Circuitos Microelectronicos - Analisis y Diseño - Muhammad Rashid

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Chapter 1 Exercise Problems EX1.1 ⎛ − Eg ⎞ ni = BT 3 / 2 exp ⎜ ⎟ ⎝ 2kT ⎠

GaAs: ni = ( 2.1× 1014 ) ( 300 ) Ge: ni = (1.66 × 1013 ) ( 300 )

3/ 2

3/ 2

⎛ ⎞ −1.4 ⎟ or ni = 1.8 × 106 cm −3 exp ⎜ ⎜ 2 ( 86 × 10−6 ) ( 300 ) ⎟ ⎝ ⎠

⎛ ⎞ −0.66 ⎟ or ni = 2.40 × 1013 cm −3 exp ⎜ ⎜ 2 ( 86 × 10−6 ) ( 300 ) ⎟ ⎝ ⎠

EX1.2 (a) majority carrier: holes, po = 1017 cm −3 minority carrier: electrons, n 2 (1.5 × 10 no = i = 1017 po

)

10 2

= 2.25 × 103 cm −3

(b) majority carrier: electrons, no = 5 × 1015 cm −3 minority carrier: holes, n 2 (1.5 × 10 ) = 4.5 × 104 cm −3 po = i = 5 × 1015 no 10 2

EX1.3 For n-type, drift current density J ≅ eμn nE or 200 = (1.6 × 10−19 ) ( 7000 ) (1016 ) E which yields E = 17.9 V / cm

EX1.4 Diffusion current density due to holes: dp J p = −eD p dx ⎛ −1 ⎞ ⎛ −x ⎞ = −eD p (1016 ) ⎜ ⎟ exp ⎜ ⎟ ⎜L ⎟ ⎜L ⎟ ⎝ p⎠ ⎝ p⎠ (a) At x = 0

(1.6 ×10 ) (10 ) (10 ) = 16 A / cm = −19

Jp

16

2

10−3 −3 (b) At x = 10 cm

⎛ −10−3 ⎞ J p = 16 exp ⎜ −3 ⎟ = 5.89 A / cm 2 ⎝ 10 ⎠

EX1.5 ⎡N N Vbi = VT ln ⎢ a 2 d ⎣ ni

⎡ (1016 )(1017 ) ⎤ ⎤ ⎢ ⎥ or Vbi = 1.23 V = 0.026 ln ( ) ⎥ 6 2 ⎥ ⎢ ⎦ ⎣ (1.8 × 10 ) ⎦

EX1.6

⎛ V ⎞ C j = C jo ⎜1 + R ⎟ ⎝ Vbi ⎠ and

−1/ 2

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⎡N N ⎤ Vbi = VT ln ⎢ a 2 d ⎥ ⎣ ni ⎦ ⎡ (1017 )(1016 ) ⎤ ⎥ = 0.757 V = ( 0.026 ) ln ⎢ ⎢ (1.5 × 1010 )2 ⎥ ⎣ ⎦ 5 ⎞ ⎛ Then 0.8 = C jo ⎜ 1 + ⎟ ⎝ 0.757 ⎠ or C jo = 2.21 pF

−1/ 2

= C jo ( 7.61)

−1/ 2

EX1.7

⎡ ⎛v ⎞ ⎤ iD = I S ⎢exp ⎜ D ⎟ − 1⎥ ⎝ VT ⎠ ⎥⎦ ⎣⎢ ⎡ ⎛ v ⎞ ⎤ so 10−3 = (10−13 ) ⎢ exp ⎜ D ⎟ − 1⎥ ⎝ 0.026 ⎠ ⎦ ⎣

⎡ 10−3 ⎤ Solving for the diode voltage, we find vD = ( 0.026 ) ln ⎢ −13 + 1⎥ ⎣10 ⎦ or vD ≅ ( 0.026 ) ln (1010 )

which yields vD = 0.599 V EX1.8 ⎛V ⎞ VPS = I D R + VD and I D ≅ I S exp ⎜ D ⎟ ⎝ VT ⎠ ( 4 − VD ) so 4 = I D ( 4 ×103 ) + VD ⇒ I D = 4 ×103 and ⎛ V ⎞ I D = (10 −12 ) exp ⎜ D ⎟ ⎝ 0.026 ⎠ By trial and error, we find I D ≅ 0.864 mA and VD ≅ 0.535 V

EX1.9

(a)

ID =

(b)

ID =

Then R =

VPS − Vγ R VPS − Vγ R

5 − 0.7 ⇒ I D = 1.08 mA 4 VPS − Vγ ⇒R= ID =

8 − 0.7 = 6.79 kΩ 1.075

(c)

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ID(mA) Diode curve 1.25 1.08

Load lines (b) (a)

0

0.7

2

4 VD(v)

6

8

EX1.10 PSpice analysis EX1.11

Quiescent diode current I DQ =

VPS − Vγ

=

10 − 0.7 = 0.465 mA 20

R Time-varying diode current: V 0.026 We find that rd = T = = 0.0559 kΩ I DQ 0.465 Then id =

vI 0.2sin ω t (V ) = ⋅ or id = 9.97sin ω t ( μ A) rd + R 0.0559 + 20 ( kΩ )

EX1.12 ⎛I ⎞ ⎛ 1.2 × 10−3 ⎞ or VD = 0.6871 V For the pn junction diode, VD ≅ VT ln ⎜ D ⎟ = ( 0.026 ) ln ⎜ −15 ⎟ ⎝ 4 × 10 ⎠ ⎝ IS ⎠ The Schottky diode voltage will be smaller, so VD = 0.6871 − 0.265 = 0.4221 V ⎛V ⎞ Now I D ≅ I S exp ⎜ D ⎟ ⎝ VT ⎠ or 1.2 × 10−3 IS = ⇒ I S = 1.07 × 10−10 A 0.4221 ⎛ ⎞ exp ⎜ ⎟ ⎝ 0.026 ⎠

EX1.13 P = I ⋅ VZ ⇒ 10 = I ( 5.6 ) ⇒ I = 1.79 mA

Also I =

10 − 5.6 = 1.79 ⇒ R = 2.46 kΩ R

Test Your Understanding Exercises TYU1.1 (a) T = 400K ⎛ − Eg ⎞ Si: ni = BT 3 / 2 exp ⎜ ⎟ ⎝ 2kT ⎠ ni = ( 5.23 × 1015 ) ( 400 )

3/ 2

⎡ ⎤ −1.1 ⎥ exp ⎢ −6 ⎢⎣ 2 ( 86 × 10 ) ( 400 ) ⎥⎦

or ni = 4.76 × 1012 cm −3

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Ge: ni = (1.66 × 1015 ) ( 400 )

3/ 2

⎡ ⎤ −0.66 ⎥ exp ⎢ −6 ⎢⎣ 2 ( 86 × 10 ) ( 400 ) ⎥⎦

or ni = 9.06 × 1014 cm −3 GaAs: ni = ( 2.1× 1014 ) ( 400 )

3/ 2

⎡ ⎤ −1.4 ⎥ exp ⎢ −6 ⎢⎣ 2 ( 86 × 10 ) ( 400 ) ⎥⎦

or ni = 2.44 × 109 cm −3 (b) T = 250 K Si: ni = ( 5.23 × 1015 ) ( 250 )

3/ 2

⎡ ⎤ −1.1 ⎥ exp ⎢ ⎢⎣ 2 ( 86 × 10−6 ) ( 250 ) ⎥⎦

or ni = 1.61× 108 cm −3 Ge: ni = (1.66 × 1015 ) ( 250 )

3/ 2

⎡ ⎤ −0.66 ⎥ exp ⎢ −6 ⎢⎣ 2 ( 86 × 10 ) ( 250 ) ⎥⎦

or ni = 1.42 × 1012 cm −3 GaAs: ni = ( 2.10 × 1014 ) ( 250 )

3/ 2

⎡ ⎤ −1.4 ⎥ exp ⎢ ⎢⎣ 2 ( 86 × 10−6 ) ( 250 ) ⎥⎦

or ni = 6.02 × 103 cm −3 TYU1.2 (a) n = 5 × 1016 cm −3 , p 0 Vγ = 0 Voltage across RL + R1 = vi ⎛ RL ⎞ 1 Voltage Divider ⇒ v0 = ⎜ ⎟ vi = vi 2 ⎝ RL + R1 ⎠

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␯0 20

2.13 For vi > 0, (Vγ = 0 )

␯i

R1

 

␯0 R2

RL

a. ⎛ R2 || RL ⎞ v0 = ⎜ ⎟ vi ⎝ R2 || RL + R1 ⎠ R2 || RL = 2.2 || 6.8 = 1.66 kΩ

␯0

⎛ 1.66 ⎞ v0 = ⎜ ⎟ vi = 0.43 vi ⎝ 1.66 + 2.2 ⎠

4.3

v0 ( rms ) =

b.

v0 ( max ) 2

⇒ v0 ( rms ) = 3.04 V

2.14 3.9 ⇒ I 2 = 0.975 mA 4 20 − 3.9 IR = = 1.3417 mA 12 I Z = 1.3417 − 0.975 ⇒ I Z = 0.367 mA IL =

PT = I Z ⋅ VZ = ( 0.367 )( 3.9 ) ⇒ PT = 1.43 mW

2.15 (a) 40 − 12 = 0.233 A 120 P = ( 0.233)(12 ) = 2.8 W

IZ =

(b)

IR = 0.233 A, IL = (0.9)(0.233) = 0.21 A 12 So 0.21 = ⇒ RL = 57.1Ω RL P = ( 0.1)( 0.233)(12 ) ⇒ P = 0.28 W

(c) 2.16

Ri VI II IZ

 VZ 

V0 RL IL

VI = 6.3 V, Ri = 12Ω, VZ = 4.8

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a. 6.3 − 4.8 ⇒ 125 mA 12 I L = I I − I Z = 125 − I Z II =

25 ≤ I L ≤ 120 mA ⇒ 40 ≤ RL ≤ 192Ω

b. PZ = I Z VZ = (100 )( 4.8 ) ⇒ PZ = 480 mW PL = I LV0 = (120 )( 4.8 ) ⇒ PL = 576 mW

2.17 a. 20 − 10 ⇒ I I = 45.0 mA 222 10 IL = ⇒ I L = 26.3 mA 380 I Z = I I − I L ⇒ I Z = 18.7 mA II =

b. PZ ( max ) = 400 mW ⇒ I Z ( max ) = ⇒ I L ( min ) = I I − I Z ( max ) = 45 − 40 ⇒ I L ( min ) = 5 mA =

400 = 40 mA 10

10 RL

⇒ RL = 2 kΩ

(c)

For Ri = 175Ω I I = 57.1 mA

I L = 26.3 mA

I Z = 30.8 mA

I Z ( max ) = 40 mA ⇒ I L ( min ) = 57.1 − 40 = 17.1 mA RL =

10 ⇒ RL = 585Ω 17.1

2.18 a.

From Eq. (2-31) 500 [ 20 − 10] − 50 [15 − 10] I Z ( max ) = 15 − ( 0.9 )(10 ) − ( 0.1)( 20 ) 5000 − 250 4 I Z ( max ) = 1.1875 A I Z ( min ) = 0.11875 A =

From Eq. (2-29(b)) Ri =

20 − 10 ⇒ Ri = 8.08Ω 1187.5 + 50

b. PZ = (1.1875 )(10 ) ⇒ PZ = 11.9 W

PL = I L ( max ) V0 = ( 0.5 )(10 ) ⇒ PL = 5 W

2.19 (a)

As approximation, assume I Z ( max ) and I Z ( min ) are the same as in problem 2.18.

V0 ( max ) = V0 ( nom ) + I Z ( max ) rZ = 10 + (1.1875)(2) = 12.375 V V0 ( min ) = V0 ( nom ) + I Z ( min ) rZ = 10 + (0.11875)(2) = 10.2375 V

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% Reg =

b. 2.20 % Reg = =

12.375 − 10.2375 × 100% ⇒ % Reg = 21.4% 10

VL ( max ) − VL ( min ) VL ( nom )

× 100%

VL ( nom ) + I Z ( max ) rz − (VL ( nom ) + I Z ( min ) rz ) VL ( nom )

⎡ I Z ( max ) − I Z ( min ) ⎦⎤ ( 3) =⎣ = 0.05 6 So I Z ( max ) − I Z ( min ) = 0.1 A

6 6 = 0.012 A, I L ( min ) = = 0.006 A 500 1000 VPS ( min ) − VZ

Now I L ( max ) = Now Ri = or 280 =

I Z ( min ) + I L ( max )

15 − 6 ⇒ I Z ( min ) = 0.020 A I Z ( min ) + 0.012

Then I Z ( max ) = 0.1 + 0.02 = 0.12 A and Ri = or 280 =

VPS ( max ) − 6 0.12 + 0.006

VPS ( max ) − VZ

I Z ( max ) + I L ( min )

⇒ VPS ( max ) = 41.3 V

2.21 Using Figure 2.21 a. VPS = 20 ± 25% ⇒ 15 ≤ VPS ≤ 25 V For VPS ( min ) :

I I = I Z ( min ) + I L ( max ) = 5 + 20 = 25 mA

Ri =

b.

VPS ( min ) − VZ II

=

15 − 10 ⇒ Ri = 200Ω 25

For VPS ( max ) ⇒ I I ( max ) =

25 − 10 ⇒ I I ( max ) = 75 mA Ri

For I L ( min ) = 0 ⇒ I Z ( max ) = 75 mA

VZ 0 = VZ − I Z rZ = 10 − ( 0.025 )( 5 ) = 9.875 V V0 ( max ) = 9.875 + ( 0.075 )( 5 ) = 10.25 V0 ( min ) = 9.875 + ( 0.005 )( 5 ) = 9.90

ΔV0 = 0.35 V

c.

% Reg =

ΔV0 × 100% ⇒ % Reg = 3.5% V0 ( nom )

2.22 From Equation (2.29(a)) VPS ( min ) − VZ 24 − 16 or Ri = 18.2Ω Ri = = I Z ( min ) + I L ( max ) 40 + 400 Also Vr =

VM VM ⇒C = 2 fRC 2 fRVr

R ≅ Ri + rz = 18.2 + 2 = 20.2Ω

Then

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C=

24 ⇒ C = 9901 μ F 2 ( 60 )(1)( 20.2 )

2.23 VZ = VZ 0 + I Z rZ VZ ( nom ) = 8 V 8 = VZ 0 + ( 0.1)( 0.5 ) ⇒ VZ 0 = 7.95 V Ii =

VS ( max ) − VZ ( nom ) Ri

=

12 − 8 = 1.333 A 3

For I L = 0.2 A ⇒ I Z = 1.133 A For I L = 1 A ⇒ I Z = 0.333 A VL ( max ) = VZ 0 + I Z ( max ) rZ

= 7.95 + (1.133)( 0.5 ) = 8.5165

VL ( min ) = VZ 0 + I Z ( min ) rZ

= 7.95 + ( 0.333)( 0.5 ) = 8.1165

ΔVL = 0.4 V ΔVL 0.4 = ⇒ % Reg = 5.0% % Reg = V0 ( nom ) 8 Vr =

VM VM ⇒C = 2 fRC 2 fRVr

R = Ri + rz = 3 + 0.5 = 3.5Ω

Then C =

12 ⇒ C = 0.0357 F 2 ( 60 )( 3.5 )( 0.8 )

2.24 (a) For −10 ≤ vI ≤ 0, both diodes are conducting ⇒ vO = 0 For 0 ≤ vI ≤ 3, Zener not in breakdown, so i1 = 0, vO = 0 vI − 3 mA 20 1 ⎛ v −3⎞ vo = ⎜ I ⎟ (10 ) = vI − 1.5 20 2 ⎝ ⎠ At vI = 10 V, vo = 3.5 V, i1 = 0.35 mA For vI > 3 i1 =

␯O(V) 4 3.5

10

(b)

3.0

10

␯I(V)

For vI < 0, both diodes forward biased

0 − vI . At vI = −10 V , i1 = −1 mA 10 v −3 For vI > 3, i1 = I . At vI = 10 V , i1 = 0.35 mA 20 −i1 =

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i1(mA)

0.35

10

10 ␯I(V)

3

1

2.25 (a) 1K ␯I

␯0

1K

2K 15

V1

1 V1 = × 15 = 5 V ⇒ for vI ≤ 5.7, v0 = vI 3 For vI > 5.7 V vI − (V1 + 0.7 ) 15 − V1 V1 + = , v0 = V1 + 0.7 1 2 1 − − 15 v 0.7 (0 ) v0 − 0.7 vI − v0 + = 1 2 1 vI 15.7 0.7 ⎛ 1 1 1⎞ + + = v0 ⎜ + + ⎟ = v0 ( 2.5 ) 1 2 1 ⎝ 1 2 1⎠ 1 vI + 8.55 = v0 ( 2.5 ) ⇒ v0 = vI + 3.42 2.5 vI = 5.7 ⇒ v0 = 5.7 vI = 15 ⇒ v0 = 9.42 ␯0(V) 9.42

5.7

0

5.7

15

␯I (V)

(b) iD = 0 for 0 ≤ vI ≤ 5.7 Then for vI > 5.7 V ⎛ v ⎞ vI − ⎜ I + 3.42 ⎟ vI − vO 2.5 ⎝ ⎠ or i = 0.6vI − 3.42 For v = 15, i = 5.58 mA = iD = I D D 1 1 1

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iD(mA) 5.58

5.7

15

␯I (V)

2.26 ⎛ 20 ⎞ For D off, vo = ⎜ ⎟ (20) − 10 = 3.33 V ⎝ 30 ⎠ Then for vI ≤ 3.33 + 0.7 = 4.03 V ⇒ vo = 3.33 V For vI > 4.03, vo = vl − 0.7; For vI = 10, vo = 9.3

(a)

␯O(V) 9.3

3.33

0

(b)

4.03

10 ␯I (V)

For vI ≤ 4.03 V , iD = 0

10 − vo vo − ( −10 ) = 10 20 3 Which yields iD = vI − 0.605 20 For vI = 10, iD = 0.895 mA

For vI > 4.03, iD +

iD(mA) 0.895

0

2.27

30

4.03

10 ␯I(V)

␯O 12.5 10.7 10.7

30

␯I

30

30 − 10.7 = 0.175 A 100 + 10 v0 = i(10) + 10.7 = 12.5 V

For vI = 30 V, i =

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b.

␯O

12.5 10.7 0

30

2.28

 5

␯I

␯O R  6.8 K

Vγ = 0.6 V vI = 15sin ω t ␯O

␯O

4.4 19.4

2.29 a. Vγ = 0 0 3 V

Vγ = 0.6 0 2.4

b. Vγ = 0 20

5

Vγ = 0.6 19.4

5

2.30

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10 6.7 0 4.7 10

2.31 One possible example is shown. Ri Ii  

DZ

Vign

L

D

 VZ  14 V 

 

RADIO VRADIO

L will tend to block the transient signals Dz will limit the voltage to +14 V and −0.7 V. Power ratings depends on number of pulses per second and duration of pulse. 2.32 ␯O(V) 40 (a)

0 ␯O(V) 35

(b)

0 5

2.33 C ␯I

␯O

 Vx 

a.

For Vγ = 0 ⇒ Vx = 2.7 V

b.

For Vγ = 0.7 V ⇒ Vx = 2.0 V

2.34 C  ␯I 

  10 V 

␯O 

2.35

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20

␯O

10 VB  0

0 ␯I

10

20

␯O

13 10 3 0

VB  3 V

␯I  VB

7

20

␯O

10 7 VB  3 V

0 3 ␯ I  VB 13

2.36 For Figure P2.32(a) 10 ␯I 0 10

␯O

20

2.37 a. 10 − 0.6 ⇒ I D1 = 0.94 mA 9.5 + 0.5 V0 = I D1 ( 9.5 ) ⇒ V0 = 8.93 V

I D1 =

ID2 = 0

b. 5 − 0.6 ⇒ I D1 = 0.44 mA 9.5 + 0.5 V0 = I D1 ( 9.5 ) ⇒ V0 = 4.18 V

I D1 =

c. d. 10 =

ID2 = 0

Same as (a)

(I ) 2

( 0.5 ) + 0.6 + I ( 9.5 ) ⇒ I = 0.964 mA

V0 = I ( 9.5 ) ⇒ V0 = 9.16 V I D1 = I D 2 =

2.38 a.

I ⇒ I D1 = I D 2 = 0.482 mA 2

I = I D1 = I D 2 = 0 V0 = 10

b.

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10 = I ( 9.5 ) + 0.6 + I ( 0.5 ) ⇒ I = I D 2 = 0.94 mA

I D1 = 0

V0 = 10 − I ( 9.5 ) ⇒ V0 = 1.07 V

c. 10 = I ( 9.5 ) + 0.6 + I ( 0.5 ) + 5 ⇒ I = I D 2 = 0.44 mA

I D1 = 0

V0 = 10 − I ( 9.5 ) ⇒ V0 = 5.82 V

d. 10 = I ( 9.5 ) + 0.6 +

I ( 0.5) ⇒ I = 0.964 mA 2

I ⇒ I D1 = I D 2 = 0.482 mA 2 V0 = 10 − I ( 9.5 ) ⇒ V0 = 0.842 V

I D1 = I D 2 =

2.39 a. V1 = V2 = 0 ⇒ D1 , D2 , D3 , on V0 = 4.4 V 10 − 4.4 ⇒ I = 0.589 mA 9.5 4.4 − 0.6 = ID2 = ⇒ I D1 = I D 2 = 7.6 mA 0.5 = I D1 + I D 2 − I = 2 ( 7.6 ) − 0.589 ⇒ I D 3 = 14.6 mA

I= I D1 I D3

b. V1 = V2 = 5 V D1 and D2 on, D3 off I 10 = I ( 9.5 ) + 0.6 + ( 0.5 ) + 5 ⇒ I = 0.451 mA 2 I I D1 = I D 2 = ⇒ I D1 = I D 2 = 0.226 mA 2 I D3 = 0 V0 = 10 − I ( 9.5 ) = 10 − ( 0.451)( 9.5 ) ⇒ V0 = 5.72 V

V1 = 5 V, V2 = 0 D1 off, D2, D3 on V0 = 4.4 V

c.

10 − 4.4 ⇒ 9.5 4.4 − 0.6 = ⇒ 0.5

I= I D2

I = 0.589 mA I D 2 = 7.6 mA I D1 = 0

I D 3 = I D 2 − I = 7.6 − 0.589 ⇒ I D 3 = 7.01 mA

V1 = 5 V, V2 = 2 V D1 off, D2, D3 on V0 = 4.4 V

d.

10 − 4.4 ⇒ 9.5 4.4 − 0.6 − 2 = ⇒ 0.5

I= I D2

I = 0.589 mA I D 2 = 3.6 mA I D1 = 0

I D 3 = I D 2 − I = 3.6 − 0.589 ⇒ I D 3 = 3.01 mA

2.40 (a) D1 on, D2 off, D3 on So I D 2 = 0 Now V2 = −0.6V , I D1 =

10 − 0.6 − ( −0.6 ) R1 + R2

=

10 ⇒ I D1 = 1.25 mA 2+6

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V1 = 10 − 0.6 − (1.25 )( 2 ) ⇒ V1 = 6.9 V I R3 = I D3

−0.6 − ( −5 )

= 2.2 mA 2 = I R 3 − I D1 = 2.2 − 1.25 ⇒ I D 3 = 0.95 mA

(b) D1 on, D2 on, D3 off So I D 3 = 0 V1 = 4.4 V , I D1 =

10 − 0.6 − 4.4 5 = 6 R1

or I D1 = 0.833 mA I R2 =

4.4 − ( −5 ) R2 + R3

=

9.4 = 0.94 mA 10

I D 2 = I R 2 − I D1 = 0.94 − 0.833 ⇒ I D 2 = 0.107 mA V2 = I R 2 R3 − 5 = ( 0.94 )( 5 ) − 5 ⇒ V2 = −0.3 V

All diodes are on V1 = 4.4V , V2 = −0.6 V

(c)

I D1 = 0.5 mA =

10 − 0.6 − 4.4 ⇒ R1 = 10 k Ω R1

I R 2 = 0.5 + 0.5 = 1 mA = I R 3 = 1.5 mA =

4.4 − ( −0.6 )

−0.6 − ( −5 ) R3

R2

⇒ R2 = 5 k Ω

⇒ R3 = 2.93 k Ω

2.41 ⎛ 0.5 ⎞ For vI small, both diodes off vO = ⎜ ⎟ vI = 0.0909vI ⎝ 0.5 + 5 ⎠ When vI − vO = 0.6, D1 turns on. So we have vI − 0.0909vI = 0.6 ⇒ vI = 0.66, vO = 0.06

vI − 0.6 − vO vI − vO vO 2v − 0.6 + = which yields vO = I 5 5 0.5 12 2vI − 0.6 When vO = 0.6, D2 turns on. Then 0.6 = ⇒ vI = 3.9 V 12 v − 0.6 − vO vI − vO vO vO − 0.6 + = + Now for vI > 3.9 I 5 5 0.5 0.5 2vI + 5.4 ; For vI = 10 ⇒ vO = 1.15 V Which yields vO = 22 For D1 on

2.42

10 V

10 K

D2

D1 ␯I

␯0 D3

D4

10 K

10 K

10 V

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For vI > 0. when D1 and D4 turn off 10 − 0.7 = 0.465 mA 20 v0 = I (10 kΩ ) = 4.65 V I=

␯0

4.65 10

4.65 4.65

10

␯I

4.65

v0 = vI for − 4.65 ≤ vI ≤ 4.65

2.43 a. R1

D2

10 V

V0 D1 ID1

R1 = 5 kΩ, R2 = 10 kΩ D1 and D2 on ⇒ V0 = 0

R2

10 V

10 − 0.7 0 − ( −10 ) − = 1.86 − 1.0 5 10 = 0.86 mA

I D1 = I D1

b. R1 = 10 kΩ, R2 = 5 kΩ, D1 off, D2 on I D1 = 0 I=

10 − 0.7 − ( −10 )

= 1.287 15 V0 = IR2 − 10 ⇒ V0 = −3.57 V

2.44 If both diodes on (a) VA = −0.7 V, VO = −1.4 V I R1 = IR2 I R1 + I D1

10 − ( −0.7 )

= 1.07 mA 10 −1.4 − ( −15 ) = = 2.72 mA 5 = I R 2 ⇒ I D1 = 2.72 − 1.07

I D1 = 1.65 mA D1 off, D2 on 10 − 0.7 − ( −15 ) I R1 = I R 2 = = 1.62 mA 5 + 10 VO = I R 2 R2 − 15 = (1.62 )(10 ) − 15 ⇒ VO = 1.2 V

(b)

VA = 1.2 + 0.7 = 1.9 V ⇒ D1 off ,

I D1 = 0

2.45

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(a)

D1 on, D2 off 10 − 0.7 I D1 = = 0.93 mA 10 VO = −15 V (b) D1 on, D2 off 10 − 0.7 I D1 = = 1.86 mA 5 VO = −15 V

2.46 15 − (V0 + 0.7 )

V0 + 0.7 V0 + 10 20 20 15 0.7 0.7 1 1 1 ⎞ ⎛ ⎛ 4.0 ⎞ − − = V0 ⎜ + + ⎟ = V0 ⎜ ⎟ 10 10 20 ⎝ 10 20 20 ⎠ ⎝ 20 ⎠ V0 = 6.975 V ID =

=

V0 ⇒ I D = 0.349 mA 20

2.47 10 K V1

Va

 VD 

10 K

Vb

V2

ID 10 K

10 K

a. V1 = 15 V, V2 = 10 V Diode off Va = 7.5 V, Vb = 5 V ⇒ VD = −2.5 V ID = 0

b.

V1 = 10 V, V2 = 15 V Diode on

V2 − Vb Vb Va Va − V1 = + + ⇒ Va = Vb − 0.6 10 10 10 10 15 10 ⎛1 1⎞ ⎛1 1⎞ ⎛ 1 1⎞ + = Vb ⎜ + ⎟ + Vb ⎜ + ⎟ − 0.6 ⎜ + ⎟ 10 10 10 10 10 10 ⎝ ⎠ ⎝ ⎠ ⎝ 10 10 ⎠ ⎛ 4⎞ 2.62 = Vb ⎜ ⎟ ⇒ Vb = 6.55 V ⎝ 10 ⎠ 15 − 6.55 6.55 ID = − ⇒ I D = 0.19 mA 10 10 VD = 0.6 V

2.48

vI = 0, D1 off, D2 on

10 − 2.5 = 0.5 mA 15 vo = 10 − ( 0.5 )( 5 ) ⇒ vo = 7.5 V for 0 ≤ vI ≤ 7.5 V I=

For vI > 7.5 V , Both D1 and D2 on vI − vo vo − 2.5 vo − 10 = + or vI = vo ( 5.5 ) − 33.75 15 10 5 When vo = 10 V, D2 turns off vI = (10 )( 5.5 ) − 33.75 = 21.25 V

For vI > 21.25 V, vo = 10 V

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2.49 a.

V01 = V02 = 0

b.

V01 = 4.4 V, V02 = 3.8 V

c.

V01 = 4.4 V, V02 = 3.8 V

Logic “1” level degrades as it goes through additional logic gates. 2.50 a.

V01 = V02 = 5 V

b.

V01 = 0.6 V, V02 = 1.2 V

c.

V01 = 0.6 V, V02 = 1.2 V

Logic “0” signal degrades as it goes through additional logic gates. 2.51 (V1 AND V2 ) OR (V3 AND V4 ) 2.52 10 − 1.5 − 0.2 I= = 12 mA = 0.012 R + 10 8.3 R + 10 = = 691.7Ω 0.012 R = 681.7Ω 2.53 10 − 1.7 − VI =8 0.75 VI = 10 − 1.7 − 8 ( 0.75 ) ⇒ VI = 2.3 V I=

2.54

 VR 

 VPS 

R 2 K

VR = 1 V, I = 0.8 mA VPS = 1 + ( 0.8 )( 2 ) VPS = 2.6 V

2.55 I Ph = η eΦA

0.6 × 10−3 = (1) (1.6 × 10−19 )(1017 ) A A = 3.75 × 10−2 cm 2

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Chapter 3 Exercise Solutions EX3.1 VTN = 1 V , VGS = 3 V , VDS = 4.5 V VDS = 4.5 > VDS ( sat ) = VGS − VTN = 3 − 1 = 2 V Transistor biased in the saturation region I D = K n (VGS − VTN ) ⇒ 0.8 = K n ( 3 − 1) ⇒ K n = 0.2 mA / V 2 2

2

(a) VGS = 2 V, VDS = 4.5 V Saturation region: I D = ( 0.2 )( 2 − 1) ⇒ I D = 0.2 mA 2

(b) VGS = 3 V, VDS = 1 V Nonsaturation region: 2 I D = ( 0.2 ) ⎡ 2 ( 3 − 1)(1) − (1) ⎤ ⇒ I D = 0.6 mA ⎣ ⎦ EX3.2 VTP = −2 V , VSG = 3 V

VSD ( sat ) = VSG + VTP = 3 − 2 = 1 V

(a) (b) (c)

VSD = 0.5 V ⇒ Nonsaturation VSD = 2 V ⇒ Saturation VSD = 5 V ⇒ Saturation

EX3.3 ⎛ R2 ⎞ ⎛ 160 ⎞ VG = ⎜ ⎟ (VDD ) = ⎜ ⎟ (10 ) = 3.636 V = VGS ⎝ 160 + 280 ⎠ ⎝ R1 + R2 ⎠ I D = 0.25 ( 3.636 − 2 ) = 0.669 mA 2

VDS = 10 − ( 0.669 )(10 ) = 3.31 V P = I DVDS = ( 0.669 )( 3.31) = 2.21 mW

EX3.4 I DQ = K P (VSG + VTP )

2

1.2 = 0.4 (VSG − 1.2 ) ⇒ VSG = 2.932 V 2

⎛ R1 ⎞ 1 VSG = ⎜ ⋅ VTTN − VDD ⎟ VDD = R2 ⎝ R1 + R2 ⎠ Note K = kΩ 1 2.932 = ( 200 )(10 ) ⇒ R2 = 682 K R2

682 R1 = 200 ⇒ R1 = 283 K 682 + R1 RD =

10 − 4 =5K 1.2

EX3.5 ⎛ R2 ⎞ ⎛ 40 ⎞ VG = ⎜ ⎟ (10 ) − 5 = ⎜ ⎟ (10 ) − 5 = −1 V R + R ⎝ 40 + 60 ⎠ ⎝ 1 2 ⎠ V − ( −5 ) 2 ID = S = K n (VGS − VTN ) RS

(a)

VS = VG − VGS

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( 5 − 1) − VGS = ( 0.5 )(1) (VGS2 − 2VGS + 1) 0.5VGS2 − 3.5 = 0 VGS2 = 7 VGS = 2.646 V I D = ( 0.5 )( 2.646 − 1) ⇒ I D = 1.354 mA VDS = 10 − (1.354 )( 3) = 5.937 V 2

4 − VGS = K n (1)(VGS − VTN )

(b)

2

(1) K n = (1.05 )( 0.5 ) = 0.525

(2) K n = ( 0.95)( 0.5 ) = 0.475 (3) VTN = (1.05 )(1) = 1.05 V

(4) VTN = ( 0.95 )(1) = 0.95 V

(1)-(3)

4 − VGS = 0.525 (VGS2 − 2.1VGS + 1.1025 )

0.525VGS2 − 0.1025VGS − 3.421 = 0 VGS =

0.1025 ± 0.010506 + 7.1841 = 2.652 V 2 ( 0.525 )

I D = 0.525 ( 2.652 − 1.05 ) = 1.348 mA VDS = 10 − (1.348 )( 3) = 5.957 V (2)-(4) 4 − VGS = 0.475 (VGS2 − 1.9VGS + 0.9025 ) 2

0.475VGS2 + 0.0975VGS − 3.5713 = 0 VGS =

−0.0975 ± 0.00950625 + 6.78547 2 ( 0.475 )

VGS = 2.641 V I D = 0.475 ( 2.641 − 0.95 ) = 1.359 mA VDS = 10 − (1.359 )( 3) = 5.924 V (1)-(4) 4 −VGS = ( 0.525) (VGS2 −1.9VGS + 0.9025) 2

0.525 VGS2 + 0.0025VGS − 3.5262 = 0

VGS =

−0.0025 ± 0.00000625 + 7.40502 2 ( 0.525)

= 2.5893 V I D = ( 0.525)( 2.5893 − 0.95) = 1.411 2

VDS = 10 − I D ( 3) = 5.7678 V (2)-(3) 4 − VGS = 0.475 (VGS2 − 2.1VGS +1.1025 ) 0.475VGS2 + 0.0025VGS − 3.4763 = 0 −0.0025 ± 0.00000625 + 6.60499 2(0.475) = 2.7027

VGS = VGS

I D = (0.475)(2.7027 − 1.05) 2 = 1.2973 mA VDS = 10 − I D (3) = 6.108 V

1.297 ≤ I DQ ≤ 1.411 mA 5.768 ≤ VDS ≤ 6.108 V EX3.6

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⎛ R2 ⎞ VG = ⎜ ⎟ (10 ) − 5 ⎝ R1 + R2 ⎠ ⎛ 200 ⎞ =⎜ ⎟ (10 ) − 5 = 0.714 V ⎝ 350 ⎠ VS = 5 − I D RS = 5 − (1.2 ) I D So VSG = VS − VG = 5 − (1.2 ) I D − 0.714

= 4.286 − (1.2 ) I D

ID =

4.286 − VSG 1.2

I D = K p (VSG + VTP )

2

(

2 4.286 − VSG = (1.2 )( 0.25 ) × VSG − 2VSG ( −1) + ( −1)

2

4.286 − VSG = ( 0.3) V − 0.6VSG + 0.3

)

2 SG

2 0.3VSG + 0.4VSG − 3.986 = 0

VSG =

( 0.4 )

−0.4 ±

+ 4 ( 0.3)( 3.986 )

2

2 ( 0.3)

Must use + sign ⇒ VSG = 3.04 V I D = ( 0.25 )( 3.04 − 1) ⇒ I D = 1.04 mA 2

VSD = 10 − I D ( RS + RD ) = 10 − (1.04 )(1.2 + 4 ) ⇒ VSD = 4.59 V VSD > VSD ( sat ) , Yes

EX3.7 VSD = 10 − I DQ ( RS + RP ) VSD = 10 − K P (VSG + VTP ) ( RS + RP ) 2

Set VSD = VSG + VTP VSG + VTP = 10 − ( 0.25 )(VSG + VTP ) ( 5.2 ) 2

1.3 (VSG + VTP ) + (VSG + VTP ) − 10 = 0 2

(VSG + VTP ) =

−1 ± 1 + 4 (1.3)(10 ) 2 (1.3)

= 2.415 V

( 3.42 V ) VSD = 2.415 V ( 2.42 V ) 2 I D = ( 0.25 )( 2.415 ) = 1.46 mA VSG = 3.415 V

EX3.8 ⎛ R2 ⎞ ⎛ 240 ⎞ VG = ⎜ ⎟ (10 ) − 5 = ⎜ ⎟ (10 ) − 5 R R + ⎝ 240 + 270 ⎠ 2 ⎠ ⎝ 1 VG = −0.294 V ID =

VS − ( −5 ) RS

=

VG − VGS + 5 2 = K n (VGS − VTN ) RS

⎛ 0.08 ⎞ 2 4.706 − VGS = ⎜ ⎟ ( 4 )( 3.9 ) (VGS − 2.4VGS + 1.44 ) ⎝ 2 ⎠ 0.624VGS2 − 0.4976VGS − 3.80744 = 0

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VGS =

0.4976 ± 0.2476 + 9.50337 2 ( 0.624 )

VGS = 2.90 V 2 ⎛ 0.08 ⎞ ID = ⎜ ⎟ ( 4 )( 2.90 − 1.2 ) ⇒ I D = 0.463 mA 2 ⎝ ⎠ VDS = 10 − I D ( 3.9 + 10 ) ⇒ VDS = 3.57 V

EX3.9 10 − VSG 2 ID = and I D = K p (VSG + VTP ) RS 0.12 = ( 0.050 )(VSG − 0.8 ) VSG = 2.35 V

2

10 − 2.35 ⇒ RS = 63.75 kΩ 0.12 VSD = 8 = 20 − I D ( RS + RD ) 8 = 20 − ( 0.12 )( 63.75 ) − ( 0.12 ) RD RS =

RD = (1) (2) (3) (4)

20 − ( 0.12 )( 63.75 ) − 8

KP KP VTP VTP ID

⇒ RD = 36.25 kΩ 0.12 = ( 0.05 )(1.05 ) = 0.0525 = ( 0.05 )( 0.95 ) = 0.0475 = −0.8 (1.05 ) = −0.84 V = −0.8 ( 0.95 ) = −0.76 V 10 − VSG 2 = = K P (VSG + VTP ) RS

(1)-(3) 2 10 − VSG = ( 63.75 )( 0.0525 ) ⎡⎣VSG − 1.68VSG + 0.7056 ⎤⎦ 2 3.347VSG − 4.623VSG − 7.6384 = 0

VSG =

4.623 ± 21.372 + 102.263 2 ( 3.347 )

VSG = 2.352 V ⇒ I D ≅ 0.120 mA VSD ≈ 8.0 V (2)-(4) 2 10 − VSG = ( 63.75 )( 0.0475 ) ⎡⎣VSG − 1.52VSG + 0.5776 ⎤⎦ 2 3.028VSG − 3.603VSG − 8.251 = 0

VSG =

3.603 ± 12.9816 + 99.936 2 ( 3.028 )

VSG = 2.35 V I D ≈ 0.120 VSD ≈ 8.0 (1)-(4) 2 10 − VSG = ( 63.75 )( 0.0525 ) ⎡⎣VSG − 1.52VSG + 0.5776 ⎤⎦ 2 − 4.087VSG − 8.06685 = 0 3.347VSG

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VSG =

4.087 ± 16.7036 + 107.999 2 ( 3.347 )

VSG = 2.279 V I D = 0.121 mA VSD = 7.89 V (2)-(3) 2 10 − VSG = ( 63.75 )( 0.0475 ) ⎡⎣VSG − 1.68VSG + 0.7056 ⎤⎦ 2 − 4.0873VSG − 7.8634 = 0 3.028VSG

VSG =

4.0873 ± 16.706 + 95.242 2 ( 3.028 )

VSG = 2.422 V I D = 0.119 mA VSD = 8.11 V

Summary 0.119 ≤ I D ≤ 0.121 mA 7.89 ≤ VSD ≤ 8.11 V EX3.10 V − VGS , I D = DD RS

I D = K n (VGS − VTN )

2

10 − VGS = (10 )( 0.2 ) (VGS2 − 2VGSVTN + VTN2 ) 10 − VGS = 2VGS2 − 8VGS + 8 2VGS2 − 7VGS − 2 = 0 VGS =

7±

(7) + 4 ( 2) 2 2 ( 2) 2

Use + sign: VGS = VDS = 3.77 V 10 − 3.77 ⇒ I D = 0.623 mA 10 Power = I DVDS = ( 0.623)( 3.77 ) ⇒ Power = 2.35 mW ID =

EX3.11 (a) VI = 4 V, Driver in Non ⋅ Sat. K nD ⎡⎣ 2 (VI − VTND ) VO − VO2 ⎤⎦ = K nL [VDD − VO − VTNL ]

2

5 ⎡⎣ 2 ( 4 − 1) VD − VD2 ⎤⎦ = ( 5 − VD − 1) = ( 4 − VO ) = 16 − 8VO + VO2 2

2

6VD2 − 38VO + 16 = 0 VD =

38 ± 1444 − 384 2 ( 6)

VD = 0.454 V (b) VI = 2 V Driver: Sat

K nD [VI − VTND ] = K nL [VDD − VO − VTNL ] 2

2

5 [ 2 − 1] = [5 − VO − 1] 2

2

5 = 4 − VO ⇒ VO = 1.76 V

EX3.12 If the transistor is biased in the saturation region

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I D = K n (VGS − VTN ) = K n ( −VTN ) 2

2

I D = ( 0.25 )( 2.5 ) ⇒ I D = 1.56 mA 2

VDS = VDD − I D RS = 10 − (1.56 )( 4 ) ⇒ VDS = 3.75 VDS > VGS − VTN = −VTN 3.75 > − ( −2.5 )

Yes — biased in the saturation region Power = I DVDS = (1.56 )( 3.75 ) ⇒ Power = 5.85 mW EX3.13 (a) For VI = 5 V, Load in saturation and driver in nonsaturation. I DD = I DL K nD ⎡⎣ 2 (VI − VTND ) VO − VO2 ⎤⎦ = K nL ( −VTNL )

2

K nD ⎡ K 2 2 ( 5 − 1)( 0.25 ) − ( 0.25 ) ⎤ = 4 ⇒ nD = 2.06 ⎣ ⎦ K nL K nL (b) I DL = K nL ( −VTNL ) ⇒ 0.2 = K nL ⎡⎣ − ( −2 ) ⎤⎦ 2

2

K nL = 50 μ A / V 2 and K nD = 103 μ A / V 2

EX3.14 For M N I DN = I DP K n (VGSN − VTN ) = K p (Vscop + VTP ) 2

2

VGSN = 1 + ( 5 − 3.25 − 1) = 1.75 V = VI Vo = VDSN ( sat ) = 1.75 − 1 ⇒ Vo = 0.75 V For M P : VI = 1.75 V

VDD − VO = VSD ( sat ) = VSGP + VTP = ( 5 − 3.25 ) − 1 = 0.75 V So Vot = 5 − 0.75 ⇒ Vot = 4.25 V

EX3.15 For RD = 10 k Ω, VDD = 5 V, and Vo = 1 V 5 −1 = 0.4 mA 10 2 ⎤⎦ I D = K n ⎡⎣ 2 (VGS − VTN ) VDS − VDS ID =

2 I D = 0.4 = K n ⎡ 2 ( 5 − 1)(1) − (1) ⎤ ⇒ K n = 0.057 mA / V 2 ⎣ ⎦ P = I D ⋅ VDS = ( 0.4 )(1) ⇒ P = 0.4 mW

EX3.16 a. V1 = 5 V, V2 = 0, M 2 cutoff ⇒ I D 2 = 0 I D = K n ⎡⎣ 2 (VI − VTN ) VO − VO2 ⎤⎦ =

5 − VO RD

( 0.05 )( 30 ) ⎡⎣ 2 ( 5 − 1)V0 − V02 ⎤⎦ = 5 − V0

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1.5V02 − 13V0 + 5 = 0 V0 =

13 ±

(13) − 4 (1.5 )( 5) ⇒ V0 = 0.40 V 2 (1.5 ) 2

5 − 0.40 ⇒ I R = I D1 = 0.153 mA 30 V1 = V2 = 5 V

I R = I D1 =

b.

5 − VO = 2 K n ⎡⎣ 2 (VI − VTN ) VO − VO2 ⎤⎦ RD

{

}

5 − V0 = 2 ( 0.05 )( 30 ) ⎡⎣ 2 ( 5 − 1)V0 − V02 ⎤⎦ 3V02 − 25V0 + 5 = 0 V0 =

25 ±

( 25) − 4 ( 3)( 5 ) ⇒ V0 = 0.205 V 2 ( 3) 2

5 − 0.205 ⇒ I R = 0.160 mA 30 = I D 2 = 0.080 mA

IR = I D1

EX3.17 M 2 & M 3 watched ⇒ I Q1 = I REF 1 = 0.4 mA 0.4 = 0.3 (VGS 3 − 1) ⇒ VGS 3 = VGS 2 = 2.15 V 2

0.4 = 0.6 (VGS 1 − 1) ⇒ VGS1 = 1.82 V 2

EX3.18 2 ⎛ 0.04 ⎞ 0.1 = ⎜ ⎟ (15 )(VSGC − 0.6 ) ⎝ 2 ⎠ VSGC = 1.177 V = VSGB 2 ⎛ 0.04 ⎞ ⎛ W ⎞ 0.2 = ⎜ ⎟ ⎜ ⎟ (1.177 − 0.6 ) ⎝ 2 ⎠ ⎝ L ⎠B ⎛W ⎞ ⎜ ⎟ = 30 ⎝ L ⎠B 2 ⎛ 0.04 ⎞ 0.2 = ⎜ ⎟ ( 25 )(VSGA − 0.6 ) ⎝ 2 ⎠ VSGA = 1.23 V

EX3.19

(a)

I REF = K n 3 (VGS 3 − VTN ) = K n 4 (VGS 4 − VTN ) VGS 3 = 2 V ⇒ VGS 4 = 3 V K K 1 2 2 ( 2 − 1) = n 4 ( 3 − 1) ⇒ n 4 = K n3 K n3 4

(b)

I Q = K n 2 (VGS 2 − VTN ) But VGS 2 = VGS 3 = 2 V

2

0.1 = K n 2 ( 2 − 1) ⇒ 2

(c)

2

2

K n 2 = 0.1 mA / V 2

0.2 = K n 3 ( 2 − 1) ⇒ K n 3 = 0.2 mA / V 2 2

0.2 = K n 4 ( 3 − 1) ⇒ K n 4 = 0.05 mA / V 2 2

EX3.20

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VS 2 = 5 − 5 = 0

RS 2 =

I D 2 = K n 2 (VGS 2 − VTN 2 )

5 = 16.7 K 0.3

2

0.3 = 0.2 (VGS 2 − 1.2 ) ⇒ VGS 2 = 2.425 V ⇒ VG 2 = VGS 2 + VS = 2.425 V 2

5 − 2.425 = 25.8 K 0.1 VS 1 = VG 2 − VDSQ1 = 2.425 − 5 = −2.575 V RD1 =

RS 1 =

−2.575 − ( −5 ) 0.1

⇒ RS 1 = 24.3 K

I D1 = K n1 (VGS 1 − VTN 1 )

2

0.1 = 0.5 (VGS1 − 1.2 ) ⇒ VGS 1 = 1.647 V VG1 = VGS 1 + VS 1 = 1.647 + ( −2.575 ) ⇒ VG1 = −0.928 V 2

⎛ R2 ⎞ 1 VG1 = ⎜ ⎟ (10 ) − 5 = ⋅ RTN ⋅ (10 ) − 5 R1 ⎝ R1 + R2 ⎠ 1 −0.928 = ( 200 )(10 ) − 5 ⇒ R1 = 491 K R1 491 R2 = 200 ⇒ R2 = 337 K 491 + R2

EX3.21 VS1 = I D RS − 5 = (0.25)(16) − 5 = −1 V I DQ = K n (VGS1 − VTN ) 2 ⇒ 0.25 = 0.5(VGS 1 − 0.8) 2 ⇒ VGS 1 = 1.507 V VG1 = VGS 1 + VS 1 = 1.507 − 1 = 0.507 V ⎛ ⎞ R3 R3 (5) ⇒ R3 = 50.7 K VG1 = ⎜ ⎟ (5) ⇒ 0.507 = 500 ⎝ R1 + R2 + R3 ⎠ VS 2 = VS 1 + VDS1 = −1 + 2.5 = 1.5 V VG 2 = VS 2 + VGS = 1.5 + 1.507 = 3.007 V ⎛ R2 + R3 ⎞ ⎛ R2 + R3 ⎞ VG 2 = ⎜ (5) ⎟ (5) ⇒ 3.007 = ⎜ R R R 500 ⎟⎠ + + ⎝ 2 3 ⎠ ⎝ 1 R2 + R3 = 300.7 R2 = 300.7 − 50.7 ⇒ R2 = 250 K R1 = 500 − 250 − 50.7 ⇒ R1 = 199.3 K VD 2 = VS 2 + VDS 2 = 1.5 + 2.5 = 4 V 5−4 ⇒ RD = 4 K RD = 0.25

EX3.22 VDS ( sat ) = VGS − VP = −1.2 − ( −4.5 ) ⇒ VDS ( sat ) = 3.3 V ⎛ ( −1.2 ) ⎞ ⎛ V ⎞ I D = I DSS ⎜ 1 − GS ⎟ = 12 ⎜ 1 − ⎟⎟ ⇒ I D = 6.45 mA ⎜ VP ⎠ ⎝ ⎝ ( −4.5 ) ⎠ 2

2

EX3.23 Assume the transistor is biased in the saturation region.

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⎛ V ⎞ I D = I DSS ⎜ 1 − GS ⎟ VP ⎠ ⎝

2

2

⎛ ⎞ V 8 = 18 ⎜ 1 − GS ⎟ ⇒ VGS = −1.17 V ⇒ VS = −VGS = 1.17 ⎜ ( −3.5 ) ⎟ ⎝ ⎠ VD = 15 − ( 8 )( 0.8 ) = 8.6 VDS = 8.6 − (1.17 ) = 7.43 V VDS = 7.43 > VGS − VP = −1.17 − ( −3.5 ) = 2.33

Yes, the transistor is biased in the saturation region. EX3.24 I D = 2.5 mA ⎛ V ⎞ I D = I DSS ⎜ 1 − GS ⎟ VP ⎠ ⎝

2

2

⎛ V ⎞ 2.5 = 6 ⎜1 − GS ⎟ ⇒ VGS = −1.42 V ⎜ ( −4 ) ⎟ ⎝ ⎠ VS = I D RS − 5 = ( 2.5 )( 0.25 ) − 5 VS = −4.375 VDS = 6 ⇒ VD = 6 − 4.375 = 1.625 5 − 1625 RD = ⇒ RD = 1.35 kΩ 2.5

( 20 )

2

R1 + R2

= 2 ⇒ R1 + R2 = 200 kΩ

VG = VGS + VS = −1.42 − 4.375 = −5.795 ⎛ R2 ⎞ VG = ⎜ ⎟ ( 20 ) − 10 ⎝ R1 + R2 ⎠ ⎛ R ⎞ −5.795 = ⎜ 2 ⎟ ( 20 ) − 10 ⇒ R2 = 42.05 kΩ → 42 kΩ ⎝ 200 ⎠ R1 = 157.95 kΩ → 158 kΩ

EX3.25 VS = −VGS . I D =

0 − VS VGS = RS RS

⎛ V ⎞ I D = I DSS ⎜ 1 − GS ⎟ VP ⎠ ⎝

2

⎛ V VGS V2 ⎞ ⎛ V ⎞ = 6 ⎜ 1 − GS ⎟ = 6 ⎜ 1 − GS + GS ⎟ 1 4 ⎠ 2 16 ⎠ ⎝ ⎝ 2 0.375VGS − 4VGS + 6 = 0 2

VGS =

4 ± 16 − 4 ( 0.375 )( 6 ) 2 ( 0.375 )

VGS = 8.86 or VGS = 1.806 V  

impossible

ID =

VGS = 1.806 mA RS

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VD = I D RD − 5 = (1.81)( 0.4 ) − 5 = −4.278

VSD = VS − V0 = −1.81 − ( −4.276 ) ⇒ VSD = 2.47 V VSD ( sat ) = VP − VGS = 4 − 1.81 = 2.19 So VSD > VSD ( sat )

EX3.26 Rin = R1 & R2 =

R1 R2 = 100 kΩ R1 + R2

I DQ = 5 mA, VS = − I DQ RS = − ( 5 )(1.2 ) = −6 V

VSDQ = 12 V ⇒ VD = VS − VSDQ

= −6 − 12 = −18 V

RD =

−18 − ( −20 ) 5

⇒ RD = 0.4 kΩ 2

⎛ V ⎞ ⎛ V ⎞ I DQ = I DSS ⎜ 1 − GS ⎟ ⇒ 5 = 8 ⎜ 1 − GS ⎟ VP ⎠ 4 ⎠ ⎝ ⎝ VGS = 0.838 V

2

VG = VGS + VS = 0.838 − 6 = −5.162 ⎛ R2 ⎞ VG = ⎜ ⎟ ( −20 ) ⎝ R1 + R2 ⎠ 1 −5.162 = (100 )( −20 ) ⇒ R1 = 387 kΩ R1 R1 R2 = 100 ⇒ ( 387 ) R2 = 100 ( 387 ) + 100 R2 R1 + R2

( 387 − 100 ) R2 = (100 )( 387 ) ⇒ R2 = 135 kΩ TYU3.1 VTN = 1.2 V , VGS = 2 V (a) V DS ( sat ) = VGS − VTN = 2 − 1.2 = 0.8 V

(i) VDS = 0.4 ⇒ Nonsaturation (ii) VDS = 1 ⇒ Saturation (iii) VDS = 5 ⇒ Saturation (b) VTN = −1.2 V , VGS = 2 V V DS ( sat ) = VGS − VTN = 2 − ( −1.2 ) = 3.2 V (i) VDS = 0.4 ⇒ Nonsaturation (ii) VDS = 1 ⇒ Nonsaturation (iii) VDS = 5 ⇒ Saturation TYU3.2

(a)

W μ n Cox 2L −14 ∈ox ( 3.9 ) ( 8.85 × 10 ) Cox = = = 7.67 × 10−8 F / cm −8 tox 450 × 10 Kn =

Kn =

(100 )( 500 ) ( 7.67 ×10−8 ) ⇒ K n = 0.274 mA / V 2 2 (7)

(b) VTN = 1.2 V, VGS = 2 V

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(i)

VDS = 0.4 V ⇒ Nonsaturation

(ii)

2 I D = ( 0.274 ) ⎡ 2 ( 2 − 1.2 )( 0.4 ) − ( 0.4 ) ⎤ ⇒ I D = 0.132 mA ⎣ ⎦ VDS = 1 V ⇒ Saturation

I D = ( 0.274 )( 2 − 1.2 ) ⇒ I D = 0.175 mA 2

(iii) VDS = 5 V ⇒ Saturation I D = ( 0.274 )( 2 − 1.2 ) ⇒ I D = 0.175 mA 2

VTN = −1.2 V , VGS = 2 V

(i)

VDS = 0.4 V ⇒ Nonsaturation

(ii)

2 I D = ( 0.274 ) ⎡ 2 ( 2 + 1.2 )( 0.4 ) − ( 0.4 ) ⎤ ⇒ I D = 0.658 mA ⎣ ⎦ VDS = 1 V ⇒ Nonsaturation

2 I D = ( 0.274 ) ⎡ 2 ( 2 + 1.2 )(1) − (1) ⎤ ⇒ I D = 1.48 mA ⎣ ⎦ (iii) VDS = 5 V ⇒ Saturation

I D = ( 0.274 )( 2 + 1.2 ) ⇒ I D = 2.81 mA 2

TYU3.3 (a) VSD (sat) = VSG + VTP = 2 − 1.2 = 0.8 V (i) Non Sat (ii) Sat (iii) Sat (b) VSD (sat) = 2 + 1.2 = 3.2 V (i) Non Sat (ii) Non Sat (iii) Sat TYU3.4 (a) (3.9)(8.85 × 10−14 ) ⎛ W ⎞ ⎛ μ p Cox ⎞ KP = ⎜ ⎟ ⎜ ⎟ Cox = 350 × 10−8 ⎝ L ⎠⎝ Z ⎠ = 9.861× 10−8 KP =

(40) ⎛ ( 300 ) ( 9.861× 10 ⎜ (2) ⎜ 2 ⎝

K P = 0.296 mA / V

(b) (i)

−8

) ⎞⎟ ⎟ ⎠

2

I D = (0.296) ⎡⎣ 2(2 − 1.2)(0.4) − (0.4) 2 ⎤⎦ = 0.142 mA

(ii)

I D = (0.296) [ 2 − 1.2] ⇒ I D = 0.189 mA 2

(iii) ID = 0.189 mA 2 (i) I D = (0.296) ⎡ 2 ( 2 + 1.2 )( 0.4 ) − ( 0.4 ) ⎤ ⎣ ⎦ = 0.710 mA (ii)

2 I D = (0.296) ⎡ 2 ( 2 + 1.2 )(1) − (1) ⎤ ⎣ ⎦ =1.60 mA

(iii) I D = ( 0.296 )( 2 + 1.2 ) = 3.03 mA

2

TYU3.5

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(a)

λ = 0, VDS ( sat ) = 2.5 − 0.8 = 1.7 V

For VDS = 2 V , VDS = 10 V ⇒ Saturation Region I D = ( 0.1)( 2.5 − 0.8 ) ⇒ I D = 0.289 mA 2

(b)

λ = 0.02 V −1 I D = K n (VGS − VTN ) (1 + λVDS ) For VDS = 2 V 2

I D = ( 0.1)( 2.5 − 0.8 ) ⎡⎣1 + ( 0.02 )( 2 ) ⎦⎤ ⇒ I D = 0.300 mA VDS = 10 V 2

(c)

2 I D = ( 0.1) ⎡( 2.5 − 0.8 ) (1 + ( 0.02 )(10 ) ) ⎤ ⇒ I D = 0.347 mA ⎣ ⎦ For part (a), λ = 0 ⇒ ro = ∞

For part (b), λ = 0.02 V −1 , −1

2 2 ro = ⎡λ K n (VGS − VTN ) ⎤ = ⎡( 0.02 )( 0.1)( 2.5 − 0.8 ) ⎤ ⎣ ⎦ ⎣ ⎦

−1

or ro = 173 k Ω

TYU3.6 VTN = VTNO + γ ⎡⎣ 2φ f + VSB − 2φ f ⎤⎦ 2φ f = 0.70 V , VTNO = 1 V

(a)

VSB = 0 ⇒, VTN = 1 V

(b)

VSB = 1 V , VTN = 1 + ( 0.35 ) ⎣⎡ 0.7 + 1 − 0.7 ⎦⎤ ⇒ VTN = 1.16 V VSB = 4 V , VTN = 1 + ( 0.35 ) ⎡⎣ 0.7 + 4 − 0.7 ⎤⎦ ⇒ VTN = 1.47 V

(c)

TYU3.7 I D = K n (VGS − VTN )

2

0.4 = 0.25 (VGS − 0.8 ) ⇒ VGS = 2.06 V 2

⎛ R2 ⎞ VGS = ⎜ ⎟ VDD ⎝ R1 + R2 ⎠ ⎛ R ⎞ 2.06 = ⎜ 2 ⎟ ( 7.5 ) ⇒ R2 = 68.8 kΩ ⎝ 250 ⎠ R1 = 181.2 kΩ VDS = 4 = VDD − I D RD 7.5 − 4 RD = ⇒ RD = 8.75 kΩ 0.4 VDS > VDS ( sat ) , Yes

TYU3.8

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VS − ( −5 )

ID =

and VS = −VGS

RS

So RS =

5 − VGS 0.1

I D = K n (VGS − VTN )

2

0.1 = ( 0.080 )(VGS − 1.2 ) ⇒ VGS = 2.32 V 5 − 2.32 So RS = ⇒ RS = 26.8 kΩ 0.1 VDS = VD − VS ⇒ VD = VDS + VS = 4.5 − 2.32 2

VD = 2.18 5 − VD 5 − 2.18 RD = = ⇒ RD = 28.2 kΩ ID 0.1

VDS > VDS ( sat ) , Yes

TYU3.9 For VDS = 2.2 V 5 − 2.2 ⇒ I D = 0.56 mA ID = 5 I D = K n (VGS − VTN ) 0.56 = K n ( 2.2 − 1)

2

2

K n = 0.389 mA / V =

W μ n Cox ⋅ 2 L

W ( 389 )( 2 ) W = ⇒ = 19.4 L L ( 40 )

TYU3.10 (a) The transition point is

(

VDD − VTNL + VTND 1 + K nD / K nL

VIt = =

)

1 + K nD /K nL

(

5 − 1 + 1 1 + 0.05/ 0.01

)

1 + 0.05/ 0.01 7.236 = ⇒ VIt = 2.236 V 3.236 VOt = VIt − VTND = 2.24 − 1 ⇒ VOt = 1.24 V

(b) We may write I D = K n D (VGSD − VTND ) = ( 0.05 )( 2.236 − 1) ⇒ I D = 76.4 μ A 2

TYU3.11

2

(

VDD − VTNL + VTND 1 + K nD /K nL

VIt =

)

1 + K nD /K nL

2.5 =

(

5 − 1 + 1 1 + K nD /K nL

)

1 + K nD /K nL

2.5 + 2.5 K nD /K nL = 5 + K nD /K nL ⇒ K nD /K nL =

b.

5 − 2.5 = 1.67 ⇒ K nD /K nL = 2.78 1.5

For VI = 5, driver in nonsaturated region.

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I DD = I DL K nD ⎡⎣ 2 (VI − VTND ) VO − VO2 ⎤⎦ = K nL (VGSL − VTNL )

2

K nD 2 ⎡ 2 (VI − VTND ) VO − VO2 ⎤⎦ = [VDD − VO − VTNL ] K nL ⎣ 2.78 ⎡⎣ 2 ( 5 − 1) V0 − V02 ⎤⎦ = [5 − V0 − 1]

2

22.24V0 − 2.78V02 = ( 4 − V0 )

2

= 16 − 8V0 + V02 3.78V02 − 30.24V0 + 16 = 0 V0 =

30.24 ±

( 30.24 ) − 4 ( 3.78 )(16 ) ⇒ V0 = 0.57 V 2 ( 3.78 ) 2

TYU3.12 We have VDS = 1.2 V < VGS − VTN = −VTN = 1.8 V Transistor is biased in the nonsaturation region. VDD − VDS 5 − 1.2 2 = ⇒ I D = 0.475 mA I D = K n ⎣⎡ 2 (VGS − VTN ) VDS − VDS ⎦⎤ and I D = 8 RS 0.475 = K n ⎡ 2 ( 0 − ( −1.8 ) ) (1.2 ) − (1.2 ) ⎤ ⎣ ⎦ 0.475 = K n ( 2.88 ) ⇒ K n = 0.165 mA/V 2 2

W μ n Cox ⋅ 2 L 165 )( 2 ) W W ( = ⇒ = 9.43 35 L L Kn =

TYU3.13 (a) Transition point for the load transistor – Driver is in the saturation region. I DD = I DL K nD (VGSD − VTND ) = K nL (VGSL − VTNL ) 2

2

VDSL ( sat ) = VGSL − VTNL = −VTNL ⇒ VDSL = VDD − VOt = 2 V Then VOt = 5 − 2 = 3 V , VOt = 3 V K nD (VIt − 1) = ( −VTNL ) K nL 0.08 (VIt − 1) = 2 ⇒ VIt = 1.89 V 0.01 (b) For the driver: VOt = VIt − VTND VIt = 1.89 V , VOt = 0.89 V

TYU3.14 2 ⎤⎦ I D = K n ⎡⎣ 2 (VGS − VTN ) VDS − VDS 2 = ( 0.050 ) ⎡ 2 (10 − 0.7 )( 0.35 ) − ( 0.35 ) ⎤ ⎣ ⎦ I D = 0.319 mA

RD =

VDD − Vo 10 − 0.35 = ⇒ RD = 30.3 kΩ ID 0.319

TYU3.15 (a) Transistor biased in the nonsaturation region

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5 − 1.5 − VDS = 12 R 2 ⎤⎦ I D = K n ⎡⎣ 2 (VGS − VTN ) VDS − VDS ID =

2 ⎤⎦ 12 = 4 ⎡⎣ 2 ( 5 − 0.8 ) VDS − VDS 2 4VDS − 33.6VDS + 12 = 0 ⇒ VDS = 0.374 V

Then R =

5 − 1.5 − 0.374 ⇒ R = 261 Ω 12

TYU3.16

a.

ID =

5 − VO = K n ⎡⎣ 2 (V2 − VTN ) VO − VO2 ⎤⎦ RD

5 − ( 0.10 )

b.

2 = K n ⎡ 2 ( 5 − 1)( 0.10 ) − ( 0.10 ) ⎤ ⇒ K n = 0.248 mA / V 2 ⎣ ⎦

25 5 − V0 = 2 ( 0.248 ) ⎡⎣ 2 ( 5 − 1) V0 − V02 ⎤⎦ 25 5 − V0 = 12.4 ⎡⎣8V0 − V02 ⎤⎦ 12.4V02 − 100.2V0 + 5 = 0 V0 =

100.2 ±

(100.2 ) − 4 (12.4 )( 5 ) ⇒ V0 = 0.0502 V 2 (12.4 ) 2

TYU3.17 2 2 I DQ = K (VGS − VTN ) ⇒ 5 = 50 (VGS − 0.15 ) ⇒ VGS = 0.466 V VS = ( 0.005 )(10 ) = 0.050 V ⇒ VGG = VGS + VS = 0.466 + 0.050 ⇒ VGG = 0.516 V VD = 5 − ( 0.005 )(100 ) ⇒ VD = 4.5 V

VDS = VD − VS = 4.5 − 0.050 ⇒ VDS = 4.45 V

TYU3.18 2 ⎤⎦ I D = K ⎡⎣ 2 (VGS − VTN ) VDS − VDS 2 = 100 ⎡ 2 ( 0.7 − 0.2 )( 0.1) − ( 0.1) ⎤ ⎣ ⎦ ID = 9 μA

RD =

2.5 − 0.1 ⇒ RD = 267 kΩ 0.009

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Chapter 3 Problem Solutions 3.1 ⎛ W ⎞ ⎛ k ′ ⎞ ⎛ 10 ⎞ ⎛ 0.08 ⎞ 2 Kn = ⎜ ⎟ ⎜ n ⎟ = ⎜ ⎟⎜ ⎟ = 0.333 mA/V L 2 1.2 2 ⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠ For VDS = 0.1 V ⇒ Non Sat Bias Region

(a)

VGS = 0 ⇒ I D = 0

(b)

2 VGS = 1 V I D = 0.333 ⎡ 2 (1 − 0.8 )( 0.1) − ( 0.1) ⎤ = 0.01 mA ⎣ ⎦

(c)

VGS = 2 V

2 I D = 0.333 ⎡ 2 ( 2 − 0.8 )( 0.1) − ( 0.1) ⎤ = 0.0767 mA ⎣ ⎦

(d)

VGS = 3 V

2 I D = 0.333 ⎡ 2 ( 3 − 0.8 )( 0.1) − ( 0.1) ⎤ = 0.143 mA ⎣ ⎦

3.2 All in Sat region ⎛ 10 ⎞⎛ 0.08 ⎞ 2 Kn = ⎜ ⎟⎜ ⎟ = 0.333 mA/V ⎝ 1.2 ⎠⎝ 2 ⎠ (a) ID = 0 (b)

I D = 0.333[1 − 0.8] = 0.0133 mA

(c)

I D = 0.333[ 2 − 0.8] = 0.480 mA

(d)

I D = 0.333[3 − 0.8] = 1.61 mA

2

2

2

3.3 (a) Enhancement-mode (b) From Graph VT = 1.5 V Now 2 0.03 = K n ( 2 − 1.5 ) = 0.25 K n ⇒ K n = 0.12 0.15 = K n ( 3 − 1.5 ) = 2.25 K n

K n = 0.0666

0.39 = K n ( 4 − 1.5 ) = 6.25 K n

K n = 0.0624

0.77 = K n ( 5 − 1.5 ) = 12.25 K n

K n = 0.0629

2

2

2

From last three, K n (Avg) = 0.0640 mA/V 2 (c)

3.4 a.

iD (sat) = 0.0640(3.5 − 1.5) 2 ⇒ iD (sat) = 0.256 mA for VGS = 3.5 V iD (sat) = 0.0640(4.5 − 1.5) 2 ⇒ iD (sat) = 0.576 mA for VGS = 4.5 V VGS = 0

VDS ( sat ) = VGS − VTN = 0 − ( −2.5 ) = 2.5 V i.

VDS = 0.5 V ⇒ Biased in nonsaturation

ii.

2 I D = (1.1) ⎡ 2 ( 0 − (−2.5) )( 0.5 ) − ( 0.5 ) ⎤ ⇒ I D = 2.48 mA ⎣ ⎦ VDS = 2.5 V ⇒ Biased in saturation

I D = (1.1) ( 0 − ( −2.5 ) ) ⇒ I D = 6.88 mA 2

iii.

VDS = 5 V Same as (ii) ⇒ I D = 6.88 mA

b. VGS = 2 V VDS ( sat ) = 2 − ( −2.5 ) = 4.5 V i.

VDS = 0.5 V ⇒ Nonsaturation I D = (1.1) ⎡⎣ 2(2 − (−2.5))(0.5) − (0.5) 2 ⎤⎦ ⇒ I D = 4.68 mA

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VDS = 2.5 V ⇒ Nonsaturation

ii.

I D = (1.1) ⎡⎣ 2(2 − (−2.5))(2.5) − (2.5) 2 ⎤⎦ ⇒ I D = 17.9 mA VDS = 5 V ⇒ Saturation

iii.

I D = (1.1) ( 2 − ( −2.5 ) ) ⇒ I D = 22.3 mA 2

3.5 VDS > VGS − VTN = 0 − ( −2 ) = 2 V Biased in the saturation region k′ W 2 I D = n ⋅ (VGS − VTN ) 2 L 2 W ⎛ 0.080 ⎞ ⎛ W ⎞ = 9.375 1.5 = ⎜ ⎟ ⎜ ⎟ ⎡⎣ 0 − ( −2 ) ⎤⎦ ⇒ L ⎝ 2 ⎠⎝ L ⎠ 3.6 kn′ = μ n Cox =

μ n ∈ox tox

=

( 600 )( 3.9 ) (8.85 ×10−14 ) tox

(a)

500 A

kn′ = 41.4 μ A/V 2

(b)

250

kn′ = 82.8 μ A/V 2

(c)

100

kn′ = 207 μ A/V 2

(d)

50

kn′ = 414 μ A/V 2

(e)

25

kn′ = 828 μ A/V 2

3.7 a. Cox = Kn =

∈ox ( 3.9 ) ( 8.85 × 10 = t0 x 450 × 10−8

−14

)⇒∈

ox

t0 x

=

2.071× 10−10 tox

= 7.67 ×10−8 F/cm 2

μ n Cox W 2

⋅

L

1 64 ( 650 ) ( 7.67 ×10−8 ) ⎛⎜ ⎞⎟ 2 ⎝ 4 ⎠ K n = 0.399 mA / V 2 =

b.

VGS = VDS = 3 V ⇒ Saturation I D = K n (VGS − VTN ) = ( 0.399 )( 3 − 0.8 ) ⇒ I D = 1.93 mA 2

2

3.8 2 ⎛ ω ⎞⎛ k′ ⎞ I D = ⎜ ⎟ ⎜ n ⎟ (VGS − VTN ) 2 2 ⎝ ⎠⎝ ⎠ 2 ⎛ ω ⎞ ⎛ 0.08 ⎞ 1.25 ⎜ ⎟⎜ ⎟ ( 2.5 − 1.2 ) ⇒ ω = 23.1 μ m ⎝ 1.25 ⎠ ⎝ 2 ⎠

3.9 ∈ Cox = ox t0 x

=

( 3.9 ) (8.85 ×10−14 ) 400 × 10−8

= 8.63 × 10−8 F/cm 2

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Kn =

μ n Cox W ⋅

L 1 ⎛W ⎞ = ( 600 ) ( 8.63 × 10−8 ) ⎜ ⎟ 2 ⎝ 2.5 ⎠ 2

K n = (1.036 × 10−5 ) W I D = K n (VGS − VTN )

2

1.2 × 10 −3 = (1.036 × 10 −5 ) W ( 5 − 1) ⇒ W = 7.24 μ m 2

3.10 Biased in the saturation region in both cases. k p′ W 2 I D = ⋅ (VSG + VTP ) 2 L 2 ⎛ 0.040 ⎞⎛ W ⎞ (1) 0.225 = ⎜ ⎟⎜ ⎟ ( 3 + VTP ) 2 L ⎝ ⎠⎝ ⎠ 2 ⎛ 0.040 ⎞ ⎛ W ⎞ 1.40 = ⎜ (2) ⎟ ⎜ ⎟ ( 4 + VTP ) ⎝ 2 ⎠⎝ L ⎠ Take ratio of (2) to (1): (4 + VTP ) 2 1.40 = 6.222 = 0.225 (3 + VTP ) 2 6.222 = 2.49 =

4 + VTP ⇒ VTP = −2.33 V 3 + VTP

W 2 ⎛ 0.040 ⎞ ⎛ W ⎞ Then 0.225 = ⎜ = 25.1 ⎟ ⎜ ⎟ ( 3 − 2.33) ⇒ L ⎝ 2 ⎠⎝ L ⎠

3.11 VS = 5 V, VG = 0 ⇒ VSG = 5 V

VTP = −0.5 V ⇒ VSD ( sat ) = VSG + VTP = 5 − 0.5 = 4.5 V

a.

VD = 0 ⇒ VSD = 5 V ⇒ Biased in saturation I D = 2 ( 5 − 0.5 ) ⇒ I D = 40.5 mA 2

b.

VD = 2 V ⇒ VSD = 3 V ⇒ Nonsaturation

c.

2 I D = 2 ⎡ 2 ( 5 − 0.5 )( 3) − ( 3) ⎤ ⇒ I D = 36 mA ⎣ ⎦ VD = 4 V ⇒ VSD = 1 V ⇒ Nonsaturation

d.

2 I D = 2 ⎡ 2 ( 5 − 0.5 )(1) − (1) ⎤ ⇒ I D = 16 mA ⎣ ⎦ VD = 5 V ⇒ VSD = 0 ⇒ I D = 0

3.12 (a) (b)

Enhancement-mode From Graph VTP = + 0.5 V

0.45 = k p ( 2 − 0.5 ) = 2.25 K p ⇒ K p =

0.20

1.25 = k p ( 3 − 0.5 ) = 6.25 K p

0.20

2.45 = k p ( 4 − 0.5 ) = 12.25 K p

0.20

2

2

2

4.10 = k p ( 5 − 0.5 ) = 20.25 K p 2

(c)

0.202 Avg K p = 0.20 mA/V 2

iD (sat) = 0.20 (3.5 − 0.5) 2 = 1.8 mA iD (sat) = 0.20 (4.5 − 0.5) 2 = 3.2 mA

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3.13 VSD ( sat ) = VSG + VTP (a)

VSD ( sat ) = −1 + 2 ⇒ VSD ( sat ) = 1 V

(b)

VSD ( sat ) = 0 + 2 ⇒ VSD ( sat ) = 2 V

(c)

VSD ( sat ) = 1 + 2 ⇒ VSD ( sat ) = 3 V

ID =

(a) (b) (c)

k p′ W k ′p W 2 2 ⋅ (VSG + VTP ) = ⋅ ⋅ ⎡⎣VSD ( sat ) ⎤⎦ 2 L 2 L 2 ⎛ 0.040 ⎞ ID = ⎜ ⎟ ( 6 )(1) ⇒ I D = 0.12 mA ⎝ 2 ⎠ 2 ⎛ 0.040 ⎞ ID = ⎜ ⎟ ( 6 )( 2 ) ⇒ I D = 0.48 mA ⎝ 2 ⎠ 2 ⎛ 0.040 ⎞ ID = ⎜ ⎟ ( 6 )( 3) ⇒ I D = 1.08 mA ⎝ 2 ⎠

3.14 VSD (sat) = VSG + VTP = 3 − 0.8 = 2.2 V ⎛ 15 ⎞⎛ 0.04 ⎞ 2 KP = ⎜ ⎟⎜ ⎟ = 0.25 mA/V ⎝ 1.2 ⎠⎝ 2 ⎠

a) b) c) d) e)

2 VSD = 0.2 Non Sat I D = 0.25 ⎡ 2 ( 3 − 0.8 )( 0.2 ) − ( 0.2 ) ⎤ = 0.21 mA ⎣ ⎦ 2 VSD = 1.2 V Non Sat I D = 0.25 ⎡ 2 ( 3 − 0.8 )(1.2 ) − (1.2 ) ⎤ = 0.96 mA ⎣ ⎦ 2 VSD = 2.2 V Sat I D = 0.25(3 − 0.8) = 1.21 mA VSD = 3.2 V Sat ID = 1.21 mA VSD = 4.2 V Sat ID = 1.21 mA

3.15 k p′ = μ p Cox =

μ p ∈ox t0 x

=

( 250 )( 3.9 ) (8.85 ×10−14 ) t0 x

(a)

tox = 500Å ⇒ k ′p = 17.3 μ A/V 2

(b)

250Å ⇒ k ′p = 34.5 μ A/V 2

(c)

100Å ⇒ k ′p = 86.3 μ A/V 2

(d)

50Å ⇒ k ′p = 173 μ A/V 2

(e)

25Å ⇒ k ′p = 345 μ A/V 2

=

8.629 × 10−11 t0 x

3.16

−14 ∈ox ( 3.9 ) ( 8.85 × 10 ) = = 6.90 × 10−8 F/cm 2 Cox = −8 t0 x 500 × 10

kn′ = ( μ n Cox ) = ( 675 ) ( 6.90 × 10−8 ) ⇒ 46.6 μ A/V 2

k ′p = ( μ p Cox ) = ( 375 ) ( 6.90 × 10−8 ) ⇒ 25.9 μ A/V 2

PMOS:

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ID =

k ′p ⎛ W ⎞ 2 ⎜ ⎟ (VSG + VTP ) 2 ⎝ L ⎠p

2 ⎛ 0.0259 ⎞⎛ W ⎞ ⎛W ⎞ 0.8 = ⎜ ⎟⎜ ⎟ ( 5 − 0.6 ) ⇒ ⎜ ⎟ = 3.19 ⎝ 2 ⎠⎝ L ⎠ p ⎝ L ⎠p

L = 4 μ m ⇒ W p = 12.8 μ m ⎛ 0.0259 ⎞ 2 Kp = ⎜ ⎟ ( 3.19 ) ⇒ K p = 41.3 μ A/V = K n ⎝ 2 ⎠ Want Kn = Kp k ′p ⎛ W ⎞ kn′ ⎛ W ⎞ ⎜ ⎟ = ⎜ ⎟ = 41.3 2 ⎝ L ⎠N 2 ⎝ L ⎠p ⎛ 46.6 ⎞ ⎛ W ⎞ ⎛W ⎞ ⎜ ⎟ ⎜ ⎟ = 41.3 ⇒ ⎜ ⎟ = 1.77 2 L ⎝ ⎠ ⎝ ⎠N ⎝ L ⎠N L = 4 μ m ⇒ WN = 7.09 μ m

3.17 VGS = 2 V, I D = ( 0.2 )( 2 − 1.2 ) = 0.128 mA 1 1 r0 = = ⇒ r0 = 781 kΩ λ I D ( 0.01)( 0.128 ) 2

VGS = 4 V, I D = ( 0.2 )( 4 − 1.2 ) = 1.57 mA 1 r0 = ⇒ r = 63.7 kΩ ( 0.01)(1.57 ) 0 2

VA =

1

λ

=

1 ⇒ VA = 100 V ( 0.01)

3.18 2 2 ⎛ 0.080 ⎞ ID = ⎜ ⎟ ( 4 )( 3 − 0.8 ) = ( 0.16 )( 3 − 0.8 ) ⇒ I D = 0.774 mA 2 ⎝ ⎠ 1 1 1 ⇒λ = = ⇒ λ (max) = 0.00646 V −1 r0 = λ ID r0 I D ( 200 )( 0.774 )

VA ( min ) =

1

λ ( max )

=

1 ⇒ VA ( min ) = 155 V 0.00646

3.19 VTN = VTNO + γ ⎡⎣ 2φ f + VSB − 2φ f ⎤⎦ ΔVTN = 2 = ( 0.8 ) ⎡ 2φ f + VSB − 2 ( 0.35 ) ⎤ ⎣ ⎦ 2.5 + 0.837 = 2 ( 0.35 ) + VSB ⇒ VSB = 10.4 V

3.20 VTN = VTNo + r ⎡⎣ 2φ f + VSB − 2φ f ⎤⎦ = 0.75 + 0.6 ⎡ 2 ( 0.37 ) + 3 − 2 ( 0.37 ) ⎤ ⎣ ⎦ = 0.75 + 0.6 [1.934 − 0.860] VTN = 1.39 V

VDS (sat) = 2.5 − 1.39 = 1.11 V

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2 ⎛ 0.08 ⎞ Sat Region I D = (15 ) ⎜ ⎟ ( 2.5 − 1.39 ) ⎝ 2 ⎠ I D = 0.739 mA

(a)

2 ⎛ 0.08 ⎞ ⎡ Non-Sat I D = (15 ) ⎜ ⎟ ⎣ 2 ( 2.5 − 1.39 )( 0.25 ) − ( 0.25 ) ⎤⎦ 2 ⎝ ⎠ I D = 0.296 mA

(b)

3.21 a. VG = %ox t0 x = ( 6 × 106 )( 275 × 10−8 ) VG = 16.5 V VG =

b.

16.5 ⇒ VG = 5.5 V 3

3.22 Want VG = ( 3)( 24 ) = %ox t0 x = ( 6 × 106 ) t0 x t0 x = 1.2 ×10−5 cm = 1200 Angstroms

3.23 ⎛ R2 ⎞ ⎛ 18 ⎞ VG = ⎜ ⎟ VDD = ⎜ ⎟ (10 ) = 3.6 V ⎝ 18 + 32 ⎠ ⎝ R1 + R2 ⎠ Assume transistor biased in saturation region V V − VGS 2 = K n (VGS − VTN ) ID = S = G RS RS 3.6 − VGS = ( 0.5 )( 2 )(VGS − 0.8 ) = VGS2 − 1.6VGS + 0.64

2

VGS2 − 0.6VGS − 2.96 = 0 VGS = ID =

0.6 ±

VG − VGS RS

( 0.6 )

2

+ 4 ( 2.96 )

⇒ VGS = 2.046 V 2 3.6 − 2.046 = ⇒ I D = 0.777 mA 2

VDS = VDD − I D ( RD + RS )

= 10 − ( 0.777 )( 4 + 2 ) ⇒ VDS = 5.34 V

VDS > VDS ( sat )

3.24

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ID(mA) 4 (a)

Q-pt Q-pt

1.67

(b) 4

5 V (V) DS

VGS = 4 V VDS (sat) = 4 − 0.8 = 3.2 V

(a)

If Sat I D = 0.25 ( 4 − 0.8 ) = 2.56 2

VDS = 1.44 × Non-Sat 2 ⎤⎦ + VDS 4 = I D RD + VDS = K n RD ⎡⎣ 2 (VGS − VT ) VDS − VDS 2 ⎤⎦ + VDS 4 = ( 0.25 )(1) ⎡⎣ 2 ( 4 − 0.8 ) VDS − VDS 2 4 = 2.6VDS − 0.25VDS 2 0.25VDS − 2.6VDS + 4 = 0

2.6 ± 6.76 − 4 = 1.88 V 2 ( 0.25 )

VDS =

4 − 1.88 = 2.12 mA 1 (b) Non-Sat region 2 ⎤⎦ + VDS 5 = I D RD + VDS = K n RD ⎡⎣ 2 (VGS − VT )VDS − VDS ID =

5 = ( 0.25 )( 3) ⎡⎣ 2 ( 5 − 0.8 ) VDS − VDS2 ⎤⎦ + VDS 2 5 = 7.3VDS − 0.75VDS 2 0.75 VDS − 7.3VDS + 5 = 0

VDS =

7.3 ± 53.29 − 15 2 ( 0.75 )

VDS = 0.741 V 5 − 0.741 ID = = 1.42 mA 3

3.25 ID(mA) 2.92 (a)

Q-pt 1.25 (b) 3.5

5 V (V) SD

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VSG = VDD = 3.5

(a)

VSD ( sat ) = 3.5 − 0.8 = 2.7 V

If biased in Sat region, I D = ( 0.2 )( 3.5 − 0.8 ) = 1.46 mA VSD = 3.5 − (1.46 )(1.2 ) = 1.75 V

2

×

Biased in Non-Sat Region. 2 ⎤⎦ 3.5 = VSD + I D RD = VSD + K p RD ⎡⎣ 2 (VSG + VTP ) VSD − VSD 2 ⎤⎦ 3.5 = VSD + ( 0.2 )(1.2 ) ⎡⎣ 2 ( 3.5 − 0.8 ) VSD − VSD 2 3.5 = VSD + 1.296 VSD − 0.24 VSD 2 − 2.296 VSD + 3.5 = 0 0.24 VSD

VSD =

+2.296 ± 5.272 − 3.36 use − sign VSD = 1.90 V 2 ( 0.24 )

2 I D = ( 0.2 ) ⎡ 2 ( 3.5 − 0.8 )(1.9 ) − (1.9 ) ⎤ = 0.2 [10.26 − 3.61] ⎣ ⎦ 3.5 − 1.90 ID = = 1.33 mA 1.2 I D = 1.33 mA

VSG = VDD = 5 V VSD ( sat ) = 5 − 0.8 = 4.2 V

(b)

If Sat Region I D = ( 0.2 )( 5 − 0.8 ) = 3.53 mA, VSD < 0 2

Non-Sat Region. 2 ⎤⎦ 5 = VSD + K p RD ⎡⎣ 2 (VSG + VTP ) VSD − VSD 2 ⎤⎦ 5 = VSD + ( 0.2 )( 4 ) ⎡⎣ 2 ( 5 − 0.8 ) VSD − VSD 2 5 = VSD + 6.72 VSD − 0.8 VSD 2 0.8 VSD − 7.72 VSD + 5 = 0

VSD = ID =

7.72 ± 59.598 − 16 use − sign VSD = 0.698 V 2 ( 0.8 )

5 − 0.698 ⇒ I D = 1.08 mA 4

3.26 10 − VS 2 = K p (VSG + VTP ) RS Assume transistor biased in saturation region ⎛ R2 ⎞ VG = ⎜ ⎟ ( 20 ) − 10 ⎝ R1 + R2 ⎠ ID =

⎛ 22 ⎞ =⎜ ⎟ ( 20 ) − 10 ⇒ VG = 4.67 V ⎝ 8 + 22 ⎠ VS = VG + VSG 10 − ( 4.67 + VSG ) = (1)( 0.5 )(VSG − 2 )

2

2 − 4VSG + 4 ) 5.33 − VSG = 0.5 (VSG

2 − VSG − 3.33 = 0 0.5VSG

VSG =

1±

(1)

2

+ 4 ( 0.5 )( 3.33)

2 ( 0.5 )

⇒ VSG = 3.77 V

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VSD

10 − ( 4.67 + 3.77 )

⇒ I D = 3.12 mA 0.5 = 20 − I D ( RS + RD ) = 20 − ( 3.12 )( 0.5 + 2 ) ⇒ VSD = 12.2 V

ID =

VSD > VSD ( sat )

3.27 VG = 0, VSG = VS Assume saturation region 2 I D = 0.4 = K p (VSG + VTP ) 0.4 = ( 0.2 )(VS − 0.8 )

2

0.4 + 0.8 ⇒ VS = 2.21 V 0.2 VD = I D RD − 5 = ( 0.4 )( 5 ) − 5 = −3 V VSD = VS − VD = 2.21 − ( −3) ⇒ VSD = 5.21 V VS =

VSD > VSD ( sat )

3.28 VDD = I DQ RD + VDSQ + I DQ RS 2 ⎛ k ′ ⎞⎛ W ⎞ (1) 10 = I DQ ( 5 ) + 5 + VGS and I DQ = ⎜ n ⎟ ⎜ ⎟ (VGS − VTN ) ⎝ 2 ⎠⎝ L ⎠ 2 ⎛ 0.060 ⎞ ⎛ W ⎞ or (2) I DQ = ⎜ ⎟ ⎜ ⎟ (VGS − 1.2 ) ⎝ 2 ⎠⎝ L ⎠ Let VGS = 2.5 V

Then from (1), 10 = I DQ ( 5 ) + 5 + 2.5 ⇒ I D = 0.5 mA W 2 ⎛ 0.060 ⎞⎛ W ⎞ Then from (2), 0.5 = ⎜ = 9.86 ⎟⎜ ⎟ ( 2.5 − 1.2 ) ⇒ 2 L L ⎝ ⎠⎝ ⎠ V 2.5 I DQ RS = VGS ⇒ RS = GS = ⇒ RS = 5 k Ω I DQ 0.5 IR =

10 = ( 0.5 )( 0.05 ) = 0.025 mA R1 + R2

Then R1 + R2 =

10 = 400 k Ω 0.025

⎛ R2 ⎞ ⎛ R2 ⎞ ⎜ ⎟ (VDD ) = 2VGS ⇒ ⎜ ⎟ (10 ) = 2 ( 2.5 ) ⇒ R1 = R2 = 200 k Ω R R + ⎝ 400 ⎠ 2 ⎠ ⎝ 1

3.29 ⎛ 75 ⎞ K n = ( 25 ) ⎜ ⎟ ⇒ 0.9375 mA/V 2 ⎝ 2⎠ ⎛ 6 ⎞ VG = ⎜ ⎟ (10 ) − 5 = −2 V ⎝ 6 + 14 ⎠ (VG − VGS ) − ( −5) 2 = I D = K n (VGS − VTN ) RS −2 − VGS + 5 = ( 0.9375 )( 0.5 )(VGS − 1)

2

3 − VGS = 0.469 (VGS2 − 2VGS + 1)

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0.469 VGS2 + 0.0625 VGS − 2.53 = 0 VGS =

−0.0625 ± 0.003906 + 4.746 ⇒ VGS = 2.26 V 2 ( 0.469 )

I D = 0.9375 ( 2.26 − 1) ⇒ I D = 1.49 mA 2

VDS = 10 − (1.49 )(1.7 ) ⇒ VDS = 7.47 V

3.30 20 = I DQ RS + VSDQ + I DQ RD (1) 20 = VSG + 10 + I DQ RD ⎛ k ′p ⎞ ⎛ W ⎞ 2 I DQ = ⎜ ⎟ ⎜ ⎟ (VSG + VTP ) L 2 ⎝ ⎠⎝ ⎠ 2 ⎛ 0.040 ⎞ ⎛ W ⎞ (2) I DQ = ⎜ ⎟ ⎜ ⎟ (VSG − 2 ) ⎝ 2 ⎠⎝ L ⎠ For example, let I DQ = 0.8 mA and VSG = 4 V ⎛ 0.040 ⎞ ⎛ W Then 0.8 = ⎜ ⎟⎜ ⎝ 2 ⎠⎝ L I DQ RS = VSG ⇒ ( 0.8 ) RS

W 2 ⎞ = 10 ⎟ ( 4 − 2) ⇒ L ⎠ = 4 ⇒ RS = 5 k Ω

From (1) 20 = 4 + 10 + ( 0.8 ) RD ⇒ RD = 7.5 k Ω IR =

20 = ( 0.8 )( 0.1) ⇒ R1 + R2 = 250 k Ω R1 + R2

⎛ R1 ⎞ ⎜ ⎟ ( 20 ) = 2VSG = ( 2 )( 4 ) ⎝ R1 + R2 ⎠ R1 ( 20 ) = 8 ⇒ R1 = 100 k Ω, R2 = 150 k Ω 250

3.31 (a)

(i)

I Q = 50 = 500 (VGS − 1.2 ) ⇒ VGS = 1.516 V 2

VDS = 5 − ( −1.516 ) =⇒ VDS = 6.516 V

(ii)

I Q = 1 = ( 0.5 )(VGS − 1.2 ) ⇒ VGS = 2.61 V 2

VDS = 5 − ( −2.61) ⇒ VDS = 7.61 V (b)

(i) Same as (a) VGS = VDS = 1.516 V

(ii)

VGS = VDS = 2.61 V

3.32 I D = K n (VGS − VTN )

2

0.25 = ( 0.2 )(VGS − 0.6 )

2

0.25 + 0.6 ⇒ VGS = 1.72 V ⇒ VS = −1.72 V 0.2 VD = 9 − ( 0.25 )( 24 ) ⇒ VD = 3 V VGS =

3.33 (a)

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ID(mA)

1.0 0.808 Q-pt 0.5

3.81

RD =

10 V (V) DS

5 −1 ⇒ RD = 8 K 0.5

I DQ = 0.5 = 0.25 (VGS − 1.4 ) ⇒ VGS = 2.81 V 2

RS =

−2.81 − ( −5 )

⇒ RS = 4.38 K 0.5 Let RD = 8.2 K, RS = 4.3 K

(b) Now

−VGS − ( −5 )

5 − VGS

= I D = 0.25 (VGS − 1.4 ) 4.3 = 1.075 (VGS2 − 2.8 VGS + 1.96 )

2

1.075 VGS2 − 2.01 VGS − 2.89 = 0 VGS =

2.01 ± 4.04 + 12.427 ⇒ VGS = 2.82 V 2 (1.075 )

I D = 0.25 ( 2.82 − 1.4 ) ⇒ I D = 0.504 mA 2

VDS = 10 − ( 0.504 )( 8.2 + 4.3) ⇒ VDS = 3.70 V

(c)

If RS = 4.3 + 10% = 4.73 K

5 − VGS = 1.18 (VGS2 − 2.8VGS + 1.96 ) 1.18 VGS2 − 2.31 VGS − 2.68 = 0 VGS =

2.31 ± 5.336 + 12.65 = 2.78 V 2 (1.18 )

I D = ( 0.25 )( 2.78 − 1.4 ) ⇒ I D = 0.476 mA 2

If Rs = 4.3 − 10% = 3.87 K

5 − VGS = ( 0.9675 ) (VGS2 − 2.8VGS + 1.96 ) 0.9675VGS2 − 1.71VGS − 3.10 = 0

VGS =

1.71 ± 2.924 + 12.0 = 2.88 V 2 ( 0.9675 )

I D = ( 0.25 )( 2.88 − 1.4 ) = 0.548 mA 2

3.34 VDD = VSD + I DQ R 9 = 2.5 + ( 0.1) R ⇒ R = 65 k Ω ⎛ k ′p ⎞ ⎛ W ⎞ 2 I DQ = ⎜ ⎟ ⎜ ⎟ (VSG + VTP ) ⎝ 2 ⎠⎝ L ⎠ W 0.025 ⎞ ⎛ W ⎞ 2 =8 ( 0.1) = ⎜⎛ ⎟ ⎜ ⎟ ( 2.5 − 1.5 ) ⇒ L L ⎝ 2 ⎠⎝ ⎠ Then for L = 4 μ m, W = 32 μ m

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3.35 5 = I DQ RS + VSDQ = I DQ ( 2 ) + 2.5 I DQ = 1.25 mA IR =

10 = (1.25 )( 0.1) ⇒ R1 + R2 = 80 k Ω R1 + R2

I DQ = K p (VSG + VTP )

2

1.25 = 0.5 (VSG + 1.5 ) ⇒ 2

1.25 − 1.5 = VSG 0.5

VSG = 0.0811 V VG = VS − VSG = 2.5 − 0.0811 = 2.42 V ⎛ R2 ⎞ VG = ⎜ ⎟ (10 ) − 5 ⎝ R1 + R2 ⎠ ⎛R ⎞ 2.42 = ⎜ 2 ⎟ (10 ) − 5 ⇒ R2 = 59.4 k Ω, R1 = 20.6 k Ω ⎝ 80 ⎠

3.36 (a) ID(mA)

0.429 Q-pt

RD =

VD − ( −5 ) I DQ

5 V (V) SD

=

5−2 ⇒ RD = 12 K 0.25

2 ⎞ ⎛ k ′p ⎞ ⎟ ⎜ ⎟ (VSG + VTP ) ⎠⎝ 2 ⎠ 2 ⎛ 0.035 ⎞ 0.25 = (15 ) ⎜ ⎟ (VSG − 1.2 ) ⇒ VSG = 2.18 V ⎝ 2 ⎠ 5 − 2.18 RS = ⇒ RS = 11.3 K 0.25 VSD = 2.18 − ( −2 ) = 4.18 V (b) k ′p = 35 + 5% = 36.75 μ A/V 2

⎛W ID = ⎜ ⎝L

5 − VSG 2 ⎛ 0.03675 ⎞ I D = (15 ) ⎜ ⎟ (VSG − 1.2 ) = 2 11.3 ⎝ ⎠

2 − 2.4VSG + 1.44 ) = 5 − VSG 3.11(VSG 2 − 6.46VSG − 0.522 = 0 3.11VSG

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VSG =

6.46 ± 41.73 + 6.49 = 2.155 V 2 ( 3.11)

5 − 2.155 = 0.252 mA 11.3 = 10 − ( 0.252 )(12 + 11.3) = 4.13 V

ID = VSD

k ′p = 35 − 5% = 33.25 μ A/V 2 5 − VSG 2 ⎛ 0.03325 ⎞ I D = (15 ) ⎜ ⎟ (VSG − 1.2 ) = 2 11.3 ⎝ ⎠

2 − 2.4VSG + 1.44 ) = 5 − VSG 2.82 (VSG 2 − 5.77VSG − 0.939 = 0 2.82VSG

VSG =

5.77 ± 33.29 + 10.59 = 2.198 V 2 ( 2.82 )

5 − 2.198 = 0.248 mA 11.3 = 10 − ( 0.248 )(12 + 11.3) = 4.22 V

ID = VSD

3.37 ID =

−VSD − ( −10 ) RD

⇒5=

−6 + 10 ⇒ RD = 0.8 kΩ RD

I D = K P (VSG + VTP ) ⇒ 5 = 3 (VSG − 1.75 ) 2

VSG =

2

5 + 1.75 = 3.04 V ⇒ VG = −3.04 3

⎛ R2 ⎞ VG = ⎜ ⎟ (10 ) − 5 = −3.04 ⎝ R1 + R2 ⎠ Rin = R1 || R2 = 80 kΩ 1 ⋅ ( 80 )(10 ) = 5 − 3.04 ⇒ R1 = 408 kΩ R1 408 R2 = 80 ⇒ R2 = 99.5 kΩ 408 + R2

3.38 ⎛ 60 ⎞ K n1 = ⎜ ⎟ ( 4 ) = 120 μ A/V 2 ⎝ 2 ⎠ ⎛ 60 ⎞ K n 2 = ⎜ ⎟ (1) = 30 μ A/V 2 ⎝ 2 ⎠ For vI = 1 V , M1 Sat. region, M2 Non-sat region.

(a)

I D 2 = I D1 30 ⎡ 2 ( −VTNL )( 5 − vO ) − ( 5 − vO ) ⎤ = 120 (1 − 0.8 ) ⎣ ⎦ We find vO2 − 6.4vO + 7.16 = 0 ⇒ vO = 4.955 V 2

(b)

2

For vI = 3 V , M1 Non-sat region, M2 Sat. region. I D 2 = I D1

30 ⎣⎡ − ( −1.8 ) ⎦⎤ = 120 ⎡⎣ 2 ( 3 − 0.8 ) vO − vO2 ⎤⎦ 2

We find 4vO2 − 17.6vO + 3.24 = 0 ⇒ vO = 0.193 V (c)

For vI = 5 V , biasing same as (b)

30 ⎡⎣ − ( −1.8 ) ⎤⎦ = 120 ⎡⎣ 2 ( 5 − 0.8 ) vO − vO2 ⎤⎦ 2

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We find 4vO2 − 33.6vO + 3.24 = 0 ⇒ vO = 0.0976 V 3.39 For vI = 5 V , M1 Non-sat region, M2 Sat. region. I D1 = I D 2 2 ⎛ kn′ ⎞ ⎛ W ⎞ ⎛ kn′ ⎞ ⎛ W ⎞ 2 ⎜ 2 ⎟ ⎜ L ⎟ ⎡⎣ 2 (VGS1 − VTN 1 ) VDS 1 − VDS 1 ⎤⎦ = ⎜ 2 ⎟ ⎜ L ⎟ (VGS 2 − VTN 2 ) ⎝ ⎠ ⎝ ⎠1 ⎝ ⎠ ⎝ ⎠2 2 W 2 ⎛ ⎞ ⎡ ⎜ ⎟ ⎣ 2 ( 5 − 0.8 )( 0.15 ) − ( 0.15 ) ⎤⎦ = (1) ⎡⎣0 − ( −2 ) ⎤⎦ ⎝ L ⎠1

⎛W ⎞ which yields ⎜ ⎟ = 3.23 ⎝ L ⎠1

3.40 a.

M1 and M2 in saturation

K n1 (VGS 1 − VTN 1 ) = K n 2 (VGS 2 − VTN 2 ) K n1 = K n 2 , VTN 1 = VTN 2 ⇒ VGS1 = VGS 2 = 2.5 V, V0 = 2.5 V 2

2

I D = (15 )( 40 )( 2.5 − 0.8 ) ⇒ I D = 1.73 mA 2

b. ⎛W ⎞ ⎛W ⎞ ⎜ ⎟ > ⎜ ⎟ ⇒ VGS1 < VGS 2 ⎝ L ⎠1 ⎝ L ⎠ 2 40 (VGS1 − 0.8 ) = (15 )(VGS 2 − 0.8 ) 2

2

VGS 2 = 5 − VGS 1 1.633 (VGS 1 − 0.8 ) = ( 5 − VGS1 − 0.8 )

2.633VGS 1 = 5.506 ⇒ VGS 1 = 2.09 V

VGS 2 = 2.91 V, V0 = VGS1 = 2.91 V

I D = (15 )(15 )( 2.91 − 0.8 ) ⇒ I D = 1.0 mA 2

3.41 (a) V1 = VGS 3 = 2.5 V 2 ⎛ W ⎞ ⎛ 0.06 ⎞ I D = 0.5 = ⎜ ⎟ ⎜ ⎟ ( 2.5 − 1.2 ) ⎝ L ⎠3 ⎝ 2 ⎠ ⎛W ⎞ ⎜ ⎟ = 9.86 ⎝ L ⎠3

V2 = 6 V ⇒ VGS 2 = V2 − V1 = 6 − 2.5 = 3.5 V 2 ⎛ W ⎞ ⎛ 0.06 ⎞ ⎛W ⎞ 0.5 = ⎜ ⎟ ⎜ ⎟ ( 3.5 − 1.2 ) ⇒ ⎜ ⎟ = 3.15 ⎝ L ⎠2 ⎝ 2 ⎠ ⎝ L ⎠2 VGS1 = 10 − V2 = 10 − 6 = 4 V 2 ⎛ W ⎞ ⎛ 0.06 ⎞ ⎛W ⎞ 0.5 = ⎜ ⎟ ⎜ ⎟ ( 4 − 1.2 ) ⇒ ⎜ ⎟ = 2.13 ⎝ L ⎠1 ⎝ 2 ⎠ ⎝ L ⎠1 (b) kn′1 = 0.06 + 5% = 0.063 mA/V 2 kn′2 = k n′3 = 0.6 − 5% = 0.057 mA/V 2 2 ⎛ 0.057 ⎞ For M3: I D = ( 9.86 ) ⎜ ⎟ (V1 − 1.2 ) ⎝ 2 ⎠ 2 ⎛ 0.057 ⎞ For M2: I D = ( 3.15 ) ⎜ ⎟ (V2 − V1 − 1.2 ) 2 ⎝ ⎠

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2 ⎛ 0.063 ⎞ For M1: I D = ( 2.13) ⎜ ⎟ (10 − V2 − 1.2 ) ⎝ 2 ⎠

0.281(V1 − 1.2 ) = 0.0898 (V2 − V1 − 1.2 ) = 0.0671( 8.8 − V2 ) 2

2

2

Take square root. 0.530 (V1 − 1.2 ) = 0.300 (V2 − V1 − 1.2 ) = 0.259 ( 8.8 − V2 )   

  

(1) 0.830V1 = 0.300V2 + 0.276

(2) 0.559V2 = 0.300V1 + 2.64

From (2) ⇒ V2 = 0.537V1 + 4.72 Substitute into (1) 0.830V1 = 0.300 [ 0.537V1 + 4.72] + 0.276 = 0.161V1 + 1.69 V1 = 2.53 V Then V2 = 0.537 ( 2.53) + 4.72 V2 = 6.08 V

3.42 ML in saturation MD in nonsaturation 2 ⎛W ⎞ ⎛W ⎞ 2 ⎜ ⎟ (VGSL − VTNL ) = ⎜ ⎟ ⎣⎡ 2 (VGSD − VTND )VDSD − VDSD ⎦⎤ ⎝ L ⎠L ⎝ L ⎠D W 2 2 (1)( 5 − 0.1 − 0.8) = ⎛⎜ ⎞⎟ ⎣⎡ 2 ( 5 − 0.8)( 0.1) − ( 0.1) ⎦⎤ L ⎝ ⎠D ⎛W ⎞ 16.81 = ⎜ ⎟ [ 0.83] ⎝ L ⎠D ⎛W ⎞ ⎜ ⎟ = 20.3 ⎝ L ⎠D

3.43 ML in saturation MD in nonsaturation 2 ⎛W ⎞ ⎛W ⎞ 2 ⎜ ⎟ (VGSL − VTNL ) = ⎜ ⎟ ⎣⎡ 2 (VGSD − VTND )VDSD − VDSD ⎦⎤ ⎝ L ⎠L ⎝ L ⎠D W 2 2 (1)(1.8 ) = ⎛⎜ ⎞⎟ ⎡⎣ 2 ( 5 − 0.8 )( 0.05) − ( 0.05) ⎤⎦ ⎝ L ⎠D ⎛W ⎞ 3.24 = ⎜ ⎟ [ 0.4175] ⎝ L ⎠D ⎛W ⎞ ⎜ ⎟ = 7.76 ⎝ L ⎠D

3.44 VDD − V0 5 − 0.1 = = 0.49 mA 10 RD Transistor biased in nonsaturation I D = 0.49 ID =

2 ⎛W ⎞ = ( 0.015 ) ⎜ ⎟ ⎡ 2 ( 4.2 − 0.8 )( 0.1) − ( 0.1) ⎤ ⎣ ⎦ ⎝L⎠ W ⎛W ⎞ 0.49 = ⎜ ⎟ 0.01005 ⇒ = 48.8 L ⎝L⎠

3.45

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5 = I D RD + Vγ + VDS

5 = (12 ) RD + 1.6 + 0.2 ⇒ RD = 267 Ω 2 ⎛ k′ ⎞⎛ W ⎞ I D = ⎜ n ⎟ ⎜ ⎟ (VGS − VTN ) ⎝ 2 ⎠⎝ L ⎠ W 2 ⎛ 0.040 ⎞ ⎛ W ⎞ = 34 12 = ⎜ ⎟ ⎜ ⎟ ( 5 − 0.8 ) ⇒ 2 L L ⎝ ⎠⎝ ⎠

3.46 5 = VSD + I D RD + Vγ 5 = 0.15 + (15 ) RD + 1.6 ⇒ RD = 217 Ω ⎛ k ′p ⎞ ⎛ W ⎞ 2 I D = ⎜ ⎟ ⎜ ⎟ (VSG + VTP ) ⎝ 2 ⎠⎝ L ⎠ W 2 ⎛ 0.020 ⎞⎛ W ⎞ = 85 15 = ⎜ ⎟⎜ ⎟ ( 5 − 0.8 ) ⇒ L ⎝ 2 ⎠⎝ L ⎠

3.47 (a) VDD − VO ⎛W = 2⎜ RD ⎝L 5 − 0.2 ⎛W = 2⎜ 20 ⎝L

⎞⎛ 0.060 ⎞ 2 ⎟⎜ ⎟ ⎡⎣( 2 )(VGS − VTN ) VO − VO ⎤⎦ ⎠⎝ 2 ⎠

2 ⎞ ⎟ ( 0.030 ) ⎡⎣ 2 ( 5 − 0.8 )( 0.2 ) − ( 0.2 ) ⎤⎦ ⎠ ⎛W ⎞ ⎛W ⎞ ⎛W ⎞ 0.24 = 0.0984 ⎜ ⎟ ⇒ ⎜ ⎟ = ⎜ ⎟ = 2.44 ⎝ L ⎠ ⎝ L ⎠1 ⎝ L ⎠ 2

(b) 5 − VO ⎛ 0.06 ⎞ 2 = ( 2.44 ) ⎜ ⎟ ⎡⎣ 2 ( 5 − 0.8 ) VO − VO ⎤⎦ 20 2 ⎝ ⎠ 5 − VO = 12.30VO − 1.464VO2 1.464VO2 − 13.30VO + 5 = 0 VO =

13.30 ± 176.89 − 29.28 2 (1.464 )

VO = 0.393 V

3.48 2 ⎞ ⎛ k n′ ⎞ ⎟ ⎜ ⎟ (VGS 1 − VTN ) 2 ⎠1 ⎝ ⎠ 2 ⎞ ⎛ kn′ ⎞ ⎟ ⎜ ⎟ (VDS 2 ( sat ) ) ⎠2 ⎝ 2 ⎠ 2 ⎛ W ⎞ ⎛ 0.08 ⎞ ⎛W ⎞ ⎛W ⎞ 0.1 = ⎜ ⎟ ⎜ ⎟ ( 0.5 ) ⇒ ⎜ ⎟ = 10 = ⎜ ⎟ L L 2 ⎝ ⎠2 ⎝ ⎠ ⎝ ⎠2 ⎝ L ⎠1 ⎛ W ⎞ ⎛ 200 ⎞ ⎛ W ⎞ ⎜ ⎟ =⎜ ⎟ ⎜ ⎟ = 20 ⎝ L ⎠3 ⎝ 100 ⎠ ⎝ L ⎠ 2 M1 & M2 matched. 2 ⎛ 0.08 ⎞ Then 0.1 = (10 ) ⎜ ⎟ (VGS 1 − 0.25 ) 2 ⎝ ⎠ VGS1 = 0.75 V

⎛W I Q1 = ⎜ ⎝L ⎛W I Q1 = ⎜ ⎝L

VD1 = −0.75 + 2 = 1.25 V RD =

2.5 − 1.25 ⇒ RD = 12.5 K 0.1

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3.49 (a) 2 ⎛ W ⎞ ⎛ k ′p ⎞ I Q 2 = ⎜ ⎟ ⎜ ⎟ (VSDB ( sat ) ) ⎝ L ⎠B ⎝ 2 ⎠ 2 ⎛ W ⎞ ⎛ 0.04 ⎞ 0.25 = ⎜ ⎟ ⎜ ⎟ ( 0.8 ) ⇒ ⎝ L ⎠B ⎝ 2 ⎠

⎛W ⎞ ⎛W ⎞ ⎜ ⎟ = 19.5 = ⎜ ⎟ ⎝ L ⎠B ⎝ L ⎠A I ⎛W ⎞ KQ 2 ⎛ W ⎞ ⎜ ⎟ = ⎜ ⎟ L I ⎝ ⎠C Q 2 ⎝ L ⎠B ⎛ 100 ⎞ =⎜ ⎟ (19.5 ) = 7.81 ⎝ 250 ⎠

2 ⎛ W ⎞ ⎛ k p′ ⎞ I Q 2 = ⎜ ⎟ ⎜ ⎟ (VSGA + VTP ) ⎝ L ⎠A ⎝ 2 ⎠ 2 ⎛ 0.04 ⎞ 0.25 = (19.5 ) ⎜ ⎟ (VSGA − 0.5 ) ⎝ 2 ⎠ VSGA = 1.30 V (b) VDA = 1.3 − 4 = −2.7 V

RD =

−2.7 − ( −5 ) 0.25

⇒ RD = 9.2 K

3.50 2 ⎞ ⎛ kn′ ⎞ ⎟ ⎜ ⎟ (VDS 2 ( sat ) ) 2 ⎠2 ⎝ ⎠ 2 ⎞ ⎛ 0.06 ⎞ ⎛W ⎞ ⎛W ⎞ ⎟ ⎜ ⎟ ( 0.5 ) ⇒ ⎜ ⎟ = 53.3 = ⎜ ⎟ ⎠2 ⎝ 2 ⎠ ⎝ L ⎠2 ⎝ L ⎠1 2 ⎛ W ⎞ ⎛ k′ ⎞ I Q = ⎜ ⎟ ⎜ n ⎟ (VGS 1 − VTN ) 2 L ⎝ ⎠1 ⎝ ⎠ 2 ⎛ 0.06 ⎞ 0.4 = ( 53.3) ⎜ ⎟ (VGS 1 − 0.75 ) ⎝ 2 ⎠ VGS1 = 1.25 V VD1 = −1.25 + 4 = 2.75 V 5 − 2.75 RD = ⇒ RD = 5.625 K 0.4

⎛W IQ = ⎜ ⎝L ⎛W 0.4 = ⎜ ⎝L

3.51 VDS ( sat ) = VGS − VP So VDS > VDS ( sat ) = −VP , I D = I DSS 3.52 VDS ( sat ) = VGS − VP = VGS + 3 = VDS ( sat ) a.

VGS = 0 ⇒ I D = I DSS = 6 mA

b.

⎛ V ⎞ ⎛ −1 ⎞ I D = I DSS ⎜ 1 − GS ⎟ = 6 ⎜ 1 − ⎟ ⇒ I D = 2.67 mA V ⎝ −3 ⎠ P ⎠ ⎝

2

2

2

c. d.

⎛ −2 ⎞ I D = 6 ⎜ 1 − ⎟ ⇒ I D = 0.667 mA ⎝ −3 ⎠ ID = 0

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3.53 ⎛ V ⎞ I D = I DSS ⎜ 1 − GS ⎟ VP ⎠ ⎝ ⎛ 1 ⎞ 2.8 = I DSS ⎜1 − ⎟ ⎝ VP ⎠

2

2

⎛ 3 ⎞ 0.30 = I DSS ⎜1 − ⎟ ⎝ VP ⎠ ⎛ 1 ⎜1 − 2.8 ⎝ VP = 0.30 ⎛ 3 ⎜1 − V P ⎝

2

2

⎞ ⎟ ⎠ = 9.33 2 ⎞ ⎟ ⎠

⎛ 1 ⎜1 − V P ⎝ ⎛ 3 ⎜1 − ⎝ VP

⎞ ⎟ ⎠ = 3.055 ⎞ ⎟ ⎠ 1 9.165 = 3.055 − 1− VP VP 8.165 = 2.055 ⇒ VP = 3.97 V VP 2

1 ⎞ ⎛ 2.8 = I DSS ⎜1 − ⎟ = I DSS ( 0.560 ) ⇒ I DSS = 5.0 mA ⎝ 3.97 ⎠

3.54 VS = −VGS , VSD = VS − VDD Want VSD ≥ VSD ( sat ) = VP − VGS VS − VDD ≥ VP − VGS − VGS − VDD ≥ VP − VGS ⇒ VDD ≤ −VP

So VDD ≤ −2.5 V ⎛ V ⎞ I D = 2 = I DSS ⎜1 − GS ⎟ VP ⎠ ⎝

2

2

⎛ V ⎞ 2 = 6 ⎜ 1 − GS ⎟ ⇒ VGS = 1.06 V ⇒ VS = −1.06 V ⎝ 2.5 ⎠

3.55 I D = K n (VGS − VTN )

2

18.5 = K n ( 0.35 − VTN ) 86.2 = K n ( 0.5 − VTN )

2

2

Then

( 0.35 − VTN ) 18.5 = 0.2146 = ⇒ VTN = 0.221 V 2 86.2 ( 0.50 − VTN ) 2

18.5 = K n ( 0.35 − 0.221) ⇒ K n = 1.11 mA / V 2 2

3.56 I D = K (VGS − VTN )

2

250 = K ( 0.75 − 0.24 ) ⇒ K = 0.961 mA / V 2 2

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3.57 2

⎛ V ⎞ V V I D = I DSS ⎜ 1 − GS ⎟ = S = − GS V R RS P ⎠ S ⎝ 2

V ⎛ V ⎞ 10 ⎜ 1 − GS ⎟ = − GS 0.2 −5 ⎠ ⎝ 2 ⎛ 2V V ⎞ 2 ⎜1 + GS + GS ⎟ = −VGS 5 25 ⎠ ⎝ 2 2 9 VGS + VGS + 2 = 0 25 5 2VGS2 + 45VGS + 50 = 0 VGS =

−45 ±

ID = −

( 45 ) − 4 ( 2 )( 50 ) ⇒ VGS 2 ( 2) 2

= −1.17 V

VGS 1.17 = ⇒ I D = 5.85 mA RS 0.2

VD = 20 − ( 5.85 )( 2 ) = 8.3 V VDS = VD − VS = 8.3 − 1.17 ⇒ VDS = 7.13 V

3.58 VDS = VDD − VS 8 = 10 − VS ⇒ VS = 2 V = I D RS = ( 5 ) RS ⇒ RS = 0.4 kΩ ⎛ V ⎞ I D = I DSS ⎜ 1 − GS ⎟ ⎝ VP ⎠

2

2

⎛ −1 ⎞ 5 = I DSS ⎜1 − ⎟ Let I DSS = 10 mA ⎝ VP ⎠ 2

⎛ −1 ⎞ 5 = 10 ⎜ 1 − ⎟ ⇒ VP = −3.41 V ⎝ VP ⎠ VG = VGS + VS = −1 + 2 = 1 V ⎛ R2 ⎞ 1 VG = ⎜ ⎟ VDD = ⋅ Rin ⋅ VDD R1 ⎝ R1 + R2 ⎠ 1 1 = ( 500 )(10 ) ⇒ R1 = 5 MΩ R1 5R2 = 0.5 ⇒ R2 = 0.556 MΩ 5 + R2

3.59 ⎛ V ⎞ I D = I DSS ⎜ 1 − GS ⎟ ⎝ VP ⎠

2

2

⎛ V ⎞ 5 = 8 ⎜ 1 − GS ⎟ ⇒ VGS = 0.838 V 4 ⎠ ⎝ VSD = VDD − I D ( RS + RD ) = 20 − ( 5 )( 0.5 + 2 ) ⇒ VSD = 7.5 V

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VS = 20 − ( 5 )( 0.5 ) = 17.5 V VG = VS + VGS = 17.5 + 0.838 = 18.3 V ⎛ R2 ⎞ 1 VG = ⎜ ⎟ VDD = ⋅ Rin ⋅ VDD R1 ⎝ R1 + R2 ⎠ 1 18.3 = (100 ) ( 20 ) ⇒ R1 = 109 kΩ R1 109 R2 = 100 ⇒ R2 = 1.21 MΩ 109 + R2

3.60 ⎛ V ⎞ I D = I DSS ⎜ 1 − GS ⎟ VP ⎠ ⎝

2

2

⎛ V ⎞ 5 = 7 ⎜ 1 − GS ⎟ ⇒ VGS = 0.465 V 3 ⎠ ⎝ VSD = VDD − I D ( RS + RD )

6 = 12 − ( 5 )( 0.3 + RD ) ⇒ RD = 0.9 kΩ

VS = 12 − ( 5 )( 0.3) = 10.5 V VG = VS + VGS = 10.5 + 0.465 = 10.965 V ⎛ R2 ⎞ VG = ⎜ ⎟ VDD ⎝ R1 + R2 ⎠ ⎛ R ⎞ 10.965 = ⎜ 2 ⎟ (12 ) ⇒ R2 = 91.4 kΩ ⇒ R1 = 8.6 kΩ ⎝ 100 ⎠

3.61 ⎛ R2 ⎞ ⎛ 60 ⎞ VG = ⎜ ⎟ VDD = ⎜ ⎟ ( 20 ) ⇒ VG = 6 V ⎝ 140 + 60 ⎠ ⎝ R1 + R2 ⎠ 2

⎛ V ⎞ V V − VGS I D = I DSS ⎜ 1 − GS ⎟ = S = G V R RS P ⎠ S ⎝ ⎛

(8 )( 2 ) ⎜⎜1 − ⎝

2

VGS ⎞ ⎟ = 6 − VGS ( −4 ) ⎟⎠

⎛ V V2 ⎞ 16 ⎜ 1 + GS + GS ⎟ = 6 − VGS 2 16 ⎠ ⎝ 2 VGS + 9VGS + 10 = 0 VGS =

−9 ±

(9)

2

− 4 (10 )

2

⇒ VGS = −1.30

⎛ ( −1.30 ) ⎞ I D = 8 ⎜⎜ 1 − ⎟ ⇒ I D = 3.65 mA ( −4 ) ⎟⎠ ⎝ VDS = VDD − I D ( RS + RD ) 2

= 20 − ( 3.65 )( 2 + 2.7 )

VDS = 2.85 V VDS > VDS ( sat ) = VGS − VP = −1.30 − ( −4 ) = 2.7 V (Yes)

3.62

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VDS = VDD − I D ( RS + RD )

5 = 12 − I D ( 0.5 + 1) ⇒ I D = 4.67 mA VS = I D RS = ( 4.67 ) ( 0.5 ) ⇒ VS = 2.33 V

⎛ R2 ⎞ ⎛ 20 ⎞ VG = ⎜ ⎟ VDD = ⎜ ⎟ (12 ) ⇒ VG = 0.511 V + R R ⎝ 450 + 20 ⎠ ⎝ 1 2 ⎠ VGS = VG − VS = 0.511 − 2.33 ⇒ VGS = −1.82 V ⎛ V ⎞ I D = I DSS ⎜ 1 − GS ⎟ VP ⎠ ⎝

2

⎛ ( −1.82 ) ⎞ 4.67 = 10 ⎜⎜ 1 − ⎟ ⇒ VP = −5.75 V VP ⎟⎠ ⎝ 2

3.63 2

⎛ V ⎞ I D = I DSS ⎜ 1 − GS ⎟ , VGS = 0 VP ⎠ ⎝ I D = I DSS = 4 mA RD =

VDD − VDS 10 − 3 ⇒ RD = 1.75 kΩ = 4 ID

3.64 VSD = VDD − I D RS 10 = 20 − (1) RS ⇒ RS = 10 kΩ R1 + R2 =

VDD 20 = = 200 kΩ 0.1 I

⎛ V ⎞ I D = I DSS ⎜ 1 − GS ⎟ VP ⎠ ⎝

2

2

⎛ V ⎞ 1 = 2 ⎜1 − GS ⎟ ⇒ VGS = 0.586 V 2 ⎠ ⎝ VG = VS + VGS = 10 + 0.586 = 10.586 ⎛ R2 ⎞ VG = ⎜ ⎟ VDD ⎝ R1 + R2 ⎠ ⎛ R ⎞ 10.586 = ⎜ 2 ⎟ ( 20 ) ⇒ R2 = 106 kΩ ⎝ 200 ⎠ R1 = 94 kΩ

3.65

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VDS = VDD − I D ( RS + RD ) 2 = 3 − ( 0.040 )(10 + RD ) ⇒ RD = 15 kΩ I D = K (VGS − VTN )

2

40 = 250 (VGS − 0.20 ) ⇒ VGS = 0.60 V 2

VG = VGS + VS = 0.60 + ( 0.040 )(10 ) = 1.0 V

⎛ R2 ⎞ VG = ⎜ ⎟ VDD ⎝ R1 + R2 ⎠ ⎛ R ⎞ 1 = ⎜ 2 ⎟ ( 3) ⇒ R2 = 50 kΩ ⎝ 150 ⎠ R1 = 100 kΩ

3.66 For VO = 0.70 V ⇒ VDS = 0.70 > VDS ( sat ) = VGS − VTN 0.75 − 0.15 = 0.6 Biased in the saturation region V − VDS 3 − 0.7 ⇒ I D = 46 μ A I D = DD = 50 RD I D = K (VGS − VTN ) ⇒ 46 = K ( 0.75 − 0.15 ) ⇒ K = 128 μ A / V 2 2

2

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Chapter 4 Exercise Solutions EX4.1 g m = 2 K n (VGS − VTN ) and I D = K n (VGS − VTN )

2

0.75 = 0.5 (VGS − 0.8 ) ⇒ VGS = 2.025 V g m = 2 ( 0.5 )( 2.025 − 0.8 ) ⇒ g m = 1.22 mA / V 2

ro =

1

λ I DQ

1 (0.01)(0.75) = 133 k Ω ro = 133 k Ω =

EX4.2 Av = − g m RD g m = 2 K n I DQ = 2

( 0.5)( 0.4 ) = 0.8944 mA/V

Av = − ( 0.8944 )(10 ) = −8.94

EX4.3 (a)

⎛ R2 ⎞ ⎛ 320 ⎞ VGS = ⎜ ⎟ VDD = ⎜ ⎟ ( 5 ) = 1.905 V R + R ⎝ 520 + 320 ⎠ 2 ⎠ ⎝ 1 I DQ = 0.20 (1.905 − 0.8 ) = 0.244 mA 2

g m = 2 K n I DQ = 2

( 0.2 )( 0.244 ) = 0.442 mA/V

ro = ∞

(b)

Av = − g m RD = − ( 0.422 )(10 ) = −4.22

(c) (d)

Ri = R1 & R2 = 520 & 320 = 198 K RO = RD = 10 K

EX4.4 At transition point, I D = 1 mA I D = K n (VGSt − VTN ) = K n (VDS ( sat ) ) 2

2

1 = 0.2 (VDS ( sat ) ) ⇒ VDS ( sat ) = 2.236 V 2

5 − 2.236 + 2.236 = 3.62 V 2 5 − 3.62 RD = = 2.76 K 0.5

Want VDSQ =

0.5 = 0.2 (VGSQ − 0.8 ) ⇒ VGSQ = 2.38 V 2

⎛ R2 ⎞ 1 VGSQ = ⎜ ⎟ VDD = ( R1 & R2 ) VDD R1 ⎝ R1 + R2 ⎠ 1 So 2.38 = ( 200 )( 5 ) ⇒ R1 = 420 K and R2 = 382 K R1

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Av = − g m RD

( 0.2 )( 0.5 ) = 0.6325 mA/V Av = − ( 0.6325 )( 2.76 )

g m = 2 K n I DQ = 2 = −1.75

EX4.5 (a) ⎛ R2 ⎞ 250 ⎛ ⎞ VG = ⎜ ⎟ (10 ) − 5 = ⎜ ⎟ (10 ) − 5 = −3 V + + 250 1000 R R ⎝ ⎠ ⎝ 1 2 ⎠ (V − VGS ) − ( −5 ) 2 = K n (VGS − VTN ) ID = G 2 −3 − VGS + 5 = 2 ( 0.5 )(VGS − 0.6 ) 2 − VGS = VGS2 − 1.2VGS + 0.36 VGS2 − 0.2VGS − 1.64 = 0 VGS =

0.2 ±

( 0.04 ) + 4 (1.64 ) 2

2

= 1.385 V

I DQ = ( 0.5 )(1.385 − 0.6 ) ⇒ I DQ = 0.308 mA VDSQ = 10 − ( 0.308 )(10 + 2 ) ⇒ VDSQ = 6.30 V 2

(b) Av =

− g m RD 1 + g m RS

g m = 2 K n I DQ = 2

( 0.5 )( 0.308 )

g m = 0.7849 mA/V Av =

− ( 0.7849 )(10 )

1 + ( 0.7849 )( 2 )

⇒

Av = −3.05

EX4.6 VSDQ = 3 V and I DQ = 0.5 mA ⇒ RD = I DQ = K P (VSG − VTP

)

5−3 ⇒ RD = 4 kΩ 0.5

2

0.5 = 1(VSG − 1) ⇒ VSG = 1.71 V ⇒ VGG = 5 − 1.71 ⇒ VGG = 3.29 V 2

Av = − g m RD g m = 2 K P I DQ = 2 (1)( 0.5 ) g m = 1.414 mA/V

Av = − (1.414 )( 4 ) ⇒ Av = −5.66 Av =

v0 −vsd 0.46sin ω t = =− = −5.66 ⇒ vi = 0.0813sin ω t vi vi vi

EX4.7 a. VSG = 9 − I DQ RS , I DQ = K P (VSG − VTP VSG = 9 − ( 2 )(1.2 )(VSG − 2 )

)

2

2

2 = 9 − 2.4 (VSG − 4VSG + 4 )

2 2.4VSG − 8.6VSG + 0.6 = 0

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(8.6 ) − 4 ( 2.4 )( 0.6 ) 2 ( 2.4 ) 2 VSG = 3.51 V, I DQ = 2 ( 3.51 − 2 ) ⇒ I DQ = 4.57 mA VSDQ = 9 + 9 − I DQ (1.2 + 1) = 18 − ( 4.57 )( 2.2 ) ⇒ VSDQ = 7.95 V VSG =

2

8.6 ±

b. V0

 VSG Vi

gmVSG



 

RD

RS

CS

( 2 )( 4.57 ) = 6.046 mA/V

g m = 2 K P I DQ = 2

V0 = g mVSG RD Av = − g m RD = − ( 6.046 )(1) ⇒ Av = −6.05

EX4.8 VDSQ = VDD − I DQ RS 5 = 10 − (1.5 ) RS ⇒ RS = 3.33 kΩ I DQ = K n (VGS − VTN ) ⇒ 1.5 = (1)(VGS − 0.8 ) 2

2

⎛ R2 VGS = 2.025 V = VG − VS = VG − 5 ⇒ VG = 7.025 V = ⎜ ⎝ R1 + R2 So R2 = 281 kΩ, R1 = 119 kΩ

Neglecting RSi , Av =

⎞ R2 ⋅10 ⎟ VDD = 400 ⎠

g m ( RS & r0 )

1 + g m ( RS & r0 )

−1

−1

r0 = ⎡⎣ λ I DQ ⎤⎦ = ⎡⎣( 0.015 )(1.5 ) ⎤⎦ = 44.4 kΩ RS & r0 = 3.33 & 44.4 = 3.1 kΩ g m = 2 K n I DQ = 2 (1)(1.5 ) = 2.45 mA / V Av =

( 2.45)( 3.1) ⇒ Av = 0.884 1 + ( 2.45 )( 3.1)

EX4.9 I DQ = K P (VSG − VTP

)

2

3 = 2 (VSG − 2 ) ⇒ VSG = 3.22 V 5 − VSG 5 − 3.22 I DQ = ⇒3= ⇒ RS = 0.593 kΩ RS RS 2

−1

−1

r0 = ⎡⎣ λ I DQ ⎤⎦ = ⎡⎣( 0.02 )( 3) ⎤⎦ = 16.7 kΩ

( 2 )( 3) = 4.9 mA / V g m ( r0 & RS ) RL = ∞, Av = 1 + g m ( r0 & RS )

g m = 2 K P I DQ = 2

For

r0 & RS = 16.7 & 0.593 = 0.573 kΩ

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Av =

( 4.9 )( 0.573) ⇒ Av = 0.737 1 + ( 4.9 )( 0.573)

If Av is reduced by 10% ⇒ Av = 0.737 − 0.0737 = 0.663 g m ( r0 RS RL )

Av =

1 + g m ( r0 RS RL )

Let r0 & RS & RL = x 0.663 =

( 4.9 ) x ⇒ 0.663 = 4.9 x (1 − 0.663) 1 + ( 4.9 ) x

x = 0.402 = 0.573 RL 0.573RL = 0.402 ⇒ ( 0.573 − 0.402 ) RL = ( 0.402 )( 0.573) ⇒ RL = 1.35 kΩ RL + 0.573

EX4.10 ⎛ R2 ⎞ 9.3 ⎞ ⎛ VG = ⎜ ⎟ VDD = ⎜ ⎟ (5) ⎝ 70.7 + 9.3 ⎠ ⎝ R1 + R2 ⎠ = 0.581 V I DQ = K p (VSG − VTP

)

2

= K P (VS − VG − VTP =

)

2

5 − VS RS

Then ( 0.4 )( 5 )(VS − 0.581 − 0.8 ) = 5 − VS 2

2 (VS − 1.381) = 5 − VS 2

2 (VS2 − 2.762VS + 1.907 ) = 5 − VS 2VS2 − 4.52VS − 1.19 = 0

VS =

4.52 ±

( 4.52 ) + 4 ( 2 )(1.19 ) 2 ( 2) 2

VS = 2.50 V ⇒ I DQ = g m = 2 K P I DQ = 2 Av = =

5 − 2.5 = 0.5 mA 5

( 0.4 )( 0.5 ) = 0.894 mA / V

g m RS R1 & R2 ⋅ 1 + g m RS R1 & R2 + RSi

( 0.894 )( 5) 70.7 & 9.3 ⋅ ⇒ Av = 0.770 1 + ( 0.894 )( 5 ) 70.7 & 9.3 + 0.5

Neglecting RSi , Av = 0.817 R0 = RS

1 1 =5 = 5 & 1.12 ⇒ R0 = 0.915 kΩ 0.894 gm

EX4.11 gmVsg V0

 Vi

 

RS

Vsg

RD

RL



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V0 = g mVsg ( RD & RL ) and Vsg = Vi Av = g m ( RD & RL ) I DQ =

5 − VSG = K p (VSG − VTP RS

5 − VSG = (1)( 4 )(VSG − 0.8 )

)

2

2

2 5 − VSG = 4 (VSG − 1.6VSG + 0.64 ) 2 4VSG − 5.4VSG − 2.44 = 0

VSG =

5.4 ± (5.4) 2 + ( 4 )( 4 )( 2.44 ) 2 ( 4)

VSG = 1.71 V 5 − 1.71 I DQ = = 0.822 mA 4 g m = 2 K p I DQ = 2 (1)( 0.822 ) = 1.81 mA / V Av = (1.81)( 2 & 4 ) = (1.81)(1.33) ⇒ Av = 2.41 Rin = RS

1 1 =4 = 4 & 0.552 ⇒ Rin = 0.485 kΩ gm 1.81

EX4.12 Kn2 =

μ n Cox ⎛ W ⎞

Av = −

2

⋅ ⎜ ⎟ = ( 0.015 )( 2 ) = 0.030 mA / V 2 ⎝ L ⎠2

K n1 K = −6 ⇒ n1 = 36 Kn2 Kn2

K n1 = ( 36 )( 0.030 ) = 1.08 mA / V 2 ⎛W ⎞ ⎛W ⎞ 1.08 = ( 0.015 ) ⎜ ⎟ ⇒ ⎜ ⎟ = 72 ⎝ L ⎠1 ⎝ L ⎠1 The transition point is found from vGSt − 1 = (10 − 1) − ( 6 )( vGSt − 1) 10 − 1 + 6 + 1 = 2.29 V 1+ 6 2.29 − 1 For Q-point in middle of saturation region VGS = + 1 ⇒ VGS = 1.645 V 2 vGSt =

EX4.13 (a) Transition points: For M 2 : vOtB = VDD − VTNL = 5 − 1.2 = 3.8 V 2 2 For M 1 : K n1 ⎡( vOtA ) (1 + λ vOtA ) ⎤ = K n 2 ⎡(VTNL ) (1 + λ2 [VDD − vOtA ]) ⎤ ⎣ ⎦ ⎣ ⎦ 2 2 3 ⎤⎦ = 25 ⎡(1.2 ) (1 + ( 0.01)( 5 ) − ( 0.01) vOtA ) ⎤ + ( 0.01) vOtA 250 ⎡⎣ vOtA ⎣ ⎦ 2 3 3 2 ⎤⎦ = 1.512 − 0.0144vOtA ( 0.01) vOtA + ( 0.01) vOtA + vOtA + 0.00144vOtA − 0.512 = 0 10 ⎡⎣vOtA

which yields vOtA ≅ 0.388 V

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3.8 − 0.388 + 0.388 ⇒ VDSQ1 = 2.094 V 2 2 2 K n1 ⎡(VGS 1 − VTND ) (1 + λ1vO ) ⎤ = K n 2 ⎡(VTNL ) (1 + λ2 [VDD − vO ]) ⎤ ⎣ ⎦ ⎣ ⎦

Then middle of saturation region V0Q =

250 ⎡(VGS 1 − 0.8 ) (1 + [ 0.01][ 2.094]) ⎤ = 25 ⎡(1.2 ) (1 + [ 0.01][5 − 2.094]) ⎤ ⎣ ⎦ ⎣ ⎦ 2

2

2 10 ⎡(VGS1 − 0.8 ) = (1.0209 ) ⎤ = 1.482 ⎣ ⎦

(VGS1 − 0.8 )

2

= 0.145 ⇒ VGS1 = 1.18 V

I DQ = K n1 ⎡(VGS1 − 0.8 ) (1 + ( 0.01)( 2.094 ) ) ⎤ ⎣ ⎦ 2

b.

2 I DQ = ( 0.25 ) ⎡( 0.145 ) (1.02094 ) ⎤ ⇒ I DQ = 37.0 μ A ⎣ ⎦ − g m1 Av = = − g m1 ( r01 & r02 ) I DQ ( λ1 + λ2 )

c.

g m1 = 2 K n1 (VGS 1 − VTND ) = 2 ( 0.25 )(1.18 − 0.8 ) = 0.19 mA/V Av =

−0.19 ⇒ Av = −257 ( 0.037 )( 0.01 + 0.01)

EX4.14 Av = − g m ( ron & rop ) ron = rop =

1 = 666.7 K ( 0.015)( 0.1)

−250 = − g m ( 666.7 & 666.7 )

g m = 0.75 mA/V = 2 K n I DQ = 2 K n ( 0.1) K n = 1.406 mA/V 2 =

k n′ ⎛ W ⎜ 2⎝L

⎞ ⎛ 0.080 ⎞ ⎛ W ⎞ ⎟=⎜ ⎟⎜ ⎟ ⎠ ⎝ 2 ⎠⎝ L ⎠

⎛W ⎞ ⎜ ⎟ = 35.2 ⎝ L ⎠1

EX4.15 (a) RO = So g m1 =

1 1 ro 2 ro1 ≈ g m1 g m1 1 1 = = 0.5 mA/V R0 2

g m1 = 2 K n I D

( 0.2 ) I D

0.5 = 2

(b) Av =

g m1 ( ro1 & ro 2 )

1 + g m1 ( ro1 & ro 2 )

ro1 = ro 2 = Av =

⇒ I D = 0.3125 mA

1 = 320 K ( 0.01)( 0.3125 )

0.5 ( 320 & 320 )

1 + ( 0.5 )( 320 & 320 )

Av = 0.988

EX4.16

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(a) 1 ro1 2 K n I D + λ1 I D Av = = 1 1 λ2 I D + λ1 I D + ro 2 ro1 g m1 +

120 =

2 0.2 I D + 0.01I D 0.01I D + 0.01I D

2.4 I D − 0.01I D = 2 0.2 I D 2.39 I D = 2 0.2 ⇒ I D = 0.140 mA g m1 = 2

( 0.2 )( 0.14 ) ⇒ g m1 = 0.335 mA/V

(b) Ro = ro1 & ro 2 ro1 = ro 2 =

1 = 714 K 0.01 ( )( 0.14 )

Ro = 357 K

EX4.17 R0 = RS 2

1 gm2

g m 2 = 0.632 mA/V, RS 2 = 8 kΩ R0 = 8

1 = 8 &1.58 ⇒ R0 = 1.32 kΩ 0.632

EX4.18 a.

I DQ1 = K n1 (VGS 1 − VTN 1 )

2

1 = 1.2 (VGS 1 − 2 ) ⇒ VGS 1 = VGS 2 = 2.91 V 2

RS = 10 kΩ ⇒ VS1 = I DQ RS − 10 = (1)(10 ) − 10 = 0

⎛ ⎞ R3 VG1 = 2.91 = ⎜ ⎟ (10 ) ⎝ R1 + R2 + R3 ⎠ ⎛ R ⎞ = ⎜ 3 ⎟ (10 ) ⇒ R3 = 145.5 kΩ ⎝ 500 ⎠ VDSQ1 = 3.5 ⇒ VS 2 = 3.5 V ⇒ VG 2 = 3.5 + 2.91 ⇒ VG 2 = 6.41 ⎛ R2 + R3 ⎞ VG 2 = ⎜ ⎟ (10 ) = 6.41 ⎝ R1 + R2 + R3 ⎠ ⎛ R + R3 ⎞ =⎜ 2 ⎟ (10 ) ⎝ 500 ⎠ R2 + R3 = 320.5 = R2 + 145.5 ⇒ R2 = 175 kΩ

Then R1 + R2 + R3 = 500 = R1 + 175 + 145.5 ⇒ R1 = 179.5 kΩ Now VS 2 = 3.5 ⇒ VD 2 = VS 2 + VSDQ 2 = 3.5 + 3.5 = 7 V So RD = b.

10 − 7 ⇒ RD = 3 kΩ 1 From Example 6-18:

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Av = − g m1 RD g m1 = 2 K n1 I DQ = 2 (1.2 )(1) = 2.19 mA / V Av = − ( 2.19 )( 3) ⇒ Av = −6.57

EX4.19 VS = I DQ RS = (1.2 )( 2.7 ) = 3.24 V VD = VS + VDSQ = 3.24 + 12 = 15.24 20 − 15.24 RD = ⇒ RD = 3.97 kΩ 1.2 ⎛ V ⎞ I D = I DSS ⎜ 1 − GS ⎟ VP ⎠ ⎝

2

2

⎛ V ⎞ V 1.2 = 4 ⎜1 − GS ⎟ ⇒ GS = 0.4523 V VP P ⎠ ⎝ VGS = ( 0.4523)( −3) = −1.357 VG = VS + VGS = 3.24 − 1.357 = 1.883 V ⎛ R2 ⎞ ⎛ R2 ⎞ VG = ⎜ ⎟ ( 20 ) = ⎜ ⎟ ( 20 ) = 1.88 ⇒ R2 = 47 kΩ. R1 = 453 kΩ ⎝ 500 ⎠ ⎝ R1 + R2 ⎠ 1 1 = = 167 kΩ r0 = λ I DQ ( 0.005 )(1.2 ) 2 I DSS ⎛ VGS ⎞ 2 ( 4 ) ⎛ 1.357 ⎞ = 1− 1− ⎟ = 1.46 mA/V 3 ⎠ ( −VP ) ⎜⎝ VP ⎟⎠ 3 ⎜⎝ Av = − g m ( r0 & RD & RL ) = − (1.46 )(167 & 3.97 & 4 ) ⇒ Av = −2.87

gm =

EX4.20 a. 2

2

⎛ V ⎞ ⎛ V ⎞ V I DQ = I DSS ⎜ 1 − GS ⎟ 2 = 8 ⎜1 − GS ⎟ ⇒ GS = 0.5 V V VP P ⎠ P ⎠ ⎝ ⎝ VGS = ( 0.5 )( −3.5 ) ⇒ VGS = −1.75

Also I DQ =

−VGS − ( −10 ) RS

2=

1.75 + 10 ⇒ RS = 5.88 kΩ RS

b.  Vi

 

gmVgs Vgs 

V0 RS

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gm = r0 =

2 I DSS VP

⎛ VGS ⎜1 − ⎝ VP

⎞ 2 ( 8 ) ⎛ 1.75 ⎞ ⎟= ⎜1 − ⎟ = 2.29 mA/V 3.5 ⎠ ⎠ 3.5 ⎝

1 = 50 kΩ ( 0.01)( 2 )

Vi = Vgs + g m RS Vgs ⇒ Vgs = Av =

c. Av =

Vi 1 + g m RS

( 2.29 ) [5.88 & 50] V0 g m RS & r0 = = ⇒ Av = 0.9234 V1 1 + g m RS & r0 1 + ( 2.29 ) [5.88 & 50] g m ( RS & RL ) & ro

1 + g m ( RS & RL & ro )

= ( 0.80 )( 0.9234 ) = 0.7387

( 2.29 )( RS & RL & ro ) = 0.7387 ⇒ RS & RL & ro = 1.235 kΩ 1 + ( 2.29 )( RS & RL & ro ) RS & ro = 5.261 kΩ 5.261RL = 1.235 kΩ ⇒ RL = 1.61 kΩ 5.261 + RL

TYU4.1 g m = 2 K n (VGS − VTN ) and I D = K n (VGS − VTN ) ⇒ VGS − VTN = 2

and g m = 2 K n

I DQ Kn

I DQ Kn

= 2 K n I DQ

( 3.4 ) g2 = 1.445 mA / V Kn = m = 4 I DQ 4 ( 2) 2

K

n

=

μ n Cox W 2

⋅

L

⎛W 1.445 = ( 0.018 ) ⎜ ⎝L

⎞ W = 80.3 ⎟⇒ L ⎠ −1

2 ro = ⎡λ K n (VGS − VTN ) ⎤ = ⎡⎣λ I DQ ⎤⎦ ⎣ ⎦

−1

−1

ro = ⎡⎣( 0.015 )( 2 ) ⎤⎦ ⇒ ro = 33.3 k Ω

TYU4.2

a.

I DQ = K n (VGS − VTN )

2

0.4 = 0.5 (VGS − 2 ) ⇒ VGS = 2.894 V 2

VDSQ = VDD − I DQ RD = 10 − ( 0.4 )(10 ) ⇒ VDSQ = 6 V

b.

g m = 2 K n (VGS − VTN ) = 2 ( 0.5 )( 2.894 − 2 ) ⇒ g m = 0.894 mA/V −1

r0 = ⎡⎣ λ I DQ ⎤⎦ , λ = 0 ⇒ r0 = ∞ v Av = 0 = − g m RD = − ( 0.894 )(10 ) ⇒ Av = −8.94 vi

c.

vi = 0.4sin ω t ⇒ vds = − ( 8.94 )( 0.4 ) sin ω t vds = −3.58sin ω t

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At VDS 1 = 6 − 3.58 = 2.42 VGS1 = 2.89 + 0.4 = 3.29

VGS1 − VTN = 3.29 − 2 = 1.29 = VDS ( sat )

So VDS 1 > VGS 1 − VTN ⇒ Biased in saturation region

TYU4.3 I DQ = K n (VGS − VTN ) = ( 0.25 )( 2 − 0.8 ) ⇒ I DQ = 0.36 mA 2

a.

2

VDSQ = VDD − I DQ RD = 5 − ( 0.36 )( 5 ) ⇒ VDSQ = 3.2 V

b.

g m = 2 K n (VGS − VTN ) = 2 ( 0.25 )( 2 − 0.8 ) ⇒ g m = 0.6 mA/V,

c.

Av =

r0 = ∞

v0 = − g m RD = − ( 0.6 )( 5 ) ⇒ Av = −3.0 vi

TYU4.4 vi = vgs = 0.1sin ω t id = g m vgs = ( 0.6 )( 0.1) sin ω t

id = 0.06sin ω t mA vds = ( −3)( 0.1) sin ω t = −0.3sin ω t ( V ) Then iD = I DQ + id = 0.36 + 0.06sin ω t = iD mA vDS = VDSQ + vds = 3.2 − 0.3sin ω t = vDS

TYU4.5 a. VSDQ = VDD − I DQ RD 7 = 12 − I DQ ( 6 ) ⇒ I DQ = 0.833 mA I DQ = K P (VSG − VTP

)

2

0.833 = 2 (VSG − 1) ⇒ VSG = 1.65 V 2

g m = 2 K P (VSG − VTP ) = 2 ( 2 )(1.65 − 1) ⇒ g m = 2.58 mA/V, r0 = ∞

b.

Av =

v0 = − g m RD = − ( 2.58 )( 6 ) ⇒ Av = −15.5 vi V0

 Vi

 

gmVsg

VSG

RD



TYU4.6 I DQ = K n (VGS − VTN ) ⇒ VGS − VTN = 2

g m = 2 K n (VGS − VTN ) = 2 K n

I DQ Kn

I DQ Kn

So g m = 2 K n I DQ

TYU4.7

η=

γ 2 2φ f + vSB

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η=

(a)

0.40 2 2 ( 0.35 ) + 1

η=

(b)

⇒ η = 0.153

0.40 2 2 ( 0.35 ) + 3

⇒ η = 0.104

( 0.5 )( 0.75) = 1.22 mA / V g mb = g m η = (1.22 )( 0.153) ⇒ g mb = 0.187 mA / V g mb = (1.22 )( 0.104 ) ⇒ g mb = 0.127 mA / V g m = 2 K n I DQ = 2

For (a), For (b),

TYU4.8 I DQ = I Q = 0.5 mA W Let = 25 L K n = ( 20 )( 25 ) = 500 μ A / V 2 0.5 + 1.5 = 2.5 V ⇒ VS = −2.5 V 0.5 Av = − g m RD

VGS =

g m = 2 ( 0.5 )( 2.5 − 1.5 ) = 1 mA/V

For Av = −4.0 ⇒ RD = 4 kΩ

VD = 5 − ( 0.5 )( 4 ) = 3 V ⇒ VDSQ = 3 − ( −2.5 ) = 5.5 V

TYU4.9 a. With RG ⇒ VGS = VDS ⇒ transistor biased in sat. region I D = K n (VGS − VTN ) = K n (VDS − VTN ) 2

2

VDS = VDD − I D RD = VDD − K n RD (VDS − VTN ) VDS = 15 − ( 0.15 )(10 )(VDS − 1.8 )

2

2

2 = 15 − 1.5 (VDS − 3.6VDS + 3.24 )

2 1.5VDS − 4.4VDS − 10.14 = 0

VDS =

4.4 ±

( 4.4 )

2

+ ( 4 )(1.5 )(10.14 )

2 (1.5 )

⇒ VDSQ = 4.45 V

I DQ = ( 0.15 )( 4.45 − 1.8 ) ⇒ I DQ = 1.05 mA 2

b. Neglecting effect of RG: Av = − g m ( RD & RL ) g m = 2 K n (VGS − VTN ) = 2 ( 0.15 )( 4.45 − 1.8 ) ⇒ g m = 0.795 mA/V Av = − ( 0.795 )(10 & 5 ) ⇒ Av = −2.65 c.

RG ⇒ establishes VGS = VDS ⇒ essentially no effect on small-signal voltage gain.

TYU4.10 a. 2 I DQ = K n (VGS − VTN ) I DQ = 0.8 ( 2 − VSG ) = 2

VSG VSG = 4 RS

2 3.2 ( 4 − 4VSG + VSG ) = VSG

2 3.2VSG − 13.8VSG + 12.8 = 0

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(13.8) − 4 ( 3.2 )(12.8 ) 2 ( 3.2 ) 2 = 1.35 V ⇒ I DQ = 0.8 ( 2 − 1.35 ) ⇒ I DQ = 0.338 mA

VSG = VSG

2

13.8 ±

b. VDSQ = VDD − I DQ ( RD + RS ) 6 = 10 − ( 0.338 )( RD + 4 ) RD =

10 − ( 0.338 )( 4 ) − 6 0.338

⇒ RD = 7.83 kΩ

c. V0

 Vgs Vi

 

gmVgs RD

 RS

Vi = Vgs + g mVgs RS ⇒ Vgs =

Vi 1 + g m RS

V0 = − g mVgs RD g m = 2 K n (VGS − VTN ) = 2 ( 0.8 )( −1.35 + 2 ) = 1.04 mA/V Av =

− (1.04 )( 7.83) V0 − g m RD = = ⇒ Av = −1.58 1 + (1.04 )( 4 ) Vi 1 + g m RS

TYU4.11 a. 5 = I DQ RS + VSG and I DQ = K p (VSG + VTP ) 2 0.8 = 0.5(VSG + 0.8) 2 ⇒ VSG = 0.465 V 5 = ( 0.8 ) RS + 0.465 ⇒ RS = 5.67 kΩ VSDQ = 10 − I DQ ( RS + RD ) 3 = 10 − ( 0.8 )( 5.67 + RD ) RD =

10 − ( 0.8 )( 5.67 ) − 3 0.8

⇒ RD = 3.08 kΩ

b. V0

 Vi

 

VSG

gmVsg

r0

RD



V0 = g mVsg ( RD & r0 ) = − g mVi ( RD & r0 ) V Av = 0 = − g m ( RD & r0 ) Vi g m = 2 K p (VSG + VTP ) = 2(0.5)(0.465 + 0.8) = 1.265 mA/V 1 1 = = 62.5 kΩ r0 = λ I 0 ( 0.02 )( 0.8 ) Av = − (1.265 )( 3.08 & 62.5 ) ⇒ Av = −3.71

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TYU4.12 (a) V0 = g mVgs r0 Vi = Vgs + V0 ⇒ Vgs = Vi − V0 So V0 = g m r0 (Vi − V0 ) Av =

( 4 )( 50 ) V0 g m r0 = = ⇒ Av = 0.995 Vi 1 + g m r0 1 + ( 4 )( 50 )

 Vgs  Ix

I x + g mVgs = I x = g mVx +

Vx

Vx and Vgs = −Vx r0

Vx 1 1 ⇒ R0 = r0 = 50 ⇒ R0 ≅ 0.25 kΩ 4 r0 gm

With RS = 4 kΩ ⇒ Av =

(b)

 

r0

gmVgs

r0 || Rs = 50 || 4 = 3.7 kΩ ⇒ Av =

g m ( r0 & RS )

1 + g m ( r0 & RS )

( 4 )( 3.7 )

1 + ( 4 )( 3.7 )

⇒ Av = 0.937

TYU4.13 (a) g m = 2 K n I DQ ⇒ 2 = 2 K n ( 0.8 ) ⇒ K n = 1.25 mA / V 2 Kn = So

μ n Cox W 2

⋅

⎛W ⎞ ⇒ 1.25 = ( 0.020 ) ⎜ ⎟ L ⎝L⎠

W = 62.5 L

I DQ = K n (VGS − VTN ) ⇒ 0.8 = 1.25 (VGS − 2 ) ⇒ VGS = 2.8 V 2

2

b. −1

−1

r0 = ⎣⎡ λ I DQ ⎦⎤ = ⎡⎣( 0.01)( 0.8 ) ⎤⎦ = 125 kΩ g m ( r0 & RL ) Av = 1 + g m ( r0 & RL ) r0 & RL = 125 & 4 = 3.88 Av =

( 2 )( 3.88 ) ⇒ Av = 0.886 1 + ( 2 )( 3.88 )

R0 =

1 1 r0 = 125 ⇒ R0 ≈ 0.5 kΩ gm 2

TYU4.14

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Rin =

1 = 0.35 kΩ ⇒ g m = 2.86 mA/V gm

V0 4 RD = RD & RL = 2.4 = RD & 4 = Ii 4 + RD

( 4 − 2.4 ) RD = ( 2.4 )( 4 ) ⇒ RD = 6 kΩ g m = 2 K n I DQ 2.86 = 2 K n ( 0.5 ) ⇒ K n = 4.09 mA / V 2 I DQ = K n (VGS − VTN )

2

0.5 = 4.09 (VGS − 1) ⇒ VGS = 1.35 V ⇒ VS = −1.35 V, VD = 5 − ( 0.5 )( 6 ) = 2 V 2

VDS = VD − VS = 2 − ( −1.35 ) = 3.35 V

We have VDS = 3.35 > VGS − VTN = 1.35 − 1 = 0.35 V ⇒ Biased in the saturation region TYU4.15 μC K n1 = n ox 2 Kn2

⎛W ⋅⎜ ⎝L μ C ⎛W = n ox ⋅ ⎜ 2 ⎝L

Av = −

⎞ 2 ⎟ = ( 0.020 )( 80 ) = 1.6 mA / V ⎠1 ⎞ 2 ⎟ = ( 0.020 )(1) = 0.020 mA / V ⎠2

K n1 1.6 =− ⇒ Av = −8.94 Kn2 0.020

The transition point is determined from vGSt − VTND = VDD − VTNL −

K n1 ( vGSt − VTND ) Kn2

vGSt − 0.8 = ( 5 − 0.8 ) − ( 8.94 )( vGSt − 0.8 ) vGSt =

( 5 − 0.8) + ( 8.94 )( 0.8) + 0.8 1 + 8.94

vGSt = 1.22 V

For Q-point in middle of saturation region VGS =

1.22 − 0.8 + 0.8 ⇒ VGS = 1.01 V 2

TYU4.16 a. I DQ 2 = 2mA, VDSQ 2 = 10 V I DQ 2 ⋅ RS 2 = 10 = 2 RS 2 ⇒ RS 2 = 5 kΩ I DQ 2 = K n 2 (VGS 2 − VTN 2 )

2

2 = 1(VGS 2 − 2 ) ⇒ VSG 2 = 3.41 V ⇒ VG 2 = 3.41 V 2

10 − 3.41 ⇒ RD1 = 3.29 kΩ 2 = 10 V ⇒ VS 1 = 3.41 − 10 = −6.59 V

Then RD1 = For VDSQ1

Then RS 1 =

−6.59 − ( −10 ) 2

I D1 = K n1 (VGS 1 − VTN 1 )

⇒ RS1 = 1.71 kΩ

2

2 = 1(VGS 1 − 2 ) ⇒ VGS 1 = 3.41 V 2

⎛ R2 ⎞ R2 1 VGS1 = ⎜ = ⋅ Rin ⎟ ( 20 ) − I DQ1 RS 1 R R R R R + + ⎝ 1 2 ⎠ 1 2 1 1 3.41 = ( 200 )( 20 ) − ( 2 )(1.71) ⇒ R1 = 586 kΩ R1

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586 R2 = 200 ⇒ ( 586 − 200 ) R2 = ( 200 )( 586 ) ⇒ R2 = 304 kΩ 586 + R2 g m1 = 2 K n1 I DQ1 = 2 (1)( 2 ) ⇒ g m1 = g m 2 = 2.828 mA/V

b.

From Example 6.17 − g g R ( R & RL ) Av = m1 m 2 D1 S 2 1 + g m 2 ( RS 2 & RL ) RS 2 & RL = 5 & 4 = 2.222 kΩ − ( 2.828 )( 2.828 )( 3.29 )( 2.222 )

Av =

1 + ( 2.828 )( 2.222 )

⇒ Av = −8.03

1 1 RS 2 = 5 = 0.3536 & 5 ⇒ R0 = 0.330 kΩ gm2 2.828

R0 =

TYU4.17 From Example 6.19 g m = 3.0 mA/V, r0 = 41.7 kΩ R1 & R2 = 420 & 180 = 126 kΩ ⎛ R1 & R2 ⎞ Vgs = ⎜ ⎟ Vi ⎝ R1 & R2 + Ri ⎠ ⎛ 126 ⎞ =⎜ ⎟ Vi = 0.863Vi ⎝ 126 + 20 ⎠ − g mVgs ( r0 & RD & RL ) Av = Vi

= − ( 3.0 )( 0.863)( 41.7 & 2.7 & 4 ) = − ( 2.589 )( 41.7 & 1.61) = − ( 2.589 )(1.55 ) ⇒ Av = −4.01

TYU4.18 a. ⎛ R2 ⎞ VG1 = ⎜ ⎟ (VDD ) ⎝ R1 + R2 ⎠ ⎛ 430 ⎞ VG1 = ⎜ ⎟ ( 20 ) = 17.2 V ⎝ 430 + 70 ⎠ 2

⎛ V ⎞ ⎛ V −V ⎞ I DQ1 = I DSS ⎜ 1 − GS ⎟ = 6 ⎜ 1 − G1 S1 ⎟ V 2 ⎝ ⎠ P ⎠ ⎝ 2

2

2

20 − VS1 ⎛ 17.2 VS 1 ⎞ ⎛ VS 1 ⎞ = 6 ⎜1 − + ⎟ = 6 ⎜ 2 − 7.6 ⎟ and I DQ1 = 2 2 4 ⎝ ⎠ ⎝ ⎠ 2

20 − VS 1 ⎛V ⎞ = 6 ⎜ S1 − 7.6 ⎟ 4 2 ⎝ ⎠ 2 ⎛V ⎞ 20 − VS1 = 24 ⎜ S 1 − 7.6VS 1 + 57.76 ⎟ 4 ⎝ ⎠

Then

= 6VS21 − 182.4VS 1 + 1386.24 6VS21 − 181.4VS 1 + 1366.24 = 0 VS 1 =

181.4 ±

(181.4 ) − 4 ( 6 )(1366.24 ) 2 ( 6) 2

VS 1 = 14.2 V ⇒ VGS 1 = 17.2 − 14.2 = 3 V > VP

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So VS1 = 16.0 ⇒ VGS1 = 17.2 − 16 = 1.2 < VP − Q 20 − 16 ⇒ I DQ1 = 1 mA 4 = 20 − I DQ1 ( RS 1 + RD1 ) = 20 − (1)( 8 ) ⇒ VSDQ1 = 12 V

on I DQ1 = VSDQ1

VD1 = I DQ1 RD1 = (1)( 4 ) = 4 V = VG 2 2

⎛ V − VS 2 ⎞ ⎛ V ⎞ I DQ 2 = I DSS ⎜ 1 − GS ⎟ = 6 ⎜⎜ 1 − G 2 ⎟ VP ⎠ ( −2 ) ⎠⎟ ⎝ ⎝ 2

2

2

V V ⎛ 4 V ⎞ ⎛ V ⎞ = 6 ⎜ 1 + − S 2 ⎟ = 6 ⎜ 3 − S 2 ⎟ and I DQ 2 = S 2 = S 2 RS 2 2 ⎠ 4 ⎝ 2 2 ⎠ ⎝ Then

VS 2 ⎛ V ⎞ = 6⎜3 − S2 ⎟ 4 2 ⎠ ⎝

2

⎛ V2 ⎞ VS 2 = 24 ⎜ 9 − 3VS 2 + S 2 ⎟ 4 ⎠ ⎝ 2 6VS 2 − 73VS 2 + 216 = 0 VS 2 =

( 73)

73 ±

2

− 4 ( 6 )( 216 )

2 ( 6)

⇒ VS 2 = 7.09 V or = 5.08 V

For VS 2 = 5.08 V ⇒ VGS 2 = 4 − 5.08 = −1.08 transistor biased ON 5.08 ⇒ I DQ 2 = 1.27 mA 4 = 20 − VS 2 = 20 − 5.08 ⇒ VDS 2 = 14.9 V

I DQ 2 = VDS 2

b. Vg2 



gm2Vgs2 Vgs2

Vi

 

RD1



V0

Vsg1 gm1Vsg1

RS2

RL



Vg 2 = g m1Vsg1 RD1 = − g m1V1 RD1 V0 = g m 2Vgs 2 ( RS 2 & RL )

Vg 2 = Vgs 2 + V0 ⇒ Vgs 2 =

Vg 2 1 + g m 2 ( RS 2 & RL )

Av =

V0 − g m1 RD1 = Vi 1 + g m 2 ( RS 2 & RL )

g m1 =

2 I DSS VP

⎛ VGS ⎞ ⎜1 − ⎟ VP ⎠ ⎝

2 ( 6 ) ⎛ 1.2 ⎞ ⎜1 − ⎟ = 2.4 mA/V 2 ⎝ 2 ⎠ 2 ( 6 ) ⎛ 1.08 ⎞ gm2 = ⎜1 − ⎟ = 2.76 mA/V 2 ⎝ 2 ⎠ − ( 2.4 )( 4 ) = −2.05 = Av Then Av = 1 + ( 2.76 )( 4 )( 2 ) =

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Chapter 4 Problem Solutions 4.1 (a) ⎛ μ C ⎞⎛W ⎞ g m = 2 K n I D = 2 ⎜ n ox ⎟ ⎜ ⎟ I D ⎝ 2 ⎠⎝ L ⎠ 0.5 = 2

( 0.040 ) ⎜⎛

ω⎞

W = 3.125 ⎟ ( 0.5 ) ⇒ L ⎝L⎠

(b) 2 ⎛ μ C ⎞⎛W ⎞ I D = ⎜ n ox ⎟ ⎜ ⎟ (VGS − VTN ) L 2 ⎝ ⎠⎝ ⎠

0.5 = ( 0.04 )( 3.125 )(VGS − 0.8 ) ⇒ VGS = 2.80 V 2

4.2 ⎛ μ p Cox ⎞ ⎛ W ⎞ gm = 2 K p I D = 2 ⎜ ⎟ ⎜ ⎟ ID ⎝ 2 ⎠⎝ L ⎠

(a)

2

W ⎛ 50 ⎞ ⎛W ⎞ = 0.3125 ⎜ ⎟ = ( 20 ) ⎜ ⎟ (100 ) ⇒ L ⎝ 2⎠ ⎝L⎠ ⎛ μ p Cox ⎞ ⎛ W ⎞ 2 ID = ⎜ ⎟ ⎜ ⎟ (VSG + VTP ) 2 L ⎝ ⎠ ⎝ ⎠

(b)

100 = ( 20 )( 0.3125 )(VSG − 1.2 ) ⇒ VSG = 5.2 V 2

4.3 I D = K n (VGS − VTN ) (1 + λVDS ) 2

I D1 1 + λVDS1 3.4 1 + λ (10 ) = ⇒ = I D 2 1 + λVDS 2 3.0 1 + λ ( 5 )

3.4 [1 + 5λ ] = 3.0 [1 + 10λ ]

3.4 − 3.0 = λ ( 3 ⋅10 − ( 3.4 ) ⋅ 5 ) ⇒ λ = 0.0308 ro =

ΔVDS 5 = = 12.5 kΩ ΔI D 0.4

4.4 r0 = ID =

1 λ ID 1

λ r0

=

1

( 0.012 )(100 )

⇒ I D = 0.833 mA

4.5 2 I D = K n (VGS − VTN ) (1 + λVDS ) I D1 1 + λVDS 1 = I D 2 1 + λVDS 2 0.20 1 + λ ( 2 ) = 0.22 1 + λ ( 4 )

0.20 (1 + 4λ ) = 0.22 (1 + 2λ )

( 0.8 − 0.44 ) λ = 0.22 − 0.20 λ = 0.0556 V −1

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ro =

ΔVDS 2 = ⇒ ro = 100 K 0.02 ΔI D

4.6 (a) (i)

ro =

1 1 = = 1000 K λ I D ( 0.02 )( 0.05 )

(ii)

ro =

1 = 100 K ( 0.02 )( 0.5)

(b) ΔI D =

(i)

ΔVDS 1 = = 0.001 mA = 1.0 μ A 1000 ro

ΔI D 1.0 ⇒ 2% = 50 ID ΔI D =

(ii)

ΔVDS 1 = = 0.01 ⇒ 10 μ A ro 100

ΔI D 10 = ⇒ 2% ID 500

4.7 I D = 1.0 mA 1 1 ro = = = 100 K λ I D ( 0.01)(1) 4.8 Av = − g m ( r0 || RD ) = − (1)( 50 ||10 ) ⇒ Av = −8.33 4.9 RD =

a.

VDD − VD SQ I DQ

=

10 − 6 ⇒ RD = 8 kΩ 0.5

For VGSQ = 2 V, for example, 2 ⎛ μ C ⎞⎛W ⎞ I DQ = ⎜ n ox ⎟ ⎜ ⎟ (VGSQ − VTN ) 2 L ⎝ ⎠⎝ ⎠ W 2 ⎛W ⎞ 0.5 = ( 0.030 ) ⎜ ⎟ ( 2 − 0.8 ) ⇒ = 11.6 L ⎝L⎠ b.

g m = 2 K n I DQ = 2 K n (VGSQ − VTN )

g m = 2 ( 0.030 )(11.6 )( 2 − 0.8 ) ⇒ g m = 0.835 mA/V ro =

1

c.

=

1

⇒ r = 133 kΩ

( 0.015)( 0.50 ) o Av = − g m ( r0 & RD ) = − ( 0.835 )(133 & 8 ) ⇒ Av = −6.30

λ I DQ

4.10 2

K n vgs2 = K n ⎡⎣Vgs sinω t ⎤⎦ = K nVgs2 sin 2ω t 1 [1 − cos 2ω t ] 2 K nVgs2 So K n vgs2 = [1 − cos 2ω t ] 2 sin 2 ω t =

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K nVgs2 Ratio of signal at 2ω to that at ω :

2

⋅ cos 2ω t

2 K n (VGSQ − VTN ) Vgs ⋅ sin ω t

The coefficient of this expression is then:

Vgs

4 (VGSQ − VTN )

4.11 0.01 =

Vgs

4 (VGSQ − VTN )

So Vgs = ( 0.01)( 4 )( 3 − 1) ⇒ Vgs = 0.08 V

4.12 Vo = − g mVgs ( ro RD ) ⎛ 50 ⎞ ⋅ Vi = ⎜ ⎟ ⋅ Vi = ( 0.9615 ) Vi R1 R2 + RSi ⎝ 50 + 2 ⎠ Av = − g m ( 0.9615 ) ( ro RD ) = − (1)( 0.9615 ) ( 50 10 ) ⇒ Av = −8.01 R1 R2

Vgs =

4.13 Av = − g m ( r0 || RD ) −10 = − g m (100 || 5 ) ⇒ g m = 2.1 mA/V 4.14 a. ⎛ R2 ⎞ VG = ⎜ ⎟ VDD ⎝ R1 + R2 ⎠ ⎛ 200 ⎞ VG = ⎜ ⎟ (12 ) = 4.8 V ⎝ 200 + 300 ⎠ V − VGS 2 = K n (VGS − VTN ) ID = G RS

4.8 − VGS = (1)( 2 ) (VGS2 − 4VGS + 4 ) 2VGS2 − 7VGS + 3.2 = 0

VGS =

7±

(7)

2

− 4 ( 2 )( 3.2 )

2 ( 2)

= 2.96 V

I D = (1)( 2.96 − 2 ) ⇒ I D = 0.920 mA 2

VDS = VDD − I D ( RD + RS )

= 12 − ( 0.92 )( 3 + 2 ) ⇒ VDS = 7.4 V

(b) Vo =

− g mVg ( RD RL ) 1 + g m RS

where Vg =

R1 R2 R1 R2 + RSi

⋅ Vi =

300 200 300 200 + 2

⋅ Vi =

120 ⋅ Vi = ( 0.9836 ) Vi 120 + 2

Then

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Av =

− g m ( 0.9836 )( RD || RL ) 1 + g m RS

g m = 2 (1)( 2.96 − 2 ) = 1.92 mA / V So Av =

− (1.92 )( 0.9836 )( 3 || 3) 1 + (1.92 )( 2 )

⇒ Av = −0.585

c.

AC load line 1 1  Slope  3.5 K 3兩兩3  2 0.92

12

7.4

−1 Δi D = ⋅ Δvds 3.5 kΩ Δi D = 0.92 mA ⇒ Δvds = ( 0.92 )( 3.5 ) = 3.22 ⇒ 6.44 V peak-to-peak

4.15 a. I D Q = 3 mA ⇒ VS = I DQ RS = ( 3)( 0.5 ) = 1.5 V I DQ = K n (VGS − VTN )

2

3 = ( 2 )(VGS − 2 ) ⇒ VGS = 3.225 V 2

VG = VGS + VS = 3.225 + 1.5 = 4.725 V ⎛ R2 ⎞ 1 VG = ⎜ ⎟ VDD = ⋅ Rin ⋅ VDD R1 ⎝ R1 + R2 ⎠ 1 4.725 = ( 200 )(15 ) ⇒ R1 = 635 kΩ R1 635R2 = 200 ⇒ R2 = 292 kΩ 635 + R2 b. − g m ( RD || RL ) Av = 1 + g m RS

g m = 2 ( 2 )( 3.225 − 2 ) = 4.90 mA / V Av =

− ( 4.90 )( 2 || 5 )

1 + ( 4.90 )( 0.5 )

⇒ Av = −2.03

4.16 From Problem 4.14: I D = 0.920 mA VDS = 7.4 V g m = 1.92 mA/V ⎛ R1 || R2 ⎞ Av = − g m ( RD || RL ) ⎜ ⎟ ⎝ R1 || R2 + RSi ⎠ ⎛ 200 || 300 ⎞ = − (1.92 )( 3 || 3) ⎜ ⎟ = − ( 2.88 )( 0.9836 ) ⎝ 200 || 300 + 2 ⎠ Av = −2.83

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AC load line 1 1 Slope   3兩兩3 1.5 K 0.92

12

7.4

−1 ⋅ ΔvDS , ΔiD = 0.92 mA ⇒ ΔvDS = ( 0.92 ) (1.5 ) = 1.38 1.5 kΩ ⇒ 2.76 V peak-to-peak output voltage swing ΔiD =

4.17 (a) Av = − g m RD −15 = −2 RD ⇒ RD = 7.5 K (b) Av = −5 =

− g m RD 1 + g m RS − ( 2 )( 7.5 ) 1 + ( 2 ) RS

⇒ RS = 1 K

4.18 (a)

Av =

− g m RD 1 + g m RS

(1)

−8 =

− g m RD 1 + g m (1)

(2)

−16 = − g m RD

Then 8 =

16 ⇒ 1 + g m (1)

g m = 1 mA/V RD = 16 K

(b)

Av = −10 =

− (1)(16 )

1 + (1) RS

RS = 0.6 K

4.19 a.

AC load line 1 Slope  5K IDQ

VDS(sat)

VDSQ

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VDSQ = V + − I DQ RD − (−VGSQ ) Output Voltage Swing = VDSQ − VDS ( sat ) = ⎡⎣V + − I DQ RD + VGSQ ⎤⎦ − (VGSQ − VTN ) = V + − I DQ RD + VTN So ΔI D = I DQ =

1 1 ⎡V + − I DQ RD + VTN ⎤⎦ ⋅ ΔVDS = 5 kΩ 5 kΩ ⎣

1 ⎡5 − I DQ (10) + 1⎤⎦ 5 kΩ ⎣ = 1.2 − 2 I DQ = I DQ ⇒ I DQ = 0.4 mA = I Q

I DQ =

b.

( 0.5 )( 0.4 ) = 0.8944 mA / V

g m = 2 K n I DQ = 2 r0 =

1 1 = = 250 kΩ λ I DQ ( 0.01)( 0.4 )

Av = − g m ( RD || RL || r0 ) = − ( 0.8944 )(10 ||10 || 250 ) ⇒ Av = −4.38

4.20 (a) I DQ = K n (VGSQ − VTN )

2

0.1 = 0.85 (VGSQ − 0.8 )

2

VGSQ = 1.143 V RS =

−1.143 − ( −5 )

⇒ RS = 38.6 K

0.1

ΔV = Δ I ( RD RL ) ro

ro =

1 = 0.1( RD RL ) ro RD RL ro = 10 K =

1 = 500 K ( 0.02 )( 0.1)

40 500 RD 40 500 + RD

⇒

37.04 RD = 10 37.04 + RD

RD = 13.7 K

(b) gm = 2 Kn I D = 2

( 0.85)( 0.1)

g m = 0.583 mA/V ro = 500 K (c) Av = − g m ( RD RL ro )

= − ( 0.583) (13.7 40 500 ) = − ( 0.583)(10 )

Av = −5.83

4.21 a. 2 I D = K n (VGS − VTN ) 2 = 4 (VGS − ( −1) )

2

VGS = −0.293 V ⇒ VS = 0.293 V = I D RS = (2) RS ⇒ RS = 0.146 kΩ

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VD = VDS + VS = 6 + 0.293 = 6.293 RD =

b. Av =

10 − 6.293 ⇒ RD = 1.85 kΩ 2 − g m ( RD RL ) 1 + g m RS

g m = 2 K n (VGS − VTN ) g m = 2 ( 4 )( −0.293 + 1) = 5.66 mA/V Av =

− ( 5.66 ) (1.85 2 )

1 + ( 5.66 )( 0.146 )

⇒ Av = −2.98

c.

AC load line

1 1.85兩兩2  0.146 1  1.107 K

Slope  2 mA

10

6

Δv0 = ( ΔiD )(1.85 || 2 ) = ( 2 )(1.85 || 2 ) = 1.92 V Δvi =

1.92 = 0.645 ⇒ Vi = 0.645 V 2.98

4.22 a. VDSQ = VDD − I DQ ( RD + RS ) 2.5 = 5 − I DQ ( 2 + RS ) 2.5 I DQ = 2 + RS I DQ = K n (VGS − VTN ) = 2

−VGS −2.5 RS ⇒ VGS = − I DQ RS = 2 + RS RS

2

⎡ −2.5RS ⎤ 2.5 Kn ⎢ − VTN ⎥ = 2 2 R + + RS S ⎣ ⎦ 2

⎡ 2.5 RS ⎤ 2.5 4 ⎢1 − ⎥ = + + RS 2 2 R S ⎦ ⎣ 2

⎡ 2 + RS − 2.5 RS ⎤ 2.5 4⎢ ⎥ = 2 + RS 2 + RS ⎣ ⎦ 4

( 2 − 1.5RS ) 2 + RS

2

= 2.5

4 ( 4 − 6 RS + 2.25 RS2 ) = 2.5 ( 2 + RS ) 9 RS2 − 26.5RS + 11 = 0 RS =

26.5 ±

( 26.5) − 4 ( 9 )(11) 2 (9) 2

RS = 0.5 kΩ or 2.44 kΩ

But RS = 2.44 kΩ ⇒ VGS = −1.37 Cut off. ⇒ RS = 0.5 kΩ,

I DQ = 1.0 mA

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b. Av =

− g m ( RD || RL ) 1 + g m RS

( 4 )(1) = 4 mA / V

g m = 2 K n I DQ = 2 Av =

− ( 4 )( 2 || 2 )

1 + ( 4 )( 0.5 )

⇒ Av = −1.33

4.23 a. 5 = I DQ RS + VSDQ + I DQ RD − 5 5 = I DQ RS + 6 + I DQ (10 ) − 5 1.

I DQ =

4 RS + 10

VS = VSDQ + I DQ RD − 5 = VSGQ

2.

1 + I DQ (10 ) = VSGQ

3.

I DQ = K p (VSGQ − 2 )

2

Choose RS = 10 kΩ ⇒ I DQ =

4 = 0.20 mA 20

VSGQ = 1 + (0.2)(10) = 3 V 0.20 = K P (3 − 2) 2 ⇒ K P = 0.20 mA / V 2

b. I DQ = ( 0.20 )( 3 − 2 ) = 0.20 mA 2

( 0.2 )( 0.2 ) = 0.4 mA / V Av = − g m ( RD || RL ) = − ( 0.4 )(10 ||10 ) ⇒ Av = −2.0

g m = 2 K P I DQ = 2

c. 4 = 0.133 mA 30 = 1 + (0.133)(10) = 2.33 V

Choose RS = 20 kΩ ⇒ I DQ = VSGQ

0.133 = K p (2.33 − 2) 2 ⇒ K p = 1.22 mA / V 2 g m = 2 (1.22)(0.133) = 0.806 mA/V Av = −(0.806)(10 10) ⇒ Av = −4.03

A larger gain can be achieved. 4.24 (a) I DQ = K p (VSGQ + VTP )

2

0.25 = 0.8 (VSGQ − 0.5 )

2

VSGQ = 1.059 V 3 − 1.059 RS = ⇒ RS = 7.76 K 0.25 VD = VS − VSDQ = 1.059 − 1.5 = −0.441 V RD =

−0.441 − ( −3) 0.25

⇒ RD = 10.2 K

(b)

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Av = − g m ( RD RL )

( 0.8)( 0.25 ) = 0.8944 mA/V Av = − ( 0.8944 )(10.2 || 2 )

g m = 2 K p I DQ = 2

Av = −1.50 (c) ΔVO = ΔI ( RD || RL ) = 0.25 (10.2 || 2 ) = 0.418 So ΔVO = 0.836 peak-to-peak

4.25 I DQ = K n (VGSQ − VTN )

2

g m = 2 K n I DQ 2.2 = 2 K n ( 6 ) ⇒ K n = 0.202 mA / V 2 6 = 0.202 ( 2.8 − VTN ) ⇒ VTN = −2.65 V 2

VDSQ = 18 − I DQ ( RS + RD ) 18 − 10 = 1.33 kΩ ⇒ RS = 1.33 − RD RS + RD = 6 g m ( RD RL ) Av = − 1 + g m RS ⎛ R ⋅1 ⎞ −2.2 ⎜ D ⎟ ⎝ RD + 1 ⎠ −1 = 1 + ( 2.2 )(1.33 − RD ) 1 + 2.93 − 2.2 RD =

2.2 RD 1 + RD

( 3.93 − 2.2 RD )(1 + RD ) = 2.2 RD 3.93 + 1.73RD − 2.2 RD2 = 2.2 RD 2.2 RD2 + 0.47 RD − 3.93 = 0 RD =

−0.47 +

( 0.47 ) + 4 ( 2.2 )( 3.93) ⇒ RD = 1.23 kΩ, 2 ( 2.2 ) = 2.8 + ( 6 )( 0.1) = 3.4 V 2

RS = 0.10 kΩ

VG = VGS + VS 1 1 VG = ⋅ Rin ⋅ VDD = (100 )(18 ) = 3.4 ⇒ R1 = 529 kΩ R1 R1 529 R2 = 100 ⇒ R2 = 123 kΩ 529 + R2

4.26 a.

VSD(sat)

VSDQ

V1

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V1 = 9 + VSG , VSD ( sat ) = VSG + VTP VSDQ = =

V1 − VSD ( sat ) 2

+ VSD ( sat )

( 9 + VSG ) − (VSG + VTP ) 2

+ (VSG + VTP )

9 + 1.5 + VSG − 1.5 2 = 3.75 + VSG = 9 + VSG − I DQ RD =

VSDQ

I DQ = K p (VSG + VTP ) Set RD = 0.1RL = 2 kΩ 3.75 = 9 − I DQ ( 2 ) ⇒ I DQ = 2.625 mA 2

b. g m = 2 K p I DQ = 2 r0 =

( 2 )( 2.625) = 4.58 mA / V

1 1 = = 38.1 kΩ λ I DQ ( 0.01)( 2.625 )

Open circuit. Av = − g m ( RD || r0 ) Av = −4.58 ( 2 || 38.1) ⇒ Av = −8.70 c. With RL Av = −4.58 ( 2 || 20 || 38.1) ⇒ Av = −7.947 ⇒ Change = 8.66% 4.27 (a) I DQ = K p (VSGQ + VTP )

2

0.5 = 0.25 (VSGQ + 0.8 )

2

VSGQ = 0.614 V = VS 10 − 0.614 RS = ⇒ RS = 18.8 K 0.5 VD = VS − VSDQ = 0.614 − 3 = −2.386 V RD =

−2.386 − ( −10 ) 0.5

⇒ RD = 15.2 K

(b) Av = − g m RD

( 0.25)( 0.5 ) = 0.7071 mA/V Av = − ( 0.7071)(15.2 )

g m = 2 K p I DQ = 2 Av = −10.7

4.28 Av = − g m ( RD RL )

VDSQ = VDD − I DQ ( RS + RD )

10 = 20 − (1)( RS + RD ) ⇒ RS + RD = 10 kΩ

Let RD = 8 kΩ, RS = 2 kΩ Av = −10 = − g m ( 8 & 20 ) g m = 1.75 mA / V = 2 K n I DQ = 2 K n (1) ⇒ K n = 0.766 mA / V 2

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VS = I DQ RS = (1)( 2 ) = 2 V I DQ = K n (VGS − VTN ) ⇒ 1 = 0.766 (VGS − 2 ) ⇒ VGS = 3.14 V 2

2

VG = VGS + VS = 3.14 + 2 = 5.14 VG =

1 1 ⋅ Rin ⋅ VDD ⇒ ( 200 )( 20 ) = 5.14 ⇒ R1 = 778 kΩ R1 R1

778R2 = 200 ⇒ R2 = 269 kΩ 778 + R2

4.29 Av =

( 4 )( 50 ) g m r0 = ⇒ Av = 0.995 1 + g m r0 1 + ( 4 )( 50 )

Ro =

1 1 r0 = 50 ⇒ R0 = 0.249 kΩ 4 gm

Av = = R0 =

g m ( r0 & RS )

1 + g m ( r0 & RS ) 4 ( 2.38 )

1 + 4 ( 2.38 )

=

4 ( 50 & 2.5 )

1 + 4 ( 50 & 2.5 )

⇒ Av = 0.905

1 1 r0 RS = 50 2.5 ⇒ R0 = 0.226 kΩ 4 gm

4.30 Av = 0.98 =

g m ( RL ro )

1 + g m ( RL ro ) g m ro ⇒ g m ro = 49 1 + g m ro

Also 0.49 =

0.49 = 0.49 =

g m ( RL ro )

1 + g m ( RL ro )

⎛ Rr ⎞ gm ⎜ L o ⎟ ⎝ RL + ro ⎠ = ⎛ Rr ⎞ 1 + gm ⎜ L o ⎟ ⎝ RL + ro ⎠

g m ( RL ro )

RL + ro + g m ( RL ro )

( 49 )(1) 49 = 1 + ro + ( 49 )(1) 50 + ro

ro = 50 K g m = 0.98 mA/V

4.31 (a) Av =

( 2 )( 25) g m ro = 1 + g m ro 1 + ( 2 )( 25 )

Av = 0.98 Ro =

1 1 ro = 25 = 0.5 || 25 gm 2

Ro = 0.49 K

(b)

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Av =

g m ( ro || RL )

1 + g m ( ro || RL )

=

2 ( 25 || 2 )

1 + 2 ( 25 || 2 )

=

2 (1.852 )

1 + 2 (1.852 )

Av = 0.787

4.32 Av =

g m ( RL || ro )

1 + g m ( RL || ro )

=

5 (10 || 100 )

1 + ( 5 )(10 || 100 )

=

5 ( 9.09 )

1 + 5 ( 9.09 )

Av = 0.978 Ro =

1 1 RL ro = 10 || 100 = 0.2 || 9.0909 gm 5

Ro = 0.196 K

4.33 a. ⎛ R2 ⎞ 396 ⎛ ⎞ VG = ⎜ ⎟ VDD = ⎜ ⎟ (10 ) ⇒ VG = 2.42 V ⎝ 396 + 1240 ⎠ ⎝ R1 + R2 ⎠ 10 − (VSG + 2.42 ) 2 = K p (VSG + VTP ) I DQ = RS 2 − 4VSG + 4 ) 7.58 − VSG = ( 2 )( 4 ) (VSG

2 − 31VSG + 24.4 = 0 8VSG

VSG =

31 ±

( 31)

2

− 4 ( 8 )( 24.4 )

2 (8)

⇒ VSG = 2.78 V

I DQ = ( 2 )( 2.78 − 2 ) ⇒ I DQ = 1.21 mA 2

VSDQ = 10 − I DQ RS = 10 − (1.21)( 4 ) ⇒ VSDQ = 5.16 V

b. g m = 2 K p I DQ = 2 r0 = Av = =

( 2 )(1.21) = 3.11 mA / V

1 1 = = 41.3 kΩ λ I DQ ( 0.02 )(1.21) g m ( RS RL r0 )

1 + g m ( RS RL r0 )

3.11( 4 4 41.3)

1 + ( 3.11) ( 4 4 41.3)

⇒ Av = 0.856

 Ii Vsg Vi

 

R1兩兩R2

r0

gmVsg 

RS

RL I0

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⎛ RS r0 I 0 = − ( g mVsg ) ⎜ ⎜R r +R L ⎝ S 0 −Vi Vsg = 1 + g m ( RS RL r0 )

⎞ ⎟⎟ ⎠

Vi = I i ( R1 R2 ) Ai =

g m ( R1 R2 ) RS r0 I0 = ⋅ I i 1 + g m ( RS RL r0 ) RS r0 + RL

( 3.11) ( 396 1240 ) 4 41.3 ⋅ 1 + ( 3.11) ( 4 4 41.3) 4 41.3 + 4 ( 3.11)( 300 ) 3.647 = ⋅ ⇒ Av = 64.2 1 + ( 3.11)(1.908 ) 3.647 + 4 =

R0 =

1 1 4 41.3 ⇒ R0 = 0.295 kΩ RS r0 = 3.11 gm

4.34 g m = 2 K n I Q = 2 (1)(1) = 2 mA / V 1 1 = = 100 kΩ λ I Q ( 0.01)(1)

r0 =

g m ( r0 RL )

⇒ Av = 0.885

1 + g m ( r0 RL )

R0 =

1 1 r0 = 100 ⇒ R0 = 0.498 kΩ 2 gm

4.35 a. Av =

=

2 (100 4 )

Av =

1 + 2 (100 4 )

gm ( 4) g m RL ⇒ 0.95 = 1 + g m RL 1 + gm ( 4)

0.95 = 4 (1 − 0.95 ) g m ⇒ g m = 4.75 mA/V

⎛1 ⎞⎛W g m = 2 ⎜ μ n Cox ⎟ ⎜ ⎝2 ⎠⎝ L

⎞ ⎟ IQ ⎠

4.75 = 2

( 0.030 ) ⎜⎛

W⎞ W = 47.0 ⎟ ( 4) ⇒ L ⎝L⎠

b.

⎛1 ⎞⎛ W g m = 2 ⎜ μ n Cox ⎟⎜ 2 ⎝ ⎠⎝ L

4.75 = 2

⎞ ⎟ IQ ⎠

( 0.030 )( 60 ) I Q

⇒ I Q = 3.13 mA

4.36 2 I DQ = K n (VGS − VTN ) 5 = 5 (VGS + 2 ) ⇒ VGS = −1 V ⇒ VS = −VGS = 1 V 2

I DQ =

VS − ( −5 ) RS

⇒ RS =

g m = 2 K n I DQ = 2 r0 =

1

λ I DQ

=

1+ 5 ⇒ RS = 1.2 kΩ 5

( 5 )( 5) = 10 mA / V

1 = 20 kΩ 0.01 ( )( 5 )

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Av = = R0 =

g m ( r0 RS RL )

1 + g m ( r0 RS RL )

(10 ) ( 20 1.2 2 ) ⇒ Av = 0.878 1 + (10 ) ( 20 1.2 2 ) 1 1 || r0 || RS = || 20 ||1.2 ⇒ Ro = 91.9 Ω 10 gm

4.37 Av =

g m (10 ) g m RS ⇒ 0.90 = 1 + g m RS 1 + g m (10 )

0.90 = 10 (1 − 0.90 ) g m ⇒ g m = 0.90 mA/V 1 1 RS = 10 ⇒ R0 = 1 kΩ gm 0.90 With RL : R0 =

Av =

g m ( RS || RL )

( 0.90 )(10 || 2 ) ⇒ Av = 0.60 1 + g m ( RS || RL ) 1 + ( 0.90 )(10 || 2 ) =

4.38 R0 =

1 RS gm

Output resistance determined primarily by gm 1 = 0.2 kΩ ⇒ g m = 5 mA/V Set gm g m = 2 K n I DQ ⇒ 5 = 2 I DQ = K n (VGS − VTN )

( 4 ) I DQ

⇒ I DQ = 1.56 mA

2

1.56 = 4 (VGS + 2 ) VGS = −1.38 V, VS = −VGS = 1.38 V 2

1.38 − ( −5 )

⇒ RS = 4.09 kΩ 1.56 5 ( 4.09 2 ) g m ( RS RL ) Av = = ⇒ Av = 0.870 1 + g m ( RS RL ) 1 + 5 ( 4.09 2 )

RS =

4.39 a.

g m = 2 K p I DQ = 2

( 5)( 5 ) = 10 mA / V

1 1 = ⇒ Ro = 100 Ω g m 10 b. Open-circuit gain g m r0 But r0 = ∞ so Av = 1.0 Av = 1 + g m r0 With RL: g m RL Av = 1 + g m RL Ro =

0.50 =

10 RL ⇒ 0.50 = 10 (1 − 0.5 ) RL ⇒ RL = 0.1 kΩ 1 + 10 RL

4.40

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−1 ⋅ Δv0 RS RL

ΔiD = I DQ =

Δv0 = − I DQ ⋅ RS RL = − v0 ( min ) = −

Av = vi =

I DQ ⋅ RS RL RS + RL

I DQ RS R 1+ S RL g m ( RS RL )

1 + g m ( RS RL )

=

v0 vi

− I DQ ( RS RL ) ⎡⎣1 + g m ( RS RL ) ⎤⎦ g m ( RS RL )

vi ( min ) = −

I DQ

⎡1 + g m ( RS RL ) ⎤⎦ gm ⎣

4.41 (a) VDD = VDSQ + I DQ RS 3 = 1.5 + ( 0.25 ) RS ⇒ RS = 6 K VS = 1.5 V I DQ = K n (VGSQ − VTN ) 0.25 = 0.5 (VGSQ − 0.4 )

2

2

VGSQ = 1.107 V VG = VGSQ + VS + 1.107 + 1.5 = 2.607 V

⎛ R2 ⎞ 1 VG = ⎜ ⎟ VDD = − RL − VDD R R R + 2 ⎠ 1 ⎝ 1 1 2.607 = ( 300 )( 3) ⇒ R1 = 345.2 K ⇒ R2 = 2291 K R1 (b) g m RS Av = g m = 2 K n I DQ = 2 ( 0.5 )( 0.25 ) = 0.7071 mA/V 1 + g m RS Av =

( 0.7071)( 6 ) ⇒ Av = 0.809 1 + ( 0.7071)( 6 )

Ro =

1 1 RS = 6 = 1.414 || 6 gm ( 0.7071)

Ro = 1.14 K

4.42 Av = + g m RD = ( 5 )( 4 ) Av = 20 Ri =

1 1 10 = 10 = 0.2 ||10 5 gm

Ri = 0.196 K

4.43 a.

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VGS + I DQ RS = 5 5 − VGS 2 = K n (VGS − VTN ) I DQ = RS

5 − VGS = (10 )( 3) (VGS2 − 2VGS + 1) 30VGS2 − 59VGS + 25 = 0

VGS =

59 ±

( 59 )

2

− 4 ( 30 )( 25 )

2 ( 30 )

⇒ VGS = 1.35 V

I DQ = ( 3)(1.35 − 1) ⇒ I DQ = 0.365 mA 2

VDSQ = 10 − ( 0.365 )( 5 + 10 ) ⇒ VDSQ = 4.53 V

b. g m = 2 K n I DQ = 2 r0 =

c.

1

( 3)( 0.365 ) ⇒ g m = 2.093 mA / V

1

=

⇒r =∞

( 0 )( 0.365 ) 0 Av = g m ( RD RL ) = ( 2.093) ( 5 4 ) ⇒ Av = 4.65 λ I DQ

4.44 a. I DQ = K p (VSG + VTP )

2

0.75 = ( 0.5 )(VSG − 1) ⇒ VSG = 2.225 V 5 − 2.225 5 = I DQ RS + VSG ⇒ RS = ⇒ RS = 3.70 kΩ 0.75 VSDQ = 10 − I DQ ( RS + RD ) 6 = 10 − ( 0.75 )( 3.70 + RD ) ⇒ RD = 1.63 kΩ 2

b. Ri =

1 gm

g m = 2 K p I DQ = 2

( 0.5 )( 0.75) = 1.225 mA / V

1 ⇒ Ri = 0.816 kΩ 1.225 Ro = RD ⇒ Ro = 1.63 kΩ Ri =

c. ⎞ ⎞⎛ RS ⎟⎟ ⋅ ii ⎟ ⎜⎜ ⎠ ⎝ RS + [1/ g m ] ⎠ 3.70 ⎛ 1.63 ⎞ ⎛ ⎞ i0 = ⎜ ⎟⎜ ⎟ ii ⎝ 1.63 + 2 ⎠ ⎝ 3.70 + 0.816 ⎠ i0 = 0.368ii = i0 = 1.84sin ω t ( μ A ) ⎛ RD i0 = ⎜ ⎝ RD + RL

v0 = i0 RL = (1.84 )( 2 ) sin ω t ⇒ v0 = 3.68sin ω t ( mV )

4.45 a. 2 I DQ = K n (VGS − VTN ) 5 = 4 (VGS − 2 ) ⇒ VGS = 3.12 V V + = I DQ RD + VDSQ − VGS 2

10 = ( 5 ) RD + 12 − 3.12 ⇒ RD = 0.224 kΩ

b.

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g m = 2 K n I DQ = 2 Ri =

( 4 )( 5 ) ⇒

g m = 8.944 mA / V

1 1 = ⇒ Ri = 0.112 kΩ g m 8.94

Av = g m ( RD RL ) = ( 8.944) ( 0.224 2 ) ⇒ Av = 1.80

c.

4.46 a. 2 I DQ = K p (VSG + VTP ) 3 = 2 (VSG − 2 ) ⇒ VSG = 3.225 V V + = I DQ RS + VSG 10 − 3.225 RS = ⇒ RS = 2.26 kΩ 3 VSDQ = 20 − I DQ ( RS + RD ) 10 = 20 − ( 3)( 2.26 + RD ) ⇒ RD = 1.07 kΩ 2

b. Av = g m ( RD RL )

( 2 )( 3) = 4.90 mA / V Av = ( 4.90 )(1.07 ||10 ) ⇒ Av = 4.74

g m = 2 K p I DQ = 2

4.47 a. Av =

(W / L )D (W / L )L

= 5 ⇒ (W / L ) D = 25 ( 0.5 ) ⇒ (W / L ) D = 12.5

1 ⎛W ⎞ μ n Cox ⎜ ⎟ = ( 30 )(12.5 ) = 375 μ A/V 2 2 ⎝ L ⎠D K L = ( 30 )( 0.5 ) = 15 μ A / V 2

KD =

Transition point: vGSD − VTND = (VDD − VTNL ) −

KD ( vGSD − VTND ) KL

375 ( vGSD − 2 ) 15 vGSD (1 + 5 ) = (10 − 2 ) + 2 + 5 ( 2 ) vGSD = 3.33 V and vDSD = 1.33 V 3.33 − 2 + 2 ⇒ VGSQ = 2.67 V VGSQ = 2 vGSD − 2 = (10 − 2 ) −

␯O

8 Q-point

VDSQ 1.33

3.33

2

␯GSD

VGSQ

b.

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I DQ = K D (VGSQ − VTND )

2

I DQ = 0.375 ( 2.67 − 2 ) ⇒ I DQ = 0.166 mA 2

VDSQ =

4.48 (a)

8 − 1.33 + 1.33 ⇒ VDSQ = 4.67 V 2 Transition point: Load: VOtB = VDD − VTNL = 10 − 2 = 8 V

Driver: 2 2 K D ⎡⎢( vOt A ) (1 + λD vOt A ) ⎤⎥ = K L ⎡( −VTNL ) 1 + λL (VDD − vOt A ) ⎤ ⎣ ⎦ ⎣ ⎦ 2 3 0.5 ⎡⎣ v0t A + ( 0.02 ) v0t A ⎤⎦ = 0.1 ⎡⎣( 4 )(1 + 0.02 ) (10 − v0t A ) ⎤⎦ We have 0.01v03t A + 0.5v02t A + 0.008v0t A − 0.48 = 0

(

)

Therefore v0t A = 0.963 V 8 − 0.963 + 0.963 = 4.48 V = VDSQ 2 2 2 Then K D ⎡(VGSD − VTND ) (1 + λD vOQ ) ⎤ = K L ⎡( −VTNL ) 1 + λL (VDD − vOQ ) ⎤ ⎣ ⎦ ⎣ ⎦ 2 ⎡ ⎤ 0.5 (VGSD − 1.2 ) (1 + ( 0.02 )( 4.48 ) ) = 0.1 ⎡⎣( 4 ) (1 + 0.02 (10 − 4.48 ) ) ⎤⎦ which yields VGSD = 2.103 V ⎣ ⎦ b. 2 I DQ = K D ⎡(VGSD − VTND ) (1 + λD vOQ ) ⎤ ⎣ ⎦ 2 I DQ = 0.5 ⎡( 2.103 − 1.2 ) (1 + ( 0.02 )( 4.48 ) ) ⎤ ⎣ ⎦ So I DQ = 0.444 mA

Now v0 Q =

(

)

c. r0 D = r0 L =

1 1 = = 112.6 kΩ λ I DQ ( 0.02 )( 0.444 )

g mD = 2 K D (VGSD − VTND ) = 2 ( 0.5 )( 2.103 − 1.2 ) ⇒ g mD = 0.903 mA / V Then Av = − g mD ( r0 D r0 L ) = − ( 0.903) (112.6 112.6 ) or Av = −50.8

4.49 2 I D = K n (VGS − VTN ) , VDS = VGS I D = 0 when VDS = 1.5 V ⇒ VTN = 1.5 V 0.8 = K n ( 3 − 1.5 ) ⇒ K n = 0.356 mA / V 2 dI I go = D = = 2 K n (VDS − VTN ) = 2 ( 0.356 )( 3 − 1.5 ) ⇒ Ro = 0.936 k Ω dVDS Ro 2

4.50 a. I DQ = K nD (VGS − VTND ) = ( 0.5 ) ( 0 − ( −1) ) 2

2

I DQ = 0.5 mA I DQ = K nL (VGSL − VTNL ) = K nL (VDD − VO − VTNL ) 2

0.5 = 0.030 (10 − V0 − 1)

2

2

0.5 = 9 − V0 ⇒ V0 = 4.92 V 0.030

b.

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I DD = I DL K nD (Vi − VTND ) = K nL (VDD − Vo − VTNL ) 2

2

K nD (Vi − VTND ) = VDD − Vo − VTNL K nL K nD (Vi − VTND ) K nL

Vo = VDD − VTNL − Av =

(W / L )D (W / L )L

dVo K nD =− =− dVi K nL

Av = −

500 ⇒ Av = −4.08 30

4.51 (a) I DQ = K L (VGSL − VTNL ) = K L (VDSL − VTNL ) 2

2

I D = ( 0.1)( 4 − 1) = 0.9 mA 2

I DQ = K D (VGSD − VTND )

2

0.9 = (1)(VGSD − 1) ⇒ VGSD = 1.95 V VGG = VGSD + VDSL = 1.95 + 4 ⇒ VGG = 5.95 V 2

b. I DD = I DL K D (VGSD − VTND ) = K L (VGSL − VTNL ) 2

2

KD (VGG + Vi − Vo − VTND ) = Vo − VTNL KL ⎛ KD ⎞ Vo ⎜ 1 + ⎟= ⎜ K L ⎟⎠ ⎝ Av =

(c) RLD

KD (VGG + Vi − VTND ) + VTNL KL

KD / KL dVo = ⇒ dVi 1 + K D / K L

Av =

1 1+ KL / KD

From Problem 4.49. 1 = 2 K L (VDSL − VTNL ) =

1 = 1.67 k Ω 2 ( 0.1)( 4 − 1)

"

g m = 2 K D I DQ = 2 (1)( 0.9 ) = 1.90 mA / V Av =

g m ( RLD || RL )

1 + g m ( RLD || RL )

=

(1.90 )(1.67 || 4 )

1 + (1.90 )(1.67 || 4 )

⇒ Av = 0.691

4.52 a.

From Problem 4.51. g m ( RLD || RL ) (1.90 )(1.67 ||10 ) Av = = 1 + g m ( RLD || RL ) 1 + (1.90 )(1.67 || 10 ) Av = 0.731

b.

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R0 =

1 1 RLD = 1.67 = 0.526 ||1.67 gm 1.90

R0 = 0.40 kΩ

4.53 ⎛ 85 ⎞ K n1 = ⎜ ⎟ ( 50 ) ⇒ 2.125 mA/V 2 ⎝ 2⎠

g m1 = 2 K n1 I D1 = 2 ro1 = ro 2 =

1

λ1 I D1

=

( 2.125 )( 0.1) = 0.9220

1 = 200 K 0.05 ( )( 0.1)

1 1 = = 133.3 K λ2 I D 2 ( 0.075 )( 0.1)

Av = − g m1 ( ro1 || ro 2 ) = − ( 0.922 )( 200 ||133.3) Av = −73.7

4.54 K p1 =

k ′p ⎛ w ⎞ ⎛ 40 ⎞ 2 ⎜ ⎟ = ⎜ ⎟ ( 50 ) ⇒ 1.0 mA/V 2 ⎝L⎠ ⎝ 2 ⎠

g m1 = 2 K p1 I D1 = 2 (1)( 0.1) = 0.6325 mA/V

ro1 = ro 2 =

1

λ1 I D1

=

1 = 133.3 K 0.075 ( )( 0.1)

1 1 = = 200 K λ2 I D 2 ( 0.05 )( 0.1)

Av = − g m1 ( ro1 ro 2 ) = − ( 0.6325 )(133.3 || 200 ) Av = −50.6

4.55 (a) ⎛ 85 ⎞ K n1 = ⎜ ⎟ ( 50 ) ⇒ 2.125 mA/V 2 ⎝ 2⎠

g m1 = 2 K n1 I D1 = 2 ro1 = ro 2 =

1

λ1 I D1

=

( 2.125 )( 0.1) = 0.922 mA/V

1

( 0.05 )( 0.1)

= 200 K

1 1 = = 133.3 K λ2 I D 2 ( 0.075 )( 0.1)

(b) Ri1 =

1 1 = = 1.085 K g m1 0.922

⎛ ⎞ Ri1 1.085 ⎛ ⎞ Vgs1 = − ⎜ ⎟ Vi = − ⎜ ⎟ Vi = −0.956Vi + + 0.050 1.085 0.050 R ⎝ ⎠ i 1 ⎝ ⎠ Vgs1 = + ( 0.956 )( 0.922 )( 200 )(133.3) Av = − g m1 ( ro1 ro 2 ) ⋅ Vi

Av = 70.5 1 1 = 0.05 + ⇒ Ri = 1.135 K 0.922 g m1

(c)

Ri = 0.05 +

(d)

Ro ≈ ro1 ro 2 = 200 133.7 ⇒ Ro ≈ 80 K

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4.56 (a) g m1 = 2 K n I D1 = 2

( 2 )( 0.1) = 0.8944 mA/V

gm2 = 2 K p I D 2 = 2

( 2 )( 0.1) = 0.8944 mA/V

ro1 = ro 2 =

1

=

1

= 100 K

λ I D ( 0.1)( 0.1) (b) The small-signal equivalent circuit gm2Vsg2

Vo





rO2 gm1Vi

Vi

Vsg2 rO1





(1) g m1Vi +

Vsg 2 ro1

+ g m 2Vsg 2 +

Vsg 2 − Vo ro 2

=0

(2) Vo Vo − Vsg 2 + = g m 2Vsg 2 ro ro 2 ⎛1 1 ⎞ ⎛ 1 ⎞ + gm2 ⎟ Vo ⎜ + ⎟ = Vsg 2 ⎜ ⎝ ro ro 2 ⎠ ⎝ ro 2 ⎠ 1 1 1 ⎛ ⎞ ⎛ ⎞ Vo ⎜ + + 0.8944 ⎟ ⇒ Vsg 2 = Vo ( 0.03317 ) ⎟ = Vsg 2 ⎜ ⎝ 50 100 ⎠ ⎝ 100 ⎠ (1) ⎛ 1 1 ⎞ V g m1Vi + Vsg 2 ⎜ + g m 2 + ⎟ = o r r ro 2 o2 ⎠ ⎝ o1

1 ⎞ Vo ⎛ 1 0.8944 Vi + Vo ( 0.03317 ) ⎜ + 0.8944 + ⎟= 100 ⎠ 100 ⎝ 100 0.8944 Vi = Vo ( 0.01 − 0.03033) Vo = −44 Vi

(c) For output resistance, set Vi = 0. gm2Vsg2 RO

Ix

rO2  Vsg2

rO1

rO

 

Vx



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(1) (2)

g m 2Vsg 2 + I x =

Vsg 2 ro1

Vx Vx − Vsg 2 + ro ro 2

+ g m 2Vsg 2 +

Vsg 2 − Vx ro 2

=0

(2) ⎛ 1 1 ⎞ V Vsg 2 ⎜ + g m 2 + ⎟ = x ro 2 ⎠ ro 2 ⎝ ro1 1 ⎞ Vx ⎛ 1 + 0.8944 + Vsg 2 ⎜ ⎟= 100 100 ⎝ ⎠ 100 Vsg 2 = Vx ( 0.010936 )

(1) ⎛1 1 ⎞ ⎛ 1 ⎞ I x = Vx ⎜ + ⎟ − Vsg 2 ⎜ + gm2 ⎟ ⎝ ro ro 2 ⎠ ⎝ ro 2 ⎠ 1 1 ⎛ ⎞ ⎛ 1 ⎞ + 0.8944 ⎟ I x = Vx ⎜ + ⎟ − Vx ( 0.010936 ) ⎜ ⎝ 50 100 ⎠ ⎝ 100 ⎠ I x = Vx ( 0.03 − 0.0098905 ) V Ro = x = 49.7 K Ix

4.57 a. 2 I D1 = K1 (VGS 1 − VTN 1 ) 0.4 = 0.1(VGS 1 − 2 ) ⇒ VGS1 = 4 V VS 1 = I D1 RS 1 = ( 0.4 )( 4 ) = 1.6 V VG1 = VS 1 + VGS 1 = 1.6 + 4 = 5.6 V 2

⎛ R2 ⎞ 1 VG1 = ⎜ ⎟ VDD = ⋅ Rin ⋅ VDD R1 ⎝ R1 + R2 ⎠ 1 5.6 = ( 200 )(10 ) ⇒ R1 = 357 kΩ R1

357 R2 = 200 ⇒ R2 = 455 kΩ 357 + R2

VDS1 = VDD − I D1 ( RS1 + RD1 ) 4 = 10 − ( 0.4 )( 4 + RD1 ) ⇒ RD1 = 11 kΩ VD1 = 10 − ( 0.4 )(11) = 5.6 V I D 2 = K 2 (VSG 2 + VTP 2 )

2

2 = 1(VSG 2 − 2 ) ⇒ VSG 2 = 3.41 V VS 2 = VD1 + VSG 2 = 5.6 + 3.41 = 9.01 10 − 9.01 RS 2 = ⇒ RS 2 = 0.495 kΩ 2 VSD 2 = VDD − I D 2 ( RS 2 + RD 2 ) 5 = 10 − ( 2 )( 0.495 + RD 2 ) ⇒ RD 2 = 2.01 kΩ 2

b.

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Vi

 

V0



 Vgs1

RD1

R1兩兩R2

gm1Vgs1

Vgs2

RD2 gm2Vsg2





V0 = g m 2Vsg 2 RD 2 = ( g m 2 RD 2 ) ( g m1Vgs1 RD1 ) Vgs1 = Vi V Av = 0 = g m1 g m 2 RD1 RD 2 Vi g m1 = 2 K1 I D1 = 2

( 0.1)( 0.4 ) = 0.4 mA / V

g m 2 = 2 K 2 I D 2 = 2 (1)( 2 ) = 2.828 mA / V Av = ( 0.4 )( 2.828 )(11)( 2.01) ⇒ Av = 25.0

c.

4

2

VSD(sat)

5

10

VSD ( sat ) = VSG + VTh = VDD − I D 2 ( RD 2 + RS 2 ) 2 = VDD − K p 2 ( RD 2 + RS 2 )VSD ( sat )

So (1)( 2.01 + 0.495 )VSD2 ( sat ) + VSD ( sat ) − VDD = 0 2 2.50VSD ( sat ) + VSD ( sat ) − 10 = 0 VSD ( sat ) =

−1 ± 1 + 4 ( 2.501)(10 ) 2 ( 2.501)

VSD ( sat ) = 1.81 V 5 − 1.81 = 3.19 ⇒ Max. output swing = 6.38 V peak-to-peak

4.58 a. 10  4 mA 2.5

VSD2(sat)

10

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2 VSD 2 ( sat ) = VDD − I D 2 ( RD 2 + RS 2 ) = VDD − K p 2 ( RD 2 + RS 2 ) VSD 2 ( sat ) 2 (1)( 2 + 0.5)VSD 2 ( sat ) + VSD 2 ( sat ) − 10 = 0 2 2.5VSD 2 ( sat ) + VSD 2 ( sat ) − 10 = 0

VSD 2 ( sat ) =

−1 ± 1 + 4 ( 2.5 )(10 ) 2 ( 2.5 )

= 1.81 V

10 − 1.81 + 1.81 ⇒ VSDQ 2 = 5.91 V 2 VSDQ 2 = VDD − I DQ 2 ( RD 2 + RS 2 ) 5.91 = 10 − I DQ 2 ( 2 + 0.5 ) ⇒ I DQ 2 = 1.64 mA VS 2 = 10 − (1.64 )( 0.5 ) = 9.18 V VSDQ 2 =

I DQ 2 = K p 2 (VSG 2 + VTP 2 )

2

1.64 = (1)(VSG 2 − 2 ) ⇒ VSG 2 = 3.28 V VD1 = VS 2 − VSG 2 = 9.18 − 3.28 = 5.90 V 10 − 5.90 RD1 = ⇒ RD1 = 10.25 kΩ 0.4 2

I DQ1 = K n1 (VGS1 − VTN 1 )

2

0.4 = ( 0.1)(VGS 1 − 2 ) ⇒ VGS1 = 4 V VS 1 = ( 0.4 )(1) = 0.4 V VG1 = VS 1 + VGS 1 = 0.4 + 4 = 4.4 V 2

⎛ R2 ⎞ 1 VG1 = ⎜ ⎟ VDD = ⋅ Rm ⋅ VDD R R R + ⎝ 1 2 ⎠ 1 1 4.4 = ⋅ ( 200 )(10 ) ⇒ R1 = 455 kΩ R1

455 R2 = 200 ⇒ R2 = 357 kΩ 455 + R2 b. I DQ 2 = 1.64 mA VSDQ 2 = 10 − (1.64 )( 2 + 0.5 ) ⇒ VSDQ 2 = 5.90 V VDSQ1 = 10 − ( 0.4 )(10.25 + 1) ⇒ VDSQ1 = 5.50 V

(c) g m1 = 2 K n1 I DQ1 = 2

( 0.1)( 0.4 ) = 0.4 mA / V

g m 2 = 2 K p 2 I DQ 2 = 2 (1)(1.64 ) = 2.56 mA / V Av = g m1 g m 2 RD1 RD 2 = ( 0.4 )( 2.56 )(10.25 )( 2 ) ⇒ Av = 21.0

4.59 a. VDSQ 2 = 7 = VDD − I DQ 2 RS 2 = 10 − I DQ 2 ( 6 ) I DQ 2 = 0.5 mA I DQ 2 = K 2 (VGS 2 − VTN 2 )

2

0.5 = ( 0.2 )(VGS 2 − 0.8 ) ⇒ VGS 2 = 2.38 V VS 1 = VS 2 + VGS 2 = 3 + 2.38 = 5.38 2

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I DQ1 =

VS 1 5.38 = = 0.269 mA 20 RS1

I DQ1 = K1 (VGS 1 − VTN 1 )

2

0.269 = ( 0.2 )(VGS 1 − 0.8 ) ⇒ VGS 1 = 1.96 V VG1 = VS 1 + VGS 1 = 5.38 + 1.96 = 7.34 V 2

⎛ R2 ⎞ 1 VG1 = ⎜ ⎟ VDD = ⋅ Rin ⋅ VDD R + R R ⎝ 1 2 ⎠ 1 1 7.34 = ( 400 )(10 ) ⇒ R1 = 545 kΩ R1 545 R2 = 400 ⇒ R2 = 1.50 MΩ 545 + R2 b. I DQ1 = 0.269 mA, I DQ 2 = 0.5 mA

VDSQ1 = 10 − ( 0.269 )( 20 ) ⇒ VDSQ1 = 4.62 V

c. Av =

g m1 RS 1 g R ⋅ m2 S 2 1 + g m1 RS1 1 + g m 2 RS 2

g m1 = 2 K1 I DQ1 = 2

( 0.2 )( 0.269 ) = 0.464 mA / V

( 0.2 )( 0.5) = 0.6325 mA / V ( 0.464 )( 20 ) ( 0.6325)( 6 ) ⋅ Av = 1 + ( 0.464 )( 20 ) 1 + ( 0.6325 )( 6 ) = ( 0.9027 )( 0.7915 )

g m 2 = 2 K 2 I DQ 2 = 2

= Av = 0.714

gm1Vgs1  Vgs1

 Vgs2 



Ix RS2

RS1

R0 =

gm2Vgs2

 

Vx

1 1 RS 2 = 6 = 1.581 6 ⇒ Ro = 1.25 kΩ gm2 0.6325

4.60 (a) I DQ1 =

10 − VGS 1 2 = K n1 (VGS 1 − VTN 1 ) RS 2

10 − VGS 1 = ( 4 )(10 ) (VGS2 1 − 4VGS 1 + 4 ) 40VGS2 1 − 159VGS 1 + 150 = 0 VGS1 =

159 ±

(159 ) − 4 ( 40 )(150 ) ⇒ VGS 1 = 2.435 V 2 ( 40 ) 2

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I DQ1 = ( 4 )( 2.435 − 2 ) ⇒ I DQ1 = 0.757 mA 2

VDSQ1 = 20 − ( 0.757 )(10 ) ⇒ VDSQ1 = 12.4 V Also I DQ 2 = 0.757 mA VDSQ 2 = 20 − ( 0.757 )(10 + 5 ) ⇒ VDSQ 2 = 8.65 V g m1 = g m 2 = 2 KI DQ = 2

(b)

( 4 )( 0.757 ) ⇒ g m1 = g m 2 = 3.48 mA / V

c.  Vgs1 Vi

 

gm1Vgs1 gm2Vgs2



RG

V0

 RS1

Vgs2 RS2

RD

RL



V0 = − ( g m 2Vgs 2 ) ( RD RL )

Vgs 2 = ( − g m1Vgs1 − g m 2Vgs 2 ) ( RS1 RS 2 ) Vi = Vgs1 − Vgs 2 ⇒ Vgs1 = Vi + Vgs 2

Vgs 2 + g m 2Vgs 2 ( RS 1 RS 2 ) = − g m1 (Vi + Vgs 2 ) ( RS 1 RS 2 )

Vgs 2 + g m 2Vgs 2 ( RS 1 RS 2 ) + g m1Vgs 2 ( RS 1 Rs 2 ) = − g m1Vi ( RS1 RS 2 ) Vgs 2 = Av =

− g m1Vi ( RS 1 RS 2 )

1 + g m 2 ( RS 1 RS 2 ) + g m1 ( RS 1 RS 2 )

V0 g m1 g m 2 ( RS 1 RS 2 )( RD RL ) = Vi 1 + ( g m1 + g m 2 ) ( RS1 RS 2 )

( 3.48 ) (10 10 )( 5 2 ) Av = ⇒ Av = 2.42 1 + ( 3.48 + 3.48 ) (10 10 ) 2

4.61 a. I DQ = 3 mA VS 1 = I DQ RS − 5 = ( 3)(1.2 ) − 5 = −1.4 V I DQ = K1 (VGS − VTN )

2

3 = 2 (VGS − 1) ⇒ VGS = 2.225 V 2

VG1 = VGS + VS 1 = 2.225 − 1.4 = 0.825 V ⎛ ⎞ R3 ⎛ R3 ⎞ VG1 = ⎜ ⎟ ( 5 ) ⇒ 0.825 = ⎜ ⎟ ( 5 ) ⇒ R3 = 82.5 kΩ ⎝ 500 ⎠ ⎝ R1 + R2 + R3 ⎠ VD1 = VS1 + VDSQ1 = −1.4 + 2.5 = 1.1 V VG 2 = VD1 + VGS = 1.1 + 2.225 = 3.325 V ⎛ R2 + R3 ⎞ ⎛ R2 + R3 ⎞ VG 2 = ⎜ ⎟ ( 5 ) ⇒ 3.325 = ⎜ ⎟ ( 5) ⎝ 500 ⎠ ⎝ R1 + R2 + R3 ⎠ R2 + R3 = 332.5 ⇒ R2 = 250 kΩ

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R1 = 500 − 250 − 82.5 ⇒ R1 = 167.5 kΩ VD 2 = VD1 + VDSQ 2 = 1.1 + 2.5 = 3.6 V RD =

5 − 3.6 ⇒ RD = 0.467 kΩ 3

b. Av = − g m1 RD

( 2 )( 3) = 4.90 mA / V

g m1 = 2 K n I DQ = 2

Av = − ( 4.90 )( 0.467 ) ⇒ Av = −2.29

4.62 a. VS 1 = I DQ RS − 10 = ( 5 )( 2 ) − 10 ⇒ VS 1 = 0 I DQ = K1 (VGS 1 − VTN )

2

5 = 4 (VGS1 − 1.5 ) ⇒ VGS 1 = 2.618 V VG1 = VGS 1 + VS1 = 2.618 V = IR3 = ( 0.1) R3 ⇒ R3 = 26.2 kΩ 2

VD1 = VS1 + VDSQ1 = 0 + 3.5 = 3.5 V VG 2 = VD1 + VGS = 3.5 + 2.62 = 6.12 V = ( 0.1)( R2 + R3 ) R2 + R3 = 61.2 kΩ ⇒ R2 = 35 kΩ VD 2 = VD1 + VDSQ 2 = 3.5 + 3.5 = 7.0 V 10 − 7 RD = ⇒ RD = 0.6 kΩ 5 10 − 6.12 R1 = ⇒ R1 = 38.8 kΩ 0.1 b. Av = − g m1 RD

( 4 )( 5 ) = 8.944 mA / V

g m1 = 2 K n I DQ = 2

Av = − ( 8.944 )( 0.6 ) ⇒ Av = −5.37

4.63 a. ⎛ V ⎞ I DQ = I DSS ⎜ 1 − GS ⎟ VP ⎠ ⎝ ⎛ V ⎞ 4 = 6 ⎜⎜ 1 − GS ⎟⎟ ⎝ ( −3) ⎠

2

2

⎡ 4⎤ VGS = ( −3) ⎢1 − ⎥ ⇒ VGS = −0.551 V 6⎦ ⎣ VDSQ = VDD − I DQ RD 6 = 10 − ( 4 ) RD ⇒ RD = 1 kΩ

b.

2 I DSS ⎛ VGS ⎞ 2 ( 6 ) ⎛ −0.551 ⎞ = 1− 1− ⎟ ⇒ g m = 3.265 mA/V −3 ⎠ ( −VP ) ⎜⎝ VP ⎟⎠ 3 ⎜⎝ 1 1 = ⇒ r0 = 25 kΩ r0 = λ I DQ ( 0.01)( 4 ) gm =

c.

Av = − g m ( r0 & RD ) = − ( 3.265 )( 25 & 1) ⇒ Av = −3.14

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4.64 VGS + I DQ ( RS1 + RS 2 ) = 0 ⎛ V ⎞ I DQ = I DSS ⎜ 1 − GS ⎟ VP ⎠ ⎝

2

2

VGS

⎛ V ⎞ + I DSS ( RS1 + RS 2 ) ⎜ 1 − GS ⎟ = 0 VP ⎠ ⎝ 2

⎛ V ⎞ VGS + ( 2 )( 0.1 + 0.25 ) ⎜1 − GS ⎟ = 0 VP ⎠ ⎝ ⎛ 2 V2 ⎞ VGS + 0.7 ⎜ 1 − VGS + GS 2 ⎟ = 0 ⎜ ⎟ ⎝ ( −2 ) ( −2 ) ⎠ 0.175VGS2 + 1.7VGS + 0.7 = 0 VGS =

−1.7 ±

(1.7 ) − 4 ( 0.175)( 0.7 ) ⇒ VGS 2 ( 0.175 ) VGS ⎞ 2 ( 2 ) ⎛ −0.431 ⎞ 2

= −0.4314 V

2 I DSS ⎛ ⎜1 − ⎟= ⎜1 − ⎟ ⇒ g m = 1.569 mA/V −VP ⎝ −2 ⎠ VP ⎠ 2 ⎝ − g m ( RD RL ) − (1.569 ) ( 8 4 ) = ⇒ Av = −3.62 Av = 1 + g m RS1 1 + (1.569 )( 0.1)

gm =

Ai =

i0 ( v0 / RL ) v0 RG ⎛ 50 ⎞ = = ⋅ = ( −3.62 ) ⎜ ⎟ ⇒ Ai = −45.2 ii ( vi / RG ) vi RL ⎝ 4⎠

4.65 I DSS = 4 mA 2 V = DD = 10 V 2 = VDD − I DQ ( RS + RD )

I DQ = VDSQ VDSQ

10 = 20 − ( 4 )( RS + RD ) ⇒ RS + RD = 2.5 kΩ

VS = 2 V = I DQ RS = 4 RS ⇒ RS = 0.5 kΩ, RD = 2.0 kΩ I DQ

⎛ V ⎞ = I DSS ⎜ 1 − GS ⎟ VP ⎠ ⎝

2

2

⎛ ⎞ ⎛ V 4⎞ 4 = 8 ⎜⎜1 − GS ⎟⎟ ⇒ VGS = ( −4.2 ) ⎜⎜1 − ⎟ ⇒ VGS = −1.23 V 8 ⎠⎟ ⎝ ⎝ ( −4.2 ) ⎠ VG = VS + VGS = 2 − 1.23 ⎛ R2 ⎞ ⎛ R2 ⎞ VG = 0.77 V = ⎜ ⎟ ( 20 ) = ⎜ ⎟ ( 20 ) ⇒ R2 = 3.85 kΩ, R1 = 96.2 KΩ ⎝ 100 ⎠ ⎝ R1 + R2 ⎠

4.66 a.

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I DSS = 5 mA 2 V 12 VDSQ = DD = =6V 2 2 12 − 6 RS = ⇒ RS = 1.2 kΩ 5 I DQ =

I DQ

⎛ V ⎞ = I DSS ⎜ 1 − GS ⎟ VP ⎠ ⎝

2

2

⎛ ⎛ V ⎞ 5 ⎞ 5 = 10 ⎜⎜ 1 − GS ⎟⎟ ⇒ VGS = ( −5 ) ⎜⎜ 1 − ⎟ ⇒ VGS = −1.464 V 10 ⎟⎠ ⎝ ⎝ ( −5 ) ⎠ VG = VS + VGS = 6 − 1.464 = 4.536 V ⎛ R2 ⎞ 1 VG = ⎜ ⎟ VDD = ⋅ Rin ⋅ VDD R1 ⎝ R1 + R2 ⎠ 1 4.536 = (100 )(12 ) ⇒ R1 = 265 kΩ R1 265 R2 = 100 ⇒ R2 = 161 kΩ 265 + R2 b. ⎛ V ⎞ 2 (10 ) ⎛ −1.46 ⎞ 2I g m = DSS ⎜ 1 − GS ⎟ = ⎜1 − ⎟ ⇒ g m = 2.83 mA/V − −5 ⎠ 5 ⎝ V ( P ) ⎝ VP ⎠ r0 =

1 1 = = 20 kΩ λ I DQ ( 0.01)( 5 )

Av =

(

g m r0 Rs RL

(

)

1 + g m r0 RS RL

)

Av =

( 2.83) ( 20 1.2 0.5 ) ⇒ Av = 0.495 1 + ( 2.83) ( 20 1.2 0.5 )

R0 =

1 1 1.2 = 0.353 1.2 ⇒ R0 = 0.273 kΩ RS = 2.83 gm

4.67 a. ⎛ R2 ⎞ ⎛ 110 ⎞ VG = ⎜ ⎟ VDD = ⎜ ⎟ (10 ) = 5.5 V ⎝ 110 + 90 ⎠ ⎝ R1 + R2 ⎠ I DQ =

10 − (VG − VGS ) RS

⎛ V ⎞ = I DSS ⎜1 − GS ⎟ VP ⎠ ⎝

⎛ V ⎞ 10 − 5.5 + VGS = ( 2 )( 5 ) ⎜1 − GS ⎟ ⎝ 1.75 ⎠

2

2

4.5 + VGS = 10 (1 − 1.143VGS + 0.3265VGS2 )

3.265VGs2 − 12.43VGS + 5.5 = 0 VGS =

12.43 ±

(12.43) − 4 ( 3.265 )( 5.5) ⇒ VGS 2 ( 3.265 ) 2

= 0.511 V

2

⎛ 0.511 ⎞ I DQ = ( 2 ) ⎜1 − ⎟ ⇒ I DQ = 1.00 mA 1.75 ⎠ ⎝ VSDQ = 10 − (1.00 )( 5 ) ⇒ VSDQ = 5.0 V

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b.

2 I DSS ⎛ VGS ⎞ 2 ( 2 ) ⎛ 0.511 ⎞ ⎜1 − ⎟= ⎜1 − ⎟ ⇒ g m = 1.618 mA/V VP ⎝ VP ⎠ 1.75 ⎝ 1.75 ⎠ g m ( RS RL ) (1.618 ) ( 5 10 ) Av = = ⇒ Av = 0.844 1 + g m ( RS RL ) 1 + (1.618 ) ( 5 10 )

gm =

⎛R ⎞ i0 ( v0 / RL ) = = Av ⋅ ⎜ i ⎟ ii ( vi / Ri ) ⎝ RL ⎠ Ri = R1 R2 = 90 110 = 49.5 kΩ Ai =

⎛ 49.5 ⎞ Ai = ( 0.844 ) ⎜ ⎟ ⇒ Ai = 4.18 ⎝ 10 ⎠ c.

AC load line 1 Slope  5兩兩10 1  3.33 K

1.0

5.0

10

Δid = 1.0 mA

vsd = ( 3.33)(1.0 ) = 3.33 V Maximum swing in output voltage = 6.66 V peak-to-peak

4.68 ⎛ V ⎞ I DQ = I DSS ⎜ 1 − GS ⎟ VP ⎠ ⎝

2

2 ⎛ 4⎞ ⎛ V ⎞ 4 = 8 ⎜1 − GS ⎟ ⇒ VGS = 4 ⎜⎜ 1 − ⎟ ⇒ VGS = 1.17 V 4 ⎠ 8 ⎟⎠ ⎝ ⎝ VSDQ = VDD − I DQ ( RS + RD ) 7.5 = 20 − 4 ( RS + RD ) ⇒ RS + RD = 3.125 kΩ

⎛ VGS ⎜1 − VP ⎝ RS = 3.125 − RD − g m RD Av = 1 + g m RS gm =

2 I DSS VP

⎞ 2 ( 8 ) ⎛ 1.17 ⎞ ⎟= ⎜1 − ⎟ ⇒ g m = 2.83 mA/V 4 ⎝ 4 ⎠ ⎠

−3 (1 + g m RS ) = − g m RD 3 ⎡⎣1 + ( 2.83)( 3.125 − RD ) ⎤⎦ = ( 2.83) RD 9.844 − 2.83RD = 0.9433RD ⇒ RD = 2.61 kΩ RS = 0.516 kΩ VS = 20 − ( 4 )( 0.516 ) ⇒ VS = 17.94 V VG = VS − VGS = 17.94 − 1.17 = 16.77 V ⎛ R2 ⎞ ⎛ R2 ⎞ VG = ⎜ ⎟ VDD = ⎜ ⎟ ( 20 ) ⇒ R2 = 335 kΩ. R1 = 65 kΩ R R + ⎝ 400 ⎠ 2 ⎠ ⎝ 1

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Chapter 5 Exercise Problems EX5.1

β=

α 1−α

0.980 = 49 1 − 0.980 0.995 = 199 For α = 0.995, β = 1 − 0.995 So 49 ≤ β ≤ 199

For α = 0.980, β =

EX5.2 BVCEO =

BVCBO n

=

β

200 3

120

or BVCEO = 40.5 V

EX5.3 V − VBE ( on ) 2 − 0.7 I B = BB = ⇒ 6.5 μ A RB 200

I C = β I = (120 )( 6.5 μ A ) ⇒ I C = 0.78 mA

VCE = VCC − I C RC = 5 − ( 0.78 )( 4 ) or VCE = 1.88 V

P = I BVBE ( on ) + I CVCE = ( 0.0065 )( 0.7 ) + ( 0.78 )(1.88 ) ≅ 1.47 mW

EX5.4 V + − VEB ( on ) − VBB 5 − 0.7 − 2.8 IB = = or I B = 4.62 μ A RB 325 I C = β I B = ( 80 )( 4.615 ) ⇒ I C = 0.369 mA

VEC = 2 = 5 − ( 0.369 ) RC which yields RC = 8.13 k Ω

EX5.5

(a)

IB =

VBB − VBE ( on ) RB

=

2 − 0.7 ⇒ 5.91 μ A 220

I C = β I B = (100 )( 5.91 μ A ) ⇒ 0.591 mA I E = (1 + β ) I B = 0.597 mA

VCE = VCC − I C RC = 10 − ( 0.591)( 4 ) or VCE = 7.64 V

6.5 − 0.7 ⇒ 26.4 μ A 220 Transistor is biased in saturation mode, so

(b)

IB =

VCE = VCE ( sat ) = 0.2 V IC =

VCC − VCE ( sat ) RC

=

10 − 0.2 or I C = 2.45 mA 4

I E = I C + I B = 2.45 + 0.0264 ≅ 2.48 mA

EX5.6 For 0 ≤ VI < 0.7 V , Qn is cutoff, VO = 9 V

When Qn is biased in saturation, we have 0.2 = 9 −

(100 )(VI − 0.7 )( 4 ) 200

So, for VI ≥ 5.1 V , VO = 0.2 V EX5.7

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⇒ VI = 5.1 V

IB =

VBB − VBE ( on )

RB + (1 + β ) RE

=

8 − 0.7 30 + ( 76 )(1.2 )

or I B = 60.2 μ A I C = β I B = ( 75 )( 60.23 μ A ) ⇒ 4.52 mA

I E = (1 + β ) I B = 4.58 mA VCE = VCC − I C RC − I E RE

= 12 − ( 4.517 )( 0.4 ) − ( 4.577 )(1.2 )

or VCE = 4.70 V

EX5.8 For VC = 4 V and ICQ = 1.5 mA, V + − VC 10 − 4 RC = = ⇒ RC = 4 k Ω I CQ 1.5 ⎛ 1+ β ⎞ ⎛ 101 ⎞ IE = ⎜ ⎟ IC = ⎜ ⎟ (1.5 ) = 1.515 mA β ⎝ 100 ⎠ ⎝ ⎠ −V ( on ) − V − also I E = BE RE Then RE =

−0.7 − ( −10 ) 1.515

⇒ RE = 6.14 k Ω

EX5.9 3 − 0.7 = 0.25 I EQ = RE RE = 9.2 kΩ ⎛ 75 ⎞ I CQ = ⎜ ⎟ ( 0.25 ) = 0.2467 mA ⎝ 76 ⎠ −0.7 + VCEQ + I CQ RC − 3 = 0 RC =

3 + 0.7 − 2 ⇒ RC = 6.89 kΩ 0.2467

EX5.10 5 = I E RE + VEB ( on ) + I B RB − 2

(a)

(b)

(c)

(d)

180 ⎞ ⎛ 5 + 2 − 0.7 = I E ⎜ 2 + ⎟ I E = 0.9859 mA 41 ⎠ ⎝ I C = 0.962 mA 180 ⎞ ⎛ 6.3 = I E ⎜ 2 + ⎟ I E = 1.2725 mA 61 ⎠ ⎝ I C = 1.25 mA 180 ⎞ ⎛ 6.3 = I E ⎜ 2 + ⎟ I E = 1.6657 mA 101 ⎠ ⎝ I C = 1.64 mA 180 ⎞ ⎛ 6.3 = I E ⎜ 2 + ⎟ I E = 1.97365 mA 151 ⎠ ⎝ I C = 1.94 mA

EX5.11

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IE =

VBB − VEB ( on ) RE

⇒ RE =

4 − 0.7 1.0

or RE = 3.3 k Ω I C = α I E = ( 0.992 )(1) = 0.992 mA

I B = I E − I C = 1.0 − 0.992 or I B = 8 μ A

VCB = I C RC − VCC = ( 0.992 )(1) − 5 or VCB = −4.01 V

EX5.12 V + − (Vγ + VCE ( sat ) ) 5 − (1.5 + 0.2 ) R= = IC 15 or R = 220 Ω 15 IB = = 1 mA 15 v − VBE ( on ) 5 − 0.7 = = 4.3 k Ω RB = I 1 IB P = I BVBE (on) + I CVCE

= (1)( 0.7 ) + (15 )( 0.2 ) = 3.7 mW

EX5.13 (a) For V1 = V2 = 0, All currents are zero and VO = 5 V. (b) For V1 = 5 V, V2 = 0; IB2 = IC2 = 0, 5 − 0.7 I B1 = = 4.53 mA 0.95 5 − 0.2 I C1 = ⇒ I C1 = I R = 8 mA 0.6 VO = 0.2 V (c) For V1 = V2 = 5 V, IB1 = IB2 = 4.53 mA; IR = 8 mA, IC1 = IC2 = IR/2 = 4 mA, VO = 0.2 V EX5.14 vO = 5 − iC RC = 5 − β iB RC ΔvO = − βΔiB RC iB =

VBB + ΔvI − VBE ( on ) RB

ΔiB =

ΔvI RB

Then

ΔvO − β RC = ΔvI RB

Let β = 100, RC = 5 k Ω, RB = 100 k Ω So

Δvo − (100 )( 5 ) = = −5 ΔvI 100

Want Q-point to be vo ( Q − pt ) = 2.5 = 5 − (100 ) I BQ ( 5 ) Then I BQ = 0.005 mA, I BQ = 0.005 =

VBB − 0.7 100

so VBB = 1.2 V Also, I CQ = β I BQ = (100 )( 0.005 ) = 0.5 mA

EX5.15

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VCEQ = 2.5 = 5 − I CQ RC

5 − 2.5 = 10 k Ω 0.25 I CQ 0.25 = = = 0.002083 mA 120 β

or RC = I BQ

Then RB =

5 − 0.7 ⇒ RB = 2.06 M Ω 0.002083

EX5.16 (a) RTH = R1 & R2 = 9 & 2.25 = 1.8 k Ω

or VTH

⎛ R2 ⎞ ⎛ 2.25 ⎞ VTH = ⎜ ⎟ ⋅ VCC = ⎜ ⎟ ( 5) + R R ⎝ 9 + 2.25 ⎠ ⎝ 1 2 ⎠ = 1.0 V

I BQ =

(b)

VTH − VBE ( on )

RTH + (1 + β ) RE

=

1 − 0.7 1.8 + (151)( 0.2 )

or I BQ = 9.375 μ A I CQ = β I BQ = (150 )( 9.375 μ A ) or I CQ = 1.41 mA

I EQ = (1 + β ) I BQ ⇒ I EQ = 1.42 mA

VCEQ = 5 − I CQ RC − I EQ RE

= 5 − (1.41)(1) − (1.42 )( 0.2 )

or VCEQ = 3.31 V

For β = 75

(c) I BQ =

1 − 0.7 = 17.6 μ A 1.8 + ( 76 )( 0.2 )

I CQ = β I BQ = ( 75 )(17.6 μ A ) or I CQ = 1.32 mA

I EQ = (1 + β ) I BQ = ( 76 )(17.6 μ A )

or I EQ = 1.34 mA

VCEQ = 5 − (1.32 )(1) − (1.34 )( 0.2 )

or VCEQ = 3.41 V EX5.17 VCEQ ≅ VCC − I CQ ( RC + RE ) or 2.5 ≅ 5 − I CQ (1 + 0.2 ) which yields I CQ = 2.08 mA, I CQ 2.08 I BQ = = = 0.0139 mA 150 β RTH = ( 0.1)(1 + β ) RE = ( 0.1)(151)( 0.2 ) or RTH = 3.02 k Ω

⎛ R2 ⎞ 1 Now VTH = ⎜ ⎟ ⋅ VCC = ⋅ RTH ⋅ VCC R R R + ⎝ 1 2 ⎠ 1 1 so VTH = ( 3.02 )( 5 ) R1

We can write VTH = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE

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or

1 ( 3.02 )( 5 ) = ( 0.0139 )( 3.02 ) + 0.7 + (151)( 0.0139 )( 0.2 ) R1

We obtain R1 = 13 k Ω and then R2 = 3.93 k Ω EX5.18 β = 150, RTH = R1 || R2 5−0 I CQ = = 5 mA 1 I CQ 5 = = 0.0333 mA I BQ = β 150 I BQ =

VTH − VBE ( on ) − ( −5 ) RTH + (1 + β ) RE

Set RTH = ( 0.1)(1 + β ) RE

⎛ R2 ⎞ We have VTH = ⎜ ⎟ (10 ) − 5 ⎝ R1 + R2 ⎠

Then I BQ

⎛ R2 ⎞ ⎜ ⎟ (10 ) − 0.7 R + R2 ⎠ = 0.0333 = ⎝ 1 (1.1)(151)( 0.2 )

⎛ R2 ⎞ which yields ⎜ ⎟ = 0.1806 ⎝ R1 + R2 ⎠ ⎛ RR ⎞ Now RTH = ⎜ 1 2 ⎟ = ( 0.1)(151)( 0.2 ) = 3.02 k Ω ⎝ R1 + R2 ⎠ so R1 ( 0.1806 ) = 3.02 k Ω

We obtain R1 = 16.7 k Ω and R2 = 3.69 k Ω EX5.19 V + − V − − VECQ 5 − ( −5 ) − 5 I CQ ≅ = 4.5 + 0.5 RC + RE

so I CQ = 1 mA, and I BQ =

I CQ

β

=

1 = 0.00833 mA 120

RTH = ( 0.1)(1 + β ) RE = ( 0.1)(121)( 0.5 )

or RTH = 6.05 k Ω We can write V + = I EQ RE + VEB ( on ) + I BQ RTH + VTH We have VTH =

1 ⋅ RTH (10 ) − 5 and if we let I EQ ≅ I CQ = 1 mA, R1

then we have 5 = (1)( 0.5 ) + 0.7 + ( 0.00833)( 6.05 ) +

1 ( 6.05 )(10 ) − 5 R1

which yields R1 = 6.91 k Ω and R2 = 48.6 k Ω EX5.20 ⎛ 2⎞ 2 ⎞ ⎛ I1 = I Q ⎜ 1 + ⎟ = ( 0.25 ) ⎜1 + ⎟ = 0.2625 mA ⎝ 40 ⎠ ⎝ β⎠ − 0 − VBE ( on ) − V 0 − 0.7 − ( −5 ) = R1 = 0.2625 I1 or R1 = 16.38 k Ω

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For VCEO = 3 V , then VCO = 2.3 V ⎛ β ⎞ ⎛ 40 ⎞ I CO = ⎜ ⎟ ⋅ I Q = ⎜ ⎟ ( 0.25 ) = 0.2439 mA ⎝ 41 ⎠ ⎝ 1+ β ⎠ + V − VCO 5 − 2.3 RC = = = 11.07 k Ω I CO 0.2439

EX5.21 RTH 50 ||100 = 33.3 k Ω ⎛ 50 ⎞ VTH = VTH = ⎜ ⎟ (10 ) − 5 = −1.67 V ⎝ 50 + 100 ⎠ −1.67 − 0.7 − ( −5 ) I B1 = ⇒ 11.2 μ A 33.3 + (101)( 2 ) I C1 = 1.12 mA, I E1 = 1.13 mA

VE1 = I E1 RE1 − 5 = (1.13)( 2 ) − 5 = −2.74 V VCE1 = 3.25 V ⇒ VC1 = 0.51 V Now VE 2 = 0.51 + 0.7 = 1.21 V 5 − 1.21 = 1.90 mA ⇒ I B 2 = 18.8 μ A 2 = 1.88 mA

IE2 = IC 2

I R1 = I C1 − I B 2 = 1.12 − 0.0188 = 1.10 mA 5 − 0.51 = 4.08 k Ω 1.10 = 2.5 ⇒ VC 2 = VE 2 − VEC 2

RC1 = VEC 2

= 1.21 − 2.5 = −1.29 V RC 2 =

−1.29 − ( −5 ) 1.88

= 1.97 k Ω

EX5.22 12 = 240 k Ω = R1 + R2 + R3 0.05 Then VB1 = ( 0.5)( 2 ) + 0.7 = 1.7 V We find

1.7 = 34 k Ω 0.05 = ( 0.5 )( 2 ) + 4 + 0.7 = 5.7 V

R3 = Also VB 2

ΔVR 2 = 5.7 − 1.7 = 4 V 4 = 80 k Ω 0.05 and R1 = 240 − 80 − 34 = 126 k Ω

so R2 =

VC 2 = 1 + 4 + 4 = 9 V Then RC =

V + − VC 2 12 − 9 = = 6 kΩ I CQ 0.5

Test Your Understanding Exercises TYU5.1

α=

β 1+ β

For β = 75, α =

75 = 0.9868 76

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For β = 125, α =

125 = 0.9921 126

TYU5.2 I E = (1 + β ) I B IE 0.780 = = 81.25 then β = 80.3 I B 0.00960

so 1 + β = Now

α=

β

=

1+ β

80.25 = 0.9877 81.25

I C = α I E = ( 0.9877 )( 0.78 ) = 0.770 mA

TYU5.3

β=

α 0.990 = = 99 1 − α 1 − 0.990

Now I B =

IE 2.15 = ⇒ 21.5 μ A and I C = α I E = ( 0.990 )( 2.15 ) = 2.13 mA 1 + β 100

TYU5.4 V 150 ro = A = IC IC

For I C = 0.1 mA ⇒ ro = 1.5 M Ω For I C = 1.0 mA ⇒ ro = 150 k Ω For I C = 10 mA ⇒ ro = 15 k Ω TYU5.5 ⎛ V ⎞ I C = I O ⎜1 + CE ⎟ VA ⎠ ⎝ At VCE = 1 V , I C = 1 mA 1 ⎞ ⎛ For VA = 75 V , I C = 1 = I O ⎜1 + ⎟ ⇒ I O = 0.9868 mA 75 ⎝ ⎠ Then, at VCE = 10 V

(a)

⎛ 10 ⎞ I C = ( 0.9868 ) ⎜ 1 + ⎟ = 1.12 mA ⎝ 75 ⎠

(b) At VCE

1 ⎞ ⎛ For VA = 150 V , I C = 1 = I O ⎜ 1 + ⎟ ⇒ I O = 0.9934 mA ⎝ 150 ⎠ 10 ⎞ ⎛ = 10 V , I C = ( 0.9934 ) ⎜ 1 + ⎟ = 1.06 mA ⎝ 150 ⎠

TYU5.6

BVCEO =

BVCBO n β

so BVCBO = 3 100 ( 30 ) = 139 V

TYU5.7 (a) For VI = 0.2 V < VBE ( on ) ⇒ I B = I C = 0, VO = 5 V and P = 0

(b) IB =

For VI = 3.6 V , transistor is driven into saturation, so VI − VBE ( on ) RB

=

V + − VCE ( sat ) 5 − 0.2 3.6 − 0.7 = 4.53 mA and I C = = = 10.9 mA RC 0.440 0.64

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I C 10.9 = = 2.41 < β which shows that the transistor is indeed driven into saturation. Now, I B 4.53

Note that

P = I BVBE ( on ) + I CVCE ( sat )

= ( 4.53)( 0.7 ) + (10.9 )( 0.2 ) = 5.35 mW

TYU5.8 For VBC = 0 ⇒ VO = 0.7 V

Then I C =

I 5 − 0.7 9.77 = 9.77 mA and I B = C = = 0.195 mA 0.44 β 50

Now VI = I B RB + VBE ( on ) = ( 0.195 )( 0.64 ) + 0.7 or VI = 0.825 V Also P = I BVBE ( on ) + I CVCE

= ( 0.195 )( 0.7 ) + ( 9.77 )( 0.7 ) = 6.98 mW

TYU5.9 V + − VC 10 − 6.34 IC = = = 0.915 mA RC 4 −VBE ( on ) − V −

And I E = Now α =

RE

=

−0.7 − ( −10 ) 10

or IE = 0.930

I C 0.915 α 0.9839 = = 0.9839 and β = = = 61 I E 0.930 1 − α 1 − 0.9839

Also I B = I E − I C = 0.930 − 0.915 ⇒ 15 μ A and VCE = VC − VE = 6.34 − ( −0.7 ) = 7.04 V TYU5.10 10 − 0.7 IE = = 1.16 mA 8 1.1625 IB = = 22.8 μ A 51 I C = ( 50 )( 22.8 μ A ) = 1.14 mA VE = V + − I E RE = 10 − (1.1625 )( 8 ) = 0.7 V VC = I C RC − 10 = (1.1397 )( 4 ) − 10 = −5.44 VEC = 0.7 − ( −5.44 ) VEC = 6.14 V TYU5.11 VBB = I B RB + VBE ( on ) + I E RE

or VBB = I B RB + VBE ( on ) + (1 + β ) I B RE Then I B =

VBB − VBE ( on )

RB + (1 + β ) RE

=

2 − 0.7 10 + ( 76 )(1)

or I B = 15.1 μ A Also I C = ( 75 )(15.1 μ A ) = 1.13 mA and I E = ( 76 )(15.1 μ A ) = 1.15 mA Now VCE = VCC + VBB − I C RC − I E RE

= 8 + 2 − (1.13)( 2.5 ) − (1.15 )(1) = 6.03 V

TYU5.12 VCE = 2.5 V ⇒ VE = 2.5 V = I E RE

We have VBB = I B RB + VBE ( on ) + VE

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so I B =

VBB − VBE ( on ) − VE RB

=

5 − 0.7 − 2.5 10

or I B = 0.18 mA Then I E = (101)( 0.18 ) = 18.2 mA And RE =

2.5 = 0.138 k Ω ⇒ 138 Ω 18.18

TYU5.13 VBB = I E RE + VEB ( on ) + I B RB 2.2 = 0.0431 mA 51 ⎛ β ⎞ ⎛ 50 ⎞ and I C = ⎜ ⎟ ⋅ I E = ⎜ ⎟ ( 2.2 ) = 2.16 mA ⎝ 51 ⎠ ⎝ 1+ β ⎠ Then VBB = ( 2.2 )(1) + 0.7 + ( 0.0431)( 50 ) I E = 2.2 mA ⇒ I B =

or VBB = 5.06 V Now VEC = 5 − I E RE = 5 − ( 2.2 )(1) = 2.8 V TYU5.14 (a) For vI = 0, iB = iC = 0, vO = 12 V , P = 0

For vI = 12 V , iB =

(b) iC =

VCC − VCE ( sat ) RC

=

vI − VBE ( on ) RB

=

12 − 0.7 = 47.1 mA 0.24

12 − 0.1 = 2.38 A 5

vO = 0.1 V and P = iBVBE ( on ) + iCVCE ( sat ) = ( 0.0471)( 0.7 ) + ( 2.38 )( 0.1) = 0.271 W

TYU5.15

For VCEQ = 2.5 V , I CQ =

(a) I BQ =

I CQ

β

=

5 − 2.5 = 1.25 mA 2

1.25 ⇒ I BQ = 12.5 μ A 100

5 − 0.7 = 344 k Ω 0.0125 IBQ is independent of β .

Then RB = (b)

For VCEQ = 1 V , I C =

β=

5 −1 = 2 mA 2

IC 2 = ⇒ β = 160 I B 0.0125

For VCEQ = 4 V , I C =

5−4 = 0.5 mA 2

IC 0.5 = ⇒ β = 40 I B 0.0125 So 40 ≤ β ≤ 160

β=

TYU5.16 5 − 0.7 I BQ = = 0.005375 mA 800

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For β = 75, I CQ = β I BQ = ( 75 )( 0.005375 ) Or I CQ = 0.403 mA For β = 150, I CQ = (150 )( 0.005375 ) Or I CQ = 0.806 mA Largest I CQ ⇒ Smallest VCEQ 5 −1 = 4.96 k Ω 0.806 5−4 For β = 75, RC = = 2.48 k Ω 0.403 For β = 150, RC =

5 − 2.5 = 4.14 k Ω 0.604 = 5 − ( 0.403)( 4.14 ) = 3.33 V

For a nominal I CQ = 0.604 mA and VCEQ = 2.5 V , RC = Now for I CQ = 0.403 mA, VCEQ

For I CQ = 0.806 mA, VCEQ = 5 − ( 0.806 )( 4.14 ) = 1.66 V So, for RC = 4.14 k Ω, 1.66 ≤ VCEQ ≤ 3.33 V TYU5.17 (a) ⎛ β ⎞ ⎛ 100 ⎞ I CQ = ⎜ ⎟ ⋅ I EQ = ⎜ ⎟ (1) = 0.99 mA ⎝ 101 ⎠ ⎝1+ β ⎠ I EQ 1 = ⇒ 9.90 μ A I BQ = 1 + β 101 VB = − I BQ RB = − ( 0.0099 )( 50 ) or VB = −0.495 V

⎛ I CQ ⎞ ⎛ 0.99 × 10−3 ⎞ VBE = VT ln ⎜ ⎟ = ( 0.026 ) ln ⎜ −14 ⎟ ⎝ 3 × 10 ⎠ ⎝ IS ⎠ or VBE = 0.630 V Then VE = VB − VBE = −0.495 − 0.630 = −1.13 V VC = 10 − ( 0.99 )( 5 ) = 5.05 V

Then VCEQ = VC − VE = 5.05 − ( −1.13) = 6.18 V

(b) I EQ

1 = 0.0196 mA 1 + β 51 VB = − ( 0.0196 )( 50 ) = −0.98 V

I EQ = 1 mA, I BQ =

⎛ β I CQ = ⎜ ⎝1+ β

=

⎞ ⎛ 50 ⎞ ⎟ ⋅ I EQ = ⎜ ⎟ (1) = 0.98 mA ⎝ 51 ⎠ ⎠

⎛ 0.98 × 10−3 ⎞ VBE = ( 0.026 ) ln ⎜ = 0.629 V −14 ⎟ ⎝ 3 × 10 ⎠ VE = −0.98 − 0.629 = −1.61 V VC = 10 − ( 0.98 )( 5 ) = 5.1 V

VCEQ = VC − VE = 5.1 − ( −1.61) = 6.71 V

TYU5.18 IQ IQ IB = = 1 + β 121 and

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⎛ IQ ⎞ VB = ⎜ ⎟ ( 20 ) = I Q ( 0.165 ) ⎝ 121 ⎠ VE = I Q ( 0.165 ) + 0.7 ⎛ β ⎞ ⎛ 120 ⎞ I CQ = ⎜ ⎟ ⋅ I EQ = ⎜ ⎟ ⋅ I Q = ( 0.992 ) I Q ⎝ 121 ⎠ ⎝ 1+ β ⎠ VC = I CQ RC − 5 = ( 0.992 ) I Q ( 4 ) − 5 = 3.97 I Q − 5 VECQ = VE − VC = ⎡⎣ I Q ( 0.165 ) + 0.7 ⎤⎦ − ⎡⎣ I Q ( 3.97 ) − 5⎤⎦ = −3.805 I Q + 5.7

Then 3 = 5.7 − 3.805I Q which yields I Q = 0.710 mA

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Chapter 5 Problem Solutions 5.1 (a)

β=

iC 510 = ⇒ β = 85 6 iB

β

85 ⇒ α = 0.9884 1 + β 86 iE = (1 + β ) iB = ( 86 )( 6 ) ⇒ iE = 516 μ A

α=

(b)

=

2.65 ⇒ β = 53 0.050 53 α = ⇒ α = 0.9815 54 iE = (1 + β ) iB = ( 54 )( 0.050 ) ⇒ iE = 2.70 mA

β=

5.2 (a)

For β = 110: α =

β 1+ β

=

110 = 0.99099 111

180 = 0.99448 181 0.99099 ≤ α ≤ 0.99448

For β = 180: α =

(b)

I C = β I B = 110 ( 50 μ A ) ⇒ I C = 5.50 mA

or I C = 180 ( 50 μ A ) ⇒ I C = 9.00 mA so 5.50 ≤ I C ≤ 9.0 mA 5.3 (a)

iE

α (b)

1.12 ⇒ 9.33 μ A 120 ⎛ 121 ⎞ = (1.12 ) ⎜ ⎟ = 1.13 mA ⎝ 120 ⎠ 120 = = 0.9917 121 50 = = 2.5 mA 20 ⎛ 21 ⎞ = ⎜ ⎟ ( 50 ) = 52.5 mA ⎝ 20 ⎠ 20 = = 0.9524 21

iB =

iB iE

α 5.4 (a)

α

β=

0.9 0.95 0.98 0.99 0.995 0.999

9 19 49 99 199 999

α 1−α

(b)

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α=

β

β

1+ β 0.9524 0.9804 0.9901 0.9934 0.9955 0.9975

20 50 100 150 220 400 5.5

1.2 ⇒ 14.8 μ A 81 ⎛ 80 ⎞ I C = (1.2 ) ⎜ ⎟ = 1.185 mA ⎝ 81 ⎠ 80 α= = 0.9877 81 VC = 5 − (1.185 )( 2 ) = 2.63 V IB =

(a)

0.80 ⇒ 9.88 μ A 81 ⎛ 80 ⎞ I C = ( 0.80 ) ⎜ ⎟ = 0.790 mA ⎝ 81 ⎠ 80 α= = 0.9877 81 VC = 5 − ( 0.790 )( 2 ) = 3.42 V IB =

(b)

Yes, VC > VB so B-C junction is reverse biased in both areas.

(c) 5.6

For VC = 0, I C = IE =

IC

α

=

5 = 2.5 mA 2

2.5 ⇒ I E = 2.546 mA 0.982

5.7 (a)

IC

α VC VC

(b)

(c)

0.75 ⇒ 12.3 μ A 61 ⎛ 60 ⎞ = ( 0.75 ) ⎜ ⎟ = 0.738 mA ⎝ 61 ⎠ 60 = = 0.9836 61 = I C RC − 10 = ( 0.738 )( 5 ) − 10 = −6.31 V

IB =

1.5 ⇒ 24.6 μ A 61 ⎛ 60 ⎞ I C = (1.5 ) ⎜ ⎟ = 1.475 mA ⎝ 61 ⎠ ⎛ 60 ⎞ α = ⎜ ⎟ = 0.9836 ⎝ 61 ⎠ VC = (1.475 )( 5 ) − 10 ⇒ VC = −2.625 V IB =

Yes, VC < 0 in both cases so that B-C junction is reverse biased.

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5.8 IC = IE =

VC − ( −10 ) RC IC

α

=

=

10 − 1.2 = 1.76 mA 5

1.76 ⇒ I E = 1.774 mA 0.992

5.9 I C = I S eVRE / VT ⎛ 0.685 ⎞ = 10−13 exp ⎜ ⎟ ⇒ I C = 27.67 mA ⎝ 0.026 ⎠ ⎛1+ β ⎞ ⎛ 91 ⎞ IE = ⎜ ⎟ I C = ⎜ ⎟ ( 27.67 ) ⎝ 90 ⎠ ⎝ β ⎠ I D = 27.98 mA I 27.67 IB = C = ⇒ I B = 0.307 mA β 90

5.10 Device 1: iE = I Eo1evEB / VT ⇒ 0.5 × 10−3 = I Eo1e0.650 / 0.026 So that I EO1 = 6.94 × 10−15 A Device 2: 12.2 × 10−3 = I Eo 2 e0.650 / 0.026 Or I Eo 2 = 1.69 × 10−13 A Ratio of areas =

I Eo 2 1.69 × 10−13 = ⇒ Ratio = 24.4 I Eo1 6.94 × 10−15

5.11 (a)

ro =

VA 250 = ⇒ ro = 250 k Ω 1 IC

(b)

ro =

VA 250 = ⇒ ro = 2.50 M Ω IC 0.1

5.12 BVC E 0 =

BVC B 0 3

β

=

60 3

100

BVC E 0 = 12.9 V

5.13 BVC E 0 = 56 =

220 3

β

BVC B 0 3

β

⇒3β =

220 = 3.93 56

β = 60.6 5.14

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BVC E 0 =

BVC B 0 3

β

BVC B 0 = ( BVC E 0 ) 3 β = ( 50 ) 3 50 BVC B 0 = 184 V

5.15 (a)

IE =

−0.7 − ( −10 )

= 1.86 mA 5 ⎛ 75 ⎞ I C = (1.86 ) ⎜ ⎟ = 1.836 mA ⎝ 76 ⎠ VC = −0.7 + 4 = 3.3 V 10 − 3.3 ⇒ RC = 3.65 K 1.836 0.5 IB = = 0.00658 mA 76 VB = I B RB = ( 0.00658 )( 25 ) ⇒ VB = 0.164 V

RC =

(b)

(c)

⎛ 75 ⎞ I C = ( 0.5 ) ⎜ ⎟ = 0.493 mA ⎝ 76 ⎠ −1 − ( −5 ) RC = ⇒ RC = 8.11 K 0.493 I O = E (10 ) + 0.7 + I E ( 4 ) − 8 76 7.3 = I E ( 4 + 0.132 ) ⇒ I E = 1.767 mA

⎛ 75 ⎞ I C = (1.767 ) ⎜ ⎟ = 1.744 mA ⎝ 76 ⎠ VCE = 8 − (1.744 )( 4 ) − ⎡⎣(1.767 )( 4 ) − 8⎤⎦ = 16 − 6.972 − 7.068 ⇒ VCE = 1.96 V

(d)

⎛I ⎞ = I E (10 + 0.263 + 2 ) + 0.7 5 = I E (10 ) + ⎜ E ⎟ ( 20 ) + 0.7 + I E ( 2 ) ⎝ 76 ⎠ I E = 0.3506 mA ⇒ I B = 4.61 μ A VC = 5 − ( 0.3506 )(10 ) VC = 1.49 V

5.16 For Fig. 5.15 (a) RE = 5 + 5% = 5.25 K IE =

−0.7 − ( −10 )

= 1.77 mA 5.25 I C = 1.75 mA 10 − 3.3 = 3.83 K RC = 1.75 RE = 5 − 5% = 4.75 K IE =

−0.7 − ( −10 )

= 1.96 mA 4.75 I C = 1.93 mA 10 − 3.3 = 3.47 K RC = 1.93 So 1.75 ≤ I C ≤ 1.93 mA 3.47 ≤ RC ≤ 3.83 K For Fig. 5.15(c) RE = 4 + 5% = 4.2 K

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IB =

8 − 0.7 = 0.0222 mA 10 + ( 76 )( 4.2 )

I C = 1.66 mA

I E = 1.69 mA VCE = 16 − (1.66 )( 4 ) − (1.69 )( 4.2 ) = 16 − 6.64 − 7.098 ⇒ VCE = 2.26 V RE = 4 − 5% = 3.8 K 8 − 0.7 = 0.0244 I C = 1.83 mA IB = 10 + ( 76 )( 3.8 ) I E = 1.86 mA VCE = 16 − (1.83)( 4 ) − (1.86 )( 3.8 ) = 16 − 7.32 − 7.068 VCE = 1.61 V So 1.66 ≤ I C ≤ 1.83 mA 1.61 ≤ VCE ≤ 2.26 V

5.17 RB =

VBB − VEB 2.5 − 0.7 = ⇒ RB = 120 k Ω IB 0.015

I CQ = ( 70 )(15μ A ) ⇒ 1.05 mA RC =

VCC − VECQ I CQ

=

5 − 2.5 ⇒ 1.05

5.18 (a) VB = − I B RB ⇒ I B =

RC = 2.38 k Ω

−VB − ( −1) = 500 RB

I B = 2.0 μ A VE = −1 − 0.7 = −1.7 V IE =

VE − ( −3) RE

=

−1.7 + 3 = 0.2708 mA 4.8

IE 0.2708 = (1 + β ) = = 135.4 ⇒ β = 134.4 0.002 IB

β

⇒ α = 0.9926 1+ β I C = β I B ⇒ I C = 0.269 mA

α=

VCE = 3 − VE = 3 − ( −1.7 ) ⇒ VCE = 4.7 V

(b) 5−4 ⇒ I E = 0.5 mA 2 4 = 0.7 + I B RB + ( I B + I C ) RC − 5 IE =

I B + IC = I E I B + IC = I E

4 = 0.7 + I B (100 ) + ( 0.5 )( 8 ) − 5 I B = 0.043 ⇒

IE 0.5 = (1 + β ) = = 11.63 IB 0.043

β = 10.63, α =

β 1+ β

⇒ α = 0.9140

5.19

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VB − 0.7 − ( −5 )

VB + 4.3 3 3 V + 4.3 ⎛ ⎞ ⎛ 50 ⎞ ⎛ 50 ⎞ IC = ⎜ ⎟ I E = ⎜ B ⎟ ⎜ 51 ⎟ 3 ⎝ 51 ⎠ ⎝ ⎠⎝ ⎠ ⎛ V + 4.3 ⎞ ⎛ 50 ⎞ VC = 5 − I C RC = 5 − ⎜ B ⎜ ⎟ (10 ) 3 ⎟⎠ ⎝ 51 ⎠ ⎝ IE =

=

Now VB = VC , so VB [1 + 3.27 ] = 5 − 14.1 = −9.05 VB = −2.12 V IE =

−2.12 + 4.3 ⇒ I E = 0.727 mA 3

5.20 10 − VE 10 − 2 = ⇒ I E = 0.80 mA 10 10 VB = VE − 0.7 = 2 − 0.7 = 1.3 V IE =

IB =

VB 1.3 = ⇒ I B = 0.026 mA RB 50

I C = I E − I B = 0.80 − 0.026 ⇒ I C = 0.774 mA

β= α=

I C 0.774 = ⇒ β = 29.77 I B 0.026

β 1+ β

=

29.77 ⇒ α = 0.9675 30.77

VEC = VE − VC = VE − ( I C RC − 10 ) = 2 − ⎡⎣( 0.774 )(10 ) − 10 ⎤⎦

VEC = 4.26 V

Load line developed assuming the VB voltage can change and the RB resistor is removed.

1 mA 0.774

Q-point

IC

4.26

20

VEC

5.21 5 − 0.7 ⇒ 17.2 μ A 250 I C = (120 )( 0.0172 ) = 2.064 mA IB =

VC = ( 2.064 )(1.5 ) − 5 = −1.90 V

VEC = 5 − ( −1.90 ) ⇒ VEC = 6.90 V

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IC (mA) 6.67

Q-point 2.06

6.9

10 V

EC (V)

5.22 ⎛ 50 ⎞ I C = ⎜ ⎟ (1) = 0.98 mA ⎝ 51 ⎠ VC = I C RC − 9 = ( 0.98 )( 4.7 ) − 9 or VC = −4.39 V 1 = 0.0196 mA 51 VE = I B RB + VEB ( on ) = ( 0.0196 )( 50 ) + 0.7 or VE = 1.68 V IB =

5.23 0.5 ⎛ 50 ⎞ I C = ⎜ ⎟ ( 0.5 ) = 0.49 mA, I B = = 0.0098 mA 51 ⎝ 51 ⎠ VE = I B RB + VEB ( on ) = ( 0.0098 )( 50 ) + 0.7 or VE = 1.19 V VC = I C RC − 9 = ( 0.49 )( 4.7 ) − 9 = −6.70 V Then VEC = VE − VC = 1.19 − ( −6.7 ) == 7.89 V PQ = I CVEC + I BVEB = ( 0.49 )( 7.89 ) + ( 0.0098 )( 0.7 ) or PQ = 3.87 mW Power Dissipated = PS = I Q ( 9 − VE ) = ( 0.5 )( 9 − 1.19 )

Or PS = 3.91 mW 5.24 I ⇒ I E1 = I E 2 = 0.5 mA 2 ≈ 0.5 mA = 5 − ( 0.5 )( 4 ) ⇒ VC1 = VC 2 = 3 V

I E1 = I E 2 = I C1 = I C 2 VC1 = VC 2

5.25 (a)

RE = 0 I B =

2 − 0.7 1.3 = RB RB

⎛ 1.3 ⎞ 5 − 2 = 0.8 ⇒ RC = 3.75 K I C = ( 80 ) ⎜ ⎟ = RC ⎝ RB ⎠ RB = 130 K

(b)

0.8 ⎛ 81 ⎞ = 0.010 mA I E = 0.8 ⎜ ⎟ = 0.81 mA 80 ⎝ 80 ⎠ 2 = ( 0.010 )( RB ) + 0.7 + ( 0.81)(1) ⇒ RB = 49 K RE = 1 K

IB =

5 = ( 0.8 ) RC + 2 + ( 0.81)(1) ⇒ RC = 2.74 K

(c)

For part (a)

IB =

2 − 0.7 = 0.01 mA 130

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I C = (120 )( 0.01) ⇒ I C = 1.20 mA VCE = 5 − (1.2 )( 3.75 ) ⇒ VCE = 0.5 V 2 = I B ( 49 ) + 0.7 + (121) I B (1)

For part (b)

I B = 0.00765 mA, I E = 0.925 mA, I C = 0.918 mA VCE = 5 − ( 0.918 )( 2.74 ) − ( 0.925 )(1) ⇒ VCE = 1.56 V Including RE result in smaller changes in Q-point values. 5.26 I BQ =

a.

VCC − VBE ( on ) RB

2 = 0.0333 mA 60 β 24 − 0.7 RB = ⇒ RB = 699 kΩ 0.0333 VCC − VCEQ 24 − 12 I CQ = ⇒ RC = ⇒ RC = 6 kΩ RC 2 I BQ =

I CQ

I BQ =

b.

=

VCC − VBE ( on ) RB

24 − 0.7 699

=

= 0.0333 mA ( Unchanged )

I CQ = β I BQ = (100 )( 0.0333) ⇒ I CQ = 3.33 mA VCEQ = VCC − I CQ RC = 24 − ( 3.33)( 6 ) ⇒ VCEQ = 4.02 V VCE = VCC − I C RC = 24 − I C ( 6 )

(c) IC (mA) 4 3.33

Q-pt (␤ ⫽ 100)

Q-pt (␤ ⫽ 60)

2

4.02

12

24

5.27 a.

VB = 0 ⇒ Cutoff ⇒ I E = 0, VC = 6 V

b.

VB = 1 V, I E =

c.

VB = 2 V. Assume active-mode

VCE

1 − 0.7 ⇒ I E = 0.3 mA 1 I C ≈ I E ⇒ VC = 6 − ( 0.3)(10 ) ⇒ VC = 3 V

2 − 0.7 = I E = 1.3 mA ≈ I C 1 VC = 6 − (1.3)(10 ) = −7 V! IE =

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Transistor in saturation 2 − 0.7 ⇒ I E = 1.3 mA 1 VE = 1.3 V, VCE ( sat ) = 0.2 V IE =

VC = VE + VCE ( sat ) = 1.3 + 0.2 ⇒ VC = 1.5 V

5.28 a.

VBB = 0. ⎛ RL Cutoff V0 = ⎜ ⎝ RC + RL V0 = 3.33 V

⎞ ⎛ 10 ⎞ ⎟ VCC = ⎜ ⎟ ( 5) ⎝ 10 + 5 ⎠ ⎠

VBB = 1 V

b.

1 − 0.7 ⇒ 6 μA 50 I C = β I B = ( 75 )( 6 ) ⇒ I C = 0.45 mA IB =

5 − V0 V = IC + 0 5 10 ⎛1 1 ⎞ 1 − 0.45 = V0 ⎜ + ⎟ ⇒ V0 = 1.83 V ⎝ 5 10 ⎠

Transistor in saturation V0 = VCE ( sat ) = 0.2 V

c.

5.29 (a) β = 100 ⎛ 100 ⎞ (i) I Q = 0.1 mA I C = ⎜ ⎟ ( 0.1) = 0.0990 mA ⎝ 101 ⎠ VO = 5 − ( 0.099 )( 5 ) ⇒ VO = 4.505 V ⎛ 100 ⎞ (ii) I Q = 0.5 mA I C = ⎜ ⎟ ( 0.5 ) = 0.495 mA ⎝ 101 ⎠ VO = 5 − ( 0.495 )( 5 ) ⇒ VO = 2.525 V

(iii) I Q = 2 mA Transistor is in saturation VO = −VBE ( sat ) + VCE ( sat ) = −0.7 + 0.2 ⇒ VO = −0.5 V (b) β = 150 ⎛ 150 ⎞ (i) I Q = 0.1 mA I C = ⎜ ⎟ ( 0.1) = 0.09934 mA ⎝ 151 ⎠ VO = 5 − ( 0.09934 )( 5 ) ⇒ VO = 4.503 V 4.503 − 4.505 × 100% = −0.044% 4.503 ⎛ 150 ⎞ (ii) I Q = 0.5 mA I C = ⎜ ⎟ ( 0.5 ) = 0.4967 mA ⎝ 151 ⎠ VO = 5 − ( 0.4967 )( 5 ) ⇒ VO = 2.517 V % change =

2.517 − 2.525 × 100% = −0.32% 2.525 (iii) I Q = 2 mA Transistor in saturation Vo = −8.5 V No change % change =

5.30

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VCB = 0.5 V ⇒ VO = 0.5 V , I C =

5 − 0.5 = 0.90 mA 5

⎛ 101 ⎞ IQ = ⎜ ⎟ ( 0.90 ) ⇒ I Q = 0.909 mA ⎝ 100 ⎠

5.31 For I Q = 0, then PQ = 0 ⎛ 50 ⎞ For I Q = 0.5 mA, I C = ⎜ ⎟ ( 0.5 ) = 0.49 mA ⎝ 51 ⎠ 0.5 IB = = 0.0098 mA, VB = 0.490 V , VE = 1.19 V 51 VC = ( 0.49 )( 4.7 ) − 9 = −6.70 V ⇒ VEC = 7.89 V P ≅ I CVEC = ( 0.49 )( 7.89 ) ⇒ P = 3.87 mW

For I Q = 1.0 mA, Using the same calculations as above, we find P = 5.95 mW For I Q = 1.5 mA, P = 6.26 mW For I Q = 2 mA, P = 4.80 mW For I Q = 2.5 mA, P = 1.57 mW For I Q = 3 mA, Transistor is in saturation. 0.7 + I B ( 50 ) = 0.2 + I C ( 4.7 ) − 9 I E = IQ = I B + IC ⇒ I B = 3 − IC Then, 0.7 + ( 3 − I C ) ( 50 ) = 0.2 + I C ( 4.7 ) − 9

Which yields I C = 2.916 mA and I B = 0.084 mA P = I BVEB + I CVEC = ( 0.084 )( 0.7 ) + ( 2.916 )( 0.2 ) or P = 0.642 mW

5.32 IE =

VEE − VEB ( on ) RE

=

9 − 0.7 ⇒ I E = 2.075 mA 4

I C = α I E = ( 0.9920 ) ( 2.075 ) ⇒ I C = 2.06 mA VBC + I C RC = VCC VBC = 9 − ( 2.06 ) ( 2.2 ) ⇒ VBC = 4.47 V

5.33 I CQ = I BQ = IR2 =

VCC − VCEQ RC I CQ

β

=

=

12 − 6 = 2.73 mA 2.2

2.73 ⇒ I BQ = 0.091 mA 30

0.7 − ( −12 )

= 0.127 mA 100 I R1 = I R 2 + I BQ = 0.127 + 0.091 = 0.218 mA V1 = I R1 R1 + 0.7 = ( 0.218 )(15 ) + 0.7 ⇒ V1 = 3.97 V

5.34

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For VCE = 4.5 5 − 4.5 = 0.5 mA I CQ = 1 0.5 = 0.02 mA I BQ = 25 0.7 − ( −5 ) = 0.057 mA I R2 = 100 I R1 = I R 2 + I BQ = 0.057 + 0.02 = 0.077 mA V1 = I R1 R1 + VBE ( on ) = ( 0.077 )(15 ) + 0.7 = 1.86 V For VCE = 1.0 5 −1 = 4 mA 1 4 = = 0.16 mA 25 = 0.057 mA

I CQ = I BQ I R2

I R1 = I R 2 + I BQ = 0.057 + 0.16 = 0.217 mA

V1 = ( 0.217 )(15 ) + 0.7 ⇒ 3.96 V

So 1.86 ≤ V1 ≤ 3.96 V IC 5 4 Range of Q-pt values

0.5 0

1

4.5

5

5.35 5 − 2.5 =5K 0.5 0.5 IB = = 0.00417 mA 120 5 − 0.7 RB = = 1032 K 0.00417

RC =

(a)

IC (mA) 1.0

Q-point 0.5

2.5

(b)

5 V (V) CE

Choose RC = 5.1 K RB = 1 MΩ

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For RB = 1 MΩ + 10% = 1.1 M, RC = 5.1 k + 10% = 5.61 K 5 − 0.7 I BQ = = 3.91 μ A ⇒ I CQ = 0.469 mA 1.1 VCEQ = 2.37 V RB = 1 MΩ + 10% = 1.1M, RC = 5.1 K − 10% = 4.59 K I BQ = 3.91 μ A ⇒ I CQ = 0.469 mA VCEQ = 2.85 V RB = 1 MΩ − 10% = 0.90 MΩ RC = 5.1 k + 10% = 5.61 K 5 − 0.7 I BQ = = 4.78 μ A ⇒ I C = 0.573 mA 0.90 VCEQ = 1.78 V RB = 1 MΩ − 10% = 0.90 MΩ RC = 5.1 k − 10% = 4.59 K I BQ = 4.78 μ A ⇒ I C = 0.573 mA VCEQ = 2.37 V IC (mA) 1.09 0.891 0.573 0.469

1.78 2.37 2.85

5 V (V) CE

5.36 VE 2 = 5 − VBE 2 VE1 = 5 − VBE1 VO = VE 2 − VE1 = ( 5 − VBE 2 ) − ( 5 − VBE1 ) VO = VBE1 − VBE 2 ⎛I ⎞ We have VBE1 = VE ln ⎜ E1 ⎟ ⎝ I EO ⎠ ⎛I ⎞ VBE 2 = VT ln ⎜ E 2 ⎟ ⎝ I EO ⎠ ⎡ ⎛I ⎞ ⎛I VO = VT ⎢ln ⎜ E1 ⎟ − ln ⎜ E 2 ⎝ I EO ⎣ ⎝ I EO ⎠

⎞⎤ ⎟⎥ ⎠⎦

⎛I ⎞ ⎛ 10 ⎞ VO = VT ln ⎜ E1 ⎟ = VT ln ⎜ I ⎟ ⎝ I ⎠ ⎝ IE2 ⎠ VO =

5.37 (a)

kT ln (10 ) e

RE = 0

(120 )( 4 ) VI − 0.7 IC = β I B VO = 5 − I C ( 4 ) = 5 − (VI − 0.7 ) 200 200 When VO = 0.2, 0.2 = 5 − 2.4 VI + 1.68 ⇒ VI = 2.7 IB =

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VO(V) 5

0.2 0.7

2.7

5

VI (V)

RE = 1 K VI − 0.7 V − 0.7 IB = = I I C = β IB 200 + (121)(1) 321

(b)

(120 )( 4 )

(VI − 0.7 ) 321 When VO = 0.2 = 5 − 1.495VI + 1.047

VO = 5 −

VI = 3.91 V VO(V) 5

0.2 0.7

3.91

5

VI (V)

5.38 For 4.3 ≤ VI ≤ 5 Q is cutoff I C = 0 VO = 0 If Q reaches saturation, VO = 4.8 4.8 IC = = 1.2 mA 4 5 − 0.7 − VI 1.2 IB = = 0.015 = ⇒ VI = 1.6 80 180 So VI ≤ 1.6, VO = 4.8

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VO(V)

4.8

1.6

5.39 (a)

4.3

5

VI (V)

For VI ≥ 4.3, Q is off and VO = 0

⎛ 101 ⎞ When transistor enters saturation, 5 = ⎜ ⎟ I C (1) + 0.2 + I C ( 4 ) ⇒ I C = 0.958 mA ⎝ 100 ⎠ VO = 3.832 V I B = 0.00958 mA

⎛ 101 ⎞ 5=⎜ ⎟ ( 0.958 )(1) + 0.7 + ( 0.00958 )(180 ) + VI ⎝ 100 ⎠ VI = 5 − 0.7 − 0.9676 − 1.7244 ⇒ VI = 1.61 V For VI = 0, transistor in saturation

5 = I E (1) + 0.2 + I C ( 4 ) ⇒ 5 = I C (1) + I B (1) + 0.2 + I C ( 4 ) 5 = I E (1) + 0.7 + I B (180 ) 5 = I C (1) + I B (1) + 0.7 + I B (180 ) I E = IC + I B

4.8 = 5 I C + I B (1) 4.3 = 1I C + 181I B I B = 4.8 − 5 I C

4.3 = I C + (181)( 4.8 − 5 I C ) 904 I C = 864.5 I C = 0.956 mA VO = 3.825 V

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VO(V) 3.832 3.825

1.61

4.3

5

VI (V)

5.40 RTH = R1 & R2 = 33 & 10 = 7.67 kΩ ⎛ R2 ⎞ ⎛ 10 ⎞ VTH = ⎜ ⎟ VCC = ⎜ ⎟ (18 ) = 4.186 V R + R ⎝ 10 + 33 ⎠ ⎝ 1 2 ⎠ V − VBE ( on ) 4.186 − 0.7 I BQ = TH = RTH + (1 + β ) RE 7.67 + ( 51)(1) I BQ = 0.0594 mA I CQ = β I BQ ⇒ I CQ = 2.97 mA I EQ = 3.03 mA VCEQ = VCC − I CQ RC − I EQ RE

= 18 − ( 2.97 )( 2.2 ) − ( 3.03)(1) ⇒ VCEQ = 8.44 V

5.41 I CQ = 1.2 mA, VCEQ = 9 V , RTH = 50 k Ω 1.2 = 0.015 mA Also I B = 80 VTH = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE ⎛ R2 ⎞ 1 1 VTH = ⎜ ⎟ (VCC ) = ⋅ RTH ⋅ VCC = ( 50 )(18 ) R1 R1 ⎝ R1 + R2 ⎠ 1 Then ( 50 )(18) = ( 0.015)( 50 ) + 0.7 + ( 81)( 0.015)(1) or R1 = 338 k Ω. R1 Then

338R2 = 50 ⇒ R2 = 58.7 k Ω 338 + R2

⎛ 81 ⎞ I EQ = ⎜ ⎟ (1.2 ) = 1.215 mA ⎝ 80 ⎠ 18 = I CQ RC + VCEQ + I EQ RE

18 = (1.2 ) RC + 9 + (1.215 )(1) ⇒ RC = 6.49 k Ω

5.42

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RTH = R1 & R2 = 20 & 15 = 8.57 k Ω ⎛ R2 ⎞ ⎛ 15 ⎞ VTH = ⎜ ⎟ (VCC ) = ⎜ ⎟ (10 ) = 4.29 V ⎝ 15 + 20 ⎠ ⎝ R1 + R2 ⎠ I EQ VCC = I EQ RE + VEB ( on ) + ⋅ RTH + VTH 1+ β ⎛ 8.57 ⎞ 10 = I EQ (1) + 0.7 + I EQ ⎜ ⎟ + 4.29 ⎝ 101 ⎠ 10 − 0.7 − 4.29 5.01 Then I EQ = = ⇒ I EQ = 4.62 mA 8.57 1.085 1+ 101 I EQ ⎛ 4.62 ⎞ VB = ⋅ RTH + VTH = ⎜ ⎟ ( 8.57 ) + 4.29 or VB = 4.68 V 1+ β ⎝ 101 ⎠

5.43 (a) RTH = 42 & 58 = 24.36 K ⎛ 42 ⎞ VTH = ⎜ ⎟ ( 24 ) = 10.08 V ⎝ 100 ⎠ 10.08 − 0.7 9.38 I BQ = = ⇒ 7.30 μ A 24.36 + (126 )(10 ) 1284.36 I CQ = 0.913 mA VCEQ = 14.8 V

I EQ = 0.9202

IC (mA)

2.38

Q-point 0.913

14.8

24 V

(b) R1 + 5% = 60.9, R2 + 5% = 44.1 RTH = 25.58 K 10.08 − 0.7 9.38 I BQ = = ⇒ 7.30 μ A 25.58 + 126 (10 ) 1285.58 I CQ = 0.912 mA VCEQ = 14.81

CE (V)

VTH = 10.08

I EQ = 0.919

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R1 + 5% = 60.9, R2 − 5% = 39.90 RTH = 24.11 K 9.50 − 0.7 8.8 I BQ = = = 6.85 μ A 24.11 + (126 )(10 ) 1284.11 I CQ = 0.857 mA VCEQ = 15.37 V

VTH = 9.50

I EQ = 0.8635 mA

R1 − 5% = 55.1 K R2 + 5% = 44.1 K 10.67 − 0.7 9.97 I BQ = = = 7.76 μ A 24.50 + 1260 1284.5 I CQ = 0.970 mA I EQ = 0.978 mA VCEQ = 14.22 V R1 − 5% = 55.1 K R2 − 5% = 39.90 10.08 − 0.7 9.38 I BQ = = = 7.31 μ A 23.14 + 1260 1283.14 I CQ = 0.914 mA I EQ = 0.9211 mA VCEQ = 14.79 V

RTH = 24.50 K

RTH = 23.14 K

VTH = 10.67 V

VTH = 10.08

So we have 0.857 ≤ I CQ ≤ 0.970 mA 14.22 ≤ VCEQ ≤ 15.37 V 5.44 a. RTH = R1 & R2 = 25 & 8 = 6.06 kΩ ⎛ R2 ⎞ ⎛ 8 ⎞ VTH = ⎜ ⎟ VCC = ⎜ ⎟ ( 24 ) ⎝ 8 + 25 ⎠ ⎝ R1 + R2 ⎠ = 5.82 V V − VBE (on) 5.82 − 0.7 I BQ = TH = RTH + (1 + β ) RE 6.06 + ( 76 )(1)

I BQ = 0.0624 mA, I CQ = 4.68 mA I EQ = 4.74 VCEQ = VCC − I CQ RC − I EQ RE = 24 − ( 4.68 )( 3) − ( 4.74 )(1) VCEQ = 5.22 V

b. I BQ =

5.82 − 0.7 ⇒ I BQ = 0.0326 mA 6.06 + (151)(1)

I CQ = 4.89 mA I EQ = 4.92

VCEQ = 24 − ( 4.89 )( 3) − ( 4.92 )(1) VCEQ = 4.41 V

5.45 (a)

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I CQ ≅ I EQ = 0.4 mA 3 3 ⇒ RC = 7.5 k Ω; RE = ⇒ RE = 7.5 k Ω 0.4 0.4 9 R1 + R2 ≅ = 112.5 k Ω ( 0.2 )( 0.4 ) RC =

⎛ R2 ⎞ VTH = ⎜ ⎟ (VCC ) = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE ⎝ R1 + R2 ⎠ (112.5 − R2 ) R2 RR 0.4 , I BQ = = 0.004 mA RTH = 1 2 = 112.5 100 R1 + R2 ⎡ (112.5 − R2 ) R2 ⎤ ⎛ 9 ⎞ R2 ⎜ ⎥ + 0.7 + (101)( 0.004 )( 7.5 ) ⎟ = ( 0.004 ) ⎢ 112.5 ⎝ 112.5 ⎠ ⎣ ⎦

We obtain R2 ( 0.08 ) = 0.004 R2 − 3.56 × 10−5 R22 + 3.73 From this quadratic, we find R2 = 48 k Ω ⇒ R1 = 64.5 k Ω (b) Standard resistor values: Set RE = RC = 7.5 k Ω and R1 = 62 k Ω, R2 = 47 k Ω Now RTH = R1 & R2 = 62 & 47 = 26.7 k Ω ⎛ R2 ⎞ ⎛ 47 ⎞ VTH = ⎜ ⎟ (VCC ) = ⎜ ⎟ ( 9 ) = 3.88 V ⎝ 47 + 62 ⎠ ⎝ R1 + R2 ⎠ VTH = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE So I BQ =

3.88 − 0.7 = 0.00406 mA 26.7 + (101)( 7.5 )

Then I CQ = 0.406 mA VRC = VRE = ( 0.406 )( 7.5 ) = 3.05 V

5.46 (a) RTH = R1 & R2 = 12 & 2 = 1.714 K ⎛ R2 ⎞ ⎛ 2⎞ VTH = ⎜ ⎟ (10 ) − 5 = ⎜ ⎟ (10 ) − 5 ⇒ VTH = −3.571 V R R + ⎝ 14 ⎠ 2 ⎠ ⎝ 1 (b) VTH = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE − 5 −3.57 = I BQ (1.714 ) + 0.7 + (101) I BQ ( 0.5 ) − 5 I BQ =

5 − 0.7 − 3.571 0.729 = ⇒ 13.96 μ A 1.714 + (101)( 0.5 ) 52.21

I CQ = 1.396 mA, I EQ = 1.410 mA

VCEQ = 10 − (1.396 )( 5 ) − (1.41)( 0.5 ) ⇒ VCEQ = 2.32 V

(d)

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RE = 0.5 + 5% = 0.525 K I BQ =

RC = 5 + 5% = 5.25 K

0.729 ⇒ 13.32 μ A 1.714 + (101)( 0.525 )

I CQ = 1.332 mA

I EQ = 1.345 mA

VCEQ = 10 − (1.332 )( 5.25 ) − (1.345 )( 0.525 )

= 10 − 6.993 − 0.7061 ⇒ VCEQ = 2.30 V

RE = 0.5 + 5% = 0.525 K I CQ = 1.332 mA

RC = 5 − 5% = 4.75 K

I EQ = 1.345 mA

VCEQ = 10 − (1.332 )( 4.75 ) − (1.345 )( 0.525 )

= 10 − 6.327 − 0.7061 ⇒ VCEQ = 2.97 V

RE = 0.5 − 5% = 0.475 K I BQ =

RC = 5 + 5% = 5.25 K

0.729 ⇒ 14.67 μ A 1.714 + (101)( 0.475 )

I CQ = 1.467 mA

I EQ = 1.482 mA

VCEQ = 10 − (1.467 )( 5.25 ) − (1.482 )( 0.475 )

= 10 − 7.70175 − 0.70395 ⇒ VCEQ = 1.59 V

RE = 0.5 − 5% = 0.475 K I CQ = 1.467 mA

RC = 5 − 5% = 4.75 K

I EQ = 1.482 mA

VCEQ = 10 − (1.467 )( 4.75 ) − (1.482 )( 0.475 )

= 10 − 6.96825 − 0.70395 ⇒ VCEQ = 2.33 V

5.47 RTH = R1 & R2 = 9 & 1 = 0.91 kΩ ⎛ R2 ⎞ ⎛ 1 ⎞ VTH = ⎜ ⎟ ( −12 ) = ⎜ ⎟ ( −12 ) = −1.2 V + R R ⎝ 1+ 9 ⎠ ⎝ 1 2 ⎠ I EQ RE + VEB ( on ) + I BQ RTH + VTH = 0 I BQ =

−VTH − VEB (on) 1.2 − 0.7 = RTH + (1 + β ) RE 0.90 + ( 76 )( 0.1)

I BQ = 0.0588,

I CQ = 4.41 mA

I EQ = 4.47 mA Center of load line ⇒ VECQ = 6 V I EQ RE + VECQ + I CQ RC − 12 = 0

( 4.47 ) ( 0.1) + 6 + ( 4.41) RC

= 12 ⇒ RC = 1.26 kΩ

5.48 (a) RTH = 36 & 68 = 23.5 K ⎛ 36 ⎞ VTH = ⎜ ⎟ (10 ) = 3.46 V ⎝ 36 + 68 ⎠ 3.46 − 0.7 I BQ = = 0.00178 mA 23.5 + ( 51)( 30 ) I CQ = 0.0888 mA

I EQ = 0.0906 mA

VCE = 10 − ( 0.0888 )( 42 ) − ( 0.0906 )( 30 ) = 10 − 3.73 − 2.72 ⇒ VCE = 3.55 V

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(b) R1 = 22.7, R2 = 12 K, RC = 14 K, RE = 10 K RTH = 7.85 k VTH = 3.46 3.46 − 0.7 I BQ = = 0.00533 mA 7.85 + ( 51)(10 ) I CQ = 0.266 mA

I EQ = 0.272 mA

VCE = 10 − ( 0.266 )(14 ) − ( 0.272 )(10 ) VCE = 3.56 V

5.49 (a) ⎛ R2 ⎞ ⎛ 68 ⎞ RTH = 36 & 68 = 23.5 K VTH = ⎜ ⎟ (10 ) − 5 = ⎜ ⎟ (10 ) − 5 = 1.54 V R + R ⎝ 36 + 68 ⎠ ⎝ 1 2 ⎠ 5 = ( 51) I BQ ( 30 ) + 0.7 + I B ( 23.5 ) + 1.54 I BQ =

2.76 = 1.78 μA ⇒ I CQ = 0.0888 mA 1553.5 I EQ = 0.0906 mA

VECQ = 10 − ( 0.0906 )( 30 ) − ( 0.0888 )( 42 )

= 10 − 2.718 − 3.7296 ⇒ VECQ = 3.55 V

(b) RTH = 12 & 22.7 = 7.85 K VTH = 1.54 RE = 10 K

RC = 14 K

5 = ( 51) I BQ (10 ) + 0.7 + I B ( 7.85 ) + 1.54 I BQ =

2.76 = 5.33 μ A I CQ = 0.266 mA 517.85 I EQ = 0.272 mA

VECQ = 10 − ( 0.272 )(10 ) − ( 0.266 )(14 )

= 10 − 2.72 − 3.724 ⇒ VECQ = 3.56 V

5.50 (a) RTH = ( 0.1)(1 + β ) RE = ( 0.1)(101)( 0.5 ) = 5.05 k Ω 1 VTH = ⋅ RTH ⋅ VCC = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE R1 I BQ =

I CQ

β

=

0.8 = 0.008 mA 100

1 Then ( 5.05)(10 ) = ( 0.008)( 5.05) + 0.7 + (101)( 0.008)( 0.5) R1 or R1 = 44.1 k Ω,

44.1R2 = 5.05 ⇒ R2 = 5.70 k Ω 44.1 + R2

⎛ 101 ⎞ Now I EQ = ⎜ ⎟ ( 0.8 ) = 0.808 mA ⎝ 100 ⎠ VCC = I CQ RC + VCEQ + I EQ RE

10 = ( 0.8 ) RC + 5 + ( 0.808 )( 0.5 ) RC = 5.75 k Ω

(b)

For 75 ≤ β ≤ 150

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⎛ R2 ⎞ ⎛ 5.7 ⎞ VTH = ⎜ ⎟ (VCC ) = ⎜ ⎟ (10 ) = 1.145 V R R + ⎝ 5.7 + 44.1 ⎠ ⎝ 1 2 ⎠ VTH = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE

For β = 75, I BQ =

1.145 − 0.7 = 0.0103 mA 5.05 + ( 76 )( 0.5 )

Then I CQ = ( 75 )( 0.0103) = 0.775 mA For β = 150, I BQ =

1.145 − 0.7 = 0.00552 mA 5.05 + (151)( 0.5 )

Then I CQ = 0.829 mA % Change =

ΔI CQ

=

I CQ

0.829 − 0.775 × 100% ⇒ % Change = 6.75% 0.80

For RE = 1 k Ω

(c)

RTH = ( 0.1)(101)(1) = 10.1 k Ω VTH =

1 1 ⋅ RTH ⋅ VCC = (10.1)(10 ) = ( 0.008 )(10.1) + 0.7 + (101)( 0.008 )(1) R1 R1

which yields R1 = 63.6 k Ω

And

63.6 R2 = 10.1 ⇒ R2 = 12.0 k Ω 63.6 + R2

⎛ R2 ⎞ ⎛ 12 ⎞ Now VTH = ⎜ ⎟ (VCC ) = ⎜ ⎟ (10 ) = 1.587 V R + R ⎝ 12 + 63.6 ⎠ ⎝ 1 2 ⎠ 1.587 − 0.7 = 0.0103 mA For β = 75, I BQ = 10.1 + ( 76 )(1)

So I CQ = 0.773 mA For β = 150, I BQ =

1.587 − 0.7 = 0.00551 mA 10.1 + (151)(1)

Then I CQ = 0.826 mA % Change =

ΔI CQ I CQ

=

0.826 − 0.773 × 100% ⇒ % Change = 6.63% 0.8

5.51 VCC ≅ I CQ ( RC + RE ) + VCEQ 10 = ( 0.8 )( RC + RE ) + 5 ⇒ RC + RE = 6.25 k Ω Let RE = 0.875 k Ω Then, for bias stable RTH = ( 0.1)(121)( 0.875 ) = 10.6 k Ω I BQ =

0.8 = 0.00667 mA 120

1 (10.6 )(10 ) = ( 0.00667 )(10.6 ) + 0.7 + (121)( 0.00667 )( 0.875 ) R1

So R1 = 71.8 k Ω and

71.8R2 = 10.6 ⇒ R2 = 12.4 k Ω 71.8 + R2

10 = 0.119 mA 71.8 + 12.4 This is close to the design specification.

Then I R ≅

5.52

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= VCC − I CQ ( RC + RE )

I CQ ≈ I EQ ⇒ VCEQ

6 = 12 − I CQ ( 2 + 0.2 )

I CQ = 2.73 mA,

I BQ = 0.0218 mA

VCEQ = 6 V VTH = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE − 6 ⎛ R2 ⎞ VTH = ⎜ ⎟ (12 ) − 6, ⎝ R1 + R2 ⎠

RTH = R 1& R2

Bias stable ⇒ RTH = ( 0.1)(1 + β ) RE = ( 0.1)(126 )( 0.2 ) = 2.52 kΩ ⎛ 1⎞ VTH = ⎜ ⎟ ( RTH )(12 ) − 6 ⎝ R1 ⎠ 1 ( 2.52 )(12 ) − 6 = ( 0.0218)( 2.52 ) + 0.7 + (126 )( 0.0218)( 0.2 ) − 6 R1 1 ( 30.24 ) = 0.7549 + 0.5494 R1 R1 = 23.2 kΩ,

23.2R 2 = 2.52 23.2 + R 2

R2 = 2.83 kΩ

5.53 a.

⎛ 81 ⎞ I CQ = 1 mA. I EQ = ⎜ ⎟ (1) = 1.01 mA ⎝ 80 ⎠ VCEQ = 12 − (1)( 2 ) − (1.01)( 0.2 ) ⇒ VCEQ = 9.80 V 1 = 0.0125 mA 80 = + ( 0.1) (1 + β ) RE = ( 0.1)( 81)( 0.2 ) = 1.62 kΩ I BQ =

RTH

⎛ R2 ⎞ 1 1 VTH = ⎜ ( RTH )(12 ) − 6 = (19.44 ) − 6 ⎟ (12 ) − 6 = R + R R R ⎝ 1 2 ⎠ 1 1 VTH = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE − 6 1 (19.44 ) − 6 = ( 0.0125 )(1.62 ) + 0.7 + ( 81)( 0.0125)( 0.2 ) − 6 R1 1 (19.44 ) = 0.923 R1 R1 = 21.1 kΩ,

21.1R2 = 1.62 21.1 + R2

R2 = 1.75 kΩ

b. R1 = 22.2 kΩ or R1 = 20.0 kΩ R2 = 1.84 kΩ or R2 = 1.66 kΩ R2 ( max ) , R1 ( min )

RTH = (1.84 ) & ( 20.0 ) = 1.685 kΩ

⎛ 1.84 ⎞ VTH = ⎜ ⎟ (12 ) − 6 = −4.99 V ⎝ 1.84 + 20.0 ⎠ −4.99 − 0.7 − ( −6 ) 0.31 = = 0.0173 mA I BQ = 1.685 + ( 81)( 0.2 ) 17.89 I CQ = 1.39 mA, I EQ = 1.40 mA

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For max, RC ⇒ VCE = 12 − (1.39 )( 2 ) − (1.40 )( 0.2 ) VCE = 8.94 V R2 ( min ) , R1 ( max )

RTH = (1.66 ) & ( 22.2 ) = 1.545 kΩ

⎛ 1.66 ⎞ VTH = ⎜ ⎟ (12 ) − 6 = −5.165 V ⎝ 1.66 + 22.2 ⎠ −5.165 − 0.7 + 6 0.135 = = 0.00761 mA ⇒ I CQ = 0.609 mA, I E = 0.616 I BQ = 1.545 + ( 81)( 0.20 ) 17.745 VCEQ = 12 − ( 0.609 )( 2 ) − ( 0.616 )( 0.2 ) VCEQ = V 10.7 V

So 0.609 ≤ I C ≤ 1.39 mA 8.94 ≤ VCEQ ≤ 10.7 V 5.54 VCEQ ≅ VCC − I CQ ( RC + RE )

5 = 12 − 3 ( RC + RE ) ⇒ RC + RE = 2.33 k Ω

Let RE = 0.333 k Ω and RC = 2 k Ω Nominal value of β = 100 RTH = ( 0.1)(1 + β ) RE = ( 0.1)(101)( 0.333) = 3.36 k Ω 3 = 0.03 mA 100 1 1 = ⋅ RTH ⋅ (12 ) − 6 = ( 3.36 )(12 ) − 6 R1 R1

I BQ = VTH

Then VTH = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE − 6 1 ( 3.36 )(12 ) − 6 = ( 0.03)( 3.36 ) + 0.7 + (101)( 0.03)( 0.333) − 6 R1 which yields R1 = 22.3 k Ω and R2 = 3.96 k Ω ⎛ R2 ⎞ ⎛ 3.96 ⎞ Now VTH = ⎜ ⎟ (12 ) − 6 = ⎜ ⎟ (12 ) − 6 or VTH = −4.19 V ⎝ 3.96 + 22.3 ⎠ ⎝ R1 + R2 ⎠ For β = 75, VTH = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE − 6 I BQ =

VTH + 6 − 0.7 −4.19 + 6 − 0.7 = = 0.0387 mA ⇒ I C = 2.90 mA RTH + (1 + β ) RE 3.36 + ( 76 )( 0.333)

For β = 150, I BQ =

−4.19 + 6 − 0.7 = 0.0207 mA 3.36 + (151)( 0.333)

Then I C = 3.10 mA Specifications are met. 5.55 RTH = R1 & R2 = 3 & 12 = 2.4 k Ω ⎛ R2 ⎞ ⎛ 12 ⎞ VTH = ⎜ ⎟ VCC = ⎜ ⎟ ( 20 ) = 16 V ⎝ 12 + 3 ⎠ ⎝ R1 + R2 ⎠ (a) For β = 75

20 = (1 + β ) I BQ RE + VEB ( on ) + I BQ RTH + VTH 20 − 0.7 − 16 = I BQ ⎡⎣( 76 )( 2 ) + 2.4 ⎤⎦

So I BQ = 0.0214 mA, I CQ = 1.60 mA, I EQ = 1.62 E

VECQ = 20 − (1.6 )(1) − (1.62 )( 2 ) or VECQ = 15.16 V

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(b) For β = 100, we find I BQ = 0.0161 mA, I CQ = 1.61 mA, VECQ = 15.13 V , I EQ = 1.63 mA 5.56 I CQ = 4.8 mA → I EQ = 4.84 mA VCEQ = VCC − I CQ RC − I EQ RE 6 = 18 − ( 4.8 )( 2 ) − ( 4.84 ) RE ⇒ RE = 0.496 kΩ RTH = ( 0.1)(1 + β ) RE = ( 0.1)(121)( 0.496 ) = 6.0 kΩ

VTH = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE I BQ = 0.040 mA VTH =

1 1 ⋅ RTH ⋅VCC = ( 6.0 )(18 ) R1 R1

1 ( 6.0 )(18) = ( 0.04 )( 6.0 ) + 0.70 + (121)( 0.04 )( 0.496 ) R1 1 (108 ) = 3.34 R1 R1 = 32.3 kΩ,

32.3 R2 = 6.0 32.3 + R2

R2 = 7.37 kΩ

5.57 For nominal β = 70 2 = 0.0286 mA → I EQ = 2.03 mA 70 = VCC − I CQ RC − I EQ RE

I BQ = VCEQ

10 = 20 − ( 2 )( 4 ) − ( 2.03) RE ⇒ RE = 0.985 K RTH = ( 0.1)(1 + β ) RE = ( 0.1)( 71)( 0.985 ) = 6.99 K

VTH = I BQ RTH + VBE ( on ) + I EQ RE

1 ⋅ RTH ⋅ VCC = I BQ RTH + VBE ( on ) + I EQ RE R1 1 ( 6.99 )( 20 ) = ( 0.0286 )( 6.99 ) + 0.70 + ( 2.03)( 0.985 ) R1 1 (139.8 ) = 2.90 R1 R1 = 48.2 K,

48.2 R2 = 6.99 48.2 + R2

R2 = 8.18 K

Check: For β = 50 ⎛ 8.18 ⎞ VTH = ⎜ ⎟ ( 20 ) = 2.90 ⎝ 8.18 + 48.2 ⎠ V − VBE ( on ) 2.90 − 0.7 I BQ = TH = = 0.0384 mA RTH + (1 + β ) RE 6.99 + ( 51)( 0.985 ) I CQ = 1.92 mA

For β = 90 I BQ =

2.90 − 0.7 = 0.0228 mA 6.99 + ( 91)( 0.985 )

I CQ = 2.05 mA

Design criterion is satisfied.

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5.58 I CQ = 1 mA → I EQ = 1.017 mA VCEQ = VCC − I CQ RC − I EQ RE 5 = 15 − (1)( 5 ) − (1.017 ) RE ⇒ RE = 4.92 kΩ Bias stable: RTH = ( 0.1)(1 + β ) RE = ( 0.1)( 61)( 4.92 ) = 30.0 kΩ 1 = 0.0167 mA I BQ = 60 1 VTH = ⋅ RTH ⋅ VCC = I BQ RTH + VBE ( on ) + I EQ RE R1 1 ( 30.0 )(15 ) = ( 0.0167 )( 30.0 ) + 0.70 + (1.017 )( 4.92 ) R1 1 ( 448.5) = 6.197 R1 R1 = 72.5 kΩ,

72.5 R2 = 30.0 72.5 + R2

R2 = 51.2 kΩ

Check: For β = 45 ⎛ 51.2 ⎞ VTH = ⎜ ⎟ (15 ) = 6.21V ⎝ 51.2 + 72.5 ⎠ V − VBE ( on ) 6.21 − 0.7 I BQ = TH = = 0.0215 mA RTH + (1 + β ) RE 30 + ( 46 )( 4.92 )

I CQ = 0.967 mA,

ΔI C = 3.27% IC

Check: For β = 75 6.21 − 0.7 I BQ = = 0.0136 mA 30.0 + ( 76 )( 4.92 ) ΔI C = 2.31% IC Design criterion is satisfied. I CQ = 1.023 mA,

5.59 (a) VCC ≅ I CQ ( RC + RE ) + VCEQ 3 = ( 0.1)( 5 RE + RE ) + 1.4 ⇒ RE = 2.67 k Ω 100 = 0.833 μ A 120 = ( 0.1)(1 + β ) RE = ( 0.1)(121)( 2.67 ) = 32.3 k Ω

RC = 13.3 k Ω, I BQ = RTH

VTH =

1 1 ⋅ RTH ⋅ VCC = ( 32.3)( 3) R1 R1

= I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE

= ( 0.000833)( 32.3) + 0.7 + (121)( 0.000833)( 2.67 ) which gives R1 = 97.3 k Ω, and R2 = 48.4 k Ω

(b)

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IR ≅

3 3 = ⇒ 20.6 μ A R1 + R2 97.3 + 48.4

I CQ = 100 μ A

P = ( I CQ + I R )VCC = (100 + 20.6 )( 3) or P = 362 μW

5.60 IE =

5 − VE 5 = = 1.67 mA RE 3

RTH = R1 || R2 = ( 0.1)(1 + β ) RE = ( 0.1)(101)( 3) = 30.3 kΩ ⎛ R2 ⎞ 1 VTH = ⎜ ⎟ ( 4 ) − 2 = ⋅ RTH ⋅ ( 4 ) − 2 + R R R ⎝ 1 2 ⎠ 1 I EQ = 0.0165 mA I BQ = 1+ β 5 = I EQ RE + VEB ( on ) + I B RTH + VTH 1 5 = (1.67 )( 3) + 0.7 + ( 0.0165 )( 30.3) + ( 30.3)( 4 ) − 2 R1 0.80 =

1 ( 30.3)( 4 ) ⇒ R1 = 152 kΩ R1

152 R2 = 30.3 ⇒ R2 = 37.8 kΩ 152 + R2

5.61 a.

RTH = R1 & R2 = 10 & 20 ⇒ RTH = 6.67 kΩ ⎛ R2 ⎞ ⎛ 20 ⎞ VTH = ⎜ ⎟ (10 ) − 5 = ⎜ ⎟ (10 ) − 5 ⇒ VTH = 1.67 V ⎝ 20 + 10 ⎠ ⎝ R1 + R2 ⎠

b. 10 = (1 + β ) I BQ RE + VEB ( on ) + I BQ RTH + VTH 10 − 0.7 − 1.67 7.63 I BQ = = ⇒ I BQ = 0.0593 mA 6.67 + ( 61)( 2 ) 128.7 I CQ = 3.56 mA, I EQ =3.62 mA VE = 10 − I EQ RE = 10 − ( 3.62 )( 2 ) VE = 2.76 V VC = I CQ RC − 10 = ( 3.56 )( 2.2 ) − 10 VC = −2.17 V

5.62 V + − V − ≅ I CQ ( RC + RE ) + VECQ 20 = ( 0.5 )( RC + RE ) + 8 ⇒ ( RC + RE ) = 24 k Ω Let RE = 10 k Ω then RC = 14 k Ω Let β = 60 from previous problem. RTH = ( 0.1)(1 + β ) RE = ( 0.1)( 61)(10 )

Or RTH = 61 k Ω

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0.5 = 0.00833 mA 60 ⎛ R2 ⎞ 1 =⎜ ⎟ (10 ) − 5 = ⋅ RTH ⋅10 − 5 R1 ⎝ R1 + R2 ⎠

I BQ = VTH

Now 10 = (1 + β ) I BQ RE + VEB ( on ) + I BQ RTH + VTH 10 = ( 61)( 0.00833)(10 ) + 0.7 + ( 0.00833) ( 61) +

1 ( 61) (10 ) − 5 R1

Then R1 = 70.0 k Ω and R2 = 474 k Ω 10 10 = ⇒ 18.4 μ A R1 + R2 70 + 474 So the 40 μ A current limit is met. IR ≅

5.63 a.

RTH = R1 & R2 = 35 & 20 ⇒ RTH = 12.7 kΩ

⎛ R2 ⎞ ⎛ 20 ⎞ VTH = ⎜ ⎟ (7) − 5 = ⎜ ⎟ ( 7 ) − 5 ⇒ VTH = −2.45 V R R + ⎝ 20 + 35 ⎠ ⎝ 1 2 ⎠

b. I BQ =

=

VTH − VBE ( on ) − ( −10 ) RTH + (1 + β ) RE

−2.45 − 0.7 + 10 ⇒ I BQ = 0.136 mA 12.7 + ( 76 )( 0.5 )

I CQ = 10.2 mA, I EQ = 10.4 mA VCEQ = 20 − I CQ RC − I EQ RE

= 20 − (10.2 )( 0.8 ) − (10.4 )( 0.5 )

VCEQ = 6.64 V

c. R2 = 20 + 5% = 21 kΩ R1 = 35 − 5% = 33.25 kΩ RE = 0.5 − 5% = 0.475 kΩ RTH = R1 & R2 = 21 & 33.25 = 12.9 kΩ ⎛ R2 ⎞ VTH = ⎜ ⎟ (7) − 5 ⎝ R1 + R2 ⎠ 21 ⎛ ⎞ =⎜ ⎟ ( 7 ) − 5 = −2.29 V ⎝ 21 + 33.25 ⎠ I BQ =

−2.29 − 0.7 − ( −10 ) 12.9 + ( 76 )( 0.475 )

= 0.143 mA

I CQ = 10.7 mA, I EQ = 10.9 mA

For RC = 0.8 + 5% = 0.84 kΩ VCEQ = 20 − (10.7 )( 0.84 ) − (10.9 )( 0.475 ) ⇒ VCEQ = 5.83 V

For RC = 0.8 − 5% = 0.76 kΩ

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VCEQ = 20 − (10.7 )( 0.76 ) − (10.9 )( 0.475 ) ⇒ VCEQ = 6.69 V R2 = 20 − 5% = 19 kΩ R1 = 35 + 5% = 36.75 kΩ RE = 0.5 + 5% = 0.525 kΩ RTH = R1 & R2 = 19 & 36.75 = 12.5 kΩ

19 ⎛ ⎞ VTH = ⎜ ⎟ ( 7 ) − 5 = −2.61 V ⎝ 19 + 36.75 ⎠ −2.61 − 0.7 − ( −10 ) = 0.128 mA I BQ = 12.5 + ( 76 )( 0.525 ) I CQ = 9.58 mA, I EQ = 9.70 mA

For RC = 0.84 kΩ VCEQ = 20 − ( 9.58 )( 0.84 ) − ( 9.70 )( 0.525 ) ⇒ VCEQ = 6.86 V

For RC = 0.76 kΩ VCEQ = 20 − ( 9.58 )( 0.76 ) − ( 9.70 )( 0.525 ) ⇒ VCEQ = 7.63 V

So 9.58 ≤ I CQ ≤ 10.7 mA and 5.83 ≤ VCEQ ≤ 7.63 V 5.64 a. RTH = 500 kΩ & 500 kΩ & 70 kΩ = 250 kΩ & 70 kΩ ⇒ RTH = 54.7 kΩ 5 − VTH 3 − VTH VTH − ( −5 ) + = 500 500 70 5 3 5 1 1 ⎞ ⎛ 1 + − = VTH ⎜ + + ⎟ − 0.0554 = VTH ( 0.0183) 500 500 70 ⎝ 500 500 70 ⎠ VTH = −3.03 V

b. I BQ =

=

VTH − VBE ( on ) − ( −5 ) RTH + (1 + β ) RE

−3.03 − 0.7 + 5 54.7 + (101)( 5 )

I BQ = 0.00227 mA I CQ = 0.227 mA, I EQ = 0.229 VCEQ = 20 − ( 0.227 )( 50 ) − ( 0.229 )( 5 ) VCEQ = 7.51 V

5.65 RTH = 30 || 60 || 20 ⇒ RTH = 10 kΩ 5 − VTH 5 − VTH VTH + = 30 60 20 5 ⎞ 1 1 ⎞ ⎛ 5 ⎛ 1 ⎜ + ⎟ = VTH ⎜ + + ⎟ ⎝ 30 60 ⎠ ⎝ 30 60 20 ⎠ VTH = 2.5 V

For β = 100

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I BQ = =

VTH − VBE ( on ) − ( −5 ) RTH + (1 + β ) RE

2.5 − 0.7 + 5 10 + (101)( 0.2 )

I BQ = 0.225 mA I CQ = 22.5 mA, I EQ = 22.7 mA VCEQ = 15 − ( 22.5 )( 0.5 ) − ( 22.7 ) ( 0.2 )

VCEQ = −0.79! In saturation ⇒ VCEQ = 0.2 V VTH = I BQ RTH + VBE + I EQ RE − 5 2.5 + 5 − 0.7 = I BQ (10 ) + I EQ ( 0.2 ) 6.8 = I BQ (10 ) + I EQ ( 0.2 ) 14.8 = I CQ ( 0.5 ) + I EQ ( 0.2 ) Transistor in saturation, I EQ = I BQ + I CQ 6.8 = I BQ (10 ) + I BQ ( 0.2 ) + I CQ ( 0.2 ) 6.8 = I BQ (10.2 ) + I CQ ( 0.2 )

51× 14.8 = I BQ ( 0.2 ) + I CQ ( 0.7 ) 754.8 = I BQ (10.2 ) + I CQ ( 35.7 ) 748 = I CQ ( 35.5 ) I CQ = 21.1 mA VCEQ = 0.2 V

5.66 I CQ = 50 μ A, I BQ = 0.625 μ A, I EQ = 50.6 μ A (a) 1 = 19.8 K 0.0506 5 = ( 0.050 ) RC + 5 + ( 0.0506 )(19.8 ) − 5 RE =

RC = 80 K

RTH = R1 & R2 Design bias stable circuit. RTH = ( 0.1)( 51)(19.8 ) = 101 K

⎛ R2 ⎞ 1 VTH = ⎜ ⎟ (10 ) − 5 = ⋅ RTH ⋅ (10 ) − 5 R1 ⎝ R1 + R2 ⎠ 1 So (101)(10 ) − 5 = I BQ (101) + 0.7 + ( 0.0506 )(19.8 ) − 5 R1 1 (1010 ) = 0.0631 + 0.7 + 1 R1 R1 = 573 K

573 R2 = 101 573 + R2

R2 = 123 K (b)

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RTH = 101 K, VTH = −3.23 V

VTH = I BQ RTH + 0.7 + (121)(19.8 ) I BQ − 5

1.07 = I BQ (101 + 2395.8 ) ⇒ I BQ = 0.429 μ A I CQ = 0.0514 mA, I EQ = 0.0519 mA

VCEQ = 10 − ( 0.0514 )( 80 ) − ( 0.0519 )(19.8 ) = 10 − 4.11 − 1.03 ⇒ VCEQ = 4.86 V

5.67 (a) 2 = 2.5 K 0.8 + I EQ RE

I EQ = 0.80 mA, RE = 12 = I CQ RC + VCEQ

12 = ( 0.8 ) RC + 7 + 2 ⇒ RC = 3.75 K VTH = I BQ RTH + VBE + I EQ RE For a bias stable circuit RTH = ( 0.1)(12.1)( 2.5 ) = 30.25 K 1 1 ⋅ RTH ⋅ VCC = ( 30.25 )(12 ) = ( 0.00667 )( 30.25 ) + 0.7 + 2 R1 R1 1 ( 363) = 2.90 ⇒ R1 = 125 K R1 125 R2 = 30.25 ⇒ R2 = 39.9 K 125 + R2

(b) Let RE = 2.4 K, RC = 3.9 K R1 = 120 K R2 = 39 K Then RTH = R1 & R2 = 120 & 39 = 29.4 K ⎛ 39 ⎞ VTH = ⎜ ⎟ (12 ) = 2.94 V ⎝ 120 + 39 ⎠ 2.94 − 0.7 2.24 = I BQ = ⇒ 7.00 μ A 29.4 + (121)( 2.4 ) 319.8 I CQ = 0.841 mΑ I EQ = 0.848 mA VCEQ = 12 − ( 0.841)( 3.9 ) − ( 0.848 )( 2.4 ) = 12 − 3.28 − 2.04 VCEQ = 6.68 V

5.68 (a)

I CQ = 100 μ A, I BQ = 1.18 μ A, I EQ = 101 μ A 2 ⇒ RE = 19.8 K 0.101 9 = ( 0.101)(19.8 ) + 6 + ( 0.1) RC − 9

RE =

RC = 100 K

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Design a bias stable circuit. BTH = ( 0.1)( 86 )(19.8 ) = 170 K

⎛ R2 ⎞ 1 VTH = ⎜ ⎟ (18 ) − 9 = (170 )(18 ) − 9 + R R R 2 ⎠ 1 ⎝ 1 9 = ( 0.101)(19.8 ) + 0.7 + ( 0.00118 )(170 ) +

1 (170 )(18 ) − 9 R1

203R2 = 170 203 + R2

R1 = 203 K

R2 = 1046 K (b) β = 125 9 = (126 ) I BQ (19.8 ) + 0.7 + I BQ (170 ) + (15.07 − 9 ) 2.23 = 0.837 μ A 2664.8 = 0.1054 mA

I BQ = I EQ

I CQ = 0.1046 mA

VECQ = 18 − ( 0.1046 )(100 ) − ( 0.1054 )(19.8 ) VECQ = 5.45 V

5.69 (a) 3 ⎛ 51 ⎞ I EQ ≈ ⎜ ⎟ ( 20 ) = 20.4 mA ⇒ RE = = 0.147 K 20.4 ⎝ 50 ⎠ 6 = 0.3 K VRC = 18 − 9 − 3 = 6 V RC = 20 For a bias stable circuit. RTH = ( 0.1)( 51)( 0.147 ) = 0.750 K V + = I EQ RE + VEB + I BQ RTH + VTH 18 = 3 + 0.7 + ( 0.4 )( 0.75 ) + 14 =

1 ( 0.75 )(18) R1

1 (13.5) ⇒ R1 = 0.964 K R1

( 0.964 ) R2 0.964 + R2

= 0.75 ⇒ R2 = 3.38 K

(b) Let RE = 0.15 K, RC = 0.3 K R1 = 1.0 K, R2 = 3.3 K

⎛ 3.38 ⎞ RTH = 1/13.3 = 0.767 K, VTH = ⎜ ⎟ (18 ) = 13.8 V ⎝ 1 + 3.38 ⎠ 18 − 13.8 − 0.7 3.5 = = 0.416 mA I BQ = 0.767 + ( 51)( 0.15 ) 8.417 I CQ = 20.8 mA, I EQ = 21.2 mA

VECQ = 18 − ( 20.8 )( 0.3) − ( 21.2 )( 0.15 ) = 18 − 6.24 − 3.18 VECQ = 8.58 V

5.70

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RTH = R1 & R2 = 100 & 40 = 28.6 kΩ

⎛ R2 ⎞ ⎛ 40 ⎞ VTH = ⎜ ⎟ (10 ) = ⎜ ⎟ (10 ) = 2.86 V ⎝ 40 + 100 ⎠ ⎝ R1 + R2 ⎠ VTH − VBE ( on ) 2.86 − 0.7 I B1 = = RTH + (1 + β ) RE1 28.6 + (121) (1) I B1 = 0.0144 mA I C1 = 1.73 mA,

I E1 = 1.75 mA

10 − VB 2 = I C1 + I B 2 3 VB 2 − VBE ( on ) − ( −10 ) IE2 = 5 10 − VB 2 VB 2 − 0.7 + 10 = I C1 + 3 (121) ( 5 )

⎛1 ⎞ 10 9.3 1 − 1.73 − = VB 2 ⎜⎜ + ⎟⎟ 3 605 ⎝ 3 (121) ( 5 ) ⎠

1.588 = VB 2 ( 0.335 ) ⇒ VB 2 = 4.74 V IE2 = I B2

4.74 − 0.7 − ( −10 )

5 = 0.0232 mA

⇒ I E 2 = 2.808 mA

I C 2 = 2.785 mA

VCEQ1 = 4.74 − (1.75 ) (1) ⇒ VCEQ1 = 2.99 V VCEQ 2 = 10 − ( 4.74 − 0.7 ) ⇒ VCEQ 2 = 5.96 V

5.71 VE1 = −0.7 I R1 = VE 2

−0.7 − ( −5 )

= 0.215 mA 20 = −0.7 − 0.7 = −1.4

IE2 =

−1.4 − ( −5 ) 1

⇒ I E 2 = 3.6 mA I B 2 = 0.0444 mA I C 2 = 3.56 mA

I E1 = I R1 + I B 2 = 0.215 + 0.0444 I E1 = 0.259 mA I B1 = 0.00320 mA I C1 = 0.256 mA

5.72 Current through V − source = I E1 + I E 2 and I E1 = I E 2 = (1 + β ) I B1 = ( 51)( 8.26 ) μ A So total current = 2 ( 51)( 8.26 ) μ A=843 μ A

P − = I ⋅ V − = ( 0.843)( 5 ) ⇒ P − = 4.22 mW (From V − source) From Example 5.19, I Q = 0.413 mA ⎛ 50 ⎞ So I C 0 = ⎜ ⎟ ( 0.413) = 0.405 mA ⎝ 51 ⎠

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P + = I ⋅ V + = ( 0.405 )( 5 ) ⇒ P + = 2.03 mW

(From V + source) 5.73 RTH = R1 & R2 = 50 & 100 = 33.3 kΩ ⎛ R2 ⎞ VTH = ⎜ ⎟ (10 ) − 5 ⎝ R1 + R2 ⎠ ⎛ 100 ⎞ =⎜ ⎟ (10 ) − 5 = 1.67 V ⎝ 100 + 50 ⎠ 5 = I E1 RE1 + VEB ( on ) + I B1 RTH + VTH ⎛ 101 ⎞ I E1 = ⎜ ⎟ ( 0.8 ) = 0.808 mA ⎝ 100 ⎠ I B1 = 0.008 mA 5 = ( 0.808 ) RE1 + 0.7 + ( 0.008 )( 33.3) + 1.67 RE1 = 2.93 kΩ VE1 = 5 − ( 0.808 ) ( 2.93) = 2.63 V VC1 = VE1 − VECQ1 = 2.63 − 3.5 = −0.87 V VE 2 = −0.87 − 0.70 = −1.57 V IE2 =

−1.57 − ( −5 ) RE 2

= 0.808 ⇒ RE 2 = 4.25 kΩ

VCEQ 2 = 4 ⇒ VC 2 = −1.57 + 4 = 2.43 V 5 − 2.43 RC 2 = ⇒ RC 2 = 3.21 kΩ 0.8 I RC1 = I C1 − I B 2 = 0.8 − 0.008 = 0.792 mA RC1 =

−0.87 − ( −5 ) 0.792

⇒ RC1 = 5.21 kΩ

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Chapter 6 Exercise Problems EX6.1

I BQ =

(a)

VBB − VBE ( on ) RB

=

2 − 0.7 ⇒ 2 μA 650

Then I CQ = β I BQ = (100 )( 2 μ A ) = 0.20 mA VCEQ = VCC − I CQ RC = 5 − ( 0.2 )(15 ) = 2 V (b)

Now I CQ 0.20 gm = = = 7.69 mA / V VT 0.026 and β VT (100 )( 0.026 ) rπ = = = 13 k Ω 0.20 I CQ (c) We find ⎛ r ⎞ Av = − g m RC ⎜ π ⎟ ⎝ rπ + RB ⎠ ⎛ 13 ⎞ = − ( 7.69 )(15 ) ⎜ ⎟ = −2.26 ⎝ 13 + 650 ⎠ EX6.2 V − VBE ( on ) 0.92 − 0.7 I BQ = BB = 100 RB

or I BQ = 0.0022 mA and I CQ = β I BQ = (150 )( 0.0022 ) = 0.33 mA (a) gm = rπ = ro =

I CQ VT

β VT I CQ

=

0.33 = 12.7 mA / V 0.026

=

(150 )( 0.026 ) 0.33

= 11.8 k Ω

VA 200 = = 606 k Ω I CQ 0.33

(b) ⎛ rπ ⎞ vo = − g m vπ ( ro & RC ) and vπ = ⎜ ⎟ ⋅ vs ⎝ rπ + RB ⎠ so ⎛ rπ ⎞ v Av = o = − g m ⎜ ⎟ ( ro & RC ) vπ ⎝ rπ + RB ⎠ ⎛ 11.8 ⎞ = − (12.7 ) ⎜ ⎟ ( 606 & 15 ) ⎝ 11.8 + 100 ⎠ which yields Av = −19.6 EX6.3 (a) V − VEB ( on ) 1.145 − 0.7 I BQ = BB = 50 RB

or I BQ = 0.0089 mA Then I CQ = β I BQ = ( 90 )( 0.0089 ) = 0.801 mA

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Now gm = rπ = ro =

I CQ VT

β VT I CQ

=

0.801 = 30.8 mA / V 0.026

=

( 90 )( 0.026 ) 0.801

= 2.92 k Ω

VA 120 = = 150 k Ω I CQ 0.801

We have Vo = g mVπ ( ro & RC )

(b)

⎛ r ⎞ and Vπ = − ⎜ π ⎟ Vs ⎝ rπ + RB ⎠ so ⎛ rπ ⎞ V Av = o = − g m ⎜ ⎟ ( ro & RC ) Vs ⎝ rπ + RB ⎠ ⎛ 2.92 ⎞ = − ( 30.8 ) ⎜ ⎟ (150 & 2.5 ) ⎝ 2.92 + 50 ⎠ which yields Av = −4.18 EX6.4 Using Figure 6.23 (a) For I CQ = 0.2 mA, 7.8 < hie < 15 k Ω, 60 < h fe < 125, 6.2 × 10−4 < hre < 50 × 10−4 , 5 < hoe < 13 μ mhos

For I CQ = 5 mA, 0.7 < hie < 1.1 k Ω, 140 < h fe < 210, 1.05 × 10−4 < hre < 1.6 ×10−4 ,

(b)

22 < hoe < 35 μ mhos EX6.5 RTH = R1 & R2 = 35.2 || 5.83 = 5 k Ω ⎛ R2 ⎞ ⎛ 5.83 ⎞ VTH = ⎜ ⎟ ⋅ VCC = ⎜ ⎟ (5) R R + ⎝ 5.83 + 35.2 ⎠ 2 ⎠ ⎝ 1 or VTH = 0.7105 V Then V − VBE ( on ) 0.7105 − 0.7 I BQ = TH = RTH 5 or I BQ = 2.1 μ A

and I CQ = β I BQ = (100 )( 2.1 μ A ) = 0.21 mA VCEQ = VCC − I CQ RC = 5 − ( 0.21)(10 ) and VCEQ = 2.9 V Now gm = ro =

I CQ VT

=

0.21 = 8.08 mA 0.026

VA 100 = = 476 k Ω I CQ 0.21

And Av = − g m ( ro & RC ) = − ( 8.08 ) ( 476 & 10 ) so Av = −79.1 EX6.6

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RTH = R1 & R2 = 250 & 75 = 57.7 k Ω ⎛ R2 ⎞ ⎛ 75 ⎞ VTH = ⎜ ⎟ (VCC ) = ⎜ ⎟ (5) ⎝ 75 + 250 ⎠ ⎝ R1 + R2 ⎠ or VTH = 1.154 V I BQ =

VTH − VBE ( on )

RTH + (1 + β ) RE

=

1.154 − 0.7 57.7 + (121)( 0.6 )

or I BQ = 3.48 μ A I CQ = β I BQ = (120 )( 3.38 μ A ) = 0.418 mA (a) Now gm = rπ =

I CQ VT

=

0.418 = 16.08 mA / V 0.026

βVT (120 )( 0.026 ) = = 7.46 k Ω 0.418 I CQ

We have Vo = − g mVπ RC We find Rib = rπ + (1 + β ) RE = 7.46 + (121)( 0.6 ) or Rib = 80.1 k Ω Also R1 & R2 = 250 & 75 = 57.7 k Ω R1 & R2 & Rib = 57.7 & 80.1 = 33.54 k Ω We find ⎛ R1 & R2 & Rib ⎞ ⎛ 33.54 ⎞ Vs′ = ⎜ ⎟ ⋅ Vs = ⎜ ⎟ ⋅ Vs ⎝ 33.54 + 0.5 ⎠ ⎝ R1 & R2 & Rib + RS ⎠

or Vs′ = ( 0.985 ) Vs Now ⎡ ⎛1+ β ⎞ ⎤ ⎡ ⎛ 121 ⎞ ⎤ Vs′ = Vπ ⎢1 + ⎜ ⎟ RE ⎥ = Vπ ⎢1 + ⎜ ⎟ ( 0.6 ) ⎥ r 7.46 ⎝ ⎠ ⎣ ⎦ ⎣⎢ ⎝ π ⎠ ⎦⎥ or Vπ = ( 0.0932 ) Vs′ = ( 0.0932 )( 0.985 ) Vs

So Av =

Vo = − (16.08 )( 0.0932 )( 0.985 )( 5.6 ) Vs

or Av = −8.27 EX6.7 RTH = R1 & R2 = 15 & 85 = 12.75 k Ω ⎛ R2 ⎞ ⎛ 85 ⎞ VTH = ⎜ ⎟ ⋅ VCC = ⎜ ⎟ (12 ) R + R ⎝ 15 + 85 ⎠ 2 ⎠ ⎝ 1 or VTH = 10.2 V Now VCC = (1 + β ) I BQ RE + VEB ( on ) + I BQ RTH + VTH

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so 12 = (101) I BQ ( 0.5 ) + 0.7 + I BQ (12.75 ) + 10.2 which yields I BQ = 0.0174 mA and I CQ = β I BQ = (100 )( 0.0174 ) = 1.74 mA I EQ = (101)( 0.0174 ) = 1.76 mA VECQ = VCC − I EQ RE − I CQ RC = 12 − (1.76 )( 0.5 ) − (1.74 )( 4 ) or VECQ = 4.16 V Now rπ =

β VT I CQ

=

(100 )( 0.026 ) 1.74

= 1.49 k Ω

We have Vo = g mVπ ( RC || RL ) ⎛ V ⎞ Vs = −Vπ − ⎜ g mVπ + π ⎟ RE rπ ⎠ ⎝ Solving for Vπ and noting that β = g m rπ , we find Av =

Vo Vs

= =

− β ( RC & RL )

rπ + (1 + β ) RE

− (100 )( 4 & 2 )

1.49 + (101)( 0.5 )

or Av = −2.56 EX6.8 Dc analysis 10 − 0.7 I BQ = = 0.00439 mA 100 + (101)( 20 ) I CQ = 0.439 mA, I EQ = 0.443 mA

Now rπ =

(100 )( 0.026 )

= 5.92 k Ω 0.439 0.439 gm = = 16.88 mA / V 0.026 100 ro = = 228 k Ω 0.439 (a) Now Vo = − g mVπ ( ro & RC ) ⎛ RB & rπ ⎞ Vπ = ⎜ ⎟ ⋅ Vs ⎝ RB & rπ + RS ⎠ RB & rπ = 100 & 5.92 = 5.59 k Ω ⎛ 5.59 ⎞ Then Vπ = ⎜ ⎟ ⋅ Vs = ( 0.918 ) Vs ⎝ 5.59 + 0.5 ⎠ V Then Av = o = − (16.88 ) ( 0.918 ) ( 228 & 10 ) Vs or Av = −148 (b) Rin = RS + RB & rπ = 0.5 + 100 & 5.92 = 6.09 k Ω

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and Ro = RC & ro = 10 & 228 = 9.58 k Ω EX6.9 (a) I CQ 0.25 gm = = = 9.615 mA / V 0.026 VT ro =

VA 100 = = 400 k Ω I CQ 0.25

Av = − g m ( ro & rc ) = − ( 9.615 )( 400 & 100 ) = −769 (b) Av = − g m ( ro & rc & rL ) = − ( 9.615 )( 400 & 100 & 100 ) Av = −427 EX6.10 5 − 0.7 I BQ = = 0.00672 mA 10 + (126 )( 5 ) I CQ = 0.84 mA, I EQ = 0.847 mA

VCEQ = 10 − ( 0.84 )( 2.3) − ( 0.847 )( 5 )

or VCEQ = 3.83 V dc load line VCE ≅ (V + − V − ) − I C ( RC + RE ) or VCE = 10 − I C ( 7.3) ac load line (neglecting ro) vce = −ic ( RC & RL ) = −ic ( 2.3 & 5 ) = −ic (1.58 ) IC (mA) 1.37

0.84

AC load line

Q-point

DC load line

3.83

10 V

CE (V)

EX6.11 (a) dc load line VEC ≅ VCC − I C ( RC + RE ) = 12 − I C ( 4.5 )

ac load line vec ≅ −ic ( RE + RC & RL ) or vec = −ic ( 0.5 + 4 & 2 ) = −ic (1.83) Q-point values 12 − 0.7 − 10.2 = 0.0150 mA I BQ = 12.75 + (121)( 0.5 ) I CQ = 1.80 mA, I EQ = 1.82 mA VECQ = 3.89 V

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IC (mA) AC load line

2.67

Q-point

1.80

DC load line

3.89

12 V (V) EC

ΔiC = I CQ = 1.80 mA ΔvEC = (1.8 )(1.83) = 3.29 V

(b)

vEC ( min ) = 3.89 − 3.29 = 0.6 V

So maximum symmetrical swing = 2 × 3.29 = 6.58 V peak-to-peak EX6.12 ⎛ R2 ⎞ 1 VTH = ⎜ (a) ⎟ ⋅ VCC = ⋅ RTH ⋅ VCC R1 ⎝ R1 + R2 ⎠ RTH = ( 0.1)(1 + β ) RE = ( 0.1)(121)(1)

so RTH = 12.1 k Ω, VTH =

1 (12.1)(12 ) R1

We can write VCC = (1 + β ) I BQ RE + VEB ( on ) + I BQ RTH + VTH We have I CQ = 1.6 mA, I BQ =

1.6 = 0.0133 mA 120

Then 1 (12.1)(12 ) R1 12.1 + (121)(1)

12 − 0.7 − I BQ = 0.0133 =

which yields R1 = 15.24 k Ω Since RTH = R1 & R2 = 12.1 k Ω, we find R2 = 58.7 k Ω Also VECQ = 12 − (1.6 )( 4 ) − (1.61)(1) = 3.99 V (b)

ac load line vec = −ic ( RC & RL )

Want Δic = I CQ − 0.1 = 1.6 − 0.1 = 1.5 mA Also Δvec = 3.99 − 0.5 = 3.49 V Now

Δvec 3.49 = = 2.327 k Ω = RC & RL Δic 1.5

So 4 & RL = 2.327 k Ω which yields RL = 5.56 k Ω EX6.13

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RTH = R1 & R2 = 25 & 50 = 16.7 k Ω ⎛ R2 ⎞ ⎛ 50 ⎞ VTH = ⎜ ⎟ ⋅ VCC = ⎜ ⎟ (5) ⎝ 25 + 50 ⎠ ⎝ R1 + R2 ⎠ or VTH = 3.33 V

I BQ =

VTH − VBE ( on )

RTH + (1 + β ) RE

=

3.33 − 0.7 16.7 + (121)(1)

or I BQ = 0.0191 mA Also I CQ = (120 )( 0.0191) = 2.29 mA

Now gm = rπ = ro =

I CQ VT

=

β VT I CQ

2.29 = 88.1 mA / V 0.026

=

(120 )( 0.026 ) 2.29

= 1.36 k Ω

VA 100 = = 43.7 k Ω I CQ 2.29

(a) ⎛ R1 & R2 & Rib ⎞ Vs′ = ⎜ ⎟ ⋅ Vs ⎝ R1 & R2 & Rib + RS ⎠ Rib = rπ + (1 + β ) ( RE & ro ) = 1.36 + (121)(1 & 43.7 )

or Rib = 120 k Ω and R1 & R2 = 16.7 k Ω Then R1 & R2 & Rib = 16.7 & 120 = 14.7 k Ω Now ⎛ 14.7 ⎞ Vs′ = ⎜ ⎟ ⋅ Vs = ( 0.967 ) Vs ⎝ 14.7 + 0.5 ⎠ and ⎛V ⎞ ⎛ 1+ β Vo = ⎜ π + g mVπ ⎟ ( RE & ro ) = Vπ ⎜ ⎝ rπ ⎠ ⎝ rπ

⎞ ⎟ RE & ro ⎠

We have Vs′ = Vπ + Vo then

Vs′ ( 0.967 )Vs = ⎛ 1+ β ⎞ ⎛1+ β ⎞ 1+ ⎜ ⎟ RE & ro 1 + ⎜ ⎟ RE & ro r ⎝ π ⎠ ⎝ rπ ⎠ We then obtain

Vπ =

Av =

=

Vo = Vs

⎛1+ β ⎝ rπ

( 0.967 ) ⎜

⎞ ⎟ RE & ro ⎠

⎛1+ β ⎞ 1+ ⎜ ⎟ RE & ro ⎝ rπ ⎠ ( 0.967 )(1 + β ) RE & ro

rπ + (1 + β ) RE & ro

Now RE & ro = 1 & 43.7 = 0.978 k Ω

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Then Av =

( 0.967 )(121)( 0.978 ) = 0.956 1.36 + (121)( 0.978 )

(b) Rib = rπ + (1 + β )( RE & ro ) or Rib = 1.36 + (121)( 0.978 ) = 120 k Ω EX6.14 For RS = 0, then r Ro = RE & ro π 1+ β Using the parameters from Exercise 4.13, we obtain 1.36 Ro = 1 43.7 121 or Ro = 11.1 Ω EX6.15 (a) For I CQ = 1.25 mA and β = 100, we find I EQ = 1.26 mA and I BQ = 0.0125 mA

Now VCEQ = 10 − I EQ RE Or 4 = 10 − (1.26 ) RE which yields RE = 4.76 k Ω Then RTH = ( 0.1)(1 + β ) RE = ( 0.1)(101)( 4.76 ) or RTH = 48.1 k Ω We have ⎛ R2 ⎞ 1 VTH = ⎜ ⎟ (10 ) − 5 = ⋅ RTH (10 ) − 5 R1 ⎝ R1 + R2 ⎠ or VTH =

1 ( 481) − 5 R1

We can write I BQ =

VTH − 0.7 − ( −5 ) RTH + (1 + β ) RE

Or 1 ( 481) − 5 − 0.7 + 5 R1 0.0125 = 48.1 + (101)( 4.76 )

which yields R1 = 65.8 k Ω Since R1 & R2 = 48.1 k Ω, we obtain R2 = 178.8 k Ω (b)

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rπ = ro =

β VT I CQ

=

(100 )( 0.026 ) 1.25

= 2.08 k Ω

VA 125 = = 100 k Ω I CQ 1.25

We may note that g mVπ = g m ( I b rπ ) = β I b Also Rib = rπ + (1 + β )( RE & RL & ro ) = 2.08 + (101) ( 4.76 1 100 ) or Rib = 84.9 k Ω Now ⎛ RE ro ⎞ Io = ⎜ 1 + β ) Ib ⎜ R r + R ⎟⎟ ( L ⎠ ⎝ E o where ⎛ R1 R2 ⎞ Ib = ⎜ ⋅I ⎜ R R + R ⎟⎟ s ib ⎠ ⎝ 1 2 We can then write ⎛ RE ro ⎞ ⎛ R1 R2 I AI = o = ⎜ 1+ β )⎜ ( ⎟ ⎜R R +R I s ⎜⎝ RE ro + RL ⎟⎠ ib ⎝ 1 2 We have RE ro = 4.76 100 = 4.54 k Ω

⎞ ⎟⎟ ⎠

so 48.1 ⎞ ⎛ 4.54 ⎞ ⎛ AI = ⎜ ⎟ (101) ⎜ ⎟ 4.54 1 48.1 + + 84.9 ⎠ ⎝ ⎠ ⎝ or AI = 29.9 r 2.08 (c) Ro = RE & ro π = 4.76 100 1+ β 101

or Ro = 20.5 Ω EX6.16 (a) RTH = R1 R2 = 70 6 = 5.53 k Ω ⎛ R2 ⎞ ⎛ 6 ⎞ VTH = ⎜ ⎟ (10 ) − 5 = ⎜ ⎟ (10 ) − 5 ⎝ 70 + 6 ⎠ ⎝ R1 + R2 ⎠ or VTH = −4.2105 V We find −4.2105 − 0.7 − ( −5 ) I BQ1 = ⇒ 2.91 μ A 5.53 + (126 )( 0.2 )

and I CQ1 = β I BQ1 = (125 )( 2.91 μ A ) = 0.364 mA I EQ1 = (1 + β ) I BQ1 = 0.368 mA

At the collector of Q1, V − 0.7 − ( −5 ) 5 − VC1 = I CQ1 + C1 RC1 (1 + β )( RE 2 ) or

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V − 0.7 − ( −5 ) 5 − VC1 = 0.364 + C1 5 (126 )(1.5)

which yields VC1 = 2.99 V also VE1 = I EQ1 RE1 − 5 = ( 0.368 )( 0.2 ) − 5 or VE1 = −4.93 V Then VCEQ1 = VC1 − VE1 = 2.99 − ( −4.93) = 7.92 V We find V − 0.7 − ( −5 ) I EQ 2 = C1 = 4.86 mA 1.5 and ⎛ β ⎞ ⎛ 125 ⎞ I CQ 2 = ⎜ ⎟ ⋅ I EQ1 = ⎜ ⎟ ( 4.86 ) = 4.82 mA β + 1 ⎝ 126 ⎠ ⎝ ⎠ We find VE 2 = VC1 − 0.7 = 2.99 − 0.7 = 2.29 V and VCEQ 2 = 5 − VE 2 = 5 − 2.29 = 2.71 V (b) The small-signal transistor parameters are: rπ 1 =

β VT I CQ1

g m1 = rπ 2 = gm2 =

=

I CQ1 VT

β VT I CQ 2 I CQ 2 VT

(125 )( 0.026 ) 0.364

0.364 = 14.0 mA / V 0.026

= =

= 8.93 k Ω

(125 )( 0.026 )

=

4.82

= 0.674 k Ω

4.82 = 185 mA / V 0.026

We find Rib1 = rπ 1 + (1 + β ) RE1 = 8.93 + (126 )( 0.2 ) Or Rib1 = 34.1 k Ω and Rib 2 = rπ 2 + (1 + β )( RE 2 & RL ) = 0.674 + (126 )(1.5 & 10 ) = 165 k Ω The small-signal equivalent circuit is:

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⫹

⫹ V␲1

r␲1

gm1V␲1

V␲2

⫹ ⫺

R1兩兩R2

RC1

RE1

We can write Vo = (1 + β ) I b 2 ( RE 2 & RL ) where ⎛ RC1 ⎞ Ib2 = ⎜ ⎟ ( − g m1Vπ 1 ) ⎝ RC1 + Rib 2 ⎠ V Vπ 1 = s ⋅ rπ 1 Rib1 Then V Av = o Vs ⎛ RC1 ⎞ ⎛ − g m1rπ 1 ⎞ = (1 + β )( RE 2 & RL ) ⎜ ⎟⎜ ⎟ ⎝ RC1 + Rib 2 ⎠ ⎝ Rib1 ⎠

so ⎛ 5 ⎞ ⎛ 125 ⎞ Av = − (126 )(1.5 & 10 ) ⎜ ⎟⎜ ⎟ ⎝ 5 + 165 ⎠ ⎝ 34.1 ⎠ or Av = −17.7 (c) Ri = R1 & R2 & Rib1 = 70 & 6 & 34.1 = 4.76 k Ω ⎛ r + RC1 ⎞ ⎛ 0.676 + 5 ⎞ and Ro = RE 2 ⎜ π 2 ⎟ = 1.5 ⎜ ⎟ ⎝ 126 ⎠ ⎝ 1+ β ⎠ or Ro = 43.7 Ω Test Your Understanding Exercises TYU6.1 I CQ 0.25 = = 9.62 mA / V gm = 0.026 VT

rπ = ro =

β VT I CQ

=

(120 )( 0.026 ) 0.25

gm2V␲2

⫺

⫺ Vs

r␲2

= 12.5 k Ω

VA 150 = = 600 k Ω I CQ 0.25

TYU6.2 V V 75 ro = A ⇒ I CQ = A = I CQ ro 200 k Ω

or ICQ = 0.375 mA

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Vo

RE2

RL

TYU6.3 RC RE Resulting gain is always smaller than this value. The effect of RS is very small. Set RC = 10 RE Now 5 ≅ I C ( RC + RE ) + VCEQ

As a first approximation, Av ≅ −

or 5 = ( 0.5 )( RC + RE ) + 2.5 which yields RC + RE = 5 k Ω So 10 RE + RE = 5 or RE = 0.454 k Ω and RC = 4.54 k Ω We have I CQ 0.5 I BQ = = = 0.005 mA β 100 and RTH = ( 0.1)(1 + β ) RE = ( 0.1)(101)( 0.454 ) or RTH = 4.59 k Ω also ⎛ R2 ⎞ 1 VTH = ⎜ ⎟ ⋅ VCC = ⋅ RTH ⋅ VCC + R R R ⎝ 1 2 ⎠ 1 or 1 23 ( 4.59 )( 5 ) = R1 R1 We can write VTH = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE VTH =

or 23 = ( 0.005 )( 4.59 ) + 0.7 + (101)( 0.005 )( 0.454 ) R1 which yields R1 = 24.1 k Ω and since R1 & R2 = 4.59 k Ω we find R2 = 5.67 k Ω TYU6.4 dc analysis RTH = R1 & R2 = 15 & 85 = 12.75 k Ω ⎛ R2 ⎞ ⎛ 85 ⎞ VTH = ⎜ ⎟ ⋅ VCC = ⎜ ⎟ (12 ) ⎝ 15 + 85 ⎠ ⎝ R1 + R2 ⎠ or VTH = 10.2 V We can write 12 − 0.7 − VTH 12 − 0.7 − 10.2 I BQ = = RTH + (1 + β ) RE 12.75 + (101)( 0.5 )

or IBQ = 0.0174 mA

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and I CQ = β I BQ = 1.74 mA ac analysis Vo = h fe I b ( RC & RL ) and Ib =

so Av =

−Vs

hie + (1 + h fe ) RE −h fe ( RC & RL ) Vo = Vs hie + (1 + h fe ) RE

For ICQ = 1.74 mA, we find h fe ( max ) = 110, h fe ( min ) = 70 hie ( max ) = 2 k Ω, hie ( min ) = 1.1 k Ω We obtain −110 ( 4 & 2 ) Av ( max ) = = −2.59 1.1 + (111)( 0.5 ) and Av ( min ) =

−70 ( 4 & 2 )

2 + ( 71)( 0.5 )

= −2.49

TYU6.5

First approximation, Av ≅ −

RC RE

This predicts a low value, so set

RC =9 RE

Now VCC ≅ I CQ ( RC + RE ) + VECQ or 7.5 = ( 0.6 )( 9 RE + RE ) + 3.75 which yields RE = 0.625 k Ω and RC = 5.62 k Ω We have RTH = ( 0.1)(1 + β ) RE = ( 0.1)(101)( 0.625 ) or RTH = 6.31 k Ω Also 1 1 VTH = ⋅ RTH ⋅ VCC = ( 6.31)( 7.5 ) R1 R1 We have I CQ 0.6 I BQ = = = 0.006 mA β 100 The KVL equation around the E-B loop gives VCC = (1 + β ) I BQ RE + VEB ( on ) + I BQ RTH + VTH or 7.5 = (101)( 0.006 )( 0.625 ) + 0.7 + ( 0.006 )( 6.31) +

1 ( 6.31)( 7.5) R1

which yields R1 = 7.41 k Ω Since RTH = R1 & R2 = 6.31 k Ω, then R2 = 42.5 k Ω TYU6.6

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⎛R ⎞ − β RC = − ( 0.95 ) ⎜ C ⎟ rπ + (1 + β ) RE ⎝ RE ⎠ ⎛ 2 ⎞ or Av = − ( 0.95 ) ⎜ ⎟ = −4.75 ⎝ 0.4 ⎠ Assume rπ = 1.2 k Ω from Example 4.6. Then −β ( 2) = −4.75 1.2 + (1 + β )( 0.4 )

We have Av =

or β = 76 TYU6.7 dc analysis: By symmetry, VTH = 0 RTH = R1 & R2 = 20 & 20 = 10 k Ω We can write 0 − 0.7 − ( −5 ) I BQ = = 0.00672 mA 10 + (126 )( 5 )

I CQ = β I BQ = (125 )( 0.00672 ) = 0.84 mA

Small-signal transistor parameters: β VT (125 )( 0.026 ) rπ = = = 3.87 k Ω I CQ 0.84 gm = ro =

I CQ VT

=

0.84 = 32.3 mA / V 0.026

VA 200 = = 238 k Ω I CQ 0.84

(a) We have Vo = − g mVπ ( ro & RC & RL ) and Vπ = Vs so Av =

Vo Vs

= − g m ( ro & RC & RL )

= − ( 32.3)( 238 & 2.3 & 5 ) or Av = −50.5 (b) Ro = ro & RC = 238 & 2.3 = 2.28 k Ω TYU6.8 We find I CQ = 0.418 mA, then

⎛ 121 ⎞ VCEQ = 5 − ( 0.418 )( 5.6 ) − ⎜ ⎟ ( 0.418 )( 0.6 ) ⎝ 120 ⎠ or VCEQ = 2.41 V So ΔvCE = ( 2.41 − 0.5 ) × 2

or ΔvCE = 3.82 V peak-to-peak TYU6.9 dc load line VCE ≅ (10 + 10 ) − I CQ ( RC + RE ) or

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VCE = 20 − I C (10 + RE )

ac load line vce = −ic RC = −ic (10 ) Now ΔvCE = VCEQ − 0.7 = ΔiC (10 ) = I CQ (10 ) so VCEQ − 0.7 = I CQ (10 ) We have that VCEQ = 20 − I CQ (10 + RE ) then I CQ (10 ) + 0.7 = 20 − I CQ (10 + RE ) or I CQ ( 20 + RE ) = 19.3

(1)

The base current is found from 10 − 0.7 I BQ = 100 + (101) RE so I CQ =

(100 )( 9.3)

100 + (101) RE

Substituting into Equation (1), ⎡ (100 )( 9.3) ⎤ ⎢ ⎥ ( 20 + RE ) = 19.3 ⎢⎣100 + (101) RE ⎥⎦ which yields RE = 16.35 k Ω Then (100 )( 9.3) I CQ = = 0.531 mA 100 + (101)(16.35 ) and VCEQ ≅ 20 − ( 0.531)(10 + 16.35 ) = 6.0 V Now ΔvCE = VCEQ − 0.7 = 6 − 0.7 = 5.3 V or, maximum symmetrical swing = 2 × 5.3 = 10.6 V peak-to-peak TYU6.10 We can write 0 − 0.7 − ( −10 ) I BQ = ⇒ 6.60 μ A 100 + (131)(10 ) and I CQ = (130 )( 6.60 μ A ) = 0.857 mA Assume nominal small-signal parameters of: hie = 4 k Ω, h fe = 134 hre = 0, hoe = 12 μ S ⇒

1 = 83.3 k Ω hoe

We find ⎛ 1 ⎞ Rib = hie + (1 + h fe ) ⎜ RE RL ⎟ hoe ⎠ ⎝ = 4 + (135 )(10 ||10 || 83.3) = 641 k Ω

To find the voltage gain:

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RB & Rib 100 & 641 ⋅ Vs = ⋅ Vs 100 & 641 + 10 RB & Rib + RS

Vs′ =

or Vs′ = ( 0.896 ) Vs Also

(1 + h fe ) R′ Vo = Vs′ hie + (1 + h fe ) R′

where R ′ = RE & RL

Then Av =

1 = 10 & 10 & 83.3 = 4.72 k Ω hoe

Vo ( 0.896 )(135 )( 4.72 ) = = 0.891 Vs 4 + (135 )( 4.72 )

To find the current gain: ⎛ ⎞ 1 ⎜ RE ⎟ hoe ⎟ ⎛ RB ⎞ Io ⎜ 1 + h fe ) ⎜ Ai = = ( ⎟ ⎟ Ii ⎜ 1 ⎝ RB + Rib ⎠ + RL ⎟⎟ ⎜⎜ RE hoe ⎝ ⎠ ⎛ 10 & 83.3 ⎞ ⎛ 100 ⎞ =⎜ ⎟ (135 ) ⎜ 100 + 641 ⎟ & + 10 83.3 10 ⎝ ⎠ ⎝ ⎠ or Ai = 8.59 To find the output resistance: 1 hie + RS RB Ro = RE 1 + h fe hoe = 10 83.3

4 + 10 & 100 ⇒ 96.0 Ω 135

TYU6.11 We find RTH = R1 & R2 = 50 & 50 = 25 k Ω ⎛ R2 ⎞ ⎛1⎞ VTH = ⎜ ⎟ ⋅ VCC = ⎜ ⎟ ( 5 ) = 2.5 V ⎝ 2⎠ ⎝ R1 + R2 ⎠ Now V − VEB ( on ) − VTH I BQ = CC RTH + (1 + β ) RE =

5 − 0.7 − 2.5 = 0.00793 mA 25 + (101)( 2 )

and I CQ = (100 )( 0.00793) = 0.793 mA

The small-signal transistor parameters: I CQ 0.793 = = 30.5 mA / V gm = 0.026 VT rπ = ro =

β VT I CQ

=

(100 )( 0.026 ) 0.793

= 3.28 k Ω

VA 125 = = 158 k Ω I CQ 0.793

(a)

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Define R ′ = RE RL ro = 2 0.5 158 ≅ 0.40 k Ω Now Av =

(1 + β ) R′ (101)( 0.4 ) = rπ + (1 + β ) R ′ 3.28 + (101)( 0.4 )

or Av = 0.925 (b) Rib = rπ + (1 + β ) R ′ = 3.28 + (101)( 0.4 ) or Rib = 43.7 k Ω Also Ro = RE & ro

rπ 3.28 = 2 & 158 1+ β 101

or Ro = 32.0 Ω TYU6.12 For VECQ = 2.5, I EQ =

VCC − VECQ RE

=

5 − 2.5 = 5 mA 0.5

then ⎛ 75 ⎞ I CQ = ⎜ ⎟ ( 5 ) = 4.93 mA ⎝ 76 ⎠ 5 I BQ = = 0.0658 mA 76 Small-signal transistor parmaters: β VT ( 75 )( 0.026 ) = = 0.396 k Ω rπ = I CQ 4.93 ro =

VA 75 = = 15.2 k Ω I CQ 4.93

Define the small-signal base current into the base, then g mVπ = − β I b Now, ⎛ RE & ro ⎞ Io = ⎜ ⎟ (1 + β ) I b ⎝ RE & ro + RL ⎠ and ⎛ R1 & R2 ⎞ Ib = ⎜ ⎟ ⋅ Ii ⎝ R1 & R2 + Rib ⎠ The current gain is ⎛ RE & ro ⎞ ⎛ R1 & R2 ⎞ I AI = o = ⎜ ⎟ (1 + β ) ⎜ ⎟ I i ⎝ RE & ro + RL ⎠ ⎝ R1 & R2 + Rib ⎠ We have RE = RL = 0.5 k Ω Rib = rπ + (1 + β )( RE & RL & ro ) = 0.396 + ( 76 )( 0.5 & 0.5 & 15.2 ) = 19.1 k Ω

and RE & ro = 0.5 & 15.2 = 0.484 k Ω Then ⎛ R1 & R2 ⎞ ⎛ 0.484 ⎞ AI = 10 = ⎜ ⎟ ⎟ ( 76 ) ⎜ ⎝ 0.484 + 0.5 ⎠ ⎝ R1 & R2 + 19.1 ⎠ which yields R1 & R2 = 6.975 k Ω Now

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⎛ R2 ⎞ 1 VTH = ⎜ ⎟ ⋅ VCC = ⋅ RTH ⋅ VCC + R R R ⎝ 1 2 ⎠ 1 or 1 VTH = ( 6.975 )( 5 ) R1 We can write V − VEB ( on ) − VTH I BQ = CC RTH + (1 + β ) RE

or 0.0658 =

5 − 0.7 − VTH 6.975 + ( 76 )( 0.5 )

which yields VTH = 1.34 =

1 ( 6.975)( 5 ) R1

or R1 = 26.0 k Ω and R2 = 9.53 k Ω TYU6.13 (a) dc analysis: V − VEB ( on ) 10 − 0.7 I EQ = EE = = 0.93 mA 10 RE ⎛ β ⎞ ⎛ 100 ⎞ I CQ = ⎜ ⎟ I EQ = ⎜ ⎟ ( 0.93) = 0.921 mA ⎝ 101 ⎠ ⎝ 1+ β ⎠ VECQ = VEE − I EQ RE − I CQ RC − ( −VCC )

= 10 − ( 0.93)(10 ) − ( 0.921)( 5 ) − ( −10 )

or VECQ = 6.1 V (b) Small-signal transistor parameters: β VT (100 )( 0.026 ) rπ = = = 2.82 k Ω I CQ 0.921 I CQ

0.921 = 35.42 mA / V 0.026 VT Small-signal current gain: I o = g mVπ and Vπ = Vs also ⎛ 1 ⎞ Vs + g mVπ = Vs ⎜⎜ + g m ⎟⎟ Ii = RE rπ ⎝ RE rπ ⎠ Then g m ( RE rπ ) I g mVπ AI = o = = Ii ⎛ 1 ⎞ 1 + g m ( RE rπ ) + gm ⎟ Vπ ⎜ ⎝ RE & rπ ⎠ gm =

=

=

( 35.42 ) (10 2.82 ) 1 + ( 35.42 ) (10 2.82 )

or AI = 0.987 (c) Small-signal voltage gain:

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Vo = g mVπ RC = g mVs RC or V Av = o = g m RC = ( 35.42 )( 5 ) Vs or Av = 177 TYU6.14 (a) V − VBE ( on ) 10 − 0.7 I BQ = EE = RB + (1 + β ) RE 100 + (101)(10 )

or I BQ = 8.38 μ A and I CQ = 0.838 mA rπ = gm = ro =

β VT

=

I CQ I CQ VT

=

(100 )( 0.026 ) 0.838

= 3.10 k Ω

0.838 = 32.23 mA / V 0.026

VA ∞ = =∞ I CQ 0.838

(b) Summing currents, we have ⎛ −V ⎞ ⎛ −V − Vs ⎞ V g mVπ + π = ⎜ π ⎟ + ⎜ π ⎟ rπ ⎝ RE ⎠ ⎝ RS ⎠ or ⎡⎛ 1 + β ⎞ 1 V 1 ⎤ Vπ ⎢⎜ + ⎥=− s ⎟+ r R R RS ⎥ E S ⎦ ⎣⎢⎝ π ⎠ We can then write ⎤ V ⎡⎛ r ⎞ Vπ = − s ⎢⎜ π ⎟ RE RS ⎥ RS ⎣⎝ 1 + β ⎠ ⎦ Now Vo = − g mVπ ( RC & RL ) So

( RC RL ) ⎡⎛ rπ ⎞ R R ⎤ Vo = gm ⎢⎜ ⎟ E S⎥ Vs RS ⎣⎝ 1 + β ⎠ ⎦ 32.23 10 1 ( ) ( ) ⎡ 3.10 ⎤ = 10 1⎥ ⎢ 1 101 () ⎣ ⎦

Av =

or Av = 0.870 Now ⎛ RE Ie = ⎜ ⎝ RE + rπ ⎛ β Ic = ⎜ ⎝ 1+ β

⎞ ⎛ 10 ⎞ ⎟ ⋅ Ii = ⎜ ⎟ ⋅ I i = ( 0.763) I i ⎝ 10 + 3.10 ⎠ ⎠

⎞ ⎛ 100 ⎞ ⎟ ⋅ Ie = ⎜ ⎟ ⋅ Ie ⎝ 101 ⎠ ⎠

⎛ RC ⎞ ⎛ 10 ⎞ Io = ⎜ ⎟ ⋅ Ic = ⎜ ⎟ ⋅ I c = ( 0.909 ) I c + R R ⎝ 10 + 1 ⎠ L ⎠ ⎝ C So we have I ⎛ 100 ⎞ AI = o = ( 0.909 ) ⎜ ⎟ ( 0.763) Ii ⎝ 101 ⎠

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or AI = 0.687 (c) We have r 3.10 Ri = RE π = 10 1+ β 101 or Ri = 30.6 Ω Also Ro = RC = 10 k Ω TYU6.15 dc analysis: We can write 5 = I BQ RB + VBE ( on ) + I EQ RE or I BQ = I CQ =

5 − 0.7 4.3 = RB + (101) RE RB + (101) RE

(100 )( 4.3) RB + (101) RE

Also 5 = I CQ RC + VCEQ + I EQ RE − 5 or ⎡ ⎛ 101 ⎞ ⎤ VCEQ = 10 − I CQ ⎢ RC + ⎜ ⎟ RE ⎥ ⎝ 100 ⎠ ⎦ ⎣ ac analysis: Vo = − g mVπ ( RC RL )

and Vs = −Vπ −

Vπ ⋅ RB = −Vπ rπ

⎛ RB ⎞ ⎜1 + ⎟ rπ ⎠ ⎝

or ⎛ r ⎞ Vπ = − ⎜ π ⎟ ⋅ Vs + r R B ⎠ ⎝ π Then V β Av = o = ( RC RL ) Vs rπ + RB

where β = g m rπ For I CQ = 1 mA, rπ = Now Av = 20 =

β VT I CQ

=

(100 )( 0.026 ) 1

= 2.6 k Ω

(100 ) ( 2 2 )

2.6 + RB which yields RB = 2.4 k Ω Also (100 )( 4.3) I CQ = 1 = 2.4 + (101) RE

which yields RE = 4.23 k Ω TYU6.16 (a)

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dc analysis: For I EQ 2 = 1 mA, ⎛ 100 ⎞ I CQ 2 = ⎜ ⎟ (1) = 0.990 mA ⎝ 101 ⎠ I EQ 2 1 I EQ1 = = = 0.0099 mA 1 + β 101 and I EQ1 0.0099 I BQ1 = = = 0.000098 mA 1+ β 101 and I CQ1 = (100 )( 0.000098 ) = 0.0098 mA

and VB1 = − I BQ1 RB = − ( 0.000098 )(10 ) or VB1 = −0.00098 ≅ 0 so VE1 = −0.7 V and VE 2 = −1.4 V Now I1 = I CQ1 + I CQ 2 = 0.0098 + 0.990 ≅ 1 mA then VO = 5 − (1)( 4 ) = 1 V and VCEQ 2 = 1 − ( −1.4 ) = 2.4 V VCEQ1 = 1 − ( −0.7 ) = 1.7 V

(b) Small-signal transistor parameters: β VT (100 )( 0.026 ) = = 265 k Ω rπ 1 = 0.0098 I CQ1 g m1 = rπ 2 = gm2 =

I CQ1 VT

β VT I CQ 2 I CQ 2 VT

0.0098 = 0.377 mA 0.026

= =

(100 )( 0.026 )

=

0.990

= 2.63 k Ω

0.990 = 38.1 mA / V 0.026

ro1 = ro 2 = ∞ (c) Small-signal voltage gain: From Figure 4.73 (b) Vo = − ( g m1Vπ 1 + g m 2Vπ 2 ) RC Vs = Vπ 1 + Vπ 2 and ⎛V ⎞ ⎛1+ β ⎞ Vπ 2 = ⎜ π 1 + g m1Vπ 1 ⎟ ⋅ rπ 2 = ⎜ ⎟ ⋅ Vπ 1rπ 2 ⎝ rπ 1 ⎠ ⎝ rπ 1 ⎠

Then ⎡ ⎛1+ β Vo = − ⎢ g m1Vπ 1 + g m 2 ⎜ ⎝ rπ 1 ⎣ Also

⎤ ⎞ ⎟ ⋅ rπ 2Vπ 1 ⎥ ⋅ RC ⎠ ⎦

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⎛1+ β Vs = Vπ 1 + ⎜ ⎝ rπ 1

⎞ ⎟ ⋅ rπ 2Vπ 1 ⎠

⎡ ⎛ r ⎞⎤ = Vπ 1 ⎢1 + (1 + β ) ⎜ π 2 ⎟ ⎥ ⎝ rπ 1 ⎠ ⎦ ⎣

or Vπ 1 =

Vs ⎛r ⎞ 1 + (1 + β ) ⎜ π 2 ⎟ ⎝ rπ 1 ⎠

Now ⎡ ⎛ rπ 2 ⎞ ⎤ ⎢ g m1 + g m 2 (1 + β ) ⎜ ⎟ ⎥ ⋅ RC V ⎝ rπ 1 ⎠ ⎦ Av = o = − ⎣ Vs ⎛r ⎞ 1 + (1 + β ) ⎜ π 2 ⎟ ⎝ rπ 1 ⎠ so ⎡ ⎛ 2.63 ⎞ ⎤ ⎢ 0.377 + ( 38.1)(101) ⎜ 265 ⎟ ⎥ ( 4 ) ⎝ ⎠⎦ Av = − ⎣ ⎛ 2.63 ⎞ 1 + (101) ⎜ ⎟ ⎝ 265 ⎠ or Av = −77.0 (d) Ri = rπ 1 + (1 + β ) rπ 2 = 265 + (101)( 2.63)

or Ri = 531 k Ω TYU6.17 (a) dc analysis: For R1 + R2 + R3 = 100 k Ω I1 =

VCC 12 = = 0.12 mA R1 + R2 + R3 100

Now VE1 = I CQ 2 RE = ( 0.5 )( 0.5 ) = 0.25 V VC1 = VE1 + VCEQ1 = 0.25 + 4 = 4.25 V and VC 2 = VC1 + VCEQ 2 = 4.25 + 4 = 8.25 V So RC =

VCC − VC 2 12 − 8.25 = = 7.5 k Ω 0.5 I CQ

Also VB1 = VE1 + VBE ( on ) = 0.25 + 0.7 = 0.95 V And ⎛ ⎞ R3 R3 VB1 = ⎜ (12 ) ⎟ ⋅ VCC ⇒ 0.95 = 100 ⎝ R1 + R2 + R3 ⎠ which yields R3 = 7.92 k Ω We have VB 2 = VC1 + VBE ( on ) = 4.25 + 0.7 = 4.95 V

and

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⎛ R2 + R3 ⎞ VB 2 = ⎜ ⎟ ⋅ VCC ⎝ R1 + R2 + R3 ⎠ or ⎛ R + 7.92 ⎞ 4.95 = ⎜ 2 ⎟ (12 ) ⎝ 100 ⎠ which yields R2 = 33.3 k Ω Then R1 = 100 − 33.3 − 7.92 = 58.8 k Ω (b) Small-signal transistor parameters: (100 )( 0.026 ) βV rπ 1 = rπ 2 = T = I CQ 0.5

or rπ 1 = rπ 2 = 5.2 k Ω and I CQ 0.5 g m1 = g m 2 = = 0.026 VT or g m1 = g m 2 = 19.23 mA/V and ro1 = ro 2 = ∞ (c) Small-signal voltage gain: We have Vo = − g m 2Vπ 2 ( RC & RL ) Also Vπ 2 + g m 2Vπ 2 = g m1Vπ 1 rπ 2 or ⎛ r ⎞ Vπ 2 = g m1Vπ 1 ⎜ π 2 ⎟ and Vπ 1 = Vs ⎝ 1+ β ⎠ We find ⎛ r ⎞ V Av = o = − g m 2 ( RC RL ) g m1 ⎜ π 2 ⎟ Vs ⎝1+ β ⎠ ⎛ β ⎞ = − g m 2 ( RC RL ) ⎜ ⎟ ⎝ 1+ β ⎠ We obtain ⎛ 100 ⎞ Av = − (19.23) ( 7.5 2 ) ⎜ ⎟ ⎝ 101 ⎠ or Av = −30.1 TYU6.18 (a) dc analysis: with vs = 0 RTH = R1 R2 = 125 30 = 24.2 k Ω ⎛ R2 ⎞ ⎛ 30 ⎞ VTH = ⎜ ⎟ ⋅ VCC = ⎜ ⎟ (12 ) ⎝ 125 + 30 ⎠ ⎝ R1 + R2 ⎠ or VTH = 2.32 V

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Now VTH = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE or

I BQ =

2.32 − 0.7 = 0.0250 mA 24.2 + ( 81)( 0.5 )

and I CQ = ( 80 )( 0.025 ) = 2.00 mA Also ⎡ ⎛1+ β ⎞ ⎤ VCEQ = VCC − I CQ ⎢ RC + ⎜ ⎟ RE ⎥ ⎝ β ⎠ ⎦ ⎣ ⎡ ⎛ 81 ⎞ ⎤ = 12 − ( 2 ) ⎢ 2 + ⎜ ⎟ ( 0.5 ) ⎥ ⎣ ⎝ 80 ⎠ ⎦ or VCEQ = 6.99 V

Power dissipated in RC: 2 PRC = I CQ RC = ( 2.0 ) ( 2 ) = 8.0 mW 2

Power dissipated in RL: I LQ = 0 ⇒ PRL = 0 Power dissipated in transistor: PQ = I BQVBEQ + I CQVCEQ = ( 0.025 )( 0.7 ) + ( 2.0 )( 6.99 ) = 14.0 mW (b) With vs = 18cos ω t ( mV )

β VT

rπ =

I CQ

=

(80 )( 0.026 ) 2.0

= 1.04 k Ω

We can write

β

( RC RL )VP cos ω t rπ Power dissipated in RL: vce =

pRL = =

vce ( rms ) RL 1 1 ⋅ 2 2 × 103

2

⎤ 1 1 ⎡β = ⋅ ⋅ ⎢ ( RC RL ) VP ⎥ 2 RL ⎣ rπ ⎦

⎡ 80 ⋅⎢ ( 2 2 ) ( 0.018)⎤⎥ ⎣1.04 ⎦

2

2

or pRL = 0.479 mW Power dissipated in RC: Since RC = RL = 2 k Ω, we have pRC = 8.0 + 0.479 = 8.48 mW Power dissipated in transistor: From the text, we have 2

⎛β ⎞ ⎛V ⎞ pQ ≅ I CQVCEQ − ⎜ ⎟ ⎜ P ⎟ ( RC RL ) ⎝ rπ ⎠ ⎝ 2 ⎠ 2

2

2

80 ⎛ ⎞ ⎛ 0.018 ⎞ 3 = ( 2 × 10−3 ) ( 6.99 ) − ⎜ ⎟ ( 2 × 10 2 × 3 ⎟ ⎜ ⎝ 1.04 × 10 ⎠ ⎝ 2 ⎠

3

)

or pQ = 13.0 mW TYU6.19 (a) dc analysis:

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RTH = R1 R2 = 53.8 10 = 8.43 k Ω ⎛ R2 ⎞ ⎛ 10 ⎞ VTH = ⎜ ⎟ ⋅ VCC = ⎜ ⎟ (5) 53.8 + 10 ⎠ R R + ⎝ ⎝ 1 2 ⎠ or VTH = 0.7837 V Now 0.7837 − 0.7 I BQ = = 0.00993 mA 8.43 and I CQ = (100 )( 0.00993) = 0.993 mA

We have VCEQ = VCC − I CQ RC or 2.5 = 5 − ( 0.993) RC which yields RC = 2.52 k Ω (b) Power dissipated in RC: 2 PRC = I CQ RC = ( 0.993) ( 2.52 ) 2

or PRC = 2.48 mW Power dissipated in transistor: PQ ≅ I CQVCEQ = ( 0.993)( 2.5 ) or PQ = 2.48 mW (c) ac analysis: Maximum ac collector current: ic = ( 0.993) cos ω t ( mA ) average ac power dissipated in RC: 1 1 2 2 pRC = ( 0.993) RC = ( 0.993) ( 2.52 ) 2 2 or pRC = 1.24 mW Now pRC 1.24 Fraction = = = 0.25 PRC + PQ 2.48 + 2.48

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Chapter 6 Problem Solutions 6.1 a. gm = rπ = r0 =

I CQ VT

β VT I CQ

=

2 ⇒ g m = 76.9 mA/V 0.026

=

(180 )( 0.026 ) 2

⇒ rπ = 2.34 kΩ

VA 150 = ⇒ r0 = 75 kΩ I CQ 2

b. 0.5 ⇒ g m = 19.2 mA/V 0.026 (180 )( 0.026 ) rπ = ⇒ rπ = 9.36 kΩ 0.5 150 r0 = ⇒ r0 = 300 kΩ 0.5

gm =

6.2 (a) gm = rπ = ro =

I CQ VT

β VT I CQ

=

0.8 = 30.8 mA/V 0.026

=

(120 )( 0.026 ) 0.8

= 3.9 K

VA 120 = = 150 K I CQ 0.8

(b) 0.08 = 3.08 mA/V 0.026 (120 )( 0.026 ) rπ = = 39 K 0.08 120 ro = = 1500 K 0.08

gm =

6.3 gm = rπ = r0 =

I CQ VT

β VT I CQ

⇒ 200 = =

I CQ 0.026

⇒ I CQ = 5.2 mA

(125)( 0.026 ) 5.2

⇒ rπ = 0.625 kΩ

VA 200 = ⇒ r0 = 38.5 kΩ I CQ 5.2

6.4

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gm = rπ =

I CQ

⇒ 80 =

VT

β VT

I CQ 0.026

⇒ 1.20 =

I CQ

⇒ I CQ = 2.08 mA

β ( 0.026 ) 2.08

⇒ β = 96

6.5 (a) 2 − 0.7 = 0.0052 mA 250 I C = (120 )( 0.0052 ) = 0.624 mA I BQ =

0.624 ⇒ g m = 24 mA / V 0.026 (120 )( 0.026 ) rπ = ⇒ rπ = 5 k Ω 0.624 ro = ∞ gm =

⎛ r ⎞ ⎛ 5 ⎞ Av = − g m RC ⎜ π ⎟ = − ( 24 )( 4 ) ⎜ ⎟ ⇒ Av = −1.88 r R + ⎝ 5 + 250 ⎠ B ⎠ ⎝ π v v vS = O = O ⇒ vS = −0.426sin100t V Av −1.88

(b) (c) 6.6 gm =

I CQ VT

, 1.08 ≤ I CQ ≤ 1.32 mA

1.08 1.32 ≤ gm ≤ ⇒ 41.5 ≤ g m ≤ 50.8 mA/V 0.026 0.026 (120 )( 0.026 ) β VT rπ = ; rπ ( max ) = = 2.89 kΩ I CQ 1.08 rπ ( min ) =

(80 )( 0.026 ) 1.32

= 1.58 kΩ

1.58 ≤ rπ ≤ 2.89 kΩ

6.7 a. rπ = 5.4 =

β VT I CQ

=

(120 )( 0.026 ) I CQ

⇒ I CQ = 0.578 mA

1 1 VCEQ = VCC = ( 5 ) = 2.5 V 2 2 VCEQ = VCC − I CQ RC ⇒ 2.5 = 5.0 − ( 0.578 ) RC ⇒ RC = 4.33 kΩ I CQ

0.578 = 0.00482 mA β 120 VBB = I BQ RB + VBE ( on ) = ( 0.00482 )( 25 ) + 0.70 ⇒ VBB = 0.820 V I BQ =

=

b.

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rπ = gm = r0 =

β VT I CQ I CQ

=

0.578

= 5.40 kΩ

0.578 = 22.2 mA/V 0.026

=

VT

(120 )( 0.026 )

VA 100 = = 173 kΩ I CQ 0.578

⎛ r ⎞ V0 = − g m ( r0 RC ) Vπ , Vπ = ⎜ π ⎟ VS ⎝ rπ + RB ⎠ β ( r0 RC ) ⎛ r ⎞ Av = − g m ⎜ π ⎟ ( r0 RC ) = − rπ + RB ⎝ rπ + RB ⎠ Av = −

(120 ) ⎡⎣173

4.33⎤⎦

5.40 + 25

=−

(120 )( 4.22 ) 30.4

⇒ Av = −16.7

6.8 a. 1 VECQ = VCC = 5 V 2 VECQ = 10 − I CQ RC ⇒ 5 = 10 − ( 0.5 ) RC ⇒ RC = 10 kΩ I CQ

0.5 = 0.005 100 VEB ( on ) + I BQ RB = VBB = ( 0.70 ) + ( 0.005 )( 50 ) ⇒ VBB = 0.95 V I BQ =

=

β

b. gm = rπ = r0 =

c.

I CQ

=

VT

β VT I CQ

0.5 ⇒ g m = 19.2 mA/V 0.026

=

(100 )( 0.026 ) 0.5

⇒ rπ = 5.2 kΩ

VA ∞ = ⇒ r0 = ∞ I CQ 0.5 Av = −

(100 )(10 ) ⇒ A = −18.1 β RC =− v 5.2 + 50 rπ + RB

6.9 10 − 4 = 1.5 mA 4 1.5 I BQ = = 0.015 mA 100 (100 )( 0.026 ) rπ = = 1.73 K 1.5 5sin ω t ( mV ) v ib = be = = 2.89sin ω t ( μ A ) rπ 1.73 kΩ So I CQ =

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iB ( t ) = I BQ + iEb = 15 + 2.89sin ω t ( μ A )

iC1 ( t ) = β iB ⇒ iC1 ( t ) = 1.5 + 0.289sin ω t ( mA )

vC ( t ) = 10 − iC1 ( t ) RC = 10 − [1.5 + 0.289sin ω t ] (γ )

vC1 ( t ) = 4 − 1.156sin ω t ( v ) Av =

vC ( t )

vbe ( t )

=

−1.156 ⇒ Av = −231 0.005

6.10 vo = 1.2sin ω t ( V ) iC ( t ) RC + vo = 0 ⇒ iC ( t ) = iC ( t ) = −0.60sin ω t ( mA ) ib ( t ) =

iC ( t )

β

−1.2sin ω t 2

= −6sin ω t ( μ A )

vbe ( t ) = ib ( t ) ⋅ rπ

g m rπ = β

100 =2K 50 vbe ( t ) = −12sin ω t ( mV ) rπ =

6.11 a. I CQ ≈ I EQ VCEQ = 5 = 10 − I CQ ( RC + RE ) = 10 − I CQ (1.2 + 0.2) I CQ = 3.57 mA 3.57 = 0.0238 mA 150 R1 & R2 = RTH = ( 0.1)(1 + β ) RE I BQ =

= ( 0.1)(151)( 0.2 ) = 3.02 kΩ

VTH =

1 ⋅ RTH ⋅ (10) − 5 R1

VTH = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE − 5 1 (3.02)(10) − 5 = ( 0.0238)(3.02) + 0.7 + (151)( 0.0238)( 0.2) − 5 R1 1 ( 30.2 ) = 1.50 ⇒ R1 = 20.1 k Ω R1 20.1R2 = 3.02 ⇒ R2 = 3.55 kΩ 20.1 + R2 b. (150 )( 0.026 ) rp = = 1.09 kΩ 3.57 3.57 gm = = 137 mA/V 0.026

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V0

 V␲  

VS

r␲

gmV␲

 R1R2

RC RE

Av =

(150 )(1.2 ) 2 β RC =2 ⇒ Av = 2 5.75 rp + (1 + β ) RE 1.09 + (151)( 0.2 )

6.12 a. ⎛ R2 ⎞ ⎛ 50 ⎞ VTH = ⎜ ⎟ VCC = ⎜ ⎟ (12 ) = 10 V + R R 50 + 10 ⎠ ⎝ ⎝ 1 2 ⎠ RTH = R1 R2 = 50 10 = 8.33 kΩ I BQ =

12 − 0.7 − 10 = 0.0119 mA 8.33 + (101)(1)

I CQ = 1.19 mA, I EQ = 1.20 mA VECQ = 12 − (1.20 )(1) − (1.19 )( 2 ) VECQ = 8.42 V iC 4

1.19

8.42

12

␯EC

b. V0

 V␲ VS

 

r␲

gmV␲

 R1R2

RC RE

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rp =

(100 )( 0.026 )

1.19 V0 = g mVp RC

= 2.18 kΩ

⎛V ⎞ VS = 2 Vp − ⎜ p + g mVp ⎟ RE ⎝ rp ⎠ ⎡ rπ + (1 + β ) RE ⎤ = −Vp ⎢ ⎥ rp ⎣ ⎦ 2 (100 )( 2 ) 2 β RC = ⇒ Av = 2 1.94 Av = rp + (1 + β ) RE 2.18 + (101)(1)

c.

Approximation: Assume rp does not vary significantly.

RC = 2 kΩ ± 5% = 2.1 kΩ or 1.9 kΩ RE = 1 kΩ ± 5% = 1.05 kΩ or 0.95 kΩ

For RC ( max ) = 2.1 kΩ and RE ( min ) Av =

− (100 )( 2.1)

2.18 + (101)( 0.95 )

= −2.14

For RC ( min ) = 1.9 kΩ and RE ( max ) = 1.05 kΩ Av =

− (100 )(1.9 )

2.18 + (101)(1.05 )

= −1.76

So 1.76 ≤ Av ≤ 2.14

6.13 (a) VCC = ⎜⎜ 1+ β ⎟⎟ I CQ RE + VECQ + I CQ RC ⎛ ⎝

⎞

β

⎠

⎛ 101 ⎞ 12 = ⎜ ⎟ I CQ (1) + 6 + I CQ ( 2 ) ⎝ 100 ⎠ so that I CQ = 1.99 mA 1.99 = 0.0199 mA 100 = ( 0.1)(1 + β ) RE = ( 0.1)(101)(1) = 10.1 k Ω

I BQ = RTH

⎛ R2 ⎞ 1 1 VTH = ⎜ ⎟ VCC = ⋅ RTH ⋅ VCC = (10.1)(12 ) R R R R + 2 ⎠ 1 1 ⎝ 1 VCC = (1 + β ) I BQ RE + VEB ( on ) + I BQ RTH + VTH 121.2 12 = (101)( 0.0199)(1) + 0.7 + ( 0.0199)(10.1) + R1 which yields R1 = 13.3 k Ω and R2 = 41.6 k Ω

(b)

Av =

2 (100 )( 2 ) 2 β RC = ⇒ Av = 2 1.95 rp + (1 + β ) RE 1.31 + (101)(1)

6.14

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I CQ = 0.25 mA, I EQ = 0.2525 mA I BQ = 0.0025 mA

I BQ RB + VBE ( on ) + I EQ ( RS + RE ) − 5 = 0 ( 0.0025)( 50 ) + 0.7 + ( 0.2525)( 0.1 + RE ) = 5 RE = 16.4 kΩ

VE = − ( 0.0025 )( 50 ) − 0.7 = −0.825 V VC = VCEQ + VE = 3 − 0.825 = 2.175 V 5 − 2.175 RC = ⇒ RC = 11.3 kΩ 0.25 − β RC Av = rπ + (1 + β ) RS rπ =

(100 )( 0.026 )

= 10.4 kΩ 0.25 − (100 )(11.3) Av = ⇒ Av = −55.1 10.4 + (101)( 0.1)

Ri = RB ⎡⎣ rπ + (1 + β ) RS ⎤⎦ = 50 ⎡⎣10.4 + (101)( 0.1) ⎤⎦ Ri = 50 20.5 ⇒ Ri = 14.5 kΩ

6.15 (a) VCC > I CQ ( RC + RE ) + VCEQ

9 = I CQ ( 2.2 + 2 ) + 3.75 So that I CQ = 1.25 mA

Assume circuit is to be designed to be bias stable. RTH = R1 & R2 = ( 0.1)(1 + β ) RE = ( 0.1)(121)( 2 ) = 24.2 Ω 1.25 I BQ = = 0.01042 mA 120 1 VTH = ⋅ RTH ⋅ VCC = I BQ RTH + VBE ( on ) + I BQ (121)( RE ) R1 1 ( 24.2 )( 9 ) = ( 0.01042 )( 24.2 ) + 0.7 + ( 0.01042 )(121)( 2 ) R1 = 0.2522 + 0.7 + 2.5216 = 3.474 R1 = 62.7 K

62.7 R2 = 24.2 62.7 + R2

R2 = 39.4 K

(b)

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1.25 = 48.08 mA / V 0.026 (120 )( 0.026 ) rp = = 2.50 k Ω 1.25 100 ro = = 80 k Ω 1.25

gm =

Vo

 IS

V␲

r␲

R1R2

ro

RC

RL

gmV␲

Vo = 2 g mVp ( ro RC RL )

Vp = I S ( R1 R2 rp )

Then Rm =

Vo = 2 g m ( R1 R2 rp )( ro RC RL ) Is

Rm = 2 48.08 ( 24.2 2.5 )( 80 2.2 1) = 2 48.08 ( 2.266 )( 0.6816 )

or Rm =

Vo = −74.3 k Ω = −74.3 V / mA Is

6.16 a. I EQ = 0.80 mA, I BQ =

0.80 = 0.0121 mA 66

I CQ = 0.788 mA 0.3 ⇒ RB = 24.8 kΩ 0.0121 V − ( −5 ) 5 − 3 = ⇒ RC = 2.54 kΩ RC = C I CQ 0.788 VB = I BQ RB ⇒ RB =

b. 0.788 = 30.3 mA / V 0.026 ( 65)( 0.026) rπ = = 2.14 kΩ 0.788 75 r0 = = 95.2 kΩ 0.788 ⎛ RC r0 ⎞ i0 = ⎜ g V , V = − vS ⎜ R r + R ⎟⎟ m π π L ⎠ ⎝ C 0 gm =

Gf =

⎛ RC r0 i0 = − gm ⎜ ⎜R r +R vS L ⎝ C 0

⎞ ⎟⎟ ⎠

⎛ 2.54 95.2 ⎞ = − ( 30.3) ⎜ ⎜ 2.54 95.2 + 4 ⎟⎟ ⎝ ⎠ G f = −11.6 mA/V

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6.17 (a) I CQ = 0.8 mA ⇒ I BQ = 0.00667 mA I BQ RS + 0.7 + (121) I BQ RE − 15 = 0 ( 0.01667 )( 2.5) + 0.7 + (121)( 0.00667 ) RE = 15 RE = 17.7 K

VE = − ( 0.00667 )( 2.5 ) − 0.7 = −0.717 V VC = −0.717 + 7 = 6.283 V 15 − 6.283 RC = = 10.9 K 0.8 (120 )( 0.026 ) 0.8 gm = = 30.77 mA/V rπ = = 3.9 K 0.026 0.8 ⎛ r ⎞ vo = − g m ( RC RL ) ⋅ vπ vπ = ⎜ π ⎟ vS ⎝ rπ + RS ⎠ Av =

− β ( RC RL ) rπ + RS

=

− (120 ) (10.9 5 ) 3.9 + 2.5

Av = −64.3 (b) For RS = 0 0.7 + (121)( 0.00667 ) RE = 15 RE = 17.7 K VE = −0.7 ⇒ VC = −0.7 + 7 = 6.3 15 − 6.3 ⇒ RC = 10.9 K 0.8 − β ( RC RL ) Av = = −30.77 (10.9 5 ) rπ

RC =

Av = −105

6.18 (a) 15 = ( 81) I BQ (10 ) + 0.7 + I BQ ( 2.5 ) 15 − 0.7 I BQ = = 0.0176 mA 2.5 + ( 81)(10 ) I CQ = 1.408 mA

(80 )( 0.026 ) 1.408 = 54.15 mA/V rπ = 0.026 1.408 rπ = 1.48 K gm =

RS V0 IS  VS

 

V␲ 

r␲

gmV␲

RC

Io

RL

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Vo = − g mVσ ( RC RL ) ⇒ Av = i AI = o = iS

− β ( RC RL ) rπ + RS

=

− ( 80 ) ( 5 5 ) 1.48 + 2.5

⇒ Av = −50.3

⎛ RC ⎞ − g mVπ ⎜ ⎟ ⎝ RC + RL ⎠ = − β ⎛ RC ⎞ ⎜ ⎟ V ⎝ RC + RL ⎠ π r

AI = −40

π

vo ( t ) = ( −50.3)( 4sin ω t ) vo ( t ) = −0.201sin ω t ( V ) 4 sin ω t ( mV ) = 1.005sin ω t ( μA ) 2.5 + 1.48 io = −40.2 sin ω t ( μA ) is =

(b) 15 − 0.7 = 1.43 mA 10 ⎛ 80 ⎞ = ⎜ ⎟ (1.43) = 1.412 mA ⎝ 81 ⎠

I EQ = I CQ

gm =

(80 )( 0.026 ) 1.412 = 54.3 mA/V rπ = = 1.47 K 0.026 1.412

Av = − g m ( RC RL ) = − ( 54.3) ( 5 5 ) ⇒ Av = −136

⎛ RC ⎞ ⎛ 5 ⎞ AI = − β ⎜ ⎟ = −80 ⎜ ⎟ ⇒ AI = −40 ⎝5+5⎠ ⎝ RC + RL ⎠ vo ( t ) = ( −136 )( 4sin ω t ) ⇒ vo ( t ) = −544sin ω t ⇒ vo ( t ) = −0.544sin ω t ( V ) is ( t ) =

4sin ω t ( mV )

= 2.72sin ω t ( μA ) 1.47 k io ( t ) = ( −40 )( 2.72sin ω t ) io ( t ) = −109sin ω t ( μA )

6.19 RTH = R1 R2 = 27 15 = 9.64 K ⎛ R2 ⎞ ⎛ 15 ⎞ VTH = ⎜ ⎟ VCC = ⎜ ⎟ ( 9 ) = 3.214 V + R R 15 + 27 ⎠ ⎝ 2 ⎠ ⎝ 1 V − VBE ( on ) 3.214 − 0.7 2.514 I BQ = TH = = RTH + (1 + β ) RE 9.64 + (101)(1.2 ) 130.84

I BQ = 0.0192 mA I CQ = 1.9214 mA gm =

(100 )( 0.026 ) 1.92 = 73.9 mA/V rπ = = 1.35 K 0.026 1.92

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RS V0 IS   

VS

ro =

V␲

RTH

r␲

gmV␲



r0

100 = 52.1 K 1.92

⎛ r R Vπ = ⎜ π TH ⎜r R +R S ⎝ π TH = 1.35 9.64 = 1.184 K

(

Vo = − g mVπ r0 RC RL rπ RTH

⎛ 1.184 ⎞ Vπ = ⎜ ⎟ VS ⎝ 1.184 + 10 ⎠ = 0.1059VS

)

(

Av = − ( 73.9 ) ( 0.1059 ) 52.1 2.2 2

⎞ ⎟⎟ VS ⎠

)

= − ( 73.9 ) ( 0.1059 ) ( 52.1 1.0476 )

= − ( 73.9 ) ( 0.1059 ) (1.027 ) Av = −8.04 ⎛ ro RC − g mVπ ⎜⎜ I ⎝ ro RC + RL AI = o = Vπ IS RTH rπ

⎞ ⎟⎟ ⎠

⎛ ro RC ⎞ AI = − g m ( RTH rπ ) ⎜ ⎜ r R + R ⎟⎟ L ⎠ ⎝ o C ro RC = 52.1 2.2 = 2.11 K RTH rπ = 9.64 1.35 = 1.184 K

⎛ 2.11 ⎞ AI = − ( 73.9 ) (1.184 ) ⎜ ⎟ ⎝ 2.11 + 2 ⎠ AI = −44.9 Ri = RTH rπ = 9.64 1.35 Ri = 1.184 K

6.20 a. 0.35 = 0.00347 mA 101 VB = 2 I B RB = 2 ( 0.00347 )(10 ) ⇒ VB = 2 0.0347 V I E = 0.35 mA, I B =

VE = VB − VBE ( on ) ⇒ VE = 2 0.735 V

b.

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RC

I0

RL

VC = VCEQ + VE = 3.5 − 0.735 = 2.77 V ⎛ b ⎞ ⎛ 100 ⎞ IC = ⎜ ⎟ IE = ⎜ ⎟ ( 0.35 ) = 0.347 mA ⎝ 101 ⎠ ⎝ 1+ b ⎠ V 1 − VC 5 − 2.77 RC = = ⇒ RC = 6.43 kΩ IC 0.347 (c) ⎛ RB rp ⎞ Av = 2 g m ⎜ R r ⎜ R r + R ⎟⎟ ( C o ) S ⎠ ⎝ B π 0.347 100 gm = = 13.3 mA/V , ro = = 288 k Ω 0.026 0.347 (100 )( 0.026 ) = 7.49 k Ω rp = 0.347 RB rp = 10 7.49 = 4.28 k Ω ⎛ 4.28 ⎞ Av = 2 (13.3) ⎜ ⎟ ( 6.43 288 ) ⇒ Av = 2 81.7 ⎝ 4.28 + 0.1 ⎠ d. ⎛ RB rp ⎞ Av = 2 g m ⎜ R r ⎜ R r + R ⎟⎟ ( C 0 ) S ⎠ ⎝ B p RB rp = 10 7.49 = 4.28 kΩ ⎛ 4.28 ⎞ Av = 2 (13.3) ⎜ ⎟ ( 6.43 288 ) ⇒ Av = 2 74.9 ⎝ 4.28 + 0.5 ⎠

6.21 a. RTH = R1 R2 = 6 1.5 = 1.2 kΩ ⎛ R2 ⎞ + ⎛ 1.5 ⎞ VTH = ⎜ ⎟V = ⎜ ⎟ ( 5 ) = 1.0 V ⎝ 1.5 + 6 ⎠ ⎝ R1 + R2 ⎠ V − VBE ( on ) 1.0 − 0.7 I BQ = TH = = 0.0155 mA RTH + (1 + β ) RE 1.2 + (181)( 0.1)

I CQ = 2.80 mA, I EQ = 2.81 VCEQ = V + − I CQ RC − I EQ RE

= 5 − ( 2.8 )(1) − ( 2.81)( 0.1) ⇒ VCEQ = 1.92 V

b.

(180 )( 0.026 )

⇒ rp = 1.67 kΩ 2.80 2.80 gm = ⇒ g m = 108 mA/V, r0 =` 0.026 (c) R1 R2 rp ⎛ ⎞ Av = 2 g m ⎜ ⎟ ( RC RL ) ⎝ R1 & R2 & rp + RS ⎠ R1 R2 rp = 6 1.5 1.67 = 0.698 k V rp =

⎛ 0.698 ⎞ Av = 2 (108 ) ⎜ ⎟ (1 1.2 ) ⇒ Av = 2 45.8 ⎝ 0.698 + 0.2 ⎠

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6.22 a. 9 = I EQ RE + VEB ( on ) + I BQ RS 0.75 = 0.00926 mA I EQ = 0.75 mA, I BQ = 81 I CQ = 0.741 mA 9 = ( 0.75 ) RE + 0.7 + ( 0.00926 )( 2 ) ⇒ RE = 11.0 kΩ b. VE = 9 − ( 0.75 )(11) = 0.75 V VC = VE − VECQ = 0.75 − 7 = −6.25 V RC =

VC − ( −9 ) I CQ

=

9 − 6.25 ⇒ RC = 3.71 kΩ 0.741

c. ⎛ rp ⎞ Av = 2 g m ⎜ ⎟ ( RC || RL || r0 ) ⎝ rp + RS ⎠ (80 )( 0.026 ) rp = = 2.81 kΩ 0.741 80 r0 = = 108 kV 0.741 2 80 Av = ( 3.71||10 ||108 ) 2.81 + 2 Av = 2 43.9

d. Ri = RS + rp = 2 + 2.81 ⇒ Ri = 4.81 kΩ 6.23 I BQ =

4 − 0.7 = 0.00647 5 + (101)( 5 )

I CQ = 0.647 mA

80 ≤ h fe ≤ 120, 10 ≤ h0e ≤ 20 mS

a.

2 .45 k

Ω ≤ hie ≤ 3

.7 k Ω   low gain

high gain

RS V0 IS  VS

 

V␲ 

r␲

gmV␲

RC

Io

RL

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⎛ 1 ⎞ V0 = 2 h fe I b ⎜ RC RL ⎟ ⎝ hoe ⎠ RB ?VS R + RS Ib = B RTH + hie RTH = RB RS = 5 1 = 0.833 kΩ

High-gain ⎛ 5 ⎞ ⎜ ⎟ VS 5 +1⎠ Ib = ⎝ = 0.1838VS 0.833 + 3.7 Low-gain ⎛ 5 ⎞ ⎜ ⎟ VS 5 +1⎠ ⎝ Ib = = 0.2538VS 0.833 + 2.45 1 1 For hoe = 10 ⇒ || Rc || RL = || 4 || 4 0.010 hoe = 100 || 2 = 1.96 kΩ

1 || 4 || 4 = 50 || 2 = 1.92 kΩ 0.020 = (120 )( 0.1838 )(1.96 ) = 43.2 = ( 80 )( 0.2538 )(1.92 ) = 39.0

For hoe = 20 ⇒ Av Av

max min

39.0 ≤ Av ≤ 43.2

b. Ri = RB hie = 5 3.7 = 2.13 kV or Ri = 5 2.45 = 1.64 kΩ 1.64 ≤ Ri ≤ 2.13 kΩ R0 =

1 1 4 = 100 || 4 = 3.85 kΩ RC = 0.010 hoe

1 || 4 = 50 || 4 = 3.70 kΩ 0.020 3.70 ≤ R0 ≤ 3.85 kΩ or R0 =

6.24 VCC  10 V

RC R1 ␯o RS  1 k CC ␯s

 

R2 RE

CE

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Assume an npn transistor with b = 100 and VA = ∞. Let VCC = 10 V . 0.5 = 50 0.01 Bias at I CQ = 1 mA and let RE = 1 k Ω

Av =

For a bias stable circuit RTH = ( 0.1)(1 + b ) RE = ( 0.1)(101)(1) = 10.1 k Ω 1 1 101 VTH = ⋅ RTH ⋅ VCC = (10.1)(10 ) = R1 R1 R1 1 I BQ = = 0.01 mA 100 VTH = I BQ RTH + VBE ( on ) + (1 + b ) I BQ RE 101 = ( 0.01)(10.1) + 0.7 + (101)( 0.01)(1) R1 which yields R1 = 55.8 k Ω and R2 = 12.3 k Ω

Now rp =

(100 )( 0.026 ) 1

= 2.6 k Ω

1 = 38.46 mA/V 0.026 Vo = − g mVp RC

gm =

⎛ R1} R2 } rp ⎞ ⎛ 10.1} 2.6 ⎞ where Vp = ⎜ ⎟ ⋅ Vs = ⎜ ⎟ .Vs ⎝ 10.1} 2.6 + 1 ⎠ ⎝ R1} R2 } rp + RS ⎠ or Vp = 0.674 Vs V Then Av = o = − ( 0.674 ) g m RC = − ( 0.674 )( 38.46 ) RC = −50 Vs which yields RC = 1.93 k Ω

With this RC, the dc bias is OK. Finish Design, Set RC = 2 K

RE = 1 K

R1 = 56 K R2 = 12 K

RTH = R1 R2 = 9.88 K ⎛ R2 ⎞ ⎛ 12 ⎞ VTH = ⎜ ⎟ VCC = ⎜ ⎟ (10 ) = 1.765 V ⎝ 12 + 56 ⎠ ⎝ R1 + R2 ⎠ 1.765 − 0.7 I BQ = = 9.60 μ A 9.88 + (101)(1) I CQ = 0.9605 mA rπ = RTH

(100 )( 0.026 ) 0.9605 rπ = 2.125 K

= 2.707 K

gm =

0.9605 = 36.94 0.026

⎛ RTH & rπ ⎞ ⎛ 2.125 ⎞ Vπ = ⎜ ⎟ Vi = ⎜ ⎟ Vi = ( 0.680 ) Vi ⎝ 2.125 + 1 ⎠ ⎝ RTH & rπ + RS ⎠ Av = − ( 0.680 ) g m RC = − ( 0.680 )( 36.94 )( 2 ) = −50.2

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Design specification met. 6.25 a. I BQ =

6 − 0.7 = 0.0169 mA 10 + (101)( 3)

I CQ = 1.69 mA, I EQ = 1.71 mA VCEQ = (16 + 6 ) − (1.69 )( 6.8 ) − (1.71)( 3) VCEQ = 5.38 V

b. 1.69 ⇒ g m = 65 mA/V 0.026 (100 )( 0.026 ) rp = ⇒ rp = 1.54 kV , 1.69 (c) − β ( RC RL ) RB Rib Av = ⋅ rπ + (1 + β ) RE RB Rib + RS gm =

r0 = ∞

Rib = rπ + (1 + β ) RE = 1.54 + (101)(3) = 304.5 k Ω RB Rib = 10 304.5 = 9.68 k Ω

Then Av =

− (100 )( 6.8 & 6.8 ) ⎛ 9.68 ⎞ ⋅⎜ ⎟ ⇒ Av = −1.06 1.54 + (101)( 3) ⎝ 9.68 + 0.5 ⎠

⎛ RC ⎞ i0 = ⎜ ⎟ ( − β ib ) ⎝ RC + RL ⎠ ⎛ RB ib = ⎜ R r + + ⎝ B π (1 + β ) RE

⎞ ⎟ iS ⎠

⎞ ⎞⎛ RB ⎟⎟ ⎟ ⎜⎜ ⎠ ⎝ RB + rπ + (1 + β ) RE ⎠ ⎞ 10 ⎛ 6.8 ⎞ ⎛ = − (100 ) ⎜ ⎟ ⇒ Ai = −1.59 ⎟ ⎜⎜ ⎝ 6.8 + 6.8 ⎠ ⎝ 10 + 1.54 + (101)( 3) ⎠⎟ (d) Ris = RS + RB Rib = 0.5 + 10 304.5 = 10.2 k Ω ⎛ RC Ai = − ( β ) ⎜ ⎝ RC + RL

(e) Av = Av =

2 b ( RC RL )

rp + (1 + b ) RE

2 (100 ) ( 6.8} 6.8 ) 1.54 + (101)( 3)

⇒ Av = 2 1.12

Ai = same as ( c ) ⇒ Ai = 2 1.59

6.26

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ie 



vCE

 vbe  

vCE

gmv

ris

o



 vCE

gmv



r=

vCe 1 = g m vCe g m

⎛ 1 ⎞ So re = rp ⎜ ⎟ r0 ⎝ gm ⎠

6.27 Let b = 100, VA = ∞ VCC

RC R1 ␯o RS  100  CC ␯s

 

R2 RE

Let VCC = 2.5 V

P = ( I R + I C ) VCC ⇒ 0.12 = ( I R + I C )( 2.5 ) ⇒ I R + I C = 48 mA, Let I R = 8mA, I C = 40 mA

R1 + R2 > I BQ =

VCC 2.5 ⇒ 312.5 k Ω = IR 8

40 = 0.4 mA 100

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Let RE = 2 k Ω. For a bias stable circuit RTH = ( 0.1)(1 + b ) RE = ( 0.1)(101)( 2 ) = 20.2 k Ω 1 VTH = ⋅ RTH ⋅ VCC = I BQ RTH + VBE ( on ) + (1 + b ) I BQ RE R1 1 ( 20.2 )( 2.5 ) = ( 0.0004 )( 20.2 ) + 0.7 + (101)( 0.0004 )( 2 ) R1 which yields R1 = 64 k V and R2 = 29.5 k Ω

(100 )( 0.026 )

rπ =

0.04

Av =

= 65 k Ω

Neglect RS

Vo 2 b RC > Vs rπ + (1 + b ) RE

−10 =

2 100 RC ⇒ RC = 26.7 k Ω 65 + (101)( 2 )

With this RC , dc biasing is OK. 6.28 100 = 20. 5 Assume a sign inversion from a common-emitter is not important. Use the configuration for Figure 6.31. Let RS = 0. Need an input resistance of

Need a voltage gain of

5 × 102 3 = 25 × 103 = 25 k Ω 0.2 × 102 6 Ri = RTH Rib . Let RTH = 50 k Ω, Rib = 50 k Ω Ri =

Rib = rp + (1 + b ) RE > (1 + b ) RE

Rib 50 = = 0.495 k Ω 1 + b 101 Let RE = 0.5 k V , VCC = 10 V , I CQ = 0.2 mA For b = 100, RE =

0.2 = 0.002 mA 100 = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE

Then I BQ = VTH

1 1 ⋅ RTH ⋅ VCC = ( 50)(10) = ( 0.002)(50) + 0.7 + (101)( 0.002 )( 0.5) R R1 1 which yields R1 = 555 k Ω and R2 = 55 k Ω Now Av =

(100)( 0.026) − β RC = 13 k Ω , rπ = rπ + (1 + β ) RE 0.2

So −20 =

− (100 ) RC

13 + (101)( 0.5)

⇒ RC = 12.7 k Ω

[Note: I CQ RC = ( 0.2 )(12.7 ) = 2.54 V. So dc biasing is OK.] 6.29

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VCC  10 V

R1

RE

CC

␯s

␯o

 

R2 RC

b = 80, Av =

2 b RC rp + (1 + b ) RE

First approximation: R ( Av ) ≈ C = 10 ⇒ RC = 10 RE RE Set RC = 12 RE

VEC ≈ VCC − I C ( RC + RE ) = 10 − I C (13RE ) 1 For VEC = VCC = 5 2 5 = 10 − I C (13RE ) For I C = 0.7 mA I E = 0.709, I B = 0.00875 mA ⇒ RE = 0.55 kΩ − RC = 6.6 kΩ Bias stable ⇒ R1 R2 = RTH = ( 0.1)(1 + β ) RE = ( 0.1)(81)( 0.55 ) = 4.46 kΩ 1 10 = ( 0.709 )( 0.55 ) + 0.7 + ( 0.00875 )( 4.46 ) + ( 4.46 )(10 ) R1 8.87 =

1 ( 4.46 ) ⇒ R1 = 5.03 kΩ R1

5.03R2 = 4.46 ⇒ R2 = 39.4 kΩ 5.03 + R2 10 10 = = 0.225 mA R1 + R2 5.03 + 39.4 0.7 + 0.225 ≅ 0.925 mA from VCC source. Now rπ = Av =

(80 ) ( 0.026 ) 0.7 (80 )( 6.6 )

= 2.97 kΩ

2.97 + ( 81)( 0.55 )

= 11.1

6.30

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5V

RC

R1

CC2 ␯o

CC1

RL  10 K ␯s

 

R2

CE RE

5V

β = 120

Let I CQ = 0.35 mA, I EQ = 0.353 mA I BQ = 0.00292 mA Let RE = 2 kΩ. For VCEQ = 4 V ⇒ 10 = 4 + ( 0.35) RC + ( 0.353)( 2) RC = 15.1 kΩ, rπ = Av =

− β ( RC RL ) rπ

(120 )( 0.026 )

=−

= 8.91 kΩ 0.35 (120 ) (15.1 10 ) 8.91

Av = −81.0

For bias stable circuit: R1 R2 = RTH = ( 0.1)(1 + β ) RE = ( 0.1)(121)( 2 ) = 24.2 kΩ ⎛ R2 ⎞ 1 VTH = ⎜ ⎟ (10) − 5 = ⋅ RTH ⋅ (10 ) − 5 R1 ⎝ R1 + R2 ⎠ VTH = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE − 5 1 ( 24.2 )(10 ) − 5 = ( 0.00292 )( 24.2 ) + 0.7 + (121)( 0.00292 )( 2 ) − 5 R1 1 ( 242 ) = 1.477, R1 = 164 kΩ R1 164 R2 = 24.2 ⇒ R2 = 28.4 kΩ 164 + R2 10 = 0.052, 0.35 + 0.052 = 0.402 mA 164 + 28.4 So bias current specification is met.

6.31 From Prob. 6.12,

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RTH = R1} R2 = 10}50 = 8.33 kΩ ⎛ R2 ⎞ ⎛ 50 ⎞ VTH = ⎜ ⎟ (12 ) = ⎜ ⎟ (12 ) = 10 V ⎝ 50 + 10 ⎠ ⎝ R1 + R2 ⎠ 12 − 0.7 − 10 = 0.0119 mA I BQ = 8.33 + (101)(1) I CQ = 1.19 mA, I EQ = 1.20 mA VECQ = 12 − (1.19 )( 2 ) − (120 )(1) = 8.42 V

1.19

1

8.42

11 12

For 1 ≤ vEC ≤ 11 DvEC = 11 − 8.42 = 2.58 ⇒ Output voltage swing = 5.16 V (peak-to-peak) 6.32 I BQ =

5 − 0.7 = 0.00315 mA 50 + (101)( 0.1 + 12.9)

I CQ = 0.315 mA, I EQ = 0.319 mA VCEQ = ( 5 + 5 ) − ( 0.315 )( 6 ) − ( 0.319 )(13) VCEQ = 3.96 V

AC load line 1 Slope  6.1 K 0.315

3.96

10

1 Δv 6.1 eC For ΔiC = 0.315 − 0.05 = 0.265 ⇒ ΔvEC = 1.62

ΔiC = −

vEC ( min ) = 3.96 − 1.62 = 2.34

Output signal swing determined by current:

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Max. output swing = 3.24 V peak-to-peak 6.33 From Problem 4.18, I CQ = 1.408 mA, I EQ = 1.426 mA (a) VECQ = 30 − (1.408 )( 5 ) − (1.426 )(10 ) = 8.7 V IC (mA) AC load line 1 Slope  RC RL 1  2.5 kΩ

1.408

8.7

␯EC (V)

vEC ( max ) = 8.7 + ΔI C ⋅ ( 2.5 ) = 8.7 + (1.408 )( 2.5 ) = 12.22

Set vEC ( max ) = 12 = 8.7 + ΔI C ( 2.5 ) ⇒ ΔI C = 1.32 mA So ΔvEC (peak-to-peak) = 2(12 − 8.7) = 6.6 V (b) ΔiC (peak-to-peak) = 2(1.32) = 2.64 mA 6.34 I EQ I BQ VE VC VECQ

= 0.80 mA, I CQ = 0.792 mA = 0.00792 mA = 0.7 + ( 0.00792 )(10 ) = 0.779 V = I CQ RC − 5 = ( 0.792 )( 4 ) − 5 = 2 1.83 V = 0.779 − ( −1.83) = 2.61 V

Load line: Assume VE remains constant at ≈ 0.78 V IC (mA) AC load line 1 Slope  RC RL 1  2.5 kΩ

1.408

8.7

␯EC (V)

21 ?vec 2 kV Collector current swing = 0.792 − 0.08 = 0.712 mA Dvec = ( 0.712 )( 2 ) = 1.424 V DiC =

Output swing determined by current. Max. output swing = 2.85 V peak-to-peak 2.85 4 = 0.712 mA peak-to-peak

Swing in i0 current =

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6.35 I BQ =

6 − 0.7 = 0.0169 mA 10 + (101)( 3)

I CQ = 1.69 mA, I EQ = 1.71 mA VCEQ = (16 + 6 ) − (1.69 )( 6.8 ) − (1.71)( 3) VCEQ = 5.38 V

AC load line 1 Slope  3.4  3 1  6.4 K

1.69

5.38

22

1 DiC = 2 Dvce 6.4

For vce ( min ) = 1 V, Dvce = 5.38 − 1 = 4.38 V ⇒ DiC =

4.38 = 0.684 mA 6.4

Output swing limited by voltage: Δvce = Max. swing in output voltage = 8.76 V peak-to-peak 1 ΔiC ⇒ Δi0 = 0.342 mA 2 or Δi0 = 0.684 mA (peak-to-peak) Δi0 =

6.36 AC load line 1 Slope  1.05 K

2.65

Q-point

ICQ

VCEQ

ro =

9

100 I CQ

Neglect ro as (E) approx. dc load line VCE = 9 − I C ( 3.4 )

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ΔI C = I CQ − 0.1 ΔVCE = VCEQ − 1 Also ΔVCE = ΔI C ( RC RL ) = ΔI C (1.05 ) Or VCEQ − 1 = ( I CQ − 0.1) (1.05 )

Substituting the expression for the dc load line. ⎡⎣9 − I CQ ( 3.4 ) − 1⎤⎦ = ( I CQ − 0.1) (1.05 )

8.105 = I CQ ( 4.45 ) ⇒ I CQ = 1.821 mA VCEQ = 2.81 V 1.821 I BQ = = 0.01821 100 RTH = ( 0.1)(101)(1.2 ) = 12.12 K 1 1 VTH = ⋅ RTH ⋅ VCC = (12.12 ) ( 9 ) = ( 0.01821) (12.12 ) + 0.7 + (101)( 0.01821)(1.2 ) R1 R1 = 0.2207 + 0.7 + 2.20705 R1 = 34.9 K R2 = 18.6 K 34.9 R2 = 12.12 34.9 + R2

6.37 dc load line 5  4.55 mA 1  0.1 AC load line 1 Slope  11.2 1  0.545 K

ICQ

VCEQ

5

For maximum symmetrical swing ΔiC = I CQ − 0.25 ΔvCE = VCEQ − 0.5 and ΔiC = I CQ − 0.25 =

VCEQ − 0.5

1 ⋅ | ΔvCE | 0.545 kΩ

0.545 VCEQ = 5 − I CQ (1.1)

0.545 ( I CQ − 0.25 ) = ⎡⎣5 − I CQ (1.1) ⎤⎦ − 0.5

( 0.545 + 1.1) I CQ = 5 − 0.5 + 0.136 I CQ = 2.82 mA,

I BQ = 0.0157 mA

RTH = R1 & R2 = ( 0.1)(1 + β ) RE

= ( 0.1)(181)( 0.1) = 1.81 kΩ

VTH =

1 ⋅ RTH ⋅ V + = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE R1

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1 (1.81)( 5 ) = ( 0.0157 )(1.81) + 0.7 + (181)( 0.0157 )( 0.1) R1 1 ( 9.05 ) = 1.013 ⇒ R1 = 8.93 kΩ R1 8.93R2 = 1.81 ⇒ R2 =2.27 kΩ 8.93 + R2

6.38 I CQ = 0.647 mA , VCEQ > 10 − ( 0.647 )( 9 ) = 4.18 V DiC = I CQ = 0.647 mA So DvCE = DiC ( 4} 4 ) = ( 0.647 )( 2 ) = 1.294 V

Voltage swing is well within the voltage specification. Then DvCE = 2 (1.294 ) = 2.59 V peak-to-peak 6.39 a. RTH = R1} R2 = 10}10 = 5 kΩ ⎛ R2 ⎞ ⎛ 10 ⎞ VTH = ⎜ ⎟ (18 ) − 9 = ⎜ ⎟ (18 ) − 9 = 0 ⎝ 10 + 10 ⎠ ⎝ R1 + R2 ⎠ 0 − 0.7 − ( −9 ) = 0.0869 mA I BQ = 5 + (181)( 0.5 )

I CQ = 15.6 mA, I EQ = 15.7 mA

VCEQ = 18 − (15.7 )( 0.5 ) ⇒ VCEQ = 10.1 V

b.

AC load line 1 Slope  0.50.3 1  0.188 K

15.6

10.1

c. rπ =

18

(180 )( 0.026 )

= 0.30 kΩ 15.6 (1 + β )( RE & RL ) ⎛ R1 & R2 & Rib ⎞ Av = ⋅⎜ ⎟ rπ + (1 + β ) ( RE RL ) ⎝ R1 & R2 & Rib + RS ⎠ Rib = rπ + (1 + β )( RE & RL ) = 0.30 + (181)( 0.5 & 0.3) or Rib = 34.2 k Ω R1 & R2 & Rib = 5 & 34.2 = 4.36 k Ω Av =

(181)( 0.5 & 0.3) ⎛ 4.36 ⎞ ⋅⎜ ⎟ ⇒ Av 0.3 + (181)( 0.5 & 0.3) ⎝ 4.36 + 1 ⎠

= 0.806

d.

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Rib = rp + (1 + b ) ( RE } RL )

Rib = 0.30 + (181)( 0.188 ) ⇒ Rib = 34.3 kΩ Ro = RE

rp + R1} R2 } RS 0.3 + 5}1 ⇒ Ro = 6.18 Ω = 0.5 1+ b 181

6.40 a. RTH = R1} R2 = 10}10 = 5 kΩ ⎛ R2 ⎞ VTH = ⎜ ⎟ ( −10 ) = 2 5 V ⎝ R1 + R2 ⎠ VTH = I BQ RTH + VBE ( on ) + (1 + b ) I BQ RE − 10 I BQ =

2 5 − 0.7 − ( −10 )

5 + (121)( 2 )

= 0.0174 mA

I CQ = 2.09 mA, I EQ = 2.11 mA VCEQ = 10 − ( 2.09 )(1) − ( 2.11)( 2 ) ⇒ VCEQ = 3.69 V

b. AC load line 1 Slope  22 1  1K

2.09

3.69

c. rπ =

10

(120 )( 0.026 )

= 1.49 kΩ 2.09 (1 + β ) ( RE RL ) ⎛ R1 R2 Rib ⎞ Av = ⋅⎜ ⎟ rπ + (1 + β ) ( RE RL ) ⎜⎝ R1 R2 Rib + RS ⎟⎠ Rib = rπ + (1 + β ) ( RE RL ) = 1.49 + (121) ( 2 2) Rib = 122.5 k Ω, Av =

R1 R2 Rib = 5 122.5 = 4.80 k Ω

(121) ( 2 2) ⎛ 4.80 ⎞ ⋅⎜ ⎟ ⇒ Av 1.49 + (121) ( 2 2) ⎝ 4.80 + 5 ⎠

= 0.484

d. Rib = rπ + (1 + b ) ( RE } RL ) Rib = 1.49 + (121) ( 2} 2 ) 1 Rib = 122 kΩ Ro = RE

rπ + R1} R2 } RS 1.49 + 5}5 1 Ro = 32.4 Ω =2 1+ b 121

6.41 a.

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RTH = R1 R2 = 60 40 = 24 kΩ

⎛ R2 ⎞ ⎛ 40 ⎞ VTH = ⎜ ⎟ VCC = ⎜ ⎟ ( 5) = 2 V ⎝ 40 + 60 ⎠ ⎝ R1 + R2 ⎠ 5 − 0.7 − 2 I BQ = = 0.0130 mA 24 + ( 51)( 3) I CQ = 0.650 mA, I EQ = 0.663 mA VECQ = 5 − I EQ RE = 5 − ( 0.663)(3) ⇒ VECQ = 3.01 V

b.

1.63 AC load line 1 Slope  51  50  34 1  1.75 K

0.65

5

3.01

c.

( 50 )( 0.026 )

80 = 2 kΩ, r0 = = 123 kΩ 0.650 0.65 Define RL′ = RE RL r0 = 3 4 123 = 1.69 kΩ

rπ =

Av =

(1 + β ) RL′ rπ (1 + β ) RL′

=

( 51)(1.69 )

2 + ( 51)(1.69 )

⎛ RE r0 Ai = (1 + β ) I b ⎜ ⎜R r +R L ⎝ E 0

⇒ Av = 0.977

⎞ ⎟⎟ ⎠

⎛ RTH ⎞ Ib = I S ⎜ ⎟ ⎝ RTH + Rib ⎠ Rib = rπ + (1 + β ) RL′ = 2 + ( 51)(1.69 ) = 88.2 RE r0 = 3 r0 = 3 123 = 2.93 24 ⎞ ⎛ 2.93 ⎞ ⎛ Ai = ( 51) ⎜ ⎟⎜ ⎟ ⇒ Ai = 4.61 2.93 4 24 88.2 ⎠ + + ⎝ ⎠⎝ d. Rib = rπ + (1 + β ) RE RL r0 = 2 + ( 51)(1.69 ) ⇒ Rib = 88.2 kΩ rπ ⎛ 2⎞ RE = ⎜ ⎟ 3 = 0.0392 3 1+ β ⎝ 51 ⎠ R0 = 38.7 Ω e. Assume variations in rπ and r0 have negligible effects R1 = 60 ± 5% R1 = 63 kΩ, R1 = 57 kΩ R0 =

R2 = 40 ± 5% R2 = 42 kΩ,

R2 = 38 kΩ

RE = 3 ± 5%

RE = 3.15 kΩ, RE = 2.85 kΩ

RL = 4 ± 5%

RL = 4.2 kΩ,

RL = 3.8 kΩ

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⎛ RE r0 ⎞ ⎛ RTH ⎞ Ai = (1 + β ) ⎜ ⎜ R r + R ⎟⎟ ⎜ R + R ⎟ L ⎠ ⎝ TH ib ⎠ ⎝ E 0 Rib = rπ + (1 + β ) ( RE RL r0 )

RTH ( max ) = 25.2 kΩ, RTH ( min ) = 22.8 kΩ Rib ( max ) = 92.5 kΩ, Rib ( min ) = 84.0 kΩ RE ( max ) , RL ( min ) , Rib = 88.6 kΩ RE ( min ) , RL ( max ) , Rib = 87.4 kΩ

RE ( max ) & r0 = 3.07 kΩ

RE ( min ) & r0 = 2.79 kΩ

For RE ( min ) , RL ( max ) , RTH ( min ) 22.8 ⎞ ⎛ 2.79 ⎞ ⎛ Ai = ( 51) ⎜ ⎟⎜ ⎟ ⇒ Ai = 4.21 ⎝ 2.79 + 4.2 ⎠ ⎝ 22.8 + 87.4 ⎠ For RE ( max ) , RL ( min ) , RTH ( max ) 25.2 ⎞ ⎛ 3.07 ⎞⎛ Ai = ( 51) ⎜ ⎟⎜ ⎟ ⇒ Ai = 5.05 3.07 3.8 25.2 + + 88.6 ⎠ ⎝ ⎠⎝

6.42 (a) 0.5 = 0.00617 mA 81 VB = I BQ RB = ( 0.00617 )(10 ) ⇒ VB = 0.0617 V

I BQ =

VE = VB + 0.7 ⇒ VE = 0.7617 V

(b) ⎛ 80 ⎞ I CQ = ( 0.5 ) ⎜ ⎟ = 0.494 mA ⎝ 81 ⎠ I CQ 0.494 gm = = ⇒ g m = 19 mA / V VT 0.026

rπ = ro =

β VT I CQ

=

(80 )( 0.026 ) 0.494

⇒ rπ = 4.21 k Ω

VA 150 ⇒ ro = 304 k Ω = I CQ 0.494

(c) RS

Vs

 

IS

VS  RB V␲

r␲

ro gmV␲



Vo RL Io

For RS = 0 ⎛V ⎞ Vo = − ⎜ π + g mVπ ⎟ ( RL & ro ) ⎝ rπ ⎠

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−Vo ⎛1+ β ⎞ ⎜ ⎟ ( RL & ro ) ⎝ rπ ⎠ Now Vs + Vπ = Vo

so that Vπ =

or Vs = Vo − Vπ = Vo +

Vo ⎛ 1+ β ⎞ ⎜ ⎟ ( RL & ro ) ⎝ rπ ⎠

We find

(1 + β )( RL & ro ) (81)( 0.5 & 304 ) = rπ + (1 + β )( RL & ro ) 4.21 + ( 81)( 0.5 & 304 ) (81)( 0.5 ) ≅ ⇒ Av = 0.906 4.21 + ( 81)( 0.5 ) Rib = rπ + (1 + β )( RL & ro ) ≅ 4.21 + ( 81)( 0.5 ) = 44.7 k Ω Av =

Vo Vs

=

⎛ RB ⎞ ⎛ ro ⎞ Ib = ⎜ ⎟ ⋅ I s and I o = ⎜ ⎟ (1 + β ) I b ⎝ RB + Rib ⎠ ⎝ ro + RL ⎠ Then ⎛ RB ⎞⎛ ro ⎞ Io = (1 + β ) ⎜ ⎟⎜ ⎟ Is ⎝ RB + Rib ⎠⎝ ro + RL ⎠ ⎛ 10 ⎞ Ai ≅ ( 81) ⎜ ⎟ (1) ⇒ Ai = 14.8 ⎝ 10 + 44.7 ⎠ (d) ⎛ RB + Rib ⎞ ⎛ 10 44.7 ⎞ Vs′ = ⎜ ⋅V = ⋅ V = 0.803) Vs ⎜ R R + R ⎟⎟ s ⎜⎜ 10 44.7 + 2 ⎟⎟ s ( s ⎠ ⎝ B ib ⎝ ⎠ Then Av = ( 0.803)( 0.906 ) ⇒ Av = 0.728 Ai =

Ai = 14.8 (Unchanged)

6.43 (a) I CQ = 1.98 mA ro =

rπ =

(100 )( 0.026 ) 1.98

= 1.313 K

VA 100 = I CQ 1.98

= 50.5 K rπ + RS 1.31 + 10 ro = 50.5 ⇒ Ro = 112 Ω 1+ β 101 0.112 & 50.5 ⇒ Ro ≅ 112 Ω (b) From Equation 4.68 (1 + β ) ( ro RL ) 100 Av = ro = = 50.5 K 1.98 rπ + (1 + β ) ( ro RL ) Ro =

(i)

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RL = 0.5 K

(101) ( 50.5 0.5) 1.31 + (101) ( 50.5 0.5 ) (101)( 0.4951) ⇒ Av = 0.974 Av = 1.31 + (101)( 0.4951) Av =

(ii) RL = 5 K Av =

ro RL = 50.5 5 = 4.5495

(101)( 4.55) ⇒ Av = 0.997 1.31 + (101)( 4.55 )

6.44 5 − 0.7 I CQ = 1.293 mA = 1.303 3.3 (125 )( 0.026 ) rπ = = 2.51 K 1.293 1.293 gm = = 49.73 mA/V 0.026 (a) Rib = rπ + (1 + β ) ( RE RL ) = 2.51 + (126 ) ( 3.3 1) I EQ =

Rib = 99.2 K Ro = RE

rπ 2.51 = 3.3 = 3.3 0.01992 1+ β 126

Ro = 19.8 Ω (b) v 2sin ω t ⇒ is ( t ) = 20.2sin ω t ( μ A ) is = s = Rib 99.2 veb ( t ) = −is ( t ) rπ = ( −20.2 )( 2.51) sin ω t veb ( t ) = −50.6sin ω t ( mV )

(126 ) ( 3.3 1) (1 + β ) ( RE RL ) (126 )( 0.7674 ) = = rπ + (1 + β ) ( RE RL ) 2.51 + (126 ) ( 3.3 1) 2.51 + (126 )( 0.7674 ) Av = 0.9747 ⇒ vo ( t ) = 1.95sin ω t ( V ) v (t ) ⇒ io ( t ) = 1.95sin ω t ( mA ) io ( t ) = o Av =

RL

6.45 a. I EQ = 1 mA , VCEQ = VCC − I EQ RE

5 = 10 − (1)( RE ) ⇒ RE = 5 kΩ

1 = 0.0099 mA 101 10 = I BQ RB + VBE ( on ) + I EQ RE 10 = ( 0.0099 ) RB + 0.7 + (1)( 5 ) ⇒ RB = 434 kΩ

I BQ =

b.

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␯b ␯0

RB

rπ =

RE

(100 )( 0.026 ) 0.99 (1 + β ) RE

= 2.63 kΩ

(101)( 5 ) v0 = = = 0.995 vb rπ + (1 + β ) RE 2.63 + (101)( 5 ) ⇒ vb =

v0 4 = ⇒ vb = 4.02 V peak-to-peak at base 0.995 0.995 RS ␯b

␯S

 

RBRib

Rib = rπ + (1 + β ) RE = 508 kΩ RB Rib = 434 508 = 234 kΩ vb =

RB Rib RB Rib + RS

⋅ vS =

vb = 0.997vS ⇒ vS =

234vS 234 = vS 234 + 0.7 234.7

4.02 ⇒ vS = 4.03 V peak-to-peak 0.997

c. Rib = rπ + (1 + β ) ( RE RL )

Rib = 2.63 + (101) ( 5 1) = 86.8 kΩ RB Rib = 434 86.8 = 72.3 kΩ

⎛ 72.3 ⎞ vb = ⎜ ⎟ vS = 0.99vS = ( 0.99 )( 4.03) ⎝ 72.3 + 0.7 ⎠ vb = 3.99 V peak-to-peak

(1 + β )( RE & RL ) ⋅ vb rπ + (1 + β )( RE & RL ) (101)( 0.833) = ( 3.99 ) 2.63 + (101)( 0.833)

v0 =

v0 = 3.87 V peak-to-peak

6.46

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RTH = R1 & R2 = 40 & 60 = 24 kΩ ⎛ 60 ⎞ VTH = ⎜ ⎟ (10 ) = 6 V ⎝ 60 + 40 ⎠ 6 − 0.7 β = 75 I BQ = = 0.0131 mA 24 + ( 76 )( 5 ) I CQ = 0.984 mA

β = 150 I BQ =

6 − 0.7 = 0.00680 mA 24 + (151)( 5 )

I CQ = 1.02 mA

β = 75 rπ =

( 75 )( 0.026 )

0.984 β = 150 rπ = 3.82 kΩ

= 1.98 kΩ

β = 75 Rib = rπ + (1 + β )( RE & RL ) = 65.3 kΩ β = 150 Rib = 130 kΩ (1 + β )( RE & RL ) R1 & R2 & Rib Av = ⋅ rπ + (1 + β )( RE & RL ) R1 & R2 & Rib + RS For β = 75, R1 & R2 & Rib = 40 & 60 & 65.3 = 17.5 k Ω Av =

( 76 )( 0.833) 17.5 ⋅ ⇒ Av = 0.789 1.98 + ( 76 )( 0.833) 17.5 + 4

For β = 150, R1 & R2 & Rib = 40 & 60 & 130 = 20.3 k Ω Av =

(151)( 0.833) 20.3 ⋅ ⇒ Av = 0.811 3.82 + (151)( 0.833) 20.3 + 4

So 0.789 ≤ Av ≤ 0.811 ⎛ RE ⎞ ⎛ RTH ⎞ Ai = (1 + β ) ⎜ ⎟ ⎟⎜ ⎝ RE + RL ⎠ ⎝ RTH + Rib ⎠ β = 75 24 ⎞ ⎛ 5 ⎞⎛ Ai = ( 76 ) ⎜ ⎟⎜ ⎟ ⇒ Ai = 17.0 ⎝ 5 + 1 ⎠ ⎝ 24 + 65.3 ⎠ β = 150 ⎛ 5 ⎞ ⎛ 24 ⎞ Ai = (151) ⎜ ⎟ ⎜ ⎟ ⇒ Ai = 19.6 ⎝ 6 ⎠ ⎝ 24 + 130 ⎠ 17.0 ≤ Ai ≤ 19.6

6.47 (a)

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⎛ I ⎞ 9 = ⎜ E ⎟ (100 ) + VBE ( on ) + I E RE ⎝ 1+ β ⎠ 9 − 0.7 IE = ⎛ 100 ⎞ ⎜ ⎟ + RE ⎝ 1+ β ⎠ 8.3 = 2.803 mA ⎛ 100 ⎞ 1 + ⎜ ⎟ ⎝ 51 ⎠ 8.3 β = 200 I E = = 5.543 mA ⎛ 100 ⎞ ⎜ ⎟ +1 ⎝ 201 ⎠ 2.80 ≤ I E ≤ 5.54 mA

β = 50 I E =

VE = I E RE , β = 50, VE = 2.80 V

β = 200, VE = 5.54 V β = 50, I CQ = 2.748 mA, rπ = 0.473 K (b)

β = 200, I CQ = 5.515 mA, rπ = 0.943 K Ri = RB & ⎡⎣ rπ + (1 + β ) RE & RL ⎤⎦

β = 50 ⇒ Ri = 100 ⎡⎣ 0.473 + ( 51)(1 & 1) ⎤⎦ = 100 25.97 = 20.6 K β = 200 ⇒ Ri = 100 ⎣⎡0.943 + ( 201)(1 & 1) ⎦⎤ = 100 101.4 = 50.3 K From Fig. (4.68) (1 + β ) ( RE RL ) ⎛ Ri ⎞ Av = ⋅⎜ ⎟ rπ + (1 + β ) ( RE RL ) ⎝ Ri + RS ⎠ =

( 51) (1 1) ⎛ 20.6 ⎞ ⋅⎜ ⎟ 0.473 + ( 51) (1 1) ⎝ 20.6 + 10 ⎠

β = 50 ⇒ Av = 0.661 β = 200 ⇒ Av =

( 201) (1 1) ⎛ 50.3 ⎞ ⎜ ⎟ 0.943 + ( 201) (1 1) ⎝ 50.3 + 10 ⎠

Av = 0.826

6.48 Vo = (1 + β ) I b RL Vs Ib = rπ + (1 + β ) RL so Av =

(1 + β ) RL rπ + (1 + β ) RL

For β = 100, RL = 0.5 k Ω rπ =

(100 )( 0.026 ) 0.5

= 5.2 k Ω

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(101)( 0.5 ) = 0.9066 5.2 + (101)( 0.5 )

Then Av ( min ) =

Then β = 180, RL = 500 k Ω rπ =

(180 )( 0.026 ) 0.5

Then Av ( max ) =

= 9.36 k Ω

(181)( 500 ) = 0.9999 9.36 + (181)( 500 )

6.49 Rib

IS



Ib

V␲ ␯S

 

gmV␲  ␤Ib

r␲

 R1R2 RE

RL I0

⎛ RE ⎞ I 0 = (1+ β ) I b ⎜ ⎟ ⎝ RE + RL ⎠ ⎛ R1 & R2 ⎞ Ib = I S ⎜ ⎟ ⎝ R1 & R2 + Rib ⎠ Rib = rπ + (1 + β )( RE & RL ) VCC = 10 V, For VCEQ = 5 V

⎛1+ β ⎞ 5 = 10 − ⎜ ⎟ I CQ RE ⎝ β ⎠ β = 80, For RE = 0.5 kΩ I CQ = 9.88 mA, I EQ = 10 mA, I BQ = 0.123 mA rπ =

(80 )( 0.026 )

= 0.211 kΩ 9.88 Rib = 0.211 + ( 81)( 0.5 & 0.5 ) ⇒ Rib = 20.46 kΩ Ai =

⎛ RE ⎞ ⎛ R1 & R2 ⎞ I0 = (1 + β ) ⎜ ⎟ ⎟⎜ IS ⎝ RE + RL ⎠⎝ R1 & R2 + Rib ⎠

⎞ R1 & R2 ⎛ 1 ⎞⎛ 8 = ( 81) ⎜ ⎟ ⎜ ⎟ ⎝ 2 ⎠ ⎝ R1 & R2 + 20.46 ⎠ 0.1975 ⎡⎣ R1} R2 + 20.46 ⎤⎦ = R1} R2 R1} R2 ⇒ 5.04 kΩ

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VTH = I BQ RTH + VBE ( on ) + (1 + b ) I BQ RE 1 ( 5.04 )(10 ) = ( 0.123)( 5.04 ) + 0.7 + (10 )( 0.5 ) ⇒ R1 = 7.97 kΩ R1 7.97 R2 = 5.04 ⇒ R2 = 13.7 kΩ 7.97 + R2 Now Ro = RE

rπ 0.211 = 0.5 or Ro = 2.59 Ω 1+ b 81

(b) Rib = 0.211 + (81) ( 0.5&2) = 32.6 k Ω ⎛ 0.5 ⎞ ⎛ 5.04 ⎞ Ai = ( 81) ⎜ ⎟⎜ ⎟ = ( 81)( 0.2 )( 0.134 ) ⎝ 0.5 + 2 ⎠ ⎝ 5.04 + 32.6 ⎠ Ai = 2.17

6.50 Ri = RTH Rib where Rib = rπ + (1 + β ) RE 5 − 3.5 VCEQ = 3.5, I CQ = 0.75 mA 2 (120 )( 0.026 ) rπ = = 4.16 k Ω 0.75 Rib = 4.16 + (121) ( 2 ) = 246 k Ω Then Ri = 120 = RTH 246 ⇒ RTH = 234 k Ω 0.75 = 0.00625 mA I BQ = 120 VTH = I BQ RTH + VBE ( on ) + (1+ β ) I BQ RE 1 1 ⋅ RTH ⋅ VCC = ( 234)(5) = ( 0.00625) ( 234) + 0.7 + (121)( 0.00625 )( 2 ) R1 R1 which yields R1 = 318 k Ω and R2 = 886 k Ω

6.51 a. Let RE = 24 Ω and VCEQ = 12 VCC = 12 V ⇒ I EQ =

12 = 0.5 A 24

I CQ = 0.493 A, I BQ = 6.58 mA rπ =

( 75)( 0.026 ) 0.493

= 3.96 Ω Reb

Is

Ib

 V

VS

 

gmV  Ib

r

 R1  R2  Rrn RE

Io

RL

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⎛ RE ⎞ I 0 = (1 + β ) I b ⎜ ⎟ ⎝ RE + RL ⎠ ⎛ RTH ⎞ Ib = I S ⎜ ⎟ ⎝ RTH + Rib ⎠ Rib = rπ + (1 + β ) ( RE & RL )

= 3.96 + ( 76 )( 24 & 8 ) ⇒ Rib = 460 Ω

Ai =

⎛ RE ⎞ ⎛ RTH ⎞ I0 = (1 + β ) ⎜ ⎟ ⎟⎜ IS ⎝ RE + RL ⎠ ⎝ RTH + Rib ⎠

⎞ ⎛ 24 ⎞ ⎛ RTH 8 = ( 76) ⎜ ⎟⎜ ⎟ ⎝ 24 + 8 ⎠ ⎝ RTH + 460 ⎠ RTH 0.140 = ⇒ RTH = 74.9 Ω (Minimum value) RTH + 460 dc analysis: 1 VTH = ⋅ RTH ⋅ VCC R1 = I BQ RTH + VBE ( on ) + I EQ RE

1 ( 74.9 )( 24 ) = ( 0.00658)( 74.9 ) + 0.70 + ( 0.5 )( 24 ) R1 = 13.19 R1 = 136 Ω,

136 R2 = 74.9 ⇒ R2 = 167 Ω 136 + R2

b.

0.493

AC load line 1 Slope  248 1  6

12

24

1 ΔiC = − Δvce 6 For ΔiC = 0.493 ⇒ Δvce = ( 0.493)( 6 ) ⇒ Max. swing in output voltage for this design = 5.92 V peak-to-peak

c. R0 =

rπ 3.96 RE = 24 = 0.0521 24 ⇒ R0 = 52 mΩ 1+ β 76

6.52 The output of the emitter follower is ⎛ RL ⎞ vo = ⎜ ⎟ ⋅ vTH ⎝ RL + Ro ⎠

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Ro 

 

␯TH

␯O

RL



For vO to be within 5% for a range of RL , we have RL ( min )

RL ( min ) + Ro

= ( 0.95 )

RL ( max )

RL ( max ) + Ro

4 10 = ( 0.95 ) which yields Ro = 0.364 k Ω 4 + Ro 10 + Ro

⎛ r + R1 & R2 & RS ⎞ We have Ro = ⎜ π ⎟ RE ro 1+ β ⎝ ⎠ The first term dominates Let R1 & R2 & RS ≅ RS , then rπ + RS r +4 ⇒ 0.364 = π 1+ β 1+ β

Ro ≅

or 0.364 =

0.364 ≅

rπ β VT 4 4 + = + 1 + β 1 + β I CQ (1 + β ) 1 + β

VT 4 + I CQ 1 + β

The factor

V 4 4 4 = 0.044 to = 0.0305. We can set Ro ≅ 0.32 = T is in the range of I CQ 91 131 1+ β

Or I CQ = 0.08125 mA. To take into account other factors, set I CQ = 0.15 mA, I BQ =

0.15 = 0.00136 mA 110

5 = 33.3 k Ω 0.15 Design a bias stable circuit. ⎛ R2 ⎞ 1 VTH = ⎜ ⎟ (10) − 5 = ( RTH )(10) − 5 R1 ⎝ R1 + R2 ⎠

For VCEQ ≅ 5 V , set RE =

RTH = ( 0.1)(1 + β ) RE = ( 0.1)(111)(33.3) = 370 k Ω VTH = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE − 5

1 ( 370 )(10 ) − 5 = ( 0.00136 )( 370 ) + 0.7 + (111)( 0.00136 )( 33.3) − 5 R 1 which yields R1 = 594 k Ω and R2 = 981 k Ω

So

Now Av =

(1 + β ) ( RE RL ) ⎛ RTH Rib ⎞ ⋅⎜ ⎟ rπ + (1 + β ) ( RE RL ) ⎝ RTH Rib + RS ⎠

Rib = rπ + (1 + β ) ( RE RL )

and rπ

=

β VT I CQ

For β = 90, RL = 4 k Ω, rπ = 15.6 k Ω, Rib = 340.6 k Ω Av =

( 91)( 33.3 & 4 ) 370 & 340.6 ⋅ ⇒ Av = 0.9332 15.6 + ( 91)( 33.3 & 4 ) 370 & 340.6 + 4

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For β = 90, RL = 10 k Ω Rib = 715.4 k Ω Av =

( 91)( 33.3 & 10 ) 370 & 715.4 ⋅ ⇒ Av = 0.9625 15.6 + ( 91)( 33.3 & 10 ) 370 & 715.4 + 4

For β = 130, RL = 4 k Ω rπ = 22.5 k Ω, Rib = 490 k Ω Av =

(131)( 33.3 & 4 ) 370 & 490 ⋅ ⇒ Av = 0.9360 22.5 + (131)( 33.3 & 4 ) 370 & 490 + 4

For β = 130, RL = 10 k Ω

Rib = 1030 k Ω

(131)( 33.3 & 10 ) 370 & 1030 ⇒ Av = 0.9645 ⋅ 22.5 + (131)( 33.3 & 10 ) 370 & 1030 + 4 Now vO ( min ) = Av ( min ) .vS = 3.73sin ω t vO ( max ) = Av ( max ) .vS = 3.86sin ω t Av =

ΔvO = 3.5% vO

6.53 PAVG = iL2 ( rms ) RL ⇒ 1 = iL2 ( rms )(12 ) so iL ( rms ) = 0.289 A ⇒ iL ( peak ) = 2 ( 0.289 ) iL ( peak ) = 0.409 A

vL ( peak ) = iL ( peak ) ⋅ RL = ( 0.409 )(12 ) = 4.91 V

4.91 = 0.982 5 With RS = 10 k Ω, we will not be able to meet this voltage gain requirement. Need to insert a buffer or an Need a gain of

op-amp voltage follower (see Chapter 9) between RS and CC1 . 1 Set I EQ = 0.5 A, VCEQ = (12 − ( −12 ) ) = 8 V 3 24 = I EQ RE + VCEQ = ( 0.5 ) RE + 8 ⇒ RE = 32 Ω 50 ( 0.5 ) = 0.49 A 51 β VT ( 50 )( 0.026 ) rπ = = = 2.65 Ω I CQ 0.49 Let β = 50, I CQ =

Rib = rπ + (1 + β ) ( RE RL ) = 2.65 + ( 51) ( 32 12 ) Rib = 448 Ω Av =

(1 + β ) ( RE RL ) ( 51) ( 32 12 ) = = 0.994 rπ + (1 + β ) ( RE RL ) 2.65 + ( 51) ( 32 12 )

So gain requirement has been met.

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0.49 = 0.0098 A = 9.8 mA 50 24 ≅ 10 I B = 98 mA Let I R ≅ R1 + R2 I BQ =

So that R1 + R2 = 245 Ω VTH =

R2 ( 24 ) − 12 = I BQ RTH + VBE ( on ) + I EQ RE − 12 R1 + R2

( 0.0098) R1 R2 ⎛ R2 ⎞ + 0.7 + ( 0.5 )( 32 ) ⎜ 245 ⎟ ( 24 ) = 245 ⎝ ⎠ Now R1 = 245 − R2 So we obtain 4 × 10−5 R22 + 0.0882 R2 − 16.7 = 0 which yields R2 = 175 Ω and R1 = 70 Ω 6.54 (a) RTH = R1 R2 = 25.6 10.4 = 7.40 k Ω ⎛ R2 ⎞ ⎛ 10.4 ⎞ VTH = ⎜ ⎟ (VCC ) = ⎜ ⎟ (18 ) = 5.2 V ⎝ 10.4 + 25.6 ⎠ ⎝ R1 + R2 ⎠ VTH = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE

I BQ =

5.2 − 0.7 = 0.0117 mA 7.40 + (126 )( 3)

Then I CQ = 1.46 mA and I EQ = 1.47 mA VCEQ = VCC − I CQ RC − I EQ RE

VCEQ = 18 − (1.46 )( 4 ) − (1.47 )( 3) ⇒ VCEQ = 7.75 V

(b) rπ =

(125) ( 0.026 )

= 2.23 k Ω 1.46 1.46 gm = = 56.2 mA / V 0.026 Re Vo Ie Is

RS

RE

r␲ Ib

␤Ib

RC

RL

RTH

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rπ + RTH 2.23 + 7.40 = = 0.0764 k Ω 1+ β 126

Re = Ie =

− ( RS RE )

(R

S

RE ) + Re

⋅ Is =

− (100 3)

(100 & 3) + 0.0764

⋅ Is

or I e = − ( 0.974 ) I s ⎛ β Vo = − I c ( RC RL ) = − ⎜ ⎝ 1+ β

⎞ ⎟ I e ( RC RL ) ⎠

⎛ β ⎞ Vo ⎛ 125 ⎞ = −⎜ ⎟ ( −0.974 )( RC & RL ) = ⎜ ⎟ ( 0.974 )( 4 & 4 ) Is 1 126 ⎠ + β ⎝ ⎝ ⎠ V Then Rm = o = 1.93 k Ω = 1.93 V / mA Is

Then

(c)

(

) (

)

Vs = I s RS R E Re = I s 100 3 0.0764 = I s ( 0.0744 ) Vs 0.0744 V V which yields o = o ( 0.0744 ) = 1.93 I s Vs

or I s =

or Av =

6.55 (a)

Vo = 25.9 Vs

β ( RC RL )

Av =

rπ + R1 R2

, RL = 12 k Ω, β = 100

Let R1 & R2 = 50 k Ω, I CQ = 0.5 mA VTH = I BQ RTH + VBE ( on ) + (1+ β ) I BQ RE

(100 )( 0.026 ) 0.5 = 0.005 mA, rπ = = 5.2 k Ω 100 0.5 1 1 ⋅ RTH ⋅ VCC = ( 50 )(12 ) = ( 0.005 )( 50 ) + 0.7 + (101)( 0.005 )( 0.5 ) R1 R1

I BQ =

which yields R1 = 500 k Ω and R2 = 55.6 k Ω Av =

(b) I CQ

(100 )(12 & 12 ) 5.2 + 50

= 10.9, Design criterion is met.

= 0.5 mA, I EQ = 0.505 mA

VCEQ = 12 − ( 0.5)(12) − ( 0.505)( 0.5) ⇒ VCEQ = 5.75 V 0.5 = 19.23 mA / V 0.026 Av = (19.23) (12 12 ) ⇒ Av = 115 Av = g m ( RC RL ) , g m =

6.56 a. Emitter current

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I EQ = I CC = 0.5 mA 0.5 = 0.00495 mA 101 VE = I EQ RE = ( 0.5 )(1) ⇒ VE = 0.5 V

I BQ =

VB = VE + VBE ( on ) = 0.5 + 0.7 ⇒ VB = 1.20 V VC = VB + I BQ RB = 1.20 + ( 0.00495 )(100 ) ⇒ VC = 1.7 V

b. rπ =

(100 )( 0.026 ) = 5.25 kΩ (100 )( 0.00495 ) (100 )( 0.00495 )

gm =

= 19.0 mA/V

0.026 Ri

gmV␲

VS

RS

 

 RE V␲

r␲

V0 RB

RL



gmV␲





RE V RE  RS S

 

RS RE  V␲

r␲



Vo = − g mVπ ( RB & RL ) Vπ = −

RE & Rie ⋅ VS = − ( 0.4971) VS RE & Rie + RS

Vo = (19 )( 0.4971) VS (100 & 1) Av = 9.37

c.

VX

 

IX

gmV␲

 RE V␲

r␲



IX =

VX VX + − g mVπ , Vπ = −VX RE rπ

IX 1 1 1 = = + + gm VX Ri RE rπ or Ri = RE rπ

1 1 = 1 5.253 gm 19

Ri = 0.84 & 0.05252 ⇒ Ri = 49.4 Ω

6.57 (a)

I EQ = 1 mA, I CQ = 0.9917 mA

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VC = 5 − ( 0.9917 )( 2 ) = 3.017 V VE = −0.7 V VCEQ = 3.72 V

(b) Av = g m ( RC RL ) 0.9917 = 38.14 mA/V 0.026 Av = ( 38.14 ) ( 2 10 ) ⇒ Av = 63.6

gm =

6.58 (a) 10 − 0.7 = 0.93 mA 10 = 0.921 mA

I EQ = I CQ

VECQ = 20 − ( 0.93)(10 ) − ( 0.921)( 5 ) VECQ = 6.10 V

(b) 0.921 = 35.42 mA/V 0.026 Av = g m ( RC RL ) = ( 35.42 ) ( 5 50 )

gm =

Av = 161

6.59 (a)

I EQ = 0.93 mA, I CQ = 0.921 mA

VECQ = 6.10 V (b)

gm =

0.921 = 35.42 mA/V rπ = 2.82 K 0.026

From Eq. 6.90 ( RC RL ) ⎡ rπ R R ⎤ Av = g m ⎢1 + β E S ⎥ RS ⎣ ⎦ ( 35.42 ) ( 50 5 ) ⎡ 2.82 ⎤ = 10 0.1⎥ ⎢ 0.1 ⎣ 101 ⎦ ( 35.42 )( 4.545 ) Av = [0.0218] 0.1 Av = 35.1 6.60 (a) ⎛ 60 ⎞ I CQ = ⎜ ⎟ (1) ⇒ I CQ = 0.984 mA ⎝ 61 ⎠ ⎛ 1⎞ VCEQ = I BQ RB + VBE ( on ) = ⎜ ⎟ (100 ) + 0.7 ⎝ 61 ⎠ VCEQ = 2.34 V

(b)

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Av = g m

(R

RL ) ⎡ rπ ⎤ RS ⎥ ⎢ RS ⎣1 + β ⎦

B

0.984 = 37.85 mA/V 0.026 rπ = 1.59 K

gm =

( 37.85) (100 2 ) ⎡1.59

⎤ ⎢ 61 0.05⎥ 0.05 ⎣ ⎦ = 1484 ⎡⎣ 0.0261 0.05⎤⎦ Av = 25.4

Av =

6.61

is ( peak ) = 2.5 mA, Vo ( peak ) = 5 mV

vo 5 × 102 3 = = 2 × 103 = 2 k Ω is 2.5 × 102 6 From Problem 4.54 ⎛ RS RE ⎞ Vo ⎛ β ⎞ =⎜ ⎟⎟ ⎟ ( RC & RL ) ⎜⎜ Is ⎝ 1+ β ⎠ ⎝ RS RE + Rie ⎠ Let RC = 4 k Ω, RL = 5 k Ω, RE = 2 k Ω So we need Rm =

Now β = 120, so we have ⎛ RS RE ⎛ 120 ⎞ 2=⎜ ⎟ ( 4 5 ) ⎜⎜ ⎝ 121 ⎠ ⎝ RS RE + Rie RS RE = 0.9075 Then RS RE + Re

⎞ ⎛ RS RE ⎞ ⎟⎟ = 2.204 ⎜⎜ ⎟⎟ ⎠ ⎝ RS RE + Rie ⎠

RS RE = 50 2 = 1.923 k Ω, so that Rie = 0.196 k Ω Assume VCEQ = 3 V

VCC ≅ I CQ ( RC + RE ) + VCEQ

5 = I CQ ( 4 + 2 ) + 3 ⇒ I CQ = 0.333 mA rπ =

(120 )( 0.026 )

= 9.37 k Ω 0.333 r + RTH 9.37 + RTH ⇒ 0.196 = Rie = π 1+ β 121 which yields RTH = 14.35 k Ω

Now VTH = I BQ RTH + VBE ( on ) + I EQ RE 1 ⎛ 121 ⎞ = 0.00833 mA, I EQ = ⎜ ⎟ (1) = 1.008 mA 120 ⎝ 120 ⎠ 1 1 = ⋅ RTH ⋅ VCC = (14.35 )( 5 ) = ( 0.00833)(14.35 ) + 0.7 + (1.008 )( 2 ) R1 R1

I BQ = VTH

which yields R1 = 25.3 k Ω and R2 = 33.2 k Ω

6.62 a.

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20 − 0.7 = 1.93 mA 10 = 1.91 mA

I EQ = I CQ

VECQ = VCC + VEB ( on ) − I C RC

= 25 + 0.7 − (1.91)( 6.5 ) ⇒ VECQ = 13.3 V

b. RS

VS

 

IS

Rie h fe Ib

V0

Ie RE

hie

RC

RL

Ib

Neglect effect hoe From Problem 6-16, assume 2.45 ≤ hie ≤ 3.7 kΩ 80 ≤ h fe ≤ 120 Vo = ( h fe I b ) ( RC & RL ) Rie =

⎛ RE ⎞ hie , Ie = ⎜ ⎟ IS 1 + h fe ⎝ RE + Rie ⎠

⎛ I ⎞ VS Ib = ⎜ e ⎟ , I S = ⎜ 1+ h ⎟ RS + RE & Rie fe ⎠ ⎝ ⎛ h fe ⎞ ⎛ RE ⎞ ⎛ ⎞ 1 Av = ⎜ R & RL ) ⎜ ⎟×⎜ ⎟ ⎜ 1 + h ⎟⎟ ( C fe ⎠ ⎝ RE + Rie ⎠ ⎝ RS + RE & Rie ⎠ ⎝ High gain device: hie = 3.7 kΩ, h fe = 120

3.7 = 0.0306 kΩ 121 RE & Rie = 10 & 0.0306 = 0.0305 Rie =

10 1 ⎛ 120 ⎞ ⎛ ⎞⎛ ⎞ Av = ⎜ ⎟ ( 6.5 & 5 ) ⎜ ⎟⎜ ⎟ ⇒ Av = 2.711 . . 121 10 0 0306 1 0 0305 + + ⎝ ⎠ ⎝ ⎠⎝ ⎠ Low gain device: hie = 2.45 kΩ, h fe = 80 2.45 = 0.03025 kΩ 81 RE & Rie = 10 & 0.03025 = 0.0302 Rie =

10 1 ⎛ 80 ⎞ ⎛ ⎞⎛ ⎞ Av = ⎜ ⎟ ( 6.5 & 5 ) ⎜ ⎟⎜ ⎟ ⇒ Av = 2.70 So Av ≈ constant ⎝ 81 ⎠ ⎝ 10 + 0.03025 ⎠ ⎝ 1 + 0.0302 ⎠ 2.70 ≤ Av ≤ 2.71

c. Ri = RE & Rie We found 0.0302 ≤ Ri ≤ 0.0305 kΩ Neglecting hoe , Ro = RC = 6.5 kΩ 6.63 a.

Small-signal voltage gain

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Av = g m ( RC & RL ) ⇒ 25 = g m ( RC & 1)

For VECQ = 3 V ⇒ VC = −VECQ + VEB ( on ) = −3 + 0.7 ⇒ VC = −2.3 VCC − I CQ RC + VC = 0 ⇒ I CQ =

5 − 2.3 2.7 = = I CQ RC RC

For I CQ = 1 mA, RC = 2.7 kΩ 1 = 38.5 mA/V 0.026 Av = ( 38.5 )( 2.7 & 1) = 28.1

gm =

Design criterion satisfied and VECQ satisfied. ⎛ 101 ⎞ IE = ⎜ ⎟ (1) = 1.01 mA ⎝ 100 ⎠ VEE = I E RE + VEB ( on ) ⇒ RE =

b. rπ =

β VT I CQ

=

(100)( 0.026) 1

5 − 0.7 ⇒ RE = 4.26 kΩ 1.01

⇒ rπ = 2.6 kΩ, g m = 38.5 mA/V, ro = ∞

6.64 a. ⎛ R2 ⎞ ⎛ 20 ⎞ VTH 1 = ⎜ ⎟ VCC = ⎜ ⎟ (10 ) ⇒ VTH 1 = 2.0 V ⎝ 20 + 80 ⎠ ⎝ R1 + R2 ⎠ RTH 1 = R1 & R2 = 20 & 80 = 16 kΩ I B1 =

2 − 0.7 = 0.0111 mA 16 + (101)(1)

I C1 = 1.11 mA ⇒ g m1 = rπ 1 =

(100)( 0.026) 1.11

1.11 ⇒ g m1 = 42.74 mA/V 0.026

⇒ rπ 1 = 2.34 kΩ

∞ ⇒ r01 = ∞ 1.11 ⎛ R4 ⎞ ⎛ 15 ⎞ VTH 2 = ⎜ ⎟ VCC = ⎜ ⎟ (10 ) = 1.50 V ⎝ 15 + 85 ⎠ ⎝ R3 + R4 ⎠ RTH 2 = R3 & R4 = 15 & 85 = 12.75 kΩ

r01 =

IB2 =

1.50 − 0.70 = 0.01265 mA 12.75 + (101)( 0.5 )

I C 2 = 1.265 mA ⇒ g m 2 = rπ 2 =

(100 )( 0.026 ) 1.26

1.265 ⇒ g m2 = 48.65 mA/V 0.026

⇒ rπ 2 = 2.06 kΩ r02 = ∞

b. Av1 = − g m1 RC1 = − ( 42.7 )( 2 ) ⇒ Av1 = −85.48 Av 2 = − g m 2 ( RC 2 RL ) = − ( 48.5) ( 4 4) ⇒ Av 2 = −97.3 c.

Input resistance of 2nd stage

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Ri 2 = R3 & R4 & rπ 2 = 15 & 85 & 2.06 = 12.75 & 2.06 ⇒ Ri 2 = 1.773 kΩ ′ Av1 = − g m1 ( RC1 Ri 2 ) = − ( 42.7 ) ( 2 1.77B) Av′1 = −40.17 Overall gain: Av = ( −40.17 )( −97.3) ⇒ Av = 3909 If we had Av1 ⋅ Av 2 = ( −85.48)( −97.3) = 8317 Loading effect reduces overall gain

6.65 a. ⎛ R2 ⎞ ⎛ 12.7 ⎞ VTH 1 = ⎜ ⎟ VCC = ⎜ ⎟ (12) ⇒ VTH 1 = 1.905 V ⎝ 12.7 + 67.3 ⎠ ⎝ R1 + R2 ⎠ RTH 1 = R1 & R2 = 12.7 & 67.3 = 10.68 kΩ 1.905 − 0.70 I B1 = = 0.00477 mA 10.68 + (121)( 2 ) I C1 = 0.572 mA 0.572 g m1 = ⇒ g m1 = 22 mA/V 0.026 (120 )( 0.026 ) rπ 1 = ⇒ rπ 1 = 5.45 kΩ 0.572 ∞ r01 = ⇒ r01 = ∞ 0.572 ⎛ R4 ⎞ ⎛ 45 ⎞ VTH 2 = ⎜ ⎟ VCC = ⎜ ⎟ (12) ⇒ VTH 2 = 9.0 V ⎝ 45 + 15 ⎠ ⎝ R3 + R4 ⎠ RTH 2 = R3 & R4 = 15 & 45 = 11.25 kΩ 9.0 − 0.70 I B2 = = 0.0405 mA 11.25 + (121)(1.6) I C2 = 4.86 mA 4.86 gm2 = ⇒ g m 2 = 187 mA/V 0.026 (120 )( 0.026 ) rπ 2 = ⇒ rπ 2 = 0.642 kΩ 4.86 r02 = ∞

b. I E1 = 0.577 mA VCEQ1 = 12 − ( 0.572 ) (10 ) − ( 0.577 ) ( 2 ) ⇒ VCEQ1 = 5.13 V I E 2 = 4.90 VCEQ 2 = 12 − ( 4.90 )(1.6 ) ⇒ VCEQ 2 = 4.16 V

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Q1 AC load line 1 107.92 1  4.42 K

Slope 

0.572

5.13

12

Q2

4.86

AC load line 1 Slope  1.60.25 1  0.216 K

4.16

12

Ri 2 = R3 R4 Rib Rib = rπ 2 + (1 + β ) ( RE 2 & RL ) = 0.642 + (121) (1.6 & 0.25 ) Rib = 26.8 Ri 2 = 15 45 26.8 Ri 2 = 7.92 kΩ c. Av1 = − g m1 ( RC1 & Ri 2 ) = − ( 22 )(10 & 7.92 ) ⇒ Av 2 = −97.2

(1 + β )( RE 2 & RL ) rπ 2 + (1 + β )( RE 2 & RL ) (121)( 0.216 ) = = 0.976 0.642 + (121)( 0.216 ) Overall gain = ( −97.2 )( 0.976 ) = −94.9 Av 2 =

d. RiS = R1 R2 rπ 1 = 67.3 12.7 5.45 ⇒ RiS = 3.61 kΩ Ro =

rπ 2 + RS RE 2 where 1+ β

RS = R3 R4 RC1 = 15 45 10 ⇒ RS = 5.29 kΩ Ro =

0.642 + 5.29 1.6 ⇒ 0.049 & 1.6 ⇒ Ro = 47.6 Ω 121

e. −1 ⋅ Δvce , ΔiC = 4.86 0.216 kΩ Δvce = ( 4.86 )( 0.216 ) = 1.05 V Max. output voltage swing = 2.10 V peak-to-peak

ΔiC =

6.66 (a)

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I R1 = IR2 IC 2

5 − 2 ( 0.7 )

= 72 mA 0.050 0.7 = = 1.4 mA 0.5 ⎛ β ⎞ =⎜ ⎟ ( 72 − 1.4 ) ⇒ I C 2 = 69.9 mA ⎝ 1+ β ⎠ 69.9 = 0.699 mA 100 ⎛ β ⎞ =⎜ ⎟ (1.4 + 0.699 ) ⇒ I C1 = 2.08 mA ⎝ 1+ β ⎠

IB2 = I C1

(b)

Vs

 

 V␲1

r␲1



gm1V␲1

r␲2 0.5 k

gm2V␲2

 V2  Vo 50 Ω

Vs = Vπ 1 + Vπ 2 + Vo (1) ⎛V ⎞ V Vo = ⎜ π 2 + π 2 + g m 2Vπ 2 ⎟ ( 0.05 ) ⎝ 0.5 rπ 2 ⎠ rπ 2 =

(100 )( 0.026 )

= 0.0372 k Ω 69.9 69.9 gm2 = = 2688 mA / V 0.026 V 1 ⎛ 1 ⎞ Vo = Vπ 2 ⎜ + + 2688 ⎟ ( 0.05 ) so that (1) Vπ 2 = o 135.8 ⎝ 0.5 0.0372 ⎠ (2)

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Vπ 1 V V + g m1Vπ 1 = π 2 + π 2 rπ 1 0.5 rπ 2 rπ 1 =

(100 )( 0.026 )

= 1.25 k Ω 2.08 2.08 g m1 = = 80 mA / V 0.026 1 ⎞ ⎛ 1 ⎞ ⎛ 1 Vπ 1 ⎜ + 80 ⎟ = Vπ 2 ⎜ + ⎟ 1.25 0.5 0.0372 ⎝ ⎠ ⎝ ⎠ ⎛ V ⎞ Vπ 1 ( 80.8 ) = Vπ 2 ( 28.88 ) = ⎜ o ⎟ ( 28.88 ) or (2) Vπ 1 = Vo ( 0.00261) ⎝ 136.7 ⎠ V V Then Vs = Vo ( 0.00261) + o + Vo = Vo (1.00993) or Av = o = 0.990 Vs 136.7

(c) Rib = rπ 1 (1 + β ) [ Rx ] Ix Vx

 

 0.5 k V␲2

r␲2

gm2V␲2



Vo 50 

⎛ 1 Vπ 2 Vπ 2 1 ⎞ + = Vπ 2 ⎜ + ⎟ r 0.5 rπ 2 0.5 π2 ⎠ ⎝ Vo V − Vπ 2 = x = I x + g m 2Vπ 2 0.05 0.05

Ix =

⎛ 1 ⎞ Ix ⎜ + gm2 ⎟ Vx ⎛ 1 ⎞ ⎝ 0.05 ⎠ − I x = Vπ 2 ⎜ + gm2 ⎟ = 0.05 0.05 ⎛ 1 ⎝ ⎠ 1 ⎞ + ⎜ ⎟ r 0.5 π2 ⎠ ⎝ We find

Vx = Rx = 4.74 k Ω Ix

Then Rib = 1.25 + (101) ( 2.89 ) ⇒ Rib = 480 k Ω

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 V␲1

r␲1

gm1V␲1



r␲2 0.5 k

gm2V␲2

 V␲2 

Ix 50 

 

Vx

To find Ro: (1) (2) (3)

Ix =

Vx V − g m 2Vπ 2 − π 2 0.05 0.5 rπ 2

⎛V ⎞ ⎛ 1 ⎞ Vπ 2 = ⎜ π 1 + g m1Vπ 1 ⎟ ( 0.5 rπ 2 ) = Vπ 1 ⎜ + 80 ⎟ ( 0.5 0.0372 ) or Vπ 2 = ( 2.77 ) Vπ 1 ⎝ 1.25 ⎠ ⎝ rπ 1 ⎠ Vπ 1 + Vπ 2 + Vx = 0 ⇒ Vπ 1 + ( 2.77 ) Vπ 1 + Vx = 0

so that Vπ 1 = − ( 0.2653) Vx

and Vπ 2 = ( 2.77 ) ⎡⎣ − ( 0.2653) Vx ⎤⎦ = − ( 0.735 ) Vx Now I x =

⎛ Vx 1 ⎞ − Vπ 2 ⎜ g m 2 + ⎟ ⎜ 0.05 0.5 rπ 2 ⎟⎠ ⎝

So that I x =

⎡ ⎤ Vx Vx 1 + ( 0.735 ) Vx ⎢ 2688 + = 0.496 Ω ⎥ which yields Ro = Ix 0.05 0.5 0.0372 ⎥⎦ ⎢⎣

6.67 a. RTH = R1 R2 = 335 125 = 91.0 kΩ ⎛ R2 ⎞ VTH = ⎜ ⎟ VCC ⎝ R1 + R2 ⎠ ⎛ 125 ⎞ =⎜ ⎟ (10 ) = 2.717 V ⎝ 125 + 335 ⎠ VTH = I B1 RTH + VBE1 + VBE 2 + I E 2 RE 2

I E 2 = (1 + β ) I E1 = (1 + β ) I B1 2

I B1 =

2.717 − 1.40 91.0 + (101) (1) 2

⇒ I B1 = 0.128 μΑ

I C1 = 12.8 μΑ I C 2 = β I E1 = β (1 + β ) I B1 = (100 )(101)( 0.128 μΑ ) I C 2 = 1.29 mΑ, I E 2 = 1.31 mΑ I RC = I C 2 + I C1 = 1.29 + 0.0128 = 1.30 mΑ

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VC = 10 − I RC RC = 10 − (1.30 )( 2.2 ) = 7.14 V VE = I E 2 RE 2 = (1.30 )(1) = 1.30 V VCE 2 = 7.14 − 1.30 = 5.84 V VCE1 = VCE 2 − VBE 2 = 5.84 − 0.7 VCE1 = 5.14 V Summary: I C1 = 12.8 μΑ I C 2 = 1.29 mΑ VCE1 = 5.14 V VCE 2 = 5.84 V b. 0.0128 g m1 = = 0.492 mΑ / V 0.026 1.292 gm2 = = 49.7 mΑ / V 0.026 Rib Ib VS

V0

 V␲1 

  R1 R2

r␲1

gm1V␲1 RC

 V␲2

r␲2

gm2V␲2



V0 = − ( g m1Vπ 1 + g m 2Vπ 2 ) RC VS = Vπ 1 + Vπ 2 , Vπ 1 = VS − Vπ 2 ⎛V ⎞ Vπ 2 = ⎜ π 1 + g m1Vπ 1 ⎟ rπ 2 r ⎝ π1 ⎠ ⎛1+ β ⎞ Vπ 2 = Vπ 1 ⎜ ⎟ rπ 2 ⎝ rπ 1 ⎠ V0 = − ⎡⎣ g m1 (VS − Vπ 2 ) + g m 2Vπ 2 ⎤⎦ RC V0 = − ⎡⎣ g m1VS + ( g m 2 − g m1 ) Vπ 2 ⎤⎦ RC ⎛r ⎞ Vπ 2 = (VS − Vπ 2 )(1 + β ) ⎜ π 2 ⎟ ⎝ rπ 1 ⎠ ⎡ ⎛ r ⎞⎤ ⎛r ⎞ Vπ 2 ⎢1 + (1 + β ) ⎜ π 2 ⎟ ⎥ = VS (1 + β ) ⎜ π 2 ⎟ ⎝ rπ 1 ⎠ ⎦ ⎝ rπ 1 ⎠ ⎣

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⎧ ⎛r ⎞⎫ VS (1 + β ) ⎜ π 2 ⎟ ⎪ ⎪ ⎪ ⎝ rπ 1 ⎠ ⎪ R V0 = − ⎨ g m1VS + ( g m 2 − g m1 ) ⋅ ⎬ C ⎛r ⎞ ⎪ 1 + (1 + β ) ⎜ π 2 ⎟ ⎪ ⎪ ⎝ rπ 1 ⎠ ⎪⎭ ⎩ V Av = 0 VS 2.01 ⎞ ⎫ ⎧ ( 49.7 − 0.492 )(101) ⎛⎜ ⎟⎪ ⎪⎪ ⎝ 203 ⎠ ⎪ 2.2 = − ⎨( 0.492 ) + ⎬ ⎛ 2.01 ⎞ ⎪ ⎪ 1 + (101) ⎜ ⎟ ⎪⎩ ⎪⎭ ⎝ 203 ⎠ Av = −55.2 c. Ris = R1 R2 Rib Rib = rπ 1 + (1 + β ) rπ 2 = 203 + (101)( 2.01) = 406 kΩ Ris = 91 406 = 74.3 kΩ = Ris R0 = RC = 2.2 kΩ

6.68 R0 Ix

 V␲1

r␲1

 

ro1 gm1V␲1



Vx

VA  V␲2

r␲2



gm2V␲2

ro2

Vx Vx − VA + + g m1Vπ 1 ro 2 ro1

(1)

I x = g m 2Vπ 2 +

(2)

Vx − VA VA + g m1Vπ 1 = ro1 rπ 1 rπ 2

Vπ 2 = VA = −Vπ 1 (3) Then from (2) ⎛ 1 Vx 1 ⎞ = VA ⎜ + g m1 + ⎟ ⎜ ro1 rπ 1 rπ 2 ⎟⎠ ⎝ ro1

⎛ 1 ⎛ ⎞ Vx Vx VA 1 ⎞ 1 + − − g m1VA or I x = Vx ⎜ + ⎟ + VA ⎜ g m 2 − − g m1 ⎟ ro 2 ro1 ro1 ro1 ⎝ ro1 ro 2 ⎠ ⎝ ⎠ Solving for VA from Equation (2) and substituting into Equation (1), we find (1) I x = g m 2VA +

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1 1 + g m1 + V ro1 rπ 1 & rπ 2 Ro = x = Ix ⎞ 1 ⎛ 1 1 ⎞ 1 ⎛ 1 + gm2 ⎟ ⎜ + g m1 + ⎟+ ⎜ ro 2 ⎝ ro1 rπ 1 & rπ 2 ⎠ ro1 ⎝ rπ 1 & rπ 2 ⎠ For β = 100, VA = 100 V , I C1 = I Bias = 1 mA ro1 = ro 2 = rπ 1 = rπ 2 =

100 = 100 k Ω 1 (100 )( 0.026 ) 1

g m1 = g m 2 =

Then Ro =

= 2.6 k Ω

1 = 38.46 mA/V 0.026 1 1 + 38.46 + 100 2.6 2.6

⎞ 1 ⎛ 1 1 ⎞ 1 ⎛ 1 + 38.46 + + 38.46 ⎟⎟ ⎜ ⎟+ ⎜ 100 ⎜⎝ 100 2.6 2.6 ⎟⎠ 100 ⎜⎝ 2.6 2.6 ⎠

or Ro = 50.0 k Ω Now I C 2 = 1 mA, I Bias = 0 Replace I Bias by

I β ⋅ = C 2 , I C1 ≅ 0.01 mA β 1+ β 1+ β

IC 2

100 100 = 100 k Ω, ro1 = = 10, 000 k Ω 1 0.01 1 = 38.46 mA/V , g m1 = 0.3846 mA/V gm2 = 0.026 (100 )( 0.026 ) = 2.6 k Ω, rπ 1 = 260 k Ω rπ 2 = 1 Then Ro = 66.4 k Ω ro 2 =

6.69 a. RTH = R1 R2 = 93.7 6.3 = 5.90 k Ω ⎛ R2 ⎞ VTH = ⎜ ⎟ VCC ⎝ R1 + R2 ⎠ 6.3 ⎞ ⎛ =⎜ ⎟ (12 ) = 0.756 V ⎝ 6.3 + 93.7 ⎠ 0.756 − 0.70 I BQ = = 0.00949 mA 5.90 I CQ = 0.949 mA VCEQ = 12 − ( 0.949 )( 6 ) ⇒ VCEQ = 6.305 V

Transistor: PQ ≈ I CQVCEQ = ( 0.949 )( 6.305 ) ⇒ PQ = 5.98 mW 2 RC : PR = I CQ RC = ( 0.949 ) ( 6 ) ⇒ PR = 5.40 mW 2

b.

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2

AC load line 1 Slope  6105 1  5.68 K

0.949

6.31

12

100 r0 = = 105 kΩ 0.949 Peak signal current = 0.949 mA

V0 ( max ) = ( 5.68 )( 0.949 ) = 5.39 V

PRC =

2 2 1 V0 ( max ) 1 ⎡ ( 5.39 ) ⎤ ⋅ = ⎢ ⎥ ⇒ PRC = 2.42 mW RC 2 2 ⎢⎣ 6 ⎥⎦

6.70 (a) 10 = I BQ RB + VBE ( on ) + (1 + β ) I BQ RE 10 − 0.7 I BQ = = 0.00369 mA 100 + (121)( 20 ) I CQ = 0.443 mA, I EQ = 0.447 mA For RC : PRC = ( 0.443) (10 ) ⇒ PRC = 1.96 mW 2

For RE : PRE = ( 0.447 ) ( 20 ) ⇒ PRE = 4.0 mW 2

(b) ΔiC = 0.667 − 0.443 = 0.224 mA 1 1 2 2 ( ΔiC ) RC = ( 0.224 ) (10 ) 2 2 = 0.251 mW

Then P RC = P RC

6.71 a. I BQ =

10 − 0.7 = 0.00596 mA 50 + (151)(10 )

I CQ = 0.894 mA, I EQ = 0.90 mA

VECQ = 20 − ( 0.894 )( 5 ) − ( 0.90 )(10 ) ⇒ VECQ = 6.53 V PQ ≅ I CQVECQ = ( 0.894 )( 6.53) ⇒ PQ = 5.84 mW

2 PRC ≅ I CQ RC = ( 0.894 ) ( 5 ) ⇒ PRC = 4.0 mW 2

2 PRE ≅ I EQ RE = ( 0.90 ) (10 ) ⇒ PRE = 8.1 mW 2

b.

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AC load line 1 52 1  1.43 K

Slope  0.894

6.53

20

−1 ΔiC = ⋅ Δvec 1.43 kΩ ΔiC = 0.894 ⇒ Δvec = ( 0.894 )(1.43) = 1.28 V ⎛ 5 ⎞ Δi0 = ⎜ ⎟ ΔiC = 0.639 mA ⎝5+2⎠ 1 2 PRL = ( 0.639 ) ( 2 ) ⇒ PRL = 0.408 mW 2 1 2 PRC = ⋅ ( 0.894 − 0.639 ) ( 5 ) ⇒ PRC = 0.163 mW 2 PRE = 0 PQ = 5.84 − 0.408 − 0.163 ⇒ PQ = 5.27 mW

6.72 I BQ =

10 − 0.70 = 0.00838 mA 100 + (101)(10 )

I CQ = 0.838 mA, I EQ = 0.846 mA

VCEQ = 20 − ( 0.838 )(10 ) − ( 0.846 )(10 ) ⇒ VCEQ = 3.16 V AC load line 1 RE RLr0

Slope  0.838

3.16

20

100 r0 = = 119 kΩ 0.838 Neglecting base currents: a. RL = 1 kΩ slope =

−1 −1 = 10 1 119 0.902 kΩ

−1 ⋅ ΔVce 0.902 kΩ ΔiC = 0.838 ⇒ ΔVce = ( 0.902 )( 0.838 ) = 0.756 V ΔiC =

1 ( 0.756 ) ⇒ PRL = 0.286 mW 2 1 2

PRL =

b.

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RL = 10 kΩ slope =

−1 −1 = 10 10 119 4.80

For ΔiC = 0.838 ⇒ Δvce = ( 0.838 )( 4.80 ) = 4.02 1 ( 3.16 ) ⇒ PRL = 0.499 mW 2 10 2

Max. swing determined by voltage PRL =

6.73 a. I BQ =

10 − 0.7 = 0.00838 mA 100 + (101)(10 )

I CQ = 0.838 mA, I EQ = 0.846 mA

VCEQ = 20 − ( 0.838 )(10 ) − ( 0.846 )(10 ) ⇒ VCEQ = 3.16 V PQ ≅ I CQVCEQ = ( 0.838 )( 3.16 ) ⇒ PQ = 2.65 mW 2 PRC ≅ I CQ RC = ( 0.838 ) (10 ) ⇒ PRC = 7.02 mW 2

b. AC load line 1 RC RL 1 1   101 0.909 K

Slope  0.838

3.16

20

−1 ΔiC = ⋅ Δvce 0.909 kΩ For ΔiC = 0.838 ⇒ Δvce = ( 0.909 )( 0.838 ) = 0.762 V ⎛ RC ⎞ ⎛ 10 ⎞ Δi0 = ⎜ ⎟ ΔiC = ⎜ ⎟ ΔiC = 0.762 mA ⎝ 10 + 1 ⎠ ⎝ RC + RL ⎠ 1 2 PRL = ( 0.762 ) (1) ⇒ PRL = 0.290 mW 2 1 2 PRC = ⋅ ( 0.838 − 0.762 ) (10 ) ⇒ PRC = 0.0289 mW 2 PQ = 2.65 − 0.290 − 0.0289 ⇒ PQ = 2.33 mW

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Chapter 7 Exercise Solutions EX7.1 (a) (i) RS = RP = 4 kΩ 1 1 ω= = rS ( RS + RP ) CS CS =

1 2π f ( RS + RP )

=

1 2π ( 20 )( 4 + 4 ) × 10 3

CS = 0.995 μ F

(ii) ⎛ RP ⎞ ω rS T ( jω ) = ⎜ ⎟ ⎝ RS + RP ⎠ 1 + ω 2 rS2 rS = ( RS + RP ) CS = 7.96 × 10 −3 RP 4 = = 0.5 RS + RP 4 + 4 f = 40 Hz T ( jω ) =

( 0.5)( 2π )( 40 ) ( 7.96 × 10 −3 )

(

)

1 + ⎡⎣ 2π ( 40 ) 7.96 × 10 −3 ⎤⎦ T ( jω ) = 0.447

f = 80 Hz T ( jω ) =

2

( 0.5)( 2π )(80 ) ( 7.96 × 10 −3 )

(

)

1 + ⎡⎣ 2π ( 80 ) 7.96 × 10 −3 ⎤⎦ T ( jω ) = 0.485

2

f = 200 Hz T ( jω ) =

( 0.5 )( 2π )( 200 ) ( 7.96 × 10 −3 )

(

)

1 + ⎡⎣ 2π ( 200 ) 7.96 × 10 −3 ⎤⎦ T ( jω ) = 0.498

2

(b)

ω=

1 1 = rP ( RS & RP ) CP

CP = =

1 2π f ( RS & RP )

(

2π 500 × 10

1

3

) (10 & 10 ) × 10

3

CP = 63.7 pF

EX7.2 a.

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⎛ RP ⎞ RP RP = 0.891 = ⇒ (1 − 0.891) RP = 0.891 ⇒ RP = 8.20 kΩ 20 log10 ⎜ ⎟ = −1 ⇒ + + R R R R R S ⎠ P S P +1 ⎝ P 1 1 ⇒ CS = fL = 2π ( RS + RP ) CS 2π (100 )(1 + 8.20 ) × 10 3 CS = 0.173 μ F fH =

1 1 ⇒ CP = 6 2π ( RS & RP ) CP 2π 10 (1 || 8.20 ) × 10 3

( )

CP = 179 pF

b. rS = ( RS + RP ) CS

(

)(

rS = 1 × 10 3 + 8.20 × 10 3 0.173 × 10 −6

)

rS = 1.59 ms open-circuit time-constant rP = ( RS & RP ) CP

(

rP = (1 & 8.20 ) × 10 3 179 × 10 −12

)

rP = 0.160 μ s short-circuit time-constant

EX7.3 a.

rS = ( Ri + RS i ) CC

b. f =

1 2π rS

RTH = R1 & R2 = 2.2 & 20 = 1.98 kΩ ⎛ R2 ⎞ ⎛ 2.2 ⎞ VTH = ⎜ ⎟ VCC = ⎜ ⎟ (10 ) = 0.991 V ⎝ 2.2 + 20 ⎠ ⎝ R1 + R2 ⎠ VTH − VBE ( on ) 0.991 − 0.7 I BQ = = = 0.0132 mA RTH + (1 + β ) RE 1.98 + ( 201)( 0.1) I CQ = 2.636 mA rπ = gm =

β VT I CQ ICQ VT

= =

( 200 )( 0.026 ) 2.636

= 1.97 kΩ

2.636 = 101.4 mA/V 0.026

Rib = τ π + (1 + β ) RE = 1.97 + ( 201)( 0.1) = 22.1 kΩ RB = R1 & R2 = 1.98 kΩ Ri = RB & Rib = 1.98 & 22.1 = 1.817 kΩ rS = ( Ri + RSi ) CC

(

= (1.817 + 0.1) ×10 3

)( 47 × 10 ) −6

= 90.1 ms 1 f = ⇒ f = 1.77 Hz 2π 90.1 × 10 −3

(

)

Midband Gain

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Aν = =

− β RC Ri ⋅ rπ + (1 + β ) RE Ri + RSi − ( 200 )( 2 )

1.82 1.97 + ( 201)( 0.1) 1.82 + 0.1 ⋅

Aν = −17.2

EX7.4 I DQ = K n (VGS − VTN )

2

a. 0.8 + 2 = VGS ⇒ VGS = 3.265 V 0.5 VS − ( −5 ) VS = −3.265 ⇒ I DQ = RS

−3.265 + 5 ⇒ RS = 2.17 kΩ 0.8 5 ⇒ RD = 6.25 kΩ VD = 0 ⇒ RD = 0.8 b. rS = ( RD + RL ) CC = (10 + 6.25 ) × 10 3 × CC 1 1 f = ⇒ CC = 2π rS 2π f 16.25 × 10 3 RS =

(

CC =

(

1

2π ( 20 ) 16.25 × 10 3

)

)

⇒ CC = 0.49 μ F

EX7.5 rS = ( RL + Ro ) CC 2 1 1 f = ⇒ CC 2 = 2π rS 2π f ( RL + Ro ) ⎧⎪ rπ + ( RS & RB ) ⎫⎪ R0 = RE & r0 ⎨ ⎬ 1+ β ⎪⎩ ⎪⎭ From Example 7-5, R0 = 35.5 Ω CC 2 =

1 2π (10 ) ⎡⎣10 × 10 3 + 35.5⎤⎦

CC 2 = 1.59 μ F

EX7.6 a.

I DQ = K P (VSG + VTP )

2

1 − ( −2 ) = VSG ⇒ VSG = 3.41 V 0.5 VS = 3.41 5 − 3.41 ⇒ RS = 1.59 kΩ RS = 1 5 For VSDG = VSGQ ⇒ VD = 0 ⇒ RD = ⇒ RD = 5 kΩ 1

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rP = ( RD & RL ) CL

b.

f =

1 1 ⇒ CL = 2π rP 2π f ( RD & RL )

CL =

( )

2π 10 6

1

( 5 & 10 ) × 10 3

⇒ CL = 47.7 pF

EX7.7 (a) RTH = 5 K VTH = −3.7527 I BQ =

−3.7527 − 0.7 − ( −5) 0.54726 = 5 + (101)( 0.5) 55.5

= 0.00986 I CQ = 0.986 mA gm = 37.925 rπ = 2.637 K 5 − Vo Vo − = 1 − Vo ( 0.4 ) 5 5 Vo = 0.035 (b) 0.986 =

Rib RS  0.1 k V0  Vi

 

RTH  5 k

V␲

r␲

gmV␲



RE  0.5 k

Vo = − gmVπ ( RC & RL )

Rib = rπ + (1 + β ) RE = 2.64 + (101)( 0.5) = 53.14 K Vπ =

Vb Vb Vb = = 101 14.885 ⎛1+ β ⎞ ⎛ ⎞ 1+ ⎜ ⎟ RE 1 + ⎜ 2.637 ⎟ ( 0.5) ⎝ ⎠ r ⎝ π ⎠

⎛ RTH & Rib ⎞ ⎛ 5 & 53.14 ⎞ Vb = ⎜ ⎟ Vi = ⎜ ⎟ Vi & + R R R ⎝ 5 & 53.14 + 0.1 ⎠ ⎝ TH ib S ⎠ ⎛ 4.57 ⎞ =⎜ ⎟ Vi = 0.9786 ⎝ 4.57 + 0.1 ⎠ (37.925)( 2.5) Av = − ( 0.9786 ) = Av = −6.23 14.885 (c)

EX7.8 a.

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RC 兩兩RL  2.5 k

I BQ =

0 − 0.7 − ( −10 )

= 0.0230 mA

0.5 + (101)( 4 )

I CQ = 2.30 mA

β VT

rπ =

I CQ I CQ

gm =

rB =

VT

= =

(100 )( 0.026 ) 2.30

= 1.13 kΩ

2.30 = 88.46 mA / V 0.026

RE ( RS + rπ ) CE

RS + rπ + (1 + β ) RE

( 4 × 10 ) ( 0.5 + 1.13) C = 3

E

0.5 + 1.13 + (101)( 4 )

rB =

1 1 = = 0.7958 ms 2π f B 2π ( 200 )

rB = 16.07CE ⇒ CE =

b. rA = RE CE = 4 × 10 3

(

fA =

fβ = =

)( 49.5 × 10 ) ⇒ r −6

A

= 0.198 s

1 1 = ⇒ f A = 0.804 H Z 2π rA 2π ( 0.198 )

EX7.9 rπ =

0.796 × 10 −3 ⇒ CE = 49.5 μ F 16.07

β 0VT I CQ

=

(150 )( 0.026 ) 0.5

= 7.8 kΩ

1 2π rπ ( Cπ + Cμ ) 1

(

2π 7.8 × 10

3

) ( 2 + 0.3) × 10

−12

⇒ f β = 8.87 MH Z

EX7.10 rπ

=

fβ = =

β 0VT I CQ

=

(150)(0.026) ⇒ rπ = 3.9 kΩ 1

1 2π rπ ( Cπ + Cμ )

(

2π 3.9 × 10

1

3

) ( 4 + 0.5) (10 ) −12

⇒ f β = 9.07 MHz

fT = β 0 f β = (150 )( 9.07 ) ⇒ fT = 1.36 GHz

EX7.11 RTH = R1 & R2 = 200 & 220 = 104.8 kΩ ⎛ R2 ⎞ ⎛ 220 ⎞ VTH = ⎜ ⎟ VCC = ⎜ ⎟ (5) = 2.619 V ⎝ 200 + 220 ⎠ ⎝ R1 + R2 ⎠ 2.62 − 0.7 = 0.009316 mA I BQ = 105 + (101)(1) I CQ = 0.9316 mA

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gm = rπ =

I CQ

=

VT

β VT I CQ

0.9316 ⇒ gm = 35.83 mA/V 0.026

=

(100)(0.026) ⇒ rπ = 2.79 kΩ 0.932

a. C M = Cμ ⎡⎣1 + gm ( RC & RL ) ⎤⎦ C M = ( 2 ) ⎡⎣1 + ( 35.83 )( 2.2 & 4.7 ) ⎤⎦ ⇒ C M = 109 pF b. RB = rS & R1 & R2 = 100 & 200 & 220 = 51.17 kΩ 1 f3 dB = 2π ( RB & rπ ) ( Cπ + Cμ ) =

1 ⇒ f3dB = 0.506 MHz 2π [ 51.17 & 2.79] × 10 3 × (10 + 109 ) × 10 −12

EX7.12 fT =

gm 2π ( Cgs + Cgd )

gm = 2 K n I DQ =2

( 0.2 )( 0.4 )

= 0.5657 mA/V fT =

0.5657 × 10 −3 ⇒ fT = 333 MHz 2π ( 0.25 + 0.02 ) × 10 −12

EX7.13 dc analysis ⎛ R2 ⎞ ⎛ 166 ⎞ VG = ⎜ ⎟ VDD = ⎜ ⎟ (10 ) = 4.15 V + R R ⎝ 166 + 234 ⎠ ⎝ 1 2 ⎠ V I D = S and VS = VG − VGS RS K n (VGS − VTN ) = 2

VG − VGS RS

( 0.5)( 0.5) (VGS2 − 4VGS + 4 ) = 4.15 − VGS 0.25VGS2 − 3.15 = 0 ⇒ VGS = 3.55 V gm = 2K n (VGS − VTN ) = 2 ( 0.5 )( 3.55 − 2 ) = 1.55 mA/V Small-signal equivalent circuit. Ri  10 k

Vi

 

RG  R1兩兩R2

Cgs

V0

 Cgd Vgs

RD

RL

gmVgs 

a.

C M = Cgd (1 + gm ( RD RL ) )

C M = ( 0.1) ⎡⎣1 + (1.55) ( 4 20 )⎤⎦ ⇒ C M = 0.617 pF

b.

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1

fH =

2πτ P τ P = ( RG & Ri ) ( Cgs + C M ) RG = R1 & R2 = 234 & 166 = 97.1 kΩ RG & Ri = 97.1 & 10 = 9.07 kΩ

(

)

rP = 9.07 × 10 3 (1 + 0.617 ) × 10 −12 = 14.7 ns fH =

1

(

2π 14.7 × 10 −9

)

⇒ f H = 10.9 MHz

EX7.14 dc analysis VTH = 0, RTH = 10 kΩ I BQ =

0 − 0.7 − ( −5 )

10 + (126 )( 5 )

= 0.00672 mA

I CQ = 0.840 mA rπ = gm = r0 =

β VT I CQ I CQ VT

= =

(125)( 0.026 ) 0.840

= 3.87 kΩ

0.840 = 32.3 mA/V 0.026

VA 200 = = 238 kΩ I CQ 0.84

High-frequency equivalent circuit C␮

RS  1 k

V0

 Vi

RB  V R1兩兩R2 ␲ 

 

r␲

r0 C␲

gmV␲

RC

a. Miller Capacitance C M = Cμ (1 + gm RL′ ) RL′ = r0 & RC & RL RL′ = 238 & 2.3 & 5 = 1.565 kΩ

C M = ( 3 ) ⎡⎣1 + ( 32.3 )(1.57 ) ⎤⎦ ⇒ Cμ = 155 pF

b. Req = RS & RB & rπ = RS & R1 & R2 & rπ Req = 1 & 20 & 20 & 3.87 = 0.736 kΩ rP = Re q ( Cπ + C M )

(

)

= 0.736 × 10 3 ( 24 + 155 ) × 10 −12 = 1.314 × 10

fH =

(

−7

1

2π 1.314 × 10 −7

)

⇒ f H = 1.21 MHz

c.

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RL

( Av )M

⎡ RB & τ π ⎤ = − gm RL′ ⎢ ⎥ ⎣⎢ RB τ π + RS ⎦⎥

( Av )M

⎡ 10 & 3.87 ⎤ = − ( 32.3 )(1.565 ) ⎢ ⎥ ⇒ ( Av ) M = −37.2 ⎣ 10 & 3.87 + 1 ⎦

EX7.15 The dc analysis 10 − 0.7 = 0.00838 mA I BQ = 100 + (101)(10 ) I CQ = 0.838 mA rπ = gm =

β VT I CQ

=

(100 )( 0.026 ) 0.838

= 3.10 kΩ

I CQ

= 32.22 mA/V VT For the input ⎡⎛ r ⎞ ⎤ rPπ = ⎢⎜ π ⎟ RE RS ⎥ Cπ ⎣⎝ 1 + β ⎠ ⎦ ⎡ 3.10 ⎤ 10 1⎥ × 10 3 × 24 × 10 −12 =⎢ ⎣ 101 ⎦ = 7.13 × 10 −10 s f Hπ =

1 1 = ⇒ f Hπ = 223 MHz 2π rPπ 2π 7.13 × 10 −10

(

)

For the output rP μ = [ RC & RL ] Cμ = (10 & 1) × 10 3 × 3 × 10 −12 = 2.73 × 10 −9 1 1 fHμ = = ⇒ f H μ = 58.4 MHz 2π rP μ 2π 2.73 × 10 −9

(

( Aν )M

)

⎡ ⎛ rπ ⎞ ⎤ ⎢ RE & ⎜ ⎟ ⎥ ⎝1+ β ⎠ ⎥ = gm ( RC & RL ) ⎢ ⎢ ⎥ ⎛ rπ ⎞ ⎢ RE & ⎜ ⎟ + RS ⎥ ⎝1+ β ⎠ ⎣⎢ ⎦⎥ ⎡ ⎛ 3.1 ⎞ ⎤ ⎢ 10 & ⎜ 101 ⎟ ⎥ ⎝ ⎠ ⎥⇒ A = ( 32.22 )(10 & 1) ⎢ ( ν )M = 0.870 ⎛ 3.1 ⎞ ⎥ ⎢ ⎢ 10 & ⎜⎝ 101 ⎟⎠ + 1 ⎥ ⎣ ⎦

EX7.16 ⎛ ⎞ R3 7.92 ⎛ ⎞ VB1 = ⎜ ⎟ (12 ) = ⎜ ⎟ (12 ) = 0.9502 V + + + + R R R 58.8 33.3 7.92 ⎝ ⎠ 2 3 ⎠ ⎝ 1 Neglecting lose currents 0.9502 − 0.7 IC = = 0.50 mA 0.5 β VT (100 )( 0.026 ) rπ = = = 5.2 K IC 0.5 IC 0.5 = = 19.23 mA/V VT 0.026 From Eq (7.127(a)), gm =

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τ Pπ

= [ RS & RB1 & rπ ][Cπ 1 + C M 1 ]

RB1 = R2 & R3 = 33.3 & 7.92 = 6.398 K C M 1 = 2Cμ 1 = 6 pF

Then 3 −12 τ Pπ = [1 & 6.398 & 5.2] × 10 × [24 + 6] × 10 ⇒ 22.24 ns 1 1 = ⇒ 7.15 MHz f Hπ = 2πτ Pπ 2π ( 22.24 × 10 −9 ) From Eq (7.128(a)) 3 −12 τ P μ = [ RC & RL ] C μ 2 = ( 7.5 & 2 ) × 10 × 3 × 10 ⇒ 4.737 ns 1 1 fHμ = = = 33.6 MHz 2πτ Pμ 2π ( 4.737 × 10 −9 ) From Eq. 7.133 Av

Av

M

M

⎡ RB1 & rπ 1 ⎤ = gm 2 ( RC & RC ) ⎢ ⎥ ⎣ RB1 & rπ 1 + RS ⎦ ⎡ 6.40 & 5.2 ⎤ = (19.23)( 7.5 & 2 ) ⎢ ⎥ ⎣ 6.40 & 5.2 + 1 ⎦ ⎡ 2.869 ⎤ = (19.23)(1.579 ) ⎢ ⎣ 2.869 + 1 ⎥⎦ = 22.5

TYU7.1 a. V0 = − ( gmVπ ) RL rπ Vπ = × Vi 1 rπ + + RS sCC T (s) = =

V0 ( s ) Vi ( s )

=

− gm rπ RL rπ + RS + (1 / sCC )

− gm rπ RL ( sCC )

1 + s ( rπ + RS ) CC

gm rπ = β T (s) =

− β RL rπ + RS

⎛ s ( rπ + RS ) CC ×⎜ ⎜ 1 + s (r + R ) C π S C ⎝

⎞ ⎟⎟ ⎠

Then τ = ( rπ + RS ) CC b. 1 f3− dB = 2π ( rπ + RS ) CC f3− dB =

1

2π ⎡⎣ 2 × 10 + 1 × 10 3 ⎤⎦ ⎡⎣10 −6 ⎤⎦ ( 2 )( 50 )( 4 ) rg R T ( jω ) max = π m L = 2 +1 rπ + RS 3

⇒ f3 dB = 53.1 Hz

T ( jω ) max = 133

c.

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兩T( j)兩 133

f

53.1 Hz

TYU7.2 a. ⎛ 1 ⎞ V0 = − g mVπ ⎜ RL & ⎟ sCL ⎠ ⎝ ⎛ r ⎞ Vπ = ⎜ π ⎟ × Vi ⎝ rπ + RS ⎠ 1 ⎛ RL × ⎜ V0 ( s ) rπ sCL ⎜ T (s) = = − gm 1 Vi ( s ) rπ + RS ⎜ ⎜ RL + sC L ⎝ T (s) =

⎞ ⎟ ⎟ ⎟ ⎟ ⎠

⎞ − β RL ⎛ 1 ×⎜ ⎟ rπ + RS ⎝ 1 + sRL CL ⎠

Then τ = RL CL b. 1 1 f3− dB = = ⇒ f3dB = 3.18 MH Z 3 2π RL CL 2π 5 × 10 10 × 10 −12

(

T ( jω ) =

)(

)

gm rπ RL ( 75)(1.5)( 5) = 1.5 + 0.5 rπ + RS

T ( jω ) max = 281

c.

281

兩T( j)兩

3.18 MHz

f

TYU7.3 a. Open-circuit time constant ( CL → open ) rS = ( RS + rπ ) CC

(

)

= ( 0.25 + 2 ) × 10 3 2 × 10 −6 = 4.5 ms

Short-circuit time constant ( CC → short )

(

)(

rP = RL CL = 4 × 10 3 50 × 10 −12

)

rP = 0.2 μ s b.

Midband gain

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⎛ rπ ⎞ V0 = − gmVπ RL , Vπ = ⎜ ⎟ Vi ⎝ τ π + RS ⎠ V −g r R Av = 0 = m π L Vi rπ + RS =

− ( 65 )( 2 )( 4 )

2 + 0.25 Av = −231

c. fL =

1 1 = ⇒ f L = 35.4 Hz 2π rS 2π 4.5 × 10 −3

fH =

1 1 ⇒ f H = 0.796 MHz = 2π rP 2π 0.2 × 10 −6

(

)

(

)

TYU7.4 Computer Analysis TYU7.5 Computer Analysis TYU7.6 (100 )( 0.026 ) βV = 10.4 kΩ rπ = 0 T = 0.25 I CQ fβ =

1 2π rπ ( Cπ + C μ )

Cπ + Cμ =

1 1 = 2π f β rπ 2π 11.5 × 10 6 10.4 × 10 3

(

)(

)

Cπ + Cμ = 1.33 pF Cμ = 0.1 pF ⇒ Cπ = 1.23 pF

TYU7.7 h fe =

β0

1 + j ( f / fβ )

f β = 5 MH Z , β 0 = 100 At f = 50 MH Z h fe =

100

⎛ 50 ⎞ 1+ ⎜ ⎟ ⎝ 5 ⎠

2

⇒ h fe = 9.95

⎛ 50 ⎞ Phase = − tan −1 ⎜ ⎟ ⇒ Phase = −84.3° ⎝ 5 ⎠

TYU7.8 f 500 ⇒ f β = 4.17 MHz fβ = T = β 0 120 fβ =

1 2π rπ (Cπ + C μ )

Cπ + Cμ =

1 1 = 2π f β rπ 2π (4.167 × 10 6 )(5 × 10 3 )

Cπ + Cμ = 7.639 pF Cμ = 0.2 pF ⇒ Cπ = 7.44 pF

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TYU7.9 (a) gm = 2K n (VGS − VTN ) = 2 ( 0.4 )( 3 − 1) ⇒ gm = 1.6 mA/V gm′ = 80% of gm = 1.28 mA/V gm gm′ = 1 + gm rS 1 + gm rS =

gm gm′

⎛ gm ⎞ 1 ⎛ 1.6 ⎞ − 1⎟ = ⎜ ⎜ 1.28 − 1 ⎟ ′ g 1.6 ⎝ ⎠ ⎝ m ⎠ rS = 0.156 kΩ ⇒ rS = 156 ohms 1 gm

rS =

(b) gm = 2K n (VGS − VTN ) = 2 ( 0.4 )( 5 − 1) ⇒ gm = 3.2 mA/V gm 3.2 gm′ = = = 2.134 1 + gm rS 1 + ( 3.2 )( 0.156 ) Δgm 3.2 − 2.134 = ⇒ A 33.3% reduction 3.2 gm

TYU7.10 fT = = Cgs = =

gm

2π ( CgsT + C gdT ) gm

2π ( Cgs + Cgsp + Cgdp ) gm − Cgsp − C gdp 2π fT 0.5 × 10 −3

(

2π 500 × 10

6

)

− ( 0.01 + 0.01) × 10 −12 ⇒ Cgs = 0.139 pF

TYU7.11 fT =

gm 2π (Cgs + Cgsp + Cgdp )

Cgsp = Cgdp 2Cgsp =

gm 1× 10−3 − C gs = − 0.4 × 10−12 2π fT 2π (350 × 106 )

2Cgsp = 0.0547 pF ⇒ Cgsp = Cgdp ≅ 0.0274 pF

TYU7.12 dc analysis ⎛ 50 ⎞ VG = ⎜ ⎟ (10 ) − 5 = −2.5 ⎝ 50 + 150 ⎠ VS = VG − VGS . I D = K n (VGS − VTN ) = 2

VS − ( −5 ) RS

VG − VGS + 5 RS

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(1)( 2 ) ⎡⎣VGS2 − 1.6VGS + 0.64 ⎤⎦ = −2.5 − VGS + 5 2 − 2.2VGS − 1.22 = 0 2VGS

2.2 ±

VGS =

( 2.2 )

2

+ 4 ( 2 )(1.22 )

2 (2)

⇒ VGS = 1.505 V

gm = 2 K n (VGS − VTN ) = 2 (1)(1.505 − 0.8 ) = 1.41 mA/V

Equivalent circuit Cgd

Ri

V0

 Vi

RG  Vgs R1兩兩R2

 

Cgs

gmVgs

RD



C M = Cgd (1 + gm RD ) = ( 0.2 ) ⎡⎣1 + (1.42 )( 5 ) ⎤⎦ ⇒ C M = 1.61 pF

(a)

(b)

τ P = ( Ri & RG ) ( Cgs + CM )

rP = [ 20 & 50 & 150 ] × 10 3 × ( 2 + 1.61) × 10 −12

(

)(

)

= 13 × 10 3 3.62 × 10 −12 = 4.71 × 10 −8 s fH =

c.

1 1 = ⇒ f H = 3.38 MHz 2π rP 2π 4.71 × 10 −8

(

( Av )M ( Av )M

)

⎛ RG ⎞ = − gm RD ⎜ ⎟ ⎝ RG + RS ⎠ ⎛ 37.5 ⎞ = − (1.41)( 5 ) ⎜ ⎟ ⇒ ( Av ) M = −4.60 ⎝ 37.5 + 20 ⎠

TYU7.13 Computer Analysis

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Chapter 7 Problem Solutions 7.1 a. T (s) = T (s) =

V0 ( s ) Vi ( s )

=

1/ ( sC1 )

⎣⎡1/ ( sC1 ) ⎦⎤ + R1

1 1 + sR1C1

b.

1

兩T 兩

fH  159 Hz

f

fH =

1 1 = ⇒ f H = 159 Hz 2π R1C1 2π (103 )(10−6 )

c. V0 ( s ) = Vi ( s ) ⋅

1 1 + sR1C1

For a step function Vi ( s ) =

1 s

K K2 1 1 = 1+ V0 ( s ) = ⋅ s 1 + sR1C1 s 1 + sR1C1 = =

K1 (1 + sR1C1 ) + K 2 s s (1 + sR1C1 )

K1 + s ( K1 R1C1 + K 2 ) s (1 + sR1C1 )

K 2 = − K1 R1C1 and K1 = 1 V0 ( s ) =

− R1C1 1 + s 1 + sR1C1

=

1 1 − 1 s +s R1C1

v0 ( t ) = 1 − e− t / R1C1

7.2 a. T (s) = T (s) =

V0 ( s ) Vi ( s )

=

R2 R2 + ⎡⎣1/ ( sC2 ) ⎤⎦

sR2 C2 1 + sR2 C2

b.

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1

兩T 兩

fL  1.59 Hz

1 1 fL = = ⇒ f L = 1.59 Hz 4 2π R2 C2 2π (10 )(10 × 10−6 )

c. V0 ( s ) = Vi ( s ) ⋅

sR2 C2 1 + sR2 C2

Vi ( s ) =

1 s

V0 ( s ) =

R2 C2 1 = 1 + sR2 C2 s + 1 R2 C2

v0 ( t ) = e− t / R2C2

7.3 a. 1 sCP T (s) = = Vi ( s ) ⎛ 1 1 ⎞ RP & + ⎜ RS + ⎟ sCP ⎝ sCS ⎠ 1 RP ⋅ sCP RP 1 RP & = = sCP R + 1 1 + sRP CP P sCP Then RP T (s) = ⎛ 1 ⎞ RP + ⎜ RS + ⎟ (1 + sRP CP ) sCS ⎠ ⎝ V0 ( s )

=

RP &

RP RP CP 1 RP + RS + + + sRS RP CP CS sCS

⎛ RP T (s) = ⎜ ⎝ RP + RS b.

⎤⎞ ⎞ ⎛ ⎡ sRP RS RP C 1 ⋅ P+ + ⋅ CP ⎥ ⎟ ⎟ × ⎜⎜ 1/ ⎢1 + ⎟ ⎠ ⎝ ⎣⎢ RP + RS CS s ( RS + RP ) CS RS + RP ⎦⎥ ⎠

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−11 ⎛ ⎤⎞ 1 ⎛ 10 ⎞ ⎜ ⎡ 10 10 3 −11 ⎟ ⎢ ⎥ 1/ 1 5 10 10 T (s) = ⎜ s × + ⋅ + + × ⋅ ( ) ⎟ −6 s ( 2 × 104 ) ⋅10−6 ⎝ 10 + 10 ⎠ ⎜ ⎢⎣ 20 10 ⎥⎦ ⎟ ⎝ ⎠ 1 1 ≅ ⋅ 1 2 1+ + s ( 5 × 10−8 ) s ( 0.02 )

s = jω T ( jω ) =

1 ⋅ 2

T ( jω ) =

1 ⋅ 2

1

⎡ ⎤ 1 1 + j ⎢ω ( 5 × 10−8 ) − ⎥ ω ( 0.02 ) ⎦ ⎣ 1 1 For ω L = = = 50 4 ( RS + RR ) CS ( 2 ×10 )(10−6 ) 1

⎡ ⎤ 1 1 + j ⎢( 50 ) ( 5 × 10−8 ) − ( 50 )( 0.02 ) ⎥⎦ ⎣ 1 1 1 1 ≈ ⋅ ⇒ T ( jω ) = ⋅ 2 1− j 2 2

For

ωH =

1

( RS & RP ) CP

=

1

( 5 ×103 )(10−11 )

= 2 × 107

1 ⎡ ⎤ 1 ⎥ 1 + j ⎢( 2 × 107 )( 5 × 10−8 ) − ⎢⎣ ( 2 ×107 ) ( 0.02 ) ⎥⎦ 1 1 1 1 ⇒ T ( jω ) = ⋅ T ( jω ) ≈ ⋅ 2 1+ j 2 2 T ( jω ) =

1 ⋅ 2

In each case, T ( jω ) =

c.

RP 2 RP + RS

1

⋅

RS = RP = 10 kΩ, CS = CP = 0.1 μ F

T (s) =

⎛ ⎡ ⎤⎞ 1 ⎜ ⎢ 1 1 1 3 −7 ⎥ ⎟ ⋅ 1/ 1 + ⋅ + + × s 5 10 10 ( )( )⎥ ⎟ 2 ⎜⎜ ⎢ 2 1 s 2 × 104 (10−7 ) ⎟ ⎦⎠ ⎝ ⎣

(

)

s = jω T ( jω ) =

1 ⋅ 2

1 ⎡ ⎤ 1 1 ⎥ 1 + + j ⎢ω ( 5 × 10−4 ) − 2 ω ( 2 × 10−3 ) ⎥⎦ ⎢⎣ 1 = 500 For ω = 4 ( 2 ×10 )(10−7 )

T ( jω ) =

1 ⋅ 2

1

⎡ ⎤ 1 ⎥ 1.5 + j ⎢( 500 ) ( 5 × 10−4 ) − ⎢⎣ ( 500 ) ( 2 ×10−3 ) ⎥⎦ 1 1 = ⋅ ⇒ T ( jω ) = 0.298 2 1.5 − j ( 0.75 )

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For ω =

1

( 5 ×10 )(10 ) −7

3

= 2 × 103

⎧ ⎡ ⎤ ⎞ ⎪⎫ 1 ⎪ ⎜⎛ 1 ⎟⎬ ⎥ ⋅ ⎨1/ 1.5 + j ⎢( 2 × 103 )( 5 × 10−4 ) − 2 ⎪ ⎜ 2 × 103 )( 2 × 10−3 ) ⎥⎦ ⎟ ⎪ ⎢⎣ ( ⎠⎭ ⎩ ⎝ 1 1 = ⋅ ⇒ T ( jω ) = 0.298 2 1.5 + j ( 0.75 )

T ( jω ) =

In each case, T ( jω ) <

7.4 Circuit (a): V R2 T= 0 = Vi 1 R2 + sC1

=

=

RP 2 RP + RS

1

⋅

R2 R1 (1/ sC1 ) R2 + R1 + (1/ sC1 ) R2 R2 +

R1 1 + sR1C1

=

R2 (1 + sR1C1 ) R2 + sR1 R2 C1 + R1

or

(1 + sR1C1 ) Vo R2 = ⋅ Vi R1 + R2 1 + sR1 R2 C1

Low frequency: Vo R2 20 2 = = = Vi R1 + R2 10 + 20 3 High frequency: Vo =1 Vi

τ 1 = R1C1 = (104 )(10 × 10−6 ) = 0.10 ⇒ f1 =

1

= 1.59 Hz

2πτ 1

τ 2 = ( R1 R 2 ) C1 = (10 20 ) × 103 × (10 × 10−6 ) ⇒ τ 2 = 0.0667 ⇒ f 2 = 兩T 兩 1.0 0.67

1.59

2.39

f

Circuit (b): 1 sC2

R2 V 1 + sR2 C2 T= o = = R2 Vi 1 + R1 R2 + R1 sR + 1 sC2 2 C2 R2

⎞ ⎛ R2 ⎞ ⎛ 1 =⎜ ⎟ ⎟ ⎜⎜ ⎝ R1 + R2 ⎠ ⎝ 1 + s ( R1 R2 ) C2 ⎠⎟

Low frequency: Vo R2 20 2 = = = Vi R1 + R2 20 + 10 3

τ = ( R1 R2 ) C2 = (10 20 ) × 103 × 10 × 10−6 = 0.0667 f =

1 2πτ

= 2.39 Hz

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1 = 2.39 Hz 2πτ 2

0.67

兩T 兩

2.39

f

7.5 a. rS = ( Ri + RP ) CS = [30 + 10] × 103 × 10 × 10−6 ⇒ rS = 0.40 s

rP = ( Ri RP ) CP = ⎡⎣30 10 ⎤⎦ × 103 × 50 × 10−12 ⇒ rP = 0.375 μ s b. 1 1 fL = = ⇒ f L = 0.398 Hz 2π rS 2π ( 0.4 ) fH =

1 1 = ⇒ f H = 424 kHz 2π rP 2π ( 0.375 × 10−6 )

At midband. CS → short, CP → open Vo = I i ( Ri RP )

T ( s ) = Ri R P = 30 10 ⇒ T ( s ) = 7.5 k Ω

c.

7.5 k

兩T 兩

fH

fL

7.6 (a) T= T

1

(1 + j 2π f τ )

max

=1

At f =

2

⇒T =

(

1 1 + ( 2π f τ )

2

)

2

=

1 1 + ( 2π f τ )

2

1 1 1 ⇒T = = 2 2πτ 2 1 + (1)

⎛1⎞ = 20 log10 ⎜ ⎟ ⇒ T dB ≅ −6 dB ⎝ 2⎠ −1 Phase = 2 tan ( 2π f τ ) = −2 tan −1 (1) = −2 ( 45° ) ⇒ Phase = −90° (b) Slope = −2 ( 6dB / oct ) = −12dB / oct = −40dB / decade T

dB

Phase = −2 ( 90° ) ⇒ Phase = −180°

7.7 (a)

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−10 ( jω )

T ( jω ) =

T =

jω ⎞ jω ⎞ ⎛ ⎛ 20 ⎜ 1 + ⎟ ( 2000 ) ⎜1 + ⎟ 20 2000 ⎝ ⎠ ⎝ ⎠ ω j ⎛ ⎞ −5 × 10−3 ⎜ −4 ⎟ ⎝ ω ⎠ = 2.5 × 10 ( jω ) = jω ⎞ ⎛ jω ⎞ ⎛ jω ⎞⎛ jω ⎞ ⎛ ⎜1 + ⎟ ⎜1 + ⎟ ⎜1 + ⎟⎜ 1 + ⎟ ω 2000 20 2000 ⎝ ⎠⎝ ⎠ ⎝ ⎠⎝ ⎠ 2.5 × 10−4 (ω )

⎛ω ⎞ ⎛ ω ⎞ 1+ ⎜ ⎟ ⋅ 1+ ⎜ ⎟ ⎝ 20 ⎠ ⎝ 2000 ⎠ 2

2

兩T 兩 5  103

2.5  104

1

20

2000



(b) jω ⎞ ⎟ 10 ⎝ ⎠ T ( jω ) = jω ⎞ ⎛ 1000 ⎜ 1 + ⎟ ⎝ 1000 ⎠

(10 )(10 ) ⎛⎜1 +

T =

( 0.10 )

⎛ω ⎞ 1+ ⎜ ⎟ ⎝ 10 ⎠

⎛ ω ⎞ 1+ ⎜ ⎟ ⎝ 1000 ⎠

2

2

兩T 兩 10

0.10

10



1000

7.8 (a) T (s) =

105

=

10

( 5 + 10 )( 5 + 500 ) (10 )( 500 )

⋅

5 ⎛ 5 ⎞⎛ 5 ⎞ + 1⎟ ⎜ + 1⎟⎜ ⎝ 10 ⎠⎝ 500 ⎠

⎛ jω ⎞ ⎜ ⎟ 10 ⎝ 10 ⎠ = ⋅ 500 ⎛ jω ⎞⎛ jω ⎞ + 1⎟⎜ + 1⎟ ⎜ ⎝ 10 ⎠⎝ 500 ⎠

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兩T 兩 0.02

10

(b) (c) (d)

500

(rod/s)

Midband gain = 0.02 ω = 500 rad/s ω = 10 rad/s

7.9 (a) T ( s) =

=

T ( jω ) =

2 × 104

( S + 10 )( S + 10 ) 3

5

2 × 104

(10 )(10 ) 3

5

⋅

1 ⎛ S ⎞⎛ S ⎞ ⎜ 3 + 1 ⎟ ⎜ 5 + 1⎟ 10 10 ⎝ ⎠⎝ ⎠

2 × 10−4 ⎛ jω ⎞ ⎛ jω ⎞ ⎜ 3 + 1 ⎟ ⎜ 5 + 1⎟ ⎝ 10 ⎠ ⎝ 10 ⎠

兩T 兩 2  104

10 3

(b) (c) (d)

2 × 10

10 5

(rod/s)

−4

ω = 103 rad/s no low freq −3dB freq.

7.10 a. ⎛ r ⎞ V0 = − g mVπ RL Vπ = ⎜ π ⎟ Vi + r R S ⎠ ⎝ π ⎛ r ⎞ ⎛ 5.2 ⎞ T = g m RL ⎜ π ⎟ = ( 29 )( 6 ) ⎜ ⎟ ⎝ 5.2 + 0.5 ⎠ ⎝ rπ + RS ⎠ Tmidband = 159

b. rS = ( RS + rπ ) CC 1 1 1 fL = ⇒ rS = = ⇒ rS = 5.31 ms Open-circuit 2π rS 2π f L 2π ( 30 ) rP =

1 1 = ⇒ τ P = 0.332 μ s Short-circuit 2π f H 2π ( 480 × 103 )

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c. rS 5.31× 10−3 = ⇒ CC = 0.932 μ F ( RS + τ π ) ( 0.5 + 5.2 ) ×103

CC =

rP = RL CL rP 0.332 × 10−6 = ⇒ CL = 55.3 pF RL 6 × 103

CL =

7.11 Computer Analysis 7.12 Computer Analysis 7.13 a. RTH = R1 R2 = 10 1.5 = 1.304 kΩ ⎛ R2 ⎞ ⎛ 1.5 ⎞ VTH = ⎜ ⎟ VCC = ⎜ ⎟ (12 ) = 1.565 V ⎝ 1.5 + 10 ⎠ ⎝ R1 + R2 ⎠ 1.565 − 0.7 I BQ = = 0.0759 mA 1.30 + (101)( 0.1) I CQ = 7.585 mA rπ =

(100 )( 0.026 )

= 0.343 kΩ 7.59 7.59 = 292 mA/V gm = 0.026

Ri = R1 R2 ⎡⎣ rπ + (1 + β ) RE ⎤⎦ = 10 1.5 ⎡⎣0.343 + (101)( 0.1) ⎤⎦ = 1.30 10.44 ⇒ Ri = 1.159 kΩ r = ( RS + Ri ) CC = [ 0.5 + 1.16] × 103 × 0.1× 10−6 r = 1.659 × 10−4 s fL =

1 1 = ⇒ f L = 959 Hz 2π r 2π (1.66 × 10−4 )

b. RS

Rib

V␲ Vi

 

V0



Ib

R1兩兩R2

r␲

gmV␲



RC RE

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V0 = − ( β I b ) RC R1b = rπ + (1 + β ) RE = 0.343 + (101)( 0.1) = 10.44 kΩ ⎛ R1 R2 ⎞ Ib = ⎜ I ⎜ R R + R ⎟⎟ i ib ⎠ ⎝ 1 2 ⎛ 1.30 ⎞ =⎜ ⎟ I i = ( 0.111) I i ⎝ 1.30 + 10.4 ⎠ Vi Ii = RS + R1 R2 Rib = Ii = V0 = Vi

Vi 0.5 + (1.3) (10.44 ) Vi 1.659 β RC ( 0.111) 1.659

⇒

V0 Vi

=

(100 )(1)( 0.111) 1.659

midband

⇒

V0 Vi

= 6.69 midband

c.

6.69

兩 VV 兩 0 i

fL  959 Hz

f

7.14 I DQ = 0.5 mA ⇒ VS = ( 0.5 )( 0.5 ) = 0.25 V 0.5 + 1.5 = 3.08 V 0.2 + VS = 3.08 + 0.25 ⇒ VG = 3.33 V

I DQ = K n (VGS − VTN ) ⇒ VGS = 2

VG = VGS

⎛ R2 ⎞ 1 VG = ⎜ ⎟ VDD ⇒ 3.33 = ⋅ Rin ⋅ VDD + R R R ⎝ 1 2 ⎠ 1 1 3.33 = ( 200 )( 9 ) ⇒ R1 = 541 kΩ R1 541R2 = 200 ⇒ R2 = 317 kΩ 541 + R2 VD = VDSQ + VS = 4.5 + 0.25 = 4.75 9 − 4.75 ⇒ RD = 8.5 kΩ 0.5 1 1 1 fL = ⇒ rL = = = 7.96 ms 2π rL 2π f L 2π ( 20 )

RD =

rL = Rin ⋅ CC ⇒ CC =

rL 7.96 × 10−3 = ⇒ CC = 0.0398 μ F 200 × 103 Rin

g m = 2 ( 0.2 )( 3.08 − 1.5 ) = 0.632 mA/V Av

midband

=

( 0.632 )(8.5) g m RD = ⇒ Av = 4.08 1 + g m RS 1 + ( 0.632 )( 0.5 )

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4.08

兩A␯兩

fL  20 Hz Phase

f

fL

90 135 180

7.15 I DQ

I DQ = K n (VGS − VTN ) ⇒ VGS = 2

+ VTN =

Kn

1 + 1 = 2.414 V 0.5

VS = −2.414 V −2.414 − ( −5 )

RS =

VD = VDSQ

⇒ RS = 2.59 kΩ 1 + VS = 3 − 2.414 = 0.586 V

5 − 0.59 ⇒ RD = 4.41 kΩ 1

RD =

b. CC  Vi

 

RG

Vgs

gmVgs RD



I0

RL

RS

⎛ ⎜ RD I 0 = − ( g mVgs ) ⎜ ⎜R +R + 1 L ⎜ D sCC ⎝ Vi Vgs = 1 + g m RS I0 ( s )

Vi ( s )

=

T (s) = =

⎞ ⎟ ⎟ ⎟ ⎟ ⎠

⎡ ⎤ − gm sCC ⋅ RD ⎢ ⎥ 1 + g m RS ⎣⎢1 + s ( RD + RL ) CC ⎦⎥

I0 ( s )

Vi ( s )

s ( RD + RL ) CC − g m RD 1 ⋅ ⋅ 1 + g m RS RD + RL 1 + s ( RD + RL ) CC

c.

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fL =

1 1 1 → rL = = = 15.92 ms 2π rL 2π f L 2π (10 )

rL = ( RD + RL ) CC ⇒ CC =

rL 15.9 × 10−3 = ⇒ CC = 1.89 μ F RD + RL ( 4.41 + 4 ) × 103

7.16 a. 9 − VSG 2 = I D = K P (VSG + VTP ) RS

2 9 − VSG = ( 0.5 )(12 ) (VSG − 4VSG + 4 ) 2 6VSG − 23VSG + 15 = 0

VSG =

( 23)

23 ±

2

− 4 ( 6 )(15 )

2 ( 6)

⇒ VSG = 3 V

g m = 2 K P (VSG + VTP ) = 2 ( 0.5 )( 3 − 2 ) ⇒ g m = 1 mA/V Ro =

1 RS = 1 12 ⇒ Ro = 0.923 kΩ gm

b.

r = ( R0 + RL ) CC

c.

fL = CC =

1 2πτ

⇒r=

1 2π f L

=

1 = 7.96 ms 2π ( 20 )

7.96 × 10−3 r = ⇒ CC = 0.729 μ F Ro + RL ( 0.923 + 10 ) × 103

7.17 a. 1 = 0.00833 mA 120 R1 || R2 = ( 0.1)(1 + β )( RE ) = ( 0.1)(121)( 4 ) = 48.4 kΩ I CQ = 1 mA, I BQ =

VTH = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE 1 ⋅ RTH ⋅ VCC = ( 0.00833)( 48.4 ) + 0.7 + (121)( 0.00833)( 4 ) R1 1 ( 48.4 )(12 ) = 5.135 R1 R1 = 113 kΩ 113R2 = 48.4 ⇒ R2 = 84.7 k Ω 113 + R2 b. r R0 = π RE r0 1+ β rπ =

(120 )( 0.026 ) 1

= 3.12 kΩ

80 = 80 kΩ 1 3.12 R0 = 4 80 = 0.02579 4 80 ⇒ R0 = 25.6 Ω 121 c. r0 =

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r = ( R0 + RL ) CC 2

r = ( 0.0256 + 4 ) × 103 × 2 × 10−6 = 8.05 × 10−3 s f =

1 1 = ⇒ f = 19.8 Hz 2π r 2π ( 8.05 × 10−3 )

7.18 (a) 5 − 0.7 = 1.075 mA I CQ = 1.064 mA 4 = 10 − (1.064 )( 2 ) − (1.075 )( 4 ) = 3.57 V

I EQ = VCEQ VCEQ gm = rπ =

I CQ

=

VT

β VT

1.064 = 40.92 mA/V 0.026

=

I CQ

(100 )( 0.026 ) 1.064

= 2.44 K

(b) For CC1 , Req1 = RS + RE

rπ 2440 = 200 + 4000 1+ β 101

Req1 = 224.0r , τ 1 = Req1CC1 = 1.053 ms For CC 2 , Req 2 = RC + RL = 2 + 47 = 49 K

τ 2 = Req 2 ⋅ Cc 2 = 49 ms f1 =

(c)

1 2πτ 1

=

1

2π (1.053 × 10−3 )

⇒ f1 = 151 Hz

7.19 (a) τ H = ( RC RL ) CL = ( 2 47 ) × 103 × 10 × 10−12 = 1.918 × 10−8 s fH =

1 2πτ H

=

1

2π (1.918 × 10−8 )

⇒ f H = 8.30 MHz

(b) 1 1 + ( f .2πτ H )

2

= 0.1

2

2 ⎛ 1 ⎞ ⎜ ⎟ = 100 = 1 + ( f .2πτ H ) ⎝ 0.1 ⎠

f =

99 2πτ H

=

99

2π (1.918 × 10−8 )

f = 82.6 MHz

7.20 (a)

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5 − VSG 2 = K P (VSG + VTP ) R1 2 5 − VSG = (1)(1.2 )(VSG − 1.5 ) = (1.2 ) (VSG − 3VSG + 2.25 ) 2

2 1.2VSG − 2.6VSG − 2.3 = 0⇒ VSG = 2.84 V

I DQ = 1.8 mA VSDQ = 10 − (1.8 )(1.2 + 1.2 ) ⇒ VSDQ = 5.68 V g m = 2 K P I DQ = 2 (1)(1.8 ) = 2.683 mA / V ro = ∞ (b) Ris =

1 1 = = 0.3727 k Ω g m 2.68

Ri = 1.2 0.373 = 0.284 k Ω

For CC1 , τ s1 = ( 284 + 200 ) ( 4.7 × 10−6 ) = 2.27 ms

For CC 2 , τ s 2 = (1.2 x103 + 50 × 103 )(10−6 ) = 51.2 ms

(c) CC2 dominates, 1 1 f 3− dB = = = 3.1 Hz 2πτ s 2 2π ( 51.2 × 10−3 ) 7.21 Assume VTN = 1V , kn′ = 80μ A / V 2 , λ = 0 Neglecting RSi = 200Ω , Midband gain is: Av = g m RD

Let I DQ = 0.2 mA, VDSQ = 5V Then RD =

9−5 ⇒ RD = 20 k Ω 0.2

We need g m =

Av RD

=

10 ⎛ k′ ⎞⎛ W ⎞ = 0.5 mA / V 2 and g m = 2 K n I DQ = 2 ⎜ n ⎟ ⎜ ⎟ I DQ 20 ⎝ 2 ⎠⎝ L ⎠

W ⎛ 0.080 ⎞ ⎛ W ⎞ or 0.5 = 2 ⎜ = 7.81 ⎟ ⎜ ⎟ ( 0.2 ) ⇒ L ⎝ 2 ⎠⎝ L ⎠ Let 9 9 R1 + R2 = = = 225 k Ω ( 0.2 ) I DQ ( 0.2 )( 0.2 ) ⎛ R2 ⎞ 2 ⎛ R2 ⎞ ⎛ 0.080 ⎞ I DQ = 0.2 = ⎜ ⎟ (9 ) = ⎜ ⎟ ( 7.81)(VGS − 1) ⇒ VGS = 1.80 = ⎜ ⎟ (9) ⇒ ⎝ 2 ⎠ ⎝ 225 ⎠ ⎝ R1 + R2 ⎠ R2 = 45 k Ω, R1 = 180 k Ω RTH = R1 R2 = 180 45 = 36 k Ω

τ1 =

1 2π f1

=

1 7.96 × 10−4 = 7.958 × 10−4 s = ( RSi + RTH ) CC or CC = ⇒ 2π ( 200 ) ( 200 + 36 ×103 )

CC = 0.022 μ F

τ2 =

1 2π f 2

=

1

2π ( 3 x103 )

= 5.305 × 10−5 s = RD CL or CL =

5.31× 10−5 ⇒ CL = 2.65 nF 20 × 103

7.22 a.

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T (s) = T (s) =

R2 + (1/ sC )

R2 + (1/ sC ) + R1 1 + sR2 C 1 + s ( R1 + R2 ) C

rA = R2 C , rB = ( R1 + R2 ) C

b. 兩T 兩 1

0.0909 fA

fB

f

c. rA = R2 C = (103 )(100 × 10−12 ) = 10−7 s = rA

rB = ( R1 + R2 ) C = [10 + 1] × 103 × 100 × 10−12 = 1.1× 10−6 s = rB

fA ≈

1 1 = ⇒ f A = 1.59 MHz 2π rA 2π (10−7 )

fB ≈

1 1 = ⇒ f B = 0.145 MHz 2π rB 2π (1.1× 10−6 )

7.23 I BQ =

10 − 0.7 = 0.00997 mA 430 + ( 201)( 2.5 )

I CQ = ( 200 ) I BQ = 1.995 mA rπ =

( 200 )( 0.026 )

= 2.61 k Ω 1.99 Rib = 2.61 + ( 201)( 2.5 ) = 505 k Ω

τs =

1

=

2π f L

1 = 0.0106 s 2π (15 )

= Req CC = ( 0.5 + 505 430 ) × 103 CC = 232.7 × 103 CC

Or CC = 4.55 × 10−8 F ⇒ 45.5 nF 7.24 10 = I BQ (300) + 0.7 + ( 201) I BQ (1) ⇒ I BQ = 0.0186 mA ⇒ I CQ = 3.7126 mA rπ =

β VT I CQ

=

( 200 )( 0.026 ) 3.71

= 1.40 K

Ri = rπ + (1 + β ) RE = 1.40 + ( 201)(1) = 202.4 K Req = RS + RB Ri = 0.1 + 300 202.4 = 121 K

τ L = Req ⋅ CC fL = CC =

1 2πτ L

τL Req

⇒τL =

=

1 2π f

=

1 = 7.958 × 10−3 s 2π ( 20 )

7.958 × 10−3 ⇒ CC = 0.0658 μ F 121× 103

7.25

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RTH = R1 R2 = 1.2 1.2 = 0.6 k Ω ⎛ R2 ⎞ ⎛ 1.2 ⎞ VTH = ⎜ ⎟ (VCC ) = ⎜ ⎟ ( 5 ) = 2.5 V ⎝ 1.2 + 1.2 ⎠ ⎝ R1 + R2 ⎠ 2.5 − 0.7 = 0.319 mA I BQ = 0.6 + (101)( 0.05 ) I CQ = 31.9 mA rπ =

(100 )( 0.026 ) 31.9

= 0.0815 k Ω 1

so that f 3− dB ( CC1 ) > τ C and f = C1

C2

f 3− dB ( CC 2 ) = 25 Hz =

1 2πτ 2

, so that τ 2 =

1 = 0.006366 s = Req CC 2 2π ( 25 )

where ⎛ r + R1 R2 RS ⎞ Req = RL + RE ⎜ π ⎟ 1+ β ⎝ ⎠ ⎛ 81.5 + 600 300 ⎞ = 10 + 50 ⎜ ⎟ = 10 + 50 2.787 ⇒ 101 ⎝ ⎠ 0.00637 Req = 12.64 Ω ⇒ CC 2 = ⇒ CC 2 = 504 μ F 12.6 Rib = rπ + (1 + β ) RE Assume CC 2 an open Rib = 81.5 + (101)( 50 ) = 5132 Ω τ 1 = (100 )τ 2 = (100 )( 0.006366 ) = 0.6366 s = Req1CC1 Req1 = RS + RTH Rib = 300 + 600 5132 = 837.2 Ω

So CC1 =

0.6366 ⇒ CC1 = 760 μ F 837.2

7.26 From Problem 7.25 RTH = 0.6 K, I CQ = 31.9 mA, rπ = 81.5 Ω 1 so f 3− dB ( CC 2 )  f 3− dB ( CC1 ) 2πτ Then f 3− dB ( CC 2 ) ⇒ CC1 acts as an open circuit and for f 3− dB ( CC1 ) ⇒ CC 2 acts as a short circuit.

τ C 2  τ C1 and f =

f 3− dB ( CC1 ) = 20 Hz =

1 2πτ C1

⇒ τ C1 = 0.007958 s

Rib = rπ + (1 + β ) ( RE RL ) = 81.5 + (101) ( 50 10 ) = 923.2 Ω

τ C1 ⇒ Req1 = RS + RTH Rib = 300 + 600 923.2 = 663.7 Ω 0.007958 ⇒ CC1 = 12 μ F 663.7 = 100τ C1 = 0.7958 s

CC1 =

τC2

⎛ r + RTH ⎞ Req 2 = RL + RE ⎜ π ⎟ = 10 + 50 ⎝ 1+ β ⎠ Req 2 = 10 + 50 6.748 = 15.95 Ω 0.7958 CC 2 = ⇒ CC 2 = 0.050 F 15.95

⎛ 81.5 + 600 ⎞ ⎜ ⎟ ⎝ 101 ⎠

7.27 a.

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I D = K n (VGS − VTN ) VGS =

2

ID 0.5 + VTN = + 0.8 = 1.8 V Kn 0.5 −VGS − ( −5 )

5 − 1.8 ⇒ RS = 6.4 kΩ 0.5 0.5 VD = VDSQ + VS = 4 − 1.8 = 2.2 V 5 − 2.2 RD = ⇒ RD = 5.6 kΩ 0.5 (b) RS =

gm = 2 Kn I D = 2

=

( 0.5 )( 0.5 ) = 1 mA / V

rA = RS CS = ( 6.4 × 103 )( 5 × 10−6 )

= 3.2 × 10−2 s 1 1 = ⇒ f A = 4.97 Hz fA = 2π rA 2π ( 3.2 × 10−2 ) ⎡ 6.4 × 103 ⎤ ⎞ −6 ⎥ ( 5 × 10 ) ⎟ CS = ⎢ + 1 1 6.4 ( )( ) ⎢⎣ ⎠ ⎦⎥ = 4.32 × 10−3 s 1 1 = ⇒ f B = 36.8 Hz fB = 2π rB 2π ( 4.32 × 10−3 )

⎛ RS rB = ⎜ ⎝ 1 + g m RS

c.

g m RD (1 + sRS CS )

Av =

⎡

⎛

RS 1 + ⎝ g m RS

(1 + g m RS ) ⎢1 + s ⎜

⎞ ⎤ ⎟ CS ⎥ ⎠ ⎦

⎣ As RS becomes large g m RD ( sRS CS ) Av → ⎡ ⎛ R ⎞ ⎤ ( g m RS ) ⎢1 + s ⎜ S ⎟ CS ⎥ ⎝ g m RS ⎠ ⎦ ⎣

⎡ ⎛ 1 ⎞ ⎤ ⎟ CS ⎥ ⎣ ⎝ gm ⎠ ⎦

( g m RD ) ⎢ s ⎜ Av =

⎛ 1 1+ s ⎜ ⎝ gm

⎞ ⎟ CS ⎠

The corner frequency f B = gm = 2 Kn I D = 2 fB =

( 0.5 )( 0.5 ) = 1 mA / V

1 ⇒ f B = 31.8 Hz ⎛ 1 ⎞ −6 2π ⎜ −3 ⎟ ( 5 × 10 ) ⎝ 10 ⎠

7.28 a.

1 and the corresponding f A → 0 2π (1/ g m ) CS

fB =

RE ( RS + rπ ) CE 1 and rB = 2π rB RS + rπ + (1 + β ) RE

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For RS = 0 rB = I EQ =

RE rπ CE rπ + (1 + β ) RE

−0.7 − ( −10 ) 5

= 1.86 mA

β = 75 ⇒ I CQ = 1.84 mA β = 125 ⇒ I CQ = 1.85 mA For f B ≤ 200 Hz ⇒ rB ≥

1 = 0.796 ms 2π ( 200 )

rπ αβ so smallest rB will occur for smallest β .

β = 75; rπ =

( 75)( 0.026 )

= 1.06 kΩ 1.84 ( 5 ×103 ) (1.06 ) CE ⇒ C = 57.2 μ F 0.796 × 10−3 = E 1.06 + ( 76 )( 5 ) b. (125 )( 0.026 ) For β = 125; rπ = = 1.76 kΩ 1.85 5 × 103 ) (1.76 ) ( 57.2 × 10−6 ) ( rB = = 0.797 ms 1.76 + (126 )( 5 ) fB =

1 1 = ⇒ f B = 199.7 Hz Essentially independent of β . 2π rB 2π ( 0.797 × 10−3 )

rA = RE CE = ( 5 × 103 )( 57.2 × 10−6 ) = 0.286 sec fA =

1 1 = ⇒ f A = 0.556 Hz Independent of β . 2π rA 2π ( 0.286 )

7.29 a. Expression for the voltage gain is the same as Equation (7.66) with RS = 0. b. rA = RE CE RE rπ CE rB = rπ + (1 + β ) RE 7.30 5 − 0.7 I C = 0.99 mA = 1 mA = I E 4.3 β VT (100 )( 0.026 ) rπ = = = 2.626 K 0.99 I CQ rA = RE CE = ( 4.3 × 103 )( 5 × 10−6 ) = 2.15 ×10 −2 s rB =

( 4.3 ×103 )( 2.626 ×103 )( 5 ×10−6 ) = 1.292 ×10−4 s RE rπ CE = rπ + (1 + β ) RE 2.626 × 103 + (101) ( 4.3 × 103 )

fA =

1 1 = = 7.40 Hz 2π rA 2π ( 2.15 × 10−2 )

fB =

1 1 = = 1.23 × 103 = 1.23 kHz 2π rB 2π (1.292 × 10−4 )

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Av

w→0

=

Av

w →∞

=

(100 )( 2 ) β RC = = 0.458 rπ + (1 + β ) RE 2.626 + (101)( 4.3) β RC rπ

=

(100 )( 2 ) 2.626

= 76.2

兩A␯兩 76.2

0.458

7.40 Hz

1.23 kHz

f

7.31 rH = ( RL RC ) CL = (10 5 ) × 103 × 15 × 10−12 rH = 5 × 10−8 s 1 1 = ⇒ f H = 3.18 MHz 2π rH 2π ( 5 × 10−8 )

fH =

10 − 0.7 = 0.93 mA, I CQ = 0.921 mA 10 0.921 gm = = 35.4 mA/V 0.026 Av = g m ( RC RL ) = 35.4 ( 5 10 ) ⇒ Av = 118

I EQ =

7.32 ⎛ R2 ⎞ VG = ⎜ ⎟ VDD ⎝ R1 + R2 ⎠

ID =

⎛ 166 ⎞ =⎜ ⎟ (10 ) ⎝ 166 + 234 ⎠ = 4.15 V

VG − VGS 2 = K n (VGS − VTN ) RS

4.15 − VGS = ( 0.5 )( 0.5 ) (VGS2 − 4VGS + 4 ) 0.25VGS2 − 3.15 = 0 ⇒ VGS = 3.55 V g m = 2 K n (VGS − VTN ) = 2 ( 0.5 )( 3.55 − 2 ) g m = 1.55 mA / V 1 1 = 0.5 = 0.5 0.645 1.55 gm

R0 = RS

R0 = 0.282 kΩ r = ( R0 RL ) CL and f H =

βω ≈ f H = 5 MHz ⇒ r = CL =

r R0 RL

=

1 2π r 1

2π ( 5 × 106 )

= 3.18 × 10−8 s

3.18 × 10−8 ⇒ CL = 121 pF ( 0.282 4 ) ×103

7.33

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(a) Low-frequency RS

CC Vo

  

Vs

RB

V␲

RC

r␲



RL

gmV␲

Mid-Band RS Vo

  

Vs

RB

V␲

r␲



RC

RL

gmV␲

High-frequency RS Vo

  

Vs

RB

V␲ 

r␲

RC

RL

CL

gmV␲

(b) 兩Am兩

fL

fH

f

(c) 12 − 0.7 = 11.3 μ A 1 MΩ = 1.13 mA

I BQ = I CQ

rπ =

(100 )( 0.026 )

= 2.3 k Ω 1.13 1.13 gm = = 43.46 mA / V 0.026 Am =

⎛ R R ⎞ V0 ( midband ) = − g m ( RC RL ) ⎜⎜ B π ⎟⎟ Vs ⎝ RB rπ + RS ⎠

⎛ 1000 2.3 ⎞ = − ( 43.46 ) ( 5.1 500 ) ⎜ ⎜ 1000 2.3 + 1 ⎟⎟ ⎝ ⎠ ⎛ 2.29 ⎞ = ( 43.46 )( 5.05 ) ⎜ ⎟ ⇒ Am = 153 ⎝ 2.29 + 1 ⎠ Am dB = 43.7 dB

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fL =

1 2πτ L

, τ L = ( RS + RB rπ ) CC or τ L = (1 + 1000 2.3) × 103 (10 × 10−6 )

⇒ τ L = 3.29 × 10−2 s ⇒ f L = 4.83 Hz fH =

1 , τ H = ( RC RL ) CL ⇒ τ L = ( 5.1 500 ) × 103 (10 × 10−12 ) 2πτ H

= 5.05 × 10−8 s ⇒ f H = 3.15 MHz

7.34 a. V0

 Vi

 

Vsg

RD

gmVsg

RL

CL



⎛ 1 ⎞ V0 = ( g mVsg ) ⎜ R0 RL ⎟ sCL ⎠ ⎝ Vsg = −Vi Av ( s ) =

V0 ( s )

⎛ 1 ⎞ = − g m ⎜ RD RL ⎟ Vi ( s ) sCL ⎠ ⎝

⎡ 1 ⎢ RD RL ⋅ sC L = − gm ⎢ ⎢ 1 ⎢ RD RL + sCL ⎣⎢ Av ( s ) = − g m ( RD RL ) ⋅

b. c.

⎤ ⎥ ⎥ ⎥ ⎥ ⎦⎥

1 1 + s ( RD RL ) CL

r = ( RD RL ) CL

r = (10 20 ) × 103 × 10 × 10−12 ⇒ r = 6.67 × 10 −8 s fH =

1 1 = ⇒ f H = 2.39 MHz 2π r 2π ( 6.67 × 10−8 )

From Example 7.6, gm = 0.705 mA/V Av = g m ( RD RL ) = ( 0.705 ) (10 20 ) ⇒ Av = 4.7 7.35 Computer Analysis 7.36 Computer Analysis 7.37 Computer Analysis 7.38

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fT =

gm

2π ( Cπ + Cμ ) I CQ

gm = fT =

1 = 38.46 mA/V 0.026

=

VT

38.46 × 10−3 2π (10 + 2 ) × 10−12

fT = 510 MHz fT

fβ =

β

=

510 ⇒ f β = 4.25 MHz 120

=

5000 MHz ⇒ f β = 33.3 MHz 150 gm

7.39 fT

fβ = fT =

β

2π ( Cπ + Cμ )

0.5 = 19.23 mA/V 0.026 19.2 × 10 −3 5 × 109 = 2π ( Cπ + 0.15 ) × 10−12 gm =

Cπ + 0.15 =

19.2 × 10−3

2π (10−12 )( 5 × 109 )

= 0.612 pF

Cπ = 0.462 pF

7.40 fβ =

a.

fT

β

=

2000 MHz = 13.3 MHz = f β 150

b. h fe =

150 1 + j ( f / fβ )

h fe =

150 1 + ( f / fβ )

2

= 10

2

⎛ f ⎞ ⎛ 150 ⎞ 2 1+ ⎜ ⎟ = ⎜ = 225 ⎜ f ⎟ ⎝ 10 ⎟⎠ ⎝ β⎠ f = f β ⋅ 224 = (13.33) 224 ⇒ f = 199.6 MHz

7.41 (a) V0 = − g mVπ RL where 1 sC1

rπ 1 + srπ C1 Vπ = ⋅ Vi = ⋅ Vi rπ 1 + rb rπ + rb 1 + srπ C1 sC1 rπ

=

⎞ ⎛ r ⎞⎛ rπ 1 ⋅ Vi = ⎜ π ⎟ ⎜ ⎟ ⋅ Vi rπ + rb + srb rπ C1 ⎝ rπ + rb ⎠ ⎜⎝ 1 + s ( rb rπ ) C1 ⎟⎠

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So Av ( s ) =

V0 ( s )

⎞ ⎛ r ⎞⎛ 1 = − g m RL ⎜ π ⎟ ⎜ ⎟ ⎜ Vi ( s ) ⎝ rπ + rb ⎠ ⎝ 1 + s ( rb rπ ) C1 ⎟⎠

(100 )( 0.026 )

(b) Midband gain: rπ =

1

= 2.6 k Ω, g m =

1 = 38.46 mA / V 0.026

(i)

For rb = 100 Ω

(ii)

⎛ 2.6 ⎞ Av1 = − ( 38.46 )( 4 ) ⎜ ⎟ ⇒ Av1 = −148.1 ⎝ 2.6 + 0.1 ⎠ For rb = 500 Ω

(c) (i)

⎛ 2.6 ⎞ Av 2 = − ( 38.46 )( 4 ) ⎜ ⎟ ⇒ Av 2 = −129.0 ⎝ 2.6 + 0.5 ⎠ 1 , τ = ( rb rπ ) C1 f 3− dB = 2πτ For rb = 100 Ω

τ 1 = ( 0.1 2.6 ) × 103 ( 2.2 ×10−12 ) = 2.12 × 10−10 s ⇒ f 3− db = 751 MHz

(ii)

For rb = 500 Ω

τ 2 = ( 0.5 2.6 ) × 103 ( 2.2 ×10−12 ) = 9.23 ×10−10 s f 3− dB = 173 MHz

7.42 (b)

f = 10 kHz = 104 Z i = 200 +

(

2500 1 − j (104 )(1.333 × 10−6 ) 1 + (10

) (1.333 ×10 )

4 2

)

−6 2

= 200 + 2500 − j 33.3 = 2700 − j 33.3 f = 100 kHz = 105

(c)

Z i = 200 +

(

2500 1 − j (105 )(1.333 × 10−6 ) 1 + (10

) (1.333 ×10 )

5 2

)

−6 2

Z i = 200 + 2456 − j 327 = 2656 − j 327 f = 1 MHz = 106

(d)

Z i = 200 +

(

2500 1 − j (106 )(1.333 × 10−6 ) 1 + (10

) (1.333 ×10 )

6 2

)

−6 2

Z i = 200 + 900 − j1200 = 1100 − j1200

7.43 a.

CM = Cμ (1 + g m RL )

b. RB兩兩RS

rb V0 

RB • Vi RB  RS

 

V␲ 

r␲

C␲

CM

gmV␲

RL

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V0 = − g mVπ RL rπ

Vπ =

Let Cπ + CM = Ci 1 sCi

⎛ RB ⎞ ⋅⎜ ⎟ Vi R + RS ⎠ 1 rπ + RB RS + rb ⎝ B sC1

Av ( s ) =

Vo ( s ) Vi ( s )

1 ⎡ ⎤ rπ ⋅ ⎢ ⎥ sCi ⎢ ⎥ 1 ⎢ ⎥ rπ + ⎥ ⎛ RB ⎞ ⎢ sCi = − g m RL ⎜ ⎥ ⎟⎢ ⎝ RB + RS ⎠ ⎢ rπ ⋅ 1 ⎥ ⎢ ⎥ sCi + RB & RS + rb ⎥ ⎢ 1 ⎢ rπ + ⎥ sCi ⎢⎣ ⎥⎦ ⎤ ⎛ RB ⎞ ⎡ rπ = − g m RL ⎜ ⎥ ⎟× ⎢ ⎝ RB + RS ⎠ ⎢⎣ rπ + (1 + srπ Ci ) ( RB RS + rb ) ⎥⎦ Let Req = ( RB RS + rb )

⎛ RB Av ( s ) = − β RL ⎜ ⎝ RB + RS Av ( s ) =

c.

⎡ ⎤ ⎞ ⎢ 1 ⎥ × ⎟ ⎢ ⎥ ⎡ ⎤ ⎠ ⎢ ( rπ + Req ) 1 + s rπ Req Ci ⎥ ⎣ ⎦⎦ ⎣

(

)

− β RL ⎛ RB ⎞ 1 ⋅⎜ ⎟⋅ rπ + Req ⎝ RB + RS ⎠ 1 + s rπ Req Ci

(

fH =

(

1

)

)

2π rπ Req Ci

7.44 High Freq. ⇒ CC1 , CC 2 , CE → short circuits C␮ V0  IS

R1兩兩R2 V␲

r␲

C␲

gmV␲

RC

RL I0



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gm = fT =

I CQ VT

=

5 = 192.3 mA/V 0.026

gm

2π ( Cπ + Cμ )

⇒ 250 × 106 =

192 × 10−3 2π ( Cπ + Cμ )

Cπ + Cμ = 122.4 pF ⇒ Cμ = 5 pF, Cπ = 117.4 pF

(

CM = Cμ 1 + g m ( RC RL )

)

= 5 ⎡⎣1 + (192.3) (1 1) ⎤⎦ ⇒ CM = 485.8 pF Ci = Cπ + CM = 117 + 485 = 603 pF rπ =

( 200 )( 0.026 )

= 1.04 kΩ

5

Req = R1 R2 rπ = 5 1.04 = 0.861 kΩ r = Req ⋅ Ci

= ( 0.861× 103 )( 603 × 10−12 ) = 5.19 × 10 −7 s

1

f =

2π r

=

1

2π ( 5.19 × 10−7 )

⇒ f = 307 kHz

7.45 RTH = R1 R2 = 60 5.5 = 5.04 kΩ ⎛ R2 ⎞ ⎛ 5.5 ⎞ VTH = ⎜ ⎟ VCC = ⎜ ⎟ (15 ) = 1.26 V ⎝ 5.5 + 60 ⎠ ⎝ R1 + R2 ⎠ 1.26 − 0.7 I BQ = = 0.0222 mA 5.04 + (101)( 0.2 ) I CQ = 2.22 mA rπ =

(100 )( 0.026 )

= 1.17 kΩ 2.22 2.22 gm = = 85.4 mA/V 0.026 Lower 3 – dB frequency: rL = Req ⋅ CC1

Req = RS + R1 R2 rπ = 2 + 60 5.5 1.17 = 2.95 k Ω rL = ( 2.95 ×103 )( 0.1× 10−6 ) = 2.95 × 10−4 s fL =

1 1 = ⇒ f L = 540 Hz 2π rL 2π ( 2.95 × 10−4 )

Upper 3 – dB frequency:

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gm

fT =

2π ( Cπ + Cμ )

⇒ 400 × 106 =

85.4 × 10−3 2π ( Cπ + Cμ )

Cπ + Cμ = 34 pF; Cμ = 2 pF ⇒ Cπ = 32 pF CM = Cμ (1 + g m RC ) = 2 (1 + ( 85.4 )( 4 ) ) ⇒ CM = 685 pF Ci = Cπ + CM = 32 + 685 = 717 pF Req = RS R1 R2 rπ

= 2 60 5.5 1.17 = 0.644 kΩ

= ( 0.644 × 103 )( 717 × 10−12 )

r = Req ⋅ Ci

= 4.62 × 10−7 s fH =

1 2π r

⇒ f H = 344 kHz

7.46 RTH = R1 R2 = 600 55 = 50.38 K ⎛ R2 ⎞ ⎛ 55 ⎞ VTH = ⎜ ⎟ (15 ) = ⎜ ⎟ (15 ) = 1.2595 V ⎝ 600 + 55 ⎠ ⎝ R1 + R2 ⎠ 1.26 − 0.7 I BQ = = 0.00222 mA 50.4 + (101)( 2 ) I CQ = 0.2217 mA rπ =

(100 )( 0.026 )

= 11.73 K 0.222 0.2217 gm = = 8.527 mA/V 0.026 Lower – 3dB Fig. τ L = R e q1 Cc1 ; Req1 = RS + RTH r π = 0.50 + 50.38 11.73 = 10.0 K

τ L = (10 × 10

3

)( 0.1×10 ) = 10 −6

−3

s

τ L = R e q1 Cc1 ; Req1 = RS + RTH r π = 0.50 + 50.38 11.73 = 10.0 K

τ L = (10 × 103 )( 0.1×10−6 ) = 10−3 s 1

fL =

2πτ 2

=

1

2π (10−3 )

⇒ f L = 159 Hz

Upper – 3dB Fig. gm 8.527 × 10−3 = = 400 × 106 fT = −12 2π ( Cπ + Cμ ) 2π ( Cπ + 2 ) × 10 Cπ + Cμ = 3.393 pF ⇒ Cπ = 1.393 pF CM = Cμ (1 + g m RC ) = 2 ⎡⎣1 + ( 8.527 )( 40 ) ⎤⎦ = 684 pF CT = Cπ + CM = 1.393 + 684 = 685.4 pF Req 2 = RS RTH rπ = 0.5 50.38 11.73 = 50.38 0.480 = 0.4750 K

τ H = R eq 2 .CT fH =

= ( 0.4750 × 103 ) ( 685.4 × 10−12 ) = 3.256 × 10−7 s 1

1 = ⇒ f H = 489 KHz 2πτ H 2π ( 3.256 × 10−7 )

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7.47 fT =

gm

2π ( Cgs + Cgd )

⎛ 40 ⎞ g m = 2 K n I D , K n = (15 ) ⎜ ⎟ = 60 μ A / V 2 ⎝ 10 ⎠ gm = 2 fT =

( 60 )(100 ) = 154.9 μ A / V

154.9 × 10−6 ⇒ fT = 44.8 MHz 2π ( 0.5 + 0.05 ) × 10−12

7.48 fT =

gm

2π ( Cgs + Cgd )

g m = 2 K n (VGs − VT )

ID then g m = 2 K n I D Kn

I D = K n (VGs − VTN ) or VGs − VTN = 2

So fT =

2 Kn I D

2π ( Cgs + Cgd )

=

Kn I D

π ( Cgs + Cgd )

( 0.2 ×10 )( 20 ×10 ) −3

(a)

fT =

(b)

fT =

(c)

10 =

−6

π ( 0.5 + 0.1) × 10−12

⇒ fT = 33.6 MHz

( 0.2 ×10 )( 250 ×10 ) −3

−6

π ( 0.5 + 0.1) × 10−12

( 0.2 ×10 ) I

⇒ fT = 118.6 MHz

−3

9

D

π ( 0.5 + 0.1) × 10−12

⇒ I D = 17.8 mA

7.49 fT =

gm

2π ( Cgs + Cgd )

Cgs + Cgd = WLCox ⎛W ⎞⎛ μ C ⎞ g m = 2 K n (VGS − VTN ) = 2 ⎜ ⎟ ⎜ n ox ⎟ (VGS − VTN ) ⎝ L ⎠⎝ 2 ⎠ ⎛W ⎞ ⎜ ⎟ ( μ n Cox )(VGS − VTN ) L Then fT = ⎝ ⎠ 2π WLCox fT =

μ n (VGS − VTN ) 2π L2

(a)

fT =

(b)

fT =

450 ( 0.5 )

2π (1.2 × 10−4 )

450 ( 0.5 )

2

2π ( 0.18 × 10−4 )

⇒ fT = 2.49 GHz

⇒ fT = 111 GHz

7.50 a.

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(

CM = Cgd ′ 1 + g m ( ro RD )

)

CM = 5 ⎡⎣1 + ( 3) (15 10 ) ⎤⎦ ⇒ CM = 95 pF b. r = ( ri ) ( Cgs + CM )

r = (10 × 103 ) ( 50 + 95 ) × 10−12 = 1.45 × 10−6 s

f =

1 1 = ⇒ f = 110 kHz 2π r 2π (1.45 × 10−6 )

7.51 fT =

gm

2π ( CgsT + CgdT )

( Eq. 7.104 )

⎛ 2⎞ Let CgdT = 0 and CgsT = ⎜ ⎟ (WLCox ) ⎝ 3⎠ ⎛μ C g m = 2 K n I D = 2 ⎜ n ox ⎝ 2

⎞ ⎡W ⎤ ⎟ ⎢ L ⎥ ID ⎠⎣ ⎦

⎛1 ⎞⎛ W ⎞ 2 ⎜ μ n Cox ⎟⎜ ⎟ I D ⎝2 ⎠⎝ L ⎠ So fT = ⎛ 2⎞ 2π ⎜ ⎟ (W LCox ) ⎝ 3⎠ ⎛1 ⎞⎛ W ⎞ ⎜ μ n Cox ⎟⎜ ⎟ I D 2 3 ⎝ ⎠⎝ L ⎠ = ⋅ 2π L W Cox fT =

3 2π L

⋅

μn I D 2W Cox L

7.52 (a) g m′ =

gm 1 + g m rS

g m = 2 K n (VGS − VTN )

−8 ⎛ μ C ⎞ ⎛ W ⎞ ⎛ ( 400 ) ( 7.25 × 10 ) ⎞⎟ K n = ⎜ n ox ⎟ ⎜ ⎟ = ⎜ (10 ) ⎟ 2 ⎝ 2 ⎠ ⎝ L ⎠ ⎝⎜ ⎠

K n = 1.45 × 10−4 A / V 2

For VGS = 5 V

g m = 2 (1.45 × 10 −4 ) ( 5 − 0.65 ) = 1.26 × 10−3 A/V g m′ = ( 0.80 ) g m = 1.01× 10−3 A/V

1.01× 10−3 =

1.26 × 10−3

1 + (1.26 × 10−3 ) rS

1 + (1.26 × 10−3 ) rS = 1.25 ⇒ rS = 196 Ω

b.

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For VGS = 3 V

g m = 2 (1.45 × 10−4 ) ( 3 − 0.65 ) = 0.6815 × 10−3 A/V g m′ =

0.6815 × 10−3

1 + ( 0.6815 × 10−3 ) (196 )

g m′ = 0.60 × 10−3 A/V Re duced by ≈ 12%

7.53 a.  Vgs Ii

Ri

gmVgs



RL

I0

rS

I 0 = g mVgs and Vgs = I i Ri − g mVgs rS so Vgs = Then Ai = b.

I i Ri 1 + g m rS

I0 g R = m i I i 1 + g m rS

As an approximation, consider 

Ii

Vgs

Ri

CgsT

I0

CM

RL

gmVgs



In this case I gm 1 Ai = 0 = g m′ Ri ⋅ where CM = C gdT (1 + g m′ RL ) and g m′ = Ii 1 + g m rs 1 + sRi ( C gsT + CM ) c. As rS increases, CM decreases, so the bandwidth increases, but the current gain magnitude decreases. 7.54 ⎛ R2 ⎞ ⎛ 225 ⎞ VGS = ⎜ ⎟ VDD = ⎜ ⎟ (10 ) ⎝ 225 + 500 ⎠ ⎝ R1 + R2 ⎠ VGS = 3.10 V

g m = 2 K n (VGS − VTN ) = 2 (1)( 3.10 − 2 ) g m = 2.207 mA / V a.

CgdT

Ri

V0

Vi

 

 R1兩兩R2

Vgs

CgsT

gmVgs

RD



b.

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CM = C gdT (1 + g m RD ) = (1) ⎡⎣1 + ( 2.207 )( 5 ) ⎤⎦ CM = 12 pF

c. r = ( Ri R1 R2 ) ( CgsT + CM ) Ri R1 R2 = 1 500 225 = 1 155 = 0.9936 kΩ

r = ( 0.9936 × 103 ) ( 5 + 12 ) × 10−12 = 1.69 × 10 −8 s fH = Av =

1 2π r

=

1

2π (1.69 × 10−8 )

− g mVgs RD Vi

and Vgs =

⇒ f H = 9.42 MHz R1 R2

R1 R2 + Ri

⋅ Vi =

155 ⋅ Vi = 0.994 Vi 155 + 1

Av = − ( 2.2 )( 5 )( 0.994 ) ⇒ Av = −10.9

7.55 RTH = R1 R2 = 33 22 = 13.2 kΩ ⎛ R2 ⎞ ⎛ 22 ⎞ VTH = ⎜ ⎟ ( 5) = ⎜ ⎟ ( 5) = 2 V ⎝ 22 + 33 ⎠ ⎝ R1 + R2 ⎠ 2 − 0.7 I BQ = = 0.00261 mA 13.2 + (121)( 4 ) I CQ = 0.3138 rπ =

(120 )( 0.026 )

= 9.94 kΩ 0.3138 0.3138 gm = = 12.07 mA/V 0.026 100 r0 = = 318 kΩ 0.3138 a. gm fT = 2π ( Cπ + Cμ )

Cπ + Cμ =

gm 12.07 × 10−3 = 2π fT 2π ( 600 × 106 )

Cπ + Cμ = 3.20 pF; Cμ = 1 pF ⇒ Cπ = 2.20 pF

(

)

CM = Cμ ⎡1 + g m ro RC RL ⎤ ⎣ ⎦ = (1) ⎡1 + (12.07 ) 318 4 5 ⎤ ⎣ ⎦ CM = 27.6 pF b.

(

)

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r = Req ( Cπ + CM ) Req = R1 R2 Rs rπ = 33 22 2 rπ = 1.74 9.94 kΩ ⇒ Req = 1.48 kΩ r = (1.48 × 103 ) ( 2.20 + 27.6 ) × 10−12 r = 4.41× 10 −8 s fH =

1 2π r

1

=

(

2π ( 4.41× 10−8 )

V0 = − g mVπ ro RC RL

⇒ f H = 3.61 MHz

)

x

⎛ R1 R2 rπ ⎞ Vπ = ⎜ ⎟V ⎜ R1 R2 rπ + RS ⎟ i ⎝ ⎠ R1 R2 rπ = 33 22 9.94 = 5.67 kΩ ⎛ 5.67 ⎞ vπ = ⎜ ⎟ Vi = 0.739Vi ⎝ 5.67 + 2 ⎠ r0 RC RL = 318 4 5 = 2.18 kΩ Av = − (12.07 )( 0.739 )( 2.18 ) Av = −19.7

7.56 RTH = R1 R2 = 40 5 = 4.44 kΩ ⎛ R2 ⎞ ⎛ 5 ⎞ VTH = ⎜ ⎟ VCC = ⎜ ⎟ (10 ) = 1.111 V ⎝ 5 + 40 ⎠ ⎝ R1 + R2 ⎠ 1.111 − 0.7 = 0.00633 mA I BQ = 4.44 + (121)( 0.5 ) I CQ = 0.760 mA rπ =

(120 )( 0.026 )

= 4.11 kΩ 0.760 0.760 = 29.23 mA/V gm = 0.026 r0 = ∞ fT =

gm

2π ( Cπ + Cμ )

Cπ + Cμ =

gm 29.23 × 10−3 = 2π fT 2π ( 250 × 106 )

Cπ + Cμ = 18.6 pF; Cμ = 3 pF ⇒ Cπ = 15.6 pF

a. CM = Cμ ⎡⎣1 + g m ( RC RL ) ⎤⎦ CM = 3 ⎡⎣1 + ( 29.2 ) ( 5 2.5 ) ⎤⎦ ⇒ CM = 149 pF For upper frequency:

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rH = Req ( Cπ + CM ) Req = rπ R1 R2 RS = 4.11 40 5 0.5 Req = 0.405 kΩ rH = ( 0.405 × 103 ) (15.6 + 149 ) × 10−12 = 6.67 ×10 −8 s 1 fH = ⇒ f H = 2.39 MHz 2π rH For lower frequency: rL = Req CC1 Req = RS + R1 R2 rπ = 0.5 + 40 5 4.11 Req = 2.64 kΩ

rL = ( 2.64 × 103 )( 4.7 × 10−6 ) = 1.24 × 10−2 s

fL =

1 ⇒ f L = 12.8 Hz 2π rL

b. 兩A␯兩 39.5

fL

fH

V0 = − g mVπ ( RC RL )

f

⎛ R1 R2 rπ ⎞ Vπ = ⎜ ⎟V ⎜ R1 R2 rπ + RS ⎟ i ⎝ ⎠ ⎛ 2.135 ⎞ Vπ = ⎜ ⎟ Vi = 0.8102Vi ⎝ 2.135 + 0.5 ⎠ AV = ( 29.23)( 0.8102 ) ( 5 2.5 ) AV = 39.5

7.57 I D = K P (VSG + VTP ) = 2

9 − VSG RS

( 2 )(1.2 ) (VSG2 − 4VSG + 4 ) = 9 − VSG 2 2.4VSG − 8.6VSG + 0.6 = 0

VSG =

8.6 ±

(8.6 ) − 4 ( 2.4 )( 0.6 ) 2 ( 2.4 ) 2

VSG = 3.512 V g m = 2 K P (VSG + VTP ) = 2 ( 2 )( 3.512 − 2 ) g m = 6.049 mA / V I D = ( 2 )( 3.512 − 2 ) = 4.572 mA 1 1 ⇒ r0 = 21.9 kΩ r0 = = λ I o ( 0.01)( 4.56 ) 2

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(

CM = CgdT 1 + g m ( ro RD )

a.

)

CM = (1) ⎡⎣1 + ( 6.04 ) ( 21.9 1) ⎤⎦ ⇒ CM = 6.785 pF

b. rH = ( Ri RG ) ( CgsT + CM )

rH = ( 2 100 ) × 103 (10 + 6.78 ) × 10−12 rH = 3.29 × 10−8 s 1

fH =

→ f H = 4.84 MHz

2πτ H

V0 = − g m ( ro RD ) ⋅ Vgs ⎛ RG ⎞ ⎛ 100 ⎞ Vgs = ⎜ ⎟ Vi = ⎜ ⎟ Vi ⎝ 102 ⎠ ⎝ RG + Ri ⎠ ⎛ 100 ⎞ Av = − ( 6.04 ) ⎜ ⎟ ( 21.9 1) ⎝ 102 ⎠ Av = −5.67

7.58 ⎛ R2 ⎞ ⎛ 22 ⎞ VG = ⎜ ⎟ ( 20 ) − 10 = ⎜ ⎟ ( 20 ) − 10 + R R ⎝ 22 + 8 ⎠ 2 ⎠ ⎝ 1 VG = 4.67 V

ID =

10 − VSG − 4.67 2 = K P (VSG + VTP ) RS

2 − 4VSG + 4 ) 5.33 − VSG = (1)( 0.5 ) (VSG 2 − VSG − 3.33 = 0 0.5VSG

VSG =

1 ± 1 + 4 ( 0.5 )( 3.33) 2 ( 0.5 )

⇒ VSG = 3.77 V

g m = 2 K p (VSG + VTP ) = 2 (1)( 3.77 − 2 ) g m = 3.54 mA / V b.

(

CM = CgdT 1 + g m ( RD RL )

)

CM = ( 3) ⎡⎣1 + ( 3.54 ) ( 2 5 ) ⎤⎦ ⇒ CM = 18.2 pF a. r = Req ( CgsT + CM ) Req = Ri R1 R2 = 0.5 8 22 = 0.461 kΩ r = ( 0.461× 103 ) (15 + 18.2 ) × 10−12 = 1.53 × 10−8 s fH =

1 2π r

⇒ f H = 10.4 MHz

c. V0 = − g mVgs ( RD RL ) ⎛ R R ⎞ ⎛ 5.87 ⎞ Vgs = ⎜ 1 2 ⎟ Vi = ⎜ ⎟ Vi ⇒ Vgs = ( 0.9215 ) Vi ⎜ R1 R2 Ri ⎟ ⎝ 5.87 + 0.5 ⎠ ⎝ ⎠ Av = − ( 3.54 )( 0.9215 ) ( 2 5 ) ⇒ Av = −4.66

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7.59 ⎛ 100 ⎞ I E = 0.5 mA ⇒ I CQ = ⎜ ⎟ ( 0.5 ) = 0.495 mA ⎝ 101 ⎠ 0.495 = 19.0 mA/V gm = 0.026 (100 )( 0.026 ) = 5.25 kΩ rπ = 0.495 a. Input: From Eq. 7.107b ⎡ rπ ⎤ rPπ = ⎢ RE RS ⎥ Cπ ⎣1 + β ⎦ ⎡ 5.25 ⎤ 0.5 0.05⎥ × 103 × 10 × 10−12 =⎢ ⎣ 101 ⎦ = 2.43 × 10−10 s

1 ⇒ f H π = 656 MHz 2π rpπ Output: From Eq. 7.108b rPμ = ( RB RL ) Cμ = (100 1) × 103 × 10−12 f Hπ =

= 9.90 × 10−10 s fH μ =

1 ⇒ f H μ = 161 MHz 2π rPμ

b. RS

Vi

gmV␲

V0



 

RE

V␲

r␲

RB

RL



V0 = − g m Vπ ( RB & RL ) g mVπ +

Vπ Vπ Vi − ( −Vπ ) + + =0 rπ RE RS

⎡ 1 1 1 ⎤ −V + ⎥= i Vπ ⎢ g m + + rπ RE RS ⎦ RS ⎣ 1 1 1 ⎤ −Vi ⎡ + + = Vπ ⎢19 + 5.25 0.5 0.05 ⎥⎦ 0.05 ⎣ Vπ ( 41.19 ) = −Vi ( 20 )

Vπ = − ( 0.4856 ) Vi

V0 = − (19 )( −0.4856 )(100 & 1) Vi Av = 9.14

c. r = CL ( RL & RB ) = (15 × 10−12 ) (1 & 100 ) × 103 r = 1.485 × 10−8 s 1 → f = 10.7 MHz 2π r Since f < f H μ ⇒ 3d B freq. dominated by CL . f =

7.60

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20 − 0.7 = 1.93 mA 10 ⎛ 100 ⎞ I CQ = ⎜ ⎟ (1.93) = 1.91 mA ⎝ 101 ⎠ 1.91 gm = = 73.5 mA/V 0.026 (100 )( 0.026 ) rπ = = 1.36 kΩ 1.91 a. Input: ⎡ rπ ⎤ rPπ = ⎢ RE RS ⎥ ⋅ Cπ ⎣1 + β ⎦ I EQ =

rPπ

⎡1.36 ⎤ 10 1⎥ × 103 × 10 × 10−12 =⎢ ⎣ 101 ⎦ = 1.327 × 10−10 s

1 ⇒ f Pπ = 1.20 GHz 2π rPπ Output: rPπ = ( RC RL ) Cμ = ( 6.5 & 5 ) × 103 × 10−12 f Pπ =

rPπ = 2.826 × 10−9 s 1 f Pμ = → f Pμ = 56.3 MHz 2π rPπ b. V0 = − g m Vπ ( RC & RL ) g mVπ +

Vπ Vπ Vi − ( −Vπ ) + + =0 rπ RE RS

⎛ Vi 1 1 1 ⎞ + Vπ ⎜ g m + + ⎟=− r R R R π E S ⎠ S ⎝ 1 1 1 ⎤ −V ⎡ + + ⎥= i Vπ ⎢ 73.5 + 1.36 10 1 ⎦ (1) ⎣

Vπ ( 75.34 ) = −Vi ⇒ Vπ = − ( 0.01327 ) Vi V0 = − ( 73.5 )( −0.01327 ) ( 6.5 5 ) Vi Av = 2.76 c.

r = CL ( RL Rc ) = (15 × 10−12 ) ( 6.5 5 ) × 103 r = 4.24 × 10−8 s

f =

1

→ f = 3.75 MHz 2π r Since f < fp μ , 3dB frequency is dominated by CL.

7.61

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VGS + I D RS = 5 5 − VGS 2 ID = = K n (VGS − VTN ) RS

5 − VGS = ( 3)(10 ) (V 2GS − 2VGS + 1) 30V 2GS − 59VGS + 25 = 0

VGS =

59 ±

( 59 )

− 4 ( 30 )( 25 )

2

2 ( 30 )

⇒ VGS = 1.349 V

g m = 2 K n (VGS − VTN ) = 2 ( 3)(1.35 − 1) g m = 2.093 mA / V On the output: rPμ = ( RD RL ) Cgd T = ( 5 4 ) × 103 × 4 × 10−12 rPμ = 8.89 × 10−9 s f Pμ =

1 → f Pμ = 17.9 MHz 2π rPμ Ri

Vi

gmVgs

V0



 

RS

Vgs

RD

RL



V0 = − g mVgs ( RD RL ) g mVgs +

Vgs RS

+

Vi − ( −Vgs ) RS

=0

⎛ V 1 1⎞ + ⎟=− i Vgs ⎜ g m + RS Ri ⎠ Ri ⎝ V 1 1⎞ ⎛ Vgs ⎜ 2.093 + + ⎟ = − i 10 2 2 ⎝ ⎠ Vgs = ( 0.1857 )Vi

Av =

V0 = ( 2.093)( 0.1857 ) ( 5 4 ) Vi

Av = 0.864

7.62 dc analysis ID =

V + − VSG 2 = K P (VSG + VTP ) RS

5 − VSG = (1)( 4 )(VSG − 0.8 )

2

2 = 4 (VSG − 1.6VSG + 0.64 )

2 4VSG − 5.4VSG − 2.44 = 0

VSG =

5.4 ±

( 5.4 )

2

+ 4 ( 4 )( 2.44 )

2 ( 4)

= 1.707

g m = 2 K P (VSG + VTP ) = 2 (1)(1.707 − 0.8 ) g m = 1.81 mA / V

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Ri

gmVgs  RS

Vgs

CgsT

CgdT

RD

RL



3 ⋅ dB frequency due to CgsT : Req =

1 RS Ri gm

1 2π Req ⋅ CgsT

fA =

1 4 0.5 = 0.246 kΩ 1.81 1 fA = = 162 MHz 2π ( 246 ) ( 4 × 10−12 )

Req =

3 − dB frequency due to CgdT 1 2π ( RD RL ) CgdT

fB = =

1 2π ( 2 4 ) × 103 × 10−12

f = 119 MHz

Midband gain Ri

Vi

gmVgs

V0



 

Vgs

RS

RD

RL



1 RS gm

1 4 1.81 Vgs = ⋅ Vi = ⋅ Vi 1 1 RS + R i 4 + 0.5 gm 1.81 −

−

= −0.492Vi

V0 = − g mVgs ( RD RL )

Av = ( 0.492 )(1.81) ( 4 2 ) ⇒ Av = 1.19

7.63 rπ =

(120 )( 0.026 )

1.02 g m = 39.23 mA/V a.

= 3.059 kΩ

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Input: f H π =

1 2π rπ

rπ = ⎡ Rs R2 R3 rπ ⎤ ( Cπ + 2Cμ ) ⎣ ⎦ Req = 0.1 20.5 28.3 3.06 = 0.096 kΩ rπ = ( 96 ) (12 + 2 ( 2 ) ) × 10−12 = 1.537 × 10−9 s f Hπ =

1

2π (1.536 × 10−9 )

Output: f H μ =

= 103.6 MHz

1 2π rμ

rμ = ( RC RL ) Cμ

= (15 10 ) × 103 × 2 × 10−12 = 6.67 × 10−9 fH μ =

1

2π ( 6.67 × 10−9 )

= 23.9 MHz

b. ⎡ R2 & R3 & rπ ⎤ A = g m ( RC & RL ) ⎢ ⎥ ⎣ R2 & R3 & rπ + RS ⎦ R2 R3 rπ = 20.5 28.3 3.059 = 2.433 kΩ ⎡ 2.433 ⎤ A = ( 39.23) ( 5 10 ) ⎢ ⎥ ⇒ A = 125.6 ⎣ 2.433 + 0.1 ⎦ c. CL = 15 pF > Cμ ⇒ CL dominates frequency response.

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Chapter 8 Exercise Solutions EX8.1

VCC = 30 V, VCE = 30 − I C RC , I CVCE = 10 1 Maximum power at VCE = VCC = 15 2 10 10 2 IC = = = VCE 15 3

a.

2 So 15 = 30 − RL ⇒ RL = 22.5 Ω ⇒ Maximum Power = 10 W 3 b. VCC = 15 V, I C ,max = 2 A VCE = 15 − I C RL

0 = 15 − 2 RL ⇒ RL = 7.5 Ω ⇒ Maximum Power = (1)( 7.5 ) = 7.5 W

EX8.2 (a)

(b)

ΔT = Pθ = ( 8 )( 2.4 ) ΔT = 19.2° C ΔT = Pθ ΔT 85 = P= θ 3.7 P = 23.0 W

EX8.3 Power = iD ⋅ vDS = (1)(12 ) = 12 watts

c.

Tsink = Tamb + P ⋅θ snk − amb

Tsink = 25 + (12 )( 4 ) ⇒ Tsink = 73°C

b.

Tcase = Tsink + P ⋅θ case − snk Tcase = 73 + (12 )(1) ⇒ Tcase = 85°C

a.

Tdev = Tcase + P ⋅θ dev − case

Tdev = 85 + (12 )( 3) ⇒ Tdev = 121°C

EX8.4

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θ dev − case = PD ,max =

TJ ,max − Tamb PD,rated

=

200 − 25 = 3.5°C/W 50

TJ ,max − Tamb

θ dev − case + θ case −snk + θ snk − amb

200 − 25 ⇒ PD ,max = 29.2 W 3.5 + 0.5 + 2 = Tamb + PD ,max (θ case −snk + θ snk − amb ) = 25 + ( 29.2 )( 0.5 + 2 ) ⇒ Tcase = 98°C

= Tcase

EX8.5 I DQ =

a.

10 − 4 ⇒ I DQ = 60 mA 0.1

b.

⎛9⎞ vds = − ⎜ ⎟ ( 60 )( 0.050 ) = −2.7 V ⇒ vDS ( min ) = 4 − 2.7 = 1.3 V ⎝ 10 ⎠ So maximum swing is determined by drain-to-source voltage. VPP = 2 × ( 2.5 ) = 5.0 V

c. 1 VP2 1 ( 2.5 ) ⋅ = ⋅ ⇒ PL = 31.25 mW 2 RL 2 0.1 2

PL =

PS = VDD ⋅ I DQ = (10 )( 60 ) = 600 mW

η=

PL PS

=

31.25 ⇒ η = 5.2% 600

EX8.6 Computer Analysis EX8.7 No Exercise Problem EX8.8 a. vI = v0 + vGSn −

VBB 2

dv dvI = 1 + GSn dv0 dv0 iDn = K n ( vGSn − VTN ) vGSn =

2

iDn + VTN Kn

dvGSn dvGSn diDn = ⋅ dvo diDn dvo

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So

dvGSn = diDn

1

1 1 ⋅ ⋅ K n 2 iDn

At v0 = 0, iDn = 0.050 A So

iDn

dvGSn 1 1 1 = ⋅ ⋅ =5 diDn 2 0.2 0.050 = iL + iDp

For a small change in v0 → Δ iL = Δ iDn − ( −Δ iDp ) 1 Δ iL 2 di 1 di 1 1 1 1 = ⋅ = 0.025 or Dn = ⋅ L = ⋅ dv0 2 dv0 2 RL 2 20

So Δ iDn =

Then

dvGSn = ( 5 )( 0.025 ) = 0.125 dv0

Then

dv dvI 1 = 1 + 0.125 = 1.125 and Av = 0 = ⇒ Av = 0.889 dv0 dvI 1.125

For v0 = 5 V, iL = 0.25 A = iDn , and iDp = 0

b.

dvGSn dvGSn diDn = ⋅ dv0 diDn dv0 dvGSn = diDn

1

1 1 1 1 1 ⋅ ⋅ = ⋅ ⋅ = 2.24 0.2 2 0.25 K n 2 iDn

diDn diL 1 = = = 0.05 dv0 dv0 20 dvGSn = ( 2.24 )( 0.05 ) = 0.112 dv0 dvI = 1 + 0.112 = 1.112 dv0 Av =

dv0 1 = ⇒ Av = 0.899 dvI 1.112

EX8.9 a. 2 Rb = rπ + (1 + β ) RE′ and RE′ = a 2 RL = (10 ) ( 8 ) = 800 Ω Ri = 1.5 kΩ = RTH Rb V 18 I Q = 2CC = = 22.5 mA a RL (10 )2 ( 8 ) rπ =

(100 )( 0.026 )

= 0.116 kΩ 22.5 Rb = 0.116 + (101)( 0.8 ) = 80.9 kΩ

1.5 = RTH 80.9 =

RTH ( 80.9 )

RTH + ( 80.9 )

⇒ ( 80.9 − 1.5 ) RTH = (1.5 )( 80.9 ) ⇒ RTH = 1.53 kΩ

⎛ R2 ⎞ 1 VTH = ⎜ ⎟ VCC = ⋅ RTH ⋅ VCC R R R + 2 ⎠ 1 ⎝ 1

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22.5 = 0.225 mA β 100 V − 0.7 1 = TH ⇒ (1.53)(18 ) = ( 0.225 )(1.53) + 0.7 ⇒ R1 = 26.4 kΩ RTH R1

I BQ = I BQ

IQ

=

26.4 R2 = 1.53 26.4 + R2

( 26.4 − 1.53) R2 = (1.53)( 26.4 ) ⇒ R2 = 1.62 kΩ b. vE = 0.9VCC = ( 0.9 )(18 ) = 16.2 V iE = 0.9 I CQ = ( 0.9 )( 22.5 ) = 20.25 mA v 16.2 v0 = E = ⇒ VP = 1.62 V a 10 i0 = aiE = (10 )( 20.25 ) ⇒ I P = 203 mA PL =

1 (1.62 )( 0.203) ⇒ PL = 0.164 W 2

EX8.10 a. ⎛ IC ⎞ ⎞ ⎟ ⇒ VBE = VT ln ⎜⎜ ⎟⎟ ⎠ ⎝ I SQ ⎠ ⎛ 5 × 10−3 ⎞ = ( 0.026 ) ln ⎜ = 0.6225 V ⇒ VD1 = VD 2 = 0.6225 −13 ⎟ ⎝ 2 × 10 ⎠

⎛V I C = I SQ exp ⎜ BE ⎝ VT VBE

⎛ 0.6225 ⎞ I Bias = I D = I SD exp ⎜ ⎟ ⎝ 0.026 ⎠ ⎛ 0.6225 ⎞ = 5 × 10−13 exp ⎜ ⎟ ⎝ 0.026 ⎠ I Bias = 12.5 mA

b.

V0 = 2 V, iL =

2 = 26.7 mA 0.075

1st approximation: iCn ≅ 26.7 mA, iBn = 0.444 mA ⎛ 26.7 × 10−3 ⎞ VBE = ( 0.026 ) ln ⎜ = 0.6661 −13 ⎟ ⎝ 2 × 10 ⎠ I D = 12.5 − 0.444 = 12.056 mA ⎛ 12.056 × 10−3 ⎞ VD = ( 0.026 ) ln ⎜ ⎟ = 0.6216 −13 ⎝ 5 × 10 ⎠ 2VD = 1.243 V VEB = 2VD − VBE = 0.5769 ⎛ 0.5769 ⎞ icP = 2 × 10−13 exp ⎜ ⎟ = 0.866 mA ⎝ 0.026 ⎠ 2nd approximation: iEn = iL + iCP = 26.7 + 0.866 ≅ 27.6 mA = iEn ⎛ 60 ⎞ iCn = ⎜ ⎟ ( 27.6 ) ⇒ iCn = 27.1 mA ⎝ 61 ⎠ iBn = 0.452 mA I D = 12.5 − 0.452 ⇒ I D = 12.05 mA

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⎛ 27.1× 10−3 ⎞ ⇒ VBEn = 0.6664 V VBEn = ( 0.026 ) ln ⎜ −13 ⎟ ⎝ 2 × 10 ⎠ ⎛ 12.05 × 10−3 ⎞ VD = ( 0.026 ) ln ⎜ ⎟ = 0.6215 V −13 ⎝ 5 × 10 ⎠ 2VDD = 1.243 V VEB = 1.243 − 0.6664 ⇒ VEBp = 0.5766 V ⎛ 0.5766 ⎞ iCP = 2 × 10−13 exp ⎜ ⎟ ⇒ iCP = 0.856 mA ⎝ 0.026 ⎠

c. 10 = 133 mA 0.075 ≅ iL = 133 mA ⇒ iCN = 131 mA

V0 = 10 V, iL = iEn

iBn = 2.18 mA ⇒ I D = 12.5 − 2.18 ⇒ I D = 10.3 mA ⎛ 10.3 × 10−3 ⎞ VD = ( 0.026 ) ln ⎜ = 0.6175 −13 ⎟ ⎝ 5 × 10 ⎠ 2VDD = 1.235 V ⎛ 131× 10−3 ⎞ ⇒ VBEn = 0.7074 V VBEn = ( 0.026 ) ln ⎜ −13 ⎟ ⎝ 2 × 10 ⎠ VEBp = 1.235 − 0.7074 ⇒ VEBp = 0.5276 V ⎛ 0.5276 ⎞ iCP = 2 × 10−13 exp ⎜ ⎟ ⇒ iCP = 0.130 mA ⎝ 0.026 ⎠

EX8.11 No Exercise Problem EX8.12 a. vI = 0 = v0 , vB 3 = 0.7 V 12 − 0.7 11.3 I R1 = = ⇒ I R1 = 45.2 mA R1 0.25 If transistors are matched, then iE1 = iE 3 i iR1 = iE1 + iB 3 = iE1 + E 3 1+ β ⎛ 1 ⎞ 1⎞ ⎛ iR1 = iE1 ⎜ 1 + ⎟ = iE1 ⎜ 1 + ⎟ 1 41 β + ⎝ ⎠ ⎝ ⎠ 45.2 iE1 = ⇒ iE1 = iE 2 = 44.1 mA 1.024 i 44.1 iB1 = iB 2 = E1 = ⇒ iB1 = iB 2 = 1.08 mA 1+ β 41 b.

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For vI = 5 V ⇒ v0 = 5 V 5 ⇒ i0 = 0.625 A 8 0.625 iE 3 ≅ 0.625 A, iB 3 = ⇒ iB 3 = 15.2 mA 41 12 − 5.7 = 25.2 mA vB 3 = 5.7 V ⇒ iR1 = 0.25 10 = 0.244 mA iE1 = 25.2 − 15.2 ⇒ iE1 = 10.0 mA ⇒ iB1 = 41 vB 4 = 5 − 0.7 = 4.3 V i0 =

I R2 =

4.3 − ( −12 ) 0.25

= 65.2 mA ≅ iE 2

65.2 = 1.59 mA 41 iI = iB 2 − iB1 = 1.59 − 0.244 ⇒ iI = 1.35 mA iB 2 =

i0 625 = ⇒ AI = 463 iI 1.35 From Equation (8.54) (1 + β ) R ( 41)( 250 ) AI = = = 641 2 RL 2 (8) AI =

c.

TYU8.1

24 = 1.2 A = I D ( max ) 20 For I D = 0 ⇒ VDS ( max ) = 24 V For VDS = 0, I D ( max ) =

Maximum power when VDS = ID =

VDS ( max ) 2 I D ( max ) 2

= 12 V and

= 0.6 A ⇒ PD ( max ) = (12 )( 0.6 ) = 7.2 Watts

TYU8.2

Maximum power at center of load line Pmax = ( 0.05 )(10 ) ⇒ Pmax = 0.5 W TYU8.3

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a.

PQ = VCEQ ⋅ I CQ = ( 7.5 )( 7.5 ) PQ = 56.3 mW

1 VP2 1 ( 6.5 ) ⋅ = ⋅ ⇒ PL = 21.1 mW 2 RL 2 1 2

b.

PL =

PS = (15 )( 7.5 ) ⇒ PS = 113 mW

η=

PL PS

=

21.1 ⇒ η = 18.7% 113

PQ = 56.3 − 21.1 = 35.2 mW

TYU8.4

a.

PL =

1 VP2 20 ⋅ ⇒ VP = 2 RL PL = 2 ( 8 )( 25 ) ⇒ VP = 20 V ⇒ VCC = ⇒ VCC = 25 V 2 RL 0.8

b.

IP =

VP 20 = ⇒ I P = 2.5 A 8 RL

c.

PQ =

VCCVP VP2 − π RL 4 RL

( 25 )( 20 ) ( 20 ) PQ = − π (8) 4 (8)

2

d.

η=

= 19.9 − 12.5 ⇒ PQ = 7.4 W

π VP π 20 = ⋅ ⇒ η = 62.8% 4VCC 4 25

TYU8.5

a.

PL =

( 4) 1 VP2 ⋅ = ⇒ PL = 80 mW 2 RL 2 ( 0.1)

b.

IP =

VP 4 ⇒ I P = 40 mA = RL 0.1

c.

PQ =

VCCVP VP2 − π RL 4 RL

PQ =

( 5 )( 4 ) ( 4 ) − = 63.7 − 40 ⇒ PQ = 23.7 mW π ( 0.1) 4 ( 0.1)

2

2

d.

η=

π VP 4VCC

TYU8.6 a. 1 ⎛ 2V I CQ ≅ ⋅ ⎜ CC 2 ⎝ RL

=

π 4

⋅ ⇒ η = 62.8% 4 5

⎞ VCC 12 = = 8 mA ⎟= ⎠ RL 1.5

RTH = R1 R2

⎛ R2 ⎞ 1 VTH = ⎜ ⎟ VCC = ⋅ RTH ⋅ VCC R1 ⎝ R1 + R2 ⎠ I CQ VTH − VBE 8 = = 0.107 mA = I BQ = β 75 RTH + (1 + β ) RE

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Let RTH = (1 + β ) RE = ( 76 )( 0.1) = 7.6 kΩ 1 ⋅ ( 7.6 )(12 ) − 0.7 R1 0.107 = 7.6 + 7.6 1 ⋅ ( 91.2 ) = 2.33 ⇒ R1 = 39.1 kΩ R1 39.1R2 = 7.6 ⇒ ( 39.1 − 7.6 ) R2 = ( 7.6 )( 39.1) ⇒ R2 = 9.43 kΩ 39.1 + R2 b. 2 2 1 1 PL = ⋅ ( 0.9 I CQ ) RL = ⎡⎣( 0.9 )( 8 ) ⎤⎦ (1.5 ) ⇒ PL = 38.9 mW 2 2 PS = VCC I CQ = (12 )( 8 ) = 96 mW

PQ = PS − PL = 96 − 38.9 ⇒ PQ = 57.1 mW

η=

PL PS

=

38.9 ⇒ η = 40.5% 96

TYU8.7

I E = I E 3 + IC 4 + IC 5 = I E 3 + IC 4 + β5 I B5

= I E 3 + β 4 I B 4 + β 5 (1 + β 4 ) I B 4

I E = (1 + β 3 ) I B 3 + β 4 β 3 I B 3 + β 5 (1 + β 4 ) β 3 I B 3

If β 4 and β 5 are large, then I E ≅ β 3 β 4 β 5 I B 3 So that composite current gain is β ≅ β 3 β 4 β 5

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Chapter 8 Problem Solutions 8.1 a.

b.

i.

VD D = 80 V

Maximum power at VDS = ID = RD =

ii.

VD D 2

= 40 V

PT 25 = = 0.625 A VD S 40 80 − 40 ⇒ RD = 64 Ω 0.625 VD D = 50 V

Maximum power at VD S =

VD D

ID =

PT 25 = =1 A VDS 25

RD =

50 − 25 ⇒ RD = 25 Ω 1

2

= 25 V

8.2 a.

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PQ (max) = I C Q ⋅ So I C Q = RL =

VC C

2 2 PQ (max) VCC

=

2(20) = 1.67 A 24

VCC − (VCC / 2) 24 − 12 = ⇒ RL = 7.2 Ω I CQ 1.67

1.67 ⇒ 20.8 mA 80 24 − 0.7 RB = ⇒ RB = 1.12 kΩ 20.8 b. I CQ ⋅ RL (1.67 )( 7.2 ) = = 462 Av = g m RL = VT 0.026 IB =

I CQ

β

=

V0 (max) = 12 V ⇒ VP =

V0 (max) 12 = ⇒ VP ≅ 26 mV Av 462

8.3 a.

For maximum power delivered to the load, set VC EQ =

VC C 2

Set VC C = 25 V = VCE ( sus )

Then I Cm =

VCC 25 = RL 0.1

I Cm = 250 mA < I C ,max 25 − 12.5 = 125 mA 0.1 V PQ ( max ) = I CQ ⋅ CC = ( 0.125 )(12.5 ) 2 = 1.56 W < PD,max I CQ =

125 = 1.25 mA 100 25 − 0.7 RB = ⇒ RB = 19.4 kΩ 1.25 1 2 1 2 b. PL ( max ) = ⋅ I CQ ⋅ RL = ( 0.125 ) (100 ) ⇒ PL ( max ) = 0.781 W ( rms ) 2 2 I BQ =

8.4

Point (b): Maximum power delivered to load. Point (a): Will obtain maximum signal current output. Point (c): Will obtain maximum signal voltage output.

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8.5 a.

b. VGG = 5 V, I D = 0.25 ( 5 − 4 ) = 0.25 A, VD S = 37.5 V, P = 9.375 W 2

VGG = 6 V, I D = 0.25 ( 6 − 4 ) = 1.0 A, VD S = 30 V, P = 30 W 2

VGG = 7 V, I D = 0.25 ( 7 − 4 ) = 2.25 A, VD S = 17.5 V, P = 39.375 W 2

VGG = 8 V, I D

= 0.25 ⎡⎣ 2 ( 8 − 4 )VD S − VD2 S ⎤⎦ =

40 − VD S

10 I D = 3.71 A, P = 10.8 W VGG = 9 V, I D

⇒ VD S = 2.92

= 0.25 ⎡⎣ 2 ( 9 − 4 ) VD S − VD2S ⎤⎦ 40 − VD S

⇒ VD S = 1.88 V 10 I D = 3.81 A, P = 7.16 W c. Yes, at VGG = 7 V, P = 39.375 W > PD ,max = 35 W =

8.6 a. VDD = 25 V 2 50 − 25 = = 1.25 A 20

Set VDSQ = I DQ

I DQ = K n (VGS − VTN )

2

1.25 + 4 = VGS = 6.5 V 0.2 ⎛ R2 ⎞ VGS = ⎜ ⎟ VDD ⎝ R1 + R2 ⎠ Let R1 + R2 = 100 kΩ ⎛ R ⎞ 6.5 = ⎜ 2 ⎟ ( 50 ) ⇒ R2 = 13 kΩ ⎝ 100 ⎠ R1 = 87 kΩ

b.

PD = I DQVDSQ = (1.25 )( 25 ) ⇒ PD = 31.25 W

c. I D ,max = 2 I DQ ⇒ I D ,max = 2.5 A VDS ,max = VDD ⇒ VDS ,max = 50 V PD ,max = 31.25 W

d.

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V0 = g m RL Vi g m = 2 K n I DQ = 2

( 0.2 )(1.25) = 1 A / V

V0 = (1)( 20 )( 0.5 ) = 10 V 1 V02 1 (10 ) ⋅ = ⋅ ⇒ PL = 2.5 W 2 RL 2 20 2

PL =

PQ = 31.25 − 2.5 ⇒ PQ = 28.75 W

8.7 (a)

PD = PD ,max − ( Slope ) (T j − 25 )

(b)

60 + 25 ⇒ T j ,max = 145°C 0.5 T j ,max − Tcase 145 − 25 or θ dev − amb = ⇒ θ dev − amb = 2°C/W = 60 θ dev − amb

At PD = 0, T j ,max = (c)

PD ,max

8.8 PD ,rated =

T j ,max − Tamb

θ dev − case

or θ dev − case =

T j ,max − Tamb PD ,rated

150 − 25 = 2.5°C/W 50 − Tamb = PD (θ dev − case + θ case − amb ) =

Then Tdev

150 − 25 = PD ( 2.5 + θ case − amb ) ⇒ 125 = PD ( 2.5 + θ case − amb )

8.9 PD = I D ⋅ VDS = ( 4 )( 5 ) = 20 W Tdev − Tamb = PD (θ dev − case + θ case − snk + θ snk − amb ) Tdev − 25 = 20 (1.75 + 0.8 + 3) = 111 ⇒ Tdev = 136°C Tdev − Tcase = PD ⋅θ dev − case = ( 20 )(1.75 ) = 35

Tcase = Tdev − 35 = 136 − 35 ⇒ Tcase = 101°C Tcase − Tsink = PD ⋅θ case − snk = ( 20 )( 0.8 ) = 16°C Tsink = Tcase − 16 = 101 − 16 ⇒ Tsink = 85°C

8.10

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Tdev − Tamb = PD (θ dev − case + θ case − amb )

200 − 25 = 25 ( 3 + θ case − amb ) ⇒ θ case − amb = 4°C/W

8.11

θ dev − case = PD = =

T j ,max − Tamb PD ,rated

=

175 − 25 = 10°C/W 15

T j ,max − Tamb

θ dev − case + θ case −snk + θ snk − amb 175 − 25 ⇒ PD = 10 W 10 + 1 + 4

8.12

η=

PL PS

PS = VCC ⋅ I Q ⎛V ⎞ PL = VP ⋅ I P = ⎜ CC ⎟ ( I Q ) ⎝ 2 ⎠ 1 ⋅ VCC ⋅ I Q ⇒ η = 50% η= 2 VCC ⋅ I Q

8.13 vo ( max ) = 4.8 V iC 3 = iC 2 = vI = vo + 0.7

−0.7 − ( −5 ) 1

iL ( max ) = −4.3 mA =

so − 3.6 ≤ vI ≤ 5.5 V vo ( min ) = −4.3 V

= 4.3 mA vS ( min ) 1

8.14

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I D 3 = K (VGS 3 − VTN ) = 2

0 − VGS 3 − ( −5 ) R

12 (VGS 3 − 0.5 ) = 5 − VGS 3 2VGS2 3 − 11VGS − 2 = 0 2

VGS 3 =

11 ±

(11)

2

+ 4 (12 )( 2 )

2 (12 )

VGS 3 = VGS 2 = 1.072 V I D 3 = I D 2 = 12 (1.072 − 0.5 ) = 3.93 mA VDS 2 ( sat ) = VGS 2 − VTN = 1.072 − 0.5 = 0.572 V 2

vo ( min ) : i2 ( max ) = −3.93 = vI ( min ) = vo ( min ) + VTN vI ( min ) = −3.43 V

V0 ( min )

1 = −3.93 + 0.5

⇒ V0 ( min ) = −3.93 V

vo ( max ) = 5 − VDS ( sat ) = 5 − 0.572 vo ( max ) = 4.43 V 4.43 I D1 ( max ) = 3.93 + = 8.36 mA 1 I D1 = 8.36 = 12 (VGS 1 − 0.5 ) ⇒ VGS 1 = 1.33 V vI ( max ) = vo + VGS1 = 4.43 + 1.33 ⇒ vI ( max ) = 5.76 V 2

8.15 a. Neglect base currents. v0 ( max ) = V + − VCE (sat) = 10 − 0.2 = 9.8 V 9.8 9.8 iL (max) = I Q = = ⇒ I Q = 9.8 mA RL 1 0 − 0.7 − ( −10 )

⇒ R = 949 Ω 9.8 iE1 ( max ) = 2 I Q ⇒ iE1 ( max ) = 19.6 mA R=

iE1 ( min ) = 0 iL ( max ) = I Q = 9.8 mA iL ( min ) = − I Q = −9.8 mA

b. 2 1 1 ( iL ( max ) ) RL = 2 ( 9.8)2 (1) ⇒ PL = 48.02 mW 2 PS = I Q (V + − V − ) + I Q ( 0 − V − )

PL =

= 9.8 ( 20 ) + 9.8 (10 ) ⇒ PS = 294 mW

η=

PL PS

=

8.16 a. I Q ( min ) = R=

48.02 ⇒ η = 16.3% 294

v0 ( max ) RL

0 − 0.7 − ( −12 ) 100

=

10 ⇒ I Q ( min ) = 100 mA 0.1

⇒ R = 113 Ω

b.

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PQ1 = I Q ⋅ VCE1 = (100 )(12 ) ⇒ PQ1 = 1.2 W P (source) = 2 I Q (12 ) = 2.4 W

c.

(10 ) 1 VP2 ⋅ = = 0.5 W 2 RL 2 (100 ) 2

PL =

PS = 1.2 + 2.4 = 3.6 W

η=

PL PS

=

0.5 ⇒ η = 13.9% 3.6

8.17 I D1 = K n (VGS − VTN ) = 12 ( 0 − ( −1.8 ) ) 2

2

I D1 = 38.9 mA (a) For RL = ∞

vo ( max ) = 4.8 V

VDS ( sat ) = VGS − VTN = 1.8 V vo ( min ) = −5 + 1.8 = −3.2 V

vI = vo + 0.7 ⇒ −2.5 ≤ vI ≤ 5.5 V

For RL = 500 Ω vo ( max ) = 4.8 V

(b)

For vo < 0, vo ( min ) = −3.2 V

I 2′ =

vo −3.2 = = −6.4 mA RL 0.5

−2.5 ≤ vI ≤ 5.5 V (c) For vo = −2V , I 2′ ( max ) = −38.9 mA −2 R2 ( min ) = ⇒ RL ( min ) = 51.4 Ω −38.9

1 v2 1 ( 2) PL = ⋅ o = ⋅ ⇒ PL = 38.9 mW 2 RL 2 51.4 2

PL = 10 ( 38.9 ) = 389 mW % =

38.9 = 10% 389

8.18 + V 2 (V ) PL = P = RL RL

2

1 (V ) 1 (V ) + ⋅ , V − = −V + PS = ⋅ 2 RL 2 RL + 2

− 2

(V ) =

+ 2

So PS

η=

PL PS

RL

⇒ η = 100%

8.19 (a)

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As maximum conversion efficiency π V η = , P = 0.785 4 VCC ⎛4⎞ So V p ( max ) = ( 0.785 )( 5 ) ⎜ ⎟ ⎝π ⎠ V p ( max ) = 5 V

(b)

Maximum power dissipation occurs when V p =

(c)

Pθ ( max ) = 2=

2VCC

π

=

2 ( 5)

π

2 VCC 2 π RL

( 5)

2

π 2 RL

⇒ RL = 1.27 Ω

8.20 P=

(a)

2 1 Vp ⋅ 2 RL

2 1 Vp ⋅ ⇒ V p = 49 V ⇒ V + = 52 V, V − = −52 V 2 24 V 49 IP = P = = 2.04 A RL 24

50 =

(b)

η=

(c)

π VP ⋅

=

4 VCC

π ⎛ 49 ⎞ ⎜ ⎟ 4 ⎝ 52 ⎠

η = 74.0% 8.21 (a) VDS ≥ VDS ( sat ) = VGS − VTN = VGS VDS = 10 − Vo ( max ) and I D = I L = K n (VGS ) Vo ( max ) RL VGS =

= K n (VGS )

2

Vo ( max ) RL ⋅ K n

So 10 − Vo ( max ) = 2

⎡⎣10 − V0 ( max ) ⎤⎦ =

Vo ( max ) RL ⋅ K n

=

2

VGS

20.5 ±

( 5 )( 0.4 )

V0 ( max )

V02 ( max ) − 20.5V0 ( max ) + 100 = 0 V0 ( max ) =

Vo ( max )

V0 ( max )

100 − 20V0 ( max ) + V02 ( max ) =

iL =

2

( 20.5 ) 2

2

2

− 4 (100 )

⇒ V0 ( max ) = 8 V

8 ⇒ iL = 1.6 mA 5 i 1.6 = L = = 2 V ⇒ VI = 10 V Kn 0.4

b.

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= 3.183 V

1 (8) PL = ⋅ = 6.4 mW 2 5 20 (1.6 ) PS = = 10.2 mW 2

π

η=

PL PS

=

6.4 ⇒ η = 62.7% 10.2

8.22

vO = iL RL and iL = iD = K n ( vGS − VTN ) or iL = K n ( vGS ) and vGS = vI − vO Then 2

vO = K n RL ( vI − vO ) or vO = 2 ( vI − vO ) 2

2

2

⎛ dv ⎞ dv0 = ( 2 )( 2 )( vI − v0 ) ⎜ 1 − 0 ⎟ dvI ⎝ dvI ⎠ dv0 ⎡1 + 4 ( vI − v0 ) ⎤⎦ = 4 ( vI − v0 ) dvI ⎣ or

4 ( vI − v0 ) dv0 = dvI 1 + 4 ( vI − v0 )

For vI = 10 V, v0 = 8 V ⇒ At vI = 0, v0 = 0 ⇒

4 (10 − 8 ) dv0 dv = ⇒ 0 = 0.889 dvI 1 + 4 (10 − 8 ) dvI

dv0 =0 dvI

At vI = 1, v0 = 0.5 ⇒

dv0 = 0.667 dvI

8.23 a. ⎛i ⎞ ⎛ 5 × 10−3 ⎞ VBE = VT ln ⎜ C ⎟ = ( 0.026 ) ln ⎜ −13 ⎟ ⎝ 5 × 10 ⎠ ⎝ IS ⎠ V VBE = BB = 0.5987 V ⇒ VBB = 1.1973 V 2 PQ = iC ⋅ vCE = ( 5 )(10 ) ⇒ PQ = 50 mW

b.

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v0 = −8 V iL =

−8 ⇒ iL = −80 mA 0.1 iCp ≈ 80 mA

⎛ iCp ⎞ ⎛ 80 × 10−3 ⎞ vEB = VT ln ⎜ ⎟ = ( 0.026 ) ln ⎜ −13 ⎟ ⎝ 5 × 10 ⎠ ⎝ IS ⎠ vEB = 0.6708 V VBB − vEB + v0 = 0.5987 − 0.6708 − 8⇒ vI = −8.072 V 2 = VBB − vEB = 1.1973 − 0.6708 = 0.5265 V

vI = VBE

⎛v ⎞ ⎛ 0.5265 ⎞ iCn = I S exp ⎜ BE ⎟ = 5 × 10−13 exp ⎜ ⎟ ⇒ iCn = 0.311 mA ⎝ 0.026 ⎠ ⎝ VT ⎠ PL = iL2 RL = ( 80 ) ( 0.1) ⇒ PL = 640 mW 2

PQn = iCn ⋅ vCE = ( 0.311) (10 − ( −8 ) ) ⇒ PQn = 5.60 mW PQp = iCp ⋅ vEC = ( 80 )( 2 ) ⇒ PQp = 160 mW

8.24 (a)

iDn = K n ( vGSn − VTN )

2

V 0.5 + 2 = vGSn = 2.5 V = BB ⇒ VBB = 5.0 V 2 2 Pn = ( 0.5 )(10 ) ⇒ Pn = Pp = 5 mW

(b)

VDS = VGS − VTN ⇒ VDS = VGS − 2 VDS = 10 − vo ( max ) and v ( max ) v ( max ) iL + VTN = O +2 = O +2 Kn RL K n ( 2 )(1)

VGS =

so 10 − v0 ( max ) =

v0 ( max ) 2

+2−2 =

v0 ( max ) 2

so v0 ( max ) = 8 V iDn = iL =

8 ⇒ iDn = iL = 8 mA 1

8 + 2 ⇒ VGS = 4 V 2 V Then vI = vo + VGS − BB = 8 + 4 − 2.5 ⇒ vI = 9.5 V 2 VBB ⎞ ⎛ vSGp = vo − ⎜ vI − ⎟ = 8 − ( 9.5 − 2.5 ) 2 ⎠ ⎝ vSGp = 1 V ⇒ M p cutoff ⇒ iDp = 0

VGS =

PL = iL2 RL = ( 8 ) (1) ⇒ PL = 64 mW 2

PMn = iDn ⋅ vDS = ( 8 )(10 − 8 ) ⇒ PMn = 16 mW PMp = iDp ⋅ vSD ⇒ PMp = 0

8.25 a.

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v0 = 24 V ⇒ iL =

24 ⇒ iL ≈ iN = 3 A 8

3 ⇒ iBn = 73.2 mA 41 For iD = 25 mA ⇒ iR1 = 25 + 73.2 = 98.2 mA iBn =

⎛i VBE = VT ln ⎜ N ⎝ IS Then 98.2 =

⎞ 3 ⎛ ⎞ ⎟ = ( 0.026 ) ln ⎜ −12 ⎟ 6 10 × ⎝ ⎠ ⎠ = 0.7004 V

30 − ( 24 + 0.7 ) R1

⇒ R1 =

5.3 ⇒ R1 = 53.97 Ω 98.2

⎛ 25 × 10−3 ⎞ = 0.5759 V VD = ( 0.026 ) ln ⎜ −12 ⎟ ⎝ 6 × 10 ⎠ VEB = 2VD − VBE = 2 ( 0.5759 ) − 0.7004 = 0.4514 V ⎛V ⎞ ⎛ 0.4514 ⎞ iP = I S exp ⎜ EB ⎟ = ( 6 × 10−12 ) exp ⎜ ⎟ ⇒ iP = 0.208 mA V ⎝ 0.026 ⎠ ⎝ T ⎠ b. Neglecting base current 30 − 0.6 30 − 0.6 iD ≈ = ⇒ iD ≈ 545 mA R1 53.97 ⎛ 0.545 ⎞ VD = ( 0.026 ) ln ⎜ = 0.656 V −12 ⎟ ⎝ 6 × 10 ⎠ Approximation for iD is okay. Diodes and transistors matched ⇒ iN = iP = 545 mA

8.26 (a) I D1 = K1 (VGS 1 − VTN ) VGS1 =

2

5 +1 = 2 V 5

I D 3 = K 3 (VGS 3 − VTN )

2

200 = K 3 ( 2 − 1) ⇒ K n3 = K p 4 = 200 μ A / V 2 2

(b) vI + VSG 4 + VGS 3 − VGS1 = vO For vo large, iL = i1 = K n1 (VGS1 − VTN ) VGS1 =

iL + VTN = K n1

⎛ So vI + 2 + 2 − ⎜ ⎜ ⎝

2

vo + VTN RL K n1

⎞ vo + 1⎟ = v0 ( 0.5)( 5) ⎟⎠

v0 −3 2.5 dv 1 dv dvI 1 =1= 0 + ⋅ ⋅ 0 dvI dvI 2 2.5v0 dvI

vI = v0 +

⎡ ⎤ 1 ⎢1 + ⎥ ⎢⎣ 2 2.5v0 ⎥⎦ For vO = 5 V : 1=

dv0 dvI

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1=

⎤ dv dv0 ⎡ dv 1 ⎢1 + ⎥ = 0 (1.1414 ) ⇒ 0 = 0.876 dvI ⎢ 2 2.5 ( 5 ) ⎥ dvI dvI ⎣ ⎦

8.27 vO = vI +

I Dn + VTN Kn

VBB − VGS and VGS = 2

For vO ≈ 0, I Dn = I DQ + iL = I DQ + Then vO = vI +

I DQ + ( vO / RL ) I DQ v VBB V or vO = vI + BB − VTN − ⋅ 1+ O − VTN − 2 Kn I DQ RL 2 Kn

For vO small, vO ≅ vI + ⎡ 1 I DQ vO ⎢1 + ⋅ ⎢⎣ 2 K n Now dvO = dvI ⎡ 1 ⎢1 + ⋅ ⎣⎢ 2 So

⋅

I DQ VBB 1 v − VTN − ⋅ 1+ ⋅ O Kn 2 2 I DQ RL

I DQ V 1 ⎤ ⎥ = vI + BB − VTN − 2 I DQ RL ⎥⎦ Kn 1

1 ⎤ ⋅ ⎥ K n I DQ RL ⎦⎥

I DQ

= 0.95

1 I DQ 1 1 ⋅ ⋅ = − 1 = 0.0526 2 K n I DQ RL 0.95

For RL = 0.1 k Ω, then Or

vO RL

1 K n I DQ

= 0.01052

K n I DQ = 95.1

We can write g m = 2 K n I DQ = 190 mA/V This is the required transconductance for the output transistor. This implies a very large transistor. 8.28 Av = − g m RL

I CQ

So −12 = − g m ( 2 ) ⇒ g m = 6 mA/V=

VT

I CQ = ( 6 )( 0.026 ) ⇒ I CQ = 0.156 mA

But for maximum symmetrical swing, set I CQ =

VCC 10 = = 5 mA ⇒ Av > 12 2 RL

Maximum power to the load: 2 (10 ) 1 VCC ⋅ = ⇒ PL ( max ) = 25 mW 2 RL 2 ( 2) 2

PL ( max ) =

PS = VCC ⋅ I CQ = (10 )( 5 ) = 50 mW

So η = 50%

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5 = 0.0278 mA β 180 R1 = RTH = 6 kΩ I CQ

I BQ =

=

VTH = I BQ RTH + VBE ( on ) + (1 + β ) I BQ RE Set RE = 20 Ω

VTH = ( 0.0278 )( 6 ) + 0.7 + (181)( 0.0278 )( 0.020 ) VTH = 0.967 V 1 ⋅ RTH ⋅ VCC R1

VTH =

0.967 =

1 ( 6 )(10 ) ⇒ R1 = 62.0 kΩ R1 R2 = 6.64 kΩ

8.29 I CQ =

VCC 15 = = 15 mA 1 RL

I BQ =

15 = 0.15 mA 100

(15) 1 V2 ⇒ PL ( max ) = 112.5 mW PL ( max ) = ⋅ CC = 2 RL 2 (1) 2

Let RTH = 10 kΩ VTH = I BQ RTH + VBE + (1 + β ) I BQ RE

= ( 0.15 )(10 ) + 0.7 + (101)( 0.15 )( 0.1)

VTH = 3.715 =

1 1 ⋅ RTH ⋅ VCC = ⋅ (10 )(15 ) R1 R1

R1 = 40.4 kΩ R2 = 13.3 kΩ

8.30 ⎛ R2 ⎞ ⎛ 1.55 ⎞ VTH = ⎜ ⎟ VCC = ⎜ ⎟ (10 ) + R R ⎝ 1.55 + 0.73 ⎠ ⎝ 1 2 ⎠ = 6.80 V RTH = R1 R2 = 0.73 1.55 = 0.496 kΩ I BQ =

VTH − VBE 6.80 − 0.70 = RTH + (1 + β ) RE 0.496 + ( 26 )( 0.02 )

I BQ = 6.0 mA, I CQ = 150 mA Av = − g m RL′ and RL′ = a 2 RL = ( 3) ( 8 ) = 72 Ω 2

gm =

I CQ VT

=

150 ⇒ 5.77 A/V 0.026

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Av = − ( 5.77 )( 72 ) = −415 Vo ′ = Av ⋅ Vi = ( 415 )( 0.017 ) = 7.06 V 7.06 = 2.35 V 3 7.06 Vo = = 2.35 V 3 7.06 Vo = = 2.35 V 3 PS = I CQ ⋅ VCC = ( 0.15 )(10 ) = 1.5 W Vo =

η=

PL PS

=

0.345 ⇒ η = 23% 1.5

8.31 a.

Assuming the maximum power is being delivered, then 36 9 Vo′ ( peak ) = 36 V ⇒ Vo = = 9 V ⇒ Vrms = ⇒ Vrms = 6.36 V 4 2 36 Vo = ⇒ Vo = 25.5 V b. 2

c.

Secondary I rms =

Primary I P =

PL 2 = ⇒ I rms = 0.314 A Vrms 6.36

0.314 ⇒ I P = 78.6 mA 4

d. PS = I CQ .VCC = ( 0.15 )( 36 ) = 5.4 W 2 ⇒ η = 37% η= 5.4 8.32 a.

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⎛V ve = ⎜ π + g mVπ ⎝ rπ

⎞ ⎟ RE′ = Vπ ⎠

⎛1 ⎞ ⎜ + g m ⎟ RE′ r ⎝ π ⎠

⎛1+ β = Vπ ⎜ ⎝ rπ vi = Vπ + ve ⇒ Vπ = vi − ve

⎞ ⎟ RE′ ⎠

⎛ 1+ β ⎞ ve = (vi − ve ) ⎜ ⎟ RE′ ⎝ rπ ⎠ 1+ β ⋅ RE′ 2 ⎛n ⎞ (1 + β ) RE′ ve rπ v = = = e where RE′ = ⎜ 1 ⎟ RL vi 1 + 1 + β ⋅ R ′ rπ + (1 + β ) RE′ vi ⎝ n2 ⎠ E rπ ⎛n ⎞ ve so ve − v0 ⎜ 1 ⎟ ⎛ n1 ⎞ ⎝ n2 ⎠ ⎜ ⎟ ⎝ n2 ⎠ (1 + β ) RE′ v 1 ⋅ so 0 = vi ⎛ n1 ⎞ rπ + (1 + β ) RE′ ⎜ ⎟ ⎝ n2 ⎠ v0 =

b. I n1 1 2 1 2 RL ⋅ I P RL , a = , I CQ = P so PL = .a 2 I CQ n2 a 2 2 PS = I CQ .VCC For η = 50% : PL =

1 2 2 ⋅ a I CQ RL a 2 I R VCC VCC V PL CQ L so a 2 = = 0.5 = 2 = = ⇒ a 2 = CC I CQ ⋅ VCC I CQ ⋅ RL ( 0.1)( 50 ) 2VCC 5 PS c. 49 ( 0.026 ) r β VT R0 = π = = ⇒ R0 = 0.255 Ω 1 + β (1 + β ) I CQ ( 50 )( 0.1)

8.33 a.

With a 10:1 transformer ratio, we need a current gain of 8 through the transistor.

⎛ R1 R2 ⎞ ⎛ R1 R2 i ie = (1 + β ) ib and ib = ⎜ ii so we need e = 8 = (1 + β ) ⎜ ⎟ ⎜R R +R ⎟ ⎜R R +R ii ib ⎠ ib ⎝ 1 2 ⎝ 1 2 Rib = rπ + (1 + β ) RL′ ≈ (1 + β ) RL′ = (101)( 0.8 ) = 80.8

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⎞ ⎟⎟ where ⎠

⎛ ⎞ R1 R2 Then 8 = (101) ⎜⎜ ⎟⎟ ⎝ R1 R2 + 80.8 ⎠ R1 R2 = 0.0792 or R1 R2 = 6.95 kΩ R1 R2 + 80.8 2VCC V 12 = RL′ ⇒ I CQ = CC = = 15 mA 2 I CQ RL′ 0.8

Set

15 = 0.15 mA 100 = I BQ RTH + VBE

I BQ = VTH

1 ⋅ RTH ⋅ VCC = I BQ RTH + VBE R1 1 ( 6.95)(12 ) = ( 0.15)( 6.95) + 0.7 ⇒ R1 = 47.9 kΩ then R2 = 8.13 kΩ R1 b. I I e = 0.9 I CQ = 13.5 mA = L ⇒ I L = 135 mA a 1 2 PL = ( 0.135 ) ( 8 ) ⇒ PL = 72.9 mW 2 PS = VCC I CQ = (12 )(15 ) ⇒ PS = 180 mW

η=

PL PS

⇒ η = 40.5%

8.34 a. VP = 2 RL PL VP = 2 ( 8 )( 2 ) = 5.66 V = peak output voltage IP =

VP 5.66 = = 0.708 A = peak output current RL 8

Set Ve = 0.9VCC = aVP to minimize distortion Then a =

( 0.9 )(18) 5.66

⇒ a = 2.86

b. 1 ⎛ I P ⎞ 1 ⎛ 0.708 ⎞ = ⎜ ⎟ ⇒ I CQ = 0.275 A 0.9 ⎜⎝ a ⎟⎠ 0.9 ⎝ 2.86 ⎠ Then PQ = VCC I Q = (18 )( 0.275 ) ⇒ PQ = 4.95 W Power rating of transistor Now I CQ =

8.35 a.

Need a current gain of 8 through the transistor.

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⎛ R1 R2 ib = 8 = (1 + β ) ⎜ ⎜R R +R ii ib ⎝ 1 2

⎞ ⎟⎟ where Rib ≈ (1 + β )( 0.9 ) = 90.9 kΩ ⎠

⎛ ⎞ R1 R2 8 =⎜ ⎟ = 0.0792 or R1 R2 = 7.82 kΩ ⎜ 101 ⎝ R1 R2 + 90.9 ⎟⎠ 2VCC 12 = 0.9 kΩ ⇒ I CQ = = 13.3 mA 2 I CQ 0.9

Set

13.3 = 0.133 mA 100 1 Then ( 7.82 )(12 ) = ( 0.133)( 7.82 ) + 0.7 ⇒ R1 = 53.9 kΩ and R 2 = 9.15 kΩ R1 b. I I e = ( 0.9 ) I CQ = 12 mA = L ⇒ I L = 120 mA a 1 2 PL = ( 0.12 ) ( 8 ) ⇒ PL = 57.6 mW 2 PS = VCC I CQ = (12 )(13.3) ⇒ PS = 159.6 mW I BQ =

η=

PL 57.6 = ⇒ η = 36.1% PS 159.6

8.36 a.

All transistors are matched. ⎛1+ β ⎞ iC 3 mA = iE1 + iB 3 = ⎜ ⎟ iC + β ⎝ β ⎠ ⎛ 61 1 ⎞ 3 = ⎜ + ⎟ iC ⇒ iC = 2.90 mA ⎝ 60 60 ⎠ b. For vo = 6 V , let RL = 200 Ω. 6 = 0.03 A = 30 mA ≅ iE 3 200 30 iB 3 = = 0.492 mA 61 iE1 = 3 − 0.492 = 2.508 mA io =

2.508 ⇒ iB1 = 41.11 μ A 61 3 iE 2 ≅ 3 mA ⇒ iB 2 = ⇒ 49.18 μ A 61 iI = iB 2 − iB1 = 49.18 − 41.11 ⇒ iI = 8.07 μ A iB1 =

Current gain Ai = VBE 3

30 × 10−3 ⇒ Ai = 3.72 × 103 8.07 × 10−6 ⎛i ⎞ ⎛ 30 × 10−3 ⎞ = VT ln ⎜ E 3 ⎟ = ( 0.026 ) ln ⎜ −13 ⎟ ⎝ 5 × 10 ⎠ ⎝ IS ⎠

VBE 3 = 0.6453 V ⎛i ⎞ ⎛ 2.508 × 10−3 ⎞ VEB1 = VT ln ⎜ E1 ⎟ = ( 0.026 ) ln ⎜ ⎟ −13 ⎝ 5 × 10 ⎠ ⎝ IS ⎠ VEB1 = 0.5807 V

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vI = v0 + VBE 3 − VEB1 = 6 + 0.6453 − 0.5807 vI = 6.0646 V Voltage gain Av =

v0 6 = ⇒ Av = 0.989 vI 6.0646

8.37 a.

For i0 = 1 A, I B 3 ≅

1 ⇒ 20 mA 50

⎡10 − ( v0,max + VBE 3 ) ⎤ 10 − VEB1 = 2⎢ − 20 ⎥ R1 R1 ⎢⎣ ⎦⎥ 10 − VBE 2vo,max = + 40 If, for simplicity, we assume VEB1 = VBE 3 = 0.7 V, then R1 R1 We can then write

If we assume v0,max = 4 V, then b.

9.3 2 ( 4 ) = + 40 which yields R1 = R2 = 32.5 Ω R1 R1

9.3 ⇒ I E1 = 0.286 A = I E 2 32.5 = 10 I S1,2 , then I E 3 = I E 4 = 2.86 A

For vI = 0, I E1 =

Since I S 3,4

c. We can write ⎧ rπ 1 ⎫ ⎪ rπ 3 + R1 ⎪ 1 + β1 ⎪ 1⎪ R0 = ⎨ ⎬ 2⎪ 1 + β3 ⎪ ⎪ ⎪ ⎩ ⎭ ( 50 )( 0.026 ) βV = 0.4545 Ω Now rπ 3 = 3 T = 2.86 IC 3 rπ 1 =

β1VT I C1

=

(120 )( 0.026 ) 0.286

= 10.91 Ω

So ⎧ 10.91 ⎫ 0.4545 + 32.5 1 ⎪⎪ 121 ⎪⎪ R0 = ⎨ ⎬ 2⎪ 51 ⎪ ⎪⎭ ⎩⎪ 10.91 = 32.5 0.0902 = 0.0900 32.5 121 Then R0 =

1 ⎧ 0.4545 + 0.0900 ⎫ ⎨ ⎬ or R 0 = 0.00534 Ω 2⎩ 51 ⎭

8.38

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{

}

1 rπ 1 + (1 + β ) ⎡ R1 ( rπ 3 + (1 + β ) 2 RL ) ⎤ ⎣ ⎦ 2 iC1 ≈ 7.2 mA and iC 3 ≈ 7.2 mA Ri =

Then rπ =

( 60 )( 0.026 ) 7.2

= 0.217 kΩ

{

}

1 0.217 + ( 61) ⎡ 2 ( 0.217 + ( 61)( 0.2 ) ) ⎤ ⎣ ⎦ 2 1 = 0.217 + 61 ⎡⎣ 2 12.4 ⎤⎦ or Ri = 52.6 kΩ 2

So Ri =

{

}

8.39 a.

b. I1 = K1 (VSG + VTP ) = 2

V + − VSG R1

5 = 10 (VSG − 2 ) ⇒ VSG = 2.707 V 2

10 − 2.707 ⇒ R1 = R2 = 1.46 kΩ R1 RL = 100 Ω For a sinusoidal output signal: c. 5=

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1 (v ) 1 ( 5) PL = ⋅ o = ⋅ ⇒ PL = 125 mW 2 RL 2 0.1 2

iD 3 ≈

2

( vo ) ( 5 ) RL

=

0.1

⇒ iD 3 = 50 mA

50 + 2 = 4.236 V 10 10 − ( 4.236 + 5 ) I1 = ⇒ I D1 = 0.523 mA 1.46 0.523 VSG1 = + 2 = 2.229 V 10 vI = 5 + 4.236 − 2.229 ⇒ vI = 7.007 V

VGS 3 =

I D2 =

(VI − VGS ) − ( −10 )

= 10 (VGS − 2 )

2

1.46 17.007 − VGS = 10 (VGS2 − 4VGS + 4 ) 1.46 14.6VGS2 − 57.4VGS + 41.4 = 0 VGS =

57.4 ±

( 57.4 ) − 4 (14.6 )( 41.4 ) 2 (14.6 ) 2

VGS 2 = 2.98 V I D 2 = 10 ( 2.98 − 2 ) ⇒ I D 2 = 9.60 mA 2

VG 4 = vI − VGS 2 = 7 − 2.98 = 4.02 V VSG 4 = 5 − 4.02 = 0.98 V ⇒ I D 4 = 0

8.40 For v0 = 0 I Q = I C 3 + I C 2 + I E1 ⎛ 1+ βn ⎞ IC 3 I B3 = I E 2 = ⎜ ⎟ IC 2 = β βn ⎝ n ⎠ I C 3 = (1 + β n ) I C 2 ⎛ β ⎞ I I B 2 = I C 1 = ⎜ P ⎟ I E1 = C 2 βn ⎝ 1+ βP ⎠ ⎛ β ⎞ I C 2 = β n ⎜ P ⎟ I E1 ⎝ 1+ βP ⎠ ⎛ β ⎞ I C 3 = (1 + β n ) β n ⎜ P ⎟ I E1 ⎜1+ β ⎟ p ⎠ ⎝ ⎛ β ⎞ ⎛ β ⎞ I Q = (1 + β n ) β n ⎜ P ⎟ I E1 + β n ⎜ P ⎟ I E1 + I E1 ⎝1+ βP ⎠ ⎝ 1+ βP ⎠ ⎛ 10 ⎞ ⎛ 10 ⎞ = ( 51)( 50 ) ⎜ ⎟ I E1 + ( 50 ) ⎜ ⎟ I E1 + I E1 ⎝ 11 ⎠ ⎝ 11 ⎠ I Q = 2318.18I E1 + 45.45 I E1 + I E1 I E1 = 1.692 μ A ⇒ I C1 = 1.534 μ A ⎛ 10 ⎞ I C 2 = ( 50 ) ⎜ ⎟ (1.692 ) ⇒ I C 2 = 76.9 μ A ⎝ 11 ⎠ ⎛ 10 ⎞ I C 3 = ( 51)( 50 ) ⎜ ⎟ (1.692 ) ⇒ I C 3 = 3.92 mA ⎝ 11 ⎠

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Because of rπ 1 and Z, neglect effect of r0. Then neglecting r01, r02 and r03, we find I X = g m 3Vπ 3 + g m 2Vπ 2 + g m1Vπ 1 +

VX rπ 1 + Z

Now ⎛ r ⎞ Vπ 1 = ⎜ π 1 ⎟ VX , Vπ 2 ≅ g m1Vπ 1rπ 2 ⎝ rπ 1 + Z ⎠ and Vπ 3 = ( g m1Vπ 1 + g m 2Vπ 2 ) rπ 3 = ⎡⎣ g m1Vπ 1 + g m 2 ( g m1Vπ 1rπ 2 ) ⎤⎦ rπ 3 ⎛ r ⎞ Vπ 3 = ⎜ π 1 ⎟ [ g m1 + g m1 g m 2 rπ 2 ] rπ 3 ⋅ VX ⎝ rπ 1 + Z ⎠ ( β + β1 β 2 ) rπ 3 ⋅ VX Vπ 3 = 1 rπ 1 + Z ⎛ r ⎞ ⎛ βr ⎞ and Vπ 2 = g m1 ⎜ π 1 ⎟ rπ 2VX = ⎜ 1 π 2 ⎟ VX ⎝ rπ 1 + Z ⎠ ⎝ rπ 1 + Z ⎠ ( β + β1 β 2 ) β 3 VX ββ β1 Then I X = 1 ⋅ V X + 1 2 ⋅ VX + ⋅ VX + rπ 1 + Z rπ 1 + Z rπ 1 + Z rπ 1 + Z Then R0 = rπ 1 =

rπ 1 + Z VX = I X 1 + β1 + β1 β 2 + ( β1 + β1 β 2 ) β 3

(10 )( 0.026 )

1.534 Z = 25 kΩ Then

= 0.169 MΩ

R0 =

169 + 25 1 + (10 ) + (10 )( 50 ) + ⎡⎣10 + (10 )( 50 ) ⎤⎦ ( 50 )

R0 =

194 = 0.00746 kΩ or Ro = 7.46 Ω 26, 011

8.41 a VBB

Neglect base currents. ⎛I ⎞ = 2VD = 2VT ln ⎜ Bias ⎟ ⎝ IS ⎠ ⎛ 5 × 10−3 ⎞ = 2 ( 0.026 ) ln ⎜ ⇒ VBB = 1.281 V −13 ⎟ ⎝ 10 ⎠

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VBE1 + VEB 3 = VBB I E1 = I E 3 + I C 2 ⎛ β ⎞ I B2 = IC 3 = ⎜ P ⎟ I E 3 ⎝ 1+ βP ⎠ ⎛ β ⎞ IC 2 = β n I B 2 = β n ⎜ P ⎟ I E 3 ⎝ 1+ βP ⎠ ⎛ β I E1 = I E 3 + β n ⎜ P ⎝ 1+ βP

⎞ ⎟ IE3 ⎠

⎡ ⎛ β I E1 = I E 3 ⎢1 + β n ⎜ P ⎝ 1+ βP ⎣

⎞⎤ ⎟⎥ ⎠⎦

⎡ ⎛ 1+ βn ⎞ ⎛ 1+ βP ⎞ ⎛ βP ⎜ ⎟ I C1 = ⎜ ⎟ I C 3 ⎢1 + β n ⎜ ⎝ βP ⎠ ⎝ 1+ βP ⎝ βn ⎠ ⎣

⎞⎤ ⎟⎥ ⎠⎦

⎡I ⎤ ⎡I ⎤ VBE1 = VT ln ⎢ C1 ⎥ , VEB 3 = VT ln ⎢ C 3 ⎥ I ⎣ S ⎦ ⎣ IS ⎦

(1.01) I C1 = ⎛⎜

21 ⎞ ⎟ IC 3 ⎝ 20 ⎠

I C1

⎡ ⎛ 20 ⎞ ⎤ ⎢1 + (100 ) ⎜ 21 ⎟ ⎥ ⎝ ⎠⎦ ⎣

⎡ 21 ⎤ = I C 3 ⎢ + 100 ⎥ = 101.05 I C 3 ⎣ 20 ⎦ = 100.05 I C 3

⎛ 100.05 I C 3 ⎞ ⎛ IC 3 ⎞ VT ln ⎜ ⎟ + VT ln ⎜ ⎟ = VBB I S ⎝ ⎠ ⎝ IS ⎠ ⎛ 100.05 I C2 3 ⎞ VT ln ⎜ ⎟ = VBB I S2 ⎝ ⎠ 100.05 I C2 3 I

2 S

⎛V ⎞ = exp ⎜ BB ⎟ ⎝ VT ⎠

⎛V ⎞ exp ⎜ BB ⎟ = 0.4995 mA = I C3 100.05 ⎝ VT ⎠ Then I E 3 = 0.5245 mA Now I C1 = 100.05 I C 3 = 49.97 mA = I C1 IC 3 =

IS

⎛ 20 ⎞ I C 2 = (100 ) ⎜ ⎟ ( 0.5245 ) = 49.95 mA = I C 2 ⎝ 21 ⎠

⎛I ⎞ ⎛ 49.97 × 10−3 ⎞ VBE1 = VT ln ⎜ C1 ⎟ = 0.026 ln ⎜ ⎟ −13 ⎝ 10 ⎠ ⎝ IS ⎠ = 0.70037 ⎛I ⎞ ⎛ 0.4995 × 10−3 ⎞ VEB 3 = VT ln ⎜ C 3 ⎟ = 0.026 ln ⎜ ⎟ 10−13 ⎝ ⎠ ⎝ IS ⎠ = 0.58062 Note: VBE1 + VEB 3 = 0.70037 + 0.58062 = 1.28099 = VBB

b.

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v0 = 10 V ⇒ iE1 ≈

10 = 0.10 A = iC1 100

100 = 1 mA 100 ⎛ 4 × 10−3 ⎞ = 2 ( 0.026 ) ln ⎜ = 1.2694 V −13 ⎟ ⎝ 10 ⎠

iB1 = VBB

⎛ 0.1 ⎞ VBE1 = ( 0.026 ) ln ⎜ −13 ⎟ = 0.7184 ⎝ 10 ⎠ VEB 3 = 1.2694 − 0.7184 = 0.55099 V ⎛ 0.55099 ⎞ I C 3 = 10−13 exp ⎜ ⎟ = 0.1598 mA ⎝ 0.026 ⎠ V 2 (10 ) PL = 0 = ⇒ PL = 1 W RL 100 2

PQ1 = iC1 ⋅ vCE1 = ( 0.1)(12 − 10 ) ⇒ PQ1 = 0.2 W PQ 3 = iC 3 ⋅ vEC 3 = ( 0.1598 ) (10 − [ 0.7 − 12]) ⇒ PQ 3 = 3.40 mW iC 2 = (100 )( iC 3 ) = (100 )( 0.1598 ) = 15.98 mA

PQ 2 = iC 2 ⋅ vCE 2 = (15.98 ) (10 − [ −12]) ⇒ PQ 2 = 0.352 W

8.42 a. ⎛ 10 × 10−3 ⎞ ⇒ VBB = 1.74195 V VBB = 3 ( 0.026 ) ln ⎜ −12 ⎟ ⎝ 2 × 10 ⎠ VBE1 + VBE 2 + VEB 3 = VBB I C1 ≈

IC 2

βn

, IC 3 ≈

IC 2

β n2

⎛I ⎞ ⎛I ⎞ ⎛I ⎞ VT ln ⎜ C1 ⎟ + VT ln ⎜ C 2 ⎟ + VT ln ⎜ C 3 ⎟ = VBB ⎝ IS ⎠ ⎝ IS ⎠ ⎝ IS ⎠ ⎡ I3 ⎤ VT ln ⎢ 3C 2 3 ⎥ = VBB ⎣ βn IS ⎦ ⎛V ⎞ I C 2 = β n I S 3 exp ⎜ BB ⎟ ⎝ VT ⎠ ⎛ 1.74195 ⎞ = ( 20 ) ( 20 × 10−12 ) 3 exp ⎜ ⎟ ⎝ 0.026 ⎠ I C 2 = 0.20 A, I C1 ≈ 10 mA, I C 3 ≈ 0.5 mA ⎛ 10 × 10−3 ⎞ VBE1 = ( 0.026 ) ln ⎜ ⇒ VBE1 = 0.58065 V −12 ⎟ ⎝ 2 × 10 ⎠ ⎛ 0.2 ⎞ VBE 2 = ( 0.026 ) ln ⎜ ⇒ VBE 2 = 0.6585 V −12 ⎟ ⎝ 2 × 10 ⎠ ⎛ 0.5 × 10−3 ⎞ VEB 3 = ( 0.026 ) ln ⎜ ⇒ VEB 3 = 0.50276 V −12 ⎟ ⎝ 2 × 10 ⎠ b. 1 V2 1 V2 PL = 10 W= ⋅ 0 = ⋅ 0 ⇒ V0 ( max ) = 20 V 2 RL 2 20

For v0 ( max ) :

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v02 ( 20 ) = ⇒ PL = 20 W 20 RL 2

PL =

20 = −1 A 20 iC 5 + iC 4 + iE 3 = −io ( max ) = 1 A i0 ( max ) = −

⎞ ⎟⎟ = 1 ⎠ i ⎛ β ⎞ i ⎛ β ⎞ ⎡ 1 ⎛ 1+ β p iC 5 + C 5 ⎜ n ⎟ + C 5 ⎜ n ⎟ ⎢ ⎜ β n ⎝ 1 + β n ⎠ β n ⎝ 1 + β n ⎠ ⎣⎢ β n ⎜⎝ β p iC 5 +

iC 5 ⎛ β n ⎞ iC 4 ⎛ 1 + β p ⋅⎜ ⎜ ⎟+ β n ⎝ 1 + β n ⎠ β n ⎜⎝ β p

⎞⎤ ⎟⎟ ⎥ = 1 ⎠ ⎦⎥

1 ⎛ 20 ⎞ ⎤ ⎛ 1 ⎞ ⎡ 1 ⎛ 6 ⎞ ⎤ ⎡ iC 5 ⎢1 + ⎜ ⎟ ⎥ + ⎜ ⎟ ⎢ ⎜ ⎟ ⎥ = 1 ⎣ 20 ⎝ 21 ⎠ ⎦ ⎝ 21 ⎠ ⎣ 20 ⎝ 5 ⎠ ⎦ iC 5 (1.05048 ) = 1 iC 5 = 0.952 A iC 4 = 0.0453 A iE 3 = 0.00272 A

⎛5⎞ iC 3 = 0.00272 ⎜ ⎟ ⎝6⎠ = 0.002267 A ⎛ 2.267 × 10−3 ⎞ VEB 3 = ( 0.026 ) ln ⎜ ⎟ = 0.54206 V −12 ⎝ 2 × 10 ⎠ VBE1 + VBE 2 = 1.74195 − 0.54206 = 1.19989 ⎛ I ⎞ ⎛I ⎞ VT ln ⎜ C 2 ⎟ + VT ln ⎜ C 2 ⎟ = 1.19989 ⎝ βn IS ⎠ ⎝ IS ⎠ ⎛ 1.19989 ⎞ iC 2 = β n ⋅ I S exp ⎜ ⎟ ⎝ 0.026 ⎠ = 20 (18.83) mA iC 2 = 93.9 mA iC 2 ⎛ β n ⎞ 93.9 = 4.47 mA ⎜ ⎟= β n ⎝ 1 + β n ⎠ 21 PQ 2 = I C 2 ( 24 − ( −20 ) ) = ( 0.0939 ) ( 44 ) = 4.13 W

iC1 =

PQ 5 = ( 0.952 ) ( −10 − ( −24 ) ) = 13.3 W

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Chapter 9 Exercise Solutions EX9.1 Av = −15 = −

R2 R1

R1 = Ri = 20 K ⇒ R2 = 15 R1 = 15 ( 20 ) = 300 K

EX9.2 We can write ACL = −

R2 ⎛ R3 ⎞ R3 ⎜1 + ⎟ − R1 ⎝ R4 ⎠ R1

R1 = R1 = 10 kΩ Want ACL = −50 Set R2 = R3 = 50 kΩ ⎛ R ⎞ ACL = −50 = −5 ⎜ 1 + 3 ⎟ − 5 ⎝ R4 ⎠ R R 50 1+ 3 = 9 ⇒ 3 = 8 = R4 R4 R4 R4 = 6.25 kΩ

EX9.3 R2 1 ⋅ R1 ⎡ ⎛ R2 ⎞ ⎤ 1 ⎢1 + ⎜1 + ⎟ ⎥ R1 ⎠ ⎦ ⎣ Ad ⎝ R Ri = R1 = 25 kΩ Let 2 = x R1

We have ACL = −

x 1 1+ (1 + x ) 5 × 103 −x = x 1.0002 + 5 × 103 x ⎞ ⎛ 12 ⎜ 1.0002 + ⎟=x × 5 103 ⎠ ⎝ −12 = −

12.0024 = x − ( 2.4 × 10 −3 ) x R2 12.0024 = 12.0313 = 0.9976 25 kΩ R2 = 300.78 kΩ x=

EX9.4 ⎛R ⎞ R R R v0 = − ⎜ F vI 1 + F vI 2 + F vI 3 + F VI 4 ⎟ R2 R3 R4 ⎝ R1 ⎠

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We need

RF R R R = 7, F = 14, F = 3.5, F = 10 R1 R2 R3 R4

Set RF = 280 kΩ 280 7 280 R2 = 14 280 R3 = 3.5 280 R4 = 10

Then R1 =

= 40 kΩ = 20 kΩ = 80 kΩ = 28 kΩ

EX9.5 We may note that

R3 R R R 3 20 = = 2 and F = = 2 so that 3 = F R2 1.5 R1 10 R2 R1

Then iL =

− ( −3) − vI ⇒ iL = 2 mA = R2 1.5 kΩ

vL = iL Z L = ( 2 × 10−3 ) ( 200 ) = 0.4 V i4 =

vL 0.4 = = 0.267 mA R2 1.5 kΩ

i3 = i4 + iL = 0.267 + 2 = 2.267 mA

v0 = i3 R3 + vL = ( 2.267 × 10−3 )( 3 × 103 ) − 0.4 ⇒ v0 = 7.2 V

EX9.6 Refer to Fig. 9.24 R1 = 2 R1 = 5 kΩ Let R1 = R3 = 2.5 kΩ Set R2 = R4 Differential Gain =

v0 R2 R2 = = 100 = ⇒ R2 = R4 = 250 kΩ v1 R1 2.5 kΩ

EX9.7

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We have the general relation that ⎛ R ⎞ ⎛ [ R4 / R3 ] ⎞ R2 v0 = ⎜ 1 + 2 ⎟ ⎜ ⎟⎟ vI 2 − vI 1 ⎜ R1 ⎠ ⎝ 1 + [ R4 / R3 ] ⎠ R1 ⎝ R1 = R3 = 10 kΩ, R2 = 20 kΩ, R4 = 21 kΩ ⎛ 20 ⎞ ⎛ [ 21/10] ⎞ ⎛ 20 ⎞ v0 = ⎜ 1 + ⎟ ⎜ ⎟⎟ vI 2 − ⎜ ⎟ vI 1 ⎜ + 10 1 21/10 [ ] ⎝ ⎠⎝ ⎝ 10 ⎠ ⎠ v0 = 2.0323vI 2 − 2.0vI a. vI 1 = 1, vI 2 = −1 v0 = −2.0323 − 2.0 ⇒ v0 = −4.032 V

b.

vI 1 = vI 2 = 1 V v0 = 2.0323 − 2.0 ⇒ v0 = 0.0323 V

c.

vcm = vI 1 = vI 2 so common-mode gain v0 = 0.0323 vcm

Acm =

d. ⎛ A ⎞ C M R RdB = 20 log10 ⎜ d ⎟ ⎝ Acm ⎠ 2.0323 ⎛ 1⎞ Ad = − ( 2.0 ) ⎜ − ⎟ = 2.016 2 ⎝ 2⎠ ⎛ 2.016 ⎞ C M R RdB = 20 log10 ⎜ ⎟ = 35.9 d B ⎝ 0.0323 ⎠

EX9.8 v0 = −

R4 ⎛ 2 R2 ⎞ ⎜1 + ⎟ ( vI 1 − vI 2 ) R3 ⎝ R1 ⎠

Differential gain (magnitude) =

R4 ⎛ 2 R2 ⎞ ⎜1 + ⎟ R3 ⎝ R1 ⎠

Minimum Gain ⇒ Maximum R1 = 1 + 50 = 51 kΩ So Ad =

20 ⎛ 2 (100 ) ⎞ ⎜1 + ⎟ ⇒ Ad = 4.92 20 ⎝ 51 ⎠

Maximum Gain ⇒ Minimum R1 = 1 kΩ Ad =

20 ⎛ 2 (100 ) ⎞ ⎜1 + ⎟ ⇒ Ad = 201 20 ⎝ 1 ⎠

Range of Differential Gain = 4.92 − 201 EX9.9 Time constant = r = R1C2 = (104 )( 0.1×10−6 ) = 1 m sec −1 ×t 0 ≤ t ≤ 1 ⇒ v0 = R1 C2 At t = 1 m sec ⇒ v0 = −1 V 0 ≤ t ≤ 2 ⇒ v0 = −1 +

1 × ( t − 1) R1 C2

At t = 2 m sec ⇒ v0 = −1 +

( 2 − 1) 1

=0

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EX9.10 v0 = vI 1 + 10vI 2 − 25vI 3 − 80vI 4 From Figure 9.40, v13 input to R1, vI4 input to R2, vI1 input to RA, and vI2 input to RB. From Equation (9.94) RF R = 25 and F = 80 R1 R2 Set RF = 500 kΩ, then R1 = 20 kΩ, and R2 = 6.25 kΩ. ⎛ R ⎞⎛ R ⎞ ⎛ R ⎞⎛ R ⎞ Also ⎜1 + F ⎟⎜ P ⎟ = 1 and ⎜1 + F ⎟ ⎜ P ⎟ = 10 ⎝ RN ⎠ ⎝ RA ⎠ ⎝ RN ⎠ ⎝ RB ⎠ where RN = R1 & R2 = 20 & 6.25 = 4.76 kΩ and RP = RA & RB & RC

We find that

RA = 10 RB

Let RA = 200 kΩ, RB = 20 kΩ 500 ⎞ ⎛ RP ⎞ ⎛ RP ⎞ ⎛ Now ⎜1 + ⎟ = 1 = (106 ) ⎜ ⎟⎜ ⎟ ⎝ 4.76 ⎠ ⎝ RA ⎠ ⎝ 200 ⎠ Then RP = 1.89 kΩ RA & RB = 200 & 20 = 18.2 kΩ

So RP = 1.89 =

18.2 RC ⇒ RC = 2.11 kΩ 18.2 + RC

EX9.11 Computer Analysis TYU9.1 ACL = −

R2 −100 kΩ = ⇒ ACL = −10 10 kΩ R1

vI = 0.25 V ⇒ vo = −2.5 V i1 =

vI 0.25 = = 0.025 mA ⇒ i1 = 25 μ A R1 10 kΩ

i2 = i1 = 25 μ A Ri = R1 = 10 kΩ

TYU9.2 (a) − R2 Av = R1 + RS −100 = −4.926 19 + 1.3 −100 = −5.076 Av ( max ) = 19 + 0.7 so 4.926 ≤ Av ≤ 5.076 Av ( min ) =

(b)

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0.1 = 5.076 μ A 19 + 0.7 0.1 = 4.926 μ A i1 ( min ) = 19 + 1.3 so 4.926 ≤ i1 ≤ 5.076 μ A i1 ( max ) =

(c)

Maximum current specification is violated.

TYU9.3 v0 = Ad ( v2 − v1 )

Ad = 103

a. v2 = 0, v0 = 5 v 5 v1 = − 0 = − 3 ⇒ v1 = −5 mV Ad 10 b. v1 = 5, v0 = −10 v0 = v2 − v1 Ad −10 = v2 − 5 ⇒ v2 = 4.99 V 103 c. v1 = 0.001, v2 = −0.001 v0 = 103 ( −0.001 − 0.001) v0 = −2 V

d. v2 = 3, v0 = 3

v0 = Ad ( v2 − v1 )

v0 = v2 − v1 Ad 3 = 3 − v1 ⇒ v1 = 2.997 V 103

TYU9.4 ⎛R ⎞ R R v0 = − ⎜ 4 vI 1 + 4 vI 2 + 4 vI 3 ⎟ R2 R3 ⎝ R1 ⎠ ⎡⎛ 40 ⎞ ⎤ ⎛ 40 ⎞ ⎛ 40 ⎞ v0 = − ⎢⎜ ⎟ ( 250 ) + ⎜ ⎟ ( 200 ) + ⎜ ⎟ ( 75 ) ⎥ ⎝ 20 ⎠ ⎝ 30 ⎠ ⎣⎝ 10 ⎠ ⎦ v0 = − [1000 + 400 + 100]

v0 = −1500 μ V = −1.5 mV

TYU9.5

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vI 1 + vI 2 + vI 3 RF = ( vI 1 + vI 2 + vI 3 ) R 3 RF 1 = ⇒ R1 = R2 = R3 ≡ R = 1 M Ω R 3 1 Then RF = M Ω = 333 k Ω 3 vO =

TYU9.6 v ⎛ R ⎞ R Av = 0 = ⎜ 1 + 2 ⎟ = 5 so that 2 = 4 vI ⎝ R1 ⎠ R1 For v0 = 10 V, vI = 2 V Then i1 =

2 = 50 μ A ⇒ R1 = 40 kΩ R1

Then R2 = 160 kΩ we find i2 =

v0 − vI 10 − 2 = = 50 μ A 160 R2

TYU9.7

v0 = Aod ( v2 − v1 ) = Aod ( vI − v1 ) v0 v − vI = −v1 or v1 = vI − 0 Aod Aod i1 =

v −v v1 = i2 and i2 = 0 1 R1 R2

⎛ 1 ⎞ v −v Then v1 ⎜ ⎟ = 0 1 R2 ⎝ R1 ⎠ ⎛ 1 1 ⎞ v v1 ⎜ + ⎟ = 0 ⎝ R1 R2 ⎠ R2 ⎛ R ⎞ ⎛ R ⎞⎛ v ⎞ v0 ⎜ 1 + 2 ⎟ v1 = ⎜ 1 + 2 ⎟ ⎜ vI − 0 ⎟ R1 ⎠ R1 ⎠⎝ Aod ⎠ ⎝ ⎝ ⎛ R2 ⎞ ⎜1 + ⎟ R1 ⎠ v0 ⎝ So Av = = vI 1 ⎛ R2 ⎞ 1+ ⎜1 + ⎟ Aod ⎝ R1 ⎠

TYU9.8

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⎛ Rb For vI 2 = 0, v2 = ⎜ ⎝ Rb + Ra

⎞ ⎟ vI 1 and ⎠

⎛ R ⎞ ⎛ Rb ⎞ v0 ( vI 1 ) = ⎜ 1 + 2 ⎟ ⎜ ⎟ vI 1 R1 ⎠⎝ Rb + Ra ⎠ ⎝ ⎛ 70 ⎞⎛ 50 ⎞ = ⎜ 1 + ⎟⎜ ⎟ vI 1 5 ⎠⎝ 50 + 25 ⎠ ⎝ = 10vI 1 ⎛ Ra ⎞ For vI 1 = 0, v2 = ⎜ ⎟ vI 2 ⎝ Rb + Ra ⎠ ⎛ R ⎞ ⎛ Ra ⎞ v0 ( vI 2 ) = ⎜ 1 + 2 ⎟ ⎜ ⎟ vI 2 R1 ⎠⎝ Rb + Ra ⎠ ⎝ ⎛ 70 ⎞⎛ 25 ⎞ = ⎜ 1 + ⎟⎜ ⎟ vI 2 5 ⎠⎝ 25 + 50 ⎠ ⎝ = 5vI 2 Then v0 = v0 ( vI 1 ) + v0 ( vI 2 ) v0 = 10vI 1 + 5vI 2

TYU9.9

RS  Ri so i1 = i2 = iS = 100 μ A v0 = −iS RF

We want −10 = − (100 × 10−6 ) RF ⇒ RF = 100 kΩ TYU9.10 We want iL = 1 mA when vI = −5 V iL =

− ( −5 ) −VI −v ⇒ R2 = I = ⇒ R2 = 5 kΩ R2 i2 10−3

vL = iL Z L = (10−3 ) ( 500 ) ⇒ vL = 0.5 V i4 =

vL 0.5 = ⇒ i4 = 0.1 mA R2 5 kΩ

i3 = i4 + iL = 0.1 + 1 = 1.1 mA If op-amp is biased at ±10 V, output must be limited to ≈ 8 V. So v0 = i3 R3 + vL 8 = (1.1× 10−3 ) R3 + 0.5 ⇒ R3 = 6.82 kΩ

Let R3 = 7.0 kΩ

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Then we want

R3 RF 7 = = = 1.4 R2 R1 5

Can choose R1 = 10 kΩ and RF = 14 kΩ TYU9.11 a. v −v i1 = I 1 I 2 R1 v01 = vI 1 + i1 R2′ , v02 = vI 2 − i1 R2 and v0 = v0 =

R4 [vI 2 − i1 R2 − vI 1 − i1 R2′ ] R3

v0 =

R4 ⎡( vI 2 − vI 1 ) − i1 ( R2 + R2′ ) ⎤⎦ R3 ⎣

R4 ( v02 − v01 ) R3

⎤ ⎛ vI 2 − vI 1 ⎞ R4 ⎡ ⎢ ( vI 2 − vI 1 ) − ⎜ ⎟ ( R2 + R2′ ) ⎥ R3 ⎣ ⎝ R1 ⎠ ⎦ For common-mode input vI 2 = vI 1 ⇒ v0 = 0 ⇒ Common Gain = 0, C M R R = ∞ v0 =

b.

Ad ( min ) ⇒ R2′ min, R1 max

⎛ 20 ⎞ ⎡ 100 + 95 ⎤ Ad = ⎜ ⎟ ⎢1 + = 4.82 51 ⎦⎥ ⎝ 20 ⎠ ⎣ ⎛ 20 ⎞ ⎡ 100 + 105 ⎤ Ad ( max ) = ⎜ ⎟ ⎢1 + ⎥ = 206 1 ⎝ 20 ⎠ ⎣ ⎦ c. A CM RR = d Acm

Acm = 0 ⇒ C M R R = ∞

TYU9.12 Differential Gain =

R4 ⎛ 2 R2 ⎞ ⎜1 + ⎟ R3 ⎝ R1 ⎠

Let R3 = R4 so the difference amplifier gain is unity. Minimum Gain ⇒ Maximum R1 ⎛ 2 R2 ⎞ So ⎜⎜1 + ⎟⎟ = 2 ⎝ R1 ( max ) ⎠ We want 2 R2 = R1 ( max )

Maximum Gain ⇒ Minimum R1 ⎛ 2 R2 ⎞ So ⎜⎜1 + ⎟⎟ = 1000 or 2 R2 = 999 R1 ( min ) R 1 ( min ) ⎠ ⎝ Ω + 100 Ω = 100.1 If R2 = 50 kΩ, let R1 ( min ) = 100 Ω fixed resistor and let R1 ( max ) = 100   k

pot

Then actual differential gain is in the range of 1.999 − 1001 TYU9.13

−1 10 μ sec −10 × 10−6 ×t 0 = r r − (10 ) (10 × 10−6 ) After 10 pulses: v0 = −5 = r

End of 1st pulse: v0 =

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100 × 10−6 = 20 μ sec = r 5 r = R1C2 = 20 μ sec = 20 × 10−6

So r =

For example, C2 = 0.01× 10−6 = 0.01 μ F ⇒ R1 = 2 kΩ TYU9.14 ⎡ ⎤ + R − ΔR R + ΔR v01 = ⎢ − ⎥V ⎣⎢ ( R − ΔR ) + ( R + ΔR ) ( R + ΔR ) + ( R − ΔR ) ⎦⎥ ⎡ R − ΔR R + ΔR ⎤ + V =⎢ − 2 R ⎥⎦ ⎣ 2R ⎛ R − ΔR − R − ΔR ⎞ + =⎜ ⎟V 2R ⎝ ⎠ ⎛ ΔR ⎞ + v01 = − ⎜ ⎟V ⎝ R ⎠ For V + = 3.5 V, ΔR = 50, R = 10 × 103 ⎛ 50 ⎞ v01 = − ⎜ 4 ⎟ ( 3.5 ) = −1.75 × 10−2 ⎝ 10 ⎠

Need an amplifier with a gain of Ad =

v0 5 = ⇒ Ad = −285.7 vi −1.75 × 10−2

Use instrumentation amplifier, Fig. 9-25. Connect v01 to vI1 and ( −v01 ) to vI2. ⎛ R ⎞ ⎛ 2R ⎞ Ad = ⎜ 4 ⎟ ⎜ 1 + 2 ⎟ = 285.7 R1 ⎠ ⎝ R3 ⎠ ⎝

Let R4 = 150 kΩ, R3 = 10 kΩ Then

R2 = 9.02 R1

Let R2 = 100 kΩ, R1 = 11.1 kΩ TYU9.15 ⎡1 ⎤ + R v01 = ⎢ − ⎥V ⎣⎢ 2 R + R (1 + δ ) ⎦⎥ ⎡ R + R (1 + δ ) − 2 R ⎤ + =⎢ ⎥V ⎣⎢ 2 ( R + R (1 + δ ) ) ⎦⎥ =

Rδ ×V + 2R ( 2 + δ )

⎛δ ⎞ v01 ≈ ⎜ ⎟ V + ⎝4⎠ + V = 5 For δ = 0.01 ⎛ 0.01 ⎞ v01 = ⎜ ⎟ ( 5 ) = 0.0125 ⎝ 4 ⎠ v 5 = 400 Need a gain Ad = 0 = v01 0.0125 ⎛ R ⎞ ⎛ 2R ⎞ Use an instrumentation amplifier Ad = 400 = ⎜ 4 ⎟ ⎜1 + 2 ⎟ R1 ⎠ ⎝ R3 ⎠ ⎝ R Let R4 = 150 kΩ, R3 = 10 kΩ then 2 = 12.8 R1

Let R2 = 150 kΩ, R1 = 11.7 kΩ

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Chapter 9 Problem Solutions 9.1 (a) vO = Ad ( v2 − v1 )

(

)

1 = Ad 10−3 − ( −10−3 ) ⇒ Ad = 500

(b) 1 = 500 ( v2 − 10−3 ) = 1 + 0.5 = 500v2

v2 = 3 mV (c) 5 = 500 (1 − v1 ) ⇒ 500v1 = 495 v1 = 0.990 V (d) (e)

vO = 0 − 3 = 500 ( v2 − ( −0.5 ) )

−250 − 3 = 500v2 v2 = −0.506 V

9.2 (a) 1 ⎛ ⎞ −3 v2 = ⎜ ⎟ vI = ( 0.49975 × 10 ) ( 3) + 1 2000 ⎝ ⎠ v2 = 1.49925 × 10−3

vO = Aod ( v2 − v1 ) = ( 5 × 103 )(1.49925 × 10−3 − 0 ) vO = 7.49625 V

(b) vO = Aod ( v2 − v1 )

3 = Aod (1.49925 × 10−3 − 0 ) Aod = 2 × 103

9.3 Av = −

R2 = −12 ⇒ R2 = 12 R1 R1

Ri = R1 = 25 kΩ ⇒ R2 = (12 )( 25 ) = 300 kΩ

9.3 (a) (b)

v2 = 3.00 V

vO = Aod ( v2 − v1 ) 2.500 = Aod ( 3.010 − 3.00 ) Aod = 250

9.4

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⎛ Ri ⎞ vid = ⎜ ⎟ vI ⎝ Ri + 25 ⎠ ⎛ Ri ⎞ 0.790 = ⎜ ⎟ ( 0.80 ) ⎝ Ri + 25 ⎠ 0.9875 ( Ri + 25 ) = Ri 24.6875 = 0.0125 Ri Ri = 1975 K

9.5 200 ⎫ = −10 ⎪ 20 ⎪ and ⎬ for each case ⎪ Ri = 20 kΩ ⎪ ⎭ Av = −

9.6 a. 100 = −10 10 Ri = R1 = 10 kΩ b. 100 & 100 Av = − = −5 10 Ri = R1 = 10 kΩ c. 100 Av = − = −5 10 + 10 Ri = 10 + 10 = 20 K Av = −

9.7

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I1 =

vI 0.5 ⇒ R1 = ⇒ R1 = 5 K R1 0.1

R2 = 15 ⇒ R2 = 75 K R1

9.8 Av = −

R2 R1 Av = −10

(a) (b) (c) (d) (e) (f)

Av = −1 Av = −0.20 Av = −10 Av = −2 Av = −1

9.9 Av = −

R2 R1 R1 = 20 K, R2 = 40 K

(a) (b) (c) (d)

R1 = 20 K, R2 = 200 K R1 = 20 K, R2 = 1000 K R1 = 80 K, R2 = 20 K

9.10 Av = −

R2 = −8 ⇒ R2 = 8 R1 R1

For vI = −1, i1 =

1 = 15 μ A ⇒ R1 = 66.7 kΩ ⇒ R2 = 533.3 kΩ R1

9.11 Av = −

R2 = −30 ⇒ R2 = 30 R1 R1

Set R2 = 1 MΩ ⇒ R1 = 33.3 kΩ

9.12 a. Av =

⎛R ⎞ R2 1.05R2 ⇒ = 1.105 ⎜ 2 ⎟ R1 0.95 R1 ⎝ R1 ⎠

⎛R ⎞ 0.95R2 = 0.905 ⎜ 2 ⎟ 1.05R1 ⎝ R1 ⎠ Deviation in gain is +10.5% and − 9.5% b. ⎛R ⎞ ⎛R ⎞ 1.01R2 0.99 R2 = 1.02 ⎜ 2 ⎟ ⇒ = 0.98 ⎜ 2 ⎟ Av ⇒ 0.99 R1 1.01R1 ⎝ R1 ⎠ ⎝ R1 ⎠ Deviation in gain = ±2%

9.13 (a)

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Av =

vO −15 = = −15 vl 1

vO = −15vl ⇒ vO = −150sin ω t ( mV )

(b) i2 = i1 =

vI = 10sin ω t ( μ A ) R1

vO ⇒ iL = −37.5sin ω t ( μ A ) RL

iL =

iO = iL − i2

iO = −47.5sin ω t ( μ A )

9.14 Av = −

R2 R1 + R5

Av = −30 ± 2.5% ⇒ 29.25 ≤ Av ≤ 30.75 So

R2 R2 = 29.25 and = 30.75 R1 + 2 R1 + 1

We have 29.25 ( R1 + 2 ) = 30.75 ( R1 + 1) Which yields R1 = 18.5 k Ω and R2 = 599.6 k Ω

For vI = 25 mV , then 0.731 ≤ vO ≤ 0.769 V

9.15 vO1 = −

R2 120 , vI = − ( 0.2 ) ⇒ vO1 = −1.2 V 20 R1

R4 ⎛ −75 ⎞ , vO1 = ⎜ ⎟ ( −1.2 ) ⇒ vO = +6 V R3 ⎝ 15 ⎠ 0.2 ⇒ i1 = i2 = 10 μ A i1 = i2 = 20 v −1.2 ⇒ i3 = i4 = −80 μ A i3 = i4 = O1 = 15 R3 1st op-amp: 90 μ A into output terminal 2nd op-amp: 80 μ A out of output terminal. vO = −

9.16 (a) R2 22 =− ⇒ Av = −22 1 R1 (b) From Eq. (9.23) R2 1 1 Av = − ⋅ = −22 ⋅ 1 R1 ⎡ ⎡ ⎤ 1 ⎛ R2 ⎞ ⎤ ⎢1 + ⎢⎣1 + 104 ( 23) ⎥⎦ ⎜1 + ⎟ ⎥ R1 ⎠ ⎦ ⎣ Aod ⎝ Av = −

Av = −21.95 (c)

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Want Av = −22 ( 0.98 ) = −21.56 So − 21.56 =

1+

−22 1 1+ ( 23) Aod

1 22 ( 23) = Aod 21.56 1 ( 23) = 0.020408 ⇒ Aod = 1127 Aod

9.17 (a) R2 1 ⋅ R1 ⎡ 1 ⎛ R2 ⎞ ⎤ ⎢1 + ⎜1 + ⎟ ⎥ R1 ⎠ ⎦ ⎣ Aod ⎝ 100 1 =− ⋅ 1 25 ⎡ ⎤ ⎢⎣1 + 5 × 103 ( 5 ) ⎥⎦ Av = −3.9960

Av = −

(b)

vO = −3.9960 (1.00 ) ⇒ vO = −3.9960 V

(c)

4 − 3.9960 × 100% = 0.10% 4

(d) vO = Aod ( v2 − v1 ) = − Aod v1 − ( −3.9960 ) vO = Aod 5 × 10+3

v1 = −

v1 = 0.7992 mV

9.18 vO = Aod ( v2 − v1 ) = − Aod v1 v −5 v1 = − O = Aod 5 × 10+3 v1 = −1 mV

9.19 Av = −

R2 ⎛ R3 R3 ⎞ + ⎟ ⎜1 + R1 ⎝ R4 R2 ⎠

a. −10 = − 10 =

b.

R2 ⎛ 100 100 ⎞ + ⎜1 + ⎟ 100 ⎝ 100 R2 ⎠

2 R2 + 1 ⇒ R2 = 450 kΩ 100 2R 100 = 2 + 1 ⇒ R2 = 4.95 MΩ 100

9.20 a.

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R2 ⎛ R3 R3 ⎞ + ⎟ ⎜1 + R1 ⎝ R4 R2 ⎠ R1 = 500 kΩ Av = −

R2 ⎛ R3 R3 ⎞ + ⎟ ⎜1 + 500 ⎝ R4 R2 ⎠ Set R2 = R3 = 500 kΩ 80 =

⎛ 500 ⎞ 500 80 = 1⎜ 1 + + 1⎟ = 2 + ⇒ R4 = 6.41 kΩ R R4 ⎝ ⎠ 4 b. For vI = −0.05 V −0.05 ⇒ i1 = i2 = −0.1 μ A i1 = i2 = 500 kΩ v X = −i2 R2 = − ( −0.1× 10−6 )( 500 × 103 ) = 0.05 i4 = −

vX 0.05 =− ⇒ i4 = −7.80 μ A 6.41 R4

i3 = i2 + i4 = −0.1 − 7.80 ⇒ i3 = −7.90 μ A

9.21 (a) Av = −1000 =

− R2 −500 = R1 R1

R1 = 0.5 K

(b) − R2 ⎛ R3 R3 ⎞ + ⎟ ⎜1 + R1 ⎝ R4 R2 ⎠ −250 ⎛ 500 500 ⎞ −1250 −1000 = + ⎜1 + ⎟= R1 ⎝ 250 250 ⎠ R1 R1 = 1.25 K Av =

9.22

i1 =

vI = i2 R

⎛v ⎞ v A = −i2 R = − ⎜ I ⎟ R = −vI ⎝R⎠ v v i3 = − A = I R R

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vA vA 2v 2v − =− A = I R R R R ⎛ 2vI ⎞ vB = v A − i4 R = −vI − ⎜ ⎟ ( R ) = −3vI ⎝ R ⎠ i4 = i2 + i3 = −

( −3vI ) 3vI vB =− = R R R 2vI 3vI 5vI i6 = i4 + i5 = + = R R R v v 5 ⎛ ⎞ v0 = vB − i6 R = −3vI − ⎜ I ⎟ R ⇒ 0 = −8 vI ⎝ R ⎠ i5 = −

From Figure 9.12 ⇒ Av = −3

9.23 (a) R2 1 ⋅ R1 ⎡ 1 ⎛ R2 ⎞ ⎤ ⎢1 + ⎜1 + ⎟ ⎥ R1 ⎠ ⎦ ⎣ Aod ⎝ 50 1 =− ⋅ ⇒ Av = −4.99985 10 ⎡ 1 ⎛ 50 ⎞ ⎤ + + 1 1 ⎢ 2 × 105 ⎜ 10 ⎟ ⎥ ⎝ ⎠⎦ ⎣

Av = −

(b)

vO = − ( 4.99985 ) (100 × 10−3 ) ⇒ vO = −499.985 mV

(c)

Error =

0.5 − 0.499985 × 100% ⇒ 0.003% 0.5

9.24 a.

From Equation (9.23) R2 1 Av = − ⋅ R1 ⎡ 1 ⎛ R2 ⎞ ⎤ ⎢1 + ⎜1 + ⎟ ⎥ R1 ⎠ ⎦ ⎣ Aod ⎝ 100 1 =− ⋅ = −0.9980 100 ⎡ 1 ⎛ 100 ⎞ ⎤ + + 1 1 ⎢ 103 ⎜ 100 ⎟ ⎥ ⎝ ⎠⎦ ⎣

Then v0 = Av ⋅ vI = ( −0.9980 )( 2 ) ⇒ v0 = −1.9960 V

b.

v0 = Aod ( v A − vB ) ⎛ 1 vB v0 − vB 1 ⎞ v = ⇒ vB ⎜ + ⎟ = 0 R1 R2 R R R2 2 ⎠ ⎝ 1 v0 vB = ⎛ R2 ⎞ ⎜1 + ⎟ R1 ⎠ ⎝

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Then v0 = Aod v A −

Aod v0 ⎛ R2 ⎞ ⎜1 + ⎟ R1 ⎠ ⎝

⎡ ⎤ ⎢ ⎥ Aod ⎥ v0 ⎢1 + = Aod v A ⎢ ⎛ R ⎞⎥ 2 ⎢ ⎜1 + ⎟ ⎥ R1 ⎠ ⎥⎦ ⎢⎣ ⎝ ⎡ ⎛ R2 ⎞ ⎤ ⎢ ⎜ 1 + ⎟ + Aod ⎥ R 1 ⎠ ⎥=A v v0 ⎢ ⎝ od A ⎢ ⎛ R ⎞ ⎥ ⎢ ⎜1 + 2 ⎟ ⎥ R1 ⎠ ⎥⎦ ⎢⎣ ⎝ ⎛ R ⎞ Aod ⎜ 1 + 2 ⎟ v A R1 ⎠ ⎝ v0 = ⎛ R ⎞ Aod + ⎜ 1 + 2 ⎟ R1 ⎠ ⎝ ⎛ R2 ⎞ ⎜1 + ⎟ vA R1 ⎠ v0 = ⎝ 1 ⎛ R2 ⎞ 1+ ⎜1 + ⎟ Aod ⎝ R1 ⎠ ⎛ 10 ⎞ ⎛ vI ⎞ ⎜1 + ⎟ ⎜ ⎟ 10 ⎠ ⎝ 2 ⎠ So v0 = ⎝ = 0.9980vI 1 ⎛ 10 ⎞ 1 + 3 ⎜1 + ⎟ 10 ⎝ 10 ⎠ For vI = 2 V v0 = 1.9960 V

9.25 ii =

(a)

vl v v R = i2 = − O ⇒ O = − 2 R1 R2 vl R1

(b) ⎞ vl v 1 ⎛ R2 = i3 + O = i3 + ⎜ − ⋅ vl ⎟ R1 RL RL ⎝ R1 ⎠ v ⎛ R ⎞ Then i3 = l ⎜ 1 + 2 ⎟ R1 ⎝ RL ⎠

i2 = i1 =

9.26 ⎛ R3 R1 ⎞ + ⎛ 0.1 1 ⎞ VX .max = ⎜ 10 ⇒ VX .max = 0.09008 V ⋅V = ⎜ ⎜ R R + R ⎟⎟ ⎜ 0.1 1 + 10 ⎟⎟ ( ) 4 ⎠ ⎝ 3 1 ⎝ ⎠ R vO = 2 ⋅ VX .max R1 10 =

R2 R ( 0.09008 ) ⇒ 2 = 111 R1 R1

So R2 = 111 k Ω

9.27 (a)

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⎛R ⎞ R R vO = − ⎜ F ⋅ vI 1 + F ⋅ vI 2 + F ⋅ vI 3 ⎟ R2 R3 ⎝ R1 ⎠ ⎡⎛ 100 ⎞ ⎤ ⎛ 100 ⎞ ⎛ 100 ⎞ = − ⎢⎜ ⎟ ( 0.5 ) + ⎜ ⎟ ( 0.75 ) + ⎜ ⎟ ( 2.5 ) ⎥ 50 20 100 ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎣ ⎦

= − [1 + 3.75 + 2.5]

vO = −7.25 V

(b) ⎡⎛ 100 ⎞ ⎛ 100 ⎞ ⎛ 100 ⎞ ⎤ −2 = − ⎢⎜ ⎟ (1) + ⎜ ⎟ ( 0.8 ) + ⎜ ⎟ vI 3 ⎥ ⎝ 20 ⎠ ⎝ 100 ⎠ ⎦ ⎣⎝ 50 ⎠ 2 = 2 + 4 + vI 3

vI 3 = −4 V

9.28 vo =

R R − RF ⋅ vI 1 − F ⋅ vI 2 − F ⋅ vI 3 R1 R2 R3

= −4vI 1 − 8vI 2 − 2vI 3 RF =4 R1

RF =8 R2

RF =2 R3

Largest resistance = RF = 250 K ⇒ R1 = 62.5 K

R2 = 31.25 K

R3 = 125 K

9.29 v0 = −4vI 1 − 0.5vI 2 = − RF =4 R1

RF R vI 1 − F vI 2 R1 R2

RF = 0.5 ⇒ R1 is the smallest resistor R2

i = 100 μ A =

vI 2 = R1 R1

⇒ R1 = 20 kΩ ⇒ RF = 80 kΩ ⇒ R2 = 160 kΩ

9.30 vI 1 = ( 0.05 ) 2 sin ( 2π ft ) = 0.0707 sin ( 2π ft ) 1 1 ⇒ 10 ms f = 1 kHz ⇒ T = 3 ⇒ 1 ms vI 2 ⇒ T2 = 100 10 R R 10 10 vO = − F ⋅ vI 1 − F ⋅ vI 2 = − ⋅ vI 1 − ⋅ vI 2 R1 R2 1 5 vO = − (10 ) ( 0.0707 sin ( 2π ft ) ) − ( 2 )( ±1 V ) vO = −0.707 sin ( 2π ft ) − ( ±2 V )

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9.31 vO = −

RF R R ⋅ vI 1 − F ⋅ vI 2 − F ⋅ vI 3 R1 R2 R3

20 20 20 ⋅ vI 1 − ⋅ vI 2 − ⋅ vI 3 10 5 2 K sin ω t = −2vI 1 − 4 [ 2 + 100sin ω t ] − 0

vO = −

Set vI 1 = −4 mV

9.32 Only two inputs. ⎡R ⎤ R vO = − ⎢ F ⋅ vI 1 + F ⋅ vI 2 ⎥ R2 ⎣ R1 ⎦ 1 ⎡ ⎤ = − ⎢3vI 1 + ⋅ vI 2 ⎥ 4 ⎣ ⎦ RF RF 1 =3 = R1 R2 4 Smallest resistor = 10 K = R1 RF = 30 K

R2 = 120 K

9.33 ⎡R ⎤ R vO = − ⎢ F ⋅ vI 1 + F ⋅ vI 2 ⎥ R R 2 ⎣ 1 ⎦ R − RF −R −5 − 5sin ω t = ( 2.5sin ω t ) F ⋅ ( 2 ) ⇒ F = 2 R1 R2 R1

RF = 2.5 R2

RF = largest resistor ⇒ RF = 200 K R1 = 100 K

R2 = 80 K

9.34 a. v0 = −

RF R R R ⋅ a3 ( −5 ) − F ⋅ a2 ( −5 ) − F ⋅ a1 ( −5 ) − F ⋅ a0 ( −5 ) R3 R2 R1 R0

So v0 =

b.

RF ⎡ a3 a2 a1 a0 ⎤ + + + ( 5) 10 ⎢⎣ 2 4 8 16 ⎥⎦ R 1 v0 = 2.5 = F ⋅ ⋅ 5 ⇒ RF = 10 kΩ 10 2

c. i.

v0 =

10 1 ⋅ ⋅ 5 ⇒ v0 = 0.3125 V 10 16

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v0 =

ii.

10 ⎡ 1 1 1 1 ⎤ + + + ( 5 ) ⇒ v0 = 4.6875 V 10 ⎢⎣ 2 4 8 16 ⎥⎦

9.35 (a) 10 ⋅ vI 1 1 20 20 vO = − ⋅ vO1 − ⋅ vI 2 = − ( 20 )( −10 ) vI 1 − ( 20 ) vI 2 1 1 vO = 200vI 1 − 20vI 2 vO1 = −

(b) vI 1 = 1 + 2sin ω t ( mV ) vI 2 = −10 mV Then vO = 200 (1 + 2sin ω t ) − 20 ( −10 ) So vO = 0.4 + 0.4sin ω t (V ) 9.36 For one-input

v1 = −

v0 Aod

v −v vI 1 − v1 v1 = + 1 0 R1 R2 & R3 RF ⎡1 VI 1 1 1 ⎤ v0 = v1 ⎢ + + ⎥− & R1 R R R R RF 2 3 F ⎦ ⎣ 1 =−

v0 Aod

⎡1 1 1 ⎤ v0 + ⎢ + ⎥− & R R R R RF 2 3 F ⎦ ⎣ 1

1 1 ⎛ 1 1 ⎞ ⎪⎫ ⎪⎧ 1 = −v0 ⎨ + + ⎜ + ⎟⎬ ⎪⎩ Aod RF RF Aod ⎝ R1 R2 & R3 ⎠ ⎪⎭ =−

⎫ v0 ⎧ 1 RF 1 +1+ ⋅ ⎨ ⎬ RF ⎩ Aod Aod R1 & R2 & R3 ⎭

⎧ ⎫ ⎪ ⎪ R 1 ⎪ ⎪ v0 = − F ⋅ vI 1 ⋅ ⎨ ⎬ where RP = R1 & R2 & R3 R1 ⎛ ⎞ R 1 F ⎪ ⎪1 + 1+ ⎪ Aod ⎜⎝ RP ⎟⎠ ⎪ ⎩ ⎭

Therefore, for three-inputs v0 =

⎛R ⎞ R R −1 × ⎜ F ⋅ vI 1 + F ⋅ vI 2 + F ⋅ vI 3 ⎟ R2 R3 1 ⎛ RF ⎞ ⎝ R1 ⎠ 1+ ⎜1 + ⎟ Aod ⎝ RP ⎠

9.37

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⎛ R ⎞ R Av = 12 = ⎜ 1 + 2 ⎟ ⇒ 2 = 11 R R1 ⎝ 1 ⎠ v v 0.5 i1 = I ⇒ R1 = I = R1 i1 0.15

R1 = 3.33 K R2 = 36.7 K

9.38

⎛ 1 ⎞ vB = ⎜ ⎟ vI ⎝ 1 + 500 ⎠

a. b.

⎛ 1 ⎞ v0 = Aod ⎜ ⎟ vi ⎝ 501 ⎠ ⎛ 1 ⎞ 2.5 = Aod ⎜ ⎟ ( 5 ) ⇒ Aod = 250.5 ⎝ 501 ⎠ ⎛ 1 ⎞ v0 = 5000 ⎜ ⎟ ( 5 ) ⇒ v0 = 49.9 V ⎝ 501 ⎠

9.39 ⎛ R ⎞ Av = ⎜ 1 + 2 ⎟ R1 ⎠ ⎝ Av = 11 (a)

(b) (c) (d) (e) (f)

Av = 2 Av = 1.2 Av = 11 Av = 3 Av = 2

9.40 (a)

R2 = 1 ⇒ R1 = R2 = 20 K R1

(b)

R2 = 9 ⇒ R1 = 20 K, R2 = 180 K R1

(c)

R2 = 49 ⇒ R1 = 20 K, R2 = 980 K R1

(d)

R2 = 0 can set R2 = 20 K, R1 = ∞ (open circuit) R1

9.41 ⎛ 50 ⎞ ⎡⎛ 20 ⎞ ⎛ 40 ⎞ ⎤ v0 = ⎜ 1 + ⎟ ⎢⎜ ⎟ vI 2 + ⎜ ⎟ vI 1 ⎥ ⎝ 50 ⎠ ⎣⎝ 20 + 40 ⎠ ⎝ 20 + 40 ⎠ ⎦ v0 = 1.33vI 1 + 0.667vI 2

9.42 (a)

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vI 1 − v2 vI 2 − v2 v2 + = 20 40 10 ⎛ 100 ⎞ vO = ⎜ 1 + ⎟ v2 = 3v2 50 ⎠ ⎝ Now 2vI 1 − 2v2 + vI 2 − v2 = 4v2 ⎛v ⎞ 2vI 1 + vI 2 = 7v2 = 7 ⎜ o ⎟ ⎝3⎠ 6 3 So vO = ⋅ vI 1 + ⋅ vI 2 7 7

(b) (c)

6 3 ( 0.2 ) + ⎜⎛ ⎟⎞ ( 0.3) ⇒ vO = 0.3 V 7 7 ⎝ ⎠ ⎛6⎞ ⎛ 3⎞ vO = ⎜ ⎟ ( 0.25 ) + ⎜ ⎟ ( −0.4 ) ⇒ vO = 42.86 mV ⎝7⎠ ⎝7⎠ vO =

9.43 ⎛ R4 ⎞ v2 = ⎜ ⎟ vI ⎝ R3 + R4 ⎠ ⎛ R ⎞ ⎛ R ⎞ ⎛ R4 ⎞ vO = ⎜1 + 2 ⎟ v2 = ⎜ 1 + 2 ⎟ ⎜ ⎟ vI R1 ⎠ R1 ⎠⎝ R3 + R4 ⎠ ⎝ ⎝ Av =

vO ⎛ R2 ⎞ ⎛ R4 ⎞ = ⎜1 + ⎟ ⎜ ⎟ vI ⎝ R1 ⎠ ⎝ R3 + R4 ⎠

9.44 (a) vO ⎛ 50 x ⎞ = ⎜⎜ 1 + ⎟ vI ⎝ (1 − x ) 50 ⎟⎠ vO ⎛ x ⎞ 1− x + x = ⎜1 + ⎟= vI ⎝ 1 − x ⎠ 1− x v 1 Av = O = vI 1 − x (b) 1 ≤ Av ≤ ∞ (c) If x = 1, gain goes to infinity.

9.45 Change resister values as shown.

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i1 =

vI = i2 R

⎛v ⎞ vx = i2 2 R + vI = ⎜ I ⎟ 2 R + vI = 3vI ⎝R⎠ v x 3I i3 = = R R 4v v 3v i4 = i2 + i3 = I + I = I R R R ⎛ 4vI ⎞ v0 = i4 2 R + vx = ⎜ ⎟ 2 R + 3vI ⎝ R ⎠ v0 = 11 vI

9.46 (a) (b)

vO =1 vI From Exercise TYU9.7 ⎛ R2 ⎞ ⎜1 + ⎟ R1 ⎠ ⎝

vO = vI ⎡ 1 ⎛ R2 ⎞ ⎤ ⎢1 + ⎜1 + ⎟ ⎥ R1 ⎠ ⎦ ⎣ Aod ⎝ But R2 = 0, R1 = ∞

vO v 1 1 = = ⇒ O = 0.999993 1 1 vI 1 + vI 1+ Aod 1.5 × 105

(b)

Want

vO 1 ⇒ Aod = 99 = 0.990 = 1 vI 1+ Aod

9.47

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v0 = Aod ( vI − v0 ) ⎛ 1 ⎞ + 1⎟ v0 = vI ⎜ ⎝ Aod ⎠ v0 1 = vI ⎛ 1 ⎞ ⎜1 + ⎟ A od ⎠ ⎝ v Aod = 104 ; 0 = 0.99990 vI Aod = 103 ;

v0 = 0.9990 vI

Aod = 102 ;

v0 = 0.990 vI

Aod = 10;

v0 = 0.909 vI

9.48 ⎛ R ⎞ v0 A = ⎜ 1 + 2 ⎟ vI R1 ⎠ ⎝ ⎛ R ⎞ ⎛ R ⎞ v01 = ⎜1 + 2 ⎟ vI , v02 = − ⎜ 1 + 2 ⎟ vI R1 ⎠ R1 ⎠ ⎝ ⎝ So v01 = −v02

9.49 (a)

iL =

vI R1

(b) vO1 = iL RL + vI = iL RL + iL R1 vOI ( max ) ≅ 10 V = iL (1 + 9 ) = 10iL So iL ( max ) ≅ 1 mA Then vI ( max ) ≅ iL R1 = (1)( 9 ) ⇒ vI ( max ) ≅ 9 V

9.50 (a) ⎛ 20 ⎞ ⎛ 20 ⎞ vX = ⎜ ⎟ ⋅ vI = ⎜ ⎟ ( 6 ) = 2 ⎝ 20 + 40 ⎠ ⎝ 60 ⎠ vO = 2 V

(b) (c)

Same as (a)

⎛ 6 ⎞ vX = ⎜ ⎟ ( 6 ) = 0.666 V ⎝ 6 + 48 ⎠ ⎛ 10 ⎞ vO = ⎜ 1 + ⎟ ⋅ v X ⇒ vO = 1.33 V ⎝ 10 ⎠

9.51 a.

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Rin =

v −v v1 and 1 0 = i1 and v0 = − Aod v1 i1 RF

So i1 =

v1 − ( − Aod v1 )

Then Rin =

RF

=

v1 (1 + Aod ) RF

v1 RF = i1 1 + Aod

b. ⎛ RS ⎞ RF ⋅ i1 i1 = ⎜ ⎟ iS and v0 = − Aod ⋅ 1 + Aod ⎝ RS + Rin ⎠ ⎛ A ⎞⎛ RS ⎞ So v0 = − RF ⎜ od ⎟⎜ ⎟ iS ⎝ 1 + Aod ⎠⎝ RS + Rin ⎠ Rin =

RF 10 = = 0.009990 1 + Aod 1001

⎞ RS ⎛ 1000 ⎞ ⎛ v0 = − RF ⎜ ⎟ iS ⎟⎜ ⎝ 1001 ⎠ ⎝ RS + 0.009990 ⎠ ⎞ RS ⎛ 1000 ⎞ ⎛ Want ⎜ ⎟ ≤ 0.990 ⎟⎜ ⎝ 1001 ⎠ ⎝ RS + 0.009990 ⎠ which yields RS ≥ 1.099 kΩ

9.52

vO = iC RF , 0 ≤ iC ≤ 8 mA For vO ( max ) = 8 V, Then RF = 1 k Ω

9.53 v 10 i = I so 1 = ⇒ R = 10 kΩ R R In the ideal op-amp, R1 has no influence. ⎛ R ⎞ Output voltage: v0 = ⎜1 + 2 ⎟ vI R⎠ ⎝ v0 must remain within the bias voltages of the op-amp; the larger the R2, the smaller the range of input voltage vI in which the output is valid. 9.54 (a)

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iL =

− vI R2

10mA =

− ( −10V ) R2

R2 = 1 K R 1 = F R2 R1 R3

Also

vL = (10mA )( 0.05k ) = 0.5 V 0.5 = 0.5 mA i2 = 1 iR 3 = 10 + 0.5 = 10.5 mA v −v 13 − 0.5 Limit vo to 13V ⇒ R3 = O L = R3 = 1.19 K 10.5 iR 3 Then

RF R3 1.19 R = = = 1.19 = F 1 R1 R2 R1

For example, RF = 119 K, R1 = 100 K

(b)

From part (a), vO = 13 V when vI = −10 V

9.55 (a) i1 = i2 and i2 =

vx + iD , vx = −i2 RF R2

⎛R Then i1 = −i1 ⎜ F ⎝ R2 ⎛ R Or iD = i1 ⎜ 1 + F R2 ⎝

⎞ ⎟ + iD ⎠ ⎞ ⎟ ⎠

(b) R1 =

vI 5 = ⇒ R1 = 5 k Ω i1 1

⎛ R ⎞ R 12 = (1) ⎜ 1 + F ⎟ ⇒ F = 11 R2 ⎠ R2 ⎝ For example, R2 = 5 k Ω, RF = 55 k Ω

9.56 VX VX − vO + R2 R3

(1)

IX =

(2)

VX VX − vO + =0 R1 RF

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⎛ R ⎞ From (2) vO = VX ⎜ 1 + F ⎟ R1 ⎠ ⎝ ⎛ 1 ⎛ R ⎞ 1 ⎞ 1 Then (1) I X = VX ⎜ + ⎟ − ⋅ VX ⎜1 + F ⎟ R1 ⎠ ⎝ ⎝ R2 R3 ⎠ R3 IX R R 1 1 1 1 1 = = + − − F = − F VX R0 R2 R3 R3 R1 R3 R2 R1 R3 =

R1 R3 − R2 RF R1 R2 R3

or Ro =

R1 R2 R3 R1 R3 − R2 RF

Note: If

RF 1 = ⇒ R2 RF = R1 R3 then Ro = ∞, which corresponds to an ideal current source. R1 R3 R2

9.57 R2 R4 = =5 R1 R3 Minimum resistance seen by vI1 is R1. Set R1 = R3 = 25 kΩ Then R2 = R4 = 125 kΩ Ad =

iL =

v0 ⇒ v0 = iL RL = ( 0.5 )( 5 ) = 2.5 V RL

v0 = 5 ( vI 2 − vI 1 )

2.5 = 5 ( vI 2 − 2 ) ⇒ vI 2 = 2.5 V

9.58 vO =

R2 ( vI 2 − vI 1 ) R1

R2 R R and 2 = 4 with R2 = R4 and R1 = R3 R1 R1 R3 Differential input resistance R 20 Ri = 2 R1 ⇒ R1 = i = = 10 K 2 2 R2 (a) = 50 ⇒ R2 = R4 = 500 K R1 Ad =

R1 = R3 = 10 K

(b)

R2 = 20 ⇒ R2 = R4 = 200 K R1 R1 = R3 = 10 K

(c)

R2 = 2 ⇒ R2 = R4 = 20 K R1 R1 = R3 = 10 K

(d)

R2 = 0.5 ⇒ R2 = R4 = 5 K R1 R1 = R3 = 10 K

9.59

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We have

⎞ ⎛ R ⎞⎛ R / R ⎞ ⎛R ⎞ ⎛ R ⎞⎛ ⎛ R2 ⎞ 1 vO = ⎜ 1 + 2 ⎟ ⎜ 4 3 ⎟ vI 2 − ⎜ 2 ⎟ vI 1 or vO = ⎜ 1 + 2 ⎟ ⎜ ⎟ vI 2 − ⎜ ⎟ vI 1 R1 ⎠ ⎝ 1 + R4 / R3 ⎠ R1 ⎠ ⎝ 1 + R3 / R4 ⎠ ⎝ ⎝ R1 ⎠ ⎝ ⎝ R1 ⎠ Set R2 = 50 (1 + x ) , R1 = 50 (1 − x ) R3 = 50 (1 − x ) , R4 = 50 (1 + x )

⎡ ⎤ ⎢ ⎥ ⎡ ⎛ 1 + x ⎞⎤ 1+ x ⎞ 1 ⎥ vI 2 − ⎛⎜ vO = ⎢1 + ⎜ ⎟⎥ ⎢ ⎟ vI 1 − 1 x − 1− x ⎠ 1 x ⎛ ⎞ ⎢ ⎥ ⎝ ⎠ ⎝ ⎣ ⎦ 1+ ⎢ ⎜⎝ 1 + x ⎟⎠ ⎥ ⎣ ⎦ ⎤ ⎡1 − x + (1 + x ) ⎤ ⎡ 1+ x ⎛ 1+ x ⎞ vO = ⎢ ⎥ vI 2 − ⎜ ⎥⋅⎢ ⎟ vI 1 1− x ⎝ 1− x ⎠ ⎣ ⎦ ⎣⎢1 + x + (1 − x ) ⎥⎦ ⎛ 1+ x ⎞ ⎛1+ x ⎞ =⎜ ⎟ vI 2 − ⎜ ⎟ vI 1 ⎝1− x ⎠ ⎝ 1− x ⎠ For vI 1 = vI 2 ⇒ vO = 0 Set

R2 = 50 (1 + x )

R1 = 50 (1 − x )

R3 = 50 (1 + x ) R4 = 50 (1 − x )

⎛ ⎞ 1 ⎟ ⎛ 1+ x ⎞⎜ ⎛ 1+ x ⎞ vO = ⎜1 + ⎟ ⎜ 1 + x ⎟ vI 2 − ⎜ ⎟ vI 1 ⎝ 1− x ⎠⎜1+ ⎝ 1− x ⎠ ⎜ ⎟⎟ ⎝ 1− x ⎠ ⎛ 1+ x ⎞ = vI 2 − ⎜ ⎟ vI 1 ⎝ 1− x ⎠ vI 1 = vI 2 = vcm vO 1 + x 1 − x − (1 + x ) −2 x = 1− = = vcm 1− x 1− x 1− x Set

R2 = 50 (1 − x )

R1 = 50 (1 + x )

R3 = 50 (1 − x ) R4 = 50 (1 + x )

⎛ ⎞ 1 ⎟ ⎛ 1− x ⎞⎜ ⎛ 1− x ⎞ vO = ⎜ 1 + ⎟ ⎜ 1 − x ⎟ vI 2 − ⎜ ⎟ vI 1 ⎝ 1+ x ⎠⎜ 1+ ⎝ 1+ x ⎠ ⎜ ⎟⎟ ⎝ 1+ x ⎠ ⎛ 1− x ⎞ = ⎜1 − ⎟ vcm ⎝ 1+ x ⎠ 1 + x − (1 − x )

2x 1+ x 1+ x Worst common-mode gain Acm =

Acm =

=

−2 x 1− x

(b)

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−2 x −2 ( 0.01) = = −0.0202 1 − x 1 − 0.01 −2 ( 0.02 ) For x = 0.02, Acm = = −0.04082 1 − 0.02 −2 ( 0.05 ) For x = 0.05, Acm = = −0.1053 1 − 0.05 1 1 For this condition, set vI 2 = + , vI 1 = − ⇒ vd = 1 V 2 2 1 1 ⎡ ⎛ 1 + x ⎞ ⎤ 1 ⎡1 − x + (1 + x ) ⎤ 1 2 = Ad = ⎢1 + ⎜ ⎥= ⋅ ⎟⎥ = ⎢ 2 ⎣ ⎝ 1 − x ⎠⎦ 2 ⎣ 1− x ⎦ 2 1− x 1− x For x = 0.01,

Acm =

For x = 0.01 Ad = 1.010

C M R RdB = 20 log10

1.010 = 33.98 dB 0.0202

C M R RdB = 20 log10

1.020 = 27.96 dB 0.04082

For x = 0.02, Ad =

1 = 1.020 0.98

For x = 0.05 Ad =

1 1.0526 = 1.0526 C M R RdB = 20 log10 ≅ 20 dB 0.95 0.1053

9.60 ⎛ 10R ⎞ ⎛ 10 ⎞ vy = ⎜ ⎟ v2 = ⎜ ⎟ ( 2.65 ) ⇒ v y = vx = 2.40909 V 10R+R ⎝ ⎠ ⎝ 11 ⎠ v2 − v y 2.65 − 2.40909 i3 = i4 = = = 0.0120 mA 20 R v −v 2.50 − 2.40909 i1 = i2 = 1 x = = 0.0045455 mA 20 R vO = vx − i2 (10R ) = ( 2.40909 ) − ( 0.0045455 )( 200 ) vO = 1.50 V

9.61 iE = (1 + β )( iB ) = ( 81)( 2 ) = 162 mA =

10 R

R = 61.73 Ω

9.62 a.

From superposition: R2 v01 = − ⋅ vI 1 R1

⎛ R ⎞ ⎛ R1 ⎞ v02 = ⎜ 1 + 2 ⎟ ⎜ ⎟ vI 2 R1 ⎠ ⎝ R3 + R4 ⎠ ⎝ Setting vI 1 = vI 2 = vcm ⎡ ⎛ ⎢⎛ R ⎞ ⎜ 1 v0 = v01 + v02 = ⎢⎜ 1 + 2 ⎟ ⎜ R3 R1 ⎠ ⎜ ⎢⎝ ⎜ 1+ R ⎢ 4 ⎝ ⎣

⎤ ⎞ ⎟ R ⎥ ⎟ − 2 ⎥ vcm ⎟ R1 ⎥ ⎟ ⎥ ⎠ ⎦

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⎛ v R ⎛ R ⎞⎜ 1 Acm = 0 = 4 ⋅ ⎜ 1 + 2 ⎟ ⎜ R4 vcm R3 ⎝ R1 ⎠ ⎜ ⎜ 1+ R 3 ⎝ R4 ⎛ R2 ⎞ R2 ⎛ R4 ⎞ ⎜1 + ⎟ − ⎜1 + ⎟ R R1 ⎠ R1 ⎝ R3 ⎠ = 3⎝ ⎛ R4 ⎞ ⎜1 + ⎟ ⎝ R3 ⎠

⎞ ⎟ R ⎟− 2 ⎟ R1 ⎟ ⎠

R4 R2 − R R1 Acm = 3 ⎛ R4 ⎞ ⎜1 + ⎟ ⎝ R3 ⎠

b.

Max. Acm

Max. Acm ⇒ Min.

R4 R and Max. 2 R3 R1

47.5 52.5 − 10.5 9.5 = 4.5238 − 5.5263 ⇒ A = cm 47.5 1 + 4.5238 1+ 10.5

max

= 0.1815

9.63

vI 1 − v A v A − vB vA − v0 = + R1 + R2 Rv R2

(1)

vI 2 − vB vB − v A vB = + R1 + R2 Rv R2

(2)

⎛ R1 ⎞ ⎛ R2 ⎞ v− = ⎜ ⎟ vA + ⎜ ⎟ vI 1 ⎝ R1 + R2 ⎠ ⎝ R1 + R2 ⎠ ⎛ R1 ⎞ ⎛ R2 ⎞ v+ = ⎜ ⎟ vB + ⎜ ⎟ vI 2 ⎝ R1 + R2 ⎠ ⎝ R1 + R2 ⎠

(3) (4)

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Now v− = v+ ⇒ R1vA + R2 vI 1 = R1vB + R2 vI 2 R So that v A = vB + 2 ( vI 2 − vI 1 ) R1 ⎛ 1 v vI 1 1 1 ⎞ v = vA ⎜ + + ⎟− B − 0 R1 + R2 R R R R R R + 2 V 2 ⎠ V 2 ⎝ 1 ⎛ 1 vI 2 1 1 ⎞ v = vB ⎜ + + ⎟− A R1 + R2 R R R R RV + 2 V 2 ⎠ ⎝ 1

(1) ( 2)

Then ⎛ 1 v ⎛ R ⎞⎛ 1 vI 1 1 1 ⎞ v 1 1 ⎞ + ⎟ ( vI 2 − vI 1 ) = vB ⎜ + + ⎟ − B − 0 + ⎜ 2 ⎟⎜ + R1 + R2 R R R R R R R R R R + + R ⎝ 1 ⎠⎝ 1 V 2 ⎠ V V 2 2 ⎠ 2 2 ⎝ 1 ⎛ 1 ⎤ vI 2 R2 1 1 ⎞ 1 ⎡ = vB ⎜ + + ⎟− ⎢ vB + ( vI 2 − vI 1 ) ⎥ R1 + R2 R + R R R R R 2 V 2 ⎠ V ⎣ 1 ⎦ ⎝ 1 Subtract (2) from (1) ⎛ R ⎞⎛ 1 v 1 1 1 ⎞ 1 R2 + + ⎟ ( vI 2 − vI 1 ) − 0 + ⋅ ( vI 2 − vI 1 ) ( vI 1 − vI 2 ) = ⎜ 2 ⎟ ⎜ R1 + R2 R R R R R R R R1 + 2 2 ⎠ 2 V V ⎝ 1 ⎠⎝ 1 v0 1 1 ⎞ 1 1 R2 ⎪⎫ ⎪⎧⎛ R ⎞ ⎛ 1 = ( vI 2 − vI 1 ) ⎨⎜ 2 ⎟ ⎜ + + ⎟+ + ⋅ ⎬ R2 ⎩⎪⎝ R1 ⎠ ⎝ R1 + R2 RV R2 ⎠ R1 + R2 RV R1 ⎭⎪ ⎛ R ⎞ ⎧ R2 R R1 R ⎫ v0 = ( vI 2 − vI 1 ) ⎜ 2 ⎟ ⎨ + 2 +1+ + 2⎬ R1 + R2 RV ⎭ ⎝ R1 ⎠ ⎩ R1 + R2 RV v0 =

R2 ⎞ 2 R2 ⎛ ⎜1 + ⎟ ( vI 2 − vI 1 ) R1 ⎝ RV ⎠

9.64

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(1) (2)

i1 =

vI 1 − vI 2 ( 0.50 − 0.030sin ω t ) − ( 0.50 + 0.030sin ω t ) = 20 R1

−0.060sin ω t 20 i1 = −3sin ω t ( μ A ) vO1 = i1 R2 + vI 1 = ( −0.0030sin ω t )(115 ) + 0.50 − 0.030sin ω t vO1 = 0.50 − 0.375sin ω t =

vO 2 = vI 2 − i1 R2 = 0.50 + 0.030sin ω t − ( −0.003sin ω t )(115 ) vO 2 = 0.50 + 0.375sin ω t R4 200 ⎡ 0.50 + 0.375sin ω t − ( 0.50 − 0.375sin ω t ) ⎤⎦ ( vO 2 − vO1 ) = 50 ⎣ R3

vO =

vO = 3sin ω t ( V ) i3 =

vO 2 0.50 + 0.375sin ω t = 50 + 200 R3 + R4

i3 = 2 + 1.5sin ω t ( μ A ) i2 =

vO1 − vO ( 0.5 − 0.375sin ω t ) − ( 3sin ω t ) = 250 R3 + R4

i2 = 2 − 13.5sin ω t ( μ A )

9.65 ⎛ 40 ⎞ vOB = ⎜1 + ⎟ vI = 2.1667 sin ω t ⎝ 12 ⎠ 30 vOC = − vI = −1.25sin ω t 12

(a) (b)

(c) vO = vOB − vOC = 2.1667 sin ω t − ( −1.25sin ω t ) vO = 3.417 sin ω t

vO 3.417 = = 6.83 0.5 vI

(d) 9.66 iO =

vI R

9.67 vO R ⎛ 2R ⎞ = 4 ⎜1 + 2 ⎟ vI 2 − vI 1 R3 ⎝ R1 ⎠ 200 ⎛ 2 (115 ) ⎞ vO = ⎜1 + ⎟ ( 0.06sin ω t ) R1 ⎠ 50 ⎝ 230 For vO = 0.5 = 1.0833 ⇒ R1 = 212.3 K R1 Ad =

vO = 8 V

230 = 32.33 ⇒ R1 = 7.11 K ⇒ R1 f = 7.11 K, R1 (potentiometer) = 205.2 K R1

9.68

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⎛ 2 R2 ⎞ ⎜1 + ⎟ ( vI 2 − vI 1 ) R1 ⎠ ⎝ Set R2 = 15 K, Set R1 = 2 K + 100 k ( Rot ) R4 R3

vO =

Want

R4 ≈8 R3

Set R3 = 10 K R 4 = 75 K

Now Gain (min) =

75 ⎛ 2 (15 ) ⎞ ⎜1 + ⎟ = 9.71 10 ⎝ 102 ⎠

Gain ( max ) =

75 ⎛ 2 (15 ) ⎞ ⎜1 + ⎟ = 120 10 ⎝ 2 ⎠

9.69 For a common-mode gain, vcm = vI 1 = vI 2 Then ⎛ R ⎞ R v01 = ⎜ 1 + 2 ⎟ vcm − 2 vcm = vcm R1 ⎠ R1 ⎝ ⎛ R ⎞ R v02 = ⎜ 1 + 2 ⎟ vcm − 2 vcm = vcm R R1 1 ⎠ ⎝ From Problem 9.62 we can write R4 R4 − R3 R3′ Acm = ⎛ R4 ⎞ ⎜1 + ⎟ R3 ⎠ ⎝ R3 = R4 = 20 kΩ, R3′ = 20 kΩ ± 5% 20 1− R3′ 1 ⎛ 20 ⎞ Acm = = 1− 1 + ( 1) 2 ⎝⎜ R3′ ⎠⎟ For R3′ = 20 kΩ − 5% = 19 kΩ 1 ⎛ 20 ⎞ ⎜ 1 − ⎟ = −0.0263 2 ⎝ 19 ⎠ For R3′ = 20 kΩ + 5% = 21 kΩ Acm =

1 ⎛ 20 ⎞ ⎜ 1 − ⎟ = 0.0238 2⎝ 21 ⎠ So Acm max = 0.0263 Acm =

9.70 a. v0 =

1 ⋅ vI ( t ′ ) dt ′ R1C2 ∫ 0.5

∫ 0.5sin ω t dt = − ω v0 = 0.5 = f =

cos ω t

1 ( 0.5 ) 0.5 ⋅ = R1C2 ω 2π R1C2 f

1 1 = ⇒ f = 31.8 Hz 2π R1C2 2π ( 50 × 103 )( 0.1× 10−6 )

Output signal lags input signal by 90°

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b. i.

f =

ii.

f =

0.5

2π ( 50 × 103 )( 0.1× 10−6 )

⇒ f = 15.9 Hz

0.5

( 0.1)( 2π ) ( 50 ×103 )( 0.1×10−6 )

⇒ f = 159 Hz

9.71 −v ⋅ t 1 vI ( t ) dt = I ∫ RC RC vI = −0.2 vO = −

Now 8=

− ( −0.2 )( 2 )

(a)

RC RC = 0.05 s

(b)

14 =

( 0.2 ) t 0.05

⇒ t = 3.5 s

9.72 a. v0 = vI

− R2

1 jω C2 R1

R2 ⋅ =−

1 jω C2

⎛ 1 ⎞ R1 ⎜ R2 + ⎟ jω C2 ⎠ ⎝

v0 R 1 =− 2⋅ vI R1 1 + jω R2 C2

b.

v0 R =− 2 vI R1

c.

f =

1 2π R2 C2

9.73 a. R ( jω C1 ) v0 − R2 = =− 2 vI R + 1 1 + jω R1C1 1 jω C1 v0 R jω R1C1 =− 2⋅ vI R1 1 + jω R1C1

b.

v0 R =− 2 vI R1

c.

f =

1 2π R1C1

9.74 Assuming the Zener diode is in breakdown,

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vO = − i2 =

R2 1 ⋅ Vz = − ( 6.8 ) ⇒ vO = −6.8 V R1 1

0 − vO 0 − ( −6.8 ) = ⇒ i2 = 6.8 mA R2 1

10 − Vz 10 − 6.8 − i2 = − 6.8 ⇒ iz = −6.2 mA!!! Rs 5.6 Circuit is not in breakdown. Now 10 − 0 10 = i2 = ⇒ i2 = 1.52 mA 5.6 + 1 Rs + R1 iz =

vO = −i2 R2 = − (1.52 )(1) ⇒ vO = −1.52 V

iz = 0

9.75 ⎡ ⎤ ⎛ v ⎞ v ⎛ v ⎞ vO = −VT ln ⎜ I ⎟ = − ( 0.026 ) ln ⎢ −14 I 4 ⎥ ⇒ vO = −0.026 ln ⎜ −I 10 ⎟ I R 10 ⎠ 10 10 ⎢ ⎥ ⎝ )( ) ⎦ ⎝ s 1⎠ ⎣( For vI = 20 mV , vO = 0.497 V For vI = 2 V , vO = 0.617 V

9.76

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⎛ 333 ⎞ v0 = ⎜ ⎟ ( v01 − v02 ) = 16.65 ( v01 − v02 ) ⎝ 20 ⎠ ⎛i ⎞ v01 = −vBE1 = −VT ln ⎜ C1 ⎟ ⎝ IS ⎠ ⎛i ⎞ v02 = −vBE 2 = −VT ln ⎜ C 2 ⎟ ⎝ IS ⎠ ⎛i ⎞ ⎛i ⎞ v01 − v02 = −VT ln ⎜ C1 ⎟ = VT ln ⎜ C 2 ⎟ ⎝ iC 2 ⎠ ⎝ iC1 ⎠ v v iC 2 = 2 , iC1 = 1 R2 R1 ⎛v R ⎞ So v01 − v02 = VT ln ⎜ 2 ⋅ 1 ⎟ ⎝ R2 v1 ⎠ Then ⎛v R ⎞ v0 = (16.65 )( 0.026 ) ln ⎜ 2 ⋅ 1 ⎟ ⎝ v1 R2 ⎠ ⎛v R ⎞ v0 = 0.4329 ln ⎜ 2 ⋅ 1 ⎟ ⎝ v1 R2 ⎠ ln ( x ) = log e ( x ) = ⎣⎡ log10 ( x ) ⎦⎤ ⋅ ⎣⎡log e (10 ) ⎦⎤ = 2.3026 log10 ( x )

⎛v R ⎞ Then v0 ≅ (1.0 ) log10 ⎜ 2 ⋅ 1 ⎟ ⎝ v1 R2 ⎠

9.77 vO = − I s R evI / VT = − (10−14 )(104 ) evI / VT vO = (10

(

−10

)e

)

vI / 0.026

For vI = 0.30 V ,

vo = 1.03 × 10−5 V

For vI = 0.60 V ,

vo = 1.05 V

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Chapter 10 Exercise Solutions EX10.1 V + − VBE ( on ) 10 − 0.7 = I REF = 15 R1 I REF = 0.62 mA I REF 0.62 = 2 2 1+ 1+ β 75 I 0 = 0.604 mA I0 =

EX10.2 V + − VBE ( on ) − V − 5 − 0.7 − ( −5 ) I REF = = R1 12 I REF = 0.775 mA I 0.775 I 0 = REF = = 0.7549 mA 2 2 1+ 1+ 75 β ΔI 0 = ( 0.02 )( 0.7549 ) = 0.0151 mA and ΔI 0 = r0 =

ΔV 1 ΔVCE 2 ⇒ r0 = CE 2 ΔI 0 r0

V 4 = 265 kΩ = A ⇒ VA = ( 265 )( 0.7549 ) ⇒ VA ≅ 200 V 0.0151 I0

EX10.3 V + − 2VBE ( on ) 9 − 2 ( 0.7 ) I REF = = R1 12 I REF = 0.6333 mA I0 =

I REF 0.6333 = = 0.6331 mA 2 2 1+ 1+ β (1 + β ) 75 ( 76 )

I 0 = 0.6331 mA = I C1 I B1 = I B 2 =

I0

β

⇒ I B1 = I B 2 = 8.44 μ A

I E 3 = I B1 + I B 2 ⇒ I E 3 = 16.88 μ A I B3 =

I E3 ⇒ I B 3 = 0.222 μ A 1+ β

EX10.4 ⎛I ⎞ I 0 RE = VT ln ⎜ REF ⎟ ⎝ I0 ⎠ ⎛ I ⎞ 0.026 ⎛ 0.75 ⎞ V ln ⎜ RE = T ln ⎜ REF ⎟ = ⎟ ⇒ RE = 3.54 kΩ I 0 ⎝ I 0 ⎠ 0.025 ⎝ 0.025 ⎠ 5 − 0.7 R1 = ⇒ R1 = 5.73 kΩ 0.75 VBE1 − VBE 2 = I 0 RE = ( 0.025 )( 3.54 ) ⇒ VBE1 − VBE 2 = 88.5 mV

EX10.5

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5 − 0.7 − ( −5 )

⇒ I REF = 0.775 mA 12 ⎛I ⎞ I 0 RE = VT ln ⎜ REF ⎟ ⎝ I0 ⎠ ⎛ 0.775 ⎞ I 0 ( 6 ) = ( 0.026 ) ln ⎜ ⎟ ⇒ I 0 ≅ 16.6 μ A ⎝ I0 ⎠ I REF =

EX10.6 ⎛I ⎞ I 0 RE = VT ln ⎜ REF ⎟ ⎝ I0 ⎠ 0.026 ⎛ 0.70 ⎞ RE = ln ⎜ ⎟ ⇒ RE = 3.465 kΩ 0.025 ⎝ 0.025 ⎠ I 0.025 gm2 = 0 = ⇒ g m 2 = 0.9615 mA/V VT 0.026

β VT

rπ 2 = r02 =

I0

=

(150 )( 0.026 ) 0.025

= 156 kΩ

VA 100 = = 4000 kΩ I 0 0.025

RE′ = RE || rπ 2 = 3.47 ||156 = 3.39 kΩ R0 = r02 (1 + g m 2 RE′ ) = 4000 ⎡⎣1 + ( 0.962 )( 3.39 ) ⎤⎦ R0 = 17.04 MΩ 1 3 dI 0 = ⋅ dVC 2 = ⇒ dI 0 = 0.176 μ A R0 17, 040

EX10.7 I REF = I R + I BR + I B1 + ... + I BN I R = I 01 = I 02 = ... = I 0 N and I BR = I B1 = I B 2 = ... = I BN =

I 01

β

⎛I ⎞ ⎛ N +1⎞ I REF = I 01 + ( N + 1) ⎜ 01 ⎟ = I 01 ⎜ 1 + ⎟ β β ⎠ ⎝ ⎠ ⎝ I REF So I 01 = I 02 = ... = I 0 N = N +1 1+

β

I 01 1 = 0.90 = N +1 I REF 1+ 50 N +1 1 1+ = 50 0.9 1 ⎛ ⎞ − 1⎟ ( 50 ) N +1 = ⎜ ⎝ 0.9 ⎠ ⎛ 1 ⎞ − 1⎟ ( 50 ) − 1 N =⎜ 0.9 ⎝ ⎠ N = 4.55 ⇒ N = 4

EX10.8

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VDS ( sat ) = 1 V = VGS 2 − VTN = VGS 2 − 2 ⇒ VGS 2 = 3 V 2 2 ⎛ μ C ⎞⎛W ⎞ I O = K n 2 (VGS 2 − VTN ) = ⎜ n ox ⎟ ⎜ ⎟ (VGS 2 − VTN ) ⎝ 2 ⎠ ⎝ L ⎠2 2 ⎛W ⎞ ⎛W ⎞ 0.20 = ( 0.020 ) ⎜ ⎟ ( 3 − 2 ) ⇒ ⎜ ⎟ = 10 L ⎝ ⎠2 ⎝ L ⎠2 2 ⎛ μ C ⎞⎛W ⎞ I REF = ⎜ n ox ⎟ ⎜ ⎟ (VGS 1 − VTN ) 2 L ⎝ ⎠ ⎝ ⎠1 VGS 1 = VGS 2 2 ⎛W ⎞ ⎛W ⎞ 0.5 = ( 0.020 ) ⎜ ⎟ ( 3 − 2 ) ⇒ ⎜ ⎟ = 25 ⎝ L ⎠1 ⎝ L ⎠1

VGS 3 = V + − VGS 1 = 10 − 3 = 7 V ⎛ μ C ⎞⎛W ⎞ 2 I REF = ⎜ n ox ⎟ ⎜ ⎟ (VGS 3 − VTN ) ⎝ 2 ⎠ ⎝ L ⎠3 2 ⎛W ⎞ ⎛W ⎞ 0.5 = ( 0.020 ) ⎜ ⎟ ( 7 − 2 ) ⇒ ⎜ ⎟ = 1 L ⎝ ⎠3 ⎝ L ⎠3

EX10.9 a.

I REF = K n (VGS − VTN )

2

0.020 = 0.080 (VGS − 1)

2

VGS = 1.5 V all transistors

b. VG 4 = VGS 3 + VGS1 + V − = 1.5 + 1.5 − 5 = −2 V VS 4 = VG 4 − VGS 4 = −2 − 1.5 = −3.5 V

VD 4 ( min ) = VS 4 + VDS 4 ( sat ) and VDS 4 ( sat ) = VGS 4 − VTN = 1.5 − 1 = 0.5 V So VD 4 ( min ) = −3.5 + 0.5 ⇒ VD 4 ( min ) = −3.0 V

c. R0 = r04 + r02 (1 + g m r04 ) 1 1 r02 = r04 = = = 2500 kΩ λ I 0 ( 0.02 )( 0.020 )

g m = 2 K n (VGS − VTN ) = 2 ( 0.080 )(1.5 − 1) ⇒ g m = 0.080 mA / V R0 = 2500 + 2500 (1 + ( 0.080 )( 2500 ) ) ⇒ R0 = 505 MΩ

EX10.10 For Q2 : vDS ( min ) = VP = 2 V ⇒ VS ( min ) = vDS ( min ) − 5 = 2 − 5 ⇒ VS ( min ) = −3 V I 0 = I DSS 2 (1 + λ vDS 2 ) = 0.5 (1 + ( 0.15 )( 2 ) ) ⇒ I 0 = 0.65 mA ⎛ v ⎞ I 0 = I DSS1 ⎜ 1 − GS 1 ⎟ ⎝ VP1 ⎠

2

2

⎛ v ⎞ 0.65 = 0.80 ⎜ 1 − GS1 ⎟ −2 ⎠ ⎝ vGS 1 = 0.0986 ⇒ vGS 1 = −0.197 V −2 vGS 1 = VI − VS − 0.197 = VI − ( −3) ⇒ VI ( min ) = −3.2 V

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Vgs 2 = 0, Vgs1 = −VX

IX =

VX VX − V1 + + g m1VX r02 r01

V1 V1 − VX + = g m1VX RD r01

(1) (2)

⎛ 1 ⎞ VX ⎜ + g m1 ⎟ r ⎝ 01 ⎠ V1 = 1 1 + RD r01

⎞ 1 ⎛ 1 ⎜ + g m1 ⎟ r r IX 1 1 1 ⎠ = = + + g m1 − 01 ⎝ 01 1 1 VX R0 r02 r01 + RD r01 1 ⎡ ⎤ ⎢ ⎞ r01 ⎥ 1 ⎛ 1 ⎥ = + ⎜ + g m1 ⎟ ⎢1 − 1 1 ⎥ r02 ⎝ r01 ⎠⎢ + ⎢⎣ RD r01 ⎥⎦ 1 ⎛ ⎞ ⎞ ⎜ RD ⎟ 1 ⎛ 1 ⎟ = + ⎜ + g m1 ⎟ ⎜ r02 ⎝ r01 ⎠⎜ 1 + 1 ⎟ ⎜R ⎟ ⎝ D r01 ⎠

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For RD  r01 ⇒

⎞ 1 1 ⎛ 1 ≅ + ⎜ + g m1 ⎟ R0 r02 ⎝ r01 ⎠

For Q1: 2I g m1 = DSS 1 VP

⎛ VGS 1 ⎞ 2 ( 0.8 ) ⎛ −0.197 ⎞ ⎜1 − ⎟= ⎜1 − ⎟ −2 ⎠ 2 ⎝ VP ⎠ ⎝ g m1 = 0.721 mA/V 1 1 = = 10.3 kΩ r0 = λ I 0 ( 0.15 )( 0.65 )

1 1 1 = + + 0.721 = 0.915 ⇒ R0 = 1.09 kΩ R0 10.3 10.3

EX10.11

a.

⎛V ⎞ I REF = I S exp ⎜ EB 2 ⎟ ⎝ VT ⎠

⎛I ⎞ ⎛ 0.5 × 10−3 ⎞ VEB 2 = VT ln ⎜ REF ⎟ = ( 0.026 ) ln ⎜ ⎟ ⇒ VEB 2 = 0.521 V −12 ⎝ 10 ⎠ ⎝ IS ⎠ 5 − 0.521 R1 = ⇒ R1 = 8.96 kΩ b. 0.5 c. Combining Equations (10.79), (10.80), and (10.81), we find ⎛ VEC 2 ⎞ ⎜1 + ⎟ ⎡ VAP ⎠ ⎛ V ⎞⎤ ⎛ V ⎞ I S 0 ⎢ exp ⎜ I ⎟ ⎥ ⎜ 1 + CEo ⎟ = I REF × ⎝ ⎛ VEB 2 ⎞ ⎝ VT ⎠ ⎦ ⎝ VAN ⎠ ⎣ ⎜1 + ⎟ ⎝ VAP ⎠ 2.5 ⎞ ⎛ 1+ ⎜ ⎟ ⎡ ⎤ ⎛V ⎞ ⎛ 2.5 ⎞ ⎝ 100 ⎠ −3 = × 10−12 ⎢ exp ⎜ I ⎟ ⎥ ⎜ 1 + 0.5 10 ) ⎛ 0.521 ⎞ ⎟ ( ⎝ VT ⎠ ⎦ ⎝ 100 ⎠ ⎣ ⎜1 + ⎟ 100 ⎠ ⎝

⎛V 1.025 ×10 −12 exp ⎜ I ⎝ VT

d.

⎞ ⎛ VI −4 ⎟ = 5.098 × 10 exp ⎜ ⎠ ⎝ VT

⎞ 8 ⎟ = 4.974 × 10 ⇒ VI = 0.521 V ⎠

⎛ 1 ⎞ 1 −⎜ ⎟ − VT ⎠ −38.46 ⎝ 0.026 = = ⇒ Av = −1923 Av = 1 1 1 1 + 0.01 0.01 + + VAN VAP 100 100

EX10.12 I CQ 0.8 gm = = = 30.77 mA/V 0.026 VT r0 = r02 =

VA 80 = = 100 kΩ I CQ 0.8

a.

V0 = − g mVπ 1 ( r0 || r02 ) , Vπ 1 = Vi Av = − g m ( r0 || r02 ) = − ( 30.77 ) [100 ||100] ⇒ Av = −1538

b.

Av = − g m ( r0 || r02 || RL ) Av = −

1540 = −770 − 770 = − ( 30.77 )( 50 || RL ) ⇒ ( 50 || RL ) = 25 ⇒ RL = 50 kΩ 2

EX10.13 (a) Neglecting effect of λ and RL

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I O = I REF = K n (VIQ − VTN ) 0.40 = 0.25 (VIQ − 1)

2

2

Then VIQ = 2.265 V

r0 = r02 =

b.

1 1 = = 125 kΩ λ I 0 ( 0.02 )( 0.4 )

g m = 2 K n (VIQ − VTN ) = 2 ( 0.25 )( 2.26 − 1) = 0.632 mA/V Av = − g m ( r0 || r02 ) = − ( 0.632 )(125 ||125 ) ⇒ Av = −39.5

Av = − g m ( r0 || r02 || RL ) −

c.

39.4 = − ( 0.632 )( 62.5 || RL ) ⇒ 62.5 || RL = 31.25 ⇒ RL = 62.5 kΩ 2

TYU10.1 For I 0 = 0.75 mA ⎛ 2⎞ 2 ⎞ ⎛ I REF = I 0 ⎜ 1 + ⎟ = ( 0.75 ) ⎜1 + ⎟ ⎝ 100 ⎠ ⎝ β⎠ I REF = 0.765 mA I REF = R1 =

V + − VBE ( on ) − V − R1

5 − 0.7 − ( −5 )

0.765 R1 = 12.2 kΩ

TYU10.2 10 − ( 0.7 )( 2 ) I REF = = 0.717 mA 12 I 0 ≈ I REF = 0.717 V 100 r0 = A = ⇒ r0 = 139.5 kΩ I 0 0.717 ΔI 0 =

1 4 ΔVCE 2 = ⇒ ΔI 0 = 0.0287 mA r0 139.5

TYU10.3 I 0 = I REF ⋅

I B3 =

I0

β

1 0.50 = ⇒ I 0 = 0.4996 mA ⎛ ⎞ ⎛ 2 2 ⎞ ⎜⎜ 1 + β (1 + β ) ⎟⎟ ⎜⎜ 1 + 50 ( 51) ⎟⎟ ⎝ ⎠ ⎝ ⎠

⇒ I B 3 = 9.99 μ A

⎛1+ β ⎞ I E3 = ⎜ ⎟ I C 3 = I E 3 = 0.5096 mA ⎝ β ⎠ IE3 0.5096 IC 2 = = ⇒ I C 2 = 0.490 mA = I C1 2⎞ ⎛ 2⎞ ⎛ ⎜ 1 + ⎟ ⎜ 1 + 50 ⎟ ⎠ ⎝ β⎠ ⎝ I B1 = I B 2 =

IC 2

β

⇒ I B1 = I B 2 = 9.80 μ A

TYU10.4

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I REF =

k n′1 ⎛ W ⎞ kn′3 ⎛ W ⎞ 2 2 ⎜ ⎟ (VGS 1 − VTN 1 ) = ⎜ ⎟ ( 5 − VGS 1 − VTN 3 ) 2 ⎝ L ⎠1 2 ⎝ L ⎠3 1

⎡⎛ 40 ⎞ ⎛ 17.3 ⎞ ⎤ ⎢⎜ 38 ⎟ ⎜ 2.70 ⎟ ⎥ (VGS1 − 0.98 ) = ( 3.98 − VGS 1 ) ⇒ VGS 1 = 1.814 V ⎠⎦ ⎣⎝ ⎠ ⎝ 2 ⎛ 0.040 ⎞ I REF = ⎜ ⎟ (17.3)(1.814 − 0.98 ) ⎝ 2 ⎠ I REF = 0.241 mA 2

kn′2 ⎛ W ⎞ 2 ⎜ ⎟ (VGS 1 − VTN 2 ) 2 ⎝ L ⎠2 2 ⎛ 0.042 ⎞ =⎜ ⎟ ( 6.92 )(1.814 − 1.0 ) ⎝ 2 ⎠ I O = 0.0963 mA IO =

TYU10.5 a. From Equation (10.52), ⎛ 3 3 ⎞ ⎜ 1− ⎟ 12 × 10 + ⎜ 12 ⎟ × 1.8 VGS 1 = ( ) ⎜ 3 3 ⎟ 1+ ⎜1+ ⎟ 12 12 ⎠ ⎝ ⎛ 0.5 ⎞ ⎛ 1 − 0.5 ⎞ VGS 1 = ⎜ ⎟ (10 ) + ⎜ ⎟ × (1.8 ) ⎝ 1 + 0.5 ⎠ ⎝ 1 + 0.5 ⎠ VGS 1 = 3.93 V also VDS1 = 3.93 V I REF = (12 )( 0.020 ) [3.93 − 1.8] ⎡⎣1 + ( 0.01)( 3.93) ⎤⎦ ⇒ I REF = 1.13 mA (W / L )2 (1 + λVDS 2 ) b. I 0 = I REF × × (W / L )1 (1 + λVDS1 ) 2

⎛ 6 ⎞ ⎡1 + ( 0.01)( 2 ) ⎤⎦ ⇒ I 0 = 0.555 mA I 0 = (1.13) × ⎜ ⎟ × ⎣ ⎝ 12 ⎠ ⎡⎣1 + ( 0.01)( 3.93) ⎤⎦

c.

For VDS 2 = 6 V ⇒ I o = 0.576 mA

TYU10.6 K n1 (VGS 1 − VTN ) = K n3 (VGS 3 − VTN ) 2

2

⎛ 0.10 ⎞ VGS 1 − 2 = ⎜⎜ ⎟⎟ (VGS 3 − 2 ) ⎝ 0.25 ⎠ VGS 1 − 2 = ( 0.6325 )(VGS 3 − 2 ) VGS 3 = 10 − VGS 1 VGS 1 − 2 = ( 0.6325 )(10 − VGS 1 ) − ( 0.6325 )( 2 )

1.6325VGS1 = 7.06 ⇒ VGS 1 = 4.325 V

I REF = K n1 (VGS 1 − VTN ) = ( 0.25 ) ( 4.325 − 2 ) ⇒ I REF = 1.35 mA 2

2

I O = 3K n 2 (VGS 1 − VTN ) = 3 ( 0.25 )( 4.325 − 2 ) I O = 3I REF ⇒ I O = 4.05 mA 2

2

TYU10.7

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I REF = 0.20 = K n1 (VGS 1 − VTN ) = 0.15 (VGS 1 − 1) ⇒ VGS 1 = VGS 2 = 2.15 V 2

I O = K n 2 (VGS 2 − VTN ) = 2

I O = K n 3 (VGS 3 − VTN )

2

0.15 2 ( 2.15 − 1) ⇒ I O = 0.10 mA 2

2

0.10 = 0.15 (VGS 3 − 1) ⇒ VGS 3 = 1.82 V 2

TYU10.8 All transistors are identical ⇒ I 0 = I REF = 250 μ A I REF = K n (VGS − VTN )

2

0.25 = 0.20 (VGS − 1) ⇒ VGS = 2.12 V 2

TYU10.9 For Q1 : iD = I DSS 1 (1 + λ vDS1 ) 2

⎛ v ⎞ For Q2 : iD = I DSS 2 ⎜ 1 − GS 2 ⎟ (1 + λ vDS 2 ) VP ⎠ ⎝ vGS 2 = −vDS1 and vDS 2 = VDS − vDS 1 So 2

⎡ −v ⎤ I DSS 1 (1 + λ vDS1 ) = I DSS 2 ⎢1 − DS 1 ⎥ ⎡⎣1 + λ (VDS − vDS1 ) ⎤⎦ VP ⎦ ⎣ I DSS 1 = I DSS 2 2

⎡ v ⎤ ⎡⎣1 + ( 0.1) vDS1 ⎤⎦ = ⎢1 − DS1 ⎥ ⎡⎣1 + ( 0.1) ( 3) − ( 0.1) vDS 1 ⎤⎦ 2 ⎦ ⎣ 2 1 + 0.1vDS 1 = (1 − vDS 1 + 0.25vDS 1 ) (1.3 − 0.1vDS 1 )

3 2 This becomes 0.025vDS 1 − 0.425vDS 1 + 1.5vDS 1 − 0.3 = 0 We find vDS 1 = 0.2127 V, vDS 2 = 2.787 V, vGS 2 = −0.2127 V

iD = I DSS 1 (1 + λ vDS 1 ) = 2 ⎡⎣1 + ( 0.1)( 0.2127 ) ⎤⎦ iD = 2.04 mA R0 = r02 + r01 (1 + g m 2 r02 )

2 I DSS ⎛ vGS 2 ⎞ 2 ( 2 ) ⎛ −0.2127 ⎞ ⎜1 − ⎟= ⎜1 − ⎟ VP ⎠ −VP ⎝ 2 ⎝ −2 ⎠ g m = 1.787 mA/V 1 1 r02 = r04 = = = 5 kΩ λ I DSS ( 0.1)( 2 ) gm =

R0 = 5 + 5 ⎡⎣1 + (1.787 )( 5 ) ⎤⎦ ⇒ R0 = 54.7 kΩ

TYU10.10

a. b.

⎛ 0.1× 10−3 ⎞ ⇒ VEB 2 = 0.557 V VEB 2 = ( 0.026 ) ln ⎜ −14 ⎟ ⎝ 5 × 10 ⎠ 5 − 0.557 R1 = ⇒ R1 = 44.4 kΩ 0.1

c.

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⎛ VEC 2 ⎞ ⎜ 1+ V ⎟ ⎤ ⎞ ⎛ VCE 0 ⎞ AP ⎟ ⎜ + = × I 1 ⎟ REF ⎟⎥ ⎜ V V ⎜ EB 2 ⎟ ⎠⎦ ⎝ AN ⎠ ⎜ 1+ V ⎟ AP ⎠ ⎝ 2.5 ⎞ ⎛ ⎜ 1 + 100 ⎟ ⎡ ⎛ VI ⎞ ⎤ ⎛ 2.5 ⎞ −14 −3 5 × 10 ⎢ exp ⎜ ⎟ ⎥ ⎜ 1 + ⎟ = ( 0.1× 10 ) ⎜ 0.557 ⎟ ⎝ VT ⎠ ⎦ ⎝ 100 ⎠ ⎣ ⎜⎜ 1 + ⎟⎟ 100 ⎠ ⎝ ⎛ ⎞ ( 5.125 ×10−14 ) exp ⎜ VVI ⎟ = 1.019 ×10−4 ⎝ T⎠ ⎡ ⎛V I S 0 ⎢ exp ⎜ I ⎝ VT ⎣

⎛V exp ⎜ I ⎝ VT 1 − 0.026 Av = ⇒ 1 1 + 100 100

d.

⎞ 9 ⎟ = 1.988 × 10 ⇒ VI = 0.557 V ⎠ Av = −1923

TYU10.11 I REF = K p1 (VSG + VTP )

a.

2

0.25 = 0.20 (VSG − 1) ⇒ VSG = 2.12 V 2

b. VDSO

From Equation (10.89) ⎡1 + λP (V + − VSG ) ⎤ K (V − V )2 TN ⎦− n I = Vo = ⎣ I REF ( λn + λP ) λn + λP

5=

1 + ( 0.015 )(10 − 2.12 ) 0.030

( 0.2 )(VI − 1) 0.25 ( 0.030 )

2

−

0.15 = 1.12 − 0.8 (VI − 1) ⇒ VI = 2.10 V 2

c.

Av =

−2 K n (VI − VTN ) I REF ( λn + λP )

Av = −

2 ( 0.2 )( 2.10 − 1.0 ) 0.25 ( 0.030 )

⇒ Av = −58.7

TYU10.12

(a)

I REF = K p1 (VSG + VTP )

2

80 = 50 (VSG − 1) ⇒ VSG = 2.26 V 2

(b)

VDSo

⎡1 + λ p (V + − VSG ) ⎤ K (V − V )2 TN ⎦− n I = Vo = ⎣ λn + λ p I REF ( λn + λ p )

⎡1 + ( 0.015 )(10 − 2.26 ) ⎤⎦ ( 50 )(VI − 1) − 5= ⎣ 0.030 (80 )( 0.030 )

2

20.83 (VI − 1) = 32.2 ⇒ VI = 2.243 V 2

(c)

Av =

−2 K n (VI − VTN ) I REF ( λn + λ p )

=

−2 ( 50 )( 2.243 − 1)

(80 )( 0.030 )

⇒ Av = −51.8

TYU10.13 a.

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gm =

IC 0 0.5 = ⇒ g m = 19.2 mA/V 0.026 VT

r0 =

VAN 120 = ⇒ r0 = 240 kΩ I CQ 0.5

r02 =

VAP 80 = ⇒ r02 = 160 kΩ I CQ 0.5 Av = − g m ( r0 || r02 || RL ) = − (19.2 ) [ 240 ||160 || 50] ⇒ Av = −631

b.

TYU10.14 1 = 38.46 mA/V 0.026 (100 )( 0.026 ) rπ 1 = rπ 2 = = 2.6 K 1 80 rO1 = rO 2 = = 80 K 1 120 rO = = 120 K 1 1 80 = 0.0257 K RO1 = 2.6 38.46 For R1 = 9.3 K I C = 1mA, g m =

( RO1 + RE ) = 9.3 ( 0.0257 + 1) = 0.924 K RE′′ = 1 [ 2.6 + 0.924] = 0.779 K RO 2 = 80 ⎡⎣1 + ( 38.46 )( 0.779 ) ⎤⎦ = 2476.7 K Av = − g m ( rO || RO 2 ) = − ( 38.46 )(120 || 2476.7 ) = − ( 38.46 )(114.5 ) R ′ = R1

Av = −4404 For RL = 100 K Av = −38.46 ⎡⎣114.5 100 ⎤⎦ = −2053 For RL = 10 K

Av = −38.46 [114.5 ||10] = −354

TYU10.15 M 1 and M 2 identical ⇒ I o = I REF a. I O = K n (VI − VYN )

2

0.25 = 0.2 (VI − 1) VI = 2.12 V g m = 2 K n (VI − VTN ) = 2 ( 0.2 )( 2.12 − 1) ⇒ g m = 0.447 mA/V 2

r0 n = r0 p =

b.

1

λn I 0

=

1

( 0.01)( 0.25 )

⇒ r0 n = 400 kΩ

1 1 = ⇒ r0 p = 200 kΩ λ p I 0 ( 0.02 )( 0.25 ) Av = − g m ( r0 || r02 || RL ) Av = − ( 0.447 ) [ 400 || 200 ||100] ⇒ Av = −25.5

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Chapter 10 Problem Solutions 10.1 a.

I1 = I 2 =

0 − 2Vγ − V − R1 + R2

2Vγ + I 2 R2 = VBE + I C R3 R2 2Vγ + ( −2Vγ − V − ) = VBE + IC R3 R1 + R2

b.

⎧⎪ ⎫⎪ R2 ⎞ − ⎛ ⎨2Vγ − ( 2Vγ + V ) ⎜ ⎟ − VBE ⎬ ⎪⎩ ⎪⎭ ⎝ R1 + R2 ⎠ Vγ = VBE and R1 = R2 IC =

1 R3

IC =

1 ⎧ 1 ⎫ − ⎨2Vγ − ( 2Vγ + V ) − VBE ⎬ R3 ⎩ 2 ⎭

or I C =

c.

−V − 2 R3

I C = 2 mA =

− ( −10 ) 2 R3

I1 = I 2 = 2 mA =

⇒ R3 = 2.5 kΩ

−2 ( 0.7 ) − ( −10 ) R1 + R2

⇒ R1 + R2 = 4.3 kΩ ⇒ R1 = R2 = 2.15 kΩ

10.2 (a)

⎛I ⎞ VBE1 = VT ln ⎜ C1 ⎟ ⎝ IS ⎠

(i)

I REF = I C1 = 10 μ A,

(ii)

I REF = I C1 = 100 μ A,

(iii)

(b)

⎛ 10 × 10−6 ⎞ VBE1 = ( 0.026 ) ln ⎜ ⎟ = 0.5388 V −14 ⎝ 10 ⎠ I O = 10 μ A ⎛ 100 × 10−6 ⎞ VBE1 = ( 0.026 ) ln ⎜ ⎟ = 0.5987 V −14 ⎝ 10 ⎠ I O = 100 μ A

⎛ 10−3 ⎞ I REF = I C1 = 1 mA, VBE1 = ( 0.026 ) ln ⎜ −14 ⎟ = 0.6585 V ⎝ 10 ⎠ I O = 1 mA IO =

I REF 2 1+

β

(i)

IO =

10 ⇒ I O = 9.615 μ A VBE1 = VBE 2 2 1+ 50

⎛I ⎞ = VT ln ⎜ O ⎟ ⎝ IS ⎠ ⎛ 9.615 × 10−6 ⎞ = ( 0.026 ) ln ⎜ ⎟ −14 ⎝ 10 ⎠ = 0.5378 V

(ii)

IO =

⎛ 96.15 × 10−6 ⎞ 100 ⇒ I O = 96.15 μ A VBE1 = ( 0.026 ) ln ⎜ ⎟ −14 2 ⎝ 10 ⎠ 1+ 50 = 0.5977 V

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IO =

(iii)

10.3 I REF =

⎛ 0.9615 × 10−3 ⎞ ⇒ I O = 0.9615 mA VBE1 = ( 0.026 ) ln ⎜ ⎟ 2 10−14 ⎝ ⎠ 1+ 50 = 0.6575 V 1

V + − VBE ( on ) − V − R1

⇒ 0.250 =

3 − 0.7 − ( −3) R1

R1 = 21.2 K I REF 0.250 = ⇒ I C1 = I C 2 = 0.2419 mA 2 2 1+ 1+ 60 β = 4.03 μ A

I C1 = I C 2 = I B1 = I B 2

10.4 I REF =

V + − VBE ( on ) − V − R1

=

5 − 0.7 − ( −5 ) 18.3

I REF = 0.5082 mA I REF 0.5082 = ⇒ I C1 = I C 2 = 0.4958 mA 2 2 1+ 1+ 80 β = ( 6.198 μ A )

I C1 = I C 2 = I B1 = I B 2

10.5 (a)

I REF =

(b)

R1 =

V + − VBE ( on ) − V − R1

V + − VBE ( on ) − V −

=

or R1 =

15 − 0.7 − ( −15 )

0 − 0.7 − ( −15 )

I REF 0.5 Advantage: Requires smaller resistance. (c) For part (a): 29.3 = 0.526 mA I O ( max ) = ( 58.6 )( 0.95 ) I O ( min ) =

0.5

⇒ R1 = 58.6 k Ω

⇒ R1 = 28.6 k Ω

29.3 = 0.476 mA ( 58.6 )(1.05 )

ΔI O = 0.526 − 0.476 = 0.05 mA ⇒ ±5% For part (b): 14.3 = 0.526 mA I O ( max ) = ( 28.6 )( 0.95)

I O ( min ) =

14.3 = 0.476 mA ( 28.6 )(1.05 )

ΔI O = 0.05 mA ⇒ ±5%

10.6 a.

⎛ 2⎞ 2 ⎞ ⎛ I REF = I 0 ⎜ 1 + ⎟ = 2 ⎜ 1 + ⎟ or I REF = 2.04 mA ⎝ 100 ⎠ ⎝ β⎠ 15 − 0.7 R1 = ⇒ R1 = 7.01 kΩ 2.04

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r0 =

b.

VA 80 = = 40 kΩ 2 I0

ΔI 0 1 ⎛ 1 ⎞ = ⇒ ΔI 0 = ⎜ ⎟ ( 9.3) = 0.2325 mA ΔVCE r0 ⎝ 40 ⎠ ΔI 0 0.2325 ΔI = ⇒ 0 = 11.6% 2 I0 I0

10.7 I 0 = nI C1 I REF = I C1 + I B1 + I B 2 = I C1 +

I C1

β

+

I0

β

⎛ ⎛ 1+ n ⎞ 1 n⎞ I REF = I C1 ⎜ 1 + + ⎟ = I C1 ⎜ 1 + ⎟ β ⎠ ⎝ β β⎠ ⎝ =

I0 ⎛ 1 + n ⎞ nI REF ⎜1 + ⎟ or I 0 = n⎝ β ⎠ ⎛ 1+ n ⎞ ⎜1 + β ⎟⎠ ⎝

10.8 IO =

I REF 2 ⎞ ⎛ ⇒ I REF = ( 0.20 ) ⎜ 1 + ⎟ = 0.210 mA 2 40 ⎝ ⎠ 1+

β

5 − 0.7 4.3 = ⇒ R1 = 20.5 K R1 = I REF 0.21

10.9 a.

b.

5 − 0.7 = 0.239 mA 18 0.239 I0 = ⇒ I 0 = 0.230 mA 2 1+ 50 VA 50 r0 = = = 218 kΩ I 0 0.230

I REF =

1 ⎛ 1 ⎞ ⋅ ΔVEC = ⎜ ⎟ (1.3) = 0.00597 mA ⇒ I 0 = 0.236 mA r0 ⎝ 217 ⎠ ⎛ 1 ⎞ ΔI 0 = ⎜ ⎟ ( 3.3) = 0.01516 mA ⇒ I 0 = 0.245 mA ⎝ 217 ⎠ ΔI 0 =

c. 10.10

5 − 0.7 − ( −5 )

⇒ R1 = 9.3 kΩ

a.

I REF = 1 =

b.

I 0 = 2 I REF ⇒ I 0 = 2 mA

c.

For VEC 2 ( min ) = 0.7 ⇒ RC 2 =

R1

5 − 0.7 ⇒ RC 2 = 2.15 kΩ 2

10.11

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I O = 0.50 mA ⇒ I OA = I OB = 0.25 mA ⎛ 3⎞ 3 ⎞ ⎛ I REF = I OA ⎜ 1 + ⎟ = 0.25 ⎜ 1 + ⎟ ⎝ 60 ⎠ ⎝ β⎠ I REF = 0.2625 mA R1 =

2.5 − 0.7 ⇒ R1 = 6.86 K 0.2625

10.12

R1 =

10 − 0.7 = 37.2 K 0.25

10.13 I 2 = 2 I1 and I3 = 3I1 (a) I 2 = 1.0 mA, I 3 = 1.5 mA (b) I1 = 0.25 mA, I 3 = 0.75 mA (c) I1 = 0.167 mA, I 2 = 0.333 mA 10.14 a.

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I 0 = I C1 and I REF = I C1 + I B 3 = I C1 + VBE 2 I C1 VBE = + R2 R2 β

I E 3 = I B1 + I B 2 + I REF = I C1 + I REF −

2 I C1

β (1 + β )

+

VBE + 1 ( β ) R2

⎛ ⎞ VBE 2 = I 0 ⎜⎜1 + ⎟⎟ (1 + β ) R2 ⎝ β (1 + β ) ⎠

I REF − I0 =

IE3 1+ β

VBE + 1 ( β ) R2

⎛ ⎞ 2 ⎜⎜ 1 + β (1 + β ) ⎟⎟ ⎝ ⎠ ⎛ ⎞ 2 0.7 I REF = ( 0.70 ) ⎜⎜ 1 + ⎟⎟ + ( 81)(10 ) 80 81 ( )( ) ⎝ ⎠ I REF = 0.700216 + 0.000864

b.

I REF = 0.7011 mA =

10 − 2 ( 0.7 ) R1

⇒ R1 = 12.27 kΩ

10.15 a. I ES 1+ β = (1 + N ) I BR

I 0i = I CR and I REF = I CR + I BS = I CR + I ES = I BR + I B1 + I B 2 + ... + I BN =

(1 + N ) I CR β

Then I REF = I CR + or I 0i =

b.

(1 + N ) I CR β (1 + β )

I REF

⎛ (1 + N ) ⎞ ⎜⎜1 + ⎟⎟ ⎝ β (1 + β ) ⎠ ⎡ ⎤ 6 I REF = ( 0.5 ) ⎢1 + ⎥ = 0.5012 mA ⎣⎢ ( 50 )( 51) ⎦⎥ R1 =

5 − 2 ( 0.7 ) − ( −5 ) 0.5012

⇒ R1 = 17.16 kΩ

10.16

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⎛ ⎞ ⎡ ⎤ 2 2 I REF = I 0 ⎜⎜ 1 + ⎟⎟ = ( 0.5 ) ⎢1 + ⎥ ⇒ I REF = 0.5004 mA ⎝ β (1 + β ) ⎠ ⎣⎢ ( 50 )( 51) ⎦⎥ 5 − 2 ( 0.7 ) − ( −5 ) R1 = ⇒ R1 = 17.19 kΩ 0.5004

10.17

I 0 = I REF ⋅

1

⎛ ⎞ 2 ⎜⎜ 1 + ⎟⎟ ⎝ β (2 + β ) ⎠ For I 0 = 0.8 mA ⎛ 2 ⎞ I REF = ( 0.8 ) ⎜⎜ 1 + ⎟⎟ ⇒ I REF = 0.8024 mA ⎝ 25 ( 27 ) ⎠ 18 − 2 ( 0.7 ) ⇒ R1 = 20.69 kΩ R1 = 0.8024

10.18

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The analysis is exactly the same as in the text. We have 1 I 0 = I REF ⋅ ⎛ ⎞ 2 ⎜⎜ 1 + ⎟⎟ ⎝ β (2 + β ) ⎠ 10.19 2 = 0.0267 mA 75 1 I C1 = 1 mA, I B1 = = 0.0133 mA 75 I E 3 = I B1 + I B 2 = 0.0133 + 0.0267 = 0.04 mA I 0 = 2 mA, I B 2 =

I E3 0.04 = = 0.000526 mA 1+ β 76 = I C1 + I B 3 ⇒ I REF = 1.000526 ≈ 1 mA

I B3 = I REF R1 =

10 − 2 ( 0.7 ) I REF

=

8.6 ⇒ R1 = 8.6 kΩ 1

10.20 (a) Assuming RO ≈ rO 3 = RO =

VA V = A I O I REF

β ro3 2 100 = = 400 K 0.25

(100 )( 400 ) 2

⇒ RO = 20 MΩ

(b) RO =

ΔV ΔV 5 ⇒ ΔI O = = ΔI O 20 MΩ 20 MΩ

ΔI O = 0.25 μ A

10.21 I REF =

V + − VBE1 − V − 5 − 0.7 = 9.3 R1

I REF = 0.4624 mA VT ⎛ I REF ⎞ 0.026 ⎛ 0.4624 ⎞ ln ⎜ ln ⎜ ⎟= ⎟ 1.5 RE ⎝ I O ⎠ ⎝ IO ⎠ ⎛ 0.4624 ⎞ I O = 0.01733ln ⎜ ⎟ ⎝ IO ⎠ IO =

By trial and error I O  41.7 μ A VBE 2 = 0.7 − I O RE VBE 2 = 0.7 − ( 0.0417 )(1.5 ) VBE 2 = 0.6375 V 10.22 (a)

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I REF =

V + − VBE1 − V − 5 − 0.7 − ( −5 ) = ⇒ I REF = 93 μ A 100 R1

⎛I ⎞ ⎛ 93 × 10−3 mA ⎞ I O RE = VT ln ⎜ REF ⎟ ⇒ I O (10 ) = 0.026 ln ⎜ ⎟ IO ⎝ IO ⎠ ⎝ ⎠ By trial and error, I O ≅ 6.8 μ A Ro = ro 2 (1 + g m 2 RE′ )

Now 30 = 4.41 M Ω 6.8 0.0068 = 0.2615 mA / V gm2 = 0.026 (100 )( 0.026 ) = 382.4 k Ω rπ 2 = 0.0068 So RE′ = rπ 2 || RE = 382 || 10 = 9.74 k Ω Then Ro = 4.41 ⎡⎣1 + ( 0.262 )( 9.74 ) ⎤⎦ ⇒ Ro = 15.6 M Ω ro 2 =

VBE1 − VBE 2 = I o RE = ( 0.0068 )(10 ) ⇒ VBE1 − VBE 2 = 0.068 V

(b) 10.23

I ⋅ ΔVC R0

ΔI 0 =

R0 = r02 (1 + g m 2 RE′ ) V 80 r02 = A = = 11.76 MΩ I 0 6.8 gm2 = rπ 2 =

I 0 0.0068 = = 0.2615 mA/V VT 0.026

(80 )( 0.026 )

= 306 kΩ 0.0068 RE′ = RE & rπ 2 = 10 & 306 = 9.68 K R0 = (11.76 ) ⎡⎣1 + ( 0.2615 )( 9.68 ) ⎤⎦ = 41.54 MΩ

Now ⎛ 1 ⎞ ΔI 0 = ⎜ ⎟ ( 5 ) ⇒ ΔI 0 = 0.120 μ A ⎝ 41.54 ⎠

10.24 (a)

I REF =

5 − 0.7 − ( −5 ) R1

= 0.50

R1 = 18.6 K ⎛I ⎞ I O RE = VT ln ⎜ REF ⎟ ⎝ IO ⎠ 0.026 ⎛ 0.50 ⎞ RE = ln ⎜ ⎟ 0.050 ⎝ 0.050 ⎠ RE = 1.20 K

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(b)

RO = rc 2 [1 + RE′ g m 2 ] RE′ = RE rπ 2 rπ 2 =

( 75 )( 0.026 )

= 39 K

0.050 VA 100 ro 2 = = ⇒ 2 MΩ I O 0.05

(c)

gm2 =

0.050 = 1.923 mA/V 0.026

RE′ = 1.20 39 = 1.164 K

RO = 2 ⎡⎣1 + (1.164 )(1.923) ⎤⎦ ⇒ RO = ( 6.477 ) MΩ ΔV 5 ΔI O = = = 0.772 μ A RO 6.477 ΔI O 0.772 × 100% = × 100 = 1.54% IO 50

10.25 Let R1 = 5 k Ω, Then I REF =

12 − 0.7 − ( −12 ) 5

⇒ I REF = 4.66 mA

Now ⎛I ⎞ 0.026 ⎛ 4.66 ⎞ I O RE = VT ln ⎜ REF ⎟ ⇒ RE = ln ⎜ ⎟ ⇒ RE ≅ 1 k Ω 0.10 ⎝ 0.10 ⎠ ⎝ IO ⎠

10.26 ⎛I ⎞ VBE = VT ln ⎜ REF ⎟ ⎝ IS ⎠ ⎛ 10−3 ⎞ −15 0.7 = ( 0.026 ) ln ⎜ ⎟ ⇒ I S = 2.03 × 10 A I ⎝ S ⎠ ⎛ 2 × 10−3 ⎞ At 2 mA, VBE = ( 0.026 ) ln ⎜ −15 ⎟ ⎝ 2.03 × 10 ⎠ = 0.718 V 15 − 0.718 ⇒ R1 = 7.14 kΩ 2 ⎛ I ⎞ 0.026 ⎛ 2 ⎞ V RE = T ln ⎜ REF ⎟ = ⋅ ln ⎜ ⎟ ⇒ RE = 1.92 kΩ I 0 ⎝ I 0 ⎠ 0.050 ⎝ 0.050 ⎠ R1 =

10.27 a. 10 − 0.7 = 0.465 mA 20 Let V − = 0 I REF ≈

⎛I ⎞ VBE ≅ VT ln ⎜ REF ⎟ ⎝ IS ⎠ ⎛ 10−3 ⎞ −15 0.7 = ( 0.026 ) ln ⎜ ⎟ ⇒ I S = 2.03 ×10 A I ⎝ S ⎠ Then ⎛ 0.465 × 10−3 ⎞ = 0.680 V VBE ≅ ( 0.026 ) ln ⎜ −15 ⎟ ⎝ 2.03 × 10 ⎠

Then

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I REF ≅

b. 10.28 I REF ≈

10 − 0.680 ⇒ I REF = 0.466 mA 20 RE =

VT ⎛ I REF ln ⎜ I0 ⎝ I0

10 − 0.7 − ( −10 ) 40

⎞ 0.026 ⎛ 0.466 ⎞ ⋅ ln ⎜ ⎟= ⎟ ⇒ RE = 400Ω 0.10 ⎝ 0.10 ⎠ ⎠

= 0.4825 mA

⎛I ⎞ VBE ≅ VT ln ⎜ REF ⎟ ⎝ IS ⎠ ⎛ 10−3 ⎞ −15 0.7 = ( 0.026 ) ln ⎜ ⎟ ⇒ I S = 2.03 × 10 A ⎝ IS ⎠ Now ⎛ 0.4825 × 10−3 ⎞ VBE = ( 0.026 ) ln ⎜ = 0.681 V −15 ⎟ ⎝ 2.03 × 10 ⎠ VBE1 = 0.681 V So 10 − 0.681 − ( −10 )

⇒ I REF = 0.483 mA 40 ⎛I ⎞ I 0 RE = VT ln ⎜ REF ⎟ ⎝ I0 ⎠ ⎛ 0.483 ⎞ I 0 (12 ) = ( 0.026 ) ln ⎜ ⎟ ⎝ I0 ⎠ I REF ≅

By trial and error. ⇒ I 0 ≅ 8.7 μ A VBE 2 = VBE1 − I 0 RE = 0.681 − ( 0.0087 )(12 ) ⇒ VBE 2 = 0.5766 V

10.29 VBE1 + I REF RE1 = VBE 2 + I 0 RE 2 VBE1 − VBE 2 = I 0 RE 2 − I REF RE1 For matched transistors ⎛I ⎞ VBE1 = VT ln ⎜ REF ⎟ ⎝ IS ⎠ ⎛I ⎞ VBE 2 = VT ln ⎜ 0 ⎟ ⎝ IS ⎠ ⎛I ⎞ Then VT ln ⎜ REF ⎟ = I 0 RE 2 − I REF RE1 ⎝ I0 ⎠ Output resistance looking into the collector of Q2 is increased.

10.30

V + − VBE1 − V − 5 − 0.7 − ( −5 ) = = 0.3174 mA R1 + RE1 27.3 + 2

(a)

I REF =

(b)

I O = I REF = 0.3174 mA Using the same relation as for the widlar current source.

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RO = ro 2 ⎡⎣1 + g m 2 ( RE rπ 2 ) ⎤⎦ ro 2 = rπ 2 =

VA 80 = = 252 K I O 0.3174

gm2 =

0.3174 = 12.21 mA/V 0.026

(100 )( 0.026 )

= 8.192 K RE || rπ 2 = 2 || 8.192 = 1.608 K 0.3174 RO = 252 ⎡⎣1 + (12.21)(1.608 ) ⎤⎦ ⇒ RO = 5.2 MΩ

(c) I O = I REF =

5 − 0.7 − ( −5 )

= 0.3407 mA 27.3 V 80 RO = ro 2 = A = ⇒ RO = 235 K I O 0.3407

10.31 Assume all transistors are matched. a. 2VBE1 = VBE 3 + I 0 RE ⎛I ⎞ VBE1 = VT ln ⎜ REF ⎟ ⎝ IS ⎠ ⎛I ⎞ VBE 3 = VT ln ⎜ 0 ⎟ ⎝ IS ⎠ ⎛I ⎞ ⎛I ⎞ 2VT ln ⎜ REF ⎟ − VT ln ⎜ 0 ⎟ = I 0 RE I ⎝ S ⎠ ⎝ IS ⎠ ⎡ ⎛I ⎞ ⎛I VT ⎢ ln ⎜ REF ⎟ − ln ⎜ 0 ⎢⎣ ⎝ I S ⎠ ⎝ IS 2

⎞⎤ ⎟ ⎥ = I 0 RE ⎠ ⎥⎦

⎛ I 2 REF ⎞ VT ln ⎜ ⎟ = I 0 RE ⎝ I0 I S ⎠ b. ⎛ 0.7 ⎞ −15 VBE = 0.7 V at 1 mA ⇒ 10−3 = I S exp ⎜ ⎟ or I S = 2.03 × 10 A ⎝ 0.026 ⎠ ⎛ 0.1× 10−3 ⎞ VBE at 0.1 mA ⇒ VBE = ( 0.026 ) ln ⎜ = 0.640 V −15 ⎟ ⎝ 2.03 × 10 ⎠ 0.640 Since I 0 = I REF , then VBE = I 0 RE ⇒ RE = or RE = 6.4 kΩ 0.1

10.32 (a) I REF =

5 − 0.7 − ( −5 ) R1

= 0.80 mA

R1 = 11.6 K RE 2 =

0.026 ⎛ 0.80 ⎞ ln ⎜ ⎟ ⇒ RE 2 = 1.44 K 0.050 ⎝ 0.050 ⎠

RE 3 =

0.026 ⎛ 0.80 ⎞ ln ⎜ ⎟ ⇒ RE 2 = 4.80 K 0.020 ⎝ 0.020 ⎠

(b) VBE 2 = 0.7 − ( 0.05 )(1.44 ) ⇒ VBE 2 = 0.628 V VBE 3 = 0.7 − ( 0.02 )( 4.80 ) ⇒ VBE 3 = 0.604 V

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10.33 (a) VBE1 = VBE 2 I REF =

V + − 2VBE1 − V − R1 + R2

Now 2VBE1 + I REF R2 = VBE 3 + I O RE or I O RE = 2VBE1 − VBE 3 + I REF R2 We have ⎛I ⎞ ⎛I ⎞ VBE1 = VT ln ⎜ REF ⎟ and VBE 3 = VT ln ⎜ O ⎟ ⎝ IS ⎠ ⎝ IS ⎠ (b) Let R1 = R2 and I O = I REF ⇒ VBE1 = VBE 3 ≡ VBE Then VBE = I O RE − I REF R2 = I O ( RE − R2 ) so I REF = I O = =

V + − V − − 2 I O ( RE − R2 ) 2 R2

⎛R ⎞ V −V − IO ⎜ E ⎟ + IO 2 R2 ⎝ R2 ⎠ +

−

Then IO =

V + −V − 2 Rε

(c) Want I O = 0.5 mA So RE = 2 R2 =

5 − ( −5 ) 2 ( 0.5 )

⇒ RE = 10 k Ω

5 − 2 ( 0.7 ) − ( −5 )

0.5 Then R1 = R2 = 8.6 k Ω

= 17.2 k Ω

10.34 a. 20 − 0.7 − 0.7 = 1.55 mA 12 I 01 = 2 I REF = 3.1 mA I 02 = I REF = 1.55 mA I 03 = 3I REF = 4.65 mA b. VCE1 = − I 01 RC1 − ( −10 ) = − ( 3.1)( 2 ) + 10 ⇒ VCE1 = 3.8 V I REF =

VEC 2 = 10 − I 02 RC 2 = 10 − (1.55 )( 3) ⇒ VEC 2 = 5.35 V VEC 3 = 10 − I 03 RC 3 = 10 − ( 4.65 )(1) ⇒ VEC 3 = 5.35 V

10.35 a.

Ist approximation

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20 − 1.4 = 2.325 mA 8 ⎛ 2.32 ⎞ Now VBE − 0.7 = ( 0.026 ) ln ⎜ ⎟ ⇒ VBE = VEB = 0.722 V ⎝ 1 ⎠ Then 2nd approximation 20 − 2 ( 0.722 ) I REF ≅ = 2.32 mA 8 I 01 = 2 I REF = 4.64 mA I 02 = I REF = 2.32 mA I 03 = 3I REF = 6.96 mA b. At the edge of saturation, VCE = VBE = 0.722 V I REF ≅

RC 2 RC 3

0 − 0.722 − ( −10 )

⇒ RC1 = 2.0 kΩ 4.64 10 − 0.722 ⇒ RC 2 = 4.0 kΩ = 2.32 10 − 0.722 ⇒ RC 3 = 1.33 kΩ = 6.96

RC1 =

10.36 I C1 = I C 2 =

10 − 0.7 − 0.7 − ( −10 ) 10

= 1.86 mA

I C 3 = I C 4 = 1.86 mA

⎛ 1.86 ⎞ I C 5 ( 0.5 ) = 0.026 ln ⎜ ⎟ ⎝ IC 5 ⎠ By Trial and error. ⇒ I C 5 = 0.136 mA = I C 6 = I C 7

2 I C 3 ( 0.8 ) + VCE 3 = 10 ⇒ VCE 3 = 10 − 2 (1.86 )( 0.8 ) VCE 3 = 7.02 V 5 = VEB 6 + VCE 5 + I C 5 ( 0.5 ) − 10

VCE 5 = 5 + 10 − 0.7 − ( 0.136 )( 0.5 ) VCE 5 = 14.2 V 5 = VEC 7 + I C 7 ( 0.8 )

VEC 7 = 5 − ( 0.136 )( 0.8 ) VEC 7 = 4.89 V

10.37 I C1 = I C 2 =

10 − 0.7 − 0.7 − ( −10 ) 10

⇒ I C1 = I C 2 = 1.86 mA

I C 4 = I C 5 = 1.86 mA

⎛I ⎞ ⎛ 1.86 ⎞ I C 3 RE1 = VT ln ⎜ C1 ⎟ ⇒ I C 3 ( 0.3) = 0.026 ln ⎜ ⎟ ⎝ IC 3 ⎠ ⎝ IC 3 ⎠ By trial and error I C 3 = 0.195 mA ⎛I ⎞ ⎛ 1.86 ⎞ I C 6 RE 2 = VT ln ⎜ C 5 ⎟ ⇒ I C 6 ( 0.5 ) = 0.026 ln ⎜ ⎟ ⎝ IC 6 ⎠ ⎝ IC 6 ⎠ By trial and error I C 6 = 0.136 mA

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10.38 10 − 0.7 = 1 mA 6.3 + 3 VBE ( QR ) = 0.7 V as assumed VRER = I REF ⋅ RER = (1)( 3) = 3 V I REF =

VRE1 = 3 V ⇒ RE1 =

VRE1 3 = ⇒ RE1 = 3 kΩ I 01 1

VRE 2 = 3 V ⇒ RE 2 =

VRE 2 3 = ⇒ RE 2 = 1.5 kΩ I 02 2

VRE 3 = 3 V ⇒ RE 3 =

VRE 3 3 = ⇒ RE 3 = 0.75 kΩ I 03 4

I 01 = 1 mA I 02 = 2 mA I 03 = 4 mA

10.38 VDS 2 ( sat ) = 2 V = VGS 2 − VTN 2 = VGS 2 − 1.5 ⇒ VGS 2 = 3.5 V 2 ⎛1 ⎞⎛W ⎞ I O = ⎜ μ n Cox ⎟ ⎜ ⎟ (VGS 2 − VTN 2 ) 2 L ⎝ ⎠ ⎝ ⎠2 2 ⎛W ⎞ ⎛W ⎞ 250 = ( 20 ) ⎜ ⎟ ( 3.5 − 1.5 ) ⇒ ⎜ ⎟ = 3.125 ⎝ L ⎠2 ⎝ L ⎠2

⎛1 ⎞⎛W ⎞ I REF = ⎜ μ n Cox ⎟ ⎜ ⎟ (VGS 2 − VTN 1 ) ⎝2 ⎠ ⎝ L ⎠1 2 ⎛W ⎞ ⎛W ⎞ 100 = ( 20 ) ⎜ ⎟ ( 3.5 − 1.5 ) ⇒ ⎜ ⎟ = 1.25 L ⎝ ⎠2 ⎝ L ⎠1 Now VGS 3 = 10 − VGS 2 = 10 − 3.5 = 6.5 V 2 ⎛W ⎞ ⎛W ⎞ So 100 = ( 20 ) ⎜ ⎟ ( 6.5 − 1.5 ) ⇒ ⎜ ⎟ = 0.2 ⎝ L ⎠3 ⎝ L ⎠3

10.39 I REF =

2.5 − VGS ⎛ 0.08 ⎞ 2 =⎜ ⎟ ( 6 )(VGS − 0.5 ) 15 ⎝ 2 ⎠

2.5 − VGS = 3.6 (VGS2 − VGS + 0.25 ) 3.6VGS2 − 2.6VGS − 1.6 = 0 VGS =

2.6 ± 6.76 + 23.04 2 ( 3.6 )

VGS = 1.12 V (1.1193) 2.5 − 1.1193 ⇒ I REF = 92.0 μ A ( 92.05 ) I REF = 15 I o = 92.0 μ A VDS 2 ( sat ) = VGS − VTN = 1.1193 − 0.5 VDS 2 ( sat ) = 0.619 V

10.39 a.

From Equation (10.50),

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VGS1 = VGS 2

VGS 1 = VGS 2

⎛ ⎛ 5 ⎞ 5 ⎞ ⎜ ⎟ ⎜ 1− ⎟ 25 ⎟ 5 + ⎜ 25 ⎟ 0.5 =⎜ ( ) ( ) ⎜ ⎜ 5 ⎟ 5 ⎟ 1 1 + + ⎜ ⎟ ⎜ ⎟ 25 ⎠ 25 ⎠ ⎝ ⎝ ⎛ 0.447 ⎞ ⎛ 1 − 0.447 ⎞ =⎜ ⎟ (5) + ⎜ ⎟ ( 0.5 ) 1 0.447 + ⎝ ⎠ ⎝ 1 + 0.447 ⎠ = 1.74 V

I REF ≅ K n1 (VGS 1 − VTN ) = (18 )( 25 )(1.74 − 0.5 ) ⇒ I REF = 0.692 mA 2

2

2 ⎛1 ⎞⎛W ⎞ I O = ⎜ μ n Cox ⎟ ⎜ ⎟ (VGS 2 − VTN ) (1 + λVDS 2 ) 2 L ⎝ ⎠ ⎝ ⎠2

b.

I 0 = (18 )(15 )(1.74 − 0.5 ) ⎡⎣1 + ( 0.02 )( 2 ) ⎤⎦ = ( 415 )(104 ) ⇒ I 0 = 0.432 mA 2

I 0 = ( 415 ) ⎡⎣1 + ( 0.02 )( 4 ) ⎤⎦ ⇒ I 0 = 0.448 mA

c.

10.40 (a) 2 ⎛ 80 ⎞ ⎛ W ⎞ I REF = 50 = ⎜ ⎟ ⎜ ⎟ (VGS − 0.5 ) ⎝ 2 ⎠ ⎝ L ⎠1 2.0 − VGS I REF = 0.050 = R Design such that VDS 2 ( sat ) = 0.25 = VGS − 0.5

VGS = 0.75 V 2 − 0.75 ⇒ R = 25 K R 2 ⎛ 80 ⎞ ⎛ W ⎞ ⎛W ⎞ 50 = ⎜ ⎟ ⎜ ⎟ ( 0.75 − 0.5 ) ⇒ ⎜ ⎟ = 20 ⎝ 2 ⎠ ⎝ L ⎠1 ⎝ L ⎠1

So 0.050 =

⎛W ⎞ ⎜ ⎟ 20 50 ⎛W ⎞ ⎝ L ⎠1 I REF = ⇒ = ⇒ ⎜ ⎟ = 40 100 ⎝ L ⎠ 2 IO ⎛W ⎞ ⎛W ⎞ ⎜ ⎟ ⎜ ⎟ ⎝ L ⎠2 ⎝ L ⎠2 1 1 RO = = ⇒ RO = 667 K (b) λ I O ( 0.015 )( 0.1)

(c)

ΔI O =

ΔV 1 = ⇒ 1.5 μ A RO 666

ΔI O ⎛ 1.5 ⎞ × 100% = ⎜ ⎟ × 100% ⇒ 1.5% IO ⎝ 100 ⎠

10.41 (a)

2 ⎛ 80 ⎞ I REF = 250 = ⎜ ⎟ ( 3)(VGS − 1) 2 ⎝ ⎠ VGS = 2.44 V I O = 250 μ A at VDS 2 = VGS = 2.44 V 1 1 = = 200 K RO = λ I O ( 0.02 )( 0.25 )

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ΔI O =

(i)

ΔV 3 − 2.44 = ⇒ 2.8 μ A RO 200

I O = 252.8 μ A ΔI O =

(ii)

ΔV 4.5 − 2.44 = ⇒ 10.3 μ A RO 200

I O = 260.3 μ A ΔI O =

(iii)

ΔV 6 − 2.44 = ⇒ 17.8 μ A RO 200

I O = 267.8 μ A 4.5 ( 250 ) = 375 μ A at VDS = 2.44 V 3 1 1 RO = = = 133.3 K λ I O ( 0.02 )( 0.375 ) IO =

(b)

ΔI O =

(i)

ΔV 3 − 2.44 = ⇒ 4.20 μ A RO 133.3

I O = 379.2 μ A ΔI O =

(ii)

ΔV 4.5 − 2.44 = ⇒ 15.5 μ A RO 133.3

I O = 390.5 μ A ΔI O =

(iii)

ΔV 6 − 2.44 = ⇒ 26.7 μ A RO 133.3

I O = 401.7 μ A

10.41 VSD 2 ( sat ) = 0.25 = VSG + VTP = VSG − 0.4 ⇒ VSG 2 = 0.65 V k ′p ⎛ W ⎞ 2 I O = ⎜ ⎟ (VSG 2 + VTP ) 2 ⎝ L ⎠2 40 ⎛ W ⎞ 2 ⎛W ⎞ 25 = ⎜ ⎟ ( 0.65 − 0.4 ) ⇒ ⎜ ⎟ = 20 2 ⎝ L ⎠2 ⎝ L ⎠2 I REF = 75 μ A =

(W /L )1 ⎛W ⎞ ⋅ I O ⇒ ⎜ ⎟ = 60 (W /L )2 ⎝ L ⎠1

k ′p ⎛ W ⎞ 2 ⎜ ⎟ (VSG 3 + VTP ) 2 ⎝ L ⎠3 = 3 − 0.65 = 2.35 V

I REF = VSG 3

Then 75 =

40 ⎛ W ⎞ 2 ⎛W ⎞ ⎜ ⎟ ( 2.35 − 0.4 ) ⇒ ⎜ ⎟ = 0.986 2 ⎝ L ⎠3 ⎝ L ⎠3

10.42 (a) VGS = VTN 1 +

I REF 0.5 = 1+ =2V K n1 0.5

⎛ I I O = K n 2 ⎜ REF ⎜ K n1 ⎝

2

⎞ ⎛ I REF ⎞ ⎟⎟ = K n 2 ⎜ ⎟ ⎝ K n1 ⎠ ⎠

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⎛ 0.5 ⎞ I 0 ( max ) = ( 0.5 )(1.05 ) ⎜ ⎟ ⇒ I 0 ( max ) = 0.525 mA ⎝ 0.5 ⎠ ⎛ 0.5 ⎞ I 0 ( min ) = ( 0.5 )( 0.95 ) ⎜ ⎟ ⇒ I 0 ( min ) = 0.475 mA ⎝ 0.5 ⎠ So 0.475 ≤ I 0 ≤ 0.525 mA (b) ⎡ I ⎤ I O = K n 2 ⎢ REF + VTN 1 − VTN 2 ⎥ ⎣⎢ K n1 ⎦⎥

2

2

⎡ 0.5 ⎤ I 0 ( min ) = ( 0.5 ) ⎢ + 1 − 1.05⎥ ⇒ I 0 (min) = 0.451 mA ⎣ 0.5 ⎦ 2

⎡ 0.5 ⎤ I 0 ( max ) = ( 0.5 ) ⎢ + 1 − 0.95⎥ ⇒ I 0 (max) = 0.551 mA ⎣ 0.5 ⎦ So 0.451 ≤ I 0 ≤ 0.551 mA

10.43

(1) (2)

Ix =

Vx − VA + g mVgs 2 ro

Ix =

VA + g mVgs1 ro

Vgs1 = Vx , Vgs 2 = −VA

So (1)

Ix =

⎛1 ⎞ Vx − VA ⎜ + g m ⎟ ro ⎝ ro ⎠

(2)

Ix =

VA + g mVx ⇒ VA = ro [ I x − g mVx ] ro

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Then Ix =

⎛1 ⎞ Vx − ro ( I x − g mVx ) ⎜ + g m ⎟ ro ⎝ ro ⎠

Ix =

⎡I ⎤ Vx g − ro ⎢ x + g m I x − m ⋅ Vx − g m2 Vx ⎥ ro ro ⎣ ro ⎦

Ix =

Vx − I x − g m ro I x + g mVx + g m2 roVx ro

⎡1 ⎤ I x [ 2 + g m ro ] = Vx ⎢ + g m + g m2 ro ⎥ r ⎣ o ⎦ 1 Since g m >> ro I x [ 2 + g m ro ] ≅ Vx ( g m )(1 + g m ro ) Then

Vx 2 + g m ro = Ro = Ix g m (1 + g m ro )

Usually, g m ro >> 2, so that Ro ≅

1 gm

10.44 VDS 2 (sat) = 2 = VGS 2 − 0.8 ⇒ VGS 2 = 2.8 V I O = 200 =

60 ⎛ W ⎞ 2 ⎛W ⎞ ⎜ ⎟ ( 2.8 − 0.8 ) ⇒ ⎜ ⎟ = 1.67 2 ⎝ L ⎠2 ⎝ L ⎠2

⎛W ⎞ ⎛W ⎞ ⎜ ⎟ I REF ⎜⎝ L ⎟⎠1 0.4 ⎝ L ⎠1 ⎛W ⎞ = ⇒ = ⇒ ⎜ ⎟ = 3.33 IO 0.2 (1.67 ) ⎝ L ⎠1 ⎛W ⎞ ⎜ ⎟ L ⎝ ⎠2 VGS 3 = 6 − 2.8 = 3.2 V 2 ⎛ 60 ⎞ ⎛ W ⎞ ⎛W ⎞ I REF = 400 = ⎜ ⎟ ⎜ ⎟ ( 3.2 − 0.8 ) ⇒ ⎜ ⎟ = 2.31 ⎝ 2 ⎠ ⎝ L ⎠3 ⎝ L ⎠3

10.45 (a) 2 2 ⎛ 60 ⎞ ⎛ 60 ⎞ I REF = ⎜ ⎟ ( 20 )(VGS 1 − 0.7 ) = ⎜ ⎟ ( 3)(VGS 3 − 0.7 ) ⎝ 2 ⎠ ⎝ 2 ⎠ VGS1 + VGS 3 = 5

20 (VGS1 − 0.7 ) = 5 − VGS1 − 0.7 3 3.582VGS 1 = 6.107 ⇒ VGS1 = VGS 2 = 1.705 V 2 ⎛ 60 ⎞ I O = ⎜ ⎟ (12 )(1.705 − 0.7 ) = 363.6 μ A at VDS 2 = 1.705 V ⎝ 2 ⎠ 2 ⎛ 60 ⎞ I REF = ⎜ ⎟ ( 20 )(1.705 − 0.7 ) = 606 μ A ⎝ 2 ⎠

(b)

RO = ΔI O =

1

λ IO

=

1 = 183.4 K 0.015 ( )( 0.3636 )

ΔV 1.5 − 1.705 = ⇒ −1.12 μ A RO 183.4

I O = 362.5 μ A

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(c)

ΔI O =

ΔV 3 − 1.705 = ⇒ 7.06 μ A RO 183.4

I O = 370.7 μ A

10.46 2 2 ⎛ 50 ⎞ ⎛ 50 ⎞ I REF = ⎜ ⎟ (15 )(VSG1 − 0.5 ) = ⎜ ⎟ ( 3)(VSG 3 − 0.5 ) ⎝ 2⎠ ⎝ 2⎠ VSG1 + VSG 3 = 10 ⇒ VSG 3 = 10 − VSG1

15 (VSG1 − 0.5) = 10 − VSG1 − 0.5 3 3.236VSG1 = 10.618 ⇒ VSG1 = 3.28 V 2 ⎛ 50 ⎞ I REF = ⎜ ⎟ (15 )( 3.28 − 0.5 ) ⇒ I REF = 2.90 mA ⎝ 2⎠ I O = I REF = 2.90 mA

VSD 2 (sat) = VSG 2 + VTP = 3.28 − 0.5 ⇒ VSD 2 (sat) = 2.78 V

10.47 VSD 2 (sat) = 1.2 = VSG 2 − 0.35 ⇒ VSG 2 = 1.55 V 2 ⎛ 50 ⎞⎛ W ⎞ ⎛W ⎞ I O = 100 = ⎜ ⎟⎜ ⎟ (1.55 − 0.35 ) ⇒ ⎜ ⎟ = 2.78 ⎝ 2 ⎠⎝ L ⎠ 2 ⎝ L ⎠2 W W I REF 200 ⎛W ⎞ L1 L1 = ⇒ = ⇒ ⎜ ⎟ = 5.56 W IO 100 2.78 ⎝ L ⎠1 L 2 VSG1 + VSG 3 = 4 ⇒ VSG 3 = 2.45 V

( ) ( )

( )

2 ⎛ 50 ⎞⎛ W ⎞ ⎛W ⎞ I REF = 200 = ⎜ ⎟⎜ ⎟ ( 2.45 − 0.35 ) ⇒ ⎜ ⎟ = 1.81 ⎝ 2 ⎠⎝ L ⎠3 ⎝ L ⎠3

10.48 2 2 ⎛ 80 ⎞ ⎛ 80 ⎞ I REF = ⎜ ⎟ ( 25 )(VSG1 − 1.2 ) = ⎜ ⎟ ( 4 )(VSG 3 − 1.2 ) 2 2 ⎝ ⎠ ⎝ ⎠ 10 − VSG1 VSG1 + 2VSG 3 = 10 ⇒ VSG 3 = 2 10 − VSG1 25 − 1.2 Then (VSG1 − 1.2 ) = 4 2 3VSG1 = 6.8 ⇒ VSG1 = 2.27 V 2 ⎛ 80 ⎞ I REF = ⎜ ⎟ ( 25 )( 2.267 − 1.2 ) ⇒ I REF = I O = 1.14 mA ⎝ 2⎠ VSD 2 (sat) = VSG 2 + VTP = 2.27 − 1.2 ⇒ VSD 2 ( sat ) = 1.07 V

10.49 VSD 2 (sat) = 1.8 = VSG 2 − 1.4 ⇒ VSG 2 = 3.2 V 2 ⎛ 80 ⎞ ⎛ W ⎞ ⎛W ⎞ I O = ⎜ ⎟ ⎜ ⎟ ( 3.2 − 1.4 ) = 100 ⇒ ⎜ ⎟ = 0.772 ⎝ 2 ⎠ ⎝ L ⎠2 ⎝ L ⎠2 W I REF L1 = W IO L 2 W 200 ⎛W ⎞ L1 = ⇒ ⎜ ⎟ = 1.54 100 0.772 ⎝ L ⎠1

( ) ( ) ( )

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Assume M3 and M4 are matched. 10 − 3.2 = 3.4 V 2VSG 3 + VSG1 = 10 ⇒ VSG 3 = 2 2 ⎛ 80 ⎞ ⎛ W ⎞ I REF = 200 = ⎜ ⎟ ⎜ ⎟ ( 3.4 − 1.4 ) L 2 ⎝ ⎠ ⎝ ⎠3,4 ⎛W ⎞ ⎜ ⎟ = 1.25 ⎝ L ⎠3,4

10.50 (a) ⎛ k ′p ⎞ ⎛ W ⎞ 2 I REF = ⎜ ⎟ ⎜ ⎟ (VSG1 + VTP ) ⎝ 2 ⎠ ⎝ L ⎠1 ⎛ k ′p ⎞ ⎛ W ⎞ 2 = ⎜ ⎟ ⎜ ⎟ (VSG 3 + VTP ) L 2 ⎝ ⎠ 3 ⎝ ⎠ But VSG 3 = 3 − VSG1 So 25 (VSG1 − 0.4 ) = 5 ( 3 − VSG1 − 0.4 ) 2

2

which yields VSG1 = 1.08 V and VSG 3 = 1.92 V I REF = 20 ( 25 )(1.08 − 0.4 ) ⇒ I REF = 231 μ A 2

(W / L )2 15 = = 0.6 I REF (W / L )1 25 Then I O = ( 0.6 )( 231) = 139 μ A IO

=

(b) VDS 2 ( sat ) = 1.08 − 0.4 = 0.68 V VR = 3 − 0.68 = 2.32 = I O R then R=

2.32 ⇒ R = 16.7 k Ω 0.139

10.51 VSD 2 (sat) = 0.35 = VSG 2 − 0.4 ⇒ VSG 2 = 0.75 V 2 ⎛W ⎞ ⎛W ⎞ I O = 80 = ( 20 ) ⎜ ⎟ ( 0.75 − 0.4 ) ⇒ ⎜ ⎟ = 32.7 L ⎝ ⎠2 ⎝ L ⎠2

( ) ( )

( )

W W I REF 50 L1 L1 = ⇒ = ⇒ W = 20.4 L1 W 80 32.7 IO L 2 VSG 3 = 3 − 0.75 = 2.25

( )

2 ⎛W ⎞ ⎛W ⎞ I REF = 50 = ( 20 ) ⎜ ⎟ ( 2.25 − 0.4 ) ⇒ ⎜ ⎟ = 0.730 ⎝ L ⎠3 ⎝ L ⎠3

10.52 a.

I REF = K n (VGS − VTN )

2

100 = 100 (VGS − 2 ) ⇒ VGS = 3 V 2

For VD 4 = −3 V, I 0 = 100 μ A

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R0 = r04 + r02 (1 + g m r04 )

b.

r02 = r04 =

1 1 = = 500 kΩ λ I 0 ( 0.02 )( 0.1)

g m = 2 K n (VGS − VTN ) = 2 ( 0.1)( 3 − 2 ) = 0.2 mA / V

R0 = 500 + 500 ⎡⎣1 + ( 0.2 )( 500 ) ⎤⎦ R0 = 51 MΩ ΔI 0 =

1 6 ⋅ ΔVD 4 = ⇒ ΔI 0 = 0.118 μ A 51 R0

10.53

Vgs 4 = − I X r02

VS 6 = ( I X − g mVgs 4 ) r04 + I X r02

= ( I X + g m I X r02 ) r04 + I X r02

VS 6 = I X ⎣⎡ r02 + (1 + g m r02 ) r04 ⎦⎤ = −Vgs 6 ⎛ V − VS 6 VX 1 ⎞ I X = g mVgs 6 + X = − VS 6 ⎜ g m + ⎟ r06 r06 r06 ⎠ ⎝ ⎛ 1 ⎞ ⎜ g m + ⎟ ⎣⎡ r02 + (1 + g m r02 ) r04 ⎦⎤ r06 ⎠ ⎝ ⎧⎪ ⎛ ⎫⎪ V 1 ⎞ I X ⎨1 + ⎜ g m + ⎟ ⎡⎣ r02 + (1 + g m r02 ) r04 ⎤⎦ ⎬ = X r06 ⎠ ⎩⎪ ⎝ ⎭⎪ r06 VX = R0 = r06 + (1 + g m r06 ) ⎡⎣ r02 + (1 + g m r02 ) r04 ⎤⎦ IX IX =

VX − IX r06

I 0 ≈ I REF = 0.2 mA = 0.2 (VGS − 1)

2

VGS = 2 V

g m = 2 K n (VGS − VTN ) = 2 ( 0.2 )( 2 − 1) = 0.4 mA / V

r02 = r04 = r06 =

1

=

1

= 250 kΩ

( 0.02 )( 0.2 ) R0 = 250 + ⎡⎣1 + ( 0.4 )( 250 ) ⎤⎦ × {250 + ⎡⎣1 + ( 0.4 )( 250 ) ⎤⎦ ( 250 )} λ I0

R0 = 2575750 kΩ ⇒ R0 = 2.58 × 109 Ω

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10.54 kn′ ⎛ W ⎞ kn′ ⎛ W ⎞ 2 2 ⎜ ⎟ (VGS1 − VTN ) = ⎜ ⎟ (VGS 3 − VTN ) 2 ⎝ L ⎠1 2 ⎝ L ⎠3 k ′p ⎛ W ⎞ 2 = ⎜ ⎟ (VGS 4 + VTP ) 2 ⎝ L ⎠4 (1) 50 ( 20 )(VGS 1 − 0.5 ) = 50 ( 5 )(VGS 3 − 0.5 ) 2

2

(2) 50 ( 20 )(VGS 1 − 0.5 ) = 20 (10 )(VGS 4 − 0.5 ) 2

2

(3) VSG 4 + VGS 3 + VGS1 = 6 From (1) 4 (VGS 1 − 0.5 ) = (VGS 3 − 0.5 ) ⇒ VGS 3 = 2 (VGS 1 − 0.5 ) + 0.5 2

2

From (2) 5 (VGS1 − 0.5 ) = (VGS 4 − 0.5 ) ⇒ VSG 4 = 5 (VGS1 − 0.5 ) + 0.5 2

2

Then (3) becomes 5 (VGS1 − 0.5 ) + 0.5 + 2 (VGS 1 − 0.5 ) + 0.5 + VGS1 = 6 which yields VGS 1 = 1.36 V and VGS 3 = 2.22 V , VSG 4 = 2.42 V k′ ⎛ W ⎞ 2 2 Then I REF = n ⎜ ⎟ (VGS1 − VTN ) = 50 ( 20 )(1.36 − 0.5 ) or I REF = I O = 0.740 mA 2 ⎝ L ⎠1 VGS 1 = VGS 2 = 1.36 V VDS 2 ( sat ) = VGS 2 − VTN = 1.36 − 0.5 ⇒ VDS 2 ( sat ) = 0.86 V

10.55 VDS 2 ( sat ) = 0.5 V = VGS 2 − VTN = VGS 2 − 0.5 ⇒ VGS 2 = 1 V kn′ ⎛ W ⎞ 2 ⎜ ⎟ (VGS 2 − VTN ) 2 ⎝ L ⎠2 2 ⎛W ⎞ ⎛W ⎞ = 50 ⎜ ⎟ (1 − 0.5 ) ⇒ ⎜ ⎟ = 4 ⎝ L ⎠2 ⎝ L ⎠2

I O = 50 μ A =

VGS1 = VGS 2 = 1 V ⇒ I REF = 150 =

kn′ ⎛ W ⎞ 2 2 ⎛W ⎞ ⎛W ⎞ ⎜ ⎟ (VGS 1 − VTN ) = 50 ⎜ ⎟ (1 − 0.5 ) ⇒ ⎜ ⎟ = 12 2 ⎝ L ⎠1 ⎝ L ⎠1 ⎝ L ⎠1

VGS 3 + VSG 4 + VGS 1 = 6 2VGS 3 = 6 − 1 = 5 V ⇒ VGS 3 = 2.5 V 2 ⎛W ⎞ ⎛W ⎞ I REF = 150 = 50 ⎜ ⎟ ( 2.5 − 0.5 ) ⇒ ⎜ ⎟ = 0.75 ⎝ L ⎠3 ⎝ L ⎠3

k ′p ⎛ W ⎞ 2 ⎜ ⎟ (VSG 4 + VTP ) 2 ⎝ L ⎠4 2 ⎛W ⎞ ⎛W ⎞ 150 = 20 ⎜ ⎟ ( 2.5 − 0.5 ) ⇒ ⎜ ⎟ 1.88 L ⎝ ⎠4 ⎝ L ⎠4 I REF =

10.56 a. As a first approximation 2 I REF = 80 = 80 (VGS 1 − 1) ⇒ VGS 1 = 2 V Then VDS 1 ≅ 2 ( 2 ) = 4 V The second approximation 80 = 80 (VGS1 − 1) ⎡⎣1 + ( 0.02 )( 4 ) ⎤⎦ 80 2 Or = (VGS 1 − 1) ⇒ VGS 1 = 1.962 86.4 2

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Then I O = K n (VGS1 − VTN ) (1 + λnVGS 1 ) 2

= 80 (1.962 − 1) ⎡⎣1 + ( 0.02 )(1.962 ) ⎤⎦ Or I 0 = 76.94 μ A 2

b. From a PSpice analysis, I 0 = 77.09 μ A for VD 3 = −1 V and I 0 = 77.14 μ A for VD 3 = 3 V. The change is ΔI 0 ≈ 0.05 μ A or 0.065%. 10.57 a.

For a first approximation,

I REF = 80 = 80 (VGS 4 − 1) ⇒ VGS 4 = 2 V 2

As a second approximation I REF = 80 = 80 (VGS 4 − 1) ⎡⎣1 + ( 0.02 )( 2 ) ⎤⎦ Or VGS 4 = 1.98 V = VGS 1 2

I O = K n (VGS 2 − VTN ) (1 + λVGS 2 ) 2

To a very good approximation I 0 = 80 μ A b. From a PSpice analysis, I 0 = 80.00 μ A for VD 3 = −1 V and the output resistance is R0 = 76.9 MΩ. Then For VD = +3 V 1 4 ΔI 0 = ⋅ VD 3 = = 0.052 μ A R0 76.9 I 0 = 80.05 μ A

10.58 (a)

VDS 3 ( sat ) = VGS 3 − VTN or VGS 3 = VDS 3 ( sat ) + VTN = 0.2 + 0.8 = 1.0 k′ ⎛ W ⎞ 2 I D = n ⎜ ⎟ (VGS 3 − VTN ) 2⎝L⎠ 2 ⎛W ⎞ ⎛W ⎞ 50 = 48 ⎜ ⎟ ( 0.2 ) ⇒ ⎜ ⎟ = 26 ⎝L⎠ ⎝ L ⎠3

(b)

VGS 5 − VTN = 2 (VGS 3 − VTN ) VGS 5 = 0.8 + 2 ( 0.2 ) ⇒ VGS 5 = 1.2 V

(c)

VD1 ( min ) = 2VDS ( sat ) = 2 ( 0.2 ) ⇒ VD1 ( min ) = 0.4 V

10.59 (a) K n1 = R= =

kn′ ⎛ W ⎞ 2 ⎜ ⎟ = 50 ( 5 ) = 250 μ A / V 2 ⎝ L ⎠1 1 K n1 I D1

⎛ ⎜1 − ⎜ ⎝

(W / L )1 ⎞⎟ (W / L )2 ⎟⎠

⎛ 5 ⎞ ⎜⎜ 1 − ⎟ = ( 8.944 )( 0.6838 ) 50 ⎟⎠ ( 0.25 )( 0.05 ) ⎝ 1

R = 6.12 k Ω (b)

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V + − V − = VSD 3 ( sat ) + VGS 1 VSD 3 ( sat ) = VSG 3 + VTP I D1 = 50 = 20 ( 5 )(VSG 3 − 0.5 ) ⇒ VSG 3 = 1.207 V Then VSD 3 ( sat ) = 1.21 − 0.5 = 0.707 V 2

Also I D1 = 50 = 50 ( 5 )(VGS1 − 0.5 ) ⇒ VGS 1 = 0.9472 V 2

Then (V + − V − )

min

= 0.71 + 0.947 = 1.66 V

(c) 2 ⎛W ⎞ ⎛W ⎞ I O1 = 25 = 50 ⎜ ⎟ ( 0.947 − 0.5 ) ⇒ ⎜ ⎟ = 2.5 ⎝ L ⎠5 ⎝ L ⎠5 2 ⎛W ⎞ ⎛W ⎞ I O 2 = 75 = 20 ⎜ ⎟ (1.207 − 0.5 ) ⇒ ⎜ ⎟ = 7.5 ⎝ L ⎠6 ⎝ L ⎠6

10.60 1 ( 5 ) = 1.667 V 3 2 ⎛1 ⎞⎛W ⎞ I REF = ⎜ μ n Cox ⎟ ⎜ ⎟ (VGS 3 − VTN ) ⎝2 ⎠ ⎝ L ⎠3 2 ⎛W ⎞ ⎛W ⎞ ⎛W ⎞ ⎛W ⎞ 100 = ( 20 ) ⎜ ⎟ (1.667 − 1) ⇒ ⎜ ⎟ = ⎜ ⎟ = ⎜ ⎟ = 11.25 ⎝ L ⎠3 ⎝ L ⎠3 ⎝ L ⎠ 4 ⎝ L ⎠5 VGS 3 =

2 ⎛1 ⎞⎛W ⎞ I O1 = ⎜ μ n Cox ⎟ ⎜ ⎟ (VGS 3 − VTN ) ⎝2 ⎠ ⎝ L ⎠1 ⎛W ⎞ ⎜ ⎟ I REF ⎝ L ⎠3 Or = I 01 ⎛W ⎞ ⎜ ⎟ ⎝ L ⎠1

⎛ W ⎞ ⎛ I 01 ⎜ ⎟ =⎜ ⎝ L ⎠1 ⎝ I REF

⎞ ⎛ W ⎞ ⎛ 0.2 ⎞ ⎛W ⎞ ⎟⎜ ⎟ = ⎜ ⎟ (11.25 ) ⇒ ⎜ ⎟ = 22.5 ⎝ L ⎠1 ⎠ ⎝ L ⎠3 ⎝ 0.1 ⎠

⎛ W ⎞ ⎛ I ⎞ ⎛ W ⎞ ⎛ 0.3 ⎞ ⎛W ⎞ And ⎜ ⎟ = ⎜ 02 ⎟ ⎜ ⎟ = ⎜ ⎟ (11.25 ) ⇒ ⎜ ⎟ = 33.75 L I L 0.1 ⎝ ⎠ 2 ⎝ REF ⎠ ⎝ ⎠3 ⎝ ⎠ ⎝ L ⎠2

10.61 I REF =

24 − VSGP − VGSN R

Also I REF = 40 (1)(VGSN − 1.2 ) I REF = 18 (1)(VSGP − 1.2 )

2

2

Then 40 (VGSN − 1.2 ) = 18 (VSGP − 1.2 ) which yields VSGP =

6.325 (VGSN − 1.2 ) + 1.2 4.243

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2 Then ⎡ 0.040 (VGSN − 1.2 ) ⎤ ⋅ R = 24 − VGSN − 1.49 (VGSN − 1.2 ) − 1.2 ⎣ ⎦ which yields VGSN = 2.69 V

and VSGP = 3.43 V 24 − 3.42 − 2.69 ⇒ I REF = 89.4 μ A 200

Now I REF =

89.4 = 17.9 μ A 5 I 2 = (1.25 )( 89.4 ) = 112 μ A I 3 = ( 0.8 )( 89.4 ) = 71.5 μ A I 4 = 4 ( 89.4 ) = 358 μ A I1 =

10.61 a.

g m ( M 0 ) = 2 K n I REF gm ( M 0 ) = 2

( 0.25)( 0.2 ) ⇒ g m ( M 0 ) = 0.447 mA/V

r0 n =

1 1 = ⇒ r0 n = 250 kΩ λn I REF ( 0.02 )( 0.2 )

r0 p =

1 1 = ⇒ r0 p = 167 kΩ λ p I REF ( 0.03)( 0.2 )

b.

Av = − g m ( r0 n || r0 p ) = − ( 0.447 )( 250 ||167 ) ⇒ Av = −44.8

c.

RL = 205 ||167 = r0 n || r0 p or RL = 100 kΩ

10.62 We have VGSN = 2.69 V and VSGP = 3.43 V 10 − 2.69 − 3.43 3.88 So I REF = = ⇒ I REF = 19.4 μ A R 200 Then I1 = ( 0.2 )(19.4 ) = 3.88 μ A I 2 = (1.25 )(19.4 ) = 24.3 μ A I 3 = ( 0.8 )(19.4 ) = 15.5 μ A I 4 = 4 (19.4 ) = 77.6 μ A 10.63 I D2

W ) ( L = W ( L) (W L ) = (W L )

2

⋅ I REF =

9 ( 200 ) ⇒ I D 2 = 120 μ A 15

1

IO

4

⎛ 20 ⎞ ⋅ I D 2 = ⎜ ⎟ (120 ) ⇒ I O = 267 μ A ⎝ 9 ⎠

3

2 ⎛ 40 ⎞ I O = 266.7 = ⎜ ⎟ ( 20 )(VSG 4 − 0.6 ) 2 ⎝ ⎠ VSG 4 = 1.416 V VSD 4 (sat) = 1.416 − 0.6 ⇒ VSD 4 ( sat ) = 0.816 V

10.64 2 ⎛ 40 ⎞ ⎛ W ⎞ I REF = 50 = ⎜ ⎟ ⎜ ⎟ (VSG1 − 0.6 ) ⎝ 2 ⎠ ⎝ L ⎠1 1.75 − VSG1 I REF = = 50 R

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VSD 2 (sat) = 0.35 = VSG 2 − 0.6 ⇒ VSG 2 = 0.95 V 1.75 − 0.95 R= ⇒ R = 16 K 0.05 2 ⎛ 40 ⎞⎛ W ⎞ ⎛W ⎞ 50 = ⎜ ⎟⎜ ⎟ ( 0.95 − 0.6 ) ⇒ ⎜ ⎟ = 20.4 2 L ⎝ ⎠⎝ ⎠1 ⎝ L ⎠1

( )

W I O1 120 ⎛W ⎞ L 2 = = ⇒ = 49 50 I REF ( 20.4 ) ⎜⎝ L ⎟⎠2

( )

W I D 3 25 ⎛W ⎞ L 3 = = ⇒ ⎜ ⎟ = 10.2 I REF 50 ( 20.4 ) ⎝ L ⎠3 VDS 5 (sat) = 0.35 = VGS 5 − 0.4 ⇒ VGS 5 = 0.75 V 2 ⎛ 100 ⎞⎛ W ⎞ ⎛W ⎞ IO 2 = ⎜ ⎟⎜ ⎟ ( 0.75 − 0.4 ) = 150 ⇒ ⎜ ⎟ = 24.5 2 L ⎝ ⎠⎝ ⎠5 ⎝ L ⎠5 W I D4 I D3 25 ⎛W ⎞ L 4 = = = ⇒ ⎜ ⎟ = 4.08 I O 2 I O 2 150 24.5 ⎝ L ⎠4

( )

10.65 For vGS = 0, iD = I DSS (1 + λ vDS ) a. b. c.

VD = −5 V, vDS = 5 iD = ( 2 ) ⎡⎣1 + ( 0.05 )( 5 ) ⎤⎦ ⇒ iD = 2.5 mA VD = 0, vDS = 10 iD = ( 2 ) ⎡⎣1 + ( 0.05 )(10 ) ⎤⎦ ⇒ iD = 3 mA VD = 5 V, vDS = 15 V iD = ( 2 ) ⎡⎣1 + ( 0.05 )(15 ) ⎤⎦ ⇒ iD = 3.5 mA

10.66 ⎛ V ⎞ I 0 = I DSS ⎜ 1 − GS ⎟ VP ⎠ ⎝ ⎛ V ⎞ 2 = 4 ⎜ 1 − GS ⎟ VP ⎠ ⎝

2

2

VGS 2 = 1− = 0.293 4 VP

So VGS = ( 0.293)( −4 ) = −1.17 V VS and VS = −VGS R ( −1.17 ) −V ⇒ R = 0.586 kΩ R = GS = − 2 I0 Finish solution: See solution Then I 0 =

10.66 Completion of solution Need vDS ≥ vDS ( sat ) = vGS − VP = −1.17 − ( −4 ) vDS ≥ 2.83 V So VD ≥ vDS ( sat ) + VS = 2.83 + 1.17 ⇒ VD ≥ 4 V

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10.67 ⎛V ⎞ I REF = I S 1 exp ⎜ EB1 ⎟ ⎝ VT ⎠

a.

⎛I ⎞ ⎛ 1× 10−3 ⎞ or VEB1 = VT ln ⎜ REF ⎟ = ( 0.026 ) ln ⎜ ⇒ VEB1 = 0.5568 −13 ⎟ ⎝ 5 × 10 ⎠ ⎝ I S1 ⎠ 5 − 0.5568 R1 = ⇒ R1 = 4.44 kΩ 1 From Equations (10.79) and (10.80) and letting VCE 0 = VEC 2 = 2.5 V

b. c. 10

−12

2.5 ⎞ ⎛ ⎜ 1 + 80 ⎟ ⎛ VI ⎞ ⎡ 2.5 ⎤ −3 exp ⎜ ⎟ ⎢1 + ⎥ = 10 ⎜ 0.5568 ⎟ ⎝ VT ⎠ ⎣ 120 ⎦ ⎜⎜ 1 + ⎟⎟ 80 ⎠ ⎝

⎛V ⎞ ⎛ 1.03125 ⎞ 1.0208333 × 10−12 exp ⎜ I ⎟ = (10−3 ) ⎜ ⎟ ⎝ 1.00696 ⎠ ⎝ VT ⎠ Then VI = 0.026 ln (1.003222 × 109 )

So VI = 0.5389 V

d.

Av =

− (1/ VT )

(1/ VAN ) + (1/ VAP )

1 −38.46 Av = 0.026 = 1 1 0.00833 + 0.0125 + 120 80 Av = −1846 −

10.68 a. b. c. other.

⎛I ⎞ ⎛ 0.5 × 10−3 ⎞ VBE = VT ln ⎜ REF ⎟ = ( 0.026 ) ln ⎜ ⎟ ⇒ VBE = 0.5208 −12 ⎝ 10 ⎠ ⎝ I S1 ⎠ 5 − 0.5208 R1 = ⇒ R1 = 8.96 kΩ 0.5 Modify Eqs. 10.79 and 10.80 to apply to pnp and npn, and set the two equation equal to each

⎛ V ⎞⎛ V I CO = I SO exp ⎜ EBO ⎟⎜ 1 + ECO VAP ⎝ VT ⎠⎝

⎞ ⎛ VBE ⎞ ⎛ VCE 2 ⎞ ⎟ ⎟ = I C 2 = I S 2 exp ⎜ ⎟ ⎜1 + ⎠ ⎝ VT ⎠ ⎝ VAN ⎠

⎛ V ⎞ ⎛ 2.5 ⎞ ⎛ VBE ⎞ ⎛ 2.5 ⎞ −12 5 × 10−13 exp ⎜ EBO ⎟ ⎜ 1 + ⎟ ⎜1 + ⎟ = 10 exp ⎜ ⎟ 80 120 V V ⎠ ⎠ ⎝ T ⎠⎝ ⎝ T ⎠⎝ ⎛V ⎞ ⎛V ⎞ 5.15625 × 10−13 exp ⎜ EBO ⎟ = 1.020833 × 10−12 exp ⎜ BE ⎟ V ⎝ T ⎠ ⎝ VT ⎠ ⎛V ⎞ exp ⎜ EBO ⎟ ⎝ VT ⎠ = 1.9798 = exp ⎛ VEBO − VBE ⎞ ⎜ ⎟ VT ⎛V ⎞ ⎝ ⎠ exp ⎜ BE ⎟ V ⎝ T ⎠

VEBO = VBE + VT ln (1.9798 ) = 0.5208 + ( 0.026 ) ln (1.9798 ) VEBO = 0.5386 ⇒ VI = 5 − 0.5386 ⇒ VI = 4.461 V

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Av =

d.

− (1/ VT )

(1/ VAN ) + (1/ VAP )

1 −38.46 0.026 Av = = 1 1 0.00833 + 0.0125 + 120 80 Av = −1846 −

10.69 a. M1 and M2 matched. For I REF = I 0 , we have VSD 2 = VSG = VSG 3 = VDS 0 = 2.5 V For M1 and M3: 2 ⎛1 ⎞⎛W ⎞ I REF = ⎜ μ p Cox ⎟ ⎜ ⎟ (VSG + VTP ) (1 + λPVSD ) ⎝2 ⎠ ⎝ L ⎠1 2 ⎛W ⎞ ⎛W ⎞ ⎛W ⎞ ⎛W ⎞ 100 = 10 ⎜ ⎟ ( 2.5 − 1) ⎡⎣1 + ( 0.02 )( 2.5 ) ⎤⎦ ⇒ ⎜ ⎟ = 4.23 = ⎜ ⎟ = ⎜ ⎟ ⎝ L ⎠1 ⎝ L ⎠1 ⎝ L ⎠3 ⎝ L ⎠ 2 For M0: 2 ⎛1 ⎞⎛W ⎞ I O = ⎜ μ n Cox ⎟ ⎜ ⎟ (VGS − VTN ) (1 + λnVDS ) ⎝2 ⎠ ⎝ L ⎠0 2 ⎛W ⎞ ⎛W ⎞ 100 = 20 ⎜ ⎟ ( 2 − 1) ⎡⎣1 + ( 0.02 )( 2.5 ) ⎤⎦ ⇒ ⎜ ⎟ = 4.76 L ⎝ ⎠0 ⎝ L ⎠0 b. r0 n = r0 p =

1

λ I0

=

1

( 0.02 )( 0.1)

= 500 kΩ

⎛1 ⎞⎛W ⎞ g m = 2 K n I O = 2 ⎜ μ n Cox ⎟ ⎜ ⎟ I O 2 ⎝ ⎠ ⎝ L ⎠o

( 0.02 )( 4.76 )( 0.1)

=2

g m = 0.195 mA/V

Av = − g m ( r0 n || r0 p ) = − ( 0.195 )( 500 || 500 ) ⇒ Av = −48.8

10.70 I REF = K p1 (VSG + VTP )

a.

2

100 = 100 (VSG − 1) ⇒ VSG = 2 V 2

b.

From Eq. 10.89 ⎡1 + λ p (V + − VSG ) ⎤ K (V − V )2 TN ⎦− n I VO = ⎣ λn + λ p I REF ( λn + λ p ) ⎡1 + ( 0.02 )(10 − 2 ) ⎤⎦ 100 (VI − 1) − 5= ⎣ 0.02 + 0.02 100 ( 0.02 + 0.02 ) 2

5 = 29 −

(VI − 1)

2

(VI − 1)

2

0.04 = 0.96

VI = 1.98 V

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Av = − g m ( r0 n || r0 p )

c.

r0 n = r0 p =

1

λ I REF

=

1 = 500 kΩ 0.02 ( )( 0.1)

( 0.1)( 0.1) = 0.2 mA / V Av = − ( 0.2 )( 500 || 500 ) ⇒ Av = −50

g m = 2 K n I REF = 2

10.71 5 − 0.6 5 − 0.6 = = 0.22 mA R1 20 From Eq. 10.96 ⎛I ⎞ 0.22 −⎜ C ⎟ − VT ⎠ ⎝ 0.026 Av = = ⎛ IC I C ⎞ 0.22 + 1 + 0.22 1 + + ⎜ ⎟ 90 ⎝ VAN RL VAP ⎠ 140 RL I REF =

=

−8.4615 −8.4615 = 1 1 0.0015714 + + 0.002444 0.004016 + RL RL RL = ∞, Av = −2107

(a) (b) (c)

RL = 250 K, Av = −1056 RL = 100 K, Av = −604

10.72 I REF =

5 − 0.6 = 0.1257 mA 35

Then I CO = 2 I REF = 0.2514 mA From Eq. 10.96 −0.2514 −9.6692 0.026 Av = = 0.2514 0.2514 1 1 0.002095 + 0.0031425 + + + 120 80 RL RL Av =

(a) (b)

−9.6692 1 RL RL = ∞ Av = −1846

0.0052375 +

RL = 250 K, Av = −1047

10.73 (a) To a good approximation, output resistance is the same as the widlar current source. R0 = r02 ⎡⎣1 + g m 2 ( rπ 2 || RE ) ⎤⎦ (b)

Av = − g m 0 ( r0 || RL || R0 )

10.74 Output resistance of Wilson source

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β r03

R0 ≅

2 Then Av = − g m ( r0 || R0 ) V 80 r03 = AP = = 400 kΩ I REF 0.2 r0 =

VAN 120 = = 600 kΩ I REF 0.2

gm =

I REF 0.2 = = 7.692 mA/V VT 0.026

⎡ (80 )( 400 ) ⎤ Av = −7.69 ⎢600 ⎥ = −7.69 [ 600 ||16, 000] ⇒ Av = −4448 2 ⎢⎣ ⎥⎦

10.75 (a)

I D 2 = I D 0 = I REF = 200 μ A

For M 2 ; ro 2 =

1

λP I D 2

=

1 = 250 K 0.02 ( )( 0.2 )

⎛ 0.04 ⎞ gm2 = 2 K P I D2 = 2 ⎜ ⎟ ( 35 )( 0.2 ) ⎝ 2 ⎠ g m 2 = 0.748 mA/V 1 1 For M 0 ; r∞ = = = 333 K λn I Do ( 0.015 )( 0.2 ) ⎛ 0.08 ⎞ g mo = 2 ⎜ ⎟ ( 20 )( 0.2 ) ⇒ g mo = 0.80 mA/V ⎝ 2 ⎠ Av = − g mo ( ro 2 || roo ) = − ( 0.80 )( 250 || 333)

(b)

Av = −114.3 Want Av = −57.15 = −0.80 (142.8 || RL ) 142.8 RL 142.8 || RL = 71.375 = ⇒ RL = 143 K 142.8 + RL

(c)

10.76 Assume M1, M2 matched I REF = I D 2 = I Do = 200 μ A 1 1 ro 2 = = = 250 K λ p I D 2 ( 0.02 )( 0.2 ) roo =

1 1 = = 333 K λn I D 0 ( 0.015 )( 0.2 )

Av = − g mo ( ro 2 roo )

−100 = − g mo ( 250 333) ⇒ g mo = 0.70 mA/V ⎛ 0.08 ⎞⎛ W ⎞ g mo = 2 ⎜ ⎟⎜ ⎟ ( 0.2 ) = 0.70 ⎝ 2 ⎠⎝ L ⎠0 ⎛W ⎞ ⎜ ⎟ = 15.3 ⎝ L ⎠0

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⎛ k ′ ⎞⎛ W ⎞ ⎛ k ′p ⎞ ⎛ W ⎞ Now ⎜ n ⎟ ⎜ ⎟ = ⎜ ⎟ ⎜ ⎟ ⎝ 2 ⎠ ⎝ L ⎠0 ⎝ 2 ⎠ ⎝ L ⎠2 ⎛ 80 ⎞ ⎛ 40 ⎞⎛ W ⎞ ⎜ ⎟ (15.3) = ⎜ ⎟⎜ ⎟ ⎝ 2⎠ ⎝ 2 ⎠⎝ L ⎠ 2 ⎛W ⎞ ⎛W ⎞ ⎜ ⎟ = ⎜ ⎟ = 30.6 ⎝ L ⎠ 2 ⎝ L ⎠1

10.77

Since Vsg 3 = 0, the circuit becomes

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I x = − g m′ Vsg 2 +

Vx − Vsg 2 r02

and Vsg 2 = I x ro 3

Then ⎛ r ⎞ V I x ⎜ 1 + g m′ ro3 + o3 ⎟ = x ro 2 ⎠ ro 2 ⎝ so that ⎛ Vx r ⎞ = Ro = ro 2 ⎜1 + g m′ ro 3 + o 3 ⎟ Ix r o2 ⎠ ⎝ or Ro = ro 2 + ro 3 (1 + g m′ ro 2 ) vo = − g m1 ( ro1 Ro ) vi

Av = Now

g m1 = 2 ro1 =

( 0.050 )( 20 )( 0.10 ) = 0.632 mA / V

1 1 = = 500 k Ω λn I DQ ( 0.02 )( 0.10 )

g m′ = 2 K p I DQ = 2 ro 2 = ro 3 =

( 0.020 )(80 )( 0.1) = 0.80 mA / V

1 1 = = 500 k Ω λ p I DQ ( 0.020 )( 0.1)

Then Ro = 500 + 500 ⎡⎣1 + ( 0.8 )( 500 ) ⎤⎦ ⇒ 201 M Ω Av = − ( 0.632 ) ( 500 201000 ) ⇒ Av = −315

10.78 From Eq. 10.105 − g m2 Av = 1 1 + ro3 ro 4 ro1 ro 2 ⎛ k′ ⎞⎛W ⎞ g m = 2 ⎜ n ⎟ ⎜ ⎟ I D1 ⎝ 2 ⎠⎝ L ⎠ =2

( 0.050 ) ( 20 ) ( 0.08 )

g m = 0.5657 mA / V

ro =

1 1 = = 625 K λ I D ( 0.02 )( 0.08 )

− ( 0.5657 ) −0.3200 Av = = 1 1 2 ( 0.00000256 ) + 2 2 ( 625) ( 625) 2

Av = −62,500

10.79

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V − ( −Vπ 2 ) Vπ 2 Vπ 2 + + g m 2Vπ 2 + O rπ 2 ro1 ro 2

(1)

g m1Vi =

(2)

VO VO − ( −Vπ 2 ) + + g m 2Vπ 2 = 0 RO 3 ro 2

(1)

⎛ 1 1 1 ⎞ V + + gm2 + ⎟ + O g m1Vi = Vπ 2 ⎜ ro 2 ⎠ ro 2 ⎝ rπ 2 ro1

(2)

⎛ 1 ⎛ 1 ⎞ 1 ⎞ VO ⎜ + ⎟ + Vπ 2 ⎜ + gm2 ⎟ = 0 ⎝ RO 3 ro 2 ⎠ ⎝ ro 2 ⎠

g m >>

1 ro

(1)

⎛ 1 + β ⎞ VO g m1Vi = Vπ 2 ⎜ ⎟+ ⎝ rπ 2 ⎠ ro 2

(2)

⎛ 1 1 ⎞ VO ⎜ + ⎟ + Vπ 2 ⋅ g m 2 = 0 ⎝ RO 3 ro 2 ⎠

(3)

Vπ 2 = −

VO gm2

⎛ 1 1 ⎞ + ⎟ ⎜ ⎝ RO 3 ro 2 ⎠

Then ⎛ 1 1 ⎞⎛ 1 + β ⎞ VO + ⎟⎜ ⎜ ⎟+ ⎝ RO 3 ro 2 ⎠⎝ rπ 2 ⎠ ro 2 ⎛ 1 1 ⎞ ⎛ 1 + β ⎞ VO = −VO ⎜ + ⎟⎜ ⎟+ R r ⎝ O 3 o 2 ⎠ ⎝ β ⎠ ro 2

(1) g m1Vi = −

VO gm2

VO ⎛ 1 + β ⎞ ⎜ ⎟ RO 3 ⎝ β ⎠ ⎛ β ⎞ VO = − g m1 RO 3 ⎜ ⎟ Vi ⎝1+ β ⎠ From Equation (10.20) RO 3 ≈ β rO 3 So ≈−

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Av =

Av =

VO − g m1ro3 β 2 = 1+ β Vi

gm =

0.25 = 9.615 mA/V 0.026

ro3 =

80 = 320 K 0.25

− ( 9.615 )( 320 )(120 ) 121

2

= −366,165

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Chapter 11 Exercise Solutions EX11.1 vE = −VBE ( on ) ⇒ vE = −0.7 V I C1 = I C 2 = 0.5 mA vC1 = vC 2 = 10 − ( 0.5 )(10 ) ⇒ vC1 = vC 2 = 5 V

EX11.2 iC 2 1 = = 0.99 IQ ⎛ vd ⎞ 1 + exp ⎜ ⎟ ⎝ VT ⎠ ⎛v ⎞ 1 1 + exp ⎜ d ⎟ = 0.99 V ⎝ T⎠ ⎛v exp ⎜ d ⎝ VT

⎞ 1 −1 ⎟= 0.99 ⎠ ⎡ 1 ⎤ − 1⎥ ⇒ vd = −119.5 mV vd = VT ln ⎢ ⎣ 0.99 ⎦

EX11.3 a.

v1 = v2 = 0 ⇒ vE = 0.7 V ΔVRC = ( 0.25 )( 8 ) = 2 V ⇒ vC1 = vC 2 = −3 V ⇒ vEC1 = 3.7 V

b.

v1 = v2 = 2.5 V ⇒ vE = 3.2 V ⇒ vEC1 = 6.2 V

c.

v1 = v2 = −2.5 V ⇒ vE = −1.8 V ⇒ vEC1 = 1.2 V

EX11.4 Let I Q = 1 mA, then I CQ1 = I CQ 2 = 0.5 mA 0.5 = 19.23 mA / V 0.026 v 1 At vC 2 , Ad = c 2 = g m RC 2 vd 2 g m1 = g m 2 =

1 (19.23) RC 2 ⇒ RC 2 = 15.6 k Ω 2 v 1 At vC1 , Ad = c1 = − g m RC1 vd 2

So, 150 =

1 (19.23) RC1 ⇒ RC1 = 10.4 k Ω 2 If V + = +10 V and V − = −10 V , dc biasing is OK.

So, −100 = −

EX11.5

Acm = 1+

⎛ I Q RC ⎞ −⎜ ⎟ ⎝ 2VT ⎠ = (1 + β ) IQ RO VT β

⎡ ( 0.8 )(12 ) ⎤ −⎢ ⎥ ⎣ 2 ( 0.026 ) ⎦ = −0.0594 101)( 0.8 )(100 ) ( 1+ ( 0.026 )(100 )

vo = Acm vcm = ( −0.0594 )(1sin ω t )( mV ) = −59.4sin ω t ( μV )

EX11.6

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vd = 40 μV , vcm = 0 vo = ( 50 )( 40 ) ⇒ 2.0 mV

(a)

(b) vd = 40 μV vcm = 200 μV Assuming Acm > 0 50 Acm

60 = 20 log10

Acm = 0.05 vo = ( 50 )( 40 ) + ( 0.05 )( 200 ) vo = 2.010 mV

EX11.7

Diff. Gain Ad =

a.

I Q RC 4VT

For v1 = v2 = 5 V ⇒ Minimum collector voltage

vC 2 = 5 V ⇒

IQ

⋅ RC = 15 − 5 = 10 V or I Q RC = 20 V for max. Ad 2 20 Then Ad = ⇒ Ad ( max ) = 192 2 ( 0.026 ) If I Q = 0.5 mA, RC = 40 kΩ

b.

⎛ I Q RC ⎞ −⎜ ⎟ ⎝ 2VT ⎠ Acm = ⎡ (1 + β ) I Q R0 ⎤ ⎢1 + ⎥ VT β ⎣ ⎦ Then ⎛ 20 ⎞ − ⎜⎜ ⎟ 2 ( 0.026 ) ⎟⎠ ⎝ Acm = ⇒ Acm = −0.199 ⎡ ( 201)( 0.5 )(100 ) ⎤ ⎢1 + ( 0.026 )( 200 ) ⎥⎦ ⎣ ⎛ 192 ⎞ and C M RRdB = 20 log10 ⎜ ⎟ ⇒ C M RRdB = 59.7 dB ⎝ 0.199 ⎠

EX11.8 Rid = 2 ⎡⎣ rπ + (1 + β ) RE ⎤⎦ rπ =

β VT I CQ

=

(100 )( 0.026 ) 0.25

= 10.4 K

Rid = 2 ⎡⎣10.4 + (101)( 0.5 ) ⎤⎦ = 122 K

EX11.9 Ad =

g m RC 2 (1 + g m RE )

( 9.62 )(10 ) 2 ⎣⎡1 + ( 9.62 ) RE ⎦⎤ 1 + ( 9.62 ) RE = 4.81 ⇒ RE = 0.396 K Rid = 2 ⎡⎣ rπ + (1 + β ) RE ⎤⎦ = 2 ⎡⎣10.4 + (101)( 0.396 ) ⎤⎦ 10 =

Rid = 100.8 K

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EX11.10 10 − VGS 4 2 = K n 3 (VGS 4 − VTN ) I1 = R1 10 − VGS 4 = ( 0.1)( 80 )(VGS 4 − 0.8 )

2

10 − VGS 4 = 8 (VGS2 4 − 1.6VGS 4 + 0.64 ) 8VGS2 4 − 11.8VGS 4 − 4.88 = 0 VGS 4 =

11.8 ±

(11.8) + 4 ( 8)( 4.88) 2 (8) 2

= 1.81 V

10 − 1.81 = 0.102 mA 80 0.102 = ID2 = = 0.0512 mA 2

I1 = I Q = I D1

= K n1 (VGS 1 − VTN )

2

0.0512 = 0.050 (VGS 1 − 0.8 ) ⇒ VGS 1 = 1.81 V 2

v01 = v02 = 5 − ( 0.0512 )( 40 ) = 2.95 V Max vcm : VDS 1 ( sat ) = VGS1 − VTN

= 1.81 − 0.8 = 1.01 V vcm ( max ) = v01 − VDS 1 ( sat ) + VGS 1 = 2.95 − 1.01 + 1.81 vcm ( max )

= 3.75 V

Min vcm : VDS 4 ( sat ) = VGS 4 − VTN = 1.81 − 0.8 = 1.01 V

vcm ( min )

= VGS 1 + VDS 4 ( sat ) − 5 = 1.81 + 1.01 − 5

vcm ( min )

= −2.18 V

−2.18 ≤ vcm ≤ 3.75 V

EX11.11

(1)( 2 ) ⇒ g f ( max ) = 1 mA / V 2 ( 2) = g f RD = (1)( 5 ) ⇒ Ad = 5

g f ( max ) = Ad

K n IQ

=

EX11.12 ⎛ K Kn iD1 1 = + ⋅ vd 1 − ⎜ n ⎜ 2 IQ IQ 2 2IQ ⎝

⎞ 2 ⎟⎟ vd ⎠

Using the parameters in Example 11.11, K n = 0.5 mA / V 2 , I Q = 1 mA, then ⎛ 0.5 ⎞ 2 iD1 1 0.5 = 0.90 = + ⋅ vd 1 − ⎜⎜ ⎟⎟ vd 2 2 (1) IQ ⎝ 2 (1) ⎠ By trial and error, vd = 0.894 V EX11.13 a.

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gm =

IQ 2VT

=

0.5 = 9.615 mA/V 2 ( 0.026 )

r02 =

VA 2 125 = = 500 kΩ I C 2 0.25

r04 =

VA 4 85 = = 340 kΩ I C 4 0.25

Ad = g m ( r02 r04 ) = ( 9.615 ) ( 500 340 ) ⇒ Ad = 1946

b.

c.

(

Ad = g m r02 r04 RL

)

Ad = ( 9.615 ) ⎡⎣500 340 100 ⎤⎦ ⇒ Ad = 644 β VT (150 )( 0.026 ) rπ = = = 15.6 kΩ I CQ 0.25

Rid = 2rπ ⇒ Rid = 31.2 kΩ

d.

R0 = r02 r04 = 500 340 ⇒ R0 = 202 kΩ

EX11.14 80 80 2 2 I REF = (10 )(VGS 4 − 0.5 ) = ( 0.33)(VGS 5 − 0.5 ) 2 2 VGS 4 + 2VGS 5 = 6 1 VGS 5 = ( 6 − VGS 4 ) 2 10 1 (VGS 4 − 0.5) = ( 6 − VGS 4 ) − 0.5 0.33 2 6.0VGS 4 = 5.252 ⇒ VGS 4 = 0.8754 V 80 2 I REF = (10 )( 0.8754 − 0.5 ) = 53.37 μ A 2 Also I Q = I REF = 53.37 μ A ⎛ 80 ⎞ g m = 2kn I Q = 2 ⎜ ⎟ (10 )( 53.37 ) ⇒ 0.2066 mA/V ⎝ 2⎠ 1 1 = = 1873.7 K ro1 = ro8 = λ ID ⎛ 0.05337 ⎞ ( 0.02 ) ⎜ ⎟ 2 ⎝ ⎠ Ad = g m ( ro1 & ro8 ) = ( 0.2066 )(1873.7 & 1873.7 ) Ad = 194

EX11.15 Ad = g m ( ro 2 & RO ) 400 = g m ( 500 & 101000 ) ⇒ g m = 0.8 mA / V g m = 2 K n I DQ ⎛ 0.08 ⎞ ⎛ W ⎞ ⎛W ⎞ ⎛W ⎞ 0.8 = 2 ⎜ ⎟ ⎜ ⎟ ( 0.1) ⇒ ⎜ ⎟ = ⎜ ⎟ = 40 ⎝ 2 ⎠ ⎝ L ⎠1 ⎝ L ⎠1 ⎝ L ⎠ 2

EX11.16

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I e 6 = (1 + β ) I b 6 = I b 7 I c 7 = β I b 7 = β (1 + β ) I b 6 Ic7 = β (1 + β ) = (100 )(101) = 1.01× 104 Ib6

EX11.17 r + 300 ROA = π A 1+ β ⎛ β I CA = ⎜ ⎝1+ β ⎛ β =⎜ ⎝1+ β

⎞ ⎟ I EA ⎠ ⎞⎛ 1 ⎞ ⎟⎜ ⎟ (1) ⇒ 9.803 μ A ⎠⎝ 1 + β ⎠ (100 )( 0.026 ) rπ A = = 265.2 K 0.009803 265.2 + 300 ROA = = 5.596 K 101 r + ROA RO = π B 10 1+ β rπ B =

(100 )( 0.026 )

= 2.626 K 0.99 2.626 + 5.596 10 RO = 101 RO = 81.4 & 104 = 80.7 Ω

EX11.18 10 − 0.7 − ( −10 ) I1 = = 0.6 ⇒ R1 = 32.2 K R1 ⎛I I Q R2 = VT ln ⎜ 1 ⎜I ⎝ Q

⎞ ⎟⎟ ⎠

.

( 0.2 ) R2 = ( 0.026 ) ln ⎛⎜

0.6 ⎞ ⎟ ⇒ R2 = 143 Ω ⎝ 0.2 ⎠

I R 6 = I1 ⇒ R3 = 0 10 = I C1 RC + VCE1 − 0.7 10.7 = ( 0.1) RC + 4 ⇒ RC = 67 K vo 2 = −0.7 + 4 = 3.3 V vE 4 = 3.3 − 1.4 = 1.9 V I R4 =

vE 4 1.9 ⇒ R4 = ⇒ R4 = 3.17 K 0.6 R4

vC 3 = vO 2 − 1.4 + vCE 4 = 3.3 − 1.4 + 3 = 4.9 V I R 5 = I R 4 = 0.6 =

10 − 4.9 ⇒ R5 = 8.5 K R5

vE 5 = 4.9 − 0.7 = 4.2 V ⇒ R6 = R7 =

0 − ( −10 ) 5

4.2 − 0.7 = 5.83 K 0.6

⇒ R7 = 2 K

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EX11.19 Ri 2 = rπ 3 + (1 + β ) rπ 4

(100 )( 0.026 )

rπ 4 =

0.6

β 2VT

rπ 3 ≈

I R4

= 4.333 K

(100 ) ( 0.026 ) 2

=

0.6

= 433.3 K

Ri 2 = 433.3 + (101)( 4.333) ⇒ Ri 2 = 871 K Ri 3 = rπ 5 + (1 + β ⎣⎡ R6 + rπ 6 + (1 + β ) R7 ⎦⎤

(100 )( 0.026 )

rπ 5 =

0.6 (100 )( 0.026 )

rπ 6 =

= 4.333 K

= 0.52 K 5 Ri 3 = 4.333 + (101) ⎣⎡5.83 + 0.52 + (101)( 2 ) ⎦⎤ Ri 3 = 21.0 MΩ gm 0.1 RC Ri 2 ) g m = = 3.846 mA/V ( 2 0.026 3.846 Ad 1 = ( 67 871) = 119.64 2 ⎛I ⎞ 0.6 A2 = ⎜ R 4 ⎟ ( R5 Ri 3 ) = (8.5 21000 ) = 98.037 2 ( 0.026 ) ⎝ 2VT ⎠ Ad 1 =

A = Ad 1 ⋅ A2 = (119.64 )( 98.037 ) = 11, 729

EX11.20 fz =

1 1 = 2π RO CO 2π (10 × 106 )( 0.2 × 10−12 )

f z = 79.6 kHz fp =

1 2π Req CO

Req = 51.98 Ω from Example 11.20 1 fp = 2π ( 51.98 ) ( 0.2 × 10−12 ) f p = 15.3 GHz

TYU11.1 Vd = V1 − V2 = 2 + 0.005sin ω t − ( 0.5 − 0.005sin ω t ) ⇒ Vd = 1.5 + 0.010sin ω t ( V ) V1 + V2 2 2 + 0.005sin ω t + 0.5 − 0.005sin ω t = ⇒ Vcm = 1.25 V 2

Vcm =

TYU11.2 For v1 = v2 = +4 V ⇒ Minimum vC1 = vC 2 = 4 V IQ I C1 = I C 2 = = 1 mA 2 10 − 4 RC = ⇒ RC = 6 kΩ 1

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TYU11.3 From Equation (11.41) g R CMRR = m O ⎛ ΔRC ⎞ ⎜ ⎟ ⎝ RC ⎠

For CMRR |dB = 75 dB ⇒ CMRR = 5623.4 Then 5623.4 =

( 3.86 )(100 ) ΔRC

⋅ (10 )

Or ΔRC = 0.686 K TYU11.4 From Equation (11.49) 1 + 2 RO g m CMRR = ⎛ Δg ⎞ 2⎜ m ⎟ ⎝ gm ⎠

For CMRR |dB = 90 dB ⇒ CMRR = 31622.8 ⎛ 1 + 2 (100 )( 3.86 ) ⎞ Then 31622.8 = ⎜ ⎟ ( 3.86 ) 2Δg m ⎝ ⎠ Δg m 0.0472 = = 0.0122 ⇒ 1.22% Or Δg m = 0.0472 mA/V or 3.86 gm

TYU11.5 For v1 = v2 = 5 V ⇒ min vC1 = vC 2 = 5 V So I C1 RC = 10 − 5 = 0.25 RC ⇒ RC = 20 kΩ Ad =

I Q RC

=

4VT

( 0.5 )( 20 ) ⇒ Ad 4 ( 0.026 )

= 96.2 Let I Q = 0.5 mA

C M RRdB = 95 db ⇒ C M RR = 5.62 × 104 ⇒ Acm =

Acm

96.2 ⇒ Acm = 1.71× 10−3 5.62 × 104

⎛ I Q RC ⎞ ⎜ ⎟ ⎝ 2VT ⎠ = = 1.71× 10−3 ⎡ (1 + β ) I Q R0 ⎤ ⎢1 + ⎥ VT β ⎣ ⎦ ⎡ ( 0.5 )( 20 ) ⎤ ⎢ ⎥ ⎣ 2 ( 0.026 ) ⎦ = 1.71× 10−3 ⎡ ( 201)( 0.5 ) R0 ⎤ ⎢1 + ⎥ ⎣ ( 0.026 )( 200 ) ⎦

1 + 19.33R0 = 1.125 × 105 ⇒ R0

= 5.82 × 103 kΩ = 5.82 MΩ

We have R0 = r04 ⎡⎣1 + g m 4 ( R2 rπ 4 ) ⎤⎦ r04 =

VA 125 = = 250 kΩ I Q 0.5

gm4 = rπ 2 =

IQ VT

β VT IQ

=

0.5 = 19.23 mA/V 0.026

=

( 200 )( 0.026 ) 0.5

= 10.4 kΩ

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5.820 = 250 ⎡⎣1 + g m 4 ( R2 rπ 4 ) ⎤⎦

19.23 ( R2 rπ 2 ) = 22.28

(R

2

rπ 2 ) = 1.159 kΩ

R2 (10.4 )

R2 + 10.4

= 1.159

R2 (10.4 − 1.16 ) = (1.16 )(10.4 ) ⇒ R2 = 1.30 kΩ Let I1 = 1 mA ⎛I I Q R2 − I1 R3 = VT ln ⎜ 1 ⎜I ⎝ Q

⎞ ⎟⎟ ⎠

( 0.5)(1.304 ) − (1) R3 = ( 0.026 ) ln ⎛⎜

1 ⎞ ⎟ ⇒ R3 = 0.634 kΩ ⎝ 0.5 ⎠

If VBE ( Q3 ) ≅ 0.7 V

10 − 0.7 − ( −10 )

R1 + R3 =

1

= 19.3 ⇒ R1 ≅ 18.7 kΩ

TYU11.6 a. v0 = Ad vd + Acm vcm vd = v1 − v2 = 0.505sin ω t − 0.495sin ω t = 0.01sin ω t v +v 0.505sin ω t + 0.495sin ω t vcm = 1 2 = 2 2 = 0.50sin ω t v0 = ( 60 )( 0.01sin ω t ) + ( 0.5 )( 0.5sin ω t ) ⇒ v0 = 0.85sin ω t ( V )

b. vd = v1 − v2 = 0.5 + 0.005sin ω t − ( 0.5 − 0.005sin ω t ) = 0.01sin ω t v +v vcm = 1 2 2 0.5 + 0.005sin ω t + 0.5 − 0.005sin ω t = 2 = 0.5 v0 = ( 60 )( 0.01sin ω t ) + ( 0.5 )( 0.5 ) ⇒ v0 = 0.25 + 0.6sin ω t ( V ) TYU11.7

a. b.

I B1 = I B 2 = rπ =

β VT I CQ

IQ / 2

=

1 ⇒ I B1 = I B 2 = 6.62 μ A 151

(1 + β ) (150 )( 0.026 )

=

1

= 3.9 kΩ

Rid = 2rπ = 2 ( 3.9 ) = 7.8 kΩ Ib =

c.

Vd 10sin ω t ( mV ) = ⇒ I b = 1.28sin ω t ( μ A ) Rid 7.8 kΩ

Ricm ≅ 2 (1 + β ) R0 = 2 (151)( 50 ) ⇒ 15.1 MΩ Ib =

Vcm 3sin ω t = ⇒ I b = 0.199sin ω t ( μ A ) Ricm 15.1 MΩ

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TYU11.8 Ad = g f RD 8 = g f ( 4 ) ⇒ g f ( max ) = 2 mA / V g f ( max ) =

( 2)

2

Kn IQ 2

⎛ 4⎞ = K n ⎜ ⎟ ⇒ K n = 2 mA / V 2 ⎝ 2⎠

TYU11.9 From Example 11-10, I Q = 0.587 mA Kn2 IQ

( 0.1)( 0.587 )

⋅ (16) ⇒ Ad = 2.74 2 1 1 = ⇒ R0 = 85.2 kΩ For M 4 , R0 = λ4 I Q ( 0.02)( 0.587 ) Ad =

2

⋅ RD =

g m = 2 K n (VGS 2 − VTN ) = 2 ( 0.1)( 2.71 − 1) = 0.342 mA/V Acm =

− ( 0.342)(16) − g m RD = ⇒ Acm = −0.0923 1 + 2 g m R0 1 + 2 ( 0.342)(85.2)

⎛ 2.74 ⎞ C M RRdB = 20 log10 ⎜ ⎟ ⇒ C M RRdB = 29.4 dB ⎝ 0.0923 ⎠

TYU11.10 1 CMRR = ⎡⎣1 + 2 2 K n I Q ⋅ Ro ⎤⎦ 2 CMRRdB = 60 dB ⇒ C M RR = 1000 1 1000 = ⎡1 + 2 2 ( 0.1)( 0.2 ) ⋅ R0 ⎤ ⎦ 2⎣ 2000 = 1 + 0.4 R0 ⇒ R0 ≅ 5 MΩ TYU11.11 Ro = ro 4 + ro 2 (1 + g m 4 ro 4 )

Assume I REF = I O = 100 μ A and λ = 0.01 V −1 ro 2 = ro 4 =

1 1 = ⇒1 MΩ λ I D ( 0.01)( 0.1)

Let K n ( all devices ) = 0.1 mA / V 2

( 0.1)( 0.1) = 0.2 mA / V Ro = 1000 + 1000 (1 + ( 0.2 )(1000 ) ) ⇒ 202 M Ω

Then g m 4 = 2 K n I D = 2

Now VGS1 = VGS 2 =

ID 0.05 + VTN = + 1 = 1.707 V 0.1 Kn

VDS1 ( sat ) = VGS 1 − VTN = 1.707 − 1 = 0.707 V So vo1 ( min ) = +4 − VGS 1 + VDS 1 ( sat ) = 4 − 1.707 + 0.707

vo1 ( min ) = 3 V = 10 − I D RD = 10 − ( 0.05 ) RD ⇒ RD = 140 kΩ

For a one-sided output, the differential gain is:

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Ad =

1 g m1 RD where g m1 = 2 K n I D 2

=2

( 0.1)( 0.05) = 0.1414 mA / V

1 ( 0.1414 )(140 ) ⇒ Ad = 9.90 2 The common-mode gain is: 2 K n I Q ⋅ RD 2 ( 0.1)( 0.1) ⋅ (140 ) Acm = = ⇒ Acm = 0.0003465 1 + 2 2 K n I Q ⋅ Ro 1 + 2 2 ( 0.1)( 0.1) ⋅ ( 202000 ) Ad =

Ad ⇒ CMRRdB = 89.1 dB Acm

Then CMRRdB = 20 log10 TYU11.12 I B5 =

a.

IQ

β (1 + β )

=

0.5 ⇒ 15.3 nA (180 )(181)

So I 0 = 15.3 nA b. For a balanced condition VEC 4 = VEC 3 = VEB 3 + VEB 5 ⇒ VEC 4 = 1.4 V VCE 2 = VC 2 − VE 2 = (10 − 1.4 ) − ( −0.7 ) ⇒ VCE 2 = 9.3 V

TYU11.13 Ad = 2 g f ( r02 r04 ) gf =

IQ 4VT

=

0.2 = 1.923 mA/V 4 ( 0.026 )

r02 =

VA 2 120 = = 1200 kΩ I C 2 0.1

r04 =

VA 4 80 = = 800 kΩ I C 4 0.1

Ad = 2 (1.923) (1200 800 ) ⇒ Ad = 1846

TYU11.14 P = ( I Q + I REF ) ( 5 − ( −5 ) )

10 = ( 0.1 + I REF )(10 ) ⇒ I REF = 0.9 mA R1 =

5 − 0.7 − ( −5 ) I REF

=

9.3 ⇒ R1 = 10.3 k Ω 0.9

⎛I ⎞ I Q RE = VT ln ⎜ REF ⎟ ⎜ I ⎟ ⎝ Q ⎠ 0.026 ⎛ 0.9 ⎞ ln ⎜ RE = ⎟ ⇒ RE = 0.571 kΩ 0.1 ⎝ 0.1 ⎠ ro 2 =

VA 2 120 = ⇒ 2.4 M Ω I C 2 0.05

ro 4 =

VA 4 80 = ⇒ 1.6 M Ω I C 4 0.05

gm =

0.05 = 1.923 mA / V 0.026

(

)

(

)

Ad = g m ro 2 ro 4 RL = (1.923) 2400 1600 90 ⇒ Ad = 158

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TYU11.15 a. R0 = r02 r04 120 r02 = = 1.2 MΩ 0.1 80 r04 = = 0.8 MΩ 0.1 R0 = 1.2 0.8 ⇒ R0 = 0.48 MΩ

b. Ad ( open circuit ) = 2 g f ( r02 r04 )

(

Ad ( with load ) = 2 g f r02 r04 RL For Ad ( with load ) =

)

1 Ad ( open circuit ) ⇒ RL = ( r02 r04 ) ⇒ RL = 0.48 MΩ 2

TYU11.16 2Kn 1 Ad = 2 ⋅ I Q ( λ2 + λ4 ) =2

2 ( 0.1) 0.1

⋅

1

( 0.01 + 0.015)

⇒ Ad = 113

TYU11.17 For the MOSFET, I D = 25 − g m1 = 2 K n I D = 2

75 = 24.26 μ A 101

( 20 )( 24.26 ) = 44.05 μ A/V

For the Bipolar, I E = 100 − 25 = 75 μ A

⎛ 100 ⎞ IC = ⎜ ⎟ ( 75 ) = 74.26 μ A ⎝ 101 ⎠ (100 )( 0.026 ) rπ 2 = = 35.0 K 0.07426 0.07426 gm2 = = 2.856 mA/V 0.026 g (1 + g m 2 rπ ) ( 44.05 ) ⎡⎣1 + ( 2.856 )( 35 ) ⎤⎦ g mC = m1 = 1 + g m1rπ 1 + ( 0.04405 )( 35 ) g mC = 1.75 mA/V

TYU11.18 From Figure 11.43 80 r04 = = 160 kΩ 0.5 R0 ≅ β r04 = (150 )(160 ) kΩ ⇒ R0 = 24 MΩ

From Figure 11.44 1 1 r06 = = ⇒ r06 = 160 kΩ λ I D ( 0.0125 )( 0.5 ) 0.5 = 0.5 (VGS − 1) ⇒ VGS = 2 V 2

g m 6 = 2 K n (VGS − VTN ) = 2 ( 0.5 )( 2 − 1) = 1 mA / V r04 = 160 kΩ

R0 = ( g m 6 )( r06 )( β r04 ) = (1)(160 )(150 )(160 ) ⇒ R0 = 3.840 MΩ

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TYU11.19 From Equation (11.126) 2 (1 + β ) β VT 2 (121)(120 )( 0.026 ) ⇒ Ri = 1.51 MΩ Ri = = IQ 0.5 rπ 11 =

β VT

=

IQ

(120 )( 0.026 ) 0.5

= 6.24 kΩ

RE′ = rπ 11 & R3 = 6.24 & 0.1 = 0.0984 kΩ g m11 = r011 =

IQ VT

=

0.5 = 19.23 mA/V 0.026

VA 120 = = 240 kΩ I Q 0.5

Then RC11 = r011 (1 + g m11 RE′ )

= 240 ⎡⎣1 + (19.23)( 0.0984 ) ⎤⎦ = 694 kΩ

rπ 8 =

β VT IC 8

=

(120 )( 0.026 ) 2

= 1.56 kΩ

Rb8 = rπ 8 + (1 + β ) R4 = 1.56 + (121)( 5 ) = 607 kΩ Then RL 7 = RC11 & Rb8 = 694 & 607 = 324 kΩ

⎡ 0.5 ⎤ ⎛ IQ ⎞ Then Av = ⎜ ⎥ ( 324 ) ⇒ Av = 3115 ⎟ RL 7 = ⎢ ⎢⎣ 2 ( 0.026 ) ⎦⎥ ⎝ 2VT ⎠ ⎛r +Z ⎞ R0 = R4 || ⎜ π 8 ⎟ ⎝ 1+ β ⎠ Z = RC11 & RC 7 V 120 RC 7 = A = = 240 kΩ I Q 0.5 Z = 694 & 240 = 178 kΩ ⎛ 1.56 + 178 ⎞ R0 = 5 || ⎜ ⎟ = 5 & 1.48 ⇒ R0 = 1.14 kΩ ⎝ 121 ⎠

TYU11.20 ⎛ IQ Av = ⎜ ⎝ 2VT

⎞ ⎟ RL 7 ⎠

⎛ 0.5 ⎞ 103 = ⎜⎜ ⎟⎟ RL 7 ⇒ RL 7 = 104 kΩ ⎝ 2 ( 0.026 ) ⎠

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Chapter 11 Problem Solutions 11.1 (a) −0.7 − ( −3) RE

= 0.1 ⇒ RE = 23 K

3 − 1.5 = 0.05 ⇒ RC = 30 K RC

(b)

vCE 2 = 6 − iC 2 ( RC + 2 RE ) = 6 − iC 2 ( 76 )

(c)

vcm ( max ) ⇒ vCB 2 = 0 ⇒ vCE 2 = 0.7 V

So 0.7 = 6 − iC 2 ( 76 ) ⇒ iC 2 = 69.74 μ A

( v ( max ) − 0.7 ) − ( −3) = 2

( 0.06974 ) ⇒ vCM ( max ) = 0.908 V 23 ( min ) ⇒ VS = −3 V ⇒ vCM ( min ) = −2.3 V

CM

vCM 11.2

Ad = 180, C M RRdB = 85 dB Ad 180 = ⇒ Acm = 0.01012 Acm Acm Assume the common-mode gain is negative. v0 = Ad vd + Acm vcm C M RR = 17, 783 =

= 180vd − 0.01012vcm v0 = 180 ( 2sin ω t ) mV − ( 0.01012 )( 2sin ω t ) V v0 = 0.36sin ω t − 0.02024sin ω t

Ideal Output:

v0 = 0.360sin ω t ( V )

Actual Output:

v0 = 0.340sin ω t ( V )

11.3 a.

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I1 =

10 − 2 ( 0.7 )

IC 2 =

⇒ I1 = 1.01 mA

8.5 I1

2 1+ β (1 + β )

=

1.01 ⇒ I C 2 ≅ 1.01 mA 2 1+ (100 )(101)

⎛ 100 ⎞ ⎛ 1.01 ⎞ IC 4 = ⎜ ⎟⎜ ⎟ ⇒ I C 4 ≅ 0.50 mA ⎝ 101 ⎠ ⎝ 2 ⎠ VCE 2 = ( 0 − 0.7 ) − ( −5 ) ⇒ VCE 2 = 4.3 V VCE 4 = ⎣⎡5 − ( 0.5 )( 2 ) ⎦⎤ − ( −0.7 ) ⇒ VCE 4 = 4.7 V

b. For VCE 4 = 2.5 V ⇒ VC 4 = −0.7 + 2.5 = 1.8 V 5 − 1.8 IC 4 = ⇒ I C 4 = 1.6 mA 2 ⎛ 1+ β ⎞ ⎛ 101 ⎞ IC 2 + ⎜ ⎟ ( 2IC 4 ) = ⎜ ⎟ ( 2 )(1.6 ) ⇒ I C 2 = 3.23 mA β ⎝ 100 ⎠ ⎝ ⎠ I1 ≈ I C 2 = 3.23 mA R1 =

10 − 2 ( 0.7 ) 3.23

⇒ R1 = 2.66 kΩ

11.4 a.

Neglecting base currents 30 − 0.7 ⇒ R1 = 73.25 kΩ I1 = I 3 = 400 μ A ⇒ R1 = 0.4 VCE1 = 10 V ⇒ VC1 = 9.3 V 15 − 9.3 ⇒ RC = 28.5 kΩ RC = 0.2 b. (100 )( 0.026 ) rπ = = 13 kΩ 0.2 50 r0 ( Q3 ) = = 125 kΩ 0.4 We have Ad =

(100 )( 28.5) ⇒ Ad 2 ( rπ + RB ) 2 (13 + 10 ) β RC

=

= 62

⎧ ⎫ ⎪ ⎪ β RC ⎪ 1 ⎪ Acm = − ⎨ 2 1 r β + rπ + RB ⎪ ( ) ⎬⎪ 1+ 0 ⎪⎩ rπ + RB ⎪⎭ ⎧ ⎫ ⎪⎪ (100 )( 28.5 ) ⎪⎪ 1 =− ⎨ ⎬ ⇒ Acm = −0.113 13 + 10 ⎪ 2 (125 )(101) ⎪ 1+ 13 + 10 ⎭⎪ ⎩⎪ ⎛ 62 ⎞ C M RRdB = 20 log10 ⎜ ⎟ ⇒ C M RRdB = 54.8 dB ⎝ 0.113 ⎠ c.

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Rid = 2 ( rπ + RB ) = 2 (13 + 10 ) ⇒ Rid = 46 kΩ 1 ⎡ rπ + RB + 2 (1 + β ) r0 ⎤⎦ 2⎣ 1 = ⎡⎣13 + 10 + 2 (101)(125 ) ⎤⎦ ⇒ Ricm = 12.6 MΩ 2

Ricm =

11.5 vCM ( max ) ⇒ VCB = 0 so that vCM ( max ) = 5 −

(a)

vCM ( max ) = 3 V

IQ 2

( RC ) = 5 −

(b) Vd ⎛ I CQ ⎞ Vd ⎛ 0.25 ⎞ ⎛ 0.018 ⎞ =⎜ =⎜ ⎟⋅ ⎟⎜ ⎟ = 0.08654 mA 2 ⎝ VT ⎠ 2 ⎝ 0.026 ⎠ ⎝ 2 ⎠ = ΔI ⋅ RC = ( 0.08654 ) ( 8 ) = 0.692 V

ΔI = g m ⋅ ΔVC 2 (c)

⎛ 0.25 ⎞ ⎛ 0.010 ⎞ ΔI = ⎜ ⎟⎜ ⎟ = 0.04808 mA ⎝ 0.026 ⎠ ⎝ 2 ⎠ ΔVC 2 = ( 0.04808 )( 8 ) = 0.385 V

11.6 P = ( I1 + I C 4 ) (V + − V − )

I1 ≅ I C 4 so 1.2 = 2 I1 ( 6 ) ⇒ I1 = I C 4 = 0.1 mA R1 =

3 − 0.7 − ( −3) 0.1

⇒ R1 = 53 k Ω

For vCM = +1V ⇒ VC1 = VC 2 = 1 V ⇒ RC =

3 −1 ⇒ RC = 40 k Ω 0.05

One-sided output 1 0.05 Ad = g m RC where g m = = 1.923 mA / V 2 0.026 Then 1 Ad = (1.923)( 40 ) ⇒ Ad = 38.5 2 11.7 a. IE ( 2 ) + I E (85) − 5 2 5 − 0.7 IE = ⇒ I E = 0.050 mA 85 + 1 ⎛ β ⎞ ⎛ I E ⎞ ⎛ 100 ⎞⎛ 0.050 ⎞ I C1 = I C 2 = ⎜ ⎟⎜ ⎟ = ⎜ ⎟⎜ ⎟ ⎝ 1 + β ⎠ ⎝ 2 ⎠ ⎝ 101 ⎠⎝ 2 ⎠

0 = 0.7 +

Or I C1 = I C 2 = 0.0248 mA VCE1 = VCE 2 = ⎡⎣5 − I C1 (100 ) ⎤⎦ − ( −0.7 ) So VCE1 = VCE 2 = 3.22 V

b.

vcm ( max ) for VCB = 0 and VC = 5 − I C1 (100 ) = 2.52 V

So vcm ( max ) = 2.52 V

vcm ( min ) for Q1 and Q2 at the edge of cutoff ⇒ vcm ( min ) = −4.3 V

(c) Differential-mode half circuits

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( 0.5) 2

(8)

⎛V ⎞ vd = Vπ + ⎜ π + g mVπ ⎟ .RE′ 2 ⎝ rπ ⎠ ⎡ (1 + β ) ⎤ = Vπ ⎢1 + RE′ ⎥ rπ ⎣ ⎦ Then −

Vπ =

− ( vd / 2 )

⎡ (1 + β ) ⎤ RE′ ⎥ ⎢1 + rπ ⎣ ⎦

vo = − g mVπ RC ⇒ Ad = rπ =

β VT

=

I CQ

β RC 1 ⋅ 2 rπ + (1 + β ) RE′

(100 )( 0.026 )

= 105 k Ω RE′ = 2 k Ω

0.0248

Then Ad =

11.8 a. RE =

1 (100 )(100 ) ⇒ Ad = 16.3 ⋅ 2 105 + (101)( 2 )

For v1 = v2 = 0 and neglecting base currents −0.7 − ( −10 ) 0.15

⇒ RE = 62 kΩ

b. Ad = rπ = Ad =

v02 β RC = vd 2 ( rπ + RB )

β VT I CQ

=

(100 )( 0.026 ) 0.075

(100 )( 50 ) ⇒ Ad 2 ( 34.7 + 0.5 )

= 34.7 kΩ

= 71.0

⎡ ⎤ ⎢ ⎥ β RC ⎢ 1 ⎥ Acm = − rπ + RB ⎢ 2 RE (1 + β ) ⎥ ⎢1 + ⎥ rπ + RB ⎥⎦ ⎢⎣ ⎡

⎤ ⎥ 1 =− ⎢ ⎥ ⇒ Acm = −0.398 34.7 + 0.5 ⎢ 2 ( 62 )(101) ⎥ ⎢⎣1 + 34.7 + 0.5 ⎥⎦ 71.0 ⇒ C M RRdB = 45.0 dB C M RRdB = 20 log10 0.398 c. Rid = 2 ( rπ + RB )

(100 )( 50 ) ⎢

Rid = 2 ( 34.7 + 0.5 ) ⇒ Rid = 70.4 kΩ

Common-mode input resistance 1 Ricm = ⎡⎣ rπ + RB + 2 (1 + β ) RE ⎤⎦ 2 1 = ⎣⎡34.7 + 0.5 + 2 (101)( 62 ) ⎦⎤ ⇒ Ricm = 6.28 MΩ 2 11.9

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(a) v1 = v2 = 1 V ⇒ VE = 1.6 IE =

9 − 1.6 ⇒ 18.97 μ A 390

IE = 9.49 μ A I C1 = I C 2 = 9.39 μ A 2 vC1 = vC 2 = ( 9.39 )( 0.51) − 9 = −4.21 V (b) 9.39 gm = ⇒ 0.361 mA/V 0.026 V ΔI = g m d = ( 0.361× 10−3 ) ( 0.005 ) = 1.805 μ A 2 ΔvC = (1.805 × 10−6 )( 510 × 103 ) = 0.921 V ⇒ vC 2 = −4.21 + 0.921 ⇒ −3.29 V vC1 = −4.21 − 0.921 ⇒ −5.13 V

11.10 (a) v1 = v2 = 0 I E1 = I E 2 ≅ 6 μ A β = 60 I C1 = I C 2 = 5.90 μ A vC1 = vC 2 = ( 5.90 )( 0.360 ) − 3 = −0.875 V VEC1 = VEC 2 = +0.6 − ( −0.875 ) = 1.475 V (b) (i) 5.90 gm = ⇒ 0.227 mA/V 0.026 Ad = g m RC = ( 0.227 )( 360 ) = 81.7 Acm = 0 (ii) ( 60 )( 0.026 ) g R Ad = m C = 40.8 rπ = 2 0.0059 = 264 K Acm =

− ( 0.227 )( 360 ) = −0.0442 2 ( 61)( 4000 ) 1+ 264

11.11

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For v1 = v2 = 0.20 V I C1 = I C 2 = 0.1 mA vC1 = vC 2 = ( 0.1)( 30 ) − 10 = −7 V 0.1 gm = = 3.846 mA/V 0.026 v ΔI = g m d = ( 3.846 )( 0.008 ) ⇒ 30.77 μ A 2 ΔvC = ΔI ⋅ RC = ( 30.77 × 10−6 )( 30 × 103 ) = 0.923 V v2 ↑⇒ I C 2 ↓⇒ vC 2 ↓⇒ vC1

= −7 + 0.923 = −6.077 V

vC 2 = −7 − 0.923 = −7.923 V

11.12 RC = 50 K For v1 = v2 = 0 IE =

−0.7 − ( −10 )

75 = 0.124 mA I C1 = I C 2 = 0.0615 mA 0.0615 gm = = 2.365 mA/V 0.026 (120 )( 0.026 ) rπ = = 50.7 K 0.0615 Differential Input v V v1 = d v2 = − d 2 2 Half-circuit. V ΔR ⎞ ⎛ ΔI = + g m d ⇒ ΔvC1 = −ΔI ⎜ RC + ⎟ 2 2 ⎠ ⎝

vo = ΔvC1 − ΔvC 2

ΔR ⎞ ⎛ ΔvC 2 = +ΔI ⎜ RC − ⎟ 2 ⎠ ⎝ ΔR ⎞ ΔR ⎞ ⎛ ⎛ = −ΔI ⎜ RC + ⎟ − ΔI ⎜ RC − ⎟ 2 ⎠ 2 ⎠ ⎝ ⎝

= −2ΔIRC ⎛ V ⎞ = −2 ⎜ g m d ⎟ RC 2⎠ ⎝ Ad = − g m RC = − ( 2.365 )( 50 ) = −118.25 Common-mode input.

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⎛V ⎞ vcm = Vπ + ⎜ π + g mVπ ⎟ ( 2 RE ) ⎝ rπ ⎠ vcm Vπ = ⎛ β⎞ 1 + ⎜ 1 + ⎟ ( 2 RE ) ⎝ rπ ⎠

ΔI = g mVπ =

g m vcm β vcm = rπ + (1 + β )( 2 RE ) ⎛1+ β ⎞ 1+ ⎜ ⎟ ( 2 RE ) ⎝ rπ ⎠

ΔR ⎞ ⎛ − β ⎜ RC + ⎟ ⋅ vcm 2 ⎠ ⎝ ΔvC1 = −ΔIR1 = rπ + (1 + β )( 2 RE ) ΔvC 2

ΔR ⎞ ⎛ − β ⎜ RC − ⎟ vcm 2 ⎠ ⎝ = −ΔIR2 = rπ + (1 + β )( 2 RE )

vo = ΔvC1 − ΔvC 2

ΔR ⎞ ΔR ⎞ ⎛ ⎛ − β ⎜ RC + ⎟ vcm ⎟ vcm + β ⎜ RC − 2 ⎠ 2 ⎠ ⎝ ⎝ =

[ ]

[ ]

⎛ ΔR ⎞ −2 β ⎜ ⎟ vcm ⎝ 2 ⎠ = rπ + (1 + β )( 2 RE )

Acm =

− (120 )( 0.5 ) − βΔR = rπ + (1 + β )( 2 RE ) 50.7 + (121)( 2 )( 75 )

= −0.0032966 118.25 C M RR = = 35,870.5 0.0032966 C M R R ∫ = 91.1 dB dB

11.13 v1 = v2 = 0 IE =

−0.7 − ( −10 )

75 = 0.124 mA I C1 = I C 2 = 0.0615 mA 0.0615 gm = = 2.365 mA/V 0.026 Δg m = 0.01 gm g m1 = 2.377 mA/V g m 2 = 2.353 mA/V rπ =

(120 )( 0.026 ) 0.0615

= 50.7 K

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Vd 2 V = − g m1 d Rc 2 Vd = + gm2 Rc 2

ΔI = g m ΔvC1 ΔvC 2

vo = ΔvC1 − ΔvC 2 = − g m1

Vd V RC − g m 2 d RC 2 2

Vd RC ( g m1 + g m 2 ) 2 R −50 Ad = − C ( g m1 + g m 2 ) = ( 2.377 + 2.353) ⇒ Ad = −118.25 2 2 Common-Mode − g m1 RC vcm − g m 2 RC vcm ΔvC1 = ΔvC 2 = ⎛ 1+ β ⎞ ⎛ 1+ β ⎞ 1+ ⎜ 1+ ⎜ ⎟ ( 2 RE ) ⎟ ( 2 RE ) ⎝ rπ ⎠ ⎝ rπ ⎠ =−

− ( g m1 − g m 2 ) RC − ( 2.377 − 2.353) ( 50 ) vo = Acm = = vcm ⎛ 1+ β ⎞ ⎛ 121 ⎞ 1+ ⎜ 1+ ⎜ ⎟ ( 2 )( 75 ) ⎟ ( 2 RE ) ⎝ 50.7 ⎠ ⎝ rπ ⎠ −1.2 = ⇒ Acm = −0.003343 358.99 C M R R ∫ = 91 dB dB

11.14 (a) v1 = v2 = 0 vE = +0.7 V 5 − 0.7 IE = = 4.3 mA 1 I C1 = I C 2 = 2.132 mA vC1 = vC 2 = ( 2.132 )(1) − 5 = −2.87 V (b) v1 = 0.5, v2 = 0 Q2 on Q1 off ⎛ 120 ⎞ I C1 = 0, I C 2 = 4.3 ⎜ ⎟ mA = 4.264 mA ⎝ 121 ⎠ vC1 = −5 V vC 2 = ( 4.264 ) (1) − 5 vC 2 = −0.736 V

2.132 = 82.0 mA/V 0.026 (82.0 ) v V ΔI = g m d ΔvC = ΔI ⋅ RC = g m d ⋅ RC = ⋅ Vd (1) = 41.0Vd 2 2 2 Vd = 0.015 ⇒ Δvc = 0.615 V

(c)

vE ≈ 0.7 V

gm =

vC 2 ↓ vC1 ↑

vC1 = −2.87 + 0.615 = −2.255 V vC 2 = −2.87 − 0.615 = −3.485 V 11.15

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(a) gm =

IC 1 = = 38.46 mA/V VT 0.026

Ad =

vo 1 = = 100 vd 0.01

Ad = g m RC 100 = 38.46 RC Rc = 2.6 K

(b) With v1 = v2 = 0 vC1 = vC 2 = 10 − (1)( 2.6 ) = 7.4 V ⇒ vcm ( max ) = 7.4 V

11.16 a. i.

( v01 − v02 ) = 0

ii. I C1 = I C 2 = 1 mA v01 − v02 = ⎡⎣V + − I C1 RC1 ⎤⎦ − ⎡⎣V + − I C 2 RC 2 ⎤⎦ = I C ( RC 2 − RC1 ) = (1)( 7.9 − 8 ) ⇒ v01 − v02 = −0.1 V

b. ⎛v ⎞ I 0 = ( I S 1 + I S 2 ) exp ⎜ BE ⎟ ⎝ VT ⎠ ⎛v ⎞ 2 × 10−3 So exp ⎜ BE ⎟ = −13 −13 ⎝ VT ⎠ 10 + 1.1× 10 = 9.524 × 109 ⎛v I C1 = I S 1 exp ⎜ BE ⎝ VT

⎞ −13 9 ⎟ = (10 )( 9.524 × 10 ) ⇒ I C1 = 0.952 mA ⎠

I C 2 = (1.1× 10−13 )( 9.524 × 109 ) ⇒ I C 2 = 1.048 mA

i. v01 − v02 = I C 2 RC 2 − I C1 RC1 ⇒ v01 − v02 = (1.048 − 0.952 )( 8 ) ⇒ v01 − v02 = 0.768 V ii. v01 − v02 = (1.048 )( 7.9 ) − ( 0.952 )( 8 ) v01 − v02 = 8.279 − 7.616 ⇒ v01 − v02 = 0.663 V 11.17 From Equation (11.12(b)) IQ iC 2 = 1 + evd / VT 1 0.90 = 1 + evd / VT 1 So evd / VT = − 1 = 0.111 0.90 vd = VT ln ( 0.111) = ( 0.026 ) ln ( 0.111) ⇒ vd = −0.0571 V 11.18 From Example 11.2, we have

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vd ( max ) 1 − − vd ( max ) / 0.026 4 ( 0.026 ) 1 + e = 0.02 v ( max ) 0.5 + d 4 ( 0.026 )

0.5 +

⎡ v ( max ) ⎤ 1 0.98 ⎢ 0.5 + d ⎥= − vd ( max ) / 0.026 4 ( 0.026 ) ⎦⎥ 1 + e ⎣⎢ 0.490 + 9.423vd ( max ) =

By trial and error vd ( max ) = 23.7 mV

1 1+ e

− vd ( max ) / 0.026

11.19 a. For I1 = 1 mA, VBE3 = 0.7 V 20 − 0.7 R1 = ⇒ R1 = 19.3 kΩ 1 ⎛ I ⎞ 0.026 ⎛ 1 ⎞ V R2 = T ⋅ ln ⎜ 1 ⎟ = ⋅ ln ⎜ ⎟ ⇒ R2 = 0.599 kΩ ⎜I ⎟ 0.1 IQ ⎝ 0.1 ⎠ ⎝ Q⎠ b. (180 )( 0.026 ) rπ 4 = = 46.8 kΩ 0.1 0.1 gm = = 3.846 mA/V 0.026 100 r04 = ⇒ 1 MΩ 0.1 From Chapter 10 R0 = r04 ⎡⎣1 + g m ( RE & rπ 4 ) ⎤⎦ RE & rπ 4 = 0.599 & 46.8 = 0.591 R0 = (1) ⎣⎡1 + ( 3.846 )( 0.591) ⎦⎤ = 3.27 MΩ r01 =

100 ⇒ 2 MΩ 0.05

⎡ ⎛ r ⎞⎤ Ricm ≅ ⎣⎡(1 + β ) R0 ⎦⎤ & ⎢(1 + β ) ⎜ 01 ⎟⎥ ⎝ 2 ⎠⎦ ⎣ = ⎣⎡(181)( 3.27 ) ⎦⎤ & ⎣⎡(181)(1) ⎦⎤ = 592 & 181 ⇒ Ricm = 139 MΩ

(c)

From Eq. (11.32(b)) − g m RC Acm = 2 (1 + β ) Ro 1+ rπ + RB 0.05 = 1.923 mA / V 0.026 (180 )( 0.026 ) rπ = = 93.6 k Ω 0.05 RB = 0

gm =

Then Acm =

− (1.923)( 50 ) ⇒ Acm = −0.00760 2 (181)( 3270 ) 1+ 93.6

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11.19 For vCM = 3.5 V and a maximum peak-to-peak swing in the output voltage of 2 V, we need the quiescent collector voltage to be VC = 3.5 + 1 = 4.5 V Assume the bias is ±10 V , and I Q = 0.5 mA.

Then I C = 0.25 mA 10 − 4.5 ⇒ RC = 22 k Ω 0.25 (100 )( 0.026 ) = 10.4 k Ω In this case, rπ = 0.25 Then (100 )( 22 ) Ad = = 101 So gain specification is met. 2 (10.4 + 0.5 ) Now RC =

For CMRRdB = 80 dB ⇒ 1 ⎡ (1 + β ) I Q Ro ⎤ 1 ⎡ (101)( 0.5 ) Ro ⎤ ⎥ ⇒ Ro = 1.03 M Ω ⎢1 + ⎥ = ⎢1 + 2⎣ VT β ⎦ 2 ⎣⎢ ( 0.026 )(100 ) ⎦⎥ Need to use a Modified Widlar current source. Ro = ro ⎡⎣1 + g m ( RE1 & rπ ) ⎤⎦ CMRR = 104 =

If VA = 100V , then ro = rπ =

(100 )( 0.026 )

100 = 200 k Ω 0.5

= 5.2 k Ω 0.5 0.5 gm = = 19.23 mA / V 0.026 Then 1030 = 200 ⎡⎣1 + (19.23)( RE1 & rπ ) ⎤⎦ ⇒ RE1 & rπ = 0.216 k Ω = RE1 & 5.2 ⇒ RE1 = 225 Ω Also let RE 2 = 225 Ω and I REF ≅ 0.5 mA

11.20 (a)

RE =

−0.7 − ( −10 ) 0.25

⇒ RE = 37.2 k Ω

(b)

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⎛1+ β ⎞ Vπ 1 V V Ve + g mVπ 1 + π 2 + g mVπ 2 = e or (1) ⎜ ⎟ (Vπ 1 + Vπ 2 ) = rπ rπ RE r R E ⎝ π ⎠ ⎛ r ⎞ Vπ 1 V1 − Ve = ⇒ Vπ 1 = ⎜ π ⎟ (V1 − Ve ) + rπ RB + rπ r R B ⎠ ⎝ π Vπ 2 = V2 − Ve Then ⎛1+ β ⎝ rπ

⎞ ⎡ rπ ⎤ V (V1 − Ve ) + (V2 − Ve )⎥ = e ⎟⎢ ⎠ ⎣ rπ + RB ⎦ RE From this, we find

(1) ⎜

V1 + Ve =

rπ + RB ⋅ V2 rπ

⎡ rπ + RB r + RB ⎤ +1+ π ⎢ ⎥ rπ ⎦ ⎣ RE (1 + β )

Now Vo = − g mVπ 2 RC = − g m RC (V2 − Ve ) We have rπ =

(120 )( 0.026 ) 0.125

≅ 25 k Ω,

gm =

0.125 = 4.81 mA / V 0.026

(i) Set V1 =

Vd V and V2 = − d 2 2

Then ⎛ ⎛ 25 + 0.5 ⎞ ⎞ Vd ⎜ 1 − ⎜ 25 ⎟ ⎟ ( −0.02 ) ⎝ ⎠ ⎝ ⎠ Ve = = 2 2.026 ⎡ 25 + 0.5 25 + 0.5 ⎤ +1+ ⎢ ⎥ 25 ⎦ ⎣ ( 37.2 )(121) Vd 2

So Ve = −0.00494Vd Now V ⎛ V ⎞ Vo = − ( 4.81)( 50 ) ⎜ − d − ( −0.00494 )Vd ⎟ ⇒ Ad = o = 119 Vd ⎝ 2 ⎠

(ii) Set V1 = V2 = Vcm Then ⎛ 25 + 0.5 ⎞ Vcm ⎜ 1 + ⎟ V ( −2.02 ) 25 ⎠ ⎝ Ve = = cm 2.02567 ⎡ 25 + 0.5 25 + 0.5 ⎤ +1+ ⎢ ⎥ 25 ⎦ ⎣ ( 37.2 )(121) Ve = Vcm ( 0.9972 ) Then Vo = − ( 4.81)( 50 ) ⎡⎣Vcm − Vcm ( 0.9972 ) ⎤⎦ or Acm =

Vo = −0.673 Vcm

11.21 From Equation (11.18)

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v0 = vC 2 − vC1 = g m RC vd gm =

I CQ VT

For I Q = 2 mA, I CQ = 1 mA 1 = 38.46 mA/V 0.026 Now 2 = ( 38.46 ) RC ( 0.015 )

Then g m =

So RC = 3.47 kΩ Now VC = V + − I C RC

= 10 − (1)( 3.47 ) = 6.53 V

For VCB = 0 ⇒ vcm ( max ) = 6.53 V

11.22 The small-signal equivalent circuit is

A KVL equation: v1 = Vπ 1 − Vπ 2 + v2 v1 − v2 = Vπ 1 − Vπ 2 A KCL equation Vπ 1 V + g mVπ 1 + π 2 + g mVπ 2 = 0 rπ rπ ⎛1 ⎞ + g m ⎟ = 0 ⇒ Vπ 1 = −Vπ 2 ⎝ rπ ⎠

(Vπ 1 + Vπ 2 ) ⎜

Then v1 − v2 = 2Vπ 1 ⇒ Vπ 1 =

1 1 ( v1 − v2 ) and Vπ 2 = − ( v1 − v2 ) 2 2

At the v01 node: v01 v01 − v02 + + g mVπ 1 = 0 RC RL ⎛ 1 ⎛ 1 1 ⎞ v01 ⎜ + ⎟ − v02 ⎜ ⎝ RL ⎝ RC RL ⎠ At the v02 node:

⎞ 1 ⎟ = g m ( v2 − v1 ) ⎠ 2

(1)

v02 v02 − v01 + + g mVπ 2 = 0 RC RL ⎛ 1 ⎛ 1 ⎞ 1 1 ⎞ v02 ⎜ + ⎟ − v01 ⎜ ⎟ = g m ( v1 − v2 ) R R L ⎠ ⎝ RL ⎠ 2 ⎝ C From (1):

(2)

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⎛ R ⎞ 1 v02 = v01 ⎜ 1 + L ⎟ − g m RL ( v2 − v1 ) ⎝ RC ⎠ 2 Substituting into (2) ⎛ R ⎞⎛ 1 ⎛ 1 ⎛ 1 1 ⎞ 1 1 ⎞ + + v01 ⎜1 + L ⎟ ⎜ ⎟ − g m RL ( v2 − v1/ ) ⎜ ⎟ − v01 ⎜ R R R R R 2 ⎝ RL C ⎠⎝ C L ⎠ L ⎠ ⎝ ⎝ C

⎞ 1 ⎟ = g m ( v1 − v2 ) ⎠ 2

⎡ ⎛ RL ⎛ 1 RL ⎞⎤ 1 ⎞ 1 + 2 + + 1⎟ ⎥ v01 ⎜ ⎟ = g m ( v1 − v2 ) ⎢1 − ⎜ ⎝ RC RC RC ⎠ 2 ⎠⎦ ⎣ ⎝ RC v01 ⎛ RL ⎞ 1 ⎛ RL ⎞ ⎜2+ ⎟ = − gm ⎜ ⎟ ( v1 − v2 ) RC ⎝ RC ⎠ 2 ⎝ RC ⎠ For v1 − v2 = vd 1 − g m RL v01 Av1 = = 2 vd ⎛ RL ⎞ ⎜2+ ⎟ RC ⎠ ⎝ 1 g m RL v02 From symmetry: Av 2 = = 2 vd ⎛ RL ⎞ ⎜2+ ⎟ RC ⎠ ⎝

Then Av =

v02 − v01 g m RL = vd ⎛ RL ⎞ ⎜2+ ⎟ RC ⎠ ⎝

11.23 The small-signal equivalent circuit is

KVL equation: v1 = Vπ 1 − Vπ 2 + v2 or v1 − v2 = Vπ 1 − Vπ 2 KCL equation:

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Vπ 1 V + g mVπ 1 + g mVπ 2 + π 2 = 0 rπ rπ ⎛1 ⎞ + g m ⎟ = 0 ⇒ Vπ 1 = −Vπ 2 ⎝ rπ ⎠

(Vπ 1 + Vπ 2 ) ⎜

Then v1 − v2 = −2Vπ 2 or Vπ 2 = − Now

v0 = − g mVπ 2 ( RC & RL )

1 ( v1 − v2 ) 2

1 g m ( RC & RL )( v1 − v2 ) 2 v 1 For v1 − v2 ≡ vd ⇒ Ad = 0 = g m ( RC & RL ) vd 2 =

11.23 a. 10 − 7 ⇒ RD = 6 kΩ 0.5 I Q = I D1 + I D 2 ⇒ I Q = 1 mA RD =

b. 10 = I D ( 6 ) + VDS − VGS and VGS =

ID + VTN Kn

For I D = 0.5 mA, VGS =

0.5 + 2 = 3.12 V 0.4

and VDS = 10.12

Load line is actually nonlinear. c. Maximum common-mode voltage when M 1 and M 2 reach the transition point, or VDS ( sat ) = VGS − VTN = 3.12 = 2 = 1.12V

Then vcm = v02 − vDS ( sat ) + VGS = 7 − 1.12 + 3.12 Or vcm ( max ) = 9 V Minimum common-mode voltage, voltage across I Q becomes zero. So vcm ( min ) = −10 + 3.12 ⇒ vcm ( min ) = −6.88 V

11.24

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We have VC 2 = − g mVπ 2 RC = − g m (Vb 2 − Ve ) RC and VC1 = − g mVπ 1 RC = − g m (Vb1 − Ve ) RC Then V0 = VC 2 − VC1

= − g m (Vb 2 − Ve ) RC − ⎡⎣ − g m (Vb1 − Ve ) RC ⎤⎦ = g m RC (Vb1 − Vb 2 )

Differential gain Ad =

V0 = g m RC Vb1 − Vb 2

Common-mode gain Acm = 0 11.25 (a) vcm = 3 V ⇒ VC1 = VC 2 = 3 V 10 − 3 Then RC = ⇒ RC = 70 k Ω 0.1 (b) CMRRdB = 75 dB ⇒ CMRR = 5623 Now CMRR = 5623 =

1 ⎡ (1 + β ) I Q Ro ⎤ ⎢1 + ⎥ β VT 2⎣ ⎦

1 ⎡ (151)( 0.2 ) Ro ⎤ ⎢1 + ⎥ ⇒ Ro = 1.45 M Ω 2 ⎣⎢ (150 )( 0.026 ) ⎦⎥

Use a Widlar current source. Ro = ro [1 + g m RE′ ] Let VA of current source transistor be 100 V. 100 0.2 = 500 k Ω, g m = = 7.69 mA / V Then ro = 0.2 0.026 (150 )( 0.026 ) rπ = = 19.5 k Ω 0.2 So 1450 = 500 ⎡⎣1 + ( 7.69 ) RE′ ⎤⎦ ⇒ RE′ = 0.247 k Ω Now RE′ = RE & rπ ⇒ 0.247 = RE & 19.5 ⇒ RE = 250Ω ⎛I Then I Q RE = VT ln ⎜ REF ⎜ I ⎝ Q

⎞ ⎟⎟ ⎠ ⎛ I REF ⎞ ⎟⎟ ⇒ I REF = 1.37 mA ⎝ ( 0.2 ) ⎠

( 0.2 )( 0.250 ) = ( 0.026 ) ln ⎜⎜ Then R1 =

10 − 0.7 − ( −10 )

11.26 At terminal A. RTHA = RA & R =

1.37

⇒ R1 = 14.1 k Ω

R (1 + δ ) ⋅ R

R (1 + δ ) + R

=

R (1 + δ ) 2+δ

≅

R = 5 kΩ 2

Variation in RTH is not significant ⎛ RA ⎞ + R (1 + δ )( 5 ) 5 (1 + δ ) VTHA = ⎜ = ⎟V = R (1 + δ ) + R 2+δ ⎝ RA + R ⎠

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At terminal B. R = 5 kΩ 2 ⎛ R ⎞ + VTHB = ⎜ ⎟ V = 2.5 V ⎝R+R⎠ From Eq. (11.27) − β RC (V2 − V1 ) VO = where V2 = VTHB and V1 = VTHA 2 ( rπ + RB ) RTHB = R & R =

RB = 5 k Ω, rπ = So VO =

(120 )( 0.026 )

0.25 − (120 )( 3)(V2 − V1 ) 2 (12.5 + 5 )

= 12.5 k Ω

= −10.3 (V2 − V1 )

We can find V2 − V1 = VTHB − VTHA ⎡ 5 (1 + δ ) ⎤ VTHB − VTHA = 2.5 − ⎢ ⎥ ⎣ 2+δ ⎦ 2.5 ( 2 + δ ) − 5 (1 + δ ) 2.5δ − 5δ = = 2+δ 2+δ −2.5δ ≅ = −1.25δ 2 Then VO = − (10.3)( −1.25 ) δ = 12.9δ

So for −0.01 ≤ δ ≤ 0.01 We have −0.129 ≤ VO 2 ≤ 0.129 V 11.27 a. Rid = 2rπ rπ =

(180 )( 0.026 )

0.2 So Rid = 46.8 kΩ

= 23.4 kΩ

Assuming rμ → ∞, then

b.

Ricm ≅ ⎡⎣(1 + β ) R0 ⎤⎦ Ricm = ⎡⎣(181)(1) ⎤⎦ = 181 ⇒ Ricm = 181 MΩ

11.28 (a) 10 − 0.7 − ( −10 ) I1 = = 0.5 ⇒ R1 = 38.6 K R1 R2 =

0.026 ⎛ 0.5 ⎞ ln ⎜ ⎟ ⇒ R2 = 236 Ω 0.14 ⎝ 0.14 ⎠

(b)

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Ricm ≈ (1 + β ) Ro 0.14 = 5.385 mA/V 0.026 (180 )( 0.026 ) rπ 4 = = 33.4 K 0.14 RE′ = 33.4 0.236 = 0.234 K

Ro = ro 4 (1 + g m 4 RE′ ) g m 4 =

100 = 714 K 0.14 Ro = 714 ⎡⎣1 + ( 5.385 )( 0.234 ) ⎤⎦ ro 4 =

= 1614 K

Ricm = (181)(1614 ) ≈ 292 MΩ

(c) Acm =

− g m1 RC 2 (1 + β ) Ro 1+ rπ 1

g m1 =

rπ 1 =

0.07 = 2.692 mA/V 0.026

(180 )( 0.026 ) 0.07

− ( 2.692 )( 40 ) 2 (181)(1614 ) 1+ 66.86 Acm = −0.0123

= 66.86 K

Acm =

11.29 Ad 1 = g m1 ( R1 rπ 3 ) g m1 = rπ 3 = Ad 2 =

I Q1 / 2 VT

β VT IQ2 / 2

= 19.23I Q1 =

2 (100 )( 0.026 ) IQ 2

=

5.2 IQ 2

IQ 2 / 2 g m 3 R2 , g m3 = = 19.23I Q 2 2 VT

(19.23) I Q 2

⋅ R2 ⇒ I Q 2 R2 = 3.12 V 2 Maximum vo 2 − vo1 = ±18 mV for linearity Then 30 =

vo3 ( max ) = ( ±18 )( 30 ) mV ⇒ ±0.54 V so I Q 2 R2 = 3.12 V is OK.

From Ad 1 :

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⎛ ⎛ 5.2 ⎞ ⎞ ⎜ R1 ⎜⎜ ⎟ ⎟ I Q 2 ⎟⎠ ⎟ ⎜ ⎝ 20 = 19.23I Q1 ( R1 & rπ 3 ) = 19.23I Q1 ⎜ ⎟ ⎜ R + ⎛⎜ 5.2 ⎞⎟ ⎟ ⎜ 1 ⎜ IQ 2 ⎟ ⎟ ⎝ ⎠⎠ ⎝ 19.23I Q1 R1 ( 5.2 ) 20 = I Q 2 R1 + 5.2 I Q1

Let

2

⋅ R1 = 5V ⇒ I Q1 R1 = 10 V

Then 20 =

19.23 (10 )( 5.2 ) I Q 2 R1 + 5.2

Now I Q1 R1 = 10 ⇒ R1 =

⇒ I Q 2 R1 = 44.8 V

10 I Q1

⎛ 10 ⎞ ⎛ IQ2 ⎞ = 44.8 ⇒ ⎜ = 4.48 So I Q 2 ⎜ ⎟ ⎜ I ⎟⎟ ⎜I ⎟ ⎝ Q1 ⎠ ⎝ Q1 ⎠ Let I Q1 = 100 μ A, I Q 2 = 448 μ A Then I Q 2 R2 = 3.12 ⇒ R2 = 6.96 k Ω I Q1 R1 = 10 ⇒ R1 = 100 k Ω

11.30 a. 20 − VGS 3 2 I1 = = 0.25 (VGS 3 − 2 ) 50 20 − VGS 3 = 12.5 (VGS2 3 − 4VGS 3 + 4 ) 12.5VGS2 3 − 49VGS 3 + 30 = 0 VGS 3 =

49 ±

( 49 )

2

− 4 (12.5 )( 30 )

2 (12.5 )

⇒ VGS 3 = 3.16 V

20 − 3.16 ⇒ I1 = I Q = 0.337 mA 50 IQ = ⇒ I D1 = 0.168 mA 2

I1 = I D1

0.168 = 0.25 (VGS 1 − 2 ) ⇒ VGS1 = 2.82 V VDS 4 = −2.82 − ( −10 ) ⇒ VDS 4 = 7.18 V 2

VD1 = 10 − ( 0.168 )( 24 ) = 5.97 V

VDS1 = 5.97 − ( −2.82 ) ⇒ VDS 1 = 8.79 V

(b)

(c)

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Max vCM ⇒ VDS 1 = VDS 2 = VDS ( sat ) = VGS1 − VTN 2.82 − 2 = 0.82 V Now VD1 = 10 − ( 0.168 )( 24 ) = 5.97 V

VS ( max ) = 5.97 − VDS1 ( sat ) = 5.97 − 0.82 VS ( max ) = 5.15 V

vCM ( max ) = VS ( max ) + VGS1 = 5.15 + 2.82 vCM ( max ) = 7.97 V

vCM ( min ) = V − + VDS 4 ( sat ) + VGS 1

VDS 4 ( sat ) = VGS 4 − VTN = 3.16 − 2 = 1.16 V Then vCM ( min ) = −10 + 1.16 + 2.82 ⇒ vCM ( min ) = −6.02 V

11.31 a. I D1 = I D 2 = 120 μ A = 100 ( VGS1 − 1.2 ) ⇒ VGS 1 = VGS 2 = 2.30 V 2

For v1 = v2 = −5.4 V and VDS1 = VDS 2 = 12 V ⇒ −5.4 − 2.30 + 12 = 4.3 V = VD 10 − 4.3 ⇒ RD = 47.5 kΩ 0.12 I Q = I D1 + I D 2 ⇒ I Q = I1 = 240 μ A RD =

I1 = 240 = 200 (VGS 3 − 1.2 ) ⇒ VGS 3 = 2.30 V 2

R1 =

20 − 2.3 ⇒ R1 = 73.75 kΩ 0.24

b. r04 =

1

λ IQ

ΔI Q =

=

1

( 0.01)( 0.24 )

= 416.7 kΩ

1 5.4 ⋅ ΔVDS = ⇒ ΔI Q ≅ 13 μ A r04 416.7

11.32 (a) I Q = 160 μ A k′ ⎛ W ⎞ 2 I D = n ⎜ ⎟ (VGS − VTN ) 2⎝L⎠ 80 2 80 = ( 4 )(VOS − 0.5 ) 2 80 = 160 (Vo5 − 0.5 )

2

80 + 0.5 = 1.207 V 160 5−2 = 37.5 K VDS = 2 − ( −1.207 ) = 3.21 V RD = 0.08 (c) VDS ( sat ) = VGS − VTN = 1.207 − 0.5 = 0.707 V VGS =

Then VS = VO 2 − VDS ( sat ) = 2 − 0.707 = +1.29 V And v1 = v2 = vcm = VGS + VS = 1.207 + 1.29 vcm = 2.50 V

(b)

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11.33 vD = 5 − ( 0.2 )( 8 ) = 3.4 V VGS =

ID + VTN Kn

0.2 + 0.8 = 1.694 V 0.25 VDS ( sat ) = VGS − VTN = 1.694 − 0.8 = 0.894 V VS = VD − VDS ( sat ) = 3.4 − 0.894 = 2.506 =

vCM = VS + VGS = 2.506 + 1.694 ⇒ vCM = 4.2 V

(b) ΔvD = ΔI D ⋅ RD

ΔI D = g m ⋅

Vd 2

gm = 2 Kn I D =2

( 0.25)( 0.2 ) = 0.4472 mA/V

ΔI D = ( 0.4472 )( 0.05 ) ⇒ 22.36 μ A

ΔvD = ( 22.36 × 10−6 )( 8 × 103 ) = 0.179 V vD 2 = 3.4 + ΔvD vD 2 = 3.4 + 0.179 ⇒ vD 2 = 3.58 V (c) vd = −50 mV

ΔI D = − ( 0.4472 )( 0.025 ) ⇒ −11.18 μ A

ΔvD = − (11.18 × 10−6 )( 8 × 103 ) = −0.0894 V vD 2 = 3.4 − 0.0894 ⇒ vD 2 = 3.31 V

11.34 a. I D1 = I D 2 = 0.5 mA v01 − v02 = ⎡⎣V + − I D1 RD1 ⎤⎦ − ⎡⎣V + − I D 2 RD 2 ⎤⎦ v01 − v02 = I D 2 RD 2 − I D1 RD1 = I D ( RD 2 − RD1 )

i.

RD1 − RD 2 = 6 kΩ, v01 − v02 = 0

ii.

RD1 = 6 kΩ, RD 2 = 5.9 kΩ

v01 − v02 = ( 0.5 )( 5.9 − 6 ) ⇒ v01 − v02 = −0.05 V

b.

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K n1 = 0.4 mA / V 2 , K n 2 = 0.44 mA / V 2 VGS1 = VGS 2 I Q = ( K n1 + K n 2 )(VGS − VTN )

2

1 = ( 0.4 + 0.44 )(VGS − VTN ) ⇒ (VGS − VTN ) = 1.19 2

2

I D1 = ( 0.4 )(1.19 ) = 0.476 mA I D 2 = ( 0.44 )(1.19 ) = 0.524 mA i. RD1 = RD 2 = 6 kΩ

v01 − v02 = ( 0.524 − 0.476 )( 6 ) ⇒ v01 − v02 = 0.288 V

ii. RD1 = 6 kΩ,

RD 2 = 5.9 kΩ

v01 − v02 = ( 0.524 )( 5.9 ) − ( 0.476 )( 6 )

= 3.0916 − 2.856 ⇒ v01 − v02 = 0.236 V

11.35 (a)

From Equation (11.69)

⎛ K Kn iD 2 1 = − ⋅ vd 1 − ⎜ n ⎜ 2IQ IQ 2 2IQ ⎝ 0.90 = 0.50 −

⎞ 2 ⎟⎟ vd ⎠

⎡ 0.1 ⎤ 2 0.1 ⋅ vd 1 − ⎢ ⎥ vd 2 ( 0.25 ) ⎣⎢ 2 ( 0.25 ) ⎦⎥

+0.40 = − ( 0.4472 ) vd 1 − ( 0.2 ) vd2 0.8945 = −vd 1 − ( 0.2 ) vd2

Square both sides 0.80 = vd2 (1 − [ 0.2] vd2 )

( 0.2 ) ( vd2 ) vd2 =

2

− vd2 + 0.80 = 0

1 ± 1 − 4 ( 0.2 )( 0.80 ) 2 ( 0.2 )

= 4V 2 or 1V 2

Then vd = ± 2 V or ± 1 V But vd

max

=

IQ kn

=

0.25 = 1.58 0.1

So vd = ±1V, ⇒ vd = −1V

b.

From part (a), vd ,max = 1.58 V

11.36

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⎛i ⎞ d ⎜ D1 ⎟ ⎜I ⎟ ⎝ Q⎠= dvd

⎛ K Kn ⋅ 1− ⎜ n ⎜ 2I 2IQ ⎝ Q

) vd

vd =0

Kn 2IQ

= So linear

⎞ 2 ⎟⎟ vd + ( ⎠

Kn iD1 1 = + ⋅ vd IQ 2 2 IQ

⎡1 ⎤ ⎛K ⎞ Kn Kn 1 + ⋅ vd ( max ) − ⎢ + ⋅ vd ( max ) ⋅ 1 − ⎜ n ⎟vd2( max ) ⎥ 2 2IQ 2 IQ ⎢⎣ 2 ⎥⎦ ⎝ 2I n ⎠ Then = 0.02 Kn 1 + ⋅v 2 2 I Q d ( max ) ⎡1 ⎤ ⎡1 ⎛K Kn Kn 0.98 ⎢ + ⋅ vd ( max ) ⎥ = ⎢ + ⋅ vd ( max ) ⋅ 1 − ⎜ n ⎜ 2I 2IQ 2IQ ⎝ Q ⎣⎢ 2 ⎦⎥ ⎢⎣ 2

0.49 + 0.98

⎤ ⎞ 2 ⎟⎟ vd ( max ) ⎥ ⎥ ⎠ ⎦

⎡1 ⎤ ⎛ 0.15 ⎞ 2 0.15 0.15 ⋅ vd ( max ) = ⎢ + ⋅ vd ( max ) ⋅ 1 − ⎜⎜ ⎟⎟ vd ( max ) ⎥ 2 ( 0.2 ) 2 ( 0.2 ) ⎢2 ⎥ ⎝ 2 ( 0.2 ) ⎠ ⎣ ⎦

0.49 + 0.600 vd ( max ) = 0.50 + 0.6124 vd ( max ) ⋅ 1 − ( 0.6124 ) vd2( max ) 0.600 vd ( max ) = 0.010 + 0.6124 vd ( max ) ⋅ 1 − ( 0.6124 ) vd2( max )

By trial and error vd ( max ) ≈ 0.429 V 11.37 (b) gm = 2 K p I D = 2

( 0.05 )( 0.008696 )

= 0.0417 mA/V Vd = ( 0.0417 )( 0.05 ) = 0.002085 mA 2 ΔvD = ( 0.002085 )( 510 ) = 1.063 ΔI = g m

vD 2 ↑⇒ vD 2 = 1.063 − 4.565 = −3.502 V vD1 = −1.063 − 4.565 = −5.628 V 9 = I S RS + VSG + 1 I S = 2I D 8 = 2 K P RS (VSG + VTP ) + VSG 2

8 = ( 2 )( 0.05 )( 390 )(VSG − 0.8 ) + VSG 2

2 − 1.6VSG + 0.64 ) + VSG 8 = 39 (VSG 2 − 61.4VSG + 16.96 = 0 39VSG

VSG =

61.4 ± 3769.96 − 4 ( 39 )(16.96 ) 2 ( 39 )

= 1.217 V VS = 2.217 I S = 0.01739 mA

I D1 = I D 2 ⇒ 8.696 μ A

vD1 = vD 2 = ( 8.696 )( 0.510 ) − 9 = −4.565 V

(b)

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g m = 2 K P I DQ = 2

( 0.05 )( 0.008696 ) = 0.0417 mA/V

Vd = ( 0.0417 )( 0.05 ) = 0.002085 mA 2 ΔvD = ( 0.002085 )( 510 ) = 1.063 V ΔvD = ΔI D ⋅ RD

ΔI D = g m ⋅

v1 ↑, I D1 ↓, vD1 ↓

vD1 = −4.565 − 1.063 = −5.628 V vD 2 = −4.565 + 1.063 = −3.502 V 11.38 (a) v1 = v2 = 0 I D = K n (VSG + VTP ) ID = 6 μA

2

6 + 0.4 = VSG 30 VSG = 0.847 V VS = +0.847 V vD = I D RD − 3

= ( 6 )( 0.36 ) − 3 = −0.84 V

VSD = VS − vD = 0.847 − ( −0.84 ) vSD = 1.69 V (b) (i) Ad = g m RD g m = 2 K n I D

=2

( 30 )( 6 ) = 26.83 μ A/V

Ad = ( 26.83)( 0.36 ) ⇒ Ad = 9.66 Acm = 0 (ii) ( 26.83)( 0.36 ) g R Ad = m D = ⇒ Ad = 4.83 2 2 − ( 26.83)( 0.36 ) − g m RD Acm = = = −0.0448 1 + 2 g m RO 1 + 2 ( 26.83)( 4 )

11.39

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For v1 = v2 = −0.30 V I D1 = I D 2 = 0.1 mA VSG =

ID − VTP KP

0.1 +1 = 2 V 0.1 = vD 2 = ( 0.1)( 30 ) − 10 = −7 V =

vD1

gm = 2 K p I D = 2

( 0.1)( 0.1) = 0.2 mA/V

⎛V ⎞ ΔI D = g m ⎜ d ⎟ = ( 0.2 )( 0.1) = 0.02 mA ⎝ 2⎠ ΔvD = ( ΔI D ) RD = ( 0.02 )( 30 ) = 0.6 V vD 2 ↑⇒ vD 2 = −7 + 0.6 ⇒ vD 2 = −6.4 V vD1 = −7 − 0.6 ⇒ vD1 = −7.6 V

11.40 For v1 = v2 = 0 0 = VGS + 2 I D RS − 10 10 = VGS + 2 K n RS (VGS − VTN )

2

= VGS + 2 ( 0.15 )( 75 )(VGS − 1)

2

22.5VGS2 − 44VGS + 12.5 = 0 So VGS = 1.61 V and I D = ( 0.15 )(1.61 − 1) ⇒ 55.9 μ A 2

gm = 2 Kn I D = 2

( 0.15 )( 0.0559 )

g m = 0.1831 mA/V Use Half-circuits – Differential gain ΔR ⎞ ⎛V ⎞⎛ vD1 = − g m ⎜ d ⎟ ⎜ RD + ⎟ 2 ⎠ ⎝ 2 ⎠⎝ ΔR ⎞ ⎛V ⎞⎛ vo 2 = g m ⎜ d ⎟ ⎜ RD − ⎟ 2 ⎠ ⎝ 2 ⎠⎝ vo = vD1 − vD 2 = − g mVd RD v Ad = o = − g m RD Vd Now – Common-Mode Gain

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Vi = Vgs + g mVgs ( 2 RS ) = Vcm Vcm Vgs = 1 + g m ( 2 RS ) vD1

vD 2

ΔR ⎞ ⎛ − g m ⎜ RD + D ⎟ Vcm 2 ⎠ ⎝ = 1 + g m ( 2 RS ) ΔR ⎞ ⎛ − gm ⎜ RD − D ⎟ Vcm 2 ⎠ ⎝ = 1 + g m ( 2 RS )

vO = vD1 − vD 2 So vo = Acm =

− g m ( ΔRD ) Vcm 1 + g m ( 2 RD )

− g m ( ΔRD ) vo = Vcm 1 + g m ( 2 RS )

Then Ad = − ( 0.1831)( 50 ) = −9.16 Acm =

− ( 0.1831)( 0.5 )

1 + ( 0.1831)( 2 )( 75 )

= −0.003216

C M R R ∫ = 69.1 dB bB

11.41 a. Ad = g m ( r02 r04 ) r02 =

VA 2 150 = = 375 kΩ I C 2 0.4

r04 =

VA 4 100 = = 250 kΩ I C 4 0.4

gm =

IC 2 0.4 = = 15.38 mA/V VT 0.026

Ad = (15.38 ) ( 375 250 ) ⇒ Ad = 2307

b. RL = r02 r04 = 375 250 ⇒ RL = 150 kΩ

11.41 From 11.40 I D1 = I D 2 = 55.9 μ A g m = 0.183 mA/V

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⎛ +V ⎞ ΔvD 2 = + g m 2 ⎜ d ⎟ RD ⎝ 2 ⎠ V V vO = ΔvD1 − ΔvD 2 = − g m1 d RD − g m 2 d RD 2 2 −V −V ⎛ Δg Δg ⎞ ⎞ ⎛ vO = d ⋅ RD ( g m 2 + g m1 ) = d ⋅ RD ⎜ g m − m + ⎜ g m − m ⎟ ⎟ 2 2 2 2 ⎠⎠ ⎝ ⎝ Ad = − g m RD = − ( 0.183) ( 50 ) = −9.15 Ad : ΔvD1 = − g m1

Vd ⋅ RD 2

Δg ⎞ Δg ⎞ ⎛ ⎛ − ⎜ g m + M ⎟ RD vcm ⎜ g m − M ⎟ RD vCM 2 ⎠ 2 ⎠ ⎝ ⎝ = + 1 + g m ( 2 RS ) 1 + g m ( 2 RS )

ACM : vO = ΔvD1 − ΔvD 2 −Δg m RD vO = vcm 1 + g m ( 2 RS )

Acm = Acm =

Δg m = ( 0.01) ( 0.183) = 0.00183

− ( 0.00183) ( 50 )

1 + ( 0.183)( 2 ) ( 75 )

= −0.003216

C M R R ∫ = 69.1 dB dB

11.42 (a) v1 = v2 = 0 5 = 2 I D RS + VSG 5 = 2 K p RS (VSG + VTP ) + VSG 2

2 5 = 2 ( 0.5 )( 2 ) (VSG − 1.6VSG + 0.64 ) + VSG 2 5 = 2VSG − 2.2VSG + 1.28 2 2VSG − 2.2VSG − 3.72 = 0

VSG =

2.2 ± 4.84 + 4 ( 2 )( 3.72 ) 2 ( 2)

VSG = 2.02 V vS = 2.02 V,

5 − 2.02 = 1.49 mA 2 = I D 2 = 0.745 mA

IS = I D1

vD1 = vD 2 = ( 0.745 (1) − 5 ) ⇒ vD1 = vD 2 = −4.26 V

(b) 5 = I S RS + VSG 2 5 = ( I D1 + I D 2 ) RS + VSG 2 2 2 5 = ⎡ K p (VSG1 + VTP ) + K p (VSG 2 + VTP ) ⎤ RS + VSG 2 ⎣ ⎦ VSG1 = VSG 2 − 1

5 = ( 0.5 )( 2 ) ⎡(VSG 2 − 1.8 ) + (VSG 2 − 0.8 ) ⎤ + VSG 2 ⎣ ⎦ 2 2 5 = ⎡⎣VSG 2 − 3.6VSG 2 + 3.24 + VSG 2 − 1.6VSG 2 + 0.64 ⎤⎦ + VSG 2 2

2

2 5 = 2VSG 2 − 4.2VSG 2 + 3.88 2 2VSG 2 − 4.2VSG 2 − 1.12 = 0

VSG 2 =

4.2 ± 17.64 + 4 ( 2 ) (1.12 ) 2 ( 2)

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VSG 2 = 2.339 V VSG1 = 1.339 V vS = 2.339 V I D1 I D1 vD1 vD1

= 0.5 (1.339 − 0.8 ) = 0.1453 mA = ( 0.1453)(1) − 5 = −4.855 V

2

I D2 I D2 vD 2 vD 2

= 0.5 ( 2.339 − 0.8 ) = 1.184 mA = (1.184 ) (1) − 5 = −3.816 V

2

(c) ΔI = g m

Vd 2

gm = 2 K p I D

vS ≈ 2.02 V = 2

( 0.5 )( 0.745 )

g m = 1.22 mA/V ΔI = (1.22 )( 0.1) = 0.122 mA ΔvD = ( ΔI ) RD = ( 0.122 )(1) = 0.122 V vD 2 ↓ vD1 ↑ vD1 = −4.26 + 0.122 vD 2 = −4.26 − 0.122 vD1 = −4.138 V vD 2 = −4.382 V

11.43

a.

gf =

IQ 4VT

⇒ I Q = g f ( 4VT ) = ( 8 )( 4 )( 0.026 )

⇒ I Q = 0.832 mA

Neglecting base currents. 30 − 0.7 R1 = ⇒ R1 = 35.2 kΩ 0.832 V 100 r04 = r02 = A = = 240 kΩ b. I CQ 0.416

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gm =

I CQ VT

=

0.416 = 16 mA / V 0.026

Ad = g m ( r02 || r04 ) = 16 ( 240 || 240 ) ⇒ Ad = 1920 Rid = 2rπ , rπ =

(180 )( 0.026 ) 0.416

= 11.25 kΩ

⇒ Rid = 22.5 kΩ R0 = r02 || r04 ⇒ R0 = 120 kΩ

c. Max. common-mode voltage when VCB = 0 for Q1 and Q2 . Therefore vcm ( max ) = V + − VEB ( Q3 ) = 15 − 0.7 vcm ( max ) = 14.3 V Min. common-mode voltage when VCB = 0 for Q5 . Therefore vcm ( min ) = 0.7 + 0.7 + ( −15 ) = −13.6 V So −13.6 ≤ vcm ≤ 14.3 V 1 (1 + β )( 2 R0 ) 2 V 100 R0 = A = = 120 kΩ I Q 0.832 Ricm ≅

Ricm = (181)(120 ) ⇒ Ricm = 21.7 MΩ

11.43 (a) gm = 2 Kn I D =2

( 0.4 )(1)

g m = 1.265 mA/V v 1 Ad = o = = 10 vd 0.1 Ad = g m RD 10 = (1.265 ) RD RD = 7.91 K

(b) Quiescent v1 = v2 = 0 vD1 = vD 2 = 10 − (1)( 7.91) = 2.09 V VGS =

ID 1 + VTN = + 0.8 = 2.38 V Kn 0.4

VDS ( sat ) = 2.38 − 0.8 = 1.58 So vcm = vD − VDS ( sat ) + VGS = 2.09 − 1.58 + 2.38 vcm = 2.89 V

11.44

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g m RD 2 For vCM = 2.5 V IQ = 0.25 mA I D1 = I D 2 = 2 Ad =

Let VD1 = VD 2 = 3 V , then RD = Then 100 =

g m ( 28 )

2 k′ ⎛ W And g m = 2 n ⎜ 2⎝L

10 − 3 ⇒ RD = 28 k Ω 0.25

⇒ g m = 7.14 mA / V ⎞ ⎟ ID ⎠

⎛ 0.080 ⎞ ⎛ W ⎞ 7.14 = 2 ⎜ ⎟ ⎜ ⎟ ( 0.25 ) ⇒ ⎝ 2 ⎠⎝ L ⎠ ⎛W ⎞ ⎛W ⎞ ⎜ ⎟ = ⎜ ⎟ = 1274 (Extremely large transistors to meet the gain requirement.) ⎝ L ⎠1 ⎝ L ⎠ 2 Need ACM = 0.10 From Eq. (11.64(b)) g m RD ACM = 1 + 2 g m Ro

( 7.14 )( 28) ⇒ Ro = 140 k Ω 1 + 2 ( 7.14 ) Ro For the basic 2-transistor current source 1 1 Ro = ro = = = 200 k Ω λ I Q ( 0.01)( 0.5 ) So 0.10 =

This current source is adequate to meet common-mode gain requirement. 11.45 Not in detail, Approximation looks good. a. −V − ( −5 ) 2 I S = GS 1 and I S = 2 I D = 2 K n (VGS 1 − VTN ) RS 5 − VGS 1 2 = 2 ( 0.050 )(VGS 1 − 1) 20 5 − VGS 1 = 2 (VGS2 1 − 2VGS1 + 1) 2VGS2 1 − 3VGS 1 − 3 = 0 VGS1 =

3±

( 3)

2

+ 4 ( 2 )( 3)

2 ( 2)

⇒ VGS1 = 2.186 V

5 − 2.186 ⇒ I S = 0.141 mA 20 I I D1 = I D 2 = S ⇒ I D1 = I D 2 = 0.0704 mA 2 v02 = 5 − ( 0.0704 )( 25 ) ⇒ v02 = 3.24 V IS =

b. g m = 2 K n (VGS − VTN ) = 2 ( 0.05 )( 2.186 − 1) g m = 0.119 mA/V 1 1 r0 = = = 710 kΩ λ I DQ ( 0.02 )( 0.0704 )

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Vgs1 = v1 − VS , Vgs 2 = v2 − VS v01 v −V + g mVgs1 + 01 S = 0 RD r0

⎛ 1 V 1⎞ v01 ⎜ + ⎟ + g m ( v1 − VS ) − S = 0 r0 ⎝ RD r0 ⎠ v02 v − VS + g mVgs 2 + 02 =0 RD r0

(1)

⎛ 1 V 1⎞ v02 ⎜ + ⎟ + g m ( v2 − VS ) − S = 0 R r r0 ⎝ D 0⎠ v − V v − VS V + g mVgs 2 = S g mVgs1 + 01 S + 02 r0 r0 RS

(2)

g m ( v1 − VS ) +

v01 v02 2VS V + − + g m ( v2 − VS ) = S r0 r0 r0 RS

g m ( v1 + v2 ) +

v01 v02 + = VS r0 r0

⎧ 2 1 ⎫ ⎨2 g m + + ⎬ r RS ⎭ 0 ⎩

(3)

From (1) ⎛ 1⎞ VS ⎜ g m + ⎟ − g m v1 r 0 ⎠ v01 = ⎝ ⎛ 1 1⎞ + ⎟ ⎜ ⎝ RD r0 ⎠ Then ⎛ 1⎞ VS ⎜ g m + ⎟ − g m v1 r0 ⎠ ⎧ v 2 1 ⎫ g m ( v1 + v2 ) + ⎝ + 02 = VS ⎨2 g m + + ⎬ (3) r0 r0 RS ⎭ ⎛ 1 1⎞ ⎩ r0 ⎜ + ⎟ R r ⎝ D 0⎠

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⎛ 1 ⎛ ⎛ 1 ⎧ 1⎞ 1⎞ 1⎞ 2 1 ⎫ ⎛ 1 1⎞ g m ( v1 + v2 ) r0 ⎜ + ⎟ + VS ⎜ g m + ⎟ − g m v1 + v02 ⎜ + ⎟ = VS ⎨2 g m + + ⎬ ⋅ r0 ⎜ + ⎟ r0 ⎠ r0 RS ⎭ ⎝ RD r0 ⎠ ⎝ RD r0 ⎠ ⎝ ⎝ RD r0 ⎠ ⎩ ⎛ 1 ⎛ r ⎞ r0 ⎞ ⎛ 1⎞ 2 1 ⎞⎛ 1 ⎞ ⎪⎫ ⎪⎧⎛ g m ( v1 + v2 ) ⎜ 1 + 0 ⎟ − g m v1 + v02 ⎜ + ⎟ = VS ⎨⎜ 2 g m + + ⎟ ⎜1 + ⎟ − ⎜ gm + ⎟⎬ R R r r R R r ⎪⎩⎝ 0 0 ⎠⎪ ⎝ D ⎠ S ⎠⎝ D ⎠ ⎝ ⎝ D 0⎠ ⎭ ⎛ 1 ⎧ ⎛ r r ⎞ r r 1⎞ 2 1 2 1⎫ g m ⎜ v1 ⋅ 0 + v2 + v2 ⋅ 0 ⎟ + v02 ⎜ + ⎟ = VS ⎨2 g m + + + 2gm ⋅ 0 + + 0 − gm − ⎬ RD ⎠ r0 RS RD RD RS RD r0 ⎭ ⎝ RD ⎝ RD r0 ⎠ ⎩ ⎧⎪ ⎫ ⎛ 1 ⎛ r r ⎞ r0 ⎞ 2 1⎞ 1 1 ⎛ g m ⎜ v1 ⋅ 0 + v2 + v2 ⋅ 0 ⎟ + v02 ⎜ + ⎟ = VS ⎨2 g m + + (1 + g m r0 )⎪⎬ (4) ⎜1 + ⎟+ RD ⎠ r0 RS ⎝ RD ⎠ RD ⎪⎩ ⎝ RD ⎝ RD r0 ⎠ ⎭⎪ ⎛ 1 ⎛ 1⎞ 1⎞ Then substituting into (2), v02 ⎜ + ⎟ + g m v2 = VS ⎜ g m + ⎟ r0 ⎠ ⎝ RD r0 ⎠ ⎝ 710 ⎤ 1 ⎤ ⎡ 710 ⎡1 + v2 + v2 + v02 ⎢ + (4) Substitute numbers: ( 0.119 ) ⎢ v1 ⎥ ⎥ 25 ⎦ ⎣ 25 ⎣ 25 710 ⎦ ⎧ ⎫ 1 1 ⎛ 710 ⎞ 2 = VS ⎨0.119 + + ⎜1 + ⎟ + ⎣⎡1 + ( 0.119 )( 710 ) ⎦⎤ ⎬ 710 20 ⎝ 25 ⎠ 25 ⎩ ⎭

( 0.119 ) [ 28.4v1 + 29.4v2 ] + ( 0.0414 ) v02 = VS {0.1204 + 1.470 + 6.8392} = VS ( 8.4296 ) or VS = 0.4010v1 + 0.4150v2 + 0.00491v02 1 ⎞ 1 ⎞ ⎛ 1 ⎛ (2) Then v02 ⎜ + ⎟ + ( 0.119 ) v2 = VS ⎜ 0.119 + ⎟ 710 ⎠ ⎝ 25 710 ⎠ ⎝ v02 ( 0.0414 ) + v2 ( 0.119 ) = ( 0.1204 ) [ 0.401v1 + 0.4150v2 + 0.00491v02 ] v02 ( 0.0408 ) = ( 0.04828 ) v1 − ( 0.0690 ) v2 v02 = (1.183) v1 − (1.691) v2

vd 2 vd v2 = vcm − 2 v ⎞ v ⎞ ⎛ ⎛ So v02 = (1.183) ⎜ vcm + d ⎟ − (1.691) ⎜ vcm − d ⎟ 2 2⎠ ⎝ ⎠ ⎝ Or v02 = 1.437vd − 0.508vcm ⇒ Ad = 1.437, Acm = −0.508 Now

v1 = vcm +

⎛ 1.437 ⎞ C M R RdB = 20 log10 ⎜ ⎟ ⇒ C M R RdB = 9.03 dB ⎝ 0.508 ⎠

11.46

KVL:

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v1 = Vgs1 − Vgs 2 + v2 So v1 − v2 = Vgs1 − Vgs 2

KCL: g mVgs1 + g mVgs 2 = 0 ⇒ Vgs1 = −Vgs 2 1 1 So Vgs1 = ( v1 − v2 ) , Vgs 2 = − ( v1 − v2 ) 2 2 Now v02 v02 − v01 + = − g mVgs 2 RD RL ⎛ 1 1 ⎞ v01 = v02 ⎜ + ⎟− ⎝ RD RL ⎠ RL v01 v01 − v02 + = − g mVgs1 RD RL ⎛ 1 1 ⎞ v02 = v01 ⎜ + ⎟− R R RL L ⎠ ⎝ D ⎛ R ⎞ From (1): v01 = v02 ⎜ 1 + L ⎟ + g m RLVgs 2 R ⎝ D ⎠

(1)

(2)

Substitute into (2): ⎛ ⎛ 1 v02 R ⎞⎛ 1 1 ⎞ 1 ⎞ − g mVgs1 = v02 ⎜1 + L ⎟ ⎜ + + ⎟ + g m RL ⎜ ⎟ Vgs 2 − R R R R R RL ⎝ ⎝ D D ⎠⎝ D L ⎠ L ⎠ ⎛ ⎛ 1 R ⎞⎛ 1 ⎞ R 1 ⎞ − g m ⋅ ( v1 − v2 ) + g m ⎜ 1 + L ⎟ ⎜ ⎟ ( v1 − v2 ) = v02 ⎜ + L2 + ⎟ ⎝ RD ⎠ ⎝ 2 ⎠ ⎝ RD RD RD ⎠ 1 ⋅ g m RL v02 ⎛ v02 RL ⎞ 1 ⎛ RL ⎞ 2 2 gm ⎜ v − v = + ⇒ A = = ( ) ⎟ 1 2 ⎜ ⎟ d2 2 ⎝ RD ⎠ RD ⎝ RD ⎠ v1 − v2 ⎛ RL ⎞ ⎜2+ ⎟ RD ⎠ ⎝

1 − ⋅ g m RL v01 From symmetry Ad 1 = = 2 v1 − v2 ⎛ RL ⎞ ⎜2+ ⎟ R D ⎠ ⎝

Then Av =

v02 − v01 g m RL = v1 − v2 ⎛ RL ⎞ ⎜2+ ⎟ RD ⎠ ⎝

11.47

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v1 − v2 = Vgs1 − Vgs 2 and g mVgs1 + g mVgs 2 = 0 ⇒ Vgs1 = −Vgs 2 Then v1 − v2 = −2Vgs 2 Or Vgs 2 = −

1 ( v1 − v2 ) 2

v0 = − g mVgs 2 ( RD RL ) = Or Ad =

gm ( RD RL ) ( v1 − v2 ) 2

gm ( RD RL ) 2

11.48 From Equation (11.64(a)), Ad = We need Ad = Then 10 =

Kn IQ 2

⋅ RD

2 = 10 0.2

K n ( 0.5 )

⋅ RD or K n ⋅ RD = 20 2 If we set RD = 20 k Ω, then K n = 1 mA / V 2

For this case VD = 10 − ( 0.25 )( 20 ) = 5 V 0.25 + 1 = 1.5 V 1 VDS ( sat ) = VGS − VTN = 1.5 − 1 = 0.5 V VGS =

Then vcm ( max ) = VD − VDS ( sat ) + VGS = 5 − 0.5 + 1.5 Or vcm ( max ) = 6 V

11.49 Vd 1 = − g mVgs1 RD = − g m RD (V1 − Vs ) Vd 2 = − g mVgs 2 RD = − g m RD (V2 − Vs ) Now Vo = Vd 2 − Vd 1 = − g m RD (V2 − Vs ) − ( − g m RD (V1 − Vs ) ) Vo = g m RD (V1 − V2 ) Define V1 − V2 ≡ Vd V Then Ad = o = g m RD and Acm = 0 Vd

11.49 Ad = g m ( r02 r04 ) g m = 2 kn I DQ

=2

( 0.12 )( 0.075 )

= 0.1897 mA/V 1 1 r02 = = = 889 kΩ λn I DQ ( 0.015 )( 0.075 ) r04 =

1

λ p I DQ

=

1 = 667 kΩ 0.02 ( )( 0.075)

Ad = ( 0.1897 ) ( 889 667 ) ⇒ Ad = 72.3

11.50 (a)

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⎛ K′ ⎞⎛W K n1 = K n 2 = ⎜ n ⎟ ⎜ ⎝ 2 ⎠⎝ L VGS1 = VGS 2 =

⎞ ⎛ 0.080 ⎞ 2 ⎟=⎜ ⎟ (10 ) = 0.40 mA / V ⎠ ⎝ 2 ⎠

ID 0.1 + VTN = + 1 = 1.5 V Kn 0.4

VDS1 ( sat ) = 1.5 − 1 = 0.5 V

For vCM = +3 V ⇒ VD1 = VD 2 = vCM − VGS 1 + VDS 1 ( sat )

= 3 − 1.5 + 0.5 ⇒ VD1 = VD 2 = 2 V RD =

10 − 2 ⇒ RD = 80 k Ω 0.1

(b) 1 g m RD and g m = 2 ( 0.4 )( 0.1) = 0.4 mA / V 2 1 Then Ad = ( 0.4 )( 80 ) = 16 2 16 C M R RdB = 45 ⇒ C M R R = 177.8 = Acm Ad =

So Acm = 0.090 Acm =

g m RD 1 + 2 g m Ro

0.090 =

( 0.4 )(80 ) 1 + 2 ( 0.4 ) Ro

⇒ Ro = 443 k Ω

If we assume λ = 0.01 V −1 for the current source transistor, then 1 1 ro = = = 500 k Ω λ I Q ( 0.01)( 0.2 ) So the CMRR specification can be met by a 2-transistor current source. ⎛W ⎞ ⎛W ⎞ Let ⎜ ⎟ = ⎜ ⎟ = 1 ⎝ L ⎠3 ⎝ L ⎠ 4 IQ 0.2 ⎛ 0.080 ⎞ 2 Then K n 3 = K n 4 = ⎜ + VTN = + 1 = 3.24 V ⎟ (1) = 0.040 mA / V and VGS 3 = 2 0.04 K ⎝ ⎠ n3 For vCM = −3 V , VD 3 = −3 − VGS1 = −3 − 1.5 = −4.5 V ⇒ VDS 3 ( min ) = −4.5 − ( −10 ) = 5.5 V > VDS 3 ( sat )

So design is OK. ⎛W ⎞ On reference side: For ⎜ ⎟ ≥ 1, VGS ( max ) = 3.24 V ⎝L⎠ 20 − VGS 3 = 20 − 3.24 = 16.76 V

Then

16.67 = 5.17 ⇒ We need six transistors in series. 3.24

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20 − 3.24 = 2.793 V 6 2 ⎛ K′ ⎞⎛W ⎞ = ⎜ n ⎟ ⎜ ⎟ (VGS − VTN ) ⎝ 2 ⎠⎝ L ⎠

VGS = I REF

2 ⎛ 0.080 ⎞⎛ W ⎞ ⎛W ⎞ 0.2 = ⎜ ⎟⎜ ⎟ ( 2.793 − 1) ⇒ ⎜ ⎟ = 1.56 for each of the 6 transistors. ⎝ 2 ⎠⎝ L ⎠ ⎝L⎠

11.51

Ad =

1 g m RD 2

gm = 2 Kn I D = 2

( 0.25 )( 0.25) = 0.50 mA / V

1 ( 0.50 )( 3) = 0.75 2 From Problem 11.26 Ad =

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V1 = VA =

5 (1 + δ ) 2+δ

, V2 = VB = 2.5 V and V1 − V2 = 1.25δ

Then Vo 2 = Ad ⋅ (V1 − V2 ) = ( 0.75 )(1.25δ ) = 0.9375δ

So for −0.01 ≤ δ ≤ 0.01 −9.375 ≤ Vo 2 ≤ 9.375 mV

11.52 From previous results v −v Ad 1 = o 2 o1 = g m1 R1 = 2 K n1 I Q1 ⋅ R1 = 20 v1 − v2 and Ad 2 = Set

I Q1 R1 2

vo3 1 1 2 K n3 I Q 2 ⋅ R2 = 30 = g m 3 R2 = 2 vo 2 − vo1 2 I Q 2 R2

= 5 V and

2

= 2.5 V

Let I Q1 = I Q 2 = 0.1 mA Then R1 = 100 k Ω, R2 = 50 k Ω 2

⎛ 0.06 ⎞ ⎛ W ⎞ ⎛ 20 ⎞ ⎛W ⎞ ⎛W ⎞ Then 2 ⎜ ⎟ ⎜ ⎟ ( 0.1) = ⎜ ⎟ ⇒ ⎜ ⎟ = ⎜ ⎟ = 6.67 ⎝ 2 ⎠ ⎝ L ⎠1 ⎝ 100 ⎠ ⎝ L ⎠1 ⎝ L ⎠ 2 2

⎛ 2 ( 30 ) ⎞ ⎛ 0.060 ⎞ ⎛ W ⎞ ⎛W ⎞ ⎛W ⎞ and 2 ⎜ ⎟ ⇒ ⎜ ⎟ = ⎜ ⎟ = 240 ⎟ ⎜ ⎟ ( 0.1) = ⎜ ⎝ 2 ⎠ ⎝ L ⎠3 ⎝ L ⎠3 ⎝ L ⎠ 4 ⎝ 50 ⎠

11.53 a.

iD1

⎛ v ⎞ = I DSS ⎜ 1 − GS 1 ⎟ VP ⎠ ⎝

⎛ v ⎞ iD 2 = I DSS ⎜ 1 − GS 2 ⎟ VP ⎠ ⎝ iD1 − iD 2

2

2

⎛ v ⎞ ⎛ v ⎞ = I DSS ⎜1 − GS 1 ⎟ − I DSS ⎜ 1 − GS 2 ⎟ VP ⎠ VP ⎠ ⎝ ⎝ I DSS

=

( vGS 2 − vGS1 )

VP

I DSS

=−

VP

⋅ vd =

I DSS

( −VP )

⋅ vd

iD1 + iD 2 = I Q ⇒ iD 2 = I Q − iD1

(

iD1 − I Q − iD1

)

2

=

I DSS

( −VP )

2

⋅ vd2

iD1 − 2 iD1 ( I Q − iD1 ) + ( I Q − iD1 ) = Then iD1 ( I Q − iD1 ) =

I DSS

( −VP )

2

⋅ vd2

⎤ I 1⎡ ⎢ I Q − DSS 2 ⋅ vd2 ⎥ 2 ⎢⎣ ( −VP ) ⎦⎥

Square both sides

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2

⎤ I 1⎡ i − iD1 I Q + ⎢ I Q − DSS 2 ⋅ vd2 ⎥ = 0 4 ⎣⎢ ( −VP ) ⎦⎥ 2 D1

⎤ I ⎛ 1⎞⎡ I Q ± I − 4 ⎜ ⎟ ⎢ I Q − DSS 2 ⋅ vd2 ⎥ ⎝ 4 ⎠ ⎢⎣ ( −VP ) ⎥⎦

2

2 Q

iD1 =

2

2 2 ⎡ ⎤ 1 2 ⎢ 2 2 I Q I DSS vd ⎛ I DSS vd2 ⎞ ⎥ ⎜ ⎟ iD1 = IQ − IQ − ± + 2 2 ⎢ 2 2 ( −VP ) ⎜⎝ ( −VP ) ⎟⎠ ⎥⎦ ⎣ Use + sign

IQ

iD1

1 2 I Q I DSS 2 = + ⋅ vd 2 2 ( −VP )2

iD1 =

IQ

IQ 2

+

⎛ I ⎞ − ⎜ DSS 2 ⋅ vd2 ⎟ ⎜ ( −V ) ⎟ P ⎝ ⎠

2

2

2

2

2

2

2

2 I DSS ⎛ I DSS 1 IQ vd −⎜ ⎜ I 2 ( −VP ) IQ ⎝ Q

⎞ ⎛ vd ⎞ ⎟⎟ ⎜ ⎟ ⎠ ⎝ VP ⎠

⎞ 2 I DSS ⎛ I DSS −⎜ ⎟ ⋅ vd ⎜ I IQ ⎠ ⎝ Q

⎞ ⎛ vd ⎞ ⎟⎟ ⎜ ⎟ ⎠ ⎝ VP ⎠

⎞ 2 I DSS ⎛ I DSS −⎜ ⎟ ⋅ vd ⎜ I IQ ⎠ ⎝ Q

⎞ ⎛ vd ⎞ ⎟⎟ ⎜ ⎟ ⎠ ⎝ VP ⎠

Or iD1 1 ⎛ 1 = +⎜ I Q 2 ⎝ −2VP We had iD 2 = I Q − iD1 Then iD 2 1 ⎛ 1 = −⎜ I Q 2 ⎝ −2VP b. If iD1 = I Q , then 1 ⎛ 1 1= +⎜ 2 ⎝ −2VP

⎞ 2 I DSS ⎛ I DSS −⎜ ⎟ ⋅ vd ⎜ I IQ ⎠ ⎝ Q

2 I DSS ⎛ I DSS VP = vd −⎜ ⎜ I IQ ⎝ Q Square both sides

2

⎞ ⎛ vd ⎞ ⎟⎟ ⎜ ⎟ ⎠ ⎝ VP ⎠

2

⎞ ⎛ vd ⎞ ⎟⎟ ⎜ ⎟ ⎠ ⎝ VP ⎠

2

2

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VP

2

⎛ I DSS ⎜⎜ ⎝ IQ

vd2 =

⎡ 2I ⎛I = v ⎢ DSS − ⎜ DSS ⎜ I ⎢ IQ ⎝ Q ⎣ 2 d

2

2 2 ⎞ ⎛ vd ⎞ ⎤ ⎟⎟ ⎜ ⎟ ⎥ ⎠ ⎝ VP ⎠ ⎥⎦

2

⎞ ⎛ 1 ⎞ 2 2 2 I DSS 2 ⋅ vd + VP ⎟⎟ ⎜ ⎟ ( vd ) − IQ ⎠ ⎝ VP ⎠ ⎛ 2I 2 I DSS ± ⎜ DSS ⎜ I IQ ⎝ Q

2

⎞ ⎛I ⎟⎟ − 4 ⎜⎜ DSS ⎠ ⎝ IQ

⎛ 2I 2 ⎜ DSS ⎜ IQ ⎝

2

2

=0 2

⎞ ⎛ 1 ⎞ 2 ⎟⎟ ⎜ ⎟ (VP ) ⎠ ⎝ VP ⎠

2

⎞ ⎛ 1 ⎞ ⎟⎟ ⎜ ⎟ ⎠ ⎝ VP ⎠

2

2 ⎛ IQ ⎞ vd2 = (VP ) ⎜ ⎟ ⎝ I DSS ⎠ 1/ 2

⎛ IQ ⎞ Or vd = VP ⎜ ⎟ ⎝ I DSS ⎠ c. For vd small, IQ 1 IQ 2 I DSS + ⋅ ⋅ vd iD1 ≈ IQ 2 2 ( −VP ) gf =

diD1 d vd

vd → 0

=

2 I DSS 1 IQ ⋅ ⋅ IQ 2 ( −VP )

⎛ 1 ⎞ I Q I DSS Or ⇒ g f ( max ) = ⎜ ⎟ 2 ⎝ −VP ⎠

11.53 Ad = g m ( ro 2 Ro ) Want Ad = 400 From Example 11.15, ro 2 = 1 M Ω Assuming that g m = 0.283 mA / V for the PMOS from Example 11.15, then Ro = 285 M Ω. ⎛ k ′ ⎞⎛ W ⎞ So 400 = g m (1000 285000 ) ⇒ g m = 0.4014 mA / V = 2 ⎜ n ⎟ ⎜ ⎟ I DQ ⎝ 2 ⎠ ⎝ L ⎠1 ⎛ 0.080 ⎞ ⎛ W ⎞ ⎛W ⎞ ⎛W ⎞ 0.04028 = ⎜ ⎟ ⎜ ⎟ ( 0.1) ⇒ ⎜ ⎟ = ⎜ ⎟ = 10.1 ⎝ 2 ⎠ ⎝ L ⎠1 ⎝ L ⎠1 ⎝ L ⎠ 2

11.54 a. I Q = I D1 + I D 2 ⇒ I Q = 1 mA v0 = 7 = 10 − ( 0.5 ) RD ⇒ RD = 6 kΩ

b. ⎛ 1 ⎞ I Q ⋅ I DSS g f ( max ) = ⎜ ⎟ 2 ⎝ −VP ⎠ ⎛ 1 ⎞ (1)( 2 ) g f ( max ) = ⎜ ⎟ ⇒ g f ( max ) = 0.25 mA/V 2 ⎝ 4⎠ c. g R Ad = m D = g f ( max ) ⋅ RD 2 Ad = ( 0.25 )( 6 ) ⇒ Ad = 1.5

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11.55 a. IS =

−VGS − ( −5 ) RS

⎛ V ⎞ = ( 2 ) I DSS ⎜ 1 − GS ⎟ VP ⎠ ⎝

⎛ V ⎞ 5 − VGS = ( 2 )( 0.8 )( 20 ) ⎜⎜ 1 − GS ⎟⎟ ⎝ ( −2 ) ⎠ 1 ⎛ ⎞ 5 − VGS = ( 2 )16 ⎜1 + VGS + VGS2 ⎟ 4 ⎝ ⎠ 2 8VGS + 33VGS + 27 = 0 VGS =

2

2

−33 ± 1089 − 4 ( 8 )( 27 ) 2 (8)

= −1.125 V IS =

5 − ( −1.125 )

20 = 0.306 mA

I D1 = I D 2 = 0.153 mA vo 2 = 1.17 V (b)

11.56 Equivalent circuit and analysis is identical to that in problem 11.36. 1 ⋅ g m RL Ad 2 = 2 ⎛ RL ⎞ ⎜2+ ⎟ RD ⎠ ⎝ 1 − ⋅ g m RL Ad 1 = 2 ⎛ RL ⎞ ⎜2+ ⎟ RD ⎠ ⎝ Av =

v02 − v01 g m RL = vd ⎛ RL ⎞ ⎜2+ ⎟ R ⎝ D ⎠

11.57 (a) Ad = g m ( ro 2 ro 4 ) 0.1 = 3.846 mA/V 0.026 120 ro 2 = = 1200 K 0.1 80 ro 4 = = 800 K 0.1 Ad = ( 3.846 ) (1200 800 ) gm =

Ad = 1846

(b)

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For Ad = 923 = ( 3.846 ) (1200 800 RL ) 240 = 480 & RL =

480 RL ⇒ RL = 480 K 480 + RL

11.58 (a) ⎛ 2⎞ I Q = 250 μ A I REF = I Q ⎜ 1 + ⎟ ⎝ β⎠ 2 ⎞ ⎛ = 250 ⎜1 + ⎟ = 252.8 μ A ⎝ 180 ⎠ 5 − ( 0.7 ) − ( −5 ) R1 = ⇒ R1 = 36.8 K 0.2528 (b) 0.125 = 4.808 mA/V Ad = g m ( ro 2 ro 4 ) gm = 0.026 150 = 1200 K ro 2 = 0.125 100 = 800 K Ad = ( 4.808 ) (1200 800 ) ro 4 = 0.125 Ad = 2308

(c) Rid = 2rπ =

2 (180 )( 0.026 ) 0.125

⇒ Rid = 74.9 K

Ro = ro 2 ro 4 = 1200 800 = 480 K = Ro

(d) vcm ( max ) = 5 − 0.7 = 4.3 V vcm ( min ) = 0.7 + 0.7 − 5 = −3.6 V 11.59 a. ⎛ IQ ⎞ ⎛ 1 ⎞ I 0 = I B3 + I B 4 ≈ 2 ⎜ ⎟ ⎜ ⎟ ⎝ 2 ⎠⎝ β ⎠ I Q 0.2 I0 = = ⇒ I0 = 2 μ A β 100 b. V 100 = 1000 kΩ r02 = r04 = A = I CQ 0.1 gm =

I CQ VT

=

0.1 = 3.846 mA/V 0.026

Ad = g m ( r02 r04 ) = ( 3.846 ) (1000 1000 ) ⇒ Ad = 1923

c. Ad = g m r02 r04 RL

(

)

Ad = ( 3.846 ) (1000 1000 250 ) ⇒ Ad = 641

11.60 a.

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Ad = g m ( r02 r04 RL ) gm =

I CQ VT

=

IQ 2VT

V 125 r02 = A 2 = I CQ I CQ r04 =

VA 4 80 = I CQ I CQ

If I Q = 2 mA, then g m = 38.46 mA/V r02 = 125 kΩ, r04 = 80 kΩ So Ad = 38.46 ⎡⎣125 80 200 ⎤⎦ Or Ad = 1508 For each gain of 1000. lower the current level For I Q = 0.60 mA, I CQ = 0.30 mA 0.3 gm = = 11.54 mA/V 0.026 125 r02 = = 417 kΩ 0.3 80 r04 = = 267 kΩ 0.3 Ad = 11.54 ⎡⎣ 417 267 200 ⎤⎦ = 1036 So I Q = 0.60 mA is adequate

b. For V + = 10 V, VBE = VEB = 0.6 V For VCB = 0, vcm ( max ) = V + − 2VEB = 10 − 2 ( 0.6 ) Or vcm ( max ) = 8.8 V 11.61 a.

From symmetry.

VGS 3 = VGS 4 = VDS 3 = VDS 4 =

0.1 +1 0.1

Or VDS 3 = VDS 4 = 2 V 0.1 +1 = 2 V 0.1 = VSD 2 = VSG1 − (VDS 3 − 10 ) = 2 − ( 2 − 10 )

VSG1 = VSG 2 = VSD1

Or VSD1 = VSD 2 = 10 V

b. r0 n = r0 p =

1

λn I DQ 1

= =

1 ⇒ 1 MΩ 0.01 ( )( 0.1) 1

⇒ 0.667 MΩ

( 0.015 )( 0.1) g m = 2 K p (VSG + VTP ) = 2 ( 0.1)( 2 − 1) = 0.2 mA / V Ad = g m ( ron rop ) = ( 0.2 ) (1000 667 ) ⇒ Ad λP I DQ

= 80

(c)

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IQ

I D 2 = I D1 =

2

= 0.1 mA

1 1 = = 1000 k Ω λn I D 4 ( 0.01)( 0.1)

ro 4 =

1

ro 2 =

λP I D 2

=

1

( 0.015)( 0.1)

= 667 k Ω

Ro = ro 2 ro 4 = 667 1000 = 400 k Ω

11.62 Ad = g m ( ro 4 ro 2 ) ⎛ 0.08 ⎞ gm = 2 ⎜ ⎟ ( 2.5 )( 0.05 ) ⎝ 2 ⎠ = 0.1414 mA/V 1 ro 4 = = 1000 K ( 0.02 )( 0.05 ) ro 2 =

1 = 1333 K ( 0.015)( 0.05 )

Ad = ( 0.1414 ) (1000 1333) Ad = 80.8

11.63 R04 = r04 ⎡⎣1 + g m 4 ( R rπ 4 ) ⎤⎦ 80 = 800 K 0.1 0.1 gm4 = = 3.846 0.026 (100 )( 0.026 ) rπ 4 = 0.1 = 26 K r04 =

R rπ 4 = 1 26 = 0.963 K Assume β = 100 rπ 3 =

(100 )( 0.026 ) 0.1

= 26 kΩ

0.1 = 3.846 mA/V 0.026 R04 = 800 ⎡⎣1 + ( 3.846 )( 0.963) ⎤⎦ ⇒ 3.763 MΩ

g m3 =

⇒ R0 = 3.763MΩ Then Av = − g m ( r02 R0 ) 120 = 1200 kΩ 0.1 0.1 gm = = 3.846 mA/V 0.026 Av = − ( 3.846 ) ⎡⎣1200 3763⎤⎦ ⇒ Av = −3499 r02 =

b. For

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R = 0, r04 = Av = − g m ( r02

80 = 800 kΩ 0.1 r04 )

= − ( 3.846 ) ⎡⎣1200 800 ⎤⎦ ⇒ Av = −1846

(c)

For part (a), Ro = ( 3.763 1.2 ) = 0.910 M Ω

For part (b), Ro = (1.2 0.8 ) = 0.48 M Ω 11.64 I B5 =

IE5 I +I I +I = B3 B4 = C 3 C 4 1+ β 1+ β β (1 + β )

Now I C 3 + I C 4 ≈ I Q IQ So I B 5 ≈ β (1 + β ) I B6 =

I Q1 IE6 = 1 + β β (1 + β )

For balance, we want I B 6 = I B 5 So that I Q1 = I Q

11.65 Resistance looking into drain of M4.

Vsg 4 ≅ I X R1 I X ± g m 4Vsg 4 =

VX − Vsg 4 r04

⎡ R ⎤ V I X ⎢1 + g m 4 R1 + 1 ⎥ = X r04 ⎦ r04 ⎣ ⎡ R ⎤ Or R0 = r04 ⎢1 + g m 4 R1 + 1 ⎥ r04 ⎦ ⎣ a.

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Ad = g m 2 ( ro 2 Ro ) g m 2 = 2 K n I DQ = 2

( 0.080 )( 0.1)

= 0.179 mA / V 1 1 ro 2 = = = 667 k Ω λn I DQ ( 0.015 )( 0.1) g m 4 = 2 K P I DQ = 2

ro 4 =

1

λ p I DQ

( 0.080 )( 0.1)

= 0.179 mA / V 1 = = 500 k Ω 0.02 ( )( 0.1)

1 ⎤ ⎡ R0 = 500 ⎢1 + ( 0.179 )(1) + ⎥ = 590.5 kΩ 500 ⎣ ⎦ Ad = ( 0.179 ) ⎡⎣667 590.5⎤⎦ ⇒ Ad = 56.06

b. When R1 = 0, R0 = r04 = 500 kΩ Ad = ( 0.179 ) ⎡⎣667 500 ⎤⎦ ⇒ Ad = 51.15 (c)

For part (a), Ro = ro 2 Ro = 667 590.5 ⇒ Ro = 313 k Ω

For part (b), Ro = ro 2 ro 4 = 667 500 ⇒ Ro = 286 kΩ 11.66 Let β = 100, VA = 100 V

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ro 2 =

VA 100 = = 1000 k Ω I CQ 0.1

Ro 4 = ro 4 [1 + g m RE′ ] where RE′ = rπ RE Now rπ =

(100 )( 0.026 )

= 26 k Ω 0.1 0.1 gm = = 3.846 mA / V 0.026 RE′ = 26 1 = 0.963 k Ω Then Ro 4 = 1000 ⎡⎣1 + ( 3.846 )( 0.963) ⎤⎦ = 4704 k Ω

Ad = g m ( ro 2 Ro 4 ) = 3.846 (1000 4704 ) ⇒ Ad = 3172

11.67 (a) For Q2, Q4

Vx − Vπ 4 V + g m 2Vπ 2 + g m 4Vπ 4 + x ro 2 ro 4

(1)

Ix =

(2)

g m 2Vπ 2 +

(3)

Vπ 4 = −Vπ 2

From (2)

Vx − Vπ 4 V = π4 ro 2 rπ 4 rπ 2

⎡ 1 ⎤ Vx 1 = Vπ 4 ⎢ + + gm2 ⎥ ro 2 r r r ⎣⎢ π 4 π 2 o 2 ⎦⎥

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Now ⎞ ⎛ 120 ⎞ ⎟=⎜ ⎟ ( 0.5 ) = 0.496 mA ⎠ ⎝ 121 ⎠ ⎛ 120 ⎞ ⎛ IQ ⎞ ⎛ 1 ⎞ ⎛ β ⎞ ⎟ ⇒ I C 2 = 0.0041 mA = ⎜ ⎟⎜ ⎟⎜ ⎟ = ( 0.5 ) ⎜⎜ 2 ⎟ ⎝ 2 ⎠⎝ 1+ β ⎠⎝ 1+ β ⎠ ⎝ (121) ⎠

⎛ β IC 4 = ⎜ ⎝ 1+ β IC 2

⎞ ⎛ IQ ⎟⎜ ⎠⎝ 2

So rπ 2 =

(120 )( 0.026 )

= 761 k Ω 0.0041 0.0041 gm2 = = 0.158 mA/V 0.026 100 ro 2 = ⇒ 24.4 M Ω 0.0041 (120 ) ( 0.026 ) rπ 4 = = 6.29 k Ω 0.496 0.496 gm4 = = 19.08 mA / V 0.026 100 ro 4 = = 202 k Ω 0.496 Now ⎡ ⎤ Vx Vx 1 1 = Vπ 4 ⎢ + + 0.158⎥ ⇒ which yields Vπ 4 = ro 2 ( 0.318) ro 2 ⎣⎢ 6.29 761 24400 ⎦⎥

From (1), ⎛ V V 1 ⎞ I x = x + x + Vπ 4 ⎜ g m 4 − g m 2 − ⎟ ro 2 ro 4 ro 2 ⎠ ⎝ 1 ⎞⎤ ⎡ ⎛ ⎜ 19.08 − 0.158 − ⎟ Ix ⎢ 1 V 1 24400 ⎠ ⎥ ⎥ which yields Ro 2 = x = 135 k Ω =⎢ + +⎝ Vx ⎢ 24400 202 Ix ( 0.318)( 24400 ) ⎥ ⎢ ⎥ ⎣ ⎦ 80 Now ro 6 = = 160 k Ω 0.5 Then Ro = Ro 2 ro 6 = 135 160 ⇒ Ro = 73.2 k Ω

(b) Ad = g mc Ro where g mc =

Δi vd / 2

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vd 2

Δi = g m1Vπ 1 + g m 3Vπ 3 and Vπ 1 + Vπ 3 = ⎛V ⎞ Also ⎜ π 1 + g m1Vπ 1 ⎟ rπ 3 = Vπ 3 ⎝ rπ 1 ⎠ ⎛1+ β ⎞ So Vπ 1 ⎜ ⎟ rπ 3 = Vπ 3 ⎝ rπ 1 ⎠ ⎛ 121 ⎞ Or Vπ 1 ⎜ ⎟ ( 6.29 ) = Vπ 3 ≅ Vπ 1 ⎝ 761 ⎠ v v Then 2Vπ 1 = d ⇒ Vπ 1 = d 2 4

⎛v ⎞ ⎛v ⎞ So Δi = ( g m1 + g m 3 ) Vπ 1 = ( 0.158 + 19.08 ) ⎜ d ⎟ = 9.62 ⎜ d ⎟ ⎝ 4⎠ ⎝ 2⎠ Δi = 9.62 ⇒ Ad = ( 9.62 )( 73.2 ) ⇒ Ad = 704 So g mc = vd / 2 Now Rid = 2 Ri where Ri = rπ 1 + (1 + β ) rπ 3 Ri = 761 + (121)( 6.29 ) = 1522 k Ω Then Rid = 3.044 M Ω

11.69 (a) Ad = 100 = g m ( ro 2 ro 4 ) Let I Q = 0.5 mA 1 1 ro 2 = = = 200 k Ω λn I D ( 0.02 )( 0.25 ) ro 4 =

1 1 = = 160 k Ω λP I D ( 0.025 )( 0.25 )

Then 100 = g m ( 200 160 ) ⇒ g m = 1.125 mA / V ⎛ K′ ⎞⎛W gm = 2 ⎜ n ⎟ ⎜ ⎝ 2 ⎠⎝ L

⎞ ⎟ ID ⎠

⎛ 0.080 ⎞ ⎛ W ⎞ ⎛W ⎞ 1.125 = 2 ⎜ ⎟ ⎜ ⎟ ( 0.25 ) ⇒ ⎜ ⎟ = 31.6 ⎝ 2 ⎠⎝ L ⎠ ⎝ L ⎠n ⎛W ⎞ ⎛W ⎞ Now ⎜ ⎟ somewhat arbitrary. Let ⎜ ⎟ = 31.6 L ⎝ ⎠P ⎝ L ⎠P

11.70

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Ad = g m ( ro 2 ro 4 )

P = ( I Q + I REF ) (V + − V − ) Let I Q = I REF Then 0.5 = 2 I Q ( 3 − ( −3) ) ⇒ I Q = I REF = 0.0417 mA ro 2 =

1 1 = = 3205 k Ω λn I D ( 0.015 )( 0.0208 )

ro 4 =

1 1 = = 2404 k Ω λP I D ( 0.02 )( 0.0208 )

Then Ad = 80 = g m ( 3205 2404 ) ⇒ g m = 0.0582 mA/V ⎛ k ′ ⎞⎛ W ⎞ gm = 2 ⎜ n ⎟ ⎜ ⎟ I D ⎝ 2 ⎠ ⎝ L ⎠n ⎛ 0.080 ⎞⎛ W ⎞ ⎛W ⎞ 0.0582 = 2 ⎜ ⎟⎜ ⎟ ( 0.0208 ) ⇒ ⎜ ⎟ = 1.02 ⎝ 2 ⎠⎝ L ⎠ n ⎝ L ⎠n

11.71 Ad = g m ( ro 2 Ro ) ≈ g m ro 2 1 ro 2 = λn I D =

1 = 666.7 K ( 0.015)( 0.1)

Ad = 400 = g m ( 666.7 ) g m = 0.60 mA/V ⎛ k′ ⎞⎛ W = 2 ⎜ n ⎟⎜ ⎝ 2 ⎠⎝ L

⎞ ⎟ ID ⎠

⎛ 0.08 ⎞ ⎛ W ⎞ 0.60 = 2 ⎜ ⎟ ⎜ ⎟ ( 0.1) ⎝ 2 ⎠⎝ L ⎠ ⎛W ⎞ 0.090 = 0.004 ⎜ ⎟ ⎝L⎠ ⎛W ⎞ ⎛W ⎞ ⎜ ⎟ = ⎜ ⎟ = 22.5 ⎝ L ⎠1 ⎝ L ⎠ 2

11.72

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Ad = g m ( Ro 4 Ro 6 ) where Ro 4 = ro 4 + ro 2 [1 + g m 4 ro 4 ]

Ro 6 = ro 6 + ro8 [1 + g m 6 ro 6 ] We have 1 = 1667 k Ω ro 2 = ro 4 = ( 0.015 )( 0.040 ) ro 6 = ro8 =

1 = 1250 k Ω 0.02 ( )( 0.040 )

⎛ 0.060 ⎞ gm4 = 2 ⎜ ⎟ (15 )( 0.040 ) = 0.268 mA/V ⎝ 2 ⎠ ⎛ 0.025 ⎞ gm6 = 2 ⎜ ⎟ (10 )( 0.040 ) = 0.141 mA/V ⎝ 2 ⎠ Then Ro 4 = 1667 + 1667 ⎡⎣1 + ( 0.268 )(1667 ) ⎤⎦ ⇒ 748 M Ω

Ro 6 = 1250 + 1250 ⎡⎣1 + ( 0.141)(1250 ) ⎤⎦ ⇒ 222.8 M Ω

(a) Ro = Ro 4 Ro 6 = 748 222.8 ⇒

Ro = 172 M Ω

(b) Ad = g m 4 ( Ro 4 Ro 6 ) = ( 0.268 )(172000 ) ⇒ Ad = 46096 11.73 Ad = g m ( ro 2 ro 4 ) ro 2 = ro 4 = =

1 λ ID 1

( 0.02 )( 0.1)

gm = 2 Kn I D = 2

= 500 K

( 0.5)( 0.1)

= 0.4472 mA/V Ad = ( 0.4472 ) ( 500 500 ) ⇒ Ad = 112 Ro = ro 2 ro 4 = 500 500 ⇒ Ro = 250 K

11.74 (a) I DP = K p (VSG + VTP )

2

0.4 + 1 = VSG 3 = 1.894 V 0.5 I DN = K n (VGS − VTN )

2

0.4 + 1 = VGS 1 = 1.894 V 0.5 VDS1 ( sat ) = VGS1 − VTN = 1.894 − 1 = 0.894 V V + = VSG 3 + VDS1 ( sat ) − VGS 1 + vCM V + = 1.894 + 0.894 − 1.894 + 4 ⇒ V + = 4.89 V = −V − (b)

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Ad = g m ( ro 2 ro 4 ) ro 2 = ro 4 =

1

λ ID

gm = 2 Kn I D

1

=

= 166.7 K

( 0.015 )( 0.4 ) = 2 ( 0.5 )( 0.4 ) = 0.8944 mA/V

Ad = ( 0.8944 ) (166.7 166.7 ) ⇒ Ad = 74.5

11.75 (a) For vcm = +2V ⇒ V + = 2.7 V If I Q is a 2-transistor current source, V − = vcm − 0.7 − 0.7 V − = −3.4 V ⇒ V + = −V − = 3.4 V (b) 100 = 1000 K Ad = g m ( ro 2 ro 4 ) ro 2 = 0.1 60 = 600 K ro 4 = 0.1 0.1 = 3.846 mA/V gm = 0.026 Ad = ( 3.846 ) (1000 600 ) ⇒ Ad = 1442

11.76 (a) (b)

V + = −V − = 3.4 V

75 = 1250 K 0.06 40 = 666.7 K ro 4 = 0.06 0.06 = 2.308 mA/V gm = 0.026 Ad = ( 2.308 ) (1250 666.7 ) ro 2 =

Ad = 1004

11.77 g m1 = 2 K n I Bias1 = 2 gm2 = rπ 2 =

I CQ VT

β VT I CQ

= =

( 0.2 )( 0.25 ) = 0.447 mA/V

0.75 = 28.85 mA/V 0.026

(120 )( 0.026 ) 0.75

= 4.16 kΩ

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i0 = g m1Vgs1 + g m 2Vπ 2 Vπ 2 = g m1Vgs1rπ 2 and vi = Vgs1 + Vπ 2 i0 = Vgs1 ( g m1 + g m 2 ⋅ g m1rπ 2 )

vi = Vgs1 + g m1Vgs1rπ 2 and Vgs1 = i0 = vi ⋅ g mC =

g m1 (1 + β )

vi 1 + g m1rπ 2

1 + g m1rπ 2

( 0.447 )(121) i0 g m1 (1 + β ) = = vi 1 + g m1rπ 2 1 + ( 0.447 )( 4.16 )

⇒ g mC = 18.9 mA/V

11.78 r0 ( M 2 ) = r0 ( Q2 ) =

1

λn I DQ

=

1

( 0.01)( 0.2 )

= 500 kΩ

VA 80 = = 400 kΩ I CQ 0.2

g m ( M 2 ) = 2 K n I DQ = 2

( 0.2 )( 0.2 )

= 0.4 mA/V Ad = g m ( M 2 ) ⎡⎣ r0 ( M 2 ) r0 ( Q2 ) ⎤⎦ = 0.4 ⎡⎣500 400 ⎤⎦ ⇒ Ad = 88.9 If the IQ current source is ideal, Acm = 0 and C M RRdB = ∞

11.79 a.

b.

Assume RL is capacitively coupled. Then

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I CQ + I DQ = I Q I DQ =

VBE 0.7 = = 0.0875 mA R1 8

I CQ = 0.9 − 0.0875 = 0.8125 mA g m1 = 2 K P I DQ = 2 (1)( 0.0875 ) ⇒ g m1 = 0.592 mA/V gm2 = rπ 2 =

I CQ VT

β VT I CQ

=

0.8125 ⇒ g m 2 = 31.25 mA/V 0.026

=

(100 )( 0.026 ) 0.8125

⇒ rπ 2 = 3.2 kΩ

c. V0 = ( − g m1Vsg − g m 2Vπ 2 ) RL Vi + Vsg = V0 ⇒ Vsg = V0 − Vi Vπ 2 = ( g m1Vsg ) ( R1 rπ 2 )

V0 = − ⎡⎣ g m1Vsg + g m 2 g m1Vsg ( R1 rπ 2 ) ⎤⎦ RL V0 = − (V0 − Vi ) ⎣⎡ g m1 + g m 2 g m1 ( R1 rπ 2 ) ⎦⎤ RL

⎡ g m1 + g m 2 g m1 ( R1 rπ 2 ) ⎦⎤ RL V0 = ⎣ Vi 1 + ⎡ g m1 + g m 2 g m1 ( R1 rπ 2 ) ⎤ RL ⎣ ⎦ We find Av =

g m1 + g m 2 g m1 ( R1 rπ 2 ) = 0.592 + ( 31.25 )( 0.592 ) ( 8 3.2 ) = 42.88 Then Av =

( 42.88 )( RL ) 1 + ( 42.88 )( RL )

11.80 a. I DQ I CQ

Assume RL is capacitively coupled. 0.7 = = 0.0875 mA 8 = 1.2 − 0.0875 = 1.11 mA

g m1 = 2 K p I DQ = 2 (1)( 0.0875 ) ⇒ g m1 = 0.592 mA/V gm2 = rπ 2 =

I CQ VT

β VT I CQ

= =

1.11 ⇒ g m 2 = 42.7 mA/V 0.026

(100 )( 0.026 ) 1.11

⇒ rπ 2 = 2.34 kΩ

b.

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Vsg = VX I X = g m 2Vπ 2 + g m1Vsg

(g

V

m1 sg

)(R

1

rπ 2 ) = Vπ 2

I X = VX ⎡⎣ g m1 + g m 2 g m1 ( R1 rπ 2 ) ⎤⎦ V 1 R0 = X = IX g m1 + g m 2 g m1 ( R1 rπ 2 ) =

1

0.592 + ( 0.592 )( 42.7 ) ( 8 2.34 )

⇒ R0 = 21.6 Ω

11.81 (a)

(1)

g m 2Vπ +

(2)

g m 2Vπ +

Then Vπ =

Vo − ( −Vπ ) ro 2 Vo − ( −Vπ ) ro 2

=0 = g m1Vi +

⎛ 1 1⎞ −Vπ −Vπ + or 0 = g m1Vi − Vπ ⎜ + ⎟ ro1 rπ ⎝ ro1 rπ ⎠

g m1Vi ⎛ 1 1⎞ ⎜ + ⎟ ⎝ ro1 rπ ⎠

From (1)

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⎛ Vo 1 ⎞ =0 ⎜ g m 2 + ⎟ Vπ + ro 2 ⎠ ro 2 ⎝ ⎛ 1 ⎞ ⎜ gm2 + ⎟ ro 2 ⎠ ⎛ 1 ⎞ Vo = −ro 2 ⎜ g m 2 + ⎟ Vπ = −ro 2 g m1Vi ⎝ ro 2 ⎠ ⎛ 1 1⎞ ⎝ ⎜ + ⎟ ⎝ ro1 rπ ⎠ ⎛ 1 ⎞ − g m1ro 2 ⎜ g m 2 + ⎟ ro 2 ⎠ V ⎝ Av = o = Vi ⎛ 1 1⎞ ⎜ + ⎟ ⎝ ro1 rπ ⎠ Now

( 0.25)( 0.025 ) = 0.158 mA / V

g m1 = 2 K n I Q = 2 gm2 = ro1 =

VT 1

λ IQ

= =

0.025 = 0.9615 mA / V 0.026 1

( 0.02 )( 0.025)

= 2000 k Ω

VA 50 = = 2000 k Ω I Q 0.025

ro 2 = rπ =

IQ

β VT IQ

=

(100 )( 0.026 ) 0.025

= 104 k Ω

Then 1 ⎞ ⎛ − ( 0.158 )( 2000 ) ⎜ 0.9615 + ⎟ 2000 ⎝ ⎠ ⇒ A = −30039 Av = v 1 ⎞ ⎛ 1 + ⎜ ⎟ ⎝ 2000 104 ⎠ To find Ro; set Vi = 0 ⇒ g m1Vi = 0

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I x = g m 2Vπ +

Vx − ( −Vπ )

Vπ = − I x ( ro1 rπ )

ro 2

Then ⎛ V 1 ⎞ I x = ⎜ g m 2 + ⎟ ( − I x ) ( ro1 rπ ) + x ro 2 ⎠ ro 2 ⎝ Combining terms, Ro =

⎡ ⎛ Vx 1 ⎞⎤ = ro 2 ⎢1 + ( ro1 rπ ) ⎜ g m 2 + ⎟ ⎥ Ix ro 2 ⎠ ⎦ ⎝ ⎣

⎡ 1 ⎞⎤ ⎛ = 2000 ⎢1 + ( 2000 104 ) ⎜ 0.9615 + ⎟ ⎥ ⇒ Ro = 192.2 M Ω 2000 ⎝ ⎠⎦ ⎣

(b)

Vo − ( −Vgs 3 )

(1)

g m 3Vgs 3 +

(2)

g m 3Vgs 3 +

(3)

−Vgs 3 − ( −Vπ 2 ) ( −Vπ 2 ) Vπ 2 + g m 2Vπ 2 + = g m1Vi + rπ 2 ro 2 ro1

From (2), Vπ 2 =

ro3 Vo − ( −Vgs 3 ) ro3

=0 = g m 2Vπ 2 +

−Vgs 3 − ( −Vπ 2 ) ro 2

⎛ 1 ⎞ Vgs 3 or 0 = Vπ 2 ⎜ g m 2 + ⎟ − r ro 2 o2 ⎠ ⎝

Vgs 3 ⎛ 1 ⎞ ro 2 ⎜ g m 2 + ⎟ ro 2 ⎠ ⎝

Then (3)

Vgs 3 ⎛ 1 1 1⎞ + gm2 + + ⎟ = g m1Vi + Vπ 2 ⎜ ro 2 ro1 ⎠ ro 2 ⎝ rπ 2

or

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Vgs 3 ⎡ 1 1 1⎤ + ⎥ = g m1Vi + ⎢ + gm2 + ro 2 ro1 ⎦ ro 2 ⎛ 1 ⎞ r ro 2 ⎜ g m 2 + ⎟ ⎣ π 2 ro 2 ⎠ ⎝ Vgs 3 Vgs 3 1 1 ⎤ ⎡ 1 + 0.9615 + + = 0.9615Vi + ⎢ ⎥ 1 ⎞ ⎣104 2000 2000 ⎦ 2000 ⎛ 2000 ⎜ 0.9615 + ⎟ 2000 ⎝ ⎠ Then Vgs 3 = 1.83 × 105 Vi Vgs 3

⎛ −V 1 ⎞ 1 ⎞ ⎛ 5 From (1), ⎜ g m 3 + ⎟ Vgs 3 = o or Vo = −2000 ⎜ 0.158 + ⎟ (1.83 ×10 ) Vi r r 2000 ⎝ ⎠ o3 ⎠ o3 ⎝ V Av = o = −5.80 × 107 Vi

To find Ro

Vx − ( −Vgs 3 )

(1)

I x = g m 3Vgs 3 +

(2)

g m 3Vgs 3 +

(3)

Vπ 2 = − I x ( ro1 rπ 2 )

ro3

Vx − ( −Vgs 3 ) ro 3

= g m 2Vπ 2 +

−Vgs 3 − ( −Vπ 2 ) ro 2

⎛ 1 ⎞ V From (1) I x = Vgs 3 ⎜ g m 3 + ⎟ + x ro 3 ⎠ ro3 ⎝ 1 ⎞ Vx ⎛ I x = Vgs 3 ⎜ 0.158 + ⎟+ 2000 ⎠ 2000 ⎝ V Ix − x 2000 So Vgs 3 = 0.1585

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From (2), ⎡ ⎛ 1 1 ⎤ V 1 ⎞ + ⎥ + x = Vπ 2 ⎜ g m 2 + ⎟ Vgs 3 ⎢ g m 3 + ro 3 ro 2 ⎦ ro 3 ro 2 ⎠ ⎣ ⎝ 1 1 ⎤ Vx 1 ⎞ ⎡ ⎛ + + = Vπ 2 ⎜ 0.9615 + Vgs 3 ⎢ 0.158 + ⎟ ⎥ 2000 2000 ⎦ 2000 2000 ⎠ ⎣ ⎝ Vx ⎡ I − Vx / 2000 ⎤ Then ⎢ x ⎥ ( 0.159 ) + 2000 = − I x ( 2000 104 ) ( 0.962 ) ⎣ 0.1585 ⎦ V We find Ro = x = 6.09 × 1010 Ω Ix

11.82 Assume emitter of Q1 is capacitively coupled to signal ground. ⎛ 80 ⎞ I CQ = 0.2 ⎜ ⎟ = 0.1975 mA ⎝ 81 ⎠ 0.2 I DQ = = 0.00247 mA 81 (80 )( 0.026 ) rπ = = 10.5 k Ω 0.1975 0.1975 g m ( Q1 ) = = 7.60 mA / V 0.026 gm ( M1 ) = 2 K n I D = 2

g m ( M 1 ) = 0.0445 mA / V

( 0.2 )( 0.00247 )

Vi = Vgs + Vπ and Vπ = g m ( M 1 ) Vgs rπ or Vgs =

Vπ g m ( M 1 ) rπ

⎛ 1 Then Vi = Vπ ⎜⎜ 1 + g M ( m 1 ) rπ ⎝

⎞ Vi ⎟⎟ or Vπ = ⎛ ⎞ 1 ⎠ ⎜⎜ 1 + g ( M ) r ⎟⎟ m 1 π ⎠ ⎝ − g m ( Q1 ) RC V Vo = − g m ( Q1 ) Vπ RC ⇒ Av = o = Vi ⎛ ⎞ 1 ⎜⎜1 + ⎟⎟ ⎝ g m ( M 1 ) rπ ⎠ − ( 7.60 )( 20 ) Then Av = ⇒ Av = −48.4 ⎛ ⎞ 1 ⎜⎜ 1 + ⎟⎟ ⎝ ( 0.0445 )(10.5 ) ⎠

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11.83 Using the results from Chapter 4 for the emitter-follower: ⎡ rπ 9 + r07 R011 ⎤ ⎢ rπ 8 + ⎥ 1+ β ⎥ R0 = R4 || ⎢ ⎢ ⎥ 1+ β ⎢ ⎥ ⎣ ⎦ β VT (100 )( 0.026 ) rπ 8 = = = 2.6 kΩ 1 IC8 IC 8

IC 9 ≈ rπ 9 =

β

=

1 = 0.01 mA 100

(100 )( 0.026 )

= 260 kΩ 0.01 V 100 r07 = A = = 500 kΩ I Q 0.2

r011 =

VA 100 = = 500 kΩ I Q 0.2

R011 = r011 [1 + g m RE′ ] , g m = rπ 11 =

(100 ) ( 0.026 )

0.2 = 7.69 0.026

= 13 kΩ 0.2 RE′ = 0.2 13 = 0.197 kΩ R011 = 500 ⎡⎣1 + ( 7.69 )( 0.197 ) ⎤⎦ = 1257 kΩ Then 260 + 500 1257 ⎤ ⎡ ⎢ 2.6 + ⎥ 101 ⎥ R0 = 5 || ⎢ 101 ⎢ ⎥ ⎢⎣ ⎥⎦ = 5 0.0863 ⇒ R0 = 0.0848 K ⇒ 84.8 Ω

11.84 Ri = rπ 1 + (1 + β ) rπ 2 rπ 2 =

(100 )( 0.026 ) 0.5

= 5.2 kΩ

(100 )( 0.026 ) (100 ) ( 0.026 ) = = 520 kΩ 0.5 ( 0.5 /100 ) Ri = 520 + (101)( 5.2 ) ⇒ Ri ≅ 1.05 MΩ (100 )( 0.026 ) rπ 3 + 50 2

rπ 1 =

R0 = 5 R0 = 5

101

, rπ 3 =

1

= 2.6 kΩ

2.6 + 50 = 5 0.521 ⇒ R0 = 0.472 kΩ 101

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⎛V ⎞ V0 = − ⎜ π 3 + g m 3Vπ 3 ⎟ ( 5 ) ⎝ rπ 3 ⎠ ⎛1+ β ⎞ V0 = −Vπ 3 ⎜ ⎟ ( 5) ⎝ rπ 3 ⎠

(1)

(V − V ) Vπ 3 = g m 2Vπ 2 + 0 π 3 50 rπ 3 ⎛ 1 1 ⎞ V g m 2Vπ 2 = Vπ 3 ⎜ + ⎟− 0 ⎝ rπ 3 50 ⎠ 50 ⎛V ⎞ Vπ 2 = ⎜ π 1 + g m1Vπ 1 ⎟ rπ 2 ⎝ rπ 1 ⎠ ⎛ 1+ β = Vπ 1 ⎜ ⎝ rπ 1

⎞ ⎟ rπ 2 ⎠

and Vin = Vπ 1 + Vπ 2 gm2 =

(2)

(3)

(4)

0.5 = 19.23 mA/V 0.026

Then ⎛ 101 ⎞ V0 = −Vπ 3 ⎜ ⎟ ( 5 ) ⇒ Vπ 3 = −V0 ( 0.005149 ) ⎝ 2.6 ⎠ And 1 ⎞ V ⎛ 1 + ⎟− 0 19.23Vπ 2 = −V0 ( 0.005149 ) ⎜ ⎝ 2.6 50 ⎠ 50 = −V0 ( 0.02208 )

(1)

(2)

Or Vπ 2 = −V0 ( 0.001148 )

And Vπ 1 = Vin − Vπ 2 = Vin + V0 ( 0.001148 )

(4)

So ⎛ 101 ⎞ −V0 ( 0.001148 ) = ⎡⎣Vin + V0 ( 0.001148 ) ⎤⎦ ⎜ ⎟ ( 5.2 ) ⎝ 520 ⎠ −V0 ( 0.001148 ) − V0 ( 0.001159 ) = Vin (1.01) ⇒ Av =

(3) V0 = −438 Vin

11.85

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I2 =

5 = 1 mA 5

1 + 0.8 = 2.21 V 0.5 2.21 − ( −5 ) I1 = = 0.206 mA 35

VGS 2 =

V0 = ( g m 2Vgs 2 ) ( R2 r02 )

Vgs 2 = ( g m1Vsg1 ) ( r01 R1 ) − V0 and Vsg1 = −Vin So Vgs 2 = − ( g m1Vin ) ( r01 R1 ) − V0 Then V0 = g m 2 ( R2 r02 ) ⎡⎣ − ( g m1Vin ) ( r01 R1 ) − V0 ⎤⎦ Av =

V0 − g m 2 ( R2 r02 ) g m1 ( r01 R1 ) = Vin 1 + g m 2 ( R2 r02 )

gm2 = 2 Kn2 I D 2 = 2

( 0.5 )(1) = 1.414 mA / V

g m1 = 2 K p1 I D1 = 2

( 0.2 )( 0.206 ) = 0.406 mA / V

r01 =

1

λ1 I D1

r02 =

=

1

( 0.01)( 0.206 )

= 485 kΩ

1 1 = = 100 kΩ λ2 I D 2 ( 0.01)(1)

R2 r02 = 5 100 = 4.76 kΩ R1 r01 = 35 485 = 32.6 kΩ Then Av =

− (1.414 )( 4.76 )( 0.406 )( 32.6 ) 1 + (1.414 )( 4.76 )

So ⇒ Av = −11.5

Output Resistance—From the results for a source follower in Chapter 6. 1 1 R0 = R2 r02 = 5 100 gm2 1.414 = 0.707 4.76 So R0 = 0.616 kΩ

11.86 a.

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R2 =

5 ⇒ R2 = 10 kΩ 0.5 I D2 0.5 − VTP 2 = + 1 = 2.41 V K p2 0.25

VSG 2 = R1 =

5 − ( −2.41) 0.1

⇒ R1 = 74.1 kΩ

b.

V0 = − ( g m 2Vsg 2 ) ( r02 R2 )

Vsg 2 = V0 − ⎡⎣ − ( g m1Vgs1 ) ( r01 R1 ) ⎤⎦ and Vgs1 = Vin Av =

V0 − ( g m 2 ) ( r02 R2 ) ( g m1 ) ( r01 R1 ) = Vin 1 + ( g m 2 ) ( r02 R2 )

g m1 = 2 K n1 I D1 = 2 gm2 = 2 K p 2 I D 2 = 2 r01 = r02 =

1

λ1 I D1

=

( 0.1)( 0.1) = 0.2 mA / V ( 0.25)( 0.5) = 0.707 mA / V

1

( 0.01)( 0.1)

= 1000 kΩ

1 1 = = 200 kΩ λ2 I D 2 ( 0.01)( 0.5 )

r02 R2 = 200 10 = 9.52 kΩ r01 R1 = 1000 74.1 = 69.0 kΩ Then Av =

− ( 0.707 )( 9.52 )( 0.2 )( 69 ) 1 + ( 0.707 )( 9.52 )

So ⇒ Av = −12.0 R0 =

1 1 10 200 R2 r02 = 0.707 gm2

= 1.414 9.52 Or R0 = 1.23 kΩ

11.87 a. I C 2 = 0.25 mA 5−2 R= ⇒ R = 12 kΩ 0.25 v − VBE ( on ) 2 − 0.7 I C 3 = 02 ⇒ RE1 = ⇒ RE1 = 2.6 kΩ RE1 0.5 RC =

5 − v03 5 − 3 = ⇒ RC = 4 kΩ IC 3 0.5

⎡ v03 − VBE ( on ) ⎤⎦ − ( −5 ) IC 4 = ⎣ RE 2 RE 2 =

3 − 0.7 + 5 ⇒ RE 2 = 2.43 kΩ 3

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b.

Input resistance to base of Q3, Ri 3 = rπ 3 + (1 + β ) RE1 rπ 3 =

(100 )( 0.026 )

= 5.2 kΩ 0.5 Ri 3 = 5.2 + (101)( 2.6 ) = 267.8 kΩ v 1 Ad 1 = 02 = g m 2 ( R Ri 3 ) vd 2 0.25 = 9.62 mA/V 0.026 1 Ad 1 = ( 9.62 ) (12 267.8 ) ⇒ Ad 1 = 55.2 2 − β ( RC Ri 4 ) v Now 03 = v02 rπ 3 + (1 + β ) RE1 gm2 =

where Ri 4 = rπ 4 + (1 + β ) RE 2 and

(1 + β ) RE 2 v0 = v03 rπ 4 + (1 + β ) RE 2 (100 )( 0.026 )

rπ 4 =

= 0.867 kΩ 3 (101)( 2.43) v0 = = 0.9965 v03 0.867 + (101)( 2.43)

Ri 4 = 0.867 + (101)( 2.43) = 246.3 kΩ rπ 3 = 5.2 kΩ So

v03 − (100 ) ( 4 246.3) = = −1.47 5.2 + (101)( 2.6 ) v02 v0 = ( 55.2 )( 0.9965 )( −1.47 ) ⇒ Ad = −80.9 vd Using Equation (11.32b) − g m 2 ( R Ri 3 )

So Ad =

c. Acm1 =

1+ rπ 2 =

2 (1 + β ) R0 rπ 2

(100 )( 0.026 )

= 10.4 kΩ 0.25 − ( 9.62 ) (12 267.8 ) = −0.0569 = Acm1 Acm1 = 2 (101)(100 ) 1+ 10.4 ⎛ v0 ⎞⎛ v03 ⎞ Then Acm = ⎜ ⎟⎜ ⎟ ⋅ Acm1 ⎝ v03 ⎠⎝ v02 ⎠

= ( 0.9965 )( −1.47 )( −0.0569 ) ⇒ Acm = 0.08335

⎛ 80.9 ⎞ C M RRdB = 20 log10 ⎜ ⎟ ⇒ C M RRdB = 59.7 dB ⎝ 0.08335 ⎠

11.88 a. RC1 =

10 − v01 10 − 2 = ⇒ RC1 = 80 kΩ I C1 0.1

RC 2 =

10 − v04 10 − 6 = ⇒ RC 2 = 20 kΩ IC 4 0.2

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b. Ad 1 =

v01 − v02 = − g m1 ( RC1 rπ 3 ) vd

0.1 = 3.846 mA/V 0.026 (180 )( 0.026 ) rπ 3 = = 23.4 kΩ 0.2 Ad 1 = − ( 3.846 ) ( 80 23.4 ) ⇒ Ad 1 = −69.6

g m1 =

Ad 2 =

v04 1 = g m 4 RC 2 v01 − v02 2

0.2 = 7.692 mA/V 0.026 1 Ad 2 = ( 7.692 )( 20 ) = 76.9 2 Then Ad = ( 76.9 )( −69.6 ) ⇒ Ad = −5352 gm4 =

11.89 a.

Neglect the effect of r0 in determining the differential-mode gain. v02 1 Ad 1 = = g m 2 ( RC Ri 3 ) where Ri 3 = rπ 3 + (1 + β ) RE vd 2 A2 =

I1 =

− β RC 2 rπ 3 + (1 + β ) RE

12 − 0.7 − ( −12 ) R1

=

23.3 = 1.94 mA ≈ I C 5 12

1 ⋅ (1.94 ) gm2 = 2 = 37.3 mA/V 0.026 ( 200 )( 0.026 ) rπ 3 = IC 3 1 (1.94 )(8) = 4.24 V 2 4.24 − 0.7 = 1.07 mA IC 3 = 3.3 ( 200 )( 0.026 ) = 4.86 kΩ rπ 3 = 1.07 Ri 3 = 4.86 + ( 201)( 3.3) = 668 kΩ

v02 = 12 −

Ad 1 =

1 ( 37.3) ⎡⎣8 668⎤⎦ = 147.4 2

Then Ad = Ad 1 ⋅ A2 = (147.4 )( −1.197 ) ⇒ Ad = −176 R0 = r05 = Acm1 =

− g m 2 ( RC Ri 3 ) 1+

rπ 2 =

VA 80 = = 41.2 kΩ I C 5 1.94

2 (1 + β ) R0 rπ 2

( 200 )( 0.026 ) 1 ⋅ (1.94 ) 2

= 5.36 kΩ

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Acm1 =

− ( 37.3) ( 8 668 )

2 ( 201)( 41.2 ) 1+ 5.36 A2 = −1.197

= −0.09539

Acm = ( −0.09539 )( −1.197 ) ⇒ Acm = 0.114

b. vd = v1 − v2 = 2.015sin ω t − 1.985sin ω t vd = 0.03sin ω t ( V ) v +v vcm = 1 2 = 2.0sin ω t 2 v03 = Ad vd + Acm vcm = ( −176 )( 0.03) + ( 0.114 )( 2 ) Or v03 = −5.052sin ω t Ideal, Acm = 0 So v03 = Ad vd = ( −176 )( 0.03) v03 = −5.28sin ω t

c. Rid = 2rπ 2 = 2 ( 5.36 ) ⇒ Rid = 10.72 kΩ 2 Ricm ≅ 2 (1 + β ) R0 (1 + β ) r0 VA 80 = = 82.5 kΩ 1 IC 2 ⋅ (1.94 ) 2 2 Ricm = ⎣⎡ 2 ( 201)( 41.2 ) ⎦⎤ ⎣⎡( 201)( 82.5 ) ⎦⎤

r0 =

= 16.6 MΩ 16.6 MΩ So ⇒ Ricm = 4.15 MΩ

11.90 a. 24 − VGS 4 2 = kn (VGS 4 − VTh ) I1 = R1 24 − VGS 4 = ( 55 )( 0.2 )(VGS 4 − 2 )

2

24 − VGS 4 = 11 (VGS2 4 − 4VGS 4 + 4 ) 11VGS2 4 − 43VGS 4 + 20 = 0 VGS 4 =

43 ±

( 43)

2

− 4 (11)( 20 )

2 (11)

= 3.37 V

24 − 3.37 = 0.375 mA = I Q 55 ⎛ 0.375 ⎞ v02 = 12 − ⎜ ⎟ ( 40 ) = 4.5 V ⎝ 2 ⎠ v02 − VGS 3 2 = I D 3 = kn (VGS 3 − VTh ) R5 I1 =

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4.5 − VGS 3 = ( 0.2 ) ( 6 ) (VGS2 3 − 4VGS 3 + 4 ) 1.2VGS2 3 − 3.8VGS 3 + 0.3 = 0 VGS 3 = I D3 =

3.8 ±

( 3.8)

2

− 4 (1.2 ) ( 0.3)

2 (1.2 )

= 3.09 V

4.5 − 3.09 = 0.235 mA 6

gm2 = 2 Kn I D2 = 2

( 0.2 ) ⎛⎜ ⎝

0.375 ⎞ ⎟ 2 ⎠

= 0.387 mA/V 1 1 g m 2 RD = ( 0.387 )( 40 ) ⇒ Ad 1 = 7.74 2 2 − g m 3 RD 2 A2 = 1 + g m 3 R5

Ad 1 =

g m3 = 2 K n I D3 = 2

( 0.2 )( 0.235 )

= 0.434 mA/V A2 =

− ( 0.434 ) ( 4 )

1 + ( 0.434 )( 6 )

= −0.482

So Ad = Ad 1 ⋅ A2 = ( 7.74 ) ( −0.482 ) ⇒ Ad = −3.73 R0 = r05 = Acm1 =

1 1 = = 133 kΩ λ I Q ( 0.02 )( 0.375 )

− ( 0.387 ) ( 40 ) − g m 2 RD = 1 + 2 g m 2 R0 1 + 2 ( 0.387 ) (133)

= −0.149

Acm = ( −0.149 )( −0.482 ) ⇒ Acm = 0.0718

b. vd = v1 − v2 = 0.3sin ω t v +v vcm = 1 2 = 2sin ω t 2 v03 = Ad vd + Acm vcm = ( −3.73)( 0.3) + ( 0.0718 )( 2 ) ⇒ v03 = −0.975sin ω t ( V ) Ideal, Acm = 0 v03 = Ad vd = ( −3.73)( 0.3)

Or ⇒ v03 = −1.12sin ω t ( V ) 11.91 The low-frequency, one-sided differential gain is

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⎛ r ⎞ v02 1 = g m RC ⎜ π ⎟ vd 2 ⎝ rπ + RB ⎠ 1 ⋅ β RC = 2 rπ + RB

Av 2 =

rπ =

(100 )( 0.026 ) 0.5

= 5.2 kΩ

1 ⋅ (100 )(10 ) ⇒ Av 2 = 87.7 Av 2 = 2 5.2 + 0.5 CM = Cμ (1 + g m RC ) 0.5 = 19.23 mA/V 0.026 CM = 2 ⎡⎣1 + (19.23)(10 ) ⎤⎦ ⇒ CM = 387 pF gm =

1 2π ⎡⎣ rπ RB ⎤⎦ ( Cπ + CM ) 1 = So ⇒ f H = 883 kHz 3 2π ⎣⎡5.2 0.5⎦⎤ × 10 × ( 8 + 387 ) × 10−12

fH =

11.92 From Equation (11.117), f Z =

a.

1 1 = 2π R0 C0 2π ( 5 × 106 )( 0.8 × 10−12 )

Or f Z = 39.8 kHz b. From Problem 11.69, f H = 883 kHz. From Equation (11.116(b)), the low-frequency common- mode gain is − g m RC Acm = ⎡⎛ RB ⎞ 2 (1 + β ) R0 ⎤ ⎢⎜ 1 + ⎥ ⎟+ rπ ⎠ rπ ⎣⎝ ⎦ rπ = 5.2 kΩ, g m = 19.23 mA/V So Acm =

− (19.23)(10 )

⎡⎛ 0.5 ⎞ 2 (101) ( 5 × 106 ) ⎤ ⎢⎜ 1 + ⎥ ⎟+ 5.2 × 103 ⎢⎣⎝ 5.2 ⎠ ⎥⎦ −4 = −9.9 × 10

⎛ 87.7 ⎞ C M RRdB = 20 log10 ⎜ = 98.9 dB −4 ⎟ ⎝ 9.9 × 10 ⎠

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11.93 a.

From Equation (7.72), fT =

gm

2π ( Cπ + Cμ )

1 = 38.46 mA/V 0.026 38.46 × 10−3 Then 800 × 106 = 2π ( Cπ + Cμ ) gm =

Or Cπ + Cμ = 7.65 × 10−12 F = 7.65 pF And Cπ = 6.65 pF CM = Cμ (1 + g m RC ) = 1 ⎡⎣1 + ( 38.46 )(10 ) ⎤⎦ = 386 pF 1 fH = 2π ⎡⎣ rπ RB ⎤⎦ ( Cπ + CM ) rπ =

(120 )( 0.026 ) 1

= 3.12 kΩ

1 2π ⎡⎣3.12 1⎤⎦ × 10 × ( 6.65 + 386 ) × 10−12 Or f H = 535 kHz fH =

b.

3

From Equation (11.140), f Z =

1 1 = 2π R0 C0 2π (10 × 106 )(10−12 )

Or f Z = 15.9 kHz 11.94 The differential-mode half circuit is:

⎛v ⎞ g m ⎜ d ⎟ RC ⎝ 2⎠ or Av = v02 = rπ ⎛1+ β ⎞ 1+ ⎜ R ⎟ E r ⎝ π ⎠ rπ =

(100 )( 0.026 ) 0.5

⎛1⎞ ⎜ ⎟ β RC ⎝ 2⎠ + (1 + β ) RE

= 5.2 kΩ

⎛1⎞ ⎜ ⎟ (100 )(10 ) 500 2 = Av = ⎝ ⎠ 5.2 + (101) RE 5.2 + (101) RE

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a.

For RE = 0.1 kΩ : Av = 32.7

b.

For RE = 0.25 kΩ : Av = 16.4

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Chapter 12 Exercise Solutions EX12.1 A 1 + Aβ A A ⇒ Aβ = −1 1 + Aβ = Af Af Af =

a.

β=

1 1 1 1 − = − = 0.05 − 0.0001 ⇒ β = 0.0499 Af A 20 104

Af

b.

=

20

(1/ β ) (1/ 0.0499 )

=

20 = 0.998 20.040

EX12.2 A 1 + Aβ 1 1 1 1 − = − 6 = 0.01 − 10−6 β= Af A 100 10

Af =

β = 0.009999 dA ⎛ Af ⎞ dA = ⋅ Af (1 + β A) A ⎜⎝ A ⎟⎠ A dAf ⎛ 100 ⎞ dAf = ⎜ 6 ⎟ ( 20 ) % ⇒ = 0.002% Af Af ⎝ 10 ⎠ dAf

=

1

⋅

EX12.3 Bandwidth = ω H (1 + β A0 ) ⎛A ⎞ ⎛ 105 ⎞ = ω H ⎜ 0 ⎟ = ( 2π )(10 ) ⎜ ⎟ ⎝ 100 ⎠ ⎝ At ⎠

ω = ( 2π ) (104 ) rad / sec ⇒ f = 10 kHz EX12.4 (a) vOA = A1 A2 vi + A2 vn = (100 )(10 ) vi + (10 ) vn So 1000vi S = = 100 i No 10vn Ni (b) A1 A2 A2 vOC = vi + vn 1 + β Α1 Α2 1 + β Α1 Α2

=

105 10 v + vn 5 i 1 + ( 0.001)10 1 + ( 0.001) (105 )

So 103 vi S = = 104 i N o 0.1vn Ni

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EX12.5 a. Vε = VS − V fb = 100 − 99 = 1 m V 5 ⇒ Av = 5000 V/V 0.001 V fb 0.099 V fb = β V0 ⇒ β = = ⇒ β = 0.0198 V/V V0 5 V0 = AvVε ⇒ Av =

Avf =

Av 1 + β Av

=

5000 ⇒ Avf = 50 V/V 1 + ( 0.0198 )( 5000 )

Rif = Ri (1 + β Av ) = ( 5 ) ⎡⎣1 + ( 0.0198 )( 5000 ) ⎤⎦ ⇒ Rif = 500 kΩ

b.

R0 f =

R0 4 = ⇒ R0 f ⇒ 40 Ω 1 + β Av 1 + ( 0.0198 )( 5000 )

EX12.6 a. I ε = I S − I fb = 100 − 99 = 1 μ A Ai =

β= Aif =

I0 5 = ⇒ Ai = 5000 A/A I ε 0.001 I fb I0

=

0.099 ⇒ β = 0.0198 A/A 5

Ai 5000 ⇒ Aif = 50 A/A = 1 + Ai β 1 + ( 5000 )( 0.0198 ) Rif =

b.

Ri 5 = ⇒ Rif ⇒ 50 Ω 1 + β Ai 1 + ( 0.0198 )( 5000 )

R0 f = (1 + β Ai ) R0 = ⎡⎣1 + ( 0.0198 )( 5000 ) ⎤⎦ ( 4 ) ⇒ R0 f = 400 kΩ

EX12.7 Avf =

Avf =

Av 104 = ⇒ Avf = 3.9984 Av 104 1+ 1+ 1 + ( R2 / R1 ) 1 + ( 30 /10 ) 105 = 3.99984 105 1+ 1 + ( 30 /10 )

3.99984 − 3.9984 × 100% ⇒ 0.0360% 3.9984

EX12.8 Use a non inverting op-amp. R R 1 + 2 = 15 ⇒ 2 = 14 R1 R1

Let

β=

R2 = 140 K R1 = 10 K 1 = 0.066667 R2 1+ R1

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Input resistance. Rif = 5 ( 0.06667 ) ( 5 × 103 ) ≅ 1.67 MΩ Rof =

50

( 0.066667 ) ( 5 × 103 )

≡ 0.15Ω

EX12.9 ⎛ ⎞ ⎟ ⎞⎜ RE ⎜ ⎟ ⋅ ii ⎟ ⎠ ⎜ R + rk ⎟ ⎜ E 1+ h ⎟ FE ⎠ ⎝ 80 0.026 ( )( ) rk = = 4.16 k Ω 0.5 Then rk 4.16 = = 0.0514 k Ω 1 + hFE 81 Then we want ⎞ io RE ⎛ 80 ⎞ ⎛ = 0.95 = ⎜ ⎟ ⎜ ⎟ ii ⎝ 81 ⎠ ⎝ RE + 0.0514 ⎠ ⎛ h io = ⎜ FE ⎝ 1 + hFE

or ⎛ ⎞ RE ⎜ ⎟ = 0.9619 ⎝ RE + 0.0514 ⎠ which yields RE ( min ) = 1.30 k Ω and ⎛ 81 ⎞ V + = I E RE + 0.7 = ⎜ ⎟ ( 0.5 )(1.3) + 0.7 ⇒ V + ( min ) = 1.36 V ⎝ 80 ⎠

EX12.10 Use the configuration shown in figure 12.20. RS = 500 Ω, RL = 200 Ω

Let 1 +

RF = 15 R1

For example, let R1 = 2 K RF = 28 K

EX12.11 a. ⎛ R2 ⎞ ⎛ 20 ⎞ VG = ⎜ ⎟ (10 ) − 5 = ⎜ ⎟ (10 ) − 5 ⎝ 20 + 30 ⎠ ⎝ R1 + R2 ⎠ VG = −1 V VS = −1 − VGS ID =

VS − ( −5 ) RS

= K n (VGS − VTN )

2

( −1) − VGS + 5 = (1.5)( 0.4 )(VGS − 2 )

2

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4 − VGS = 0.6 (V 2GS − 4VGS + 4 )

0.6V 2GS − 1.4VGS − 1.6 = 0 VGS =

1.4 ±

(1.4 )

2

+ 4 ( 0.6 )(1.6 )

2 ( 0.6 )

VGS = 3.174 V

g m = 2 K n (VGS − VTN ) = 2 (1.5 )( 3.17 − 2 ) g m = 3.52 mA / V

⎛ RD ⎞ ⎛ 2 ⎞ I0 = − ⎜ ⎟ ( g mVgs ) = − ⎜ ⎟ ( 3.52 ) Vgs ⎝ 2+2⎠ ⎝ RD + RL ⎠ I 0 = −1.76Vgs Vi = Vgs + g mVgs RS ⇒ Vgs =

Vgs =

Vi 1 + g m RS

Vi = ( 0.4153) Vi 1 + ( 3.52 )( 0.4 )

I 0 = − (1.76 )( 0.4153)Vi ⇒ Agf =

I0 = −0.731 mA / V Vi

b. For K n = 1 mA / V 2 From dc analysis: 4 − VGS = (1)( 0.4 )(VGS − 2 )

2

4 − VGS = 0.4 (V 2GS − 4VGS + 4 ) 0.4V 2GS − 0.6VGS − 2.4 = 0

VGS =

0.6 ±

( 0.6 ) + 4 ( 0.4 )( 2.4 ) 2 ( 0.4 ) 2

VGS = 3.31V g m = 2 (1)( 3.31 − 2 ) = 2.623 mA / V ⎛ 2 ⎞ I0 = − ⎜ ⎟ ( 2.623) Vgs = −1.311Vgs ⎝ 2+2⎠ Vi Vgs = = Vi ( 0.488 ) 1 + ( 2.623)( 0.4 )

I 0 = − (1.31)( 0.488 )Vi

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Agf =

I0 = −0.6398 mA / V Vi

% change =

0.731 − 0.6398 ⇒ 12.5% 0.731

EX12.12 Use the circuit with the configuration shown in Figure 12.27. The LED replaces RL . 1 ⇒ RE = 100 Ω RE

Agf = 10 mS = 10 × 10−3 =

EX12.13 dc analysis: 10 − V0 V0 = ID + 4.7 47 + 20 I D = K n (VGS − VTN )

(1)

2

(2)

⎛ 20 ⎞ VGS = ⎜ (3) ⎟ V0 = 0.2985V0 ⎝ 20 + 47 ⎠ 2.13 − V0 ( 0.213) = I D + ( 0.0149 ) V0 (1) I D = 2.13 − V0 ( 0.2279 )

From (2): 2.13 − V0 ( 0.2279 ) = 1 ⎡⎣( 0.2985V0 ) − 1.5⎤⎦

2

2.13 − V0 ( 0.2279 ) = 0.0891V0 − 0.8955V0 + 2.25 2

0.0891V0 − 0.6676V0 + 0.12 = 0 2

V0 =

0.6676 ±

( 0.6676 ) − 4 ( 0.0891)( 0.12 ) 2 ( 0.0891) 2

V0 = 7.31 V 10 − 7.31 7.31 − = 0.572 − 0.109 4.7 67 I D = 0.463 mA ID =

VGS =

a.

0.463 + 1.5 = 2.18 1 g m = 2 K n (VGS − VTN ) = 2 (1)( 2.18 − 1.5 ) ⇒ g m = 1.36 mA/V

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V0 − Vgs V0 + g mVgs + = 0 (1) RD RF

Vgs − Vi RS

+

Vgs − V0 RF

=0

(2)

⎛ 1 1 ⎞ V0 Vi + + Vgs ⎜ ⎟= ⎝ RS RF ⎠ RF RS V 1 ⎞ V ⎛ 1 Vgs ⎜ + ⎟ = 0 + i ⎝ 20 47 ⎠ 47 20 Vgs ( 0.0713) = V0 ( 0.0213) + Vi ( 0.050 ) Vgs = V0 ( 0.299 ) + Vi ( 0.701)

From (1): V0 V Vgs + (1.36 ) Vgs + 0 − =0 4.7 47 47 V0 ( 0.213) + (1.36 ) Vgs + V0 ( 0.213) − Vgs ( 0.0213) = 0 V0 ( 0.234 ) + Vgs (1.34 ) = 0 V0 ( 0.234 ) + (1.34 ) ⎡⎣(V0 )( 0.299 ) + (Vi )( 0.701) ⎤⎦ = 0 V0 ( 0.635 ) + Vi ( 0.939 ) = 0 ⇒ Avf =

V0 = −1.48 Vi

b. For K n = 1.5 mA / V 2 From dc analysis: 2.13 − V0 ( 0.2279 ) = 1.5 ⎡⎣( 0.2985V0 ) − 1.5⎤⎦

2

= 1.5 ⎡⎣0.0891V0 − 0.8955V0 + 2.25⎤⎦ 2

= 0.1337V0 − 1.343V0 + 3.375 2

0.1337V0 − 1.115V0 + 1.245 = 0 2

V0 = V0 =

1.115 ±

(1.115) − 4 ( 0.1337 )(1.245 ) 2 ( 0.1337 ) 2

1.115 ± 0.7597 ⇒ V0 = 7.01 V 2 ( 0.1337 )

10 − 7.01 7.01 − = 0.636 − 0.105 4.7 67 I D = 0.531 mA ID =

0.531 + 1.5 = 2.09 1.5 g m = 2 K n (VGS − VTN ) = 2 (1.5 )( 2.09 − 1.5 )

VGS =

= 1.77 mA/V

From ac analysis: V0 V Vgs + (1.77 )Vgs + 0 − =0 4.7 47 47

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V0 ( 0.213) + (1.77 ) Vgs + V0 ( 0.0213) − Vgs ( 0.0213) = 0 V0 ( 0.234 ) + Vgs (1.75 ) = 0 V0 ( 0.234 ) + (1.75 ) ⎡⎣V0 ( 0.299 ) + Vi ( 0.701) ⎤⎦ = 0 V0 ( 0.757 ) + Vi (1.23) = 0 ⇒ Avf =

% change =

V0 = −1.62 Vi

1.62 − 1.48 ⇒ 9.46% 1.48

EX12.14 a. Input resistance.

V0 − Vgs V0 + g mVgs + = 0 (1) RD RF Ii =

Vgs − V0 RF

(2)

So V0 = Vgs − I i RF ⎛ 1 ⎛ 1 ⎞ 1 ⎞ V0 ⎜ + ⎟ + Vgs ⎜ g m − ⎟ = 0 (1) RF ⎠ ⎝ RD RF ⎠ ⎝ 1 ⎞ 1 ⎞ ⎛ 1 ⎛ ⎡⎣Vgs − I i ( 47 ) ⎤⎦ ⎜ + ⎟ + Vgs ⎜1.36 − ⎟ = 0 47 ⎠ ⎝ 4.7 47 ⎠ ⎝

⎡⎣Vgs − I i ( 47 ) ⎤⎦ ( 0.234 ) + Vgs (1.34 ) = 0 Vgs Vgs (1.57 ) = I i (11.0 ) ⇒ Rif = = 7.0 kΩ Ii

Output Resistance.

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IX =

VX VX + g mVgs + RD RS + RF

⎛ RS ⎞ ⎛ 20 ⎞ Vgs = ⎜ ⎟ VX = ⎜ ⎟ VX = 0.2985VX ⎝ 20 + 47 ⎠ ⎝ RS + RF ⎠ V VX I X = X + (1.36)(0.2985)VX + 4.7 20 + 47 I X = VX [ 0.213 + 0.406 + 0.0149] R0 f =

b.

VX = 1.58 kΩ IX

From part (a)

1 ⎞ ⎛ ⎡⎣Vgs − I i ( 47 ) ⎤⎦ ( 0.234 ) + Vgs ⎜ 1.77 − ⎟ = 0 47 ⎠ ⎝ Vgs Vgs (1.98 ) = I i (11) ⇒ Rif = = 5.56 kΩ Ii

IX =

VX VX + (1.77 )( 0.2985 ) VX + 4.7 20 + 47

I X = VX [ 0.213 + 0.528 + 0.0149] ⇒ R0 f =

VX = 1.32 kΩ IX

EX12.15 Use the circuit with the configuration shown in Figure 12.41 Let RF = 10 K EX12.16 ⎛ 5.5 ⎞ VTH = ⎜ ⎟ (10 ) = 0.9735 V ⎝ 5.5 + 51 ⎠ RTH = 5.5 || 51 = 4.965 kΩ

I BQ =

0.973 − 0.7 = 0.00217 mA 4.96 + (121)(1)

I CQ = 0.2605 mA rπ = 11.98 kΩ, g m = 10.02 mA / V Req = RS || R1 || R2 || rπ

= (10 ) || 51|| 5.5 || 12 = 2.598 kΩ

From Equation (12.99(b)): ⎛ ⎞ Req T = ( g m RC ) ⎜ ⎜ R + R + R ⎟⎟ F eq ⎠ ⎝ C 2.598 ⎛ ⎞ = (10 )(10 ) ⎜ ⎟ ⇒ T = 2.75 10 82 2.598 + + ⎝ ⎠ EX12.17 Computer Analysis EX12.18

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Vε = −Vt , V0 = AvVε = − AvVt ⎛ R1 || Ri ⎞ ⎛ R1 || Ri ⎞ Vr = ⎜ ⎟ V0 = − ⎜ ⎟ ( AvVt ) R || R + R 2 ⎠ ⎝ 1 i ⎝ R1 || Ri + R2 ⎠ T =−

⎛ R1 || Ri ⎞ Vr = Av ⎜ ⎟ Vt ⎝ R1 || Ri + R2 ⎠

or T=

Av R 1+ 2 R1 Ri

EX12.19 ⎛ f′ ⎞ Phase = −180° = −3 tan −1 ⎜ 5 ⎟ ⎝ 10 ⎠ ⎛ f′ ⎞ or tan −1 ⎜ 5 ⎟ = 60° ⇒ f ′ = 1.732 × 105 Hz ⎝ 10 ⎠ β (100 ) T ( f ′) = 1 = 3 2 ⎡ f′ ⎞ ⎤ ⎛ ⎢ 1+ ⎜ 5 ⎟ ⎥ ⎢ ⎝ 10 ⎠ ⎥⎦ ⎣ β (100 ) ⇒ β = 0.08 = 3 ⎡ 1 + (1.732 )2 ⎤ ⎣⎢ ⎦⎥

EX12.20 Afo =

1000 = 7.092 1 + ( 0.140 )(1000 )

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( 0.140 )(1000 )

T =1=

⎛ f ⎞ ⎛ f ⎞ 1 + ⎜ 3 ⎟. 1 + ⎜ 4 ⎟ 10 ⎝ ⎠ ⎝ 5 × 10 ⎠

2

⎛ f ⎞ 1+ ⎜ 6 ⎟ ⎝ 10 ⎠

2

By trial and error, at f = 7.65 × 104 Hz

( 0.140 )(1000 ) 2 2 1 + ( 76.5 ) 1 + (1.53) 1 + ( 0.0765 ) ( 0.140 )(1000 ) = = 1.00 76.5 ( )(1.828 )(1.0 )

T =

2

Then φ = − ⎡⎣ tan −1 (76.5) + tan −1 (1.53) + tan −1 (0.0765) ⎤⎦ = − [89.25 + 56.83 + 4.37 ] = −150.45

Phase Marge = 180 − 150.45 = 29.5° EX12.21 The new loop gain function is 105 1 T ′( f ) = × ⎛ f ⎞⎛ f ⎞ ⎛1 + j ⋅ f ⎞ ⎛1 + j ⋅ f ⎞ ⎟⎜ ⎟ ⎜1 + j ⋅ ⎟⎜ 1 + j ⋅ ⎟ ⎜ 107 ⎠ ⎝ 5 × 108 ⎠ f PD ⎠ ⎝ 5 × 105 ⎠ ⎝ ⎝ ⎛ f ⎞ f ⎞ ⎪⎧ ⎛ f ⎞ ⎛ f ⎞ ⎪⎫ −1 ⎛ Phase = − ⎨ tan −1 ⎜ + tan −1 ⎜ 7 ⎟ + tan −1 ⎜ ⎟ + tan ⎜ 5 ⎟ 8 ⎟⎬ ⎝ 5 × 10 ⎠ ⎝ 10 ⎠ ⎝ 5 × 10 ⎠ ⎪⎭ ⎪⎩ ⎝ f PD ⎠ For a phase margin 45° ⇒ Phase = −135°, the poles are far apart so this will occur at approximately f ′ = 5 × 105 Hz. Then 105

T ′( f ) = 1 =

2

⎛ 5 × 105 ⎞ 1+ ⎜ ⎟ × 1+1× 1× 1 ⎝ f PD ⎠ 105

1=

(1.414 )

⎛ 5 × 105 ⎞ 1+ ⎜ ⎟ ⎝ f PD ⎠

2

2

⎛ 5 × 105 ⎞ 9 1+ ⎜ ⎟ = 5 × 10 ⎝ f PD ⎠

f PD ≅

5 × 105 5 × 109

⇒ f PD = 7.07 Hz

EX12.22 Af ( 0 ) =

A0 1 + β A0

=

2 × 105

1 + ( 0.05 ) ( 2 × 105 )

⇒ Af (0) ≅ 20

f C = f PD (1 + β A0 ) = 100 ⎡⎣1 + ( 0.05 ) ( 2 × 105 ) ⎤⎦ ⇒ fC ≅ 1 MHz

EX12.23 Phase Margin = 45° ⇒ Phase = −135° This will occur at approximately f ′ = 107 Hz

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105

T ′( f ) = 1 =

2

⎛ 107 ⎞ 1+ ⎜ ⎟ × 2× 1 ⎝ f PD ⎠ 2

⎛ 107 ⎞ 9 1+ ⎜ ⎟ = 5 × 10 f ⎝ PD ⎠ 107 ⇒ f PD = 141 Hz f PD ≅ 5 × 109

TYU12.1 A Af = 1 + Aβ A f + A f Aβ = A

Af = A (1 − Af β ) A=

Af 1 − Af β

=

80 1 − ( 80 )( 0.0120 )

A = 2000

TYU12.2 dAf dA ⎛ Af ⎞ dA 1 = ⋅ =⎜ ⎟. Af (1 + β A) A ⎝ A ⎠ A dA dAf = A Af

⎛ A ⋅⎜ ⎜A ⎝ f

⎞ ⎛ 5 × 105 ⎞ dA = ±5% ⎟⎟ = ( 0.001) ⎜ ⎟⇒ A ⎝ 100 ⎠ ⎠

TYU12.3 Af ⋅ f H = A0 ⋅ f1 Af =

6 A0 ⋅ f1 (10 ) ( 8 ) ⇒ Af ( 0 ) = 32 = fH 250 × 103

TYU12.4 Vε = VS − V fb = 100 − 99 = 1 mV Ag =

β= Agf =

I 0 5 mA = ⇒ Ag = 5 A/V Vε 1 mV V fb

=

I0

99 mV ⇒ β = 19.8 V/A 5 mA

Ag 1 + β Ag

=

5 ⇒ Agf = 0.05 A/V = 50 mA/V 1 + (19.8 )( 5 )

TYU12.5

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I ε = I S − I fb = 100 − 99 = 1 μ A Az =

β= Azf =

V0 5V = ⇒ Az = 5 × 106 V/A Iε 1 μ A I fb

=

V0

99 μ A ⇒ β = 1.98 × 10−5 A/V 5V

Az 1 + β Az

TYU12.6 rπ =

a. gm =

I CQ

=

VT

=

5 × 106

1 + (1.98 × 10−5 )( 5 × 106 )

⇒ Azf = 5 × 104 V/A = 50 V/mA

hFEVT (100 )( 0.026 ) = = 5.2 kΩ I CQ 0.5

0.5 = 19.23 mA / V 0.026

⎛1 ⎞ ⎛ 1 ⎞ + 19.23 ⎟ ( 2 ) ⎜ + g m ⎟ RE ⎜ r 5.2 ⎠ ⎠ = ⎝ Avf = ⎝ π ⎛1 ⎞ ⎛ 1 ⎞ + 19.23 ⎟ ( 2 ) 1 + ⎜ + g m ⎟ RE 1 + ⎜ ⎝ 5.2 ⎠ ⎝ rπ ⎠

(19.42 )( 2 ) ⇒ Avf = 0.97490 1 + (19.42 )( 2 ) = rπ + (1 + hFE ) RE = 5.2 + (101)( 2 ) ⇒ Rif =

Rif

rπ 5.2 =2 ⇒ R0 f = 0.0502 kΩ ⇒ 50.2 Ω 1 + hFE 101

R0 f = RE

b.

= 207.2 kΩ

hFE = 150 ⇒ rπ = 7.8 kΩ, g m = 19.23mA/V

⎛ 1 ⎞ + 19.23 ⎟ ( 2 ) ⎜ (19.36 )( 2 ) 7.8 ⎠ Avf = ⎝ = 1 + (19.36 )( 2 ) ⎛ 1 ⎞ 1+ ⎜ + 19.23 ⎟ ( 2 ) 7.8 ⎝ ⎠ Avf = 0.97482 ⇒ 0.0082% change in Avf Rif = 7.8 + (101)( 2 ) = 209.8 kΩ ⇒ 1.25% change in Rif R0 f = RE

rπ 7.8 =2 = 2 0.0517 1 + hFE 151

R0 f = 50.36 Ω ⇒ 0.319% change in R0 f

TYU12.7

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V0 = ( g mVgs ) RS Vi = Vgs + g m RSVgs Vi Vgs = 1 + g m RS g m = 2 K n I DQ = 2

( 0.2 )( 0.25 ) = 0.447 mA / V

Avf =

V0 g m RS = Vi 1 + g m RS

Avf =

( 0.447 )( 5 ) ⇒ Avf 1 + ( 0.447 )( 5 )

= 0.691

Rif = ∞

I X = g mVgs =

VX RS

Vgs = −VX ⎛ 1 ⎞ I X = VX ⎜ g m + ⎟ RS ⎠ ⎝ 1 1 R0 f = RS = 5 gm 0.447 R0 f = 1.55 kΩ

TYU12.8 Computer Analysis TYU12.9 Computer Analysis TYU12.10

a.

Agf =

b.

Agf =

hFE ⋅ Ag

1 + ( hFE Ag ) RE

=

( 200 ) (103 ) ⇒ Agf 1 + ( 200 ) (103 )(103 )

( 200 ) (104 ) ⇒ Agf 1 + ( 200 ) (104 )(103 )

= 10−3 A/V=1 mA/V

≅ 1 mA/V

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Percent change is negligible. ( 4.5 × 10−7 % )

( 200 ) (104 ) ( 200 ) (103 ) − 1 + ( 200 ) (104 )(103 ) 1 + ( 200 ) (103 )(103 ) 10−3

=

− =

( 200 ) (104 ) ⎡⎣1 + ( 200 ) (104 )(103 )⎤⎦

(10 )( 2 ×10 )( 2 ×10 ) −3

9

8

(200)(103 )[1 + (200)(104 )(103 )] (10−3 )(2 × 109 )(2 × 108 ) 200 × 104 − 200 × 103

(10−3 )( 2 ×109 )( 2 ×108 )

TYU12.11 T = Ai β =

T ( f1 )

Ai 0 β

= 4.5 × 10−9

(10 ) ( 0.01) 5

=

⎛ ⎛ f ⎞ f ⎞ ⎜ 1 + j ⋅ ⎟ 1 + j ⋅ ⎜ 10 ⎟ ⎝ ⎠ f1 ⎠ ⎝ 3 10 =1= 2 ⎛ f′⎞ 1+ ⎜ E ⎟ ⎝ 10 ⎠

⎛ f′⎞ 1 + ⎜ E ⎟ = 106 ⎝ 10 ⎠ 2

f E′ = 10 106 − 1 ⇒ f E′ ≈ 104 Hz

⎛ 104 ⎞ ⎛ f′⎞ Phase = φ = − tan −1 ⎜ E ⎟ = − tan −1 ⎜ ⎟ ⎝ 10 ⎠ ⎝ 10 ⎠ = − tan −1 (103 )

φ ≈ 90° Phase Margin = 180 − 90 ⇒ Phase Margin = 90° TYU12.12 Αι 0 β ⎛ f ⎞⎛ f ⎞ ⎜ 1+ j ⋅ ⎟ ⎜1 + j ⋅ ⎟ f1 ⎠ ⎝ f2 ⎠ ⎝ ⎡ ⎛ f ⎞ ⎛ f ⎞⎤ Phase = − ⎢ tan −1 ⎜ ⎟ + tan −1 ⎜ ⎟ ⎥ ⎝ f1 ⎠ ⎝ f2 ⎠⎦ ⎣ T = Ai β =

Phase Margin = 60° ⇒ Phase = −120° ⎡ ⎛ f ⎞ ⎛ f ⎞⎤ −120° = − ⎢ tan −1 ⎜ 4 ⎟ + tan −1 ⎜ 5 ⎟ ⎥ 10 10 ⎠ ⎦ ⎝ ⎠ ⎝ ⎣ At f ′ = 7.66 × 104 Hz, Phase = − ⎡⎣ tan −1 ( 7.66 ) + tan −1 ( 0.766 ) ⎤⎦ = − [82.56 + 37.45] = −120°

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(10 ) β 5

T ( f ′) = 1 =

1+ ( 7.66 ) × 1 + ( 0.766 ) 2

(10 ) β

2

5

1=

( 7.725 )(1.26 )

⇒ β = 9.73× 10−5

TYU12.13 Phase Margin = 60° ⇒ Phase = −120° ⎛ f′ ⎞ Phase = −120° = −3 tan −1 ⎜ 5 ⎟ ⎝ 10 ⎠ ⎛ f′ ⎞ tan −1 ⎜ 5 ⎟ = 40° ⇒ f ′ = 0.839 × 105 Hz ⎝ 10 ⎠ β (100 ) T ( f ′) = 1 = 3 2 ⎡ f′ ⎞ ⎤ ⎛ ⎢ 1+ ⎜ 5 ⎟ ⎥ ⎢ ⎝ 10 ⎠ ⎥⎦ ⎣ β (100 ) = ⇒ β = 0.0222 3 ⎡ 1 + ( 0.839 )2 ⎤ ⎣⎢ ⎦⎥

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Chapter 12 Problem Solutions 12.1 (a) Af =

A 1 + Aβ

120 =

5 × 105 1 + (5 × 105 ) β

120 + (120)(5 × 105 ) β = 5 × 105 β = 0.008331 (b) 5 × 103 120 = 1 + (5 × 103 ) β 120 + (120)(5 × 103 ) β = 5 × 103

β = 0.008133 12.2

Af =

(a)

A 1 + Aβ

β = 0.15

T = Aβ T =∞ (i) A = 80 dB ⇒ A = 104 ⇒ T = 1.5 × 103 (ii) (iii) T = 15 1 (i) Af = = 6.667

β

(ii)

Af = 6.662

(iii)

Af = 6.25

(b) (i) (ii) (iii) (i)

T =∞ T = 2.5 × 103 T = 25 1 Af = = 4.00

(ii)

Af = 3.9984

(iii)

Af = 3.846

β

12.3 (a) A 1 ≅ = 125 1 + Aβ β β = 0.0080 (b) Af =

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Af = (125)(0.9975) = 124.6875 124.6875 =

A 1 + (0.008) A

124.6875 [1 + (0.008) A] = A 124.6875 = A [1 − 0.9975] A = 49,875

12.4 (a)

100 =

104 1 + (104 ) β

β = 9.9 × 10−3 (b)

⎛ Af ⎞ dA =⎜ ⎟ ⎝ A⎠ A Af

dAf

dAf ⎛ 100 ⎞ = ⎜ 4 ⎟ (−0.10) = −0.001 ⇒ = −0.10% Af ⎝ 10 ⎠

12.5

⎛ Af ⎞ dA =⎜ ⎟ ⎝ A⎠ A Af

dAf

−0.001 = Af = 500 500 =

Af 5 × 104

(−0.10)

5 × 104 1 + (5 × 104 ) β

β = 1.98 ×10−3 12.6 (a) For Fig. P12.6(a) vo1 vo 200 10 = = vi 1 + 200 β1 vo1 1 + 10 β1

⎛ 200 ⎞ ⎛ 10 ⎞ Af = ⎜ ⎟⎜ ⎟ = 40 ⎝ 1 + 200 β1 ⎠ ⎝ 1 + 10 β1 ⎠ 50 = (1 + 200 β1 )(1 + 10 β1 ) = 1 + 210 β1 + 2000 β12 2000 β12 + 210 β1 − 49 = 0 −210 ± 44100 + 4(2000)(49) 2(2000) β1 = 0.1126 For Fig P12.6(b) (200)(10) 40 = 1 + (200)(10) β 2

β1 =

β 2 = 0.0245 Fig. P12.6(a) (b)

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A1 = 180 180 10 ⎛ ⎞⎛ ⎞ Af = ⎜ ⎟ ⎜ ⎟ = (8.4634)(4.704) ⎝ 1 + (180)(0.1126) ⎠ ⎝ 1 + (10)(0.1126) ⎠ Af = 39.81 dAf Af

=

39.81 − 40 ⇒ −0.475% 40

Fig. P12.6(b) (180)(10) = 39.91 Af = 1 + (180)(10)(0.0245) dAf 39.91 − 40 = ⇒ −0.225% 40 Af (c)

Fig. P12.6(b) is a better feedback circuit.

12.7 (a)

VO = (−10)(−15)(−20)Vε = −3000Vε

Vε = β VO + VS So VO = −3000( β VO + VS ) We find V −3000 Avf = O = VS 1 + 3000 β

For Avf = −120 = (b) Then Avf =

−3000 ⇒ β = 0.008 1 + 3000 β

Now VO = (−9)(−13.5)(−18)Vε = −2187Vε −2187 −2187 = = −118.24 1 + 2187 β 1 + 2187(0.008)

% change =

120 − 118.24 × 100 ⇒ 1.47% change 120

12.8 (105 )(4) = (50) f B ⇒ f B = 8 kHz 12.9 (a)

(50) f3− dB = (105 )(4) ⇒ f 3− dB = 8 kHz

(b)

(10) f3− dB = (105 )(4) ⇒ f3− dB = 40 kHz

12.10 (50)(20 × 103 ) = 5A0 so A0 = 2 × 105 12.11 Low freq. Af =

A0 1 + A0 β

5000 ⇒ β = 0.0098 1 + (5000) β Freq. response 100 =

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Af =

A 1 + Aβ

5000 f f ⎞ ⎛ ⎞⎛ ⎜1 + j f ⎟⎜ 1 + j f ⎟ 1 ⎠⎝ 2 ⎠ = ⎝ (5000)(0.0098) 1+ f ⎞⎛ f ⎞ ⎛ ⎜ 1 + j f ⎟⎜ 1 + j f ⎟ 1 ⎠⎝ 2 ⎠ ⎝ 5000 f ⎞⎛ f ⎞ ⎛ ⎜ 1 + j f ⎟ ⎜ 1 + j f ⎟ + 49 1 ⎠⎝ 2 ⎠ ⎝ 5000 = f f ⎛ jf ⎞ ⎛ jf 1+ j + j + ⎜ ⎟⎜ f1 f 2 ⎝ f1 ⎠ ⎝ f 2 =

=

⎞ ⎟ + 49 ⎠

5000 f f ⎛ jf ⎞ ⎛ jf ⎞ 50 + j + j + ⎜ ⎟ ⎜ ⎟ f1 f 2 ⎝ f1 ⎠ ⎝ f 2 ⎠

Also Af =

Af 0 f ⎞⎛ f ⎞ ⎛ ⎜1 + j f ⎟ ⎜ 1 + j f ⎟ A ⎠⎝ B ⎠ ⎝

=

100 f f ⎛ f ⎞⎛ f ⎞ 1+ j j +j + j fA f B ⎜⎝ f A ⎟⎠ ⎜⎝ f B ⎟⎠

So 100 100 = 1 ⎛ jf ⎞ ⎛ jf ⎞ f f ⎛ f ⎞⎛ f ⎞ f f +j + ⎜ j ⎟⎜ j ⎟ 1 + j +j + ⎜ ⎟⎜ ⎟ 1+ j fA f B ⎝ f A ⎠⎝ f B ⎠ 50 f1 50 f 2 50 ⎝ f1 ⎠ ⎝ f 2 ⎠ Then 1 1 1 1 + = + f A f B 50 f1 50 f 2

and

1 1 = f A f B 50 f1 f 2

f1 = 10 and f 2 = 2000 1 1 1 1 + = + = 0.002 + 0.000010 = 0.002010 f A f B 50(10) 50(2000) and f 1 1 1 = ⇒ = B f A f B (50)(10)(2000) f A 106

Then

fB 1 + = 0.002010 6 fB 10

10−6 f B2 + 1 = 2.01 + 10−3 f B 10−6 f B2 − 2.01× 10−3 f B + 1 = 0 fB =

2.01× 10−3 ± 4.0401× 10 −6 − 4(10−6 )(1) 2(10−6 )

fB =

2.01× 10−3 ± 2.0025 × 10−4 2(10−6 )

+ sign

f B = 1.105 × 103 Hz

+ sign

f A = 9.05 × 102 Hz

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12.12 (a)

Fig. P12.6(a) ⎡ ⎤ ⎛ 200 ⎞ ⎢ ⎥ ⎜ f ⎟ ⎢ ⎥ ⎜⎜ 1 + j ⎟⎟ f1 ⎠ 10 ⎤ ⎢ ⎥⎡ ⎝ Af = ⎢ ⎢ ⎥ ⎥ 200 + 1 (10)(0.1126) ⎛ ⎞ ⎣ ⎦ ⎢1 + ⎜ ⎟ (0.1126) ⎥ ⎢ ⎜ 1+ j f ⎟ ⎥ ⎢⎣ ⎜⎝ ⎥⎦ f1 ⎟⎠ 200 ⎡ ⎤ =⎢ ⎥ (4.704) f ⎞ ⎛ ⎢ ⎜1 + j ⎟ + 22.52 ⎥ f1 ⎠ ⎣⎢ ⎝ ⎦⎥ =

940.73 23.52 + j

f f1

=

940.73 1 ⋅ f 23.52 1 + j (23.52) f1

40 f −3dB = (23.52)(100) ⇒ 2.352 kHz jf 1+ (23.52) f1 Fig P12.6(b) (200)(10) f 1+ j f1 2000 = Af = f (0.0245)(200)(10) 1+ 1 + j + 49 f f1 1+ j f1 =

=

(b)

2000 1 ⋅ f −3dB = (50)(100) ⇒ 5 KHz 50 1 + j f (50) f1 Overall feedback ⇒ wider bandwidth.

12.13 v0 = A1 A2 vi + A1vn v0 = (100)vi + (1)vn = (100)(10) + (1)(1) ⇒

S0 1000 = = 1000 N0 1

12.14 (a)

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(b)

Circuit (b) – less distortion 12.15 (a) Low input R ⇒ Shunt input Low output R ⇒ Shunt output Or a Shunt-Shunt circuit (b) High input R ⇒ Series input High output R ⇒ Series output Or a series-Series circuit (c) Shunt-Series circuit (d) Series-Shunt circuit 12.16 (a)

Ri (max) = Ri (1 + T ) = 10(1 + 104 ) ⇒ Ri (max) ≅ 105 k Ω

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Ri 10 = ≅ 10−3 k Ω 1 + T 1 + 104 Or Ri (min) = 1Ω Ri (min) =

Ro (max) = Ro (1 + T ) = 1(1 + 104 ) ⇒ Ro (max) ≅ 104 k Ω

(b)

Ro 1 = ≅ 10−4 k Ω 1 + T 1 + 104 Or Ro (min) = 0.1Ω Ro (min) =

12.17 io Series output = current signal and Shunt input = current vi signal. Also, Shunt output = voltage signal and Series input = voltage signal. Two possible solutions are shown.

Overall Transconductance Amplifier, Ag =

12.18 ⎛ R1 || Ri ⎞ Vε = − ⎜ ⎟ ⋅ Vx ⎝ R1 || Ri + R2 ⎠

12.19 Vε = Vi − V fb = 50 − 48 = 2mV Av =

β=

Vo 5 = = 2.5 × 103 V/V Vε 0.002 V fb Vo

=

0.048 = 0.0096 V/V 5

V 5 Arf = o = = 100 V/V Vi 0.05

12.20

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⎛ R ⎞ R Avf ≈ ⎜1 + 2 ⎟ = 20 ⇒ 2 = 19 R R1 ⎝ 1 ⎠ vd = iS Ri

vS − vd (vs − vd ) − v0 + R1 R2

(1)

v0 − A0 L vd v0 − (vs − vd ) + =0 R0 R2

(2)

is =

⎛ 1 (v − v ) 1 ⎞ A v v0 ⎜ + ⎟ = 0 L d + S d R0 R2 ⎝ R0 R2 ⎠ A0 L vd (vS − vd ) + R0 R2 v0 = ⎛ 1 1 ⎞ ⎜ + ⎟ ⎝ R0 R2 ⎠

From (1): 1 ⎡ A0 L vd (vS − vd ) ⎤ ⋅⎢ + R2 ⎥⎦ vS − vd vS − vd R2 ⎣ R0 + − iS = R1 R2 ⎛ 1 1 ⎞ ⎜ + ⎟ ⎝ R0 R2 ⎠ A0 L 1 1 ⎞ ⎛ ⎛ − ⎜ 1 ⎜ 1 R0 R2 R2 ⎟ 1 1 ⎟ − vd ⎜ + − + iS = vS ⎜ + R ⎜ R1 R2 ⎜ R1 R2 1 + R2 ⎟ 1+ 2 ⎜ ⎟ ⎜ R R0 0 ⎠ ⎝ ⎝ vd = iS Ri

⎞ ⎟ ⎟ ⎟ ⎟ ⎠

⎧ ⎡⎛ 1 ⎡⎛ 1 1 ⎞ ⎛ R ⎞ A0 L 1 ⎤ ⎫ 1 ⎞ ⎛ R2 ⎞ 1 − ⎥⎪ ⎪ Ri ⎢⎜ + ⎟ ⎜ 1 + 2 ⎟ + ⎢ ⎜ + ⎟ ⎜1 + ⎟ − R R R R R ⎪ ⎝ 1 2 ⎠⎝ 0 ⎠ 0 2 ⎦⎪ ⎢ ⎝ R1 R2 ⎠ ⎝ R0 ⎠ R2 iS ⎨1 + ⎣ ⎬ = vS ⎢ R R ⎪ ⎪ 1+ 2 1+ 2 ⎢ R0 R0 ⎪ ⎪ ⎢⎣ ⎩ ⎭ ⎧⎪ R ⎡1 R 1 ⎡ 1 R2 1 1 A0 L ⎤ ⎫⎪ 1⎤ iS ⎨1 + 2 + Ri ⎢ + 2 ⋅ + + ⎥ ⎬ = vS ⎢ + ⋅ + ⎥ ⎪⎩ R0 ⎣ R1 R1 R0 R0 R0 ⎦ ⎪⎭ ⎣ R1 R1 R0 R0 ⎦ ⎧⎪ ⎡R ⎛ R ⎞ ⎤ ⎪⎫ ⎡ R ⎛ R ⎞⎤ iS ⎨ R0 + R2 + Ri ⎢ 0 + ⎜1 + 2 ⎟ + A0 L ⎥ ⎬ = vS ⎢ 0 + ⎜ 1 + 2 ⎟ ⎥ (1) R1 ⎠ R1 ⎠ ⎦ ⎪⎩ ⎣ R1 ⎝ ⎦ ⎪⎭ ⎣ R1 ⎝ Let R2 = 190 kΩ , R1 = 10 kΩ ⎧ ⎡ 0.1 ⎤⎫ ⎡ 0.1 ⎤ iS ⎨0.1 + 190 + 100 ⋅ ⎢ + 20 + 105 ⎥ ⎬ = vS ⎢ + 20 ⎥ ⎣ 10 ⎦⎭ ⎣ 10 ⎦ ⎩ 7 iS (1.000219 × 10 ) = vS (20.01) vS ≅ 5 × 105 kΩ ⇒ Rif ≅ 500 MΩ iS Output Resistance Rif =

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⎤ ⎥ ⎥ ⎥ ⎥ ⎥⎦

IX =

VX − A0 L vd VX + R0 R2 + R1 || Ri

vd =

− R1 || Ri ⋅ VX R1 || Ri + R2

A0 L ⋅ R1 || Ri IX 1 1 1 = = + + VX R0 f R0 R0 ( R1 || Ri + R2 ) R2 + R1 || Ri R1 || Ri = 10 ||100 = 9.09

1 1 105 ⎛ 9.09 ⎞ 1 = + ⋅⎜ ⎟+ R0 f 0.1 0.1 ⎝ 9.09 + 190 ⎠ 190 + 9.09 = 10 + 4.566 × 104 + 0.00502 R0 f = 2.19 × 10−5 kΩ ⇒ R0 f = 0.0219 Ω

12.21 a.

vS − vd v0 − (vS − vd ) v = and vd = 0 R1 R2 A

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⎛ 1 vS vS v 1 ⎞ + = 0 + vd ⎜ + ⎟ R1 R2 R2 R R ⎝ 1 2 ⎠ v0 v0 ⎛ 1 1 ⎞ + ⎜ + ⎟ R2 A ⎝ R1 R2 ⎠ ⎛ 1 1 ⎞ v ⎡ 1 ⎛ R ⎞⎤ vS ⎜ + ⎟ = 0 ⎢1 + ⎜ 1 + 2 ⎟ ⎥ R1 ⎠ ⎦ ⎝ R1 R2 ⎠ R2 ⎣ A ⎝ =

⎛ R2 ⎞ ⎜1 + ⎟ R1 ⎠ v0 = ⎝ vS 1⎛ R ⎞ 1 + ⎜1 + 2 ⎟ A⎝ R1 ⎠

which can be written as v A Avf = 0 = vS ⎡ ⎛ R ⎞⎤ 1 + ⎢ A / ⎜1 + 2 ⎟ ⎥ R1 ⎠ ⎦ ⎣ ⎝ 1 b. β= R 1+ 2 R1 20 =

c.

105 1 + (105 ) β

105 −1 So β = 20 5 ⇒ β = 0.04999 10 R2 1 R 1 Then = −1 = − 1 ⇒ 2 = 19.004 R1 β 0.04999 R1 A → 9 × 104 9 × 104 Af = = 19.99956 1 + (9 × 104 )(0.04999)

d.

ΔAf Af

=

ΔAf −4.444 × 10−4 = −2.222 × 10 −3 % ⇒ = −0.005% 20 Af

12.22 Aif = Rif =

Ai 1000 = = 90.9 A/A 1 + β i Ai 1 + (1000)(0.01) Ri 1 + β i Ai

=

1 ⇒ Rif = 90.9 Ω 1 + (0.01)(1000)

Rof = Ro (1 + β i Ai ) = 10 [1 + (0.01)(1000) ] ⇒ Rof = 110 kΩ

12.23

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I ε = I i − I fb = 50 − 47.5 = 2.5 μ A Ai =

βi =

Io 5 = = 2000 A/A I ε 0.0025 I fb Io

=

0.0475 = 0.0095 A/A 5

I 5 Aif = o = = 100 A/A I i 0.05

12.24 a.

Assume that V1 is at virtual ground. V0 = − I fb RF Now I fb = I 0 +

I fb RF V0 = I0 − R3 R3

I fb = I S − I ε

and I 0 = Ai I ε =

I0 Ai

so I0 Ai From above ⎛ R ⎞ I fb ⎜ 1 + F ⎟ = I 0 R3 ⎠ ⎝ I fb = I S −

⎛ I 0 ⎞ ⎛ RF ⎞ ⎜ I S − ⎟ ⎜1 + ⎟ = I0 Ai ⎠ ⎝ R3 ⎠ ⎝ ⎡ ⎛ R ⎞ 1 ⎛ R ⎞⎤ I S ⎜ 1 + F ⎟ = I 0 ⎢1 + ⎜ 1 + F ⎟ ⎥ R3 ⎠ R3 ⎠ ⎦ ⎝ ⎣ Ai ⎝ or

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Aif =

⎛ RF ⎞ ⎜1 + ⎟ R3 ⎠ ⎝ = ⎡ 1 ⎛ RF ⎞ ⎤ ⎢1 + ⎜ 1 + ⎟⎥ R3 ⎠ ⎦ ⎣ Ai ⎝ Ai = = Aif Ai 1+ ⎛ RF ⎞ ⎜1 + ⎟ R3 ⎠ ⎝

I0 IS

b.

βi =

c.

25 =

1 ⎛ RF ⎞ ⎜1 + ⎟ R3 ⎠ ⎝

105 1 + (105 ) β i

105 −1 so β i = 25 5 ⇒ β i = 0.03999 10 RF R 1 1 = −1 = − 1 ⇒ F = 24.0 so R3 β i R3 0.03999

Ai = 105 − (0.15)(105 ) = 8.5 × 104

d.

so Aif = so

ΔAif

Aif

8.5 × 104 = 24.9989 1 + (8.5 × 104 )(0.03999) =−

1.10 × 10−3 = −4.41× 10 −5 ⇒ −4.41× 10 −3 % 25

12.24 b. V0 = ( g m1 Vgs + g m 2 Vπ ) RL (1) Vi = Vgs + V0

(2)

⎛ 1 + hFE ⎞ V1 Vπ V + g m 2 Vπ + 1 = 0 ⇒ Vπ ⎜ =0 ⎟+ rπ RE 2 ⎝ rπ ⎠ RE 2 Vπ V −V = g m1 Vgs + 1 π = 0 rπ RD1 or ⎡V ⎤ V V1 = RD1 ⎢ π + π − g m1 Vgs ⎥ (4) ⎣ rπ RD1 ⎦

Then ⎛ 1 + hFE Vπ ⎜ ⎝ rπ

(3)

⎞ RD1 ⎟+ ⎠ RE 2

⎧⎪⎛ 1 + hFE Vπ ⎨⎜ ⎩⎪⎝ rπ Let

⎡ Vπ Vπ ⎤ − g m1 Vgs ⎥ = 0 (3) ⎢ + ⎣ rπ RD1 ⎦ ⎫ ⎞ RD1 ⎛ RD1 ⎞ 1 ⎪ + ⎬ = g m1 ⎜ ⎟+ ⎟ Vgs R r R ⎪ E2 π E2 ⎭ ⎝ RE 2 ⎠ ⎠

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Vπ ⋅

⎛R ⎞ 1 = g m1 ⎜ D1 ⎟ Vgs Req ⎝ RE 2 ⎠

⎛R ⎞ so Vπ = g m1 Req ⎜ D1 ⎟ Vgs ⎝ RE 2 ⎠ Then ⎡ ⎛R ⎞ ⎤ V0 = ⎢ g m1 Vgs + g m1 g m 2 Req ⎜ D1 ⎟ Vgs ⎥ RL ⎝ RE 2 ⎠ ⎦ ⎣ so ⎡ ⎛ R ⎞⎤ V0 = g m1 RL ⎢1 + g m 2 Req ⎜ D1 ⎟ ⎥ (Vi − V0 ) ⎝ RE 2 ⎠ ⎦ ⎣ so ⎡ ⎛ R ⎞⎤ g m1 RL ⎢1 + g m 2 Req ⎜ D1 ⎟ ⎥ V ⎝ RE 2 ⎠ ⎦ ⎣ Av = 0 = Vi ⎡ ⎛ R ⎞⎤ 1 + g m1 RL ⎢1 + g m 2 Req ⎜ D1 ⎟ ⎥ ⎝ RE 2 ⎠ ⎦ ⎣ c. Set Vi = 0

I X + g m1 Vgs + g m 2Vπ =

(1)

VX RL

Vgs = −VX

From part (b), we have ⎛R ⎞ ⎛R ⎞ Vπ = g m1 Req ⎜ D1 ⎟ Vgs = − g m1 Req ⎜ D1 ⎟ VX R ⎝ E2 ⎠ ⎝ RE 2 ⎠ Then ⎛R ⎞ IX 1 1 = = + g m1 g m 2 Req ⎜ D1 ⎟ VX R0 RL ⎝ RE 2 ⎠ or R0 = RL

1 g m1

1 ⎛R ⎞ g m1 g m 2 Req ⎜ D1 ⎟ ⎝ RE 2 ⎠

12.25

I S = I ε + I fb , V1 = I ε Ri I fb = I 0 +

V0 and V0 = V1 − I fb RF R3

I 0 = Ai I ε ⇒ I ε =

I0 Ai

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Now I fb = Ai I ε +

1 (V1 − I fb RF ) R3

⎡ R ⎤ V I fb ⎢1 + F ⎥ = Ai Iε + 1 R R3 3 ⎦ ⎣ I fb = I S − I e ⎡ R ⎤ V ( I S − I e ) ⎢1 + F ⎥ = Ai I e + 1 R3 ⎣ R3 ⎦ ⎡⎛ R ⎞ ⎤ V ⎡ R ⎤ I S ⎢1 + F ⎥ = I e ⎢⎜ 1 + F ⎟ + Ai ⎥ + 1 R3 ⎠ ⎣ R3 ⎦ ⎣⎝ ⎦ R3 V Iε = 1 Ri ⎤ 1 ⎪⎫ ⎤ ⎪⎧ 1 ⎡⎛ RF ⎞ ⎟ + Ai ⎥ + ⎬ ⎥ = V1 ⎨ ⋅ ⎢⎜ 1 + R3 ⎠ ⎦ ⎦ R3 ⎪⎭ ⎩⎪ Ri ⎣⎝ ⎛ RF ⎞ ⎜1 + ⎟ R3 ⎠ V1 ⎝ Rif = = I S ⎧⎪ 1 ⎡⎛ RF ⎞ ⎤ 1 ⎫⎪ ⎨ ⋅ ⎢⎜ 1 + ⎟ + Ai ⎥ + ⎬ R3 ⎠ ⎦ R3 ⎭⎪ ⎩⎪ Ri ⎣⎝ ⎡ R I S ⎢1 + F ⎣ R3

The 1/ R3 term in the denominator will be negligible. Using the results of Problem 12.15: 25 Rif = ⎧1 ⎡ 5 ⎫ ⎨ ⎣ (25) + 10 ⎤⎦ ⎬ 2 ⎩ ⎭ Rif ≅ 5 × 10−4 kΩ ⇒ Rif = 0.5 Ω Output Resistance (Let Z L = 0)

IX =

VX VX + Ai I ε + R3 RF + Ri

Iε =

VX RF + Ri

so A + 1 RF IX 1 1 , = = + i = 24 VX R0 f R3 RF + Ri R3 Let RF = 240 kΩ, R3 = 10 k Ω 1 1 105 + 1 = + Rof 10 240 + 2

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so R0 f ≈

RF + Ri 240 + 2 ⇒ R0 f ≈ 2.42 × 10−3 kΩ or R0 f ≈ 2.42 Ω = 5 10 + 1 Ai + 1

12.26 Agf =

Ag 5 = = 0.2 A/V 1 + β z Ag 1 + (4.8)(5)

Rif = Ri [1 + β z Ag ] = (10) [1 + (4.8)(5) ] = 250 kΩ

Rof = Ro [1 + β z Ag ] = (10) [1 + (4.8)(5) ] = 250 kΩ

12.27 Vε = Vi − V fb = 40 − 38 = 2.0 mV I o 8 mA = = 4 A/V Vε 2 mV

Ag =

V fb

βz =

I0

Agf =

=

38 mV = 4.75 V/A 8 mA

I0 8 mA = = 0.2 A/V Vi 40 mV

12.28

IE =

V − Vε (1 + hFE ) ⋅ I0 = S hFE RE

Also I 0 = hFE ( AgVε ) so Vε =

I0 hFE Ag

Then V I0 1 + hFE ⋅ I0 = S − hFE RE hFE Ag RE ⎡1 + hFE ⎤ VS 1 + ⎢ ⎥ I0 = h h A R RE ⎥ FE g E ⎦ ⎣⎢ FE ⎡ Ag (1 + hFE ) RE + 1 ⎤ VS ⎢ ⎥ I0 = hFE Ag RE RE ⎥⎦ ⎣⎢ I0 1 = VS RE

⎡ ⎤ hFE Ag RE hFE Ag I0 ⋅⎢ ≈ ⎥⇒ ⎢⎣1 + Ag (1 + hFE ) RE ⎥⎦ VS 1 + (hFE Ag ) RE

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b.

Bz = RE

c.

10 =

5 × 105 1 + (5 × 105 ) β z

5 × 105 −1 β z = 10 5 ⇒ β z = RE = 0.099998 kΩ 5 × 10 d. If Ag → 5.5 × 105 then

Agf = ΔAgf

Agf

5.5 × 105 = 10.0000182 1 + (5.5 × 105 )(0.099998) =

1.82 × 10−5 ⇒ 1.82 × 10−4 % 10

12.29

I E = (1 + hFE ) AgVε , I E =

Now (1 + hFE ) Ag I S Ri =

Vε − I S and Vε = I S Ri , Vε = VS − Vε = VS − I S Ri RE

1 ⋅ (VS − I S Ri ) − I S RE

⎡ ⎤ V Ri + 1⎥ I S = S ⎢(1 + hFE ) Ag Ri + R RE E ⎣ ⎦ ⎡ ⎤ VS R = RE ⎢(1 + hFE ) Ag Ri + i + 1⎥ IS RE ⎦ ⎣ From Problem 12.16: (1 + hFE ) Ag ≈ hFE Ag = 5 × 105 mS

Rif =

RE ≈ 0.1 kΩ 20 ⎤ ⎡ + 1⎥ so Rif = (0.1) ⎢(5 × 105 )(20) + 0.1 ⎣ ⎦ or Rif = 106 kΩ

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Vπ = AgVε rπ I X = g mVπ +

VX − (−Vε ) R0

(1)

Vε = −( I X + AgVε )( RE || Ri )

(2)

or Vε ⎡⎣1 + Ag ( Rε || Ri ) ⎤⎦ = − I X ( RE || Ri ) Now: V V I X = g m Ag rπ Vε + X + ε (1) R0 R0 ⎛ 1 ⎞ ⎡ − I X ( RE || Ri ) ⎤ VX I X = ⎜ g m Ag rπ + ⎟ ⎢ ⎥+ R0 ⎠ ⎣⎢1 + Ag ( RE || Ri ) ⎦⎥ R0 ⎝ V R0 f = X IX ⎧⎪ ⎛ 1 ⎞ ⎡ ( RE || Ri ) ⎤ ⎫⎪ = R0 ⎨1 + ⎜ g m Ag rπ + ⎟ ⎢ ⎥⎬ R0 ⎠ ⎣⎢1 + Ag ( RE || Ri ) ⎦⎥ ⎭⎪ ⎩⎪ ⎝ g m rπ Ag = hFE Ag = 5 × 105 mS

Let hFE = 100 so Ag = 5× 103 mS RE || Ri = 0.1 || 20 ≈ 0.1 kΩ Then ⎤ ⎫⎪ 1 ⎞⎡ 0.1 ⎪⎧ ⎛ R0 f = 50 ⎨1 + ⎜ 5 × 105 + ⎟ ⎢ ⎥⎬ 3 50 ⎠ ⎣1 + (5 × 10 )(0.1) ⎦ ⎪⎭ ⎪⎩ ⎝ or R0 f = 5.04 MΩ

12.30 Azf =

Az 1 + β g Az

=

5 = 0.2 V/μA 1 + (4.8)(5)

Rif =

Ri 1 = ⇒ Rif = 40 Ω 1 + β g Az 1 + (4.8)(5)

Rof =

Ro 1 = ⇒ Rof = 40 Ω 1 + β g Az 1 + (4.8)(5)

12.31

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I ε = I i − I fb = 40 − 38 = 2 μA Az =

Vo 8 V = =4 Iε 2 μA

βg = Azf =

I fb Vo

=

38 μA = 4.75 8 V

Vo 8 V = = 0.2 I i 40 μA

12.32 a.

Assuming V1 is at virtual ground (−V0 ) = − I fb RF and (−V0 ) = − Az Iε ⇒ I ε =

V0 Az

I fb = I S − I ε ⎛V ⎞ So V0 = ( I S − I ε ) RF = I S RF − ⎜ 0 ⎟ RF ⎝ Az ⎠ ⎡ R ⎤ V0 ⎢1 + F ⎥ = I S RF ⎣ Az ⎦ V RF AR so Azf = 0 = = z F I S ⎡ RF ⎤ Az + RF ⎢1 + A ⎥ z ⎦ ⎣ Az Az = or Azf = ⎛ 1 ⎞ 1 + Az β g 1 + Az ⎜ ⎟ ⎝ RF ⎠ 1 RF

b.

βg =

c.

5 × 104 =

5 × 106 1 + (5 × 106 ) β g

5 × 106 −1 4 β g = 5 × 10 6 ⇒ β g = 1.98 × 10−5 5 × 10 1 RF = ⇒ RF = 50.5 kΩ

βg

d.

Az = (0.9)(5 × 106 ) = 4.5 × 106

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Azf = ΔAzf Azf

4.5 × 106 = 4.994 × 104 1 + (4.5 × 106 )(1.98 × 10−5 ) =−

55.4939 = −1.11× 10−3 ⇒ −0.111% 5 × 104

12.33

V1 = I ε Ri , − V0 = − Az I ε ⇒ V0 = Ax I ε I fb = I S − I ε and −V0 = V1 − I fb RF

− Az I ε = V1 − ( I S − I ε ) RF ⎛V ⎞ ⎛V ⎞ − Az ⎜ 1 ⎟ = V1 − I S RF + ⎜ 1 ⎟ RF ⎝ Ri ⎠ ⎝ Ri ⎠ ⎡ A R ⎤ I S RF = V1 ⎢1 + z + F ⎥ ⎣ Ri Ri ⎦ V RF Rif = 1 = I S ⎡ Az RF ⎤ ⎢1 + R + R ⎥ i i ⎦ ⎣ Or, using the results from problem 12.18. 50.5 × 103 Rif = ⎡ 5 × 106 50.5 × 103 ⎤ ⎢1 + 10 × 103 + 10 × 103 ⎥ ⎣ ⎦ =

50.5 × 103 ⇒ Rif = 99.79 Ω [1 + 500 + 5.05]

12.34 Assume I CQ = 0.2 mA Then rπ =

(100)(0.026) = 13 k Ω 0.2

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(1)

⎛ 1 1 Vi − VA VA VA − Vo V V 1 ⎞ = + ⇒ i + o = VA ⎜ + + ⎟ Ri R1 R2 Ri R2 ⎝ Ri R1 R2 ⎠

Now Vi Vo ⎛1 1 1⎞ + = VA ⎜ + + ⎟ ⇒ Vi + Vo = VA (12) 10 10 ⎝ 10 1 10 ⎠ 1 or VA = (Vi + Vo ) 12 ⎛ AvVε − Vo ⎞ Vo − VA (2) ⎜ ⎟ (1 + hFE ) = R2 ⎝ Ro + rπ ⎠ where Vε = Vi − VA Then ⎛ Av (Vi − VA ) − Vo ⎞ Vo − VA ⎜ ⎟ (1 + hFE ) = R0 + rπ R2 ⎝ ⎠ we find AvVi (1 + hFE ) Vo (1 + hFE ) Vo AvVA (1 + hFE ) VA − − = − Ro + rπ Ro + rπ R2 Ro + rπ R2 Then (5 × 103 )(101)Vi Vo (101) Vo ⎛ (5 × 103 )(101) 1 ⎞ − − =⎜ − ⎟ VA 14 14 10 ⎝ 14 10 ⎠ Rearranging terms, we find V Avf = o =10.97 Vi ⎛ Vi ⎞ Vi Vi = =⎜ ⎟ Ri I i ⎛ Vi − VA ⎞ ⎝ Vi − VA ⎠ ⎜ ⎟ ⎝ Ri ⎠ 1 1 VA = (Vi + Vo ) = (Vi + 10.97Vi ) = 0.9975Vi 12 12 Then 1 ⎛ ⎞ Rif = ⎜ ⎟ (10 k Ω) ⇒ Rif = 4 M Ω ⎝ 1 − 0.9975 ⎠ To find the output resistance: Rif =

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Ix +

( AvVε )(1 + hFE ) Vx = Ro + rπ R2 + R1 || Ri

⎛ R1 || Ri ⎞ Vε = − ⎜ ⎟ ⋅ Vx ⎝ R1 || Ri + R2 ⎠ Now R1 || Ri = 1 || 10 = 0.909 Then Vε = −0.0833Vx Now 3 1 ⎪⎧ ⎡ (5 × 10 )(0.0833) + 1 ⎤ ⎪⎫ I x = Vx ⎨ ⎢ ⎬ ⎥ (101) + 1 + 13 10 + 0.909 ⎭⎪ ⎦ ⎩⎪ ⎣

= Vx {3.012 × 103 + 0.0917}

Or Vx = Rof = 3.32 × 10−4 k Ω ⇒ Rof = 0.332 Ω Ix 12.35 a. Neglecting base currents I C 2 = 0.5 mA, VC 2 = 12 − (0.5)(22.6) = 0.7 V I C1 = 0.5 mA ⇒ v0 = 0

Then I C 3 = 2 mA b.

rπ 1 = rπ 2 =

hFE ⋅ VT (100)(0.026) = = 5.2 kΩ I C1 0.5

0.5 = 19.23 mA / V 0.026 (100)(0.026) rπ 3 = = 1.3 kΩ 2 2 g m3 = = 76.92 mA / V 0.026 g m1 = g m 2 =

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Vπ 1 V + g m1Vπ 1 + g m1Vπ 2 + π 2 = 0 rπ 1 rπ 1

⎛ 1 ⎞ (Vπ 1 + Vπ 2 ) ⎜ + g m1 ⎟ = 0 ⇒ Vπ 1 = −Vπ 2 ⎝ rπ 1 ⎠ Vπ 1 ( RS + rπ 1 ) − Vπ 2 + Vb 2 Vi = rπ 1

(1)

⎛ R ⎞ or Vi = Vπ 1 ⎜ 1 + S ⎟ − Vπ 2 + Vb 2 ⎝ rπ 1 ⎠ But Vπ 2 = −Vπ 1 so ⎛ R ⎞ Vi = Vπ 1 ⎜ 2 + S ⎟ + Vb 2 (2) rπ 1 ⎠ ⎝ V02 V −V + g m1Vπ 2 + 02 0 = 0 RC rπ 3

(3)

Vπ 3 V V −V + g m 3Vπ 3 = 0 + 0 b 2 rπ 3 RL R2 Vπ 3 = V02 − V0 so ⎛ 1 + hFE (V02 − V0 ) ⎜ ⎝ rπ 3

⎞ ⎛ 1 1 ⎞ V + ⎟ − b2 ⎟ = V0 ⎜ R R R2 ⎝ L 2 ⎠ ⎠

(4)

Vb 2 − V0 Vb 2 Vπ 2 + + = 0 (5) R2 R1 rπ 1 Substitute numbers into (2), (3), (4) and (5): 1 ⎞ ⎛ Vi = −Vπ 2 ⎜ 2 + ⎟ + Vb 2 5.2 ⎠ ⎝ Vi = −Vπ 2 (2.192) + Vb 2 (2) 1 ⎞ ⎛ 1 ⎛ 1 ⎞ + V02 ⎜ ⎟ + (19.23)Vπ 2 − V0 ⎜ ⎟ = 0 ⎝ 22.6 1.3 ⎠ ⎝ 1.3 ⎠ V02 (0.8135) + (19.23)Vπ 2 − (0.7692)V0 = 0 (3)

⎛ 101 ⎞ ⎛ 101 1 1 ⎞ ⎛ 1 ⎞ V02 ⎜ + + ⎟ − Vb 2 ⎜ ⎟ ⎟ = V0 ⎜ ⎝ 1.3 ⎠ ⎝ 1.3 4 50 ⎠ ⎝ 50 ⎠ V02 (77.69) = V0 (77.96) − Vb 2 (0.02) (4) ⎛ 1 1⎞ ⎛ 1 ⎞ ⎛ 1 ⎞ Vb 2 ⎜ + ⎟ − V0 ⎜ ⎟ + Vπ 2 ⎜ ⎟=0 ⎝ 50 10 ⎠ ⎝ 50 ⎠ ⎝ 5.2 ⎠ Vb 2 (0.120) − V0 (0.020) + Vπ 2 (0.1923) = 0

(5)

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From (2): Vb 2 = Vi + Vπ 2 (2.192). Substitute in (4) and (5) to obtain: V02 (77.69) = V0 (77.96) − [Vi + Vπ 2 (2.192)](0.02) (4′) [Vi + Vπ 2 (2.192)](0.120) − V0 (0.020) + Vπ 2 (0.1923) = 0 (5′) So we now have the following three equations: V02 (0.8135) + (19.23)Vπ 2 − (0.7692)V0 = 0 (3) V02 (77.69) = V0 (77.96) − Vi (0.02) − Vπ 2 (0.04384) (4′) (0.120)Vi + Vπ 2 (0.4553) − V0 (0.020) = 0 (5′) From (3): V02 = V0 (0.9455) − Vπ 2 (23.64). Substitute for V02 in (4′) to obtain: (77.69)[Vo (0.9455) − Vπ 2 (23.64)] = V0 (77.96) − Vi (0.02) − Vπ 2 (0.04384) or 0 = V0 (4.504) − Vi (0.02) + Vπ 2 (1836.5) Next, solve (5′) for Vπ 2 : (0.120)Vi + Vπ 2 (0.4553) − V0 (0.020) = 0 Vπ 2 = V0 (0.04393) − Vi (0.2636) Finally, 0 = V0 (4.504) − Vi (0.02) + (1836.5)[V0 (0.04393) − Vi (0.2636)] 0 = V0 (85.18) − Vi (484.12) So V 484.12 ⇒ Avf = 5.68 Avf = 0 = Vi 85.18

12.36 a.

RTH = R1 || R2 = 400 || 75 = 63.2 kΩ

⎛ R2 ⎞ ⎛ 75 ⎞ VTH = ⎜ ⎟ VCC = ⎜ ⎟ (10) = 1.579 V + R R 75 + 400 ⎠ ⎝ ⎝ 1 2 ⎠ 1.579 − 0.7 I BQ1 = = 0.007106 mA 63.2 + (121)(0.5) I CQ1 = 0.853 mA

VC1 = 10 − (0.853)(8.8) = 2.49 V 2.49 − 0.7 = 0.497 mA 3.6 = 10 − (0.497)(13) = 3.54 V

IC 2 ≈ VC 2

IC 3 ≈

3.54 − 0.7 = 2.03 mA 1.4

Then

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(120)(0.026) = 3.66 kΩ 0.853 0.853 g m1 = = 32.81 mA / V 0.026 (120)(0.026) rπ 2 = = 6.28 kΩ 0.497 0.497 gm2 = = 19.12 mA / V 0.026 (120)(0.026) rπ 3 = = 1.54 kΩ 2.03 2.03 g m3 = = 78.08 mA / V 0.026 b. rπ 1 =

Vi = Vπ 1 + Vε 1 ⇒ Vε 1 = Vi − Vπ 1

(1)

Vπ 1 V V −V + g m1Vπ 1 = ε 1 + ε 1 0 rπ 1 RE1 RF

(2)

Vπ 2 = −( g m1Vπ 1 )( RC1 || rπ 2 )

(3)

g m 2Vπ 2 +

Vπ 3 + V0 Vπ 3 + =0 RC 2 rπ 3

(4)

Vπ 3 V V −V + g m 3Vπ 3 = 0 + 0 ε 1 (5) rπ 3 RE 3 RF Substitute numbers in (2), (3), (4) and (5): 1⎞ V ⎛ 1 ⎞ ⎛ 1 Vπ 1 ⎜ + 32.81⎟ = (Vi − Vπ 1 ) ⎜ + ⎟− 0 ⎝ 3.66 ⎠ ⎝ 0.5 10 ⎠ 10 (2) or Vπ 1 (35.18) = Vi (2.10) − V0 (0.10) Vπ 2 = −(32.81)Vπ 1 (88 || 6.28) or Vπ 2 = −Vπ 1 (120.2) (3) (19.12)Vπ 2 +

Vπ 3 V0 Vπ 3 + + =0 13 13 1.54

or Vπ 2 (19.12) + Vπ 3 (0.7263) + V0 (0.07692) = 0 (4) 1 ⎞ V −V ⎛ 1 ⎞ ⎛ 1 Vπ 3 ⎜ + 78.08 ⎟ = V0 ⎜ + ⎟ − i π1 10 ⎝ 1.54 ⎠ ⎝ 1.4 10 ⎠ or Vπ 3 (78.73) = V0 (0.8143) − Vi (0.10) + Vπ 1 (0.10) (5) Now substituting Vπ 2 = −Vπ 1 (120.2) in (4): (19.12)[−Vπ 1 (120.2)] + Vπ 3 (0.7263) + V0 (0.07692) = 0

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or −Vπ 1 (2298.2) + Vπ 3 (0.7263) + V0 (0.07692) = 0 Then Vπ 3 = Vπ 1 (3164.3) − V0 (0.1059) Substituting Vπ 3 = Vπ 1 (3164.3) − V0 (0.1059) in (5): (78.73)[Vπ 1 (3164.3) − V0 (0.1059)] = V0 (0.8143) − Vi (0.10) + Vπ 1 (0.10) or Vπ 1 (2.49 × 105 ) − V0 (9.152) = −Vi (0.10) Then Vπ 1 = V0 (3.674 × 10−5 ) − Vi (4.014 × 10−7 ) Now substituting Vπ 1 = V0 (3.674 × 10−5 ) −Vi (4.014 × 10−7 ) in (2):

(35.18)[V0 (3.674) ×10−5 ) − Vi (4.014 ×10 −7 )] = Vi (2.10) − V0 (0.10) or V0 (0.1013) = Vi (2.10) So

V0 = 20.7 Vi Rif =

c. I RB1 = I b1 =

Vi and I i = I RB1 + I b1 Ii

Vi RB1

Vπ 1 rπ 1

Now Vπ 1 = (20.7Vi )(3.674 × 10−5 ) − Vi (4.014 × 10−7 ) Vπ 1 = Vi (7.60 × 10−4 ) Then Vi Rif = Vi V (7.60 × 10−4 ) + i 63.2 3.66 1 = 0.01582 + 2.077 × 10−4

or Rif = 62.4 kΩ d.

To determine R0 f :

Equation (1) is modified to Vπ 1 + Ve1 = 0 (Vi = 0) Equation (5) is modified to: Vπ 3 (78.73) + I X = V0 (0.8143) + Vπ 1 (0.10) (5) Now Vπ 1 (35.18) = −V0 (0.10) (2) Vπ 2 = −Vπ 1 (120.2) (3) Vπ 2 (19.12) + Vπ 3 (0.7263) + V0 (0.07692) = 0 (4) Now Vπ 1 = −V0 (0.002843) so

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Vπ 2 = −(−V0 )(0.002843)(120.2) Vπ 2 = V0 (0.3417) Then V0 (0.3417)(19.12) + Vπ 3 + (0.7263) + V0 (0.07692) = 0 or Vπ 3 = −V0 (9.101) (4) So then −V0 (9.101)(78.73) + I X = V0 (0.8143) + (0.10)(−V0 )(0.002843) or I X = V0 (717.3) (5) or V R0 f = 0 = 0.00139 kΩ ⇒ R0 f = 1.39 Ω IX

12.37 (a)

(1)

Vi − VA V −V V + g m1Vπ 1 = A + A O rπ 1 RE RF

(2)

VB V + g m1Vπ 1 + B = 0 RC1 rπ 2

(3)

VC V − Vo + g m 2Vπ 2 + C =0 RC 2 rπ 3

(4)

g m 3Vπ 3 +

VC − VO VO − VA = rπ 3 RF

Vπ 1 = Vi − VA Vπ 2 = VB Vπ 3 = VC − VO

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⎛ 1 ⎞ V V −V + g m1 ⎟ = A + A O r R RF E ⎝ π1 ⎠

(1)

(Vi − VA ) ⎜

(2)

⎛ 1 1 ⎞ VB ⎜ + ⎟ + g m1 (Vi − VA ) = 0 R r ⎝ C1 π 2 ⎠

(3)

⎛ 1 VO 1 ⎞ VC ⎜ + =0 ⎟ + g m 2VB − rπ 3 ⎝ RC 2 rπ 3 ⎠

(4)

(VC − VO ) ⎜ g m3 +

(1) (2) (3) (4)

(Vi − VA )(555.5) = VA (20) + (VA − VO )(0.8333) VB (5.109) + 550(Vi − VA ) = 0 VC (3.257) + 178VB − VO (1.718) = 0 (VC − VO )(173.7) = (VO − VA )(0.8333)

(1) (2) (3) (4)

Vi (555.5) + VO (0.8333) = VA (576.3) VB (5.109) + 550Vi − VA (550) = 0 VC (3.257) + 178VB − VO (1.718) = 0 VC (173.7) + VA (0.8333) = VO (174.5)

⎛

1 ⎞ VO − VA ⎟= rπ 3 ⎠ RF

⎝ (100)(0.026) 14.3 rπ 1 = = 0.182 K g m1 = = 550 mA/V 14.3 0.026 (100)(0.026) 4.62 rπ 2 = = 0.563 K g m 2 = = 178 mA/V 4.62 0.026 (100)(0.026) 4.47 rπ 3 = = 0.582 K g m 3 = = 172 mA/V 4.47 0.026 V −V 1 ⎞ V + 550 ⎟ = A + A O (1) (Vi − VA ) ⎛⎜ 1.2 ⎝ 0.182 ⎠ 0.05 1 1 ⎛ ⎞ (2) VB ⎜ + ⎟ + (550)(Vi − VA ) = 0 ⎝ 0.3 0.563 ⎠ VO 1 ⎞ ⎛ 1 VC ⎜ + =0 (3) ⎟ + 178 VB − 0.65 0.582 0.582 ⎝ ⎠ 1 ⎞ VO − VA (4) (VC − VO ) ⎛⎜172 + ⎟= 0.582 1.2 ⎝ ⎠

VB = VA (107.7) − Vi (107.7) From (2) From (4) VC = VO (1.0046) − VA (0.004797) Substitute into (3) (3.257) [VO (1.0046) − VA (0.004797) ] +(178)[VA (107.7) − Vi (107.7)] − Vo (1.718) = 0 VO (3.272) − VA (0.01562) + VA (19170.6) − Vi (19170.6) − VO (1.718) = 0 VA (19170.6) = Vi (19170.6) − VO (1.554) VA = Vi (1.00) − VO (0.00008106) Substitute into (1)

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Vi (555.5) + Vo (0.8333) = (576.3) [Vi (1.00) − Vo (0.00008106) ] = Vi (576.3) − Vo (0.0467) Vo (0.880) = Vi (20.8) Vo = Avf = 23.6 Vi Ideal R + RE 1.2 + 0.05 = = 25.0 Avf = F 0.05 RE Rif =

(b)

Vi V V − VA and I i = π 1 = i rπ 1 rπ 1 Ii

We have VA = Vi (1.00) − Vo (0.00008106) = Vi (1.00) − (23.6)Vi (0.00008106) VA = Vi (0.99809) Vi (1 − 0.99809) = Vi (0.01051) 0.182 Vi Rif = ⇒ Rif = 95.1 K Vi (0.01051)

Then I i =

To find Rof , set Vi = 0 I X + g m 3Vπ 3 +

Vπ 3 Vx − VA = rπ 3 RF

Vπ 3 = VC − VX I X + (VC − VX )( g m 3 +

V − VA 1 )= X rπ 3 RF

For Vi = 0, we have VC = VX (1.0046) − VA (0.004797) VA (576.3) = VX (0.8333) VA = VX (0.001446) VC = VX (1.0046) − VX (0.001446)(0.004797) VC = VX (1.0046) 1 ⎞ VX (1 − 0.004797) ⎛ I X + VX (1.0046 − 1.0) ⎜172 + ⎟= 0.582 1.2 ⎝ ⎠ I X + VX (0.7991) = VX (0.8293) I X = VX (0.03024) Rof =

VX = 33.1 K IX

12.38 RD1 = RE1 = 8 K VGG = VGS + ( I D + I C ) RL ID =

10 − VD 8

IC =

10 − (VD + 0.7) 8

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VGG =

4.5 = 36 =

ID + VTN + ( I D + I C ) RL kn

(10 − VD )

⎛ 10 − VD ⎞ ⎛ 9.3 − VD ⎞ +1+ ⎜ ⎟ (1.8) + ⎜ ⎟ (1.8) 8(0.4) 8 ⎠ ⎝ 8 ⎠ ⎝

8 3.2

10 − VD + 8 + 18 − 1.8VD + 16.74 − 1.8VD

36 = 4.472 10 − VD + 42.74 − 3.6VD 3.6VD − 6.74 = 4.472 10 − VD 12.96VD2 − 48.528VD + 45.4276 = 19.999(10 − VD ) 12.96VD2 − 28.528VD − 154.6 = 0 28.528 ± 813.847 + 4(12.96)(154.6) 2(12.96) VD = 4.726 V VD =

10 − 4.726 = 0.6593 mA 8 9.3 − 4.726 IC = 8 = 0.5718 mA 0.6593 VGS = + 1 = 2.284 V 0.4 VO = 4.5 − 2.284 = 2.216 V ID =

VDS = VD − VO = 4.726 − 2.216 = 2.51V

(V ( sat ) = 1.28 V ) DS

Also VO = ( I D + I C ) RL = (0.6593 + 0.5718)(1.8) = 2.216 V OK VECQ 2 = (4.726 + 0.7) − 2.216 = 3.21 V (b)

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(1)

Vi = Vgs + Vo

(2)

V VA + g m1Vgs = π RD1 rπ 2

(3)

VB Vπ + + g m 2Vπ = 0 RE 2 rπ 2 Vπ = VB − VA

Vgs = Vi − Vo

(2)

VA V − VA + g m1 (Vi − Vo ) = B RD1 rπ 2

(3)

VB VB − VA + + g m 2 (VB − VA ) = 0 RE 2 rπ 2

(4)

g m1Vgs + g m 2Vπ =

VO RL

g m1 (Vi − Vo ) + g m 2 (VB − VA ) =

VO RL

g m1 = 2 K n I D1 = 2 (0.4)(0.6593) = 1.027 mA/V gm2 = rπ 2 =

I C 2 0.5718 = = 21.99 mA/V VT 0.026

(100)(0.026) = 4.547 K 0.5718

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(2) (3)

VA V − VA + (1.027)(Vi − Vo ) = B 8 4.547 VB VB − VA + + (21.99)(VB − VA ) = 0 8 4.547 0.125VA + 1.027Vi − 1.027Vo = 0.2199VB − 0.2199VA

(2) (3) 0.125VB + 0.2199VB − 0.2199VA + 21.99VB − 21.99VA = 0 (2) 0.3449VA − 0.2199VB = −1.027Vi + 1.027Vo (3) 22.21VA = 22.335VB VA = 1.0056VB Then (0.3449)(1.0056)VB − 0.2199VB = −1.027Vi + 1.027Vo 0.1269VB = −1.027Vi + 1.027Vo VB = −8.093Vi + 8.093Vo VA = −8.138Vi + 8.138Vo Then (4)

(1.027)(Vi − Vo ) + (21.99) [ −8.093Vi + 8.093Vo + 8.138Vi − 8.138Vo ] =

Vo 1.8

1.027Vi − 1.027Vo − 177.965Vi + 177.965Vo + 178.955Vi − 178.955Vo = 0.5556Vo 2.017Vi − 2.5726Vo = 0 Vo = 0.784 = AV Vi

Set Vi = 0 I X + g m 2Vπ + g m1Vgs =

VX RL

I X + g m 2 (VB − VA ) + g m1 (−VX ) =

VX RL

VB = 8.093VX VA = 8.138VX I X + (21.99) [8.093VX − 8.138VX ] + 1.027( −VX ) = VX (0.5556) I X = 2.572VX VX = Rof = 0.389 K IX

12.39 a.

g m = 2 K n I DQ = 2 (0.5)(0.5) = 1 mA / V

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V0 = ( g mVgs ) RS

Vi = Vgs + V0 so Vgs = Vi − V0 Then V0 = g m RS (Vi − V0 ) Av =

g m RS 1(2) = ⇒ Av = 0.667 1 + g m RS 1 + (1)(2)

To determine R0 f I X + g mVgs =

VX and Vgs = −VX RS

IX 1 1 = = gm + VX R0 f RS so R0 f =

1 1 RS = 2 ⇒ R0 f = 0.667 kΩ 1 gm

For K n = 0.8 mA / V 2

b.

g m = 2 (0.8)(0.5) = 1.265 mA / V Av =

(1.265)(2) = 0.7167 1 + (1.265)(2)

ΔAf

=

Af

0.7167 − 0.667 ⇒ 7.45% increase 0.667

1 || 2 = 0.7905 || 2 1.265 = 0.5666 kΩ

R0 f = R0 f

ΔR0 f R0 f

=

0.5666 − 0.667 ⇒ 15.05% decrease 0.667

12.40 dc analysis: RTH 1 = 150 || 47 = 35.8 kΩ , ⎛ 47 ⎞ VTH 1 = ⎜ ⎟ (25) = 5.96 V ⎝ 47 + 150 ⎠ RTH 2 = 33 || 47 = 19.4 kΩ , ⎛ 33 ⎞ VTH 2 = ⎜ ⎟ (25) = 10.3 V ⎝ 33 + 47 ⎠ 5.96 − 0.7 I B1 = = 0.0187 mA 35.8 + (51)(4.8) I C1 = (50)(0.0187) = 0.935 mA 10.3 − 0.7 = 0.03705 mA 19.4 + (51)(4.7) = (50)(0.03705) = 1.85 mA

I B2 = IC 2

(50)(0.026) = 1.39 kΩ; 0.935 (50)(0.026) = 0.703 kΩ rπ 2 = 1.85 rπ 1 =

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0.935 = 35.96 mA / V 0.026 1.85 = = 71.15 mA / V 0.026

g m1 = gm2

VS = Vπ 1 + Ve

(1)

Vπ 1 V V − V0 + g m1Vπ 1 = e + e rπ 1 R1 RF

(2)

g m1Vπ 1 +

Vπ 2 Vπ 2 Vπ 2 + + =0 RC1 RB 2 rπ 2

(3)

V0 V0 − Ve (4) + =0 RC 2 RF Substitute numerical values in (2), (3) and (4): Ve = VS − Vπ 1 (1) g m 2Vπ 2 +

Vπ 1 1 ⎞ ⎛ 1 ⎛ 1 ⎞ + (35.96)Vπ 1 = (VS − Vπ 1 ) ⎜ + ⎟ − V0 ⎜ ⎟ 1.39 ⎝ 0.1 4.7 ⎠ ⎝ 4.7 ⎠ or Vπ 1 (46.89) = VS (10.213) − V0 (0.2128) (2) 1 1 ⎞ ⎛1 + (35.96)Vπ 1 + Vπ 2 ⎜ + ⎟=0 ⎝ 10 19.4 0.703 ⎠ or (35.96)Vπ 1 + Vπ 2 (1.574) = 0

(3)

1 ⎞ ⎛ 1 ⎛ 1 ⎞ + (71.15)Vπ 2 + V0 ⎜ ⎟ − (VS − Vπ 1 ) ⎜ ⎟=0 ⎝ 4.7 4.7 ⎠ ⎝ 4.7 ⎠ or (71.15)Vπ 2 + V0 (0.4255) − VS (0.2128) + Vπ 1 (0.2128) = 0 (4) From (3): Vπ 2 = −Vπ 1 (22.85) Then substitute in (4): −(71.15)Vπ 1 (22.85) + V0 (0.4255) − VS (0.2128) + Vπ 1 (0.2128) = 0 or −Vπ 1 (1625.6) + V0 (0.4255) − VS (0.2128) = 0 From (2): Vπ 1 = VS (0.2178) − V0 (0.004538) Then −(1625.6)[VS (0.2178) − V0 (0.004538)] + V0 (0.4255) − VS (0.2128) = 0 or −VS (354.3) + V0 (7.802) = 0 Finally V ⇒ 0 = 45.4 VS

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12.41 For example, use a z-stage amplifier. Each stage shown in Fig. 12.29. 12.42

For M 3 : K n 3 =

kn′ 2

⎛W ⎞ ⎛W ⎞ ⋅ ⎜ ⎟ Let ⎜ ⎟ = 25 L ⎝ ⎠3 ⎝ L ⎠3

⎛ 0.080 ⎞ 2 Then K n 3 = ⎜ ⎟ (25) = 1 mA / V ⎝ 2 ⎠ Want vo = 0 for vi = 0, so that

I D 3 = I Q 2 = 0.4 = 1 ⋅ (VGS 3 − VTN ) 2 0.4 + 2 = 2.63 V 1 For VGS 3 = 2.63V ⇒ VG 3 = 12 − I D 2 RD Or 2.63 = 12 − (0.1) RD ⇒ RD = 93.7 k Ω

Then VGS 3 =

g m 3 = 2 K n 3 I D 3 = 2 (1)(0.4) = 1.26 mA / V ⎛ R1 ⎞ ⎛ 10 ⎞ VA = ⎜ ⎟ (Vo ) = ⎜ ⎟ (Vo ) = 0.0769Vo + R R ⎝ 120 + 10 ⎠ 2 ⎠ ⎝ 1 (Small amount of feedback) (1) Vi = Vgs1 − Vgs 2 + VA

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g mVgs1 + g mVgs 2 = 0 ⇒ Vgs1 = −Vgs 2

(2) Then

Vi = −2Vgs 2 + VA ⇒ Vgs 2 =

1 (VA − Vi ) 2

Vgs 2 = 0.0384 Vo − 0.5Vi

(3)

VB = − g mVgs 2 RD = − g m RD [0.03846Vo − 0.5Vi ]

(4)

Vgs 3 = VB − Vo and Vo = g m 3Vgs 3 [ RL || ( R1 + R2 )]

So Vo = g m 3 [ RL || ( R1 + R2 ) ] (VB − Vo ) Then Vo = g m 3 ⎡⎣ RL || ( R1 + R2 ) ⎤⎦ ⎡⎣ − g m RD ( 0.03846Vo − 0.5Vi ) − Vo ⎤⎦ Or Vo ⎡⎣1 + g m 3 [ RL || ( R1 + R2 )][ g m RD (0.03846) + 1]⎤⎦ = g m 3 [ RL || ( R1 + R2 ) ][ 0.5 g m RD ] ⋅ Vi Now RL || ( R1 + R2 ) = 4 ||130 = 3.88 So Vo ⎡⎣1 + (1.26)(3.88) [ g m (93.7)(0.03846) + 1]⎤⎦ = (1.26)(3.88)(0.5) g m (93.7)Vi Rearranging terms, we find Vo 229 g m = = 10 ⇒ g m = 1.11 mA / V Vi 5.89 + 17.6 g m We have ⎛ k ′ ⎞⎛ W gm = 2 ⎜ n ⎟ ⎜ ⎝ 2 ⎠⎝ L

⎞ ⎛ 0.080 ⎞ ⎛ W ⎟ I D ⇒ 1.11 = 2 ⎜ ⎟⎜ ⎠ ⎝ 2 ⎠⎝ L

⎞ ⎛W ⎞ ⎛W ⎞ ⎟ (0.1) ⇒ ⎜ ⎟ = ⎜ ⎟ = 77 ⎠ ⎝ L ⎠1 ⎝ L ⎠ 2

12.43 Assuming an ideal op-amp, then from Equation (12.58) I0 R 20 = 1+ 1 = = 100 IS R2 0.2 Then

R1 = 99 R2

For example, set R2 = 5 kΩ and R1 = 495 kΩ 12.44 ⎛ h ⎞ ⎛ 100 ⎞ I C1 = ⎜ FE ⎟ I E1 = ⎜ ⎟ (0.2) = 0.198 mA ⎝ 101 ⎠ ⎝ 1 + hFE ⎠ VC1 = 10 − (0.198)(40) = 2.08 V

(a)

2.08 − 0.7 = 1.38 mA 1 ⎛ 100 ⎞ =⎜ ⎟ (1.38) = 1.37 mA ⎝ 101 ⎠

IE2 = IC 2

For Q1 : (100)(0.026) = 13.1 k Ω 0.198 0.198 = = 7.62 mA / V 0.026

rπ 1 = g m1

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For Q2 : (100)(0.026) = 1.90 k Ω 1.37 1.37 = = 52.7 mA / V 0.026

rπ 2 = gm2

(b)

IS =

Vπ 1 Vπ 1 Vπ 1 − Ve + + RS rπ 1 RF

g m1Vπ 1 +

Vπ 2 + Ve Vπ 2 + =0 RC1 rπ 2

(1) (2)

Vπ 2 V V −V (3) + g m 2Vπ 2 = e + e π 1 rπ 2 RE RF Substitute numerical values in (1), (2), and (3): 1 1⎞ ⎛1 ⎛1⎞ + ⎟ − Ve ⎜ ⎟ I S = Vπ 1 ⎜ + 10 13.1 10 ⎝ ⎠ ⎝ 10 ⎠ I S = Vπ 1 (0.2763) − Ve (0.10) (1) 1 ⎞ ⎛ 1 ⎛ 1 ⎞ (7.62)Vπ 1 + Vπ 2 ⎜ + ⎟ + Ve ⎜ ⎟ = 0 ⎝ 40 1.90 ⎠ ⎝ 40 ⎠ (7.62)Vπ 1 + Vπ 2 (0.5513) + Ve (0.025) = 0

(2)

⎛ 1 ⎞ ⎛1 1 ⎞ ⎛ 1⎞ + 52.7 ⎟ = Ve ⎜ + ⎟ − Vπ 1 ⎜ ⎟ Vπ 2 ⎜ ⎝ 1.90 ⎠ ⎝ 1 10 ⎠ ⎝ 10 ⎠ (3) Vπ 2 (53.23) = Ve (1.10) − Vπ 1 (0.10) From (3), Ve = Vπ 2 (48.39) + Vπ 1 (0.0909) Substituting into (1), I S = Vπ 1 (0.2763) − (0.10) [Vπ 2 (48.39) + Vπ 1 (0.0909)]

or I S = Vπ 1 (0.2672) − Vπ 2 (4.839) (1′) and substituting into (2), (7.62)Vπ 1 + Vπ 2 (0.5513) + ( 0.025 ) ⎡⎣Vπ 2 ( 48.39 ) + Vπ 1 ( 0.0909 ) ⎤⎦ = 0 or (7.622)Vπ 1 + Vπ 2 (1.761) = 0 ⇒ Vπ 1 = −Vπ 2 (0.2310) Then substituting (2′) into (1′), we obtain I S = (0.2672)(−Vπ 2 )(0.2310) − Vπ 2 (4.839) or I S = −Vπ 2 (4.901) Now

(2′)

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⎛ RC 2 ⎞ I O = − g m 2Vπ 2 ⎜ ⎟ ⎝ RC 2 + RL ⎠ ⎛ 2 ⎞ = −(52.7) ⎜ ⎟ Vπ 2 = −(42.16)Vπ 2 ⎝ 2 + 0.5 ⎠ Then ⎛ −IS ⎞ I O = −(42.16) ⎜ ⎟ ⎝ 4.901 ⎠ or I Aif = O = 8.60 Is

(c)

Ri =

Vπ 1 and Ri = RS || Rif IS

We had Vπ 1 = −Vπ 2 (0.2310) and I S = −Vπ 2 (4.901) so ⎛ −Vπ 1 ⎞ IS = − ⎜ ⎟ (4.901) = Vπ 1 (21.22) ⎝ 0.2310 ⎠ Then V 1 = 0.04713 Ri = π 1 = IS 21.22 Finally 10 Rif 0.04713 = ⇒ Rif = 47.4 Ω 10 + Rif 12.45 (a) Vπ 1 V −V + π 1 e2 RF RS RB1 rπ 1

(1)

Ii =

(2)

g m1Vπ 1 +

(3)

Vπ 2 V V −V + g m 2Vπ 2 = e 2 + e 2 π 1 rπ 2 RE 2 RF

(4)

⎛ RC 2 ⎞ I o = −( g m 2Vπ 2 ) ⎜ ⎟ ⎝ RC 2 + RL ⎠

VC1 V + π2 = 0 RC1 RB 2 rπ 2

Now (1)′

Ii =

Vπ 1 V − e2 RS RB1 rπ 1 RF RF

⎛ ⎞ RF ⎟V − I R So Ve 2 = ⎜ ⎜ RS RB1 rπ 1 RF ⎟ π 1 i F ⎝ ⎠ Now, from (2) V + Ve 2 Vπ 2 g m1Vπ 1 + π 2 + =0 Rc1 RB 2 rπ 2

(2)′

⎛ Vπ 2 Ve 2 1 ⎞ + =0 ⎜ g m1 + ⎟ Vπ 1 + r R R R r C1 RB 2 π2 ⎠ ⎝ C1 B2 π 2

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Also (3)′

⎛ ⎛ 1 Vπ 1 1 ⎞ 1 ⎞ = Ve 2 ⎜ + ⎜ gm2 + ⎟ Vπ 2 + ⎟ rπ 2 ⎠ RF ⎝ RE 2 RF ⎠ ⎝

And ⎛ I ⎞ ⎛ R + RL ⎞ Vπ 2 = − ⎜ O ⎟ ⎜ C 2 ⎟ ⎝ g m 2 ⎠ ⎝ RC 2 ⎠ Substitute (1)′ into (2)′ and (3)′

(4)′

⎡ ⎤ RF ⎢ − Vπ 1 − I i RF ⎥ = 0 ⎢ RS RB1 rπ 1 RF ⎥ ⎣ ⎦ ⎡ ⎤ ⎛ Vπ 1 ⎛ 1 RF 1 ⎞ 1 ⎞ =⎜ + − Vπ 1 − I i RF ⎥ (3)″ ⎜ gm2 + ⎟ Vπ 2 + ⎟⎢ rπ 2 ⎠ RF ⎝ RE 2 RF ⎠ ⎢ RS RB1 rπ 1 RF ⎥ ⎝ ⎣ ⎦ Solve for Vπ 1 from (2)″ and substitute into (3)″. Also use Equation (4)′.

(2)″

⎛ Vπ 2 1 ⎞ 1 + ⎜ g m1 + ⎟ Vπ 1 + r R R R r C1 RB 2 π2 ⎠ ⎝ C1 B2 π 2

(b)

RB1 = R1 R2 = 20 80 = 16 K

⎛ 20 ⎞ VTH 1 = ⎜ ⎟ (10) = 2 V ⎝ 100 ⎠ 2 − 0.7 I BQ1 = = 0.0111 mA 16 + (101)(1) I CQ1 = 1.11 mA RTH 2 = 15 85 = 12.75 K ⎛ 15 ⎞ VTH 2 = ⎜ ⎟ (10) = 1.5 V ⎝ 100 ⎠ 1.5 − 0.7 I BQ 2 = = 0.01265 mA 12.75 + (101)(0.5) I CQ 2 = 1.265 mA 1.11 = 42.69 mA/V 0.026 1.265 gm2 = = 48.65 mA/V 0.026 (100)(0.026) rπ 1 = = 2.34 K 1.11 (100)(0.026) rπ 2 = = 2.06 K 1.265 Now RC1 RB 2 = 2 12.75 = 1.729 K

g m1 =

RS RB1 rπ 1 RF ≅ RB1 rπ 1 RF = 16 2.34 10 = 1.695 K RC1 RB 2 rπ 2 = 1.729 2.06 = 0.940 K

Now (2)″

(3)″

Vπ 2 1 ⎞ 1 ⎡ 10 ⎛ ⎤ + ⋅ Vπ 1 − I i (10) ⎥ ⎜ 42.69 + ⎟ Vπ 1 + 2.06 ⎠ 0.940 1.729 ⎣⎢1.695 ⎝ ⎦ 46.587Vπ 1 + 1.064Vπ 2 − 5.784 I i = 0 Vπ 1 ⎛ 1 1 ⎞ ⎡ 10 ⎛ 101 ⎞ ⎤ =⎜ + ⎟⎢ ⋅ Vπ 1 − I i (10) ⎥ ⎜ ⎟ Vπ 2 + 2.06 10 0.5 10 1.695 ⎝ ⎠ ⎝ ⎠⎣ ⎦ 49.03Vπ 2 = 12.29Vπ 1 − 21I i

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From (2)″ Vπ 1 = (0.1242) I i − (0.02284)Vπ 2 Then 49.03Vπ 2 = 12.29 [ (0.1242) I i − (0.02284)Vπ 2 ] − 21I i (3)″ 49.31Vπ 2 = −19.47 I i

From (4)′

⎛ I ⎞⎛ 4 + 4 ⎞ Vπ 2 = − ⎜ o ⎟ ⎜ ⎟ = −(0.0411) I o ⎝ 48.65 ⎠ ⎝ 4 ⎠

Then (49.31) [ −(0.0411) I o ] = −19.47 I i Io = Ai = 9.61 Ii

12.46 a.

RTH = 13.5 || 38.3 = 9.98 kΩ ⎛ 13.5 ⎞ VTH = ⎜ ⎟ (10) = 2.606 V ⎝ 13.5 + 38.3 ⎠ (120)(2.606 − 0.7) I C1 = = 1.75 mA 9.98 + (121)(1)

VC1 = 10 − (1.75 )( 3) = 4.75 V 4.75 − 0.7 = 0.50 mA 8.1 (120)(0.026) rπ 1 = = 1.78 kΩ 1.75 1.75 g m1 = = 67.31 mA / V 0.026 (120)(0.026) rπ 2 = = 6.24 kΩ 0.50 0.50 gm2 = = 19.23 mA / V 0.026 b. IC 2 ≈

VS − Vπ 1 Vπ 1 V −V = + π 1 e2 RS RB || rπ 1 RF g m1Vπ 1 +

Vπ 2 + Ve 2 Vπ 2 + =0 RC1 rπ 2

Vπ 2 V V −V + g m 2Vπ 2 = ε 2 + ε 2 π 1 rπ 2 RE 2 RF and

(1) (2) (3)

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V0 = −( g m 2Vπ 2 ) RC 2 (4) Substitute numerical values in (1), (2), and (3) VS Vπ 1 ⎡ 1 1 ⎤ Ve 2 = Vπ 1 ⎢ + + ⎥− 0.6 ⎣ 0.6 9.98 ||1.78 1.2 ⎦ 1.2 VS (1.67) = Vπ 1 (4.011) − Ve 2 (0.8333)

(1)

1 ⎞ Ve 2 ⎛1 (67.31)Vπ 1 + Vπ 2 ⎜ + =0 ⎟+ ⎝ 3 6.24 ⎠ 3 or Vπ 1 (67.31) + Vπ 2 (0.4936) + Ve 2 (0.3333) = 0

(2)

V V ⎛ 1 ⎞ V Vπ 1 ⎜ + 19.23 ⎟ = e 2 + e 2 − π 2 ⎝ 6.24 ⎠ 8.1 1.2 1.2 or Vπ 2 (19.39) = Ve 2 (0.9568) − Vπ 1 (0.8333) (3) From (1) Ve 2 = Vπ 1 (4.813) − VS (2.00) Then Vπ 1 (67.31) + Vπ 2 (0.4936) + (0.3333)[Vπ 1 (4.813) − VS (2.00)] = 0 or Vπ 1 (68.91) + Vπ 2 (0.4936) − VS (0.6666) = 0 (2′) and Vπ 2 (19.39) = (0.9568) [Vπ 1 (4.813) − VS (2.00) ] − Vπ 1 (0.8333)

or Vπ 2 (19.39) = Vπ 1 (3.772) − VS (1.914) (3′) We find Vπ 1 = VS (0.009673) − Vπ 2 (0.007163) Then Vπ 2 (19.39) = (3.772)[VS (0.009673) − Vπ 2 (0.007163)] − VS (1.914) Vπ 2 (19.42) = VS ( −1.878) or Vπ 2 = −VS (0.09670) so that V0 = −(19.23)(4)(−VS )(0.09670) Then V0 = 7.44 VS 12.47 Using the circuit from Problem 12.32, we have Rif =

Vπ 1 . IS

VS − Vπ 1 RS From Problem 12.32 Vπ 1 = VS (0.009673) − Vπ 2 (0.007163)

Where I S =

= VS (0.009673) − (0.007163)(−VS )(0.09670) = VS (0.01037)

So Rif =

VS (0.01037) ⋅ (0.6) = 0.00629 kΩ VS − VS (0.01037)

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or Rif = 6.29 Ω 12.48 RTH = 1.4 ||17.9 = 1.298 kΩ ⎛ 1.4 ⎞ VTH = ⎜ ⎟ (10) = 0.7254 V ⎝ 1.4 + 17.9 ⎠ 0.7254 − 0.7 = 0.0196 mA I B1 = 1.298 I C1 = (50)(0.0196) = 0.98 mA Neglecting dc base currents, VB 2 = 10 − (0.98)(7) = 3.14 V

3.14 − 0.7 = 3.25 mA 0.25 + 0.5 ⎛ 50 ⎞ I C 2 = ⎜ ⎟ (3.25) = 3.19 mA ⎝ 51 ⎠ (50)(0.026) rπ 1 = = 1.33 kΩ 0.98 0.98 g m1 = = 37.7 mA / V 0.026 (50)(0.026) rπ 2 = = 0.408 kΩ 3.19 3.19 gm2 = = 123 mA / V 0.026 IE2 =

IS =

Vπ 1 V −V + π1 1 R1 || R2 || rπ 1 RF

g m1Vπ 1 +

Vπ 2 Vπ 2 + Ve 2 + =0 rπ 2 RC1

Vπ 2 V −V + g m 2Vπ 2 = e 2 1 rπ 2 RE1

(1) (2) (3)

Ve 2 − Vπ 1 V −V V (4) = 1 + 1 π1 RE1 RE 2 RF Enter numerical values in (1), (2), (3) and (4): Vπ 1 V −V IS = + π1 1 17.9 ||1.4 ||1.33 5 or

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I S = Vπ 1 (1.722) − V1 (0.20) (1) (37.7)Vπ 1 +

Vπ 2 V + Ve 2 + π2 =0 0.408 7

or Vπ 1 (37.7) + Vπ 2 (2.594) + Ve 2 (0.1429) = 0 Vπ 2 V −V + (123)Vπ 2 = e 2 1 0.408 0.25 or Vπ 2 (125.5) = Ve 2 (4) − V1 (4)

(2)

(3)

Ve 2 − V1 V −V V = 1 + 1 π1 0.25 0.50 5 or Ve 2 (4) = V1 (6.20) − Vπ 1 (0.20) (4) From (4): Ve 2 = V1 (1.55) − Vπ 1 (0.05) Then substituting in (3): Vπ 2 (125.5) = (4)[V1 (1.55) − Vπ 1 (0.05)] − V1 (4) or Vπ 2 (125.5) = V1 (2.20) − Vπ 1 (0.20) (3′) and substituting in (2): Vπ 1 ( 37.7 ) + Vπ 2 ( 2.594 ) + ( 0.1429 ) ⎡⎣V1 (1.55 ) − Vπ 1 ( 0.05 ) ⎤⎦ = 0 or Vπ 1 (37.69) + Vπ 2 (2.594) + V1 (0.2215) = 0 Now V1 = −Vπ 1 (170.16) − Vπ 2 (11.71) Then substituting in (1): I S = Vπ 1 (1.722) − (0.20)[ −Vπ 1 (170.16) − Vπ 2 (11.71)] or I S = Vπ 1 (35.75) + Vπ 2 (2.342) and substituting in (3′): Vπ 2 (125.5) = (2.20)[ −Vπ 1 (170.16) − Vπ 2 (11.71)] − Vπ 1 (0.20) or Vπ 2 (151.3) = −Vπ 1 (374.55) so that Vπ 1 = −Vπ 2 (0.4040)

Then I S = (35.75)[−Vπ 2 (0.4040)] + Vπ 2 (2.342) I S = −Vπ 2 (12.10) ⎛ RC 2 ⎞ I 0 = −( g m 2Vπ 2 ) ⎜ ⎟ ⎝ RC 2 + RL ⎠ ⎛ 2.2 ⎞ = −(123) ⎜ ⎟ Vπ 2 = −(64.43)Vπ 2 ⎝ 2.2 + 2 ⎠ or Vπ 2 = −(0.01552) I 0 Then I0 I 1 = ⇒ 0 = 5.33 I S (0.01552)(12.10) IS

12.49

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For example, use the circuit shown in Figure P12.30 12.50 rπ 1 = 6.24 kΩ, rπ 2 = 3.12 kΩ, rπ 3 = 1.56 kΩ g m1 = 19.23mA / V , g m 2 = 38.46 mA / V, g m 3 = 76.92 mA / V

VS = Vπ 1 + Ve1

(1)

Vπ 1 V V −V + g m1Vπ 1 = e1 + e1 e 3 rπ 1 RE1 RF Vπ 2 = − g m1Vπ 1 ( RC1 || rπ 2 ) g m 2Vπ 2 +

(2)

(3)

Vπ 3 + Ve 3 Vπ 3 + =0 RC 2 rπ 3

(4)

Vπ 3 V V −V (5) + g m 3Vπ 3 = e 3 + e 3 e1 rπ 3 RE 2 RF Enter numerical values in (2)-(5): Vπ 1 1 ⎞ ⎛ 1 ⎛ 1 ⎞ + (19.23)Vπ 1 = Ve1 ⎜ + ⎟ − Ve 3 ⎜ ⎟ 6.24 0.1 0.8 ⎝ ⎠ ⎝ 0.8 ⎠ or Vπ 1 (19.39) = Ve1 (11.25) − Ve 3 (1.25) (2) Vπ 2 = −(19.23)Vπ 1 (5 || 3.12) = −(36.94)Vπ 1 (3) 1 ⎞ ⎛1 ⎛1⎞ (38.46)Vπ 2 + Vπ 3 ⎜ + ⎟ + Ve 3 ⎜ ⎟ = 0 ⎝ 2 1.56 ⎠ ⎝ 2⎠ or Vπ 2 (38.46) + Vπ 3 (1.141) + Ve 3 (0.5) = 0

(4)

1 ⎞ ⎛ 1 ⎞ ⎛ 1 ⎛ 1 ⎞ Vπ 3 ⎜ + 76.92 ⎟ = Ve 3 ⎜ + ⎟ − Ve 3 ⎜ ⎟ ⎝ 1.56 ⎠ ⎝ 0.1 0.8 ⎠ ⎝ 0.8 ⎠ or Vπ 3 (77.56) = Ve 3 (11.25) − Ve1 (1.25) (5) From (1) Vπ 1 = VS − Ve1 Then (VS − Ve1 )(19.39) = Ve1 (11.25) − Ve3 (1.25) or VS (19.39) = Ve1 (30.64) − Ve 3 (1.25) (2′) Vπ 2 = −VS (36.94) + Ve1 (36.94)

(3′)

(38.46)[−VS (36.94) + Ve1 (36.94)] + Vπ 3 (1.141) + Ve 3 (0.5) = 0

(4′)

From (5): Ve 3 = Vπ 3 (6.894) + Ve1 (0.1111) Then VS (19.39) = Ve1 (30.64) − (1.25)[Vπ 3 (6.894) + Ve1 (0.1111)]

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or VS (19.39) = Ve1 (30.50) − Vπ 3 (8.6175) (2″) and −VS (1420.7) + Ve1 (1420.7) + Vπ 3 (1.141) + (0.5)[Vπ 3 (6.894) + Ve1 (0.1111)] = 0 or −VS (1420.7) + Ve1 (1420.76) + Vπ 3 (4.588) = 0 (4″) From (2″): Ve1 = VS (0.6357) + Vπ 3 (0.2825) Then substituting in (4″): −VS (1420.7) + (1420.76)[VS (0.6357) + Vπ 3 (0.2825)] + Vπ 3 (4.588) = 0 −VS (517.5) + Vπ 3 (405.95) = 0 Now I 0 = g m 3Vπ 3 = 76.92Vπ 3 or Vπ 3 = I 0 (0.0130) Then −VS (517.5) + I 0 (0.0130)(405.95) = 0 or I0 = 98.06 mA / V VS

12.52 (100)(0.026) = 5.2 kΩ 0.5 0.5 g m1 = g m 2 = = 19.23 mA / V 0.026 (100)(0.026) rπ 3 = = 1.3 kΩ 2 2 g m3 = = 76.92 mA / V 0.026 rπ 1 = rπ 2 =

Vπ 1 V + g m1Vπ 1 + g m 2Vπ 2 + π 2 = 0 rπ 1 rπ 2

(1)

Since rπ 1 = rπ 2 and g m1 = g m 2 , then Vπ 1 = −Vπ 2 VS = Vπ 1 − Vπ 2 + Ve 3 = −2Vπ 2 + Ve 3 (2) g m 2Vπ 2 +

Vπ 3 Vπ 3 + Ve 3 + =0 rπ 3 RC 2

Vπ 3 V V + g m 3Vπ 3 = e3 + π 2 rπ 3 RF rπ 2

(3)

(4)

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⎛ RC 3 ⎞ I0 = − ⎜ ⎟ ( g m 3Vπ 3 ) ⎝ RC 3 + RL ⎠ From (2): Ve 3 = VS + 2Vπ 2

(19.23) Vπ 2 +

(5)

Vπ 3 Vπ 3 1 + + (VS + 2Vπ 2 ) = 0 1.3 18.6 18.6

or (19.23)Vπ 2 + (0.8230)Vπ 3 + (0.05376)VS = 0 (3′) V ⎛ 1 ⎞ ⎛1⎞ + 76.92 ⎟ = ⎜ ⎟ (VS + 2Vπ 2 ) + π 2 Vπ 3 ⎜ 1.3 10 5.2 ⎝ ⎠ ⎝ ⎠ or (77.69)Vπ 3 = (0.3923)Vπ 2 + (0.1)VS (4′) ⎛ 2 ⎞ I0 = − ⎜ (5′) ⎟ (76.92)Vπ 3 = −(51.28)Vπ 3 ⎝ 2 +1⎠ From (3′): Vπ 2 = −(0.04255)Vπ 3 − (0.002780)VS Then (77.69)Vπ 3 = (0.3923)[ −(0.04255)Vπ 3 − (0.002780)VS ] + (0.1)VS

(77.71)Vπ 3 = (0.0989)VS or Vπ 3 = (0.001273)VS so that I 0 = −(51.28)(0.001273)VS or I0 = −(0.0653) mA/V VS

12.52 A=

5 × 103 2

f ⎞⎛ f ⎞ ⎛ ⎜1 + j 4 ⎟ ⎜1 + j 5 ⎟ 10 10 ⎝ ⎠⎝ ⎠ f ⎛ ⎞ ⎛ f ⎞ Phase = φ = − tan −1 ⎜ 4 ⎟ − 2 tan −1 ⎜ 5 ⎟ ⎝ 10 ⎠ ⎝ 10 ⎠ By trial and error, when f = 1.095 × 105 Hz , φ ≅ 180°

For T = 1 at f = 1.095 × 105 Hz, 1=

β ( 5 × 103 ) ⎛ f ⎞ 1+ ⎜ 4 ⎟ ⎝ 10 ⎠

2

⎡ ⎛ f ⎞ ⋅ ⎢1 + ⎜ 5 ⎟ ⎢⎣ ⎝ 10 ⎠

2

⎤ ⎥ ⎥⎦

⇒1=

β ( 5 × 103 )

(10.996 )( 2.199 )

⇒ β = 4.84 × 10−3

12.53 Use the basic circuit shown in Figure 12.27.

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For the ideal case I0 1 = Vi RE we want I0 = 10−3 A/V = 1 mA/V Vi Set RE = 1 kΩ Since the op-amp has a finite gain, finite input resistance, and finite output resistance, the closed-loop gain is slightly less than the ideal. RE will need to be slightly decreased to increase the gain. 12.54 dc analysis I E RE + VEB (on) + I B RB + VCC = 0 5 − 0.7 = 0.0343 100 + (51)(0.5) I C = (50)(0.0343) = 1.71 mA IB =

(50)(0.026) = 0.760 kΩ 1.71 1.71 gm = = 65.77 mA/V 0.026 a.

Then rπ =

To determine Rif : IS +

Vπ V − (−Vπ ) + 0 =0 RB || rπ RF

(1)

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V0 V0 − (−Vπ ) (2) + RC RF Now from (2): V ⎛1 1 ⎞ (65.77)Vπ − π = V0 ⎜ + ⎟ 10 ⎝ 1 10 ⎠ (65.67)Vπ = V0 (1.10) or V0 = (59.7)Vπ and from (1): Vπ V V + π + 0 =0 IS + 100 0.760 10 10 g mVπ =

I S + Vπ (0.8543) + (0.1)(59.7)Vπ = 0 I S = −Vπ (6.824) Now (−Vπ ) ⇒ Rif = 147 Ω Rif = IS

To determine R0 f :

IX =

VX VX + − g mVπ RC RF + RB rπ

(3)

⎛ −( RB rπ ) ⎞ Vπ = ⎜⎜ ⎟⎟ (VX ) (4) ⎝ ( RB rπ ) + RF ⎠ Now ⎛ −(100 0.760) ⎞ Vπ = ⎜⎜ ⎟⎟ (VX ) = −(0.07014)VX ⎝ (100 0.760) + 10 ⎠ so 1 ⎛1 ⎞ I X = VX ⎜ + + (65.77)(0.07014) ⎟ 1 10.754 ⎝ ⎠ VX ⇒ R0 f = 175 Ω R0 f = IX b. From part (a), we find I Vπ = − S 6.824 then ⎛ −IS ⎞ V0 = (59.7) ⎜ ⎟ ⎝ 6.824 ⎠ or

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V0 = −8.75 kΩ VS

c.

If capacitance is finite, a phase shift will be introduced.

12.55 dc analysis: VGS = VDS ID =

VDD − VGS = K n (VGS − VTN ) 2 RD

10 − VGS = (0.20)(8)(VGS − 2) 2 10 − VGS = 1.6(VGS2 − 4VGS + 4) 1.6VGS2 − 5.4VGS − 3.6 = 0 5.4 ± (5.4) 2 + 4(1.6)(3.6) = 3.95 V 2(1.6) 10 − 3.95 ID = = 0.756 mA 8 g m = 2 K n I D = 2 (0.2)(0.756) ⇒ g m = 0.778 mA/V

VGS =

a. Vgs − VS RS

+

Vgs − V0 RF

=0

⎛ 1 1 ⎞ VS V0 Vgs ⎜ + + ⎟= R R RS RF F ⎠ ⎝ S V0 − Vgs V0 + g mVgs + =0 RD RF

(1)

⎛ 1 ⎛ 1 ⎞ 1 ⎞ V0 ⎜ + − g m ⎟ (2) ⎟ = Vgs ⎜ ⎝ RD RF ⎠ ⎝ RF ⎠ So from (1): 1 ⎞ VS V0 ⎛1 Vgs ⎜ + + ⎟= ⎝ 10 100 ⎠ 10 100 or Vgs (0.11) = VS (0.10) + V0 (0.010) Vgs = VS (0.909) + V0 (0.0909)

Then from (2):

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⎛1 1 ⎞ ⎛ 1 ⎞ V0 ⎜ + − 0.778 ⎟ ⎟ = Vgs ⎜ ⎝ 8 100 ⎠ ⎝ 100 ⎠ V0 (0.135) = Vgs (−0.768) = (−0.768)[VS (0.909) + V0 (0.0909)] V0 (0.2048) = −VS (0.6981) so V Av = 0 = −3.41 VS

b. We have Vgs = VS (0.909) + V0 (0.0909) = VS (0.909) + (0.0909) − (3.41VS ) = 0.599VS Now V V0 (−3.41VS ) RS Azf = 0 = = I S VS − Vgs VS − 0.599VS RS or (−3.41)(10) Azf = ⇒ Azf = −85.0 V/ma 0.401 Vgs Vgs 0.401VS 0.599VS c. Rif = = = ⋅ (10) ⇒ Rif = 14.9 kΩ 0.401VS IS RS d.

IX =

VX VX + g mVgs + RD RS + RF

⎛ RS ⎞ ⎛ 10 ⎞ Vgs = ⎜ ⎟ VX = ⎜ ⎟ VX R + R ⎝ 10 + 100 ⎠ F ⎠ ⎝ S = (0.0909)VX 1 ⎤ ⎡1 I X = VX ⎢ + (0.778)(0.0909) + 10 + 100 ⎥⎦ ⎣8 IX 1 = = 0.2048 ⇒ R0 f = 4.88 kΩ VX R0 f

12.56 As g m → ∞,

V0 − RF −100 = = = −10 VS RS 10

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To be within 10% of ideal, V0 = −10(0.9) = −9 VS From Problem 12.41, we had Vgs = VS (0.909) + V0 (0.0909) = VS (0.909) + ( −9VS )(0.0909) = 0.0909VS Also from Problem 12.41, we had V0 (0.135) = Vgs (0.010 − g m )

or (−9VS )(0.135) = (0.0909)VS (0.010 − g m ) −1.215 = 0.000909 − 0.0909 g m or g m = 13.36 mA/V 12.57 dc analysis RTH = 24 || 150 = 20.7 kΩ ⎛ 24 ⎞ VTH = ⎜ ⎟ (12) = 1.655 V ⎝ 24 + 150 ⎠ 1.655 − 0.7 I BQ = = 0.00556 mA 20.7 + (151)(1)

so I CQ = 0.834 mA (150)(0.026) = 4.68 kΩ 0.834 0.834 gm = = 32.08 mA / V 0.026

rπ =

VS − Vπ Vπ V − V0 = + π RS RB || rπ RF

(1)

V0 V0 − Vπ + = 0 (2) RC RF From (1): ⎡1 VS 1 1 ⎤ V0 = Vπ ⎢ + + ⎥− R RF 5 5 20.7 || 4.68 ⎣ F ⎦ g mVπ +

or

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⎛ 1 ⎞ V0 VS (0.20) = Vπ ⎜ 0.4620 + ⎟− R RF ⎝ F ⎠ From (2): ⎛ ⎛1 1 ⎞ 1 ⎞ ⎜ 32.08 − ⎟ Vπ + V0 ⎜ + ⎟=0 R ⎝ ⎝ 6 RF ⎠ F ⎠

so ⎛ 1 ⎞ −V0 ⎜ 0.1667 + ⎟ R ⎝ F ⎠ (2) Vπ = ⎛ 1 ⎞ ⎜ 32.08 − ⎟ RF ⎠ ⎝ Then ⎡ ⎛ 1 ⎢ −V0 ⎜ 0.1667 + RF ⎛ 1 ⎞⎢ ⎝ VS (0.20) = ⎜ 0.4620 + ⎟⎢ R ⎛ 1 ⎞ ⎝ F ⎠ ⎢ ⎜ 32.08 − ⎟ RF ⎠ ⎢⎣ ⎝

⎞⎤ ⎟⎥ ⎠ ⎥ − V0 ⎥ RF ⎥ ⎥⎦ Neglect the RF in the denominator term. Now V0 V = −5 ⇒ VS = − 0 = −V0 (0.20) 5 VS ⎡ −V ( 0.1667 RF + 1) ⎤ −V0 (0.20)(0.20) RF = (0.4620 RF + 1) ⎢ 0 ⎥ − V0 32.08RF ⎣ ⎦ 2 −1.283RF = −(0.4620 RF + 1)(0.1667 RF + 1) − 32.08 RF 1.206 RF2 − 32.71RF − 1 = 0 32.71 ± (32.71) 2 + 4(1.206)(1) 2(1.206) so that RF = 27.2 kΩ RF =

12.58 dc analysis RTH = 4 || 15 = 3.16 kΩ = RB ⎛ 4 ⎞ VTH = ⎜ ⎟12 = 2.526 V ⎝ 4 + 15 ⎠ 2.526 − 0.7 I BQ = = 0.00251 3.16 + (181)(4) I CQ = 0.452 mA (180)(0.026) = 10.4 kΩ 0.452 0.452 gm = = 17.4 mA/V 0.026

rπ =

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Vi − Vπ 1 Vπ 1 V −V = + π1 0 RS RB || rπ RF

(1)

g mVπ 1 +

Vπ 2 =0 RC || RB || rπ

(2)

g mVπ 2 +

Vπ 3 =0 RC || RB || rπ

(3)

g mVπ 3 +

V0 V0 V0 − Vπ 1 + + =0 RC RL RF

(4)

Now RC || RB || rπ = 8 || 3.16 ||10.4 = 1.86 kΩ RB || rπ = 3.16 ||10.4 = 2.42 kΩ Now substituting in (2): V (17.4)Vπ 1 + π 2 = 0 or Vπ 2 = −(32.36)Vπ 1 1.86 and substituting in (3): V (17.4)Vπ 2 + π 3 = 0 1.86 V (17.4)[−(32.36)Vπ 1 ] + π 3 = 0 1.86 or Vπ 3 = (1047.3)Vπ 1 Substitute numerical values in (1): ⎛1 Vi 1 1 ⎞ V0 = Vπ 1 ⎜ + + ⎟− R RF 10 10 2.42 ⎝ F ⎠ or ⎛ 1 ⎞ V0 Vi (0.10) = Vπ 1 ⎜ 0.513 + ⎟− R RF ⎝ F ⎠ Substitute numerical values in (4): ⎛ 1 1 1 ⎞ Vπ 1 =0 (17.4)(1047.3)Vπ 1 + V0 ⎜ + + ⎟− ⎝ 8 4 RF ⎠ RF ⎛ ⎛ 1 ⎞ 1 ⎞ Vπ 1 ⎜ 1.822 × 104 − ⎟ + V0 ⎜ 0.375 + ⎟=0 RF ⎠ RF ⎠ ⎝ ⎝ ⎛ 1 ⎞ −V0 ⎜ 0.375 + ⎟ RF ⎠ ⎝ Vπ 1 = 1 1.822 × 104 − RF so that

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⎡ ⎛ 1 ⎞⎤ ⎢ −V0 ⎜ 0.375 + ⎟⎥ RF ⎠ ⎥ V0 ⎛ 1 ⎞⎢ ⎝ − Vi (0.10) = ⎜ 0.513 + ⎟ RF ⎠ ⎢ 1.822 × 104 − 1 ⎥ RF ⎝ ⎢ ⎥ RF ⎥ ⎣⎢ ⎦ V V We have 0 = −80 or Vi = − 0 80 Vi ⎡ ⎛ 1 ⎞⎤ ⎢ − ⎜ 0.375 + ⎟⎥ R (0.10) ⎛ 1 ⎞⎢ ⎝ 1 F ⎠ ⎥ − = ⎜ 0.513 + − ⎟⎢ ⎥ 1 80 R R 4 ⎝ F ⎠ 1.822 × 10 − F ⎢ ⎥ RF ⎥ ⎢⎣ ⎦ Neglect that 1/ RF term in the denominator. (0.513RF + 1)(0.375 RF + 1) −(0.00125 RF ) = − −1 1.822 × 104 RF 22.775 RF2 = (0.513RF + 1)(0.375 RF + 1) + 1.822 × 10 4 RF We find 22.58 RF2 − 1.822 × 104 RF − 1 = 0 RF =

1.822 × 104 ± (1.822 ×10 4 ) 2 + 4(22.58)(1) 2(22.58)

or RF = 0.807 MΩ 12.59

a.

VS − (−Vd ) −Vd − V1 = RS RF

or ⎛ 1 1 ⎞ VS V1 + + =0 Vd ⎜ ⎟+ ⎝ RS RF ⎠ RS RF V0 − V1 V1 V1 − (−Vd ) = + R1 R2 RF or ⎛ 1 V0 1 1 ⎞ Vd = V1 ⎜ + + ⎟+ R1 ⎝ R1 R2 RF ⎠ RF

(1)

(2)

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V0 A0 L Substitute numerical values in (1) and (2): V0 ⎛ 1 1 ⎞ VS V1 ⋅⎜ + ⎟ + + = 0 104 ⎝ 5 10 ⎠ 5 10 or V0 (0.3 × 10−4 ) + VS (0.20) + V1 (0.10) = 0

and V0 = A0 LVd or Vd =

(1)

V0 ⎛ 1 1 1⎞ V ⎛1⎞ = V1 ⎜ + + ⎟ + 04 ⋅ ⎜ ⎟ 50 ⎝ 50 10 10 ⎠ 10 ⎝ 10 ⎠ or V0 (0.02 − 10−5 ) = V1 (0.22) (2) ⎛ 0.02 − 10−5 ⎞ Then V1 = V0 ⎜ ⎟ ⎝ 0.22 ⎠ and

⎡ ⎛ 0.02 − 10−5 ⎞ ⎤ V0 (0.3 × 10−4 ) + VS (0.20) + (0.10) ⎢V0 ⎜ ⎟⎥ = 0 ⎣ ⎝ 0.22 ⎠ ⎦ V0 ⎡⎣0.3 × 10−4 − 0.4545 × 10−5 + 0.00909⎤⎦ + VS (0.20) = 0

Then b.

V0 V −0.20 = ⇒ 0 = −21.94 −3 VS 9.115 × 10 VS Rif =

Now Vd =

−Vd −V ⋅ R = d S VS − (−Vd ) VS + Vd RS

V0 21.94VS =− A0 L 104

(21.94 × 10−4 )(5) 1 − 21.94 × 10−4 or Rif = 1.099 × 10−2 kΩ ⇒ Rif = 10.99 Ω

Then Rif =

c. Because of the A0LVd source, R0 f = 0 12.60 For example, use the circuit shown in Figure 12.41 12.61 Break the loop

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It = Iε

Now Ai I t +

V0 V0 + =0 R1 RF + RS Ri

⎛ RS ⎞ V0 Ir = ⎜ ⎟⋅ ⎝ RS + Ri ⎠ RF + RS Ri ⎛ R + Ri ⎞ or V0 = I r ⎜ S ⎟ ⋅ ( RF + RS Ri ) ⎝ RS ⎠ Then ⎛ 1 ⎞ ⎡ ⎛ RS + Ri ⎞ ⎤ 1 Ai I t + ⎜⎜ + ⎟⎟ × ⎢ I r ⎜ ⎟ ( RF + RS Ri ) ⎥ = 0 ⎦ ⎝ R1 RF + RS Ri ⎠ ⎣ ⎝ RS ⎠ Ai I T = − r ⇒T = It ⎡1 ⎤ ⎛ RS + Ri ⎞ 1 ⎢ + ⎥⎜ ⎟ ( RF + RS Ri ) R R R R RS ⎠ + F S i ⎦⎝ ⎣ 1

12.62

Vπ 1 V V −V + g m1Vπ 1 = ε 1 + ε 1 0 rπ 1 RE1 RF g m1Vπ 1 +

Vr RC1 rπ 2

(1)

= 0 ⇒ Vr = −( g m1Vπ 1 )( RC1 rπ 2 )

(2)

Vπ 2 = Vt so that g m 2Vt +

Vπ 3 + V0 Vπ 3 + = 0 (3) RC 2 rπ 3

Vπ 3 V V −V + g m 3Vπ 3 = 0 + 0 ε 1 (4) rπ 3 RE 3 RF From (4): ⎛ 1 ⎛ 1 ⎞ V 1 ⎞ V0 ⎜ + + g m3 ⎟ + ε 1 ⎟ = Vπ 3 ⎜ ⎝ RE 3 RF ⎠ ⎝ rπ 3 ⎠ RF

But Vε 1 = −Vπ 1 ⎛ 1 ⎞ V + g m3 ⎟ − π 1 Vπ 3 ⎜ ⎝ rπ 3 ⎠ RF so V0 = ⎛ 1 1 ⎞ + ⎜ ⎟ R R F ⎠ ⎝ E3

Then

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⎡⎛ 1 ⎞ ⎛ 1 1 ⎞⎤ + Vπ 1 ⎢⎜ + g m1 ⎟ − ⎜ ⎟⎥ = ⎠ ⎝ RE1 RF ⎠ ⎦ ⎣⎝ rπ 1

⎛ 1 ⎞ V −Vπ 3 ⎜ + g m3 ⎟ + π 1 r ⎝ π3 ⎠ RF ⎛ 1 1 ⎞ RF ⋅ ⎜ + ⎟ ⎝ RE 3 RF ⎠

(1′)

and ⎛ 1 ⎞ V + g m3 ⎟ − π 1 Vπ 3 ⎜ ⎛ 1 1 ⎞ ⎝ rπ 3 ⎠ RF = 0 + g m 2Vt + Vπ 3 ⎜ ⎟+ ⎛ 1 1 ⎞ ⎝ RC 2 rπ 3 ⎠ + RC 2 ⋅ ⎜ ⎟ R R F ⎠ ⎝ E3

(3′)

From (3′), solve for Vπ 3 and substitute into (1′). Then from (1′), solve for Vπ 1 and substitute into (2). Then T =−

Vr . Vt

12.63

Vr Vr Vr − Ve + + =0 RS rπ 1 RF g m1Vt +

(1)

Vπ 2 + Ve Vπ 2 + = 0 (2) RC1 rπ 2

Vπ 2 V V − Vr + g m 2Vπ 2 = e + e (3) rπ 2 RE RF Using the parameters from Problem 12.29, we obtain 1 1⎞ V ⎛1 Vr ⎜ + + ⎟− e = 0 ⎝ 10 15.8 10 ⎠ 10 or Vr (0.2633) = Ve (0.10) (1) 1 ⎞ Ve ⎛ 1 =0 (7.62)Vt + Vπ 2 ⎜ + ⎟+ ⎝ 40 2.28 ⎠ 40 or Vt (7.62) + Vπ 2 (0.4636) + Ve (0.025) = 0

(2)

⎛ 1 ⎞ ⎛1 1 ⎞ V Vπ 2 ⎜ + 52.7 ⎟ = Ve ⎜ + ⎟ − r 2.28 ⎝ ⎠ ⎝ 1 10 ⎠ 10 or Vπ 2 (53.14) = Ve (1.10) − Vr (0.10) Then Vπ 2 = Ve (0.0207) − Vr (0.001882) (3) Substituting in (2): Vt (7.62) + (0.4636)[Ve (0.0207) − Vr (0.001882)] + Ve (0.025) = 0

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or Vt (7.62) + Ve (0.03460) − Vr (0.0008725) = 0 From (1) Ve = Vr (2.633) Then Vt (7.62) + Vr (2.633)(0.03460) − Vr (0.0008725) = 0 Vt (7.62) + Vr (0.09023) = 0 Vr = −84.45 Vt Now V T = − r ⇒ T = 84.45 Vt

or

12.64

Vπ = −Vt

g mVπ =

V0 V0 + RC RF + RB rπ

(1)

and ⎛ RB rπ ⎞ (2) Vr = ⎜ V ⎜ R r + R ⎟⎟ 0 F ⎠ ⎝ B π Now ⎛1 ⎞ 1 (65.77)Vπ = V0 ⎜ + ⎜ 1 10 + 100 0.760 ⎟⎟ ⎝ ⎠ or (65.77)Vπ = V0 (1.0930) and ⎛ 0.754 ⎞ Vr = ⎜ ⎟ V0 = (0.07011)V0 ⎝ 10 + 0.754 ⎠ so V0 = (14.26)Vr Then (65.77)(−Vt ) = (14.26)Vr (1.0930) Vr = −4.22 so that T = 4.22 Vt

12.64 Want f1 = 12 MHz for a phase margin of 45° TdB ( 0 ) = 80 dB ⇒ T ( 0 ) = 104

Then

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T ( 0)

T(f)=

⎛ f ⎜1 + j f PD ⎝ Set f = f1 and T

⎞⎛ f ⎞ ⎟⎜ 1 + j 6 ⎟ 12 × 10 ⎠ ⎠⎝ =1

So 104

T =1=

2

⎛ 12 × 106 ⎞ 1+ ⎜ ⎟ ⋅ 2 ⎝ f PD ⎠ which yields 12 × 106 104 = ⇒ f PD = 1.70 kHz f PD 2

12.65 a.

⎛ f ⎞ ⎛ f ⎞ − 2 tan −1 ⎜ 4 ⎟ φ = − tan −1 ⎜ 2 ⎟ 5 10 × ⎝ ⎠ ⎝ 10 ⎠

or ⎛ f ⎞ ⎛ f ⎞ −180 = − tan −1 ⎜ 180 2 ⎟ − 2 tan −1 ⎜ 1804 ⎟ ⇒ f180 ≈ 1.05 × 104 Hz ⎝ 5 × 10 ⎠ ⎝ 10 ⎠ β (105 ) T ( f180 ) = 1 = b. 2 2 ⎛ 1.05 × 104 ⎞ ⎡ ⎛ 1.05 ×104 ⎞ ⎤ 1+ ⎜ 1 + ⎢ ⎥ ⎟ ⎜ ⎟ 2 4 ⎝ 5 × 10 ⎠ ⎢⎣ ⎝ 10 ⎠ ⎥⎦ β (105 ) or 1= (21.02)(2.105)

β = 4.42 × 10−4 12.65 A0 = 80 dB ⇒ A0 = 104 A0 Af ( 0 ) = 1 + β A0 or 5 =

104

1 + β (104 )

⇒ β ≈ 0.2

Then T ( 0 ) = β A0 = 0.2 × 104 Inserting a dominate pole ⎛ f ⎞ −1 ⎛ f ⎞ −1 ⎛ f ⎞ φ = − tan −1 ⎜ ⎟ − tan ⎜ 6 ⎟ − tan ⎜ 7 ⎟ 10 f ⎝ ⎠ ⎝ 10 ⎠ ⎝ PD ⎠ If we want a phase margin of 45°, then ⎛ f ⎞ ⎛ f ⎞ −135° ≈ −90° − tan −1 ⎜ 6 ⎟ − tan −1 ⎜ 7 ⎟ 10 ⎝ ⎠ ⎝ 10 ⎠ By trial and error, f ≈ 0.845 MHz Then 0.2 × 104 1 T =1= × 2 2 2 ⎛ 0.845 × 106 ⎞ ⎛ 0.845 ⎞ ⎛ 0.845 ⎞ + 1 + 1+ ⎜ 1 ⎜ ⎟ ⎟ ⎜ ⎟ ⎝ 10 ⎠ f PD ⎝ 1 ⎠ ⎝ ⎠

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0.845 × 106 0.2 ×104 ≈ f PD (1.309 )(1.0036 )

so f PD = 555 Hz 12.66

β (103 )

T = β Av =

(a)

f ⎞ ⎛ ⎜1 + j 4 ⎟ 10 ⎠ ⎝ f f φ = −2 tan −1 4 − tan −1 5 10 10 Set φ = −180°

2

f ⎞ ⎛ ⎜1 + j 5 ⎟ 10 ⎠ ⎝

By trial and error, f = 4.58 ×10 4 Hz Set T = 1 at f = 4.58 ×10 4 Hz

(b)

β (103 )

1=

2

⎛ 4 2 ⎞ 4 2 ⎜ 1 + ⎜⎛ 4.58 ×10 ⎟⎞ ⎟ 1 + ⎜⎛ 4.58 ×10 ⎟⎞ 4 5 ⎜ ⎝ 10 ⎠ ⎟⎠ ⎝ 10 ⎠ ⎝ β (103 ) ⇒ β = 0.02417 1= (21.976)(1.10) 103 = 39.7 1 + (10 )(0.02417)

(c)

Avfo =

(d)

Wait T < 1 at f = 4.58 × 104 Hz, so system is stable for smaller values of β .

3

12.67 ⎛ f ⎞ ⎛ f ⎞ ⎛ f ⎞ − tan −1 ⎜ − tan −1 ⎜ 5 ⎟ 4 ⎟ 4 ⎟ ⎝ 10 ⎠ ⎝ 5 × 10 ⎠ ⎝ 10 ⎠ 4 At f = 8.1× 10 Hz, φ = −180.28°

φ = − tan −1 ⎜

Determine T ( f ) at this frequency. T = β (103 ) ×

=

a.

1 ⎛ 8.1× 104 ⎞ 1+ ⎜ ⎟ 4 ⎝ 10 ⎠

2

×

1 ⎛ 8.1× 104 ⎞ 1+ ⎜ 4 ⎟ ⎝ 5 × 10 ⎠

2

×

1 ⎛ 8.1× 104 ⎞ 1+ ⎜ ⎟ 5 ⎝ 10 ⎠

2

β (103 ) (8.161)(1.904)(1.287) For β = 0.005

T ( f ) = 0.250 < 1 ⇒ Stable

b.

For β = 0.05

T ( f ) = 2.50 > 1 ⇒ Unstable

12.68 (b)Phase margin = 80° ⇒ φ = −100° ⎛ f ⎞ ⎛ f ⎞ − tan −1 ⎜ 3 ⎟ 4 ⎟ ⎝ 10 ⎠ ⎝ 5 × 10 ⎠ 3 By trial and error, f = 1.16 × 10 Hz

φ = −100 = −2 tan −1 ⎜

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Then

β (5 × 103 )

T =1=

2

⎛ 3 2 ⎞ 3 2 ⎜ 1 + ⎛⎜ 1.16 × 10 ⎞⎟ ⎟ ⋅ 1 + ⎛⎜ 1.16 × 10 ⎞⎟ 3 4 ⎜ ⎝ 10 ⎠ ⎟⎠ ⎝ 5 × 10 ⎠ ⎝ β (5 × 103 ) = ⇒ β = 4.7 × 10−4 (2.35)(1.00)

12.69

For β = 0.005,

c.

T ( f ) = 1(0 dB) at f ≈ 2.10 × 104 Hz

Then 4 4 ⎛ 2.10 × 104 ⎞ −1 ⎛ 2.10 × 10 ⎞ −1 ⎛ 2.10 × 10 ⎞ ⎟ − tan ⎜ ⎟ − tan ⎜ ⎟ 3 4 5 ⎝ 10 ⎠ ⎝ 10 ⎠ ⎝ 10 ⎠

φ = − tan −1 ⎜ or

= −87.27 − 64.54 − 11.86

φ = −163.7 System is stable. Phase margin = 16.3° For β = 0.05, T ( f ) = 1 (0 dB) at f ≈ 6.44 × 104 Hz

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Then 4 4 ⎛ 6.44 × 104 ⎞ −1 ⎛ 6.44 × 10 ⎞ −1 ⎛ 6.44 × 10 ⎞ − − tan tan ⎟ ⎜ ⎟ ⎜ ⎟ 3 4 5 ⎝ 10 ⎠ ⎝ 10 ⎠ ⎝ 10 ⎠

φ = − tan −1 ⎜ or

= −89.11 − 81.17 − 32.78

φ = −203.1° ⇒ System is unstable. 12.70

β (105 )

T = Aβ =

f ⎞⎛ f ⎞⎛ f ⎞ ⎛ 1 + j 5 ⎟ ⎜1 + j ⎜1 + j ⎟ 4 ⎟⎜ 5 × 10 ⎠ ⎝ 10 ⎠ ⎝ 5 × 105 ⎠ ⎝ Phase Margin = 60° ⇒ φ = −120° So ⎛ f ⎞ ⎛ f ⎞ ⎛ f ⎞ −120 = − tan −1 ⎜ − tan −1 ⎜ 5 ⎟ − tan −1 ⎜ 4 ⎟ 5 ⎟ ⎝ 5 × 10 ⎠ ⎝ 10 ⎠ ⎝ 5 × 10 ⎠ By trial and error, at f = 105 Hz , φ ≅ −120° Then β (105 ) T = 1= 2 2 2 ⎛ 105 ⎞ ⎛ 105 ⎞ ⎛ 105 ⎞ ⋅ 1+ ⎜ 5 ⎟ ⋅ 1+ ⎜ 1+ ⎜ 4 ⎟ 5 ⎟ ⎝ 5 × 10 ⎠ ⎝ 10 ⎠ ⎝ 5 × 10 ⎠ 1=

β (105 )

( 2.236 )(1.414 )(1.02 )

⇒ β = 3.22 × 10−5

12.71 (a)

100 =

105 ⇒ β = 9.99 × 10−3 1 + (105 ) β

T = 1 = β Av =

1=

(9.99 × 10−3 )(105 ) ⎛ f ⎞ 1+ ⎜ 3 ⎟ ⎝ 10 ⎠ 999 2

2

⎛ f ⎞ 1+ ⎜ 5 ⎟ ⎝ 10 ⎠

⎛ f ⎞ ⎛ f ⎞ 1+ ⎜ 3 ⎟ 1+ ⎜ 5 ⎟ ⎝ 10 ⎠ ⎝ 10 ⎠ 5 f = 3.08 × 10 Hz

2

2

Phase f f − tan −1 5 3 10 10 5 5 3.08 ×10 −1 3.08 × 10 = − tan −1 − tan 103 105 = −89.81 − 72.01 φ = −161.8 Stable (b) Phase Margin = 180 − 161.8 = 18.2°

φ = − tan −1

12.72

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φ = − tan −1

(a)

For φ = −180°

f f f − tan −1 6 − tan −1 4 5 × 10 10 5 × 107

By trial and error, f180 = 7.25 × 106 Hz T =

(b)

=

(0.10)(105 ) ⎛ 7.25 × 106 ⎞ 1+ ⎜ ⎟ 4 ⎝ 5 × 10 ⎠

2

⎛ 7.25 ×106 ⎞ 1+ ⎜ ⎟ 6 ⎝ 10 ⎠

2

⎛ 7.25 ×106 ⎞ 1+ ⎜ ⎟ 7 ⎝ 5 × 10 ⎠

2

104 (145)(7.319)(1.01)

T = 9.33 System is unstable T =1=

104 ⎛ f ⎞ 1+ ⎜ 4 ⎟ ⎝ 5 × 10 ⎠

2

⎛ f ⎞ 1+ ⎜ 6 ⎟ ⎝ 10 ⎠

2

⎛ f ⎞ 1+ ⎜ 7 ⎟ ⎝ 5 × 10 ⎠

2

f = 2.14 ×107 Hz 7 7 ⎛ 2.14 × 107 ⎞ −1 ⎛ 2.14 × 10 ⎞ −1 ⎛ 2.14 × 10 ⎞ ⎟ − tan ⎜ ⎟ − tan ⎜ ⎟ 4 6 7 ⎝ 5 × 10 ⎠ ⎝ 10 ⎠ ⎝ 5 × 10 ⎠

φ = − tan −1 ⎜

= −89.87 − 87.32 − 23.17 φ = −200.4°

For f = 7.25 × 106 Hz

(c)

(0.0010)(105 )

T =

2

⎛ 7.25 × 106 ⎞ ⎛ 7.25 ×106 ⎞ 1+ ⎜ ⎟ 1+ ⎜ ⎟ 4 6 ⎝ 5 × 10 ⎠ ⎝ 10 ⎠ 100 = (145)(7.319)(1.01)

2

⎛ 7.25 ×106 ⎞ 1+ ⎜ ⎟ 7 ⎝ 5 × 10 ⎠

2

T = 0.0933 System is stable. T =1=

100 ⎛ f ⎞ 1+ ⎜ 4 ⎟ ⎝ 5 × 10 ⎠

2

⎛ f ⎞ 1+ ⎜ 6 ⎟ ⎝ 10 ⎠

2

⎛ f ⎞ 1+ ⎜ 7 ⎟ ⎝ 5 × 10 ⎠

2

f = 2.13 ×106 Hz 6 6 2.13 × 106 −1 2.13 × 10 −1 2.13 × 10 − tan − tan 5 × 104 106 5 × 107 = −88.66 − 64.85 − 2.44 φ = −155.9°

φ = − tan −1

12.73 (a)

f180 f − 2 tan −1 1805 5 × 103 10 = 1.05 × 105 Hz

φ = −180 = − tan −1 f180

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(0.0045)(2 × 103 )

T =

(b)

2 2 ⎛ 1.05 × 105 ⎞ ⎡ ⎛ 1.05 × 105 ⎞ ⎤ 1+ ⎜ ⎢1 + ⎜ ⎟ ⎥ 3 ⎟ 5 ⎝ 5 × 10 ⎠ ⎢⎣ ⎝ 10 ⎠ ⎥⎦ 9 = (21.02)(2.1025)

0 T = f180 = 0.204

System is stable 9

T =1=

⎛ f ⎞ 1+ ⎜ 3 ⎟ ⎝ 5 × 10 ⎠

2

⎡ ⎛ f ⎞2 ⎤ ⎢1 + ⎜ 5 ⎟ ⎥ ⎣⎢ ⎝ 10 ⎠ ⎦⎥

f = 3.88 × 104 Hz 3.88 × 104 3.88 ×10 4 − 2 tan −1 3 5 × 10 105 = −82.66 − 42.41 φ = −125.1°

φ = − tan −1

(c)

T =

=

(0.15)(2 × 103 ) ⎛ 1.05 × 105 ⎞ 1+ ⎜ 3 ⎟ ⎝ 5 × 10 ⎠

2

⎡ ⎛ 1.05 × 105 ⎞ 2 ⎤ ⎢1 + ⎜ ⎟ ⎥ 5 ⎢⎣ ⎝ 10 ⎠ ⎥⎦

300 (21.02)(2.1025)

T = 6.79 System is unstable T = 1=

300 ⎛ f ⎞ 1+ ⎜ 3 ⎟ ⎝ 5 × 10 ⎠

2

⎡ ⎛ f ⎞2 ⎤ ⎢1 + ⎜ 5 ⎟ ⎥ ⎢⎣ ⎝ 10 ⎠ ⎥⎦

f = 2.33 × 105 Hz 5 2.33 × 105 −1 2.33 × 10 − 2 tan 5 × 103 105 = −88.77 − 133.54

φ = − tan −1

φ = −222.3° 12.74 Phase Margin = 45° ⇒ φ = −135° φ = −135° ⎛ f ⎞ ⎛ f ⎞ ⎛ f ⎞ ⎛ f ⎞ = − tan −1 ⎜ 3 ⎟ − tan −1 ⎜ 4 ⎟ − tan −1 ⎜ 5 ⎟ − tan −1 ⎜ 6 ⎟ ⎝ 10 ⎠ ⎝ 10 ⎠ ⎝ 10 ⎠ ⎝ 10 ⎠ 4 At f = 10 Hz, φ = −135.6°

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T =1

1

= β (103 ) ×

2

1

×

⎛ 104 ⎞ ⎛ 104 ⎞ 1+ ⎜ 3 ⎟ 1+ ⎜ 4 ⎟ ⎝ 10 ⎠ ⎝ 10 ⎠ 1 1 × × 2 4 2 ⎛ 10 ⎞ ⎛ 104 ⎞ 1+ ⎜ 5 ⎟ 1+ ⎜ 6 ⎟ ⎝ 10 ⎠ ⎝ 10 ⎠

1=

2

×

β (103 ) (10.05)(1.414)(1.005)(1.00)

or

β = 0.01428 12.75 1 1 × f ⎛ ⎞ f ⎞ ⎛1 + j 3 ⎟ ⎜1 + j ⎟ ⎜ 300 × 10 ⎠ f PD ⎠ ⎝ ⎝ 1 1 × × f f ⎛ ⎞ ⎛ ⎞ ⎜1 + j ⎜1 + j ⎟ 6 ⎟ 2 × 10 ⎠ ⎝ 25 × 106 ⎠ ⎝ Phase Margin = 45° ⇒ φ = −135° at f = 300 kHz T = 5000 ×

3 ⎛ 300 × 103 ⎞ −1 ⎛ 300 × 10 ⎞ −135° = − tan −1 ⎜ −0−0 ⎟ − tan ⎜ 3 ⎟ ⎝ 300 × 10 ⎠ ⎝ f PD ⎠ = −90° − 45° Now T = 1@ f = 300 kHz

T =1≈

5000 2

⎛ 300 × 103 ⎞ 1+ ⎜ ⎟ ⋅ 2 ⋅1 ⋅1 ⎝ f PD ⎠ 2

⎛ 300 × 103 ⎞ ⎛ 5000 ⎞ 1+ ⎜ ⎟ =⎜ ⎟ ⎝ f PD ⎠ ⎝ 2 ⎠ f PD ≈

12.76 (a)

2

300 × 103 2 ⇒ f PD = 84.8 Hz 5000

At f = 106 Hz,

6 106 −1 10 2 tan − 104 106 = −89.4 − 90 ≈ −180°

φ = − tan −1

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T =

=

103 ⎛ 106 ⎞ 1+ ⎜ 4 ⎟ ⎝ 10 ⎠

2

⎡ ⎛ 106 ⎞ 2 ⎤ ⎢1 + ⎜ 6 ⎟ ⎥ ⎢⎣ ⎝ 10 ⎠ ⎥⎦

103 =5 (100)(2)

T > 1 at f = f180 = 106 Hz, System is unstable.

103

T=

(b)

⎛ f ⎞⎛ f ⎞⎛ f ⎞ ⎜1 + j ⎟⎜ 1 + j 4 ⎟ ⎜ 1 + j 6 ⎟ f PD ⎠ ⎝ 10 ⎠ ⎝ 10 ⎠ ⎝ Phase margin = 45° ⇒ φ = −135

φ = −135 = − tan −1

f f PD

− tan −1

2

f f − 2 tan −1 6 4 10 10

4

f ≅ 10 Hz T =1=

1≅

103 ⎛ 104 ⎞ 1+ ⎜ ⎟ ⎝ f PD ⎠

2

⎛ 104 ⎞ 1+ ⎜ 4 ⎟ ⎝ 10 ⎠

2

⎡ ⎛ 104 ⎞ 2 ⎤ ⎢1 + ⎜ 6 ⎟ ⎥ 10 ⎠ ⎥ ⎣⎢ ⎝ ⎦

103

⎛ 104 ⎞ ⎜ ⎟ (1.414)(1.0) ⎝ f PD ⎠ (104 )(1.414) f PD = 103 f PD = 14.14 Hz

12.77 (a)

⎛ f180 ⎞ ⎛ f ⎞ ⎛ f ⎞ − tan −1 ⎜ 180 4 ⎟ − tan −1 ⎜ 1805 ⎟ 4 ⎟ ⎝ 10 ⎠ ⎝ 5 × 10 ⎠ ⎝ 10 ⎠

φ = −180 = − tan −1 ⎜

f180 ≅ 8.06 × 104 Hz

(b)

500

T =

2

⎛ 8.06 × 10 ⎞ ⎛ 8.06 × 104 ⎞ 1+ ⎜ 1 + ⎟ ⎜ ⎟ 4 4 ⎝ 10 ⎠ ⎝ 5 × 10 ⎠ 500 = (8.122)(1.897)(1.284) 4

2

⎛ 8.06 ×104 ⎞ 1+ ⎜ ⎟ 5 ⎝ 10 ⎠

2

T = 25.3 500 ⎛ ⎞ f ⎛ f ⎞⎛ f ⎞⎛ f ⎞ 1+ j 5 ⎟ ⎜1 + j ⎟⎜ 1 + j 4 ⎟ ⎜ 1 + j 4 ⎟⎜ × 10 5 10 10 f ⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ PD ⎠ Phase Margin = 60° ⇒ φ = −120°

(c)

T=

−120 = − tan −1

f f f f − tan −1 4 − tan −1 − tan −1 5 10 5 × 104 10 f PD

Assume tan −1

f f PD

≅ 90°

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Then f ≅ 4.2 × 103 Hz 500

T =1=

2

⎛ 4.2 × 103 ⎞ ⎛ 4.2 × 103 ⎞ 1+ ⎜ ⎟ 1+ ⎜ ⎟ 4 ⎝ 10 ⎠ ⎝ f PD ⎠ 500

1=

2

⎛ 4.2 × 103 ⎞ 1+ ⎜ 4 ⎟ ⎝ 5 × 10 ⎠

2

⎛ 4.2 × 103 ⎞ 1+ ⎜ ⎟ 5 ⎝ 10 ⎠

2

⎛ 4.2 × 103 ⎞ 1+ ⎜ ⎟ (1.085)(1.004)(1.0) ⎝ f PD ⎠

4.2 × 103 500 ≅ (1.0846)(1.0035)(1.0) f PD f PD = 9.14 Hz

12.78 50 =

(a) T=

104 ⇒ β = 0.0199 1 + (104 ) β

(0.0199)(104 ) ⎛ f ⎞⎛ f ⎞ ⎜1 + j ⎟⎜ 1 + j 5 ⎟ 10 ⎠ f PD ⎠ ⎝ ⎝

Phase margin = 45° ⇒ φ = −135° −135 = − tan −1

f f PD

− tan −1

f 105

5

f = 10 Hz T =1=

(0.0199)(104 ) ⎛ 105 ⎞ 1+ ⎜ ⎟ ⎝ f PD ⎠

2

⎛ 105 ⎞ 1+ ⎜ 5 ⎟ ⎝ 10 ⎠

2

105 (0.0199)(104 ) = f PD 1.414 f PD = 711 Hz 20 =

(b)

T=

104 ⇒ β = 0.0499 1 + (104 ) β

(0.0499)(104 ) f ⎞⎛ f ⎞ ⎛ ⎜1 + j ⎟⎜ 1 + j 5 ⎟ 711 ⎠⎝ 10 ⎠ ⎝

T =1=

(0.0499)(104 ) ⎛ f ⎞ 1+ ⎜ ⎟ ⎝ 711 ⎠

2

⎛ f ⎞ 1+ ⎜ 5 ⎟ ⎝ 10 ⎠

2

f = 1.76 × 105 Hz 5 ⎛ 1.76 × 105 ⎞ −1 ⎛ 1.76 × 10 ⎞ ⎟ − tan ⎜ ⎟ 5 ⎝ 711 ⎠ ⎝ 10 ⎠

φ = − tan −1 ⎜

= −89.77 − 60.40

φ = −150.2 Phase Margin = 180 − 150.2 = 29.8°

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2

12.79 (a) 20 =

AO = 100 dB ⇒ AO = 105

105 ⇒ β = 0.04999 1 + (105 ) β

(0.04999)(105 ) ⎛ f ⎞⎛ f ⎞⎛ f ⎞ ⎜1 + j ⎟⎜ 1 + j 6 ⎟⎜ 1 + j 7 ⎟ f 10 10 ⎝ ⎠⎝ ⎠ PD ⎠ ⎝ Phase Margin = 45° ⇒ φ = −135° T=

−135 = − tan −1

f f f − tan −1 6 − tan −1 7 10 10 f PD

f ≈ 106 Hz T =1=

1=

(0.04999)(105 ) ⎛ 106 ⎞ 1+ ⎜ ⎟ ⎝ f PD ⎠

2

⎛ 106 ⎞ 1+ ⎜ 6 ⎟ ⎝ 10 ⎠

2

⎛ 106 ⎞ 1+ ⎜ 7 ⎟ ⎝ 10 ⎠

2

(0.04999)(105 ) 2

⎛ 106 ⎞ 1+ ⎜ ⎟ (1.414)(1.005) ⎝ f PD ⎠

106 (0.04999)(105 ) = f PD (1.414)(1.005) f PD = 2.84 Hz

(b) T =1=

5=

105 ⇒ β = 0.19999 1 + (105 ) β (0.19999)(105 )

⎛ f ⎞ 1+ ⎜ ⎟ ⎝ 284 ⎠

2

⎛ f ⎞ 1+ ⎜ 6 ⎟ ⎝ 10 ⎠

2

⎛ f ⎞ 1+ ⎜ 7 ⎟ ⎝ 10 ⎠

2

f = 2.25 × 106 Hz 6 6 ⎛ 2.25 × 106 ⎞ −1 ⎛ 2.25 × 10 ⎞ −1 ⎛ 2.25 × 10 ⎞ ⎟ − tan ⎜ ⎟ − tan ⎜ ⎟ 6 7 ⎝ 284 ⎠ ⎝ 10 ⎠ ⎝ 10 ⎠

φ = − tan −1 ⎜

= −89.99 − 66.04 − 12.68

φ = −168.7 Phase Margin = 180 − 168.7 = 11.3° 12.80 a. T( f ) =

T (0) = 100 dB ⇒ T (0) = 105 105 f ⎞⎛ f ⎞⎛ f ⎛ ⎞ 1+ j ⎜1 + j ⎟ ⎜1 + j ⎟ 6 ⎟⎜ 10 ⎠ ⎝ 5 × 10 ⎠ ⎝ 10 × 106 ⎠ ⎝

T =1= = 105 ×

1 ⎛ f ⎞ 1+ ⎜ ⎟ ⎝ 10 ⎠

2

×

1 ⎛ f ⎞ 1+ ⎜ 6 ⎟ ⎝ 5 × 10 ⎠

2

×

1

f ⎛ ⎞ 1+ ⎜ 6 ⎟ ⎝ 10 × 10 ⎠

2

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By trial and error f = 0.976 MHz ⎛ 0.976 × 106 ⎞ −1 ⎛ 0.976 ⎞ −1 ⎛ 0.976 ⎞ ⎟ − tan ⎜ ⎟ − tan ⎜ ⎟ 10 5 ⎝ ⎠ ⎝ 10 ⎠ ⎝ ⎠

φ = − tan −1 ⎜

= −90° − 11.05° − 5.574° = −106.6° Phase Margin = 180° − 106.6° = 73.4°

b.

f P′1 ∝

1 10 75 so = CF f P′1 20

or f P′1 = 2.67 Hz Now T =1= = 105 ×

×

1 ⎛ f ⎞ 1+ ⎜ ⎟ ⎝ 2.67 ⎠

2

×

1 ⎛ f ⎞ 1+ ⎜ 6 ⎟ ⎝ 5 × 10 ⎠

2

1 2

f ⎛ ⎞ 1+ ⎜ 6 ⎟ ⎝ 10 × 10 ⎠ By trial and error f ≈ 2.66 ×105 Hz then ⎛ 2.66 × 105 ⎞ −1 ⎛ 0.266 ⎞ −1 ⎛ 0.266 ⎞ φ = − tan −1 ⎜ ⎟ − tan ⎜ ⎟ − tan ⎜ ⎟ ⎝ 5 ⎠ ⎝ 10 ⎠ ⎝ 2.67 ⎠ = −90° − 3.045° − 1.524° = −94.57° Phase Margin = 180° − 94.57° = 85.4°

12.81 1 where τ = ( Ro1 Ri 2 ) Ci 2πτ = (500 1000) × 103 × 2 × 10−12 ⇒ τ = 6.67 × 10−7 s

(a)

f 3− dB =

Then 1 ⇒ f 3− dB = 239 kHz 2π (6.67 × 10−7 ) (b) For 1 1 f PD = 10 Hz, τ = = = 0.0159 s 2π f PD 2π (10)

f 3− dB =

Then τ = ( Ro1 Ri 2 ) ( Ci + CM )

0.0159 = ( 500 1000 ) × 103 × ( Ci + CM )

or ( Ci + CM ) = 4.77 ×10−8 = 2 x10−12 + CM ⇒ CM = 0.0477 μ F 12.82 Assuming a phase margin of 45°.

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f ⎛ f ⎞ ⎛ ⎞ −135° ≈ −90° − tan −1 ⎜ − tan −1 ⎜ 6 ⎟ 6 ⎟ ⎝ 2 × 10 ⎠ ⎝ 25 × 10 ⎠ By trial and error, f ≈ 1.74 MHz Then T =1 1

= 5000 ×

2

⎛ 1.74 × 106 ⎞ 1+ ⎜ ⎟ f PD ⎠ ⎝ 1 1 × × 2 2 ⎛ 1.74 ⎞ ⎛ 1.74 ⎞ 1+ ⎜ 1+ ⎜ ⎟ ⎟ ⎝ 2 ⎠ ⎝ 25 ⎠

or 1.74 × 106 5000 ≈ f PD (1.325)(1.0024) so f PD = 462 Hz 12.83 20 =

105 ⇒ β = 0.04999 1 + (105 ) β

Phase Margin = 60° ⇒ φ = −120° f

−120 = − tan −1

Assume tan −1

f PD

− tan −1

f f − tan −1 6 10 5 × 107

f = 90° f PD

f ≅ 30° ⇒ f = 5.77 × 105 Hz 106 (0.04999)(105 ) T = 1= 2 2 2 ⎛ 5.77 × 105 ⎞ ⎛ 5.77 × 105 ⎞ ⎛ 5.77 ×105 ⎞ 1+ ⎜ 1 1 + + ⎟ ⎜ ⎟ ⎜ ⎟ 6 7 f PD ⎝ 10 ⎠ ⎝ 5 × 10 ⎠ ⎝ ⎠

tan −1

1=

4.999 × 103 2

⎛ 5.77 × 105 ⎞ 1+ ⎜ ⎟ (1.155)(1.0) f PD ⎝ ⎠

5.77 × 105 4.999 ×103  (1.155)(1.0) f PD f PD = 133.3 Hz

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Chapter 13 Exercise Solutions EX13.1 I C1 = I C 2 ≅ 9.5 μ A 9.5 μ A I B1 = I B 2 = = 0.0475 μ A ⇒ I B1 = I B 2 = 47.5 nA 200 EX13.2 5 − 0.6 − 0.6 − (−5) = 0.22 mA I REF = 40 I C17 = I C13 B = 0.75I REF = (0.75)(0.22) = 0.165 mA 0.165 (0.165)(0.1) + 0.6 + I C16 = 200 50 = 0.000825 + 0.01233 I C16 = 13.2 μ A EX13.3

⎛V ⎞ 0.18 ×10−3 = 10−14 exp ⎜ D ⎟ ⎝ VT ⎠ ⎛ 0.18 × 10−3 ⎞ VD = VT ln ⎜ ⎟ −14 ⎝ 10 ⎠ ⎛ 0.18 × 10−3 ⎞ = (0.026) ln ⎜ ⎟ −14 ⎝ 10 ⎠ VD = 0.6140 VBB = 2VDD ≅ 1.228 V

⎛ V /2 ⎞ I C14 = I C 20 = I S exp ⎜ BB ⎟ ⎝ VT ⎠ ⎛ 0.6140 ⎞ = 3 × 10−14 exp ⎜ ⎟ ⎝ 0.026 ⎠ I C14 = I C 20 = 0.541 mA

EX13.4 100 ro 6 = ⇒ 10.5 MΩ 0.0095 Then, using results from Example 13.4

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Ract1 = ro 6 [1 + g m 6 ( R2 || rπ 6 ) ] = 10.5 [1 + (0.365)(1|| 547) ] = 14.3 MΩ V 100 ro 4 = A = ⇒ 10.5 MΩ I C 4 0.0095 ⎛ 9.5 ⎞ Ad = − ⎜ ⎟ (10.5 ||14.3 || 4.07 ) = −889 ⎝ 0.026 ⎠

EX13.5 100 = 556 K 0.18 Ri 3 = rπ 22 + (1 + β P )[ R19 || R20 ] = 7.22 + (51)(556 ||111) ⇒ 4.73 MΩ 100 Ract 2 = = 185 K 0.54 100 Ro17 = = 185 K 0.54 −(200)(201)(50)(185 || 4730 ||185) Av 2 = 4070[50 + [9.63 + (201)(0.1)]] −182358786.9 = 32450.1 Av 2 = −562 R19 = ro13 A =

EX13.6 100 100 ro17 = = 185 K ro13 B = = 185 K 0.54 0.54 RC17 = 185 [1 + (20.8)(0.1|| 9.63) ] RC17 = 566 K 7.22 + 566 ||185 = 2.88 K Re 22 = 51 100 = 556 K RC19 = 0.18 0.65 + 2.88 || 556 Re 20 = = 0.0689 51 = 68.9 Ω RO = 22 + 68.9 = 90.9 Ω EX13.7 Ci = C1 (1+ | A2 |) = 30(1 + 562) = 16890 pF Ri 2 = 4.07 MΩ Ro1 = Ract1 || ro 4 = 14.3 ||10.5 = 6.05 MΩ Then Req = Ro1 || Ri 2 = 6.05 || 4.07 = 2.43 MΩ

Then f PD =

1 1 = 6 2π Req Ci 2π (2.43 × 10 )(16890 × 10−12 )

= 3.88 Hz

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EX13.8 For M 5 , M 6 ⎛ 40 ⎞ K P = (12.5) ⎜ ⎟ = 250 μ A / V 2 ⎝ 2 ⎠ 5 + 5 − VSG 5 0.25(VSG 5 − 0.5) 2 = 225 2 V 56.25(VSG 0.25) − + = 10 − VSG 5 SG 5 5 2 56.25VSG 5 − 55.25VSG 5 + 4.0625 = 0

55.25 ± 3052.5625 − 914.0625 2(56.25) = 0.902 V

VSG 5 = VSG 5

10 − 0.902 ⇒ 40.4 μ A 225 Current in M 1 − M 4 = 20.2 μ A I set = I Q =

EX13.9 ⎛ 40 ⎞ K P1 = K P 2 = (12.5) ⎜ ⎟ = 250 μ A/V 2 ⎝ 2 ⎠ From Exercise Ex 13.8, I D1 = I D 2 = 20.2 μ A ro 4 = ro 2 =

1 1 = = 2.48 MΩ λ I D (0.02)(0.0202)

Ad = 2 K P1 I Q (ro 2 || ro 4 ) = 2(0.25)(0.0404)(2480 || 2480) Ad = 176 ⎛ 80 ⎞ K n 7 = (12.5) ⎜ ⎟ = 500 μ A / V 2 ⎝ 2⎠ g m 7 = 2 K n 7 I Q = 2 (0.50)(0.0404) = 0.284 mA/V ro 7 = ro8 =

1

λ ID7

=

1 ⇒ 1.24 MΩ (0.02)(0.0404)

Av 2 = g m 7 (ro 7 || ro8 ) = (0.284)(1240 || 1240) = 176 AV = Ad ⋅ AV 2 = (176)(176) = 30,976

EX13.10 (a) ⎛ k ′ ⎞⎛ W ⎞ ⎛ I Q1 ⎞ g m1 = 2 ⎜ n ⎟ ⎜ ⎟ ⎜ ⎟ ⎝ 2 ⎠ ⎝ L ⎠1 ⎝ 2 ⎠

⎛ 0.08 ⎞ ⎛ 0.25 ⎞ =2 ⎜ ⎟ (20) ⎜ ⎟ ⎝ 2 ⎠ ⎝ 2 ⎠ = 0.6325 mA/V

1 = 800 K (0.01)(0.25/ 2) 1 ro 4 = = 533.3 K (0.015)(0.25/ 2) ro 2 =

Ad 1 = g m1 (ro 2 || ro 4 ) = (0.6325)(800 || 533.3) Ad 1 = 202.4

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⎛ k′ ⎞⎛ W ⎞ g m5 = 2 ⎜ n ⎟ ⎜ ⎟ IQ 2 ⎝ 2 ⎠ ⎝ L ⎠5

⎛ 0.04 ⎞ =2 ⎜ ⎟ (80)(0.25) ⎝ 2 ⎠ = 1.265 mA/V

1 = 266.7 K (0.015)(0.25) 1 ro9 = = 400 K (0.01)(0.25) ro 5 =

A2 = − g m 5 (ro 5 || ro 9 ) = −(1, 265)(266.7 || 400) A2 = −202.4

Overall gain (assuming A3 = 1) A = Ad 1 ⋅ A2 = (202.4)(−202.4) = −40,966 EX13.11 I D1 = I D 2 = 25 μ A g m1 = g m8 = 2

k ′p ⎛ W ⎜ 2 ⎝L

⎞ ⎛ 40 ⎞ ⎟ I DQ = 2 ⎜ ⎟ (25)(25) ⇒ g m1 = g m8 = 224 μ A / V ⎠ ⎝ 2 ⎠

⎛ k′ ⎞⎛ W gm6 = 2 ⎜ n ⎟ ⎜ ⎝ 2 ⎠⎝ L

⎞ ⎛ 80 ⎞ ⎟ I DQ = 2 ⎜ ⎟ (25)(25) ⇒ g m 6 = 316 μ A / V ⎠ ⎝ 2⎠ 1 1 = = 2 MΩ ro1 = ro 6 = ro8 = ro10 = λ I D (0.02)(25)

ro 4 =

1

λ ID4

=

1 =1 MΩ (0.02)(50)

Ro8 = g m8 (ro8 ro10 ) = (224)(2)(2) = 896 M Ω Ro 6 = g m 6 (ro 6 ) ( ro 4 || ro1 ) = 316(2) (1|| 2 ) = 421 M Ω

Then Ad = g m1 ( Ro 6 Ro8 ) = 224 ( 421 896 ) ⇒ Ad = 64,158 EX13.12 V + − V − = VEB1 + VEB 6 + VBE 7 + I1 R1 = 0.6 + 0.6 + 0.6 + (0.236)(8) = 3.69 V So V + = −V − = 1.845 V EX13.13 Ad = 2 K P I Q 5 ( Ri 2 ) = 2(1)(0.2)(26) Ad = 16.4

EX13.14

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rπ 13 =

β VT I C13

=

(200)(0.026) 0.20

= 26 kΩ Ri 2 = rπ 13 + (1 + β ) RE13 = 26 + 201(1) = 227 kΩ r010 = r012 = g m12 = rπ 12 =

1

λ I D10

=

1 = 500 kΩ (0.02)(0.1)

VA 50 = = 500 kΩ I C12 0.1 I C12 0.1 = = 3.85 mA / V VT 0.026

β VT I C12

=

(200)(0.026) = 52 kΩ 0.1

Ract1 = r012 [1 + g m12 (rπ 12 || R5 )] = 500[1 + (3.85)(52 || 0.5)] = 1453 kΩ Ad = 2 K n I Q 5 ⋅ ( ro10 || Ract1 || R12 ) = 2(0.6)(0.2) ⋅ (500 ||1453 || 227) = (0.490)(141) ⇒ Ad = 69.1

EX13.15 From Example 13.15, f PD = 265 Hz Av = Adi ⋅ A2 = (16.4)(1923) = 31,537 fT = f PD ⋅ Av = (265)(31,537) = 8.36 MHz

TYU13.1 Computer Analysis TYU13.2 Computer Analysis TYU13.3 Vi N (max) = V + − VEB (on) = 15 − 0.6 = 14.4 V Vi N (min) ≅ 4VBE (on) + V + = 4(0.6) − 15 = −12.6 V −12.6 ≤ Vi N (cm) ≤ 14.4 V TYU13.4

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V0 (max) ≅ V + − 2VBE (on) = 15 − 2(0.6)

a.

V0 (max) = 13.8 V V0 (min)

= 3VBE (on) + V − = 3(0.6) − 15

V0 (min)

≅ −13.2 V −13.2 ≤ V0 ≤ 13.8 V V0 (max) = 5 − 1.2 = 3.8 V

b.

V0 (min) ≅ 3VBE + V − = 3(0.6) − 5 = −3.2 V −3.2 ≤ V0 ≤ 3.8 V

TYU13.5 15 − 2(0.6) − (−15) = 0.72 mA I REF ≅ 40 ⎛I ⎞ ⎛ 0.72 × 10−3 ⎞ VBE = VT ln ⎜ REF ⎟ = (0.026) ln ⎜ ⎟ −14 ⎝ 10 ⎠ ⎝ IS ⎠ = 0.650 V

So I REF =

30 − 2(0.65) ⇒ I REF = 0.718 mA 40 VBE11 = 0.650 V

⎛I ⎞ I C10 R4 = VT ln ⎜ REF ⎟ ⎝ I C10 ⎠ ⎛ 0.718 ⎞ I C10 (5) = (0.026) ln ⎜ ⎟ ⎝ I C10 ⎠ By trial and error: I C10 = 18.9 μ A

VBE10 = VBE11 − I C10 R4 = 0.650 − (0.0189)(5) ⇒ VBE10 ≅ 0.556 V I C10 18.9 = = 9.45 μ A 2 2 ⎛I ⎞ ⎛ 9.45 × 10−6 ⎞ = VT ln ⎜ C 6 ⎟ = (0.026)ln ⎜ ⎟ ⇒ VBE 6 = 0.537 V −14 ⎝ 10 ⎠ ⎝ IS ⎠

IC 6 ≅ VBE 6

TYU13.6 10 − 0.6 − 0.6 − (−10) I REF = ⇒ I REF = 0.47 mA 40 ⎛I ⎞ I C10 R4 = VT ln ⎜ REF ⎟ ⎝ I C10 ⎠ ⎛ 0.47 ⎞ I C10 (5) = (0.026) ln ⎜ ⎟ ⎝ I C10 ⎠ By trial and error: ⇒ I C10 ≅ 17.2 μ A IC 6 ≅ I C13 B

I C10 ⇒ I C 6 = 8.6 μ A 2 = (0.75) I REF ⇒ I C13 B = 0.353 mA

I C13 A = (0.25) I REF ⇒ I C13 A = 0.118 mA

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TYU13.7

R0 = R6 + RE14 RE14 =

rπ 14 + R0 d R013 A

1 + βn The diode resistance can be found as ⎛V ⎞ I D = I S exp ⎜ D ⎟ ⎝ VT ⎠ ⎛ 1 1 ∂I D = = IS ⎜ rd ∂VD ⎝ VT

⎞ ⎛ VD ⎟ ⋅ exp ⎜ ⎠ ⎝ VT

⎞ ID ⎟= ⎠ VT

or rd =

VT V 0.026 = T = ⇒ 144 Ω I D I C13 A 0.18

RE 22 =

rπ 22 + R017 R013 B 1+ βP

R013 B = r013 B = 92.6 kΩ R017 = r017 ⎡⎣1 + g m17 ( R8 rπ 17 ) ⎦⎤ = 283 kΩ

From previous calculations RE 22 = 1.51 kΩ R0 d = 2rd + RE 22 = 2(0.144) + 1.51 = 1.80 kΩ R013 A = r013 A = 278 kΩ βV (200)(0.026) rπ 14 = n T = = 1.04 kΩ I C14 5 1.04 + 1.8 || 278 ⇒ 14.1 Ω 201 R0 = R6 + RE14 = 27 + 14.1 ⇒ R0 ≅ 41 Ω RE14 =

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TYU13.8 For M 6 we have VSG 5 = VSG 6 = 1.06 V So VSD 6 (sat) = 1.06 − 0.5 = 0.56 V For M 1 and M 2 ID =

IQ

= K P (VSG1 + VTP )

2

2 0.0397 = 0.125(VSG1 − 0.5) 2 ⇒ VSG1 = 0.898 V 2 So maximum input voltage = V + − VSD 6 (sat) − VSG1 = 5 − 0.56 − 0.898 ⇒ Vi N (max) = 3.54 V

For M 3, K p = (6.25)(20) = 125 μ A / V 2 I D3 =

IQ

=

39.7 μA 2

2 39.7 = 125(VGS 3 − VTN ) 2 2 VTN = 0.5V ⇒ VGS 3 = 0.898 V VSD1 (sat) = 0.898 − 0.5 = 0.398 V Vi N (min) = V − + VGS 3 + VSD1 (sat) − VSG1 = −5 + 0.898 + 0.398 − 0.898 Vi N (min) = −4.60 V

−4.60 ≤ Vi N (cm) ≤ 3.54 V TYU13.9 V0 (max) = V + − VSD 8 (sat) VSG 8 = VSG 5 = 1.06 V VSD 8 (sat) = 1.06 − 0.5 = 0.56 V V0 (max) = 5 − 0.56 = 4.44 V V0 (min) = V − + VDS 7 (sat) VGS 7 = 1.06 ⇒ VDS 7 (sat) = 1.06 − 0.5 = 0.56 V0 (min) = −5 + 0.56 = −4.44 −4.44 ≤ V0 ≤ 4.44 V

TYU13.10 (a) For M 5, K p 5 = 125 μ A / V 2

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V + − V − − VSG 5 Rset

K p 5 (VSG 5 + VTP ) 2 =

5 + 5 − VSG 5 100 2 12.5(VSG 5 − VSG 5 + 0.25) = 10 − VSG 5 0.125(VSG 5 − 0.5) 2 =

2 12.5VSG 5 − 11.5VSG 5 − 6.875 = 0

11.5 ± (11.5) 2 + 4(12.5)(6.875) 2(12.5) = 1.33 V

VSG 5 = VSG 5 Then

10 − 1.33 ⇒ 86.7 μ A 100 IQ = ID4 = = 43.35 μ A 2

I REF = I Q = I D 8 = I D 7 = I D1 = I D 2 = I D 3

K p1 = K p 2 = 125μ A / V 2

(b)

ro 2 = ro 4 =

I

λ ID

=

1 = 1153 k Ω (0.02)(0.04335)

Input stage gain Ad = 2 K p1 I Q ⋅ (ro 2 ro 4 ) = 2(0.125)(0.0867) ⋅ (1153 1153) ⇒ Ad = 84.9

Transconductance of M 7 g m 7 = 2 K n 7 I D 7 = 2 (0.250)(0.0867) = 0.294 mA / V ro 7 = ro8 =

1

λ I D7

=

1 = 577 k Ω (0.02)(0.0867)

Second stage gain Av 2 = g m 7 (ro 7 ro8 ) = (0.294)(577 577) ⇒ Av 2 = 84.8 Overall gain = Ad ⋅ Av 2 = (84.9)(84.8) ⇒ A = 7, 200

TYU13.11 (a) Ad = Bg m1 ( ro 6 ro8 ) ⎛ k ′ ⎞⎛ W g m1 = 2 ⎜ n ⎟ ⎜ ⎝ 2 ⎠⎝ L g m1 = 400 μ A / V

ro 6 = ro8 =

1

λP I D 6

=

⎞ ⎛ 80 ⎞ ⎟ I D1 = 2 ⎜ ⎟ (20)(50) ⎠ ⎝ 2⎠

1 = 0.333 M Ω (0.02)(150)

1 1 = = 0.333 M Ω λn I D 8 (0.02)(150)

Ad = 3(400) ( 0.333 || 0.333) ⇒ Ad = 200

(b)

f PD =

1 2π Ro (CL + CP )

where Ro = ro 6 || ro8 = 0.333 || 0.333 M Ω

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f PD =

1 ⇒ f PD = 477 kHz 2π ( 0.333 || 0.333) × 106 × 2 × 10−12

f PD ⋅ Ad = (477 × 103 )(200) ⇒ 95.5 MHz

TYU13.12 (a) From Exercise TYU 13.11, g m1 = 400 μ A / V ro 6 = ro8 = ro10 = ro12 = 0.333 M Ω ⎛ k′ g m10 = 2 ⎜ P ⎝ 2

⎞⎛W ⎞ ⎛ 40 ⎞ ⎟ ⎜ L ⎟ I D10 = 2 ⎜ 2 ⎟ (20)(150) ⇒ g m10 = 490 μ A / V ⎝ ⎠ ⎠⎝ ⎠

⎛ k ′ ⎞⎛ W ⎞ ⎛ 80 ⎞ g m12 = 2 ⎜ n ⎟ ⎜ ⎟ I D12 = 2 ⎜ ⎟ (20)(150) ⇒ g m12 = 693 μ A / V ⎝ 2⎠ ⎝ 2 ⎠⎝ L ⎠ Ro10 = g m10 (ro10 ro 6 ) = (490)(0.333)(0.333) = 54.4 M Ω Ro12 = g m12 (ro12 ro8 ) = (693)(0.333)(0.333) = 77.0 M Ω Ad = Bg m1 ( Ro10 || Ro12 ) = 3(400)(54.4 || 77.0) ⇒ Ad = 38, 254 Ro = Ro10 || Ro12 = 54.4 || 77.0 = 31.9 M Ω

(b)

1 = 2.50 kHz 2π (31.9 × 106 )(2 × 10−12 ) f PD ⋅ Ad = (2.5 × 103 )(38, 254) ⇒ 95.6 MHz f PD =

TYU13.13 (a) Ad = g m1 ( Ro 6 || Ro8 )

From Example 13.10, g m1 = 316 μ A / V , Ro8 = 316 M Ω Now Ro 6 = g m 6 (ro 6 )(ro 4 || ro1 ) ro1 = 1 M Ω, ro 4 = 0.5 M Ω I 50 gm6 = C 6 = ⇒ 1.923 mA / V VT 0.026 ro 6 =

VA6 80 = = 1.6 M Ω I C 6 50

Then Ro 6 = (1.923)(1600)(0.5 || 1) = 1026 M Ω Ad = (316)(1026 || 316) ⇒ Ad = 76,343 (b)

1 ⇒ f PD = 329 Hz 2π (316 || 1026) × 106 × 2 × 10−12 f PD ⋅ Ad = (329)(76,343) ⇒ 25.1 MHz f PD =

TYU13.14 For Q7 and R1 VSG = VBE 7 + I1 R1 = 0.6 + I1 (5) For M 8 : I 2 = K p (VSG + VTP ) 2 I 2 = 0.3(VSG − 1.4) 2 By trial and error:

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VSG = 2.54 V I1 = I 2 = 0.388 mA

TYU13.15 For J 6 biased in the saturation region ⇒ I C 3 = I DSS = 300 μ A Q1 , Q2 , Q3 are matched ⇒ I C1 = I C 2 = I C 3 = 300 μ A

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Chapter 13 Problem Solutions 13.1

Computer Simulation

13.2

Computer Simulation

13.3

(

Ad = g m1 ro 2 ro 4 Ri 6

(a)

)

g m1 =

I C1 20 = ⇒ 0.769 mA / V VT 0.026

ro 2 =

VA 2 80 = = 4 MΩ I C 2 20

ro 4 =

VA 4 80 = = 4 MΩ I C 2 20

Ri 6 = rπ 6 + (1 + β n ) ⎡⎣ R1 rπ 7 ⎤⎦ (120)(0.026) rπ 7 = = 15.6 k Ω 0.2 V (on) 0.6 = = 0.030 mA I C 6 ≅ BE 20 R1 (120)(0.026) = 104 k Ω 0.030

rπ 6 =

Then Ri 6 = 104 + (121) ⎡⎣ 20 15.6 ⎤⎦ ⇒ 1.16 M Ω Then

(

)

Ad = 769 4 4 1.16 ⇒ Ad = 565

Now ⎛ R1 Vo = − I c 7 ro 7 = −( β n I b 7 )ro 7 = − β n ro 7 ⎜ ⎝ R1 + rπ 7

⎞ ⎟ Ic6 ⎠

⎛ R1 ⎞ Vo1 = − β n (1 + β n )ro 7 ⎜ ⎟ I b 6 and I b 6 = + R r Ri 6 ⎝ 1 π7 ⎠ Then − β n (1 + β n )ro 7 ⎛ R1 ⎞ V Av 2 = o = ⎜ ⎟ Vo1 Ri 6 ⎝ R1 + rπ 7 ⎠ ro 7 =

VA 80 = = 400 k Ω I C 7 0.2

So Av 2 =

−(120)(121)(400) ⎛ 20 ⎞ ⎜ ⎟ ⇒ Av 2 = −2813 1160 ⎝ 20 + 15.6 ⎠

Overall gain = Ad ⋅ Av 2 = (565)(−2813) ⇒ A = −1.59 ×106

(b)

Rid = 2rπ 1 and rπ 1 =

(80)(0.026) = 104 k Ω 0.020

Rid = 208 k Ω

(c)

f PD =

1 and CM = (10)(1 + 2813) = 28,140 pF 2π Req CM

Req = ro 2 ro 4 Ri 6 = 4 4 1.16 = 0.734 M Ω f PD =

1 = 7.71 Hz 2π (0.734 × 10 )(28,140 × 10−12 ) 6

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Gain-Bandwidth Product = (7.71)(1.59 × 106 ) ⇒ 12.3 MHz

13.4 a. b.

Q3 acts as the protection device. Same as part (a).

13.5 If we assume VBE (on) = 0.7 V, then Vin = 0.7 + 0.7 + 50 + 5 So breakdown voltage ≈ 56.4 V. 13.6 (a)

I REF =

15 − 0.6 − 0.6 − (−15) = 0.50 ⇒ R5 = 57.6 k Ω R5

⎛I ⎞ I C10 R4 = VT ln ⎜ REF ⎟ ⎝ I C10 ⎠ 0.026 ⎛ 0.50 ⎞ R4 = ln ⎜ ⎟ ⇒ R4 = 2.44 k Ω 0.030 ⎝ 0.030 ⎠ 5 − 0.6 − 0.6 − (−5) ⇒ I REF = 0.153 mA 57.6 ⎛ 0.153 ⎞ I C10 (2.44) = (0.026) ln ⎜ ⎟ ⎝ I C10 ⎠

(b)

I REF =

By trial and error, I C10 ≅ 21.1 μ A

13.7 (a)

I REF ≅ 0.50 mA

⎛I ⎞ ⎛ 0.50 × 10−3 ⎞ VBE = VT ln ⎜ REF ⎟ = (0.026) ln ⎜ ⎟ ⇒ VBE11 = 0.641V = VEB12 −14 ⎝ 10 ⎠ ⎝ IS ⎠ Then 15 − 0.641 − 0.641 − (−15) ⇒ R5 = 57.4 k Ω R5 = 0.50 0.026 ⎛ 0.50 ⎞ ln ⎜ R4 = ⎟ ⇒ R4 = 2.44 k Ω 0.030 ⎝ 0.030 ⎠ ⎛ 0.030 × 10−3 ⎞ VBE10 = 0.026 ln ⎜ ⎟ ⇒ VBE10 = 0.567 V −14 ⎝ 10 ⎠

(b)

From Problem 13.6, I REF ≅ 0.15 mA

⎛ 0.15 × 10−3 ⎞ VBE11 = VEB12 = 0.026 ln ⎜ ⎟ = 0.609 V −14 ⎝ 10 ⎠ 5 − 0.609 − 0.609 − (−5) ⇒ I REF = 0.153 mA Then I REF = 57.4 Then I C10 ≅ 21.1 μ A from Problem 13.6

13.8 5 − 0.6 − 0.6 − (−5) ⇒ I REF = 0.22 mA 40 ⎛I ⎞ I C10 R4 = VT ln ⎜ REF ⎟ ⎝ I C10 ⎠

a.

I REF =

⎛ 0.22 ⎞ I C10 (5) = (0.026) ln ⎜ ⎟ ⎝ I C10 ⎠

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By trial and error; I C10 ≅ 14.2 μ A I C10 ⇒ I C 6 = 7.10 μ A 2 = 0.75 I REF ⇒ I C17 = 0.165 mA

IC 6 ≅ I C17

I C13 A = 0.25I REF ⇒ I C13 A = 0.055 mA

(b) Using Example 13.4 rπ 17 = 31.5 kΩ RE′ = 50 [31.5 + (201)(0.1)] = 50 51.6 = 25.4 kΩ rπ 16 =

β nVT I C16

and

0.165 (0.165)(0.1) + 0.6 + = 0.0132 mA 200 50 rπ 16 = 394 kΩ Then Ri 2 = 394 + (201)(25.4) ⇒ 5.5 MΩ rπ 6 = 732 kΩ 0.00710 gm6 = = 0.273 mA / V 0.026 50 r06 = = 7.04 MΩ 0.0071 Then Ract1 = 7.04[1 + (0.273)(1 732)] = 8.96 MΩ 50 r04 = = 7.04 MΩ 0.0071 Then ⎛ 7.1 ⎞ Ad = − ⎜ ⎟ (7.04 8.96 5.5) ⎝ 0.026 ⎠ or Ad = −627 Gain of differential amp stage I C16 =

Using Example 13.5, and neglecting the input resistance to the output stage: V 50 Ract 2 = A = = 303 kΩ I C13 B 0.165 Av 2 =

−(200)(201)(50)(303) (303)

(5500)[50 + 31.5 + (201)(0.1)] or Av 2 = −545 Gain of second stage

13.9 I C10 = 19 μ A From Equation (13.6) ⎡ β 2 + 2β P + 2 ⎤ ⎡ (10) 2 + 2(10) + 2 ⎤ = 2 I C10 = 2 I ⎢ P2 I ⎥ ⎢ ⎥ 2 ⎣ (10) + 3(10) + 2 ⎦ ⎣ β P + 3β P + 2 ⎦ ⎡122 ⎤ = 2I ⎢ ⎥ ⎣132 ⎦

So ⎛ 132 ⎞ 2 I = (19) ⎜ ⎟ = 20.56 μ A ⎝ 122 ⎠ I C 2 = I = 10.28 μ A

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IC 9 =

I B9

2I

=

20.56 ⇒ I C 9 = 17.13 μ A 2⎞ ⎛ + 1 ⎜ ⎟ ⎝ 10 ⎠

⎛ 2 ⎞ ⎜1 + ⎟ β ⎝ P ⎠ I 17.13 = C9 = ⇒ I B 9 = 1.713 μ A βP 10

IB4 =

10.28 I = ⇒ I B 4 = 0.9345 μ A (1 + β P ) 11

⎛ β IC 4 = I ⎜ P ⎝1+ βP

⎞ ⎛ 10 ⎞ ⎟ = (10.28) ⎜ ⎟ ⇒ I C 4 = 9.345 μ A 11 ⎠ ⎝ ⎠

13.10 VB 5 − V − = VBE (on) + I C 5 (1) = 0.6 + (0.0095)(1) = 0.6095 0.6095 IC 7 = ⇒ I C 7 = 12.2 μ A 50 I C 8 = I C 9 = 19 μ A I REF = 0.72 mA I E13 = I REF = 0.72 mA I C14 = 138 μ A Power = (V + − V − ) [ I C 7 + I C 8 + I C 9 + I REF + I E13 + I C14 ] = 30[0.0122 + 0.019 + 0.019 + 0.72 + 0.72 + 0.138] ⇒ Power = 48.8 mW Current supplied by V + and V − = I C 7 + I C 8 + I C 9 + I REF + I E13 + I C14 = 1.63 mA

13.11 (a)

vcm (min) = −15 + 0.6 + 0.6 + 0.6 + 0.6 = −12.6 V

vcm (max) = +15 − .6 = 14.4 V So − 12.6 ≤ vcm ≤ 14.4 V

(b)

vcm (min) = −5 + 4(0.6) = −2.6 V

vcm (max) = 5 − 0.6 = 4.4 V So − 2.6 ≤ vcm ≤ 4.4 V 13.12 If v0 = V − = −15 V , the base voltage of Q14 is pulled low, and Q18 and Q19 are effectively cut off. As a first approximation 0.6 I C14 = = 22.2 mA 0.027 22.2 I B14 = = 0.111 mA 200 Then I C15 = I C13 A − I B14 = 0.18 − 0.111 = 0.069 mA Now ⎛I ⎞ VBE15 = VT ln ⎜ C15 ⎟ ⎝ 15 ⎠ ⎛ 0.069 × 10−3 ⎞ = (0.026) ln ⎜ ⎟ −14 ⎝ 10 ⎠ = 0.589 V

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As a second approximation 0.589 I C14 = ⇒ I C14 = 21.8 mA 0.027 21.8 I B14 = = 0.109 mA 200 and I C15 = 0.18 − 0.109 ⇒ I C15 = 0.071 mA 13.13 a. Neglecting base currents: I D = I BIAS Then ⎛I ⎞ VBB = 2VD = 2VT ln ⎜ D ⎟ ⎝ IS ⎠ ⎛ 0.25 × 10−3 ⎞ = 2(0.026) ln ⎜ −14 ⎟ ⎝ 2 × 10 ⎠

or VBB = 1.2089 V ⎛V / 2⎞ I CN = I CP = I S exp ⎜ BB ⎟ ⎝ VT ⎠ ⎛ 1.2089 ⎞ = 5 × 10−14 exp ⎜ ⎟ ⎝ 2(0.026) ⎠ So I CN = I CP = 0.625 mA

For vI = 5 V, v0 ≅ 5 V

b.

5 = 1.25 mA 4 As a first approximation I CN ≈ iL = 1.25 mA iL ≅

⎛ 1.25 × 10−3 ⎞ VBEN = (0.026) ln ⎜ = 0.6225 V −14 ⎟ ⎝ 5 × 10 ⎠ Neglecting base currents, VBB = 1.2089 V

Then VEBP = 1.2089 − 0.6225 = 0.5864 V ⎛ 0.5864 ⎞ I CP = 5 × 10−14 exp ⎜ ⎟ ⇒ I CP = 0.312 mA ⎝ 0.026 ⎠ As a second approximation, I CN = iL + I CP = 1.25 + 0.31 ⇒ I CN ≅ 1.56 mA

13.14 R1 + R2 =

VBB 1.157 = = 64.28 kΩ (0.1) I BIAS 0.018

⎛I VBE = VT ln ⎜ C ⎝ IS

⎞ ⎛ (0.9) I BIAS ⎞ ⎟ = (0.026) ln ⎜ ⎟ IS ⎠ ⎝ ⎠

⎛ 0.162 × 10−3 ⎞ = (0.026) ln ⎜ ⎟ −14 ⎝ 10 ⎠

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VBE = 0.6112 V ⎛ R2 ⎞ VBE = ⎜ ⎟ VBB ⎝ R1 + R2 ⎠ ⎛ R ⎞ 0.6112 = ⎜ 2 ⎟ (1.157) ⎝ 64.28 ⎠ So R2 = 33.96 kΩ

Then R1 = 30.32 kΩ 13.15

(

Ad = − g m ro 4 ro 6 Ri 2

(a)

)

From example 13.4 9.5 gm = = 365 μ A / V , ro 4 = 5.26 M Ω 0.026 Now ro 6 = ro 4 = 5.26 M Ω Assuming R8 = 0, we find Ri 2 = rπ 16 + (1 + β n ) RE′ = 329 + (201) ( 50 9.63) ⇒ 1.95 M Ω

Then

(

)

Ad = −(365) 5.26 5.26 1.95 ⇒ Ad = −409

(b) Av 2 =

From Equation (13.20),

(

− β n (1 + β n ) R9 Ract 2 Ri 3 R017

{

}

)

Ri 2 R9 + ⎡⎣ rπ 17 + (1 + β n ) Rg ⎤⎦

For Rg = 0, Ri 2 = 1.95 M Ω Using the results of Example 13.5 Av 2 =

(

−200(201)(50) 92.6 4050 92.6 (1950){50 + 9.63}

)⇒A

v2

= −792

13.16 Let I C10 = 40 μ A, then I C1 = I C 2 = 20 μ A. Using Example 13.5, Ri 2 = 4.07 MΩ (200)(0.026) = 260 kΩ 0.020 0.020 gm6 = = 0.769 mA/V 0.026 50 r06 = ⇒ 2.5 MΩ 0.02 Then Ract1 = 2.5[1 + (0.769)(1 260)] = 4.42 MΩ 50 ⇒ 2.5 MΩ r06 = 0.02 Then rπ 6 =

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⎛ I CQ ⎞ Ad = − ⎜ ⎟ (r04 Ract1 Ri 2 ) ⎝ VT ⎠ ⎛ 20 ⎞ = −⎜ ⎟ (2.5 4.42 4.07) ⎝ 0.026 ⎠ So Ad = −882

13.17 From Problem 13.8 I1 = I 2 = 7.10 μ A, I C17 = 0.165 mA, I C13 A = 0.055 mA I C16 ≈ I B17 +

I E17 R8 + VBE17 R9

=

0.165 (0.165)(0.1) + 0.6 + 200 50

= 0.000825 + 0.01233

I C16 = 0.0132 mA (200)(0.026) = 31.5 K 0.165 RE1 = R9 [ rπ 17 + (1 + β ) R8 ] = 50 [31.5 + (201)(0.1)]

rπ 17 =

= 50 51.6 = 25.4 K rπ 16 =

(200)(0.026) = 394 K 0.0132

Then Ri 2 = rπ 16 + (1 + β ) RE1 = 394 + (201)(25.4) ⇒ 5.50 MΩ Now (200)(0.026) rπ 6 = = 732 K 0.0071 0.0071 gm6 = = 0.273 mA/V 0.026 50 ro 6 = ⇒ 7.04 MΩ 0.0071 Ract1 = ro 6 [1 + g m 6 ( R rπ 6 )] = 7.04[1 + (0.273)(1 732)] = 8.96 MΩ 50 ro 4 = ⇒ 7.04 MΩ 0.0071 Then Ad = − g m1 (ro 4 Ract1 Ri 2 ) ⎛ 7.10 ⎞ = −⎜ ⎟ (7.04 8.96 5.5) ⎝ 0.026 ⎠ Ad = −627

50 50 ⇒ 303 K Ro17 = = 303 K 0.165 0.165 From Eq. (13.20), assuming Ri 3 → ∞

Now Ract 2 =

Av 2 ≅ − =

β (1 + β ) R9 ( Ract 2 R017 ) Ri 2 { R9 + [rπ 17 + (1 + β ) R8 ]} −(200)(201)(50)(303 303)

(5500)[50 + 31.5 + (201)(0.1)] Av 2 = −545

=

−3.045 × 108 5.588 × 105

Overall gain Av = (−627)(−545) = 341, 715

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13.18 Using results from 13.17 ⎛ 100 ⎞ Ri 2 = 5.50 MΩ, Ract1 ⎜ ⎟ [1 + (0.273)(1 732)] ⇒ 17.93 MΩ ⎝ 0.0071 ⎠ 100 ⇒ 14.08 MΩ ro 4 = 0.0071 ⎛ 7.10 ⎞ Ad = − ⎜ ⎟ (14.08 17.93 5.50) ⎝ 0.026 ⎠ Ad = −885 Now 100 100 Ract 2 = = 606 K Ro17 = = 606 K 0.165 0.165 −(200)(201)(50)(606 606) −6.09 × 108 Av 2 = = (5500)[50 + 31.5 + (201)(0.1)] 5.588 × 105 Av 2 = −1090 Overall gain Av = (−885)(−1090) = 964, 650 13.19 Now Re14 =

rπ 14 + R01 and R0 = R6 + Re14 1+ βP

Assume series resistance of Q18 and Q19 is small. Then R01 = r013 A Re 22

where Re 22 =

rπ 22 + R017 r013 B 1+ βP

and R017 = r017 [1 + g m17 ( R8 rπ 17 )] Using results from Example 13.6, rπ 17 = 9.63 kΩ rπ 22 = 7.22 kΩ g m17 = 20.8 mA/V r017 = 92.6 kΩ Then R017 = 92.6[1 + (20.8)(0.1 9.63)] = 283 kΩ 50 r013 B = = 92.6 kΩ 0.54 Then 7.22 + 283 92.6 = 1.51 kΩ Re 22 = 51 R01 = r013 A Re 22 = 278 1.51 = 1.50 kΩ (50)(0.026) = 0.65 kΩ rπ 14 = 2 Then 0.65 + 1.50 Re14 = = 0.0422 kΩ 51 or Re14 = 42.2 Ω Then R0 = 42.2 + 27 ⇒ R0 = 69.2 Ω 13.20

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⎡ ⎛ r Rid = 2 ⎢ rπ 1 + (1 + β n ) ⎜ π 3 ⎝ 1+ βP ⎣ β n = 200, β P = 10 (a) I C1 = 9.5 μ A

⎞⎤ ⎟⎥ ⎠⎦

(200)(0.026) = 547 K 0.0095 (10)(0.026) rπ 3 = = 27.4 K 0.0095 Then (201)(27.4) ⎤ ⎡ Rid = 2 ⎢547 + ⎥ 11 ⎣ ⎦ Rid ⇒ 2.095 MΩ rπ 1 =

(b) I C1 = 7.10 μ A (200)(0.026) rπ 1 = = 732 K 0.0071 (10)(0.026) rπ 3 = = 36.6 K 0.0071 (201)(36.6) ⎤ ⎡ Rid = 2 ⎢ 732 + ⎥ 11 ⎣ ⎦ Rid ⇒ 2.80 MΩ 13.21 We can write A0 ⎛ f ⎞⎛ f ⎞ ⎜1 + j ⎟⎜ 1 + j ⎟ f PD ⎠⎝ f1 ⎠ ⎝ 181, 260 = f ⎞⎛ f ⎞ ⎛ ⎜1 + j ⎟ ⎜1 + j ⎟ 10.7 ⎠ ⎝ f1 ⎠ ⎝ Phase: ⎛ f ⎞ −1 ⎛ f ⎞ φ = − tan −1 ⎜ ⎟ − tan ⎜ ⎟ ⎝ 10.7 ⎠ ⎝ f1 ⎠ A( f ) =

For a Phase margin = 70°, φ = −110° So ⎛ f ⎞ −1 ⎛ f ⎞ −110° = − tan −1 ⎜ ⎟ − tan ⎜ ⎟ ⎝ 10.7 ⎠ ⎝ f1 ⎠ Assuming f  10.7, we have ⎛ f ⎞ f tan −1 ⎜ ⎟ = 20° ⇒ = 0.364 f1 ⎝ f1 ⎠ At this frequency, A( f ) = 1, so

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1=

=

181, 260 2

⎛ f ⎞ 2 1+ ⎜ ⎟ ⋅ 1 + (0.364) ⎝ 10.7 ⎠ 170,327 2

⎛ f ⎞ 1+ ⎜ ⎟ ⎝ 10.7 ⎠ f = 170,327 ⇒ f = 1.82 MHz or 10.7 Then, second pole at f f1 = ⇒ f1 = 5 MHz 0.364

13.22 a.

Original g m1 and g m 2

⎛W ⎞⎛ μ C ⎞ K p1 = K p 2 = ⎜ ⎟⎜ P ox ⎟ = (12.5)(10) ⎝ L ⎠⎝ 2 ⎠ = 125 μ A / V 2 So ⎛ IQ ⎞ g m1 = g m 2 = 2 K p1 ⎜ ⎟ ⎝ 2⎠

= 2 (0.125)(10) = 0.09975 mA/V

⎛W ⎞ If ⎜ ⎟ is increased to 50, then ⎝L⎠ K p1 = K p 2 = (50)(10) = 500 μ A / V 2

So g m1 = g m 2 = 2 (0.5)(0.0199) = 0.1995 mA/V b. Gain of first stage Ad = g m1 (r02 r04 ) = (0.1995)(5025 5025)

or Ad = 501 Voltage gain of second stage remains the same, or Av 2 = 251 Then Av = Ad ⋅ Av 2 = (501)(251) or Ad = 125, 751 13.24 a.

K p = (10)(20) = 200 μ A / V 2 = 0.2 mA / V 2

I REF = I SET

10 − VSG − (−10) 200 = k P (VSG − 1.5) 2 =

2 20 − VSG = (0.2)(200)(VSG − 3VSG + 2.25) 2 40VSG − 119VSG + 70 = 0

VSG =

119 ± (119) 2 − 4(40)(70) ⇒ VSG = 2.17 V 2(40)

Then

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20 − 2.17 ⇒ I REF = 89.2 μ A 200 M 5 , M 6 , M 8 matched transistors so that I REF =

I Q = I D 7 = I REF = 89.2 μ A

b. Small-signal voltage gain of input stage: Ad = 2 K p1 I Q ⋅ ( ro 2 ro 4 ) 1 = 1.12 MΩ ⎛ 89.2 ⎞ (0.02) ⎜ ⎟ ⎝ 2 ⎠ 1 1 r04 = = = 2.24 MΩ λn I D ⎛ 89.2 ⎞ (0.01) ⎜ ⎟ ⎝ 2 ⎠ Then Ad = 2(200)(89.2) ⋅ (1.12 2.24) r02 =

1 = λP I D

or Ad = 141 Small-signal voltage gain of second stage: Av 2 = g m 7 (r07 r08 ) K n 7 = (20)(20) = 400 μ A / V 2 So g m 7 = 2 K n 7 I D 7 = 2 (0.4)(0.0892) = 0.378 mA/V 1 1 r08 = = = 561 kΩ λP I D 7 (0.02)(0.0892) r07 =

1

λn I D 7

=

1 = 1121 kΩ (0.01)(0.0892)

So Av 2 = (0.378)(1121 561) ⇒ Av 2 = 141 Then overall voltage gain Av = Ad ⋅ Av 2 = (141)(141) ⇒ Av = 19,881 13.25 Small-signal voltage gain of input stage: Ad = 2 K p1 I Q ⋅ ( ro 2 ro 4 ) K p1 = (10)(10) = 100 μ A / V 2 1 1 = = 1000 kΩ I ⎛ ⎞ ⎛ 0.2 ⎞ λP ⎜ Q ⎟ (0.01) ⎜ ⎟ ⎝ 2 ⎠ ⎝ 2⎠ 1 1 r04 = = = 2000 kΩ ⎛ IQ ⎞ ⎛ 0.2 ⎞ (0.005) λn ⎜ ⎟ ⎜ ⎟ ⎝ 2 ⎠ ⎝ 2⎠ r02 =

Then Ad = 2(0.1)(0.2) ⋅ (1000 2000) or Ad = 133 Small-signal voltage gain of second stage: Av 2 = g m 7 ( r07 r08 ) K n 7 = (20)(20) = 400 μ A / V 2 So

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g m 7 = 2 K n 7 I D 7 = 2 (0.4)(0.2) = 0.566 mA/V 1 1 r08 = = = 500 kΩ λP I D 7 (0.01)(0.2) r07 =

1

λn I D 7

=

1 = 1000 kΩ (0.005)(0.2)

So Av 2 = (0.566)(1000 500) ⇒ Av 2 = 189 Then overall voltage gain is Av = Ad ⋅ Av 2 = (133)(189) ⇒ Av = 25,137 13.26 f PD =

1 2π Req Ci

where Req = r04 r02 and Ci = C1 (1 + Av 2 ) We can find that Av 2 = 251 and r04 = r02 = 5.025 MΩ Now Req = 5.025 5.025 = 2.51 MΩ and Ci = 12(1 + 251) = 3024 pF So 1 f PD = 2π (2.51× 106 )(3024 × 10−12 ) or f PD = 21.0 Hz 13.27 f PD =

1 2π Req Ci

where Req = r04 r02 From Problem 13.22, r02 = 1.12 MΩ, r04 = 2.24 MΩ and Av 2 = 141 So 1 8= 2π (1.12 2.24) × 106 × Ci or Ci = 2.66 × 10−8 = C1 (1 + Av 2 ) = C1 (142) or C1 = 188 pF 13.28 R0 = r07 r08 We can find that r07 = r08 = 2.52 MΩ Then R0 = 2.52 2.52 or R0 = 1.26 MΩ

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13.29 a.

V0 = ( g m1Vgs1 )(r01 r02 ) VI = Vgs1 + V0

Then V0 = g m1 (r01 r02 )(VI − V0 ) or Av =

g m1 (r01 r02 ) 1 + g m1 (r01 r02 )

b.

I X + g m1Vgs1 =

VX VX and Vgs1 = −VX + r02 r01

R0 =

1 r r g m1 01 02

13.30 (a)

(b)

2 ⎛ 80 ⎞ I Q 2 = ⎜ ⎟ (20) [1.1737 − 0.7 ] ⎝ 2⎠ I Q 2 = 180 μ A 2 ⎛ 80 ⎞ I D 6 = ⎜ ⎟ (25) (VGS 6 − 0.7 ) = 25 ⇒ VGS 6 = 0.8581 V ⎝ 2⎠ 2 ⎛ 40 ⎞ I D 7 = ⎜ ⎟ (50) (VSG 7 − 0.7 ) = 25 ⇒ VSG 7 = 0.8581 V 2 ⎝ ⎠

Set VSG 8 P = VGS 8 N = 0.8581 V ⎛ 40 ⎞ ⎛ W ⎞ ⎛W ⎞ 180 = ⎜ ⎟ ⎜ ⎟ (0.8581 − 0.7) 2 ⇒ ⎜ ⎟ = 360 ⎝ 2 ⎠ ⎝ L ⎠8 P ⎝ L ⎠8 P ⎛ 80 ⎞ ⎛ W ⎞ ⎛W ⎞ 180 = ⎜ ⎟ ⎜ ⎟ (0.8581 − 0.7) 2 ⇒ ⎜ ⎟ = 180 ⎝ 2 ⎠ ⎝ L ⎠8 N ⎝ L ⎠8 N

13.31

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VGS11 ⇒

2 ⎛ 80 ⎞ 200 = ⎜ ⎟ (20) (VGS 11 − 0.7 ) ⎝ 2⎠ VGS 11 = 1.20 V

Let M 12 = 2 transistors in series. Than 5 − 1.20 = 1.90 V 2 2 ⎛ 80 ⎞⎛ W ⎞ ⎛W ⎞ ⎛W ⎞ 200 = ⎜ ⎟⎜ ⎟ (1.90 − 0.7 ) ⇒ ⎜ ⎟ = ⎜ ⎟ = 3.47 L L 2 ⎝ ⎠⎝ ⎠12 ⎝ ⎠12 A ⎝ L ⎠12 B

VGS12 =

13.32 (a) 2 ⎛ 80 ⎞ I Q 2 = 250μ A = ⎜ ⎟ (5) (VGS 8 − 0.7 ) ⎝ 2⎠ ⇒ VGS 8 = 1.818 V 1.818 ⇒ VGS 6 = VSG 7 = = 0.909 V 2 ⎛ 80 ⎞ I D 6 = I D 7 = ⎜ ⎟ (25)(0.909 − 0.7) 2 = 43.7 μ A ⎝ 2⎠ (b)

⎛ 80 ⎞ ⎛ 250 ⎞ g m1 = 2 ⎜ ⎟ (15) ⎜ ⎟ ⇒ 0.5477 mA/V ⎝ 2⎠ ⎝ 2 ⎠ 1 ro 2 = = 800 K ( 0.01)( 0.125)

r04 =

1

= 533.3K

( 0.015)( 0.125) Ad 1 = g m1 ( ro 2 ro 4 ) = ( 0.5477 ) ( 800

533.3)

Ad 1 = 175 Second stage:

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A2 = − g m 5 (ro 5 ro 9 ) ⎛ 40 ⎞ g m 5 = 2 ⎜ ⎟ (80)(250) ⇒ 1.265 mA/V ⎝ 2 ⎠ 1 r05 = = 266.7 K (0.015)(0.25) 1 r09 = = 400 K (0.01)(0.25) A2 = −(1.265)(266.7 400) A2 = −202 Assume the gain of the output stage ≈ 1, then Av = Ad 1 ⋅ A2 = (175)(−202) Av = −35,350

13.33 (a)

Ad = g m1 ( Ro 6 Ro8 )

g m1 = 2 K n I DQ = 2 (0.5)(0.025) ⇒ 224 μ A / V g m1 = g m8 g m 6 = 2 (0.5)(0.025) ⇒ 224 μ A / V ro1 = ro 6 = ro8 = ro10 = ro 4 =

1 1 = = 2.67 M Ω λ I DQ (0.015)(25)

1 1 = ⇒ 1.33 M Ω λ I D 4 ( 0.015 )( 50 )

Now Ro8 = g m8 (ro8 ro10 ) = (224)(2.67)(2.67) = 1597 M Ω Ro 6 = g m 6 (ro 6 )(ro 4 ro1 ) = (224)(2.67)(2.67 1.33) ⇒ Ro 6 = 531 M Ω Then Ad = (224)(531 1597) ⇒ Ad = 89, 264 (b)

Ro = Ro 6 Ro8 = 531 1597 ⇒ Ro = 398 M Ω

(c)

f PD =

1 1 = ⇒ f PD = 80 Hz 2π Ro CL 2π ( 398 × 106 )( 5 × 10−12 )

GBW = (89, 264)(80) ⇒ GBW = 7.14 MHz

13.34 (a) ro1 = ro8 = ro10 = ro 6 = ro 4 =

1

λn I D

=

1 1 = = 2 MΩ λ p I D (0.02)(25)

1 = 2.67 M Ω (0.015)(25)

1 1 = = 1.33 M Ω λn I D 4 (0.015)(50)

⎛ 35 ⎞ ⎛ W ⎞ ⎛W ⎞ g m1 = 2 ⎜ ⎟ ⎜ ⎟ (25) = 41.8 ⎜ ⎟ = g m8 ⎝ 2 ⎠ ⎝ L ⎠1 ⎝ L ⎠1 ⎛ 80 ⎞⎛ W ⎞ ⎛W ⎞ g m 6 = 2 ⎜ ⎟⎜ ⎟ (25) = 63.2 ⎜ ⎟ ⎝ 2 ⎠⎝ L ⎠6 ⎝ L ⎠6 Ro = Ro 6 Ro8 = [ g m 6 (ro 6 )(ro 4 ro1 )] [ g m8 (ro8 ro10 )]

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⎛W ⎞ ⎛W ⎞ Define X 1 = ⎜ ⎟ and X 6 = ⎜ ⎟ L ⎝ ⎠1 ⎝ L ⎠6 Then Ro = ⎣⎡ 63.2 X 6 ( 2.67 ) (1.33 2 ) ⎦⎤ ⎡⎣ 41.8 X 1 ( 2 )( 2 ) ⎤⎦ 22,539 X 1 X 6 = 134.8 X 6 167.2 X 1 = 134.8 X 6 + 167.2 X 1 Ad = g m1 Ro

⎛ 22,539 X 1 X 6 ⎞ = (41.8 X 1 ) ⎜ ⎟ 134.8 167.2 X + X 6 1 ⎠ ⎝ = 10, 000

1 ⎛W ⎞ ⎛W ⎞ Now X 6 = ⎜ ⎟ = ⎜ ⎟ = 0.674 X 1 L 2.2 ⎝ ⎠6 ⎝ L ⎠1 We then find ⎛W ⎞ ⎛W ⎞ X 12 = ⎜ ⎟ = 4.06 = ⎜ ⎟ ⎝ L ⎠1 ⎝ L ⎠p

and ⎛W ⎞ ⎜ ⎟ = 1.85 ⎝ L ⎠n 13.35 Let V + = 5V , V − = −5V P = IT (10) = 3 ⇒ IT = 0.3 mA ⇒ I REF = 0.1 mA = 100 μ A 1 ro1 = ro8 = ro10 = = 1 MΩ (0.02)(50) 1 ro 6 = = 1.33 MΩ (0.015)(50) 1 ro 4 = = 0.667 M Ω (0.015)(100) ⎛ 35 ⎞ ⎛ W ⎞ g m1 = 2 ⎜ ⎟ ⎜ ⎟ (50) = 59.2 X 1 = g m8 ⎝ 2 ⎠ ⎝ L ⎠1 ⎛W ⎞ where X 1 = ⎜ ⎟ ⎝ L ⎠1 Assume all width-to-length ratios are the same. ⎛ 80 ⎞ ⎛ W ⎞ g m 6 = 2 ⎜ ⎟ ⎜ ⎟ (50) = 89.4 X 1 ⎝ 2 ⎠⎝ L ⎠ Now Ro = Ro 6 Ro8 = ⎡⎣ g m 6 ( ro 6 ) ( ro 4 ro1 ) ⎤⎦ ⎡⎣ g m8 ( ro8 ro10 ) ⎤⎦

= ⎡⎣89.4 X 1 (1.33) ( 0.667 1) ⎤⎦ ⎡⎣59.2 X 1 (1)(1) ⎤⎦ ( 47.6 X 1 )( 59.2 X 1 ) = [ 47.6 X 1 ] [59.2 X 1 ] = 47.6 X 1 + 59.2 X 1

So Ro = 26.4 X 1 Now Ad = g m1 Ro = ( 59.2 X 1 )( 26.4 X 1 ) = 25, 000 So that X 12 =

W = 16 for all transistors L

13.36

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(a)

Ad = Bg m1 (ro 6 ro8 ) ro 6 = ro8 =

1 1 = = 0.741 M Ω λ I DQ (0.015)(90)

⎛ k ′ ⎞⎛ W ⎞ g m1 = 2 ⎜ n ⎟ ⎜ ⎟ I D1 = 2 (500)(30) = 245 μ A / V ⎝ 2 ⎠⎝ L ⎠ Ad = (3)(245)(0.741 0.741) ⇒ Ad = 272

(b)

Ro = ro 6 ro8 = 0.741 0.741 ⇒ Ro = 371 k Ω

(c)

f PD =

1 1 = ⇒ f PD = 85.8 kHz 2π Ro C 2π (371× 103 )(5 × 10−12 )

GBW = (272)(85.8 × 103 ) ⇒ GBW = 23.3 MHz

13.37 (a)

1 = 0.5 M Ω (0.02)(2.5)(40) 1 ro8 = = 0.667 M Ω (0.015)(2.5)(40) Ad = Bg m1 ( ro 6 ro8 )

ro 6 =

400 = (2.5) g m1 ( 0.5 0.667 ) ⇒ g m1 = 560 μ A / V

⎛ 80 ⎞ ⎛ W ⎞ ⎛W ⎞ g m1 = 560 = 2 ⎜ ⎟ ⎜ ⎟ (40) ⇒ ⎜ ⎟ = 49 ⎝ 2 ⎠⎝ L ⎠ ⎝L⎠ Assume all (W/L) ratios are the same except for ⎛W ⎞ ⎛W ⎞ M 5 and M 6 . ⎜ ⎟ = ⎜ ⎟ = 122.5 ⎝ L ⎠5 ⎝ L ⎠ 6

(b) Assume the bias voltages are V + = 5V , V − = −5V .

⎛W ⎞ ⎛W ⎞ Assume ⎜ ⎟ = ⎜ ⎟ = 49 ⎝ L ⎠ A ⎝ L ⎠B ⎛ 80 ⎞ I Q = ⎜ ⎟ (49)(VGSA − 0.5) 2 = 80 ⇒ VGSA = 0.702 V ⎝ 2⎠ Then ⎛ 80 ⎞ ⎛ W ⎞ I REF = 80 = ⎜ ⎟ ⎜ ⎟ (VGSC − 0.5) 2 ⎝ 2 ⎠ ⎝ L ⎠C For four transistors

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10 − 0.702 = 2.325 V 4 ⎛ 80 ⎞ ⎛ W ⎞ ⎛W ⎞ 80 = ⎜ ⎟ ⎜ ⎟ (2.325 − 0.5) 2 ⇒ ⎜ ⎟ = 0.60 ⎝ 2 ⎠ ⎝ L ⎠C ⎝ L ⎠C

VGSC =

(c)

f 3− dB =

1 2π Ro C

Ro = 0.5 0.667 = 0.286 M Ω

1 = 185 kHz 2π (286 × 103 )(3 × 10−12 ) GBW = (400)(185 × 103 ) ⇒ 74 MHz f 3− dB =

13.38 (a) From previous results, we can write Ro10 = g m10 (ro10 ro 6 ) Ro12 = g m12 (ro12 ro8 ) Ad = Bg m1 ( Ro10 Ro12 ) Now ro10 = ro 6 =

1 1 = = 0.5 M Ω (0.02)(2.5)(40) λP B ( I Q / 2 )

ro12 = ro8 =

1 1 = = 0.667 M Ω (0.015)(2.5)(40) λn B ( I Q / 2 )

Assume all transistors have the same width-to-length ratios except for M 5 and M 6 . ⎛W Let ⎜ ⎝L Then

⎞ 2 ⎟= X ⎠

⎛ k ′p ⎞ ⎛ W ⎞ ⎛ 35 ⎞ g m10 = 2 ⎜ ⎟ ⎜ ⎟ ( I DQ10 ) = 2 ⎜ ⎟ X 2 (2.5)(40) ⎝ 2⎠ ⎝ 2 ⎠ ⎝ L ⎠10 = 83.67 X ⎛ k′ ⎞⎛ W ⎞ ⎛ 80 ⎞ g m12 = 2 ⎜ n ⎟ ⎜ ⎟ ( I DQ12 ) = 2 ⎜ ⎟ X 2 (2.5)(40) ⎝ 2⎠ ⎝ 2 ⎠ ⎝ L ⎠12 = 126.5 X ⎛ 80 ⎞ g m1 = 2 ⎜ ⎟ X 2 (40) = 80 X ⎝ 2⎠ Then Ro10 = (83.67 X )(0.5)(0.5) = 20.9 X M Ω

Ro12 = (126.5 X )(0.667)(0.667) = 56.3 X M Ω We want 20, 000 = (2.5)(80 X )[20.9 X 56.3 X ] ⎡ (20.9 X )(56.3 X ) ⎤ 2 = 200 X ⎢ ⎥ = 3048 X ⎣ 20.9 X + 56.3 X ⎦

Then ⎛W ⎞ X 2 = 6.56 = ⎜ ⎟ ⎝L⎠ Then ⎛W ⎞ ⎛W ⎞ ⎜ ⎟ = ⎜ ⎟ = (2.5)(6.56) = 16.4 ⎝ L ⎠ 6 ⎝ L ⎠5

(b)

Assume bias voltages are V + = 5V , V − = −5V

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⎛W ⎞ ⎛W ⎞ Assume ⎜ ⎟ = ⎜ ⎟ = 6.56 ⎝ L ⎠ A ⎝ L ⎠B ⎛ 80 ⎞ I Q = 80 = ⎜ ⎟ (6.56)(VGSA − 0.5) 2 ⇒ VGSA = 1.052 V ⎝ 2⎠ Need 5 transistors in series 10 − 1.052 VGSC = = 1.79 V 5 Then ⎛ 80 ⎞ ⎛ W ⎞ ⎛W ⎞ I REF = 80 = ⎜ ⎟ ⎜ ⎟ (1.79 − 0.5) 2 ⇒ ⎜ ⎟ = 1.20 ⎝ 2 ⎠ ⎝ L ⎠C ⎝ L ⎠C

(c)

f 3− dB =

1 where Ro = Ro10 Ro12 2π Ro C

Now Ro10 = 20.9 6.56 = 53.5 M Ω Ro12 = 56.3 6.56 = 144 M Ω Then Ro = 53.5 144 = 39 M Ω 1 = 1.36 kHz 2π (39 × 106 )(3 × 10−12 ) GBW = (20, 000)(1.36 x103 ) ⇒ GBW = 27.2 MHz f 3− dB =

13.39 Ad = g m ( M 2 ) ⋅ ⎡⎣ ro 2 ( M 2 ) ro 2 (Q2 ) ⎤⎦ ⎛ 40 ⎞ g m ( M 2 ) = 2 ⎜ ⎟ (25)(100) = 447 μ A / V ⎝ 2 ⎠ 1 1 ro 2 ( M 2 ) = = = 500 k Ω λ I DQ (0.02)(0.1) ro 2 (Q2 ) =

VA 120 = = 1200 k Ω I CQ 0.1

Then Ad = 447(0.5 1.2) ⇒ Ad = 158 13.40

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Ad = g m ( M 2 ) ⋅ ⎡⎣ ro 2 ( M 2 ) ro 2 (Q2 ) ⎤⎦

⎛ 80 ⎞ g m ( M 2 ) = 2 ⎜ ⎟ (25)(100) = 632 μ A / V ⎝ 2⎠ 1 1 ro 2 ( M 2 ) = = = 667 k Ω λ I DQ (0.015)(0.1) VA 80 = = 800 k Ω I CQ 0.1

ro 2 (Q2 ) =

Ad = (632) ( 0.667 0.80 ) ⇒ Ad = 230

13.41 (a)

I REF = 200 μ A Ad = g m1 ( Ro 6 Ro8 )

K n = K p = 0.5 mA / V 2

λn = λ p = 0.015 V −1

where Ro8 = g m8 (ro8 ro10 ) Ro 6 = g m 6 (ro 6 ) ( ro 4 ro1 ) Now g m8 = 2 K P I D 8 = 2 (0.5)(0.1) = 0.447 mA/V 1 1 ro8 = = = 667 k Ω λP I D 8 (0.015)(0.1) ro10 =

1 = 667 k Ω λP I D 8

gm6 =

IC 6 0.1 = = 3.846 mA/V 0.026 VT

ro 6 =

VA 80 = = 800 k Ω I C 6 0.1

ro 4 =

1 1 = = 333 k Ω λn I D 4 (0.015)(0.2)

ro1 =

1

λ p I D1

=

1 = 667 k Ω (0.015)(0.1)

g m1 = 2 K P I D1 = 2 (0.5)(0.1) = 0.447 mA/V So Ro8 = (0.447)(667)(667) ⇒ 198.9 M Ω Ro 6 = (3.846)(800)(333 667) ⇒ 683.4 M Ω

Then Ad = 447(198.9 683.4) ⇒ Ad = 68,865 13.42 Assume biased at V + = 10V , V − = −10V . P = 3I REF (20) = 10 ⇒ I REF = 167 μ A Ad = g m1 ( Ro 6 Ro8 ) = 25, 000 kn′ = 80 μ A / V 2 , k ′p = 35 μ A / V 2

λn = 0.015V −1 , λ p = 0.02 V −1 ⎛W ⎞ ⎛W ⎞ Assume ⎜ ⎟ = 2.2 ⎜ ⎟ L ⎝ ⎠p ⎝ L ⎠n

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Ro8 = g m8 ( ro8 ro10 ) Ro 6 = g m 6 (ro 6 )(ro 4 ro1 ) ro8 = ro10 =

1

λP I D 8 1

λP I D 8

=

1 = 0.60 M Ω (0.02)(83.3)

= 0.60 M Ω

⎛ k ′p ⎞ ⎛ W ⎞ ⎛ 35 ⎞ g m8 = 2 ⎜ ⎟ ⎜ ⎟ I D 8 = 2 ⎜ ⎟ (2.2) X 2 (83.3) ⎝ 2⎠ ⎝ 2 ⎠ ⎝ L ⎠8 = 113.3 X ⎛W ⎞ where X 2 = ⎜ ⎟ ⎝ L ⎠n VA 80 ro 6 = = = 0.960 M Ω I C 6 83.3

ro 4 = ro1 = gm6 =

1

λn I D 4

=

1 = 0.40 M Ω (0.015)(167)

1 1 = = 0.60 M Ω λ p I D1 (0.02)(83.3) IC 6 83.3 = = 3204 μ A / V 0.026 VT

⎛ k p′ ⎞ ⎛ W ⎞ ⎛ 35 ⎞ g m1 = 2 ⎜ ⎟ ⎜ ⎟ I D1 = 2 ⎜ ⎟ (2.2) X 2 (83.3) ⎝ 2⎠ ⎝ 2 ⎠ ⎝ L ⎠1 = 113.3 X Now Ro 6 = (3204)(0.960) ⎡⎣0.40 0.60 ⎤⎦ = 738 M Ω Ro8 = (113.3 X )(0.60)(0.60) = 40.8 X M Ω Then Ad = 25, 000 = (113.3 X ) ⎡⎣ 738 40.8 X ⎤⎦ ⎡ 30,110 X ⎤ = (113.3 X ) ⎢ ⎥ ⎣ 738 + 40.8 X ⎦ which yields X = 2.48 or ⎛W ⎞ X 2 = 6.16 = ⎜ ⎟ ⎝ L ⎠n

and ⎛W ⎞ ⎜ ⎟ + (2.2)(6.16) = 12.3 ⎝ L ⎠P 13.43 For vcm (max), assume VCB (Q5 ) = 0. Then VS = 15 − 0.6 − 0.6 = 13.8 V 0.236 I D 9 = I D10 = = 0.118 mA 2 Using parameters given in Example 13.11 I 0.118 VSG = D 9 − VTP = + 1.4 = 2.17 V 0.20 KP Then vcm (max) = 13.8 − 2.17 ⇒ vcm (max) = 11.6 V

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For vcm (min) , assume VSD ( M 9 ) = VSD ( sat ) = VSG + VTP = 2.17 − 1.4 = 0.77 V Now VD10 = I D10 (0.5) + 0.6 + I D10 (0.5) − 15 = 0.118 + 0.6 − 15 ⇒ VD10 = −14.28 V Then vcm (min) = −14.28 + VSD (sat) − VSG = −14.28 + 0.77 − 2.17 = −15.68 V Then, common-mode voltage range −15.68 ≤ vcm ≤ 11.6 Or, assuming the input is limited to ±15 V, then −15 ≤ vcm ≤ 11.6 V 13.44 For I1 = I 2 = 300 μ A, VSG = VBE + (0.3)(8) = 0.6 + 2.4 = 3.0 V Then I 2 = K P (VSG + VTP ) 2 0.3 = K P (3 − 1.4)2 ⇒ K P = 0.117 mA / V 2 13.45 For VCB = 0 for both Q6 and Q7 , then VS = 0.6 + 0.6 + VSG + (−VS ) So 2VS = 1.2 + VSG Now I1 0.6 + I 2 R1 = VSG = + VTP and I1 = I 2 KP Also I1 = I 2 = K P (VSG + VTP ) 2 so 0.6 + (0.25)(8)(VSG − 1.4) 2 = VSG 2 − 2.8VSG + 1.96) = VSG 0.6 + 2(VSG 2 2VSG − 6.6VSG + 4.52 = 0 6.6 ± (6.6) 2 − 4(2)(4.52) = 2.33 V 2(2) Then 2VS = 1.2 + 2.33 = 3.53 and VS = 1.765 V VSG =

13.46 I C 5 = I C 4 = 300 μ A Using the parameters from Examples 13.12 and 13.13, we have βV (200)(0.026) Ri 2 = rπ 13 = n T = = 17.3 kΩ 0.3 I C13 Ad = 2 K n I Q 5 ⋅ ( Ri 2 ) = 2(0.6)(0.3) ⋅ (17.3)

or Ad = 10.38 Now

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I C13 0.3 = = 11.5 mA/V 0.026 VT

g m13 = r013 =

VA 50 = = 167 kΩ I C13 0.3

Then | Av 2 | = g m13 ⋅ r013 = (11.5)(167) or Av 2 = 1917 Overall gain: Av = (10.38)(1917) = 19,895 Assuming the resistances looking into Q4 and into the output stage are very large, we have

13.47 | Av 2 | =

β R013 rπ 13 + (1 + β ) RE13

where R013 = r013 ⎡⎣1 + g m13 ( RE13 rπ 13 ) ⎤⎦ I C13 = 300 μ A, r013 =

50 = 167 kΩ 0.3

0.3 = 11.5 mA / V 0.026 (200)(0.026) = = 17.3 kΩ 0.3

g m13 = rπ 13

So R013 = (167) ⎡⎣1 + (11.5) (1 17.3) ⎤⎦ ⇒ 1.98 MΩ Then | Av 2 | =

(200)(1980) = 1814 17.3 + (201)(1)

Now Ci = C1 (1 + Av 2 ) = 12 [1 + 1814] ⇒ Ci = 21, 780 pF 1 f PD = 2π Req Ci Req = Ri 2 r012 r010

Neglecting R3 , r010 =

1 1 = = 333 kΩ λ I D10 (0.02)(0.15)

Neglecting R5 , 50 r012 = = 333 kΩ 0.15 Ri 2 = rπ 13 + (1 + β ) RE13

= 17.3 + (201)(1) = 218 kΩ

Then f PD =

1 2π ⎡⎣ 218 333 333⎤⎦ × 103 × ( 21, 780 ) × 10−12

or f PD = 77.4 Hz Unity-Gain Bandwidth Gain of first stage:

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Ad = 2 K n I Qs ⋅ ( R12 ro12 ro10 ) = 2(0.6)(0.3) ⋅ (218 333 333) = (0.6)(218 333 333)

or Ad = 56.6 Overall gain: Av = (56.6)(1814) = 102, 672 Then unity-gain bandwidth = (77.4)(102, 672) ⇒ 7.95 MHz 13.48 Since VGS = 0 in J 6 , I REF = I DSS ⇒ I DSS = 0.8 mA 13.49 a.

Ri 2 = rπ 5 + (1 + β ) [ rπ 6 + (1 + β ) RE ]

(100)(0.026) = 13 kΩ 0.2 I 200 μ A ≅ C6 = = 2 μA 100 β

rπ 6 = IC 5

So rπ 5 =

(100)(0.026) = 1300 kΩ 0.002

Then Ri 2 = 1300 + (101) [13 + (101)(0.3) ] or Ri 2 = 5.67 MΩ Av = g m 2 ( r02 r04 Ri 2 )

b. gm2 =

2 ⋅ I D ⋅ I DSS VP

=

2 ⋅ (0.1)(0.2) 3

= 0.0943 mA / V r02 =

1 1 = = 500 kΩ λ I D (0.02)(0.1)

r04 =

VA 5.0 = = 500 kΩ I C 4 0.1

Then Av = (0.0943)[500 || 500 || 5670] or Av = 22.6 13.50 a.

Need VSD (QE ) ≥ VSD ( sat ) = VP For minimum bias ±3 V

Set VP = 3 V and VZK = 3 V I REF 2 =

VZK − VD1 R3

3 − 0.6 ⇒ R3 = 24 kΩ 0.1 Set bias in QE = I REF 2 + I Z 2 = 0.1 + 0.1 = 0.2 mA Therefore,

so that R3 =

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I DSS = 0.2 mA

b.

Neglecting base currents 12 − 0.6 I 01 = I REF 1 = 0.5 mA = R4 so that R4 = 22.8 kΩ

13.51 a. gm2

We have 2 = ⋅ I D ⋅ I DSS | VP |

=

2 ⋅ (0.5)(1) 4

= 0.354 mA/V r02 = r04 =

1

λ ID

=

1 = 100 kΩ (0.02)(0.5)

VA 100 = = 200 kΩ I D 0.5

0.5 = 19.23 mA/V 0.026 (200)(0.026) = 10.4 kΩ rπ 4 = 0.5 So R04 = r04 ⎡⎣1 + g m 4 ( rπ 4 R2 ) ⎤⎦ gm4 =

= 200 ⎡⎣1 + (19.23) (10.4 0.5 ) ⎤⎦ = 2035 kΩ

Ad = g m 2 ( r02 R04 RL )

For RL → ∞ Ad = 0.354 (100 || 2035 ) = 33.7 b.

With these parameter values, gain can never reach 500. Similarly for this part, gain can never reach 700.

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Chapter 14 Exercise Solutions EX14.1 a.

ACL =

b.

dACL ACL ACL =

−50 ⇒ ACL = −49.949 ⎡ ⎛ 1 ⎞ ⎤ ⎢1 + ⎜ 5 × 104 ⎟ ( 51) ⎥ ⎠ ⎣ ⎝ ⎦ dA 51 = 10 × ⇒ CL = 0.0102% ACL 5 × 104

−50 ⇒ ACL = −49.943 51 ⎤ ⎡ ⎢⎣1 + 4.5 × 104 ⎥⎦

EX14.2 a. For R0 = 0 1 1 1 = + (1 + 104 ) = 0.1 + 103 Ri f 10 10 ⇒ Ri f = 10−3 kΩ = 1 Ω

b.

For R0 = 10 kΩ

1 1 1 ⎡1 + 104 + 1 ⎤ 104 = + ×⎢ ⎥ ≅ 0.1 + Ri f 10 10 ⎣ 1 + 1 + 1 ⎦ 3 (10 ) Rif = 3 × 10−3 kΩ ⇒ Ri f = 3 Ω

EX14.3 ⎛ 40 ⎞ 40 (1 + 104 ) + 99 ⎜ 1 + ⎟ 1 ⎠ ⎝ Ri f = 99 1+ 1 5 4 × 10 + 4.059 × 103 ≅ 100 Ri f = 4.04 × 103 kΩ ⇒ Ri f = 4.04 MΩ

EX14.4 R 1 + 2 = 100 R1 a.

1 1 ⎡ 105 ⎤ = ⎢ ⎥ = 10 ⇒ R0 f = 0.1 Ω R0 f 100 ⎣100 ⎦

b.

1 1 ⎡ 105 ⎤ 2 = ⎢ ⎥ = 10 R0 f 10 ⎣100 ⎦

R0 f = 10−2 kΩ ⇒ R0 f = 10 Ω

EX14.5 From Equation (14.43)

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ACL ( f ) =

=

ACL 0 1+ j ⋅

f f PD ( A0 /ACL 0 ) 25

f 1+ j ⋅ ( 50 ) (104 / 25)

=

25 1+ j ⋅

f 2 × 104

f = 2 kHz

a.

v0 = 25 ⇒ v0 ( peak ) = 1.25 mV vI f = 20 kHz

b.

v0 1 = ⋅ 25 ⇒ v0 ( peak ) = 0.884 mV vI 2 c. f = 100 kHz v0 25 25 = = = 4.90 2 vI 5.099 1 + (100 / 20 ) ⇒ v0 = 0.245 mV

EX14.6 Full-scale response = 1× 5 = 5 V t=

5 ⇒ t = 2.5 μ s 2

EX14.7 a.

FPBW =

SR 0.63 × 106 = 2π V0 ( max ) 2π (1)

FPBW = 1.0 × 105 ⇒ FPBW = 100 kHz

b.

FPBW =

0.63 × 106 = 1.0 × 104 2π (10 )

⇒ FPBW = 10 kHz

EX14.8 ⎛I ⎞ ⎛ 1.85 × 10−14 ⎞ V0 S = VT ln ⎜ S 2 ⎟ = (0.026) ln ⎜ −14 ⎟ ⎝ 2 × 10 ⎠ ⎝ I S1 ⎠ ⇒ V0 S = 2.03 mV

EX14.9 We need iC1 = iC 2 , vEC 3 = vEC 4 = 0.6 V, and vCE1 = vCE 2 = 10 V By Equation (14.60(a)) ⎡ ⎛ v ⎞ ⎤ ⎛ 10 ⎞ iC1 = I S 1 ⎢exp ⎜ BE1 ⎟ ⎥ ⎜ 1 + ⎟ ⎝ VT ⎠ ⎦ ⎝ 50 ⎠ ⎣ ⎡ ⎛ v ⎞ ⎤ ⎛ 0.6 ⎞ = I S 3 ⎢ exp ⎜ EB 3 ⎟ ⎥ ⎜1 + ⎟ 50 ⎠ ⎝ VT ⎠ ⎦ ⎝ ⎣ By Equation (14.60(b))

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⎡ ⎛ v ⎞ ⎤ ⎛ 10 ⎞ iC 2 = I S 2 ⎢ exp ⎜ BE 2 ⎟ ⎥ ⎜ 1 + ⎟ ⎝ VT ⎠ ⎦ ⎝ 50 ⎠ ⎣ ⎡ ⎛ v ⎞ ⎤ ⎛ 0.6 ⎞ = I S 4 ⎢ exp ⎜ EB 4 ⎟ ⎥ ⎜ 1 + ⎟ 50 ⎠ ⎝ VT ⎠ ⎦ ⎝ ⎣ I S1 = I S 2, take the ratio: ⎛v −v ⎞ I exp ⎜ BE1 BE 2 ⎟ = S 3 VT ⎝ ⎠ IS 4 vBE1 − vBE 2 = V0 S

⎛I ⎞ = VT ln ⎜ S 3 ⎟ ⎝ IS 4 ⎠ = 0.026 ⋅ ln (1.05 )

⇒ V0 S = 1.27 m V

EX14.10 VOS = 0.020 =

I Q ⎛ ΔK n ⎞ 1 ⋅ ⋅⎜ ⎟ 2 2Kn ⎝ Kn ⎠ 1 150 ⎛ ΔK n ⎞ ⋅ ⋅ 2 2 ( 50 ) ⎜⎝ 50 ⎟⎠

⇒ ΔK n = 1.63μ A / V 2 ⇒

ΔK n 1.63 = ⇒ 3.26% 50 Kn

EX14.11 ⎛ R5 ⎞ + Want ⎜ ⎟V = 5 m V ⎝ R5 + R4 ⎠ R5  R4 so

R5 =

R5 × V + = 0.005 R4

( 0.005)(100 )

= 0.05 kΩ 10 ⇒ R5 = 50 Ω

EX14.12 R1′ = 25 1 = 0.9615 kΩ R2′ = 75 1 = 0.9868 kΩ For I Q = 100 μ A ⇒ iC1 = iC 2 = 50 μ A From Equation (14.75) ⎛ 50 × 10−6 ⎞ ( 0.026 ) ln ⎜ −14 ⎟ + ( 0.050 )( 0.9615 ) ⎝ 10 ⎠ ⎛i ⎞ = ( 0.026 ) ln ⎜ C 2 ⎟ + ( 0.050 )( 0.9868 ) ⎝ IS 4 ⎠ 0.58065 + 0.048075 ⎛i ⎞ = ( 0.026 ) ln ⎜ C 2 ⎟ + 0.04934 ⎝ IS 4 ⎠

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⎛i ⎞ ln ⎜ C 2 ⎟ = 22.284 ⎝ IS 4 ⎠ 50 × 10−6 = 4.7625 × 109 IS 4 I S 4 ≅ 1.05 × 10−14 A

EX14.13 From Equation (14.79) ⎛ R ⎞ v0 = I B1 R2 − I B 2 R3 ⎜ 1 + 2 ⎟ R1 ⎠ ⎝ For v0 = 0 ⎛ 100 ⎞ 0 = (1.1× 10−6 ) (100 kΩ ) − (1.0 × 10−6 ) R3 ⎜ 1 + ⎟ 10 ⎠ ⎝ R3 (11) = (1.1)(100 kΩ ) ⇒ R3 = 10 kΩ

TYU14.1 v1CM ( max ) = V + − VSD1 ( sat ) − VSG1 v1CM ( min ) = V − + VDS 4 ( sat ) + VSD1 ( sat ) − VSG1 We have: I REF = 100 μ A, kn′ = 80 μ A / V 2 , k ′p = 40 μ A / V 2 , ⎛W ⎜ ⎝L For

⎞ ⎟ = 25 ⎠ M1 :

2 ⎛ 40 ⎞ I D = 50 = ⎜ ⎟ ( 25 )(VSG1 + VTP ) ⎝ 2 ⎠

So 50 = 500 (VSG1 − 0.5 ) ⇒ VSG1 = 0.816 V 2

VSD1 ( sat ) = 0.816 − 0.5 = 0.316 V

Then vCM ( max ) = V + − 0.316 − 0.816 = V + − 1.13 V For M 4 : 2 ⎛ 80 ⎞ I D = 100 = ⎜ ⎟ ( 25 )(VGS 4 − VTN ) ⎝ 2⎠

So 100 = 1000 (VGS 4 − 0.5 ) ⇒ VGS 4 = 0.816 V 2

VDS 4 ( sat ) = 0.816 − 0.5 = 0.316 V

VCM ( min ) = V − + 0.316 + 0.316 − 0.816 = V − − 0.184 So V − − 0.184 ≤ vCM ≤ V + − 1.13 V

TYU14.2 vo ( max ) = V + − VSD 8 ( sat ) − VSG10 vo ( min ) = V − + VDS 4 ( sat ) + VDS 6 ( sat ) Now

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50

VSG 8 = VSG10 =

+ 0.5 = 0.816 V

( 40/ 2 )( 25) VSD 8 ( sat ) = VSD10 ( sat ) = 0.316 V So vo ( max ) = V + − 0.316 − 0.816 = V + − 1.13 Also VGS 6 =

50 + 0.5 = 0.724 V (80/ 2 )( 25)

VGS 4 =

100 + 0.5 = 0.816 V (80/ 2 )( 25)

VDS 6 ( sat ) = 0.724 − 0.5 = 0.224 V VDS 4 ( sat ) = 0.816 − 0.5 = 0.316 V So vo ( min ) = V − + 0.316 + 0.224 = V − + 0.54

Then V − + 0.54 ≤ vo ≤ V + − 1.13 V TYU14.3 500 = −25 20 Within 0.1% ⇒ −25 + ( 0.001)( 25 ) ⇒ ACL = −24.975 ACL ( ideal ) = −

−24.975 =

−25 ⎡ 26 ⎤ ⎢1 + A ⎥ 0L ⎦ ⎣

−25 26 = − 1 = 0.0010 A0 L −24.975 A0 L = 25.974

TYU14.4

ACL ( ∞ )

a.

ACL =

b.

dACL 100 = 10 × 5 = 0.01% 10 ACL

⎡ A (∞) ⎤ 1 + ⎢ CL ⎥ ⎣ A0 L ⎦ R 495 ACL ( ∞ ) = 1 + 2 = 1 + = 100 R1 5 100 ACL = ⇒ ACL = 99.90 100 1+ 5 10 ACL ( ∞ ) = 100

ACL = 99.90 − ( 0.0001)( 99.90 ) ⇒ ACL = 99.89

TYU14.5

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ACL ( ∞ ) − ACL

= 1−

ACL ( ∞ )

ACL = 1− ACL ( ∞ )

1 A (∞) 1 + CL A0 L

ACL ( ∞ ) ACL ( ∞ ) −1 A0 L A0 L So 0.001 = = A (∞) A (∞) 1 + CL 1 + CL A0 L A0 L 1+

0.001 = 0.999 ⋅ ACL ( ∞ ) =

ACL ( ∞ ) A0 L

0.001 0.001 ⋅ A0 L = ⋅ (104 ) ⇒ ACL ( ∞ ) = 10.010 0.999 0.999

or ACL = (1 − 0.001)(10.010 ) ⇒ ACL = 10.0 TYU14.6 ii ⎛ Rif ⎞ =⎜ ⎟ I1 ⎝ Ri ⎠ iI 0.1 a. = = 1× 10−5 i1 104 iI 10 = = 1× 10−3 i1 104

b.

TYU14.7 Voltage follower R2 = 0, R1 = ∞

Rif = Ri (1 + A0 L ) = 10 (1 + 5 × 105 ) ≅ 5 × 106 kΩ ⇒ Rif = 5000 MΩ

TYU14.8 f3dB =

(105 ) (10 ) ⇒ 20 kHz fT = 50 ACL 0

f max = f 3dB = V0 ( max ) =

SR 2π V0 ( max )

SR 0.8 × 106 = 2π f3dB 2π ( 20 × 103 )

⇒ V0 ( max ) = 6.37 V

TYU14.9 v0 = I B1 R3 = (10−6 )( 200 × 103 ) a. ⇒ v0 = 0.20 V

b.

R4 = R1 R2 R3 = 100 50 200 ⇒ R4 = 28.6 kΩ

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Chapter 14 Problem Solutions 14.1 Ad =

vo = −80 vi

vo (max) = 4.5 ⇒ vi (max) = 56.25 mV

So vi (max)rms =

56.25 2

= 39.77 mV

14.2

(a) 4.5 = 0.028125 mA 160 4.5 iL = = 4.5 mA 1 Output Circuit = 4.528 mA v −4.5 vi = − o = ⇒ vi = −0.05625 V A 80 (b) v 4.5 io ≈ 15 mA = o = RL RL i2 =

⇒ RL (min) = 300 Ω

14.3 (1) (2)

v2 = 12.5 mV

(3) (4) (5)

AOL = 2 × 104 v1 = 8 μ V AOL = 1000

vo = 2 V

14.4 From Eq. (14.4)

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ACL =

−15 =

− R2 / R1 1 ⎛ R2 ⎞ 1+ ⎜1 + ⎟ AoL ⎝ R1 ⎠ − R2 / R1 1 ⎛ R2 ⎞ 1+ ⎜1 + ⎟ R1 ⎠ 2 × 103 ⎝

⎛R ⎞ 15 ⎜ 2 ⎟ R R 1 ⎞ ⎛ + ⎝ 1 3⎠ = 2 15 ⎜ 1 + 3 ⎟ R1 × × 2 10 2 10 ⎝ ⎠ 15.0075 =

R2 (0.9925) R1

R2 = 15.12 R1

14.5

vI − v1 v1 − v0 v1 = + and v0 = − A0 L v1 R1 R2 Ri so that v1 = −

v0 A0 L

⎛ 1 vI v0 1 1⎞ + = v1 ⎜ + + ⎟ R1 R2 ⎝ R1 R2 Ri ⎠ So ⎡1 vI 1 ⎛ 1 1 1 ⎞⎤ = −v0 ⎢ + + ⎟⎥ ⎜ + R1 ⎣ R2 A0 L ⎝ R1 R2 Ri ⎠ ⎦ Then v0 −(1/ R1 ) = = ACL vI ⎡ 1 1 ⎛ 1 1 1 ⎞⎤ + ⎟⎥ ⎢ + ⎜ + ⎣ R2 A0 L ⎝ R1 R2 Ri ⎠ ⎦ From Equation (14.20) for RL = ∞ and R0 = 0

1 1 1 (1 + A0 L ) = + ⋅ Rif Ri R2 1

For Ri = 1 kΩ

a.

−(1/ 20) ⎡ 1 1 ⎛ 1 1 1 ⎞⎤ ⎢100 + 103 ⎜ 20 + 100 + 1 ⎟ ⎥ ⎝ ⎠⎦ ⎣ −0.05 = [0.01 + 1.06 × 10−3 ]

ACL =

or

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⇒ ACL = −4.52 1 1 1 + 103 = + ⇒ Rif = 90.8 Ω Rif 1 100

For Ri = 10 kΩ

b. ACL =

=

−(1/ 20) ⎡ 1 1 ⎛ 1 1 1 ⎞⎤ ⎢100 + 103 ⎜ 20 + 100 + 10 ⎟ ⎥ ⎝ ⎠⎦ ⎣ −0.05 [0.01 + 1.6 × 10−4 ]

or ⇒ ACL = −4.92 1 1 1 + 103 = + ⇒ Rif = 98.9 Ω Rif 10 100

For Ri = 100 kΩ

c. ACL =

=

−(1/ 20) ⎡ 1 1 ⎛ 1 1 1 ⎞⎤ ⎢100 + 103 ⎜ 20 + 100 + 100 ⎟ ⎥ ⎝ ⎠⎦ ⎣ −0.05 [0.01 + 7 × 10−5 ]

or ⇒ ACL = −4.965 1 1 1 + 103 = + ⇒ Rif = 99.8 Ω Rif 100 100

14.6

⎛ R2 ⎞ ⎜1 + ⎟ R1 ⎠ v ⎝ ACL = o = vi ⎡ 1 ⎛ R2 ⎞ ⎤ ⎢1 + ⎜1 + ⎟ ⎥ R1 ⎠ ⎦ ⎣ AOL ⎝ For the ideal: ⎛ R2 ⎞ 0.10 = 50 ⎜1 + ⎟ = R1 ⎠ 0.002 ⎝ vo (actual ) = (0.10)(1 − 0.001) = 0.0999 So

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0.0999 50 = = 49.95 1 0.002 1 + (50) AOL which yields AOL = 1000

14.7 From Equation (14.18) ⎛A 1 ⎞ − ⎜ OL − ⎟ v ⎝ Ro R2 ⎠ Avf 1 = o1 = v1 ⎛ 1 1 1 ⎞ + + ⎟ ⎜ R R R o 2 ⎠ ⎝ L Or ⎛ 5 × 103 1 ⎞ −⎜ − ⎟ 3 1 100 ⎠ ⋅ v = −(4.99999 × 10 ) ⋅ v vo1 = ⎝ 1 1 1.11 ⎛1 1 1 ⎞ ⎜ + + ⎟ ⎝ 10 1 100 ⎠ vo1 = −4.504495 × 103 ⋅ v1 Now i1 vi − v1 = ≡K v1 R1v1 Then vi − v1 = KR1v1 which yields vi v1 = KR1 + 1 Now, from Equation (14.20) 1 ⎤ ⎡ 1 + 5 × 103 + ⎥ 1 1 ⎢ 10 K= + ⎢ ⎥ 10 100 ⎢ 1 + 1 + 1 ⎥ 10 100 ⎦⎥ ⎣⎢ ⎡ 5.0011× 103 ⎤ = (0.1) + (0.01) ⎢ ⎥ = 45.15495 1.11 ⎣ ⎦ Then vi vi v1 = = ( 45.15495)(10 ) + 1 452.5495

We find vi ⎡ ⎤ vo1 = −4.504495 × 103 ⎢ ⎥ ⎣ 452.5495 ⎦ Or v Avf 1 = o1 = −9.9536 vi

For the second stage, RL = ∞

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⎛ 5 × 103 1 ⎞ −⎜ − ⎟ 1 100 ⎠ vo 2 = ⎝ ⋅ v1′ = −4.950485 × 103 ⋅ v1′ ⎛1 1 ⎞ ⎜ + ⎟ ⎝ 1 100 ⎠ ⎡ ⎤ 1 1 ⎢1 + 5 × 103 ⎥ K≡ + ⎢ ⎥ = 49.61485 10 100 ⎢ 1 + 1 ⎥ ⎢⎣ 100 ⎥⎦ vo1 vo1 vo1 v1′ = = = KR1 + 1 (49.61485)(10) + 1 497.1485 Then vo 2 −4.950485 × 103 = = −9.95776 497.1485 vo1 So v Avf = o 2 = (−9.9536)(−9.95776) ⇒ Avf = 99.12 vi

14.8 a.

v1 − vI v v −v + 1 + 1 0 =0 R3 + Ri R1 R2

(1)

⎡ 1 vI 1 1⎤ v v1 ⎢ + + ⎥= 0 + ⎣ R3 + Ri R1 R2 ⎦ R2 R3 + Ri v0 v0 − A0 L vd v0 − v1 + + =0 RL R0 R2 or ⎡1 A v 1 1⎤ v v0 ⎢ + + ⎥ = 1 + 0L d R R R R R0 0 2 ⎦ 2 ⎣ L ⎛ v −v ⎞ vd = ⎜ I 1 ⎟ ⋅ Ri ⎝ R3 + Ri ⎠ So substituting numbers: vI 1⎤ v 1 ⎡ 1 v1 ⎢ + + ⎥= 0 + ⎣10 + 20 10 40 ⎦ 40 10 + 20 or v1[0.15833] = v0 [0.025] + vI [0.03333] 1 ⎤ v (104 )vd ⎡1 1 v0 ⎢ + + ⎥= 1 + 0.5 ⎣1 0.5 40 ⎦ 40

(2)

(3)

(1)

(2)

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or v0 [3.025] = v1 [ 0.025] + ( 2 × 104 ) vd ⎛ v −v ⎞ vd = ⎜ I 1 ⎟ ⋅ 20 = 0.6667 ( vI − v1 ) ⎝ 10 + 20 ⎠ So v0 [3.025] = v1 [ 0.025] + ( 2 × 104 ) ( 0.6667 )( vI − v1 )

(3) (2)

or v0 [3.025] = 1.333 ×10 4 vI − 1.333 × 104 v1 From (1): v1 = v0 ( 0.1579 ) + vI ( 0.2105 ) Then v0 [3.025] = 1.333 × 104 vI − 1.333 × 104 ⎡⎣ v0 ( 0.1579 ) + vI ( 0.2105 ) ⎤⎦ v0 ⎡⎣ 2.1078 ×103 ⎤⎦ = vI ⎡⎣1.0524 ×104 ⎤⎦ or v ACL = 0 = 4.993 vI To find Rif : Use Equation (14.27) ⎛ 0.5 0.5 ⎞ iI ⎜ 1 + + ⎟ 1 40 ⎠ ⎝

3 ⎧⎛ 1 1 ⎞ ⎛ 0.5 0.5 ⎞ 0.5 ⎫ (10 ) vd = v1 ⎨⎜ + ⎟ ⎜ 1 + + − − ⎬ ⎟ 1 40 ⎠ (40) 2 ⎭ 40 ⎩⎝ 10 40 ⎠ ⎝ iI (1.5125) = v1{(0.125)(1.5125) − 0.0003125} − 25vd or iI (1.5125) = vI {0.18875} − 25vd Now vd = iI Ri = iI (20) and v1 = vI − iI (20) So iI (1.5125) = [vI − iI (20)] ⋅ [0.18875] − 25iI (20)

iI [505.3] = vI (0.18875) or vI = 2677 kΩ iI Now Rif = 10 + 2677 ⇒ Rif = 2.687 MΩ To determine R0 f : Using Equation (14.36) ⎡ 1 1 ⎢⎢ A0 L = ⋅ R2 R0′ f R0 ⎢ ⎢1 + R R 1 i ⎣ or R0′ f = 3.5 Ω

⎤ ⎡ ⎤ ⎥ 1 ⎢ 103 ⎥ ⎥= ⎥ ⋅⎢ ⎥ 0.5 ⎢1 + 40 ⎥ ⎥ ⎢ 10 20 ⎥ ⎣ ⎦ ⎦

Then R0 f = 1 kΩ 3.5 Ω ⇒ R0 f = 3.49 Ω

b. Using Equation (14.16) dACL dA ⎛ 5 ⎞ = (−10) ⎜ 3 ⎟ ⇒ CL = −(0.05)% ACL ACL ⎝ 10 ⎠ 14.9

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v0 − A0 L vd v0 − vI + = 0 and vd = vI − v0 R0 Ri So v0 A0 L v v − ⋅ (vI − v0 ) + 0 − I = 0 R0 R0 Ri Ri

⎡1 A ⎡1 A ⎤ 1⎤ v0 ⎢ + 0 L + ⎥ = vI ⎢ + 0 L ⎥ ⎣ R0 R0 Ri ⎦ ⎣ Ri R0 ⎦ ⎡ 1 (104 ) 1 ⎤ ⎡ 1 (104 ) ⎤ + + + v0 ⎢ ⎥ = vI ⎢ ⎥ ⎣ 0.2 0.2 100 ⎦ ⎣100 0.2 ⎦ v0 [5.000501× 104 ] = vI [5.000001× 10 4 ]

So ACL =

Set vI = 0

b. i0 =

v0 = 0.9999 vI

v0 − A0 L vd v0 + and vd = −v0 R0 Ri

⎡1 A 1⎤ i0 = v0 ⎢ + 0 L + ⎥ R R R 0 i ⎦ ⎣ 0 Then 1 1 A0 L 1 = + + R0 f R0 R0 Ri

or 1 1 (104 ) 1 = + + R0 f 0.2 0.2 100 which yields R0 f ≅ 0.02 Ω 14.10

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vI 1 − v1 vI 2 − v1 v1 − v0 + = 20 10 40 vI 1 vI 2 v0 1 1⎤ ⎡1 + + = v1 ⎢ + + ⎥ 20 10 40 ⎣ 20 10 40 ⎦ v and v0 = − A0 L v1 so that v1 = − 0 A0L Then ⎧1 1 ⎛ 7 ⎞⎫ ⋅ vI 1 (0.05) + vI 2 (0.10) = −v0 ⎨ + ⎟⎬ 3 ⎜ ⎩ 40 2 × 10 ⎝ 40 ⎠ ⎭ = −v0 [2.50875 × 10−2 ] ⇒ v0 = −1.993vI 1 − 3.986vI 2 Δv0 2 − 1.993 Δv = ⇒ 0 = 0.35% 2 v0 v0

14.11

⎛ 40 ⎞ ⎛ 4⎞ vB = ⎜ ⎟ v2 = ⎜ ⎟ v2 = 0.8v2 ⎝ 40 + 10 ⎠ ⎝ 5⎠ v1 − v A v A − v0 = 10 40 v1 v0 1 ⎞ ⎛1 + = vA ⎜ + ⎟ 10 40 ⎝ 10 40 ⎠ v1 (0.1) + v0 (0.025) = vA (0.125) v0 = A0 L vd = A0 L (vB − v A ) or v0 = A0 L [0.8v2 − v A ] v0 − 0.8v2 = −v A A0 L ⇒ v A = 0.8v2 −

(1)

(2) (3)

v0 A0 L

Then

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⎡ v ⎤ v1 (0.1) + v0 (0.025) = (0.125) ⎢0.8v2 − 0 ⎥ A 0L ⎦ ⎣ 0.125 ⎡ ⎤ v1 (0.1) − v2 (0.1) = −v0 ⎢0.025 + 103 ⎥⎦ ⎣ = −v0 [2.5125 × 10−2 ] ⇒ Ad = ⇒

v0 = 3.9801 v2 − v1

ΔAd 0.0199 = ⇒ 0.4975% Ad 4

14.12 a.

Considering the second op-amp and Equation (14.20), we have ⎡ ⎤ 1 1 1 ⎢1 + 100 ⎥ 101 = + ⋅⎢ ⎥ = 0.10 + (0.1)(11) Rif 2 10 0.1 ⎢ 1 + 1 ⎥ ⎣⎢ 0.1 ⎦⎥ So Rif 2 = 0.0109 kΩ The effective load on the first op-amp is then RL1 = 0.1 + Rif 2 = 0.1109 kΩ

Again using Equation (14.20), we have 1 1 + 100 + 1 1 1 0.1109 = 0.10 + 110.017 = + ⋅ 11.017 Rif 10 1 1 + 1 + 1 0.1109 1 so that Rif = 99.1 Ω b.

To determine R0 f :

For the first op-amp, we can write, using Equation (14.36) ⎡ ⎤ ⎡ ⎤ ⎥ 1 ⎢ 100 ⎥ 1 1 ⎢ A0 L ⎥ = ⋅⎢ ⎥ = ⋅⎢ R2 ⎥ 1 ⎢ 40 ⎥ R0 f 1 R0 ⎢ 1+ 1+ ⎢ R1 || Ri ⎥ ⎢⎣ 1 || 10 ⎥⎦ ⎣ ⎦ which yields R0 f 1 = 0.021 kΩ For the second op-amp, then ⎡ ⎢ A0 L 1 1 ⎢ = ⋅ R2 R0 f R0 ⎢ ⎢1 + ( R + R ) || R 1 0f1 i ⎣

⎤ ⎥ ⎥ ⎥ ⎥ ⎦

⎡ ⎤ ⎥ 1 ⎢ 100 ⎥ = ⋅⎢ 0.10 1 ⎢1 + ⎥ ⎢⎣ (0.121) ||10 ⎥⎦ or R0 f = 18.4 Ω

c.

To find the gain, consider the second op-amp.

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v01 − (−vd 2 ) vd 2 −vd 2 − v02 + = 0.1 Ri 0.1

(1)

v01 v02 1 1 ⎞ ⎛ 1 + vd 2 ⎜ + + ⎟=− 0.1 0.1 ⎝ 0.1 10 0.1 ⎠ or v01 (10) + vd 2 (20.1) = −v02 (10) v02 − A0 L vd 2 v02 − (−vd 2 ) + =0 R0 0.1

(2)

v02 ⎛ 100 1 ⎞ v02 − vd 2 ⎜ − =0 ⎟+ 1 ⎝ 1 0.1 ⎠ 0.1 v02 (11) − vd 2 (90) = 0 or vd 2 = v02 (0.1222) Then Equation (1) becomes v01 (10) + v02 (0.1222)(20.1) = −v02 (10) or v01 = −v02 (1.246) Now consider the first op-amp.

vI − (−vd 1 ) vd 1 −vd 1 − v01 + = 1 Ri 1

(1)

⎛1 1 1⎞ vI (1) + vd 1 ⎜ + + ⎟ = −v01 (1) ⎝ 1 10 1 ⎠ or vI (1) + vd 1 (2.1) = −v01 (1) v01 v − A0 L vd 1 v01 − (−vd 1 ) + 01 + =0 0.1109 R0 1

(2)

1 1⎞ ⎛ 1 ⎛ 100 1 ⎞ v01 ⎜ + + ⎟ − vd 1 ⎜ − ⎟=0 ⎝ 0.1109 1 1 ⎠ ⎝ 1 1⎠ v01 (11.017) − vd 1 (99) = 0 or vd 1 = v01 (0.1113) Then Equation (1) becomes

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vI (1) + v01 (0.1113)(2.1) = −v01

or vI = −v01 (1.234) We had v01 = −v02 (1.246) So vI = v02 (1.246)(1.234) or

v02 = 0.650 vI

v02 =1 vI So ratio of actual to ideal = 0.650.

d.

Ideal

14.13 (a)

For the op-amp. A0 L ⋅ f 3dB = 106

106 = 50 Hz 2 × 104 For the closed-loop amplifier. 106 f 3dB = = 40 kHz 25 (b) Open-loop amplifier. 2 × 104 2 × 104 ⇒| A | = A= 2 f ⎛ f ⎞ 1+ j 1 + f3dB ⎜ ⎟ ⎝ f3dB ⎠ f 3dB =

f = 0.25 f 3dB ⇒ A = f = 5 f 3− dB ⇒ A =

2 × 104 1 + (0.25)

2

= 1.94 × 104

2 × 104

= 3.92 × 103 1 + (5) 2 Closed-loop amplifier 25 f = 0.25 f 3dB ⇒ A = = 24.25 1 + (0.25) 2 f = 5 f 3− dB ⇒ A =

25 1 + (5) 2

= 4.90

14.14 The open loop gain can be written as A0 A0 L ( f ) = ⎛ f ⎞⎛ f ⎞ ⎜1 + j ⋅ ⎟⎜ 1 + j ⋅ ⎟ f 5 106 ⎠ × ⎝ PD ⎠ ⎝ where A0 = 2 × 105. The closed-loop response is A0 L ACL = 1 + β A0 L At low frequency, 2 × 105 100 = 1 + β (2 × 105 ) So that β = 9.995 × 10−3. Assuming the second pole is the same for both the open-loop and closed-loop, then ⎛ f ⎞ f ⎞ −1 ⎛ φ = − tan −1 ⎜ ⎟ − tan ⎜ ⎟ × f 5 106 ⎠ ⎝ ⎝ PD ⎠

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For a phase margin of 80°, φ = −100°. So ⎛ f ⎞ −100 = −90 − tan −1 ⎜ 6 ⎟ ⎝ 5 × 10 ⎠ or f = 8.816 × 105 Hz Then A0 L = 1 2 × 105

=

⎛ 8.816 × 105 ⎞ 1+ ⎜ ⎟ f PD ⎝ ⎠

2

⎛ 8.816 × 105 ⎞ 1+ ⎜ ⎟ 6 ⎝ 5 × 10 ⎠

2

or 8.816 × 105 ≅ 1.9696 × 105 f PD or f PD = 4.48 Hz 14.15 (a) 1st stage (10) f3− dB = 1 MHz ⇒ f 3− dB = 100 kHz 2nd stage (50) f3− dB = 1 MHz ⇒ f 3− dB = 20 kHz Bandwidth of overall system ≅ 20 kHz (b) If each stage has the same gain, so 2 K = 500 ⇒ K = 22.36 Then bandwidth of each stage (22.36) f 3− dB = 1 MHz ⇒ f 3− dB = 44.7 kHz 14.16

A=

Ao 1+ j

A=

f f3− dB Ao

⎛ f ⎞ 1+ ⎜ ⎟ ⎝ f3− dB ⎠

2

⇒ 200 =

5 × 104 ⎛ 104 ⎞ 1+ ⎜ ⎟ ⎝ f3− dB ⎠

2

⇒ f3− dB = 40 Hz

Then fT = (5 × 104 )(40) ⇒ fT = 2 MHz 14.17

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(5 × 104 ) f PD = 106 ⇒ f PD = 20 Hz (25) f 3− dB 106 ⇒ f3− dB = 40 kHz Av =

Avo 1+ j

f

⇒ Av =

f3− dB

25 f ⎛ ⎞ 1+ ⎜ 3 ⎟ ⎝ 40 × 10 ⎠

2

At f = 0.5 f 3− dB = 20 kHz Av =

25 1 + (0.5) 2

= 22.36

At f = 2 f 3− dB = 80 kHz Av =

25 1 + (2) 2

14.18 (20 × 103 ) ⋅ Avf

= 11.18

MAX

= 106 ⇒ Avf

MAX

= 50

14.19 From Equation (14.55), SR 10 × 106 F P BW = = 2π VP 0 2π (10) or F P BW = f max = 159 kHz 14.20 a. VP 0

Using Equation (14.55), 8 × 106 = 2π (250 × 103 )

or VP 0 = 5.09 V b.

1 1 = = 4 × 10−6 s f 250 × 103 One-fourth period = 1 μ s

Period T =

Slope =

VP 0 = SR = 8 V/μ s 1μ s ⇒ VP 0 = 8 V

14.21 For input (a), maximum output is 5 V.

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S R = 1 V/μs so

For input (b), maximum output is 2 V.

For input (c), maximum output is 0.5 V so the output is

14.22 For input (a), max v01 = 3 V.

Then v02

max

= 3(3) = 9 V

For input (b), max v01 = 1.5 V.

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Then v02

max

= 3 (1.5 ) = 4.5 V

14.23 f MAX = 20 kHz, SR = 0.8 V / μ s V po =

SR 0.8 × 106 = ⇒ V po = 6.37 V 2π f MAX 2π (20 × 103 )

14.24 ⎛V ⎞ ⎛V ⎞ I1 = I S 1 exp ⎜ BE1 ⎟ , I 2 = I S 2 exp ⎜ BE 2 ⎟ ⎝ VT ⎠ ⎝ VT ⎠ Want I1 = I 2 , so ⎛V ⎞ 5 × 10−14 (1 + x) exp ⎜ BE1 ⎟ I1 ⎝ VT ⎠ =1= I2 ⎛V ⎞ 5 × 10−14 (1 − x) exp ⎜ BE 2 ⎟ ⎝ VT ⎠ =

⎛ V −V ⎞ (1 + x) exp ⎜ BE1 BE 2 ⎟ VT (1 − x) ⎝ ⎠

Or ⎛ V − VBE1 ⎞ ⎛ VOS ⎞ 1+ x = exp ⎜ BE 2 ⎟ = exp ⎜ ⎟ 1− x V ⎝ ⎠ ⎝ VT ⎠ T ⎛ 0.0025 ⎞ = exp ⎜ ⎟ = 1.10 ⎝ 0.026 ⎠ Now 1 + x = (1 − x )(1.10) ⇒ x = 0.0476 ⇒ 4.76% 14.25 From Equation (14.62), vCE 2 ⎛ vCE1 ⎞ ⎛ ⎜1+ V ⎟ I ⎜ 1+ V AN ⎟ AN ⎜ = S3 ⋅⎜ ⎜ 1 + vEB ⎟ I S 4 ⎜ 1 + vEC 4 ⎜ V ⎟ ⎜ VAP AP ⎠ ⎝ ⎝

⎞ ⎟ ⎟ ⎟ ⎟ ⎠

For vCE 2 = 0.6 V, then vEC 4 = 5 V. We have vCE1 = 5 V so

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5 ⎞ ⎛ ⎛ 0.6 ⎞ ⎜ 1 + 80 ⎟ I S 3 ⎜ 1 + 80 ⎟ ⋅⎜ ⎜ ⎟= ⎟ ⎜⎜ 1 + 0.6 ⎟⎟ I S 4 ⎜⎜ 1 + 5 ⎟⎟ 80 ⎠ 80 ⎠ ⎝ ⎝ or I S 3 (1.0625) 2 = = 1.112 I S 4 (1.0075) 2 So I S 3 = (10−14 ) (1.112 )

or I S 3 = 1.112 × 10−14 A 14.26

By superposition: R vo (vi ) = − 2 ⋅ vi = −50vi R1 ⎛ R ⎞ vo (vos ) = ⎜ 1 + 2 ⎟ ⋅ vos = 51vos R1 ⎠ ⎝ So vo = vo ( vi ) + vo ( vos ) = −50vi + 51vos

For vi = 20 mV and vos + 2.5 mV vo = −50(0.02) + 51(0.0025) = −0.8725 V For vi = 20 mV and vos = −2.5 mV vo = −50(0.02) + 51(−0.0025) = −1.1275 V So −1.1275 ≤ vo ≤ −0.8725 V 14.27 vo = −50vi = −50 ⎡⎣ ±2.5 mV + sin ω t ( mV ) ⎤⎦ vo = [ ±0.125 − 0.25sin ω t ] ( v )

14.28

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0.5 × 10−3 = 5 × 10−8 A 104 Also i dV 1 I I = C o ⇒ Vo = ∫ Idt = ⋅ t dt C0 C Then 5 × 10−8 5= t ⇒ t = 103 s 10 × 10−6 I=

14.29 a.

⎛ 100 ⎞ | v01 | = 10 ⎜ 1 + ⎟ or | v01 | = 110 mV 10 ⎠ ⎝

Then ⎛ 50 ⎞ | v02 | = | v01 | (5) + 10 ⎜ 1 + ⎟ = (110)(5) + (10)(6) ⎝ 10 ⎠ or | v02 | = 610 mV

14.30 v0 due to vI 1 ⎞ ⎛ v0 = (0.5) ⎜1 + ⎟ = 0.9545 V ⎝ 1.1 ⎠ Wiper arm at V + = 10 V, (using superposition)

⎛ R1 || R5 ⎞ ⎛ 0.0909 ⎞ v1 = ⎜ ⎟ (10) = ⎜ ⎟ (10) + R || R R ⎝ 0.0909 + 10 ⎠ 4 ⎠ ⎝ 1 5 = 0.090 ⎛ 1⎞ Then v01 = − ⎜ ⎟ (0.090) = −0.090 ⎝ 1⎠ Wiper arm in center, v1 = 0 and v02 = 0

Wiper arm at V − = −10 V, v1 = −0.090 So v03 = 0.090 Finally, total output v0 : (from superposition) Wiper arm at V + , v0 = 0.8645 V Wiper arm in center, v0 = 0.9545 V Wiper arm at V − ,

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v0 = 1.0445 V

14.31 a. R1′ = R2′ = 0.5 || 25 = 0.490 kΩ or R1′ = R2′ = 490 Ω b.

From Equation (14.75), ⎛ 125 × 10−6 ⎞ (0.026) ln ⎜ + (0.125) R1′ −14 ⎟ ⎝ 2 × 10 ⎠ ⎛ 125 × 10−6 ⎞ = (0.026) ln ⎜ + (0.125) R2′ −14 ⎟ ⎝ 2.2 × 10 ⎠ 0.586452 + (0.125) R1′ = 0.583974 + (0.125) R2′ 0.002478 = (0.125)( R2′ − R1′)

So R2′ − R1′ = 0.0198 kΩ ⇒ 19.8 Ω Then R2 (1 − x) Rx R × Rx − 1 = 0.0198 R2 + (1 − x) Rx R1 + xRx (0.5)(1 − x)(50) (0.5)(50) x − = 0.0198 (0.5) + (1 − x)(50) (0.5) + x(50) 25(1 − x) 25 x − = 0.0198 50.5 − 50 x 0.5 + 50 x (0.5 + 50 x)(25 − 25 x) − (25 x)(50.5 − 50 x) = 0.0198 (50.5 − 50 x )(0.5 + 50 x)

25 {0.5 − 0.5 x + 50 x − 50 x 2 − 50.5 x + 50 x 2 } = 0.0198 {25.25 + 2525 x − 25 x − 2500 x 2 } 25 {0.5 − x} = 0.0198 {25.25 + 2500 x − 2500 x 2 } 0.5 − x = 0.019998 + 1.98 x − 1.98 x 2 1.98 x 2 − 2.98 x + 0.48 = 0 x=

2.98 ± (2.98) 2 − 4(1.98)(0.48) 2(1.98)

So x = 0.183 and 1 − x = 0.817 14.32 R1′ = R1 ||15 = 0.5 ||15 = 0.4839 kΩ R2′ = R2 || 35 = 0.5 || 35 = 0.4930 kΩ From Equation (14.75), ⎛i ⎞ ⎛i ⎞ (0.026) ln ⎜ C1 ⎟ + iC1 R1′ = (0.026) ln ⎜ C 2 ⎟ + iC 2 R2′ ⎝ IS3 ⎠ ⎝ IS 4 ⎠ ⎛i ⎞ (0.026) ln ⎜ C1 ⎟ = iC 2 R2′ − iC1 R1′ ⎝ iC 2 ⎠ ⎛i ⎞ ⎡ i R′ ⎤ (0.026) ln ⎜ C1 ⎟ = iC 2 R2′ ⎢1 − C1 ⋅ 1 ⎥ ′ i i R 2 ⎦ C2 ⎝ C2 ⎠ ⎣ ⎡ ⎛i ⎞ ⎛ i ⎞⎤ (0.026) ln ⎜ C1 ⎟ = iC 2 (0.4930) ⎢1 − (0.9815) ⎜ C1 ⎟ ⎥ ⎝ iC 2 ⎠ ⎝ iC 2 ⎠ ⎦ ⎣ By trial and error:

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iC1 = 252 μ A and iC 2 = 248 μ A

or iC1 = 1.0155 iC 2 14.33 From Eq. (14.79), we have ⎛ R ⎞ vo = I B1 R2 − I B 2 R3 ⎜ 1 + 2 ⎟ R1 ⎠ ⎝ I B1 = 1 μ A I B 2 = 2 μ A Setting vo = 0, we have ⎛ 200 ⎞ 0 = (10−6 )( 200 × 103 ) − ( 2 × 10−6 ) R3 ⎜ 1 + ⎟ 20 ⎠ ⎝ 200 × 10−3 R3 = ⇒ R = 9.09 K ( 2 ×10−6 ) (11) 3

14.34

1+

R2 = 80 R1 R1 = 6.329 K

V f = vI = 5sin ω t ( mV ) I1 + I B = I 2

(a)

VI = 0 ⇒ VX = 0 I 2 = I B VO = (10−6 )(500 × 103 ) ⇒ vo = 0.50 V

(b)

v − VX VX + IB = o R1 R2 ⎛ 1 ⎛ 1 v 1 ⎞ 1 ⎞ VX ⎜ + ⎟ + I B = 0 ⇒ vo = R2 ⎜ + ⎟ vI + I B R2 R1 ⎝ R1 R2 ⎠ ⎝ R1 R2 ⎠ vo = 80 ⎡⎣5sin ω t ( mV ) ⎤⎦ + (10−6 )( 500 × 103 ) vo = [ 0.5 + 0.4sin ω t ] ( v )

14.35 a.

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For I B 2 = 1 μ A, then v0 = − (10−6 )(104 ) or

v0 = −0.010 V

b.

If a 10 kΩ resistor is included in the feedback loop

Now v0 = − I B 2 (10) + I B1 (10) = 0 Circuit is compensated if I B1 = I B 2 .

14.36 From Equation (14.83), we have v0 = R2 I 0S where R2 = 40 kΩ and I 0 S = 3 μ A. Then v0 = ( 40 × 103 )( 3 × 10−6 ) or v0 = 0.12 V 14.37 a. Assume all bias currents are in the same direction and into each op-amp. v01 = I B1 (100 kΩ ) = (10−6 )(105 ) ⇒ v01 = 0.1 V Then v02 = v01 ( −5 ) + I B1 ( 50 kΩ )

= ( 0.1)( −5 ) + (10−6 )( 5 × 104 ) = −0.5 + 0.05

or v02 = −0.45 V b. Connect R3 = 10 ||100 = 9.09 kΩ resistor to noninverting terminal of first op-amp, and R3 = 10 || 50 = 8.33 kΩ resistor to noninverting terminal of second op-amp.

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14.38 a. For a constant current through a capacitor. 1 t v0 = ∫ I dt C 0 0.1× 10−6 or v0 = ⋅ t ⇒ v0 = (0.1)t 10−6 v0 = 1 V b. At t = 10 s, c. Then 100 × 10−12 v0 = ⋅ t ⇒ v0 = (10−4 )t −6 10 At t = 10 s, v0 = 1 mV 14.39 a. Assume all bias currents are into the op-amp. v01 = I B1 ( 50 kΩ ) = (10 ×10−6 )( 50 ×103 ) or v01 = v02 = 0.5 V v03 = ( −1)( v01 ) + (10 × 10−6 )( 20 × 103 )

or v03 = −0.3 V b.

RA = 10 || 50 ⇒ RA = 8.33 kΩ

RB = 20 || 20 ⇒ RB = 10 kΩ c. Assume the worst case offset current, that is, I 0 S = I B1 − I B 2 or I 0 S = I B 2 − I B1 . From Equation (14.83), v01 = R2 I 0 S = ( 50 × 103 )( 2 × 10−6 )

or v01 = v02 = 0.1 V v03 = ( −1) v01 − I 0 S R2

= ( −1)( 0.1) − ( 2 × 10−6 )( 20 × 103 )

or v03 = −0.14 V 14.40 a. Using Equation (14.79), Circuit (a), ⎛ 50 ⎞ v0 = ( 0.8 × 10−6 )( 50 × 103 ) − ( 0.8 ×10−6 )( 25 × 103 ) ⎜1 + ⎟ ⎝ 50 ⎠ or v0 = 0

Circuit (b), ⎛ 50 ⎞ v0 = ( 0.8 × 10−6 )( 50 × 103 ) − ( 0.8 × 10 −6 )(103 ) ⎜1 + ⎟ ⎝ 50 ⎠ −2 = 4 × 10 − 1.6 or v0 = −1.56 V

b. Assume I B1 = 0.7 μ A and I B 2 = 0.9 μ A, then using Equation (14.79): Circuit (a),

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⎛ 50 ⎞ v0 = ( 0.7 × 10−6 )( 50 × 103 ) − ( 0.9 × 10−6 )( 25 × 103 ) ⎜ 1 + ⎟ ⎝ 50 ⎠ = 0.035 − 0.045

or

v0 = −0.010 V

Circuit (b), ⎛ 50 ⎞ v0 = ( 0.7 × 10−6 )( 50 × 103 ) − ( 0.9 × 10−6 )(106 ) ⎜1 + ⎟ ⎝ 50 ⎠ = 0.035 − 1.8

or

v0 = −1.765 V

14.41 a.

If R = 0, ⎛ 100 ⎞ v0,max = ⎜1 + ⎟ V0 S + I B (100 kΩ) 10 ⎠ ⎝ = (11)(10 × 10−3 ) + (2 × 10−6 )(100 × 103 ) v0,max = 0.110 + 0.20 ⇒ v0,max = 0.310 V

b.

R = 10.1|| 100 = 9.17 kΩ = R

14.42 a.

⎛ Ri ⎞ ⎜ ⎟ (15) = 0.010 V ⎝ Ri + R2 ⎠

15 = 0.0006667 15 + R2 15(1 − 0.0006667) = 0.0006667 R2 Then R2 = 22.48 MΩ

b.

R1 = Ri || RF = 15 || 10 ⇒ R1 = 6 kΩ

14.43 a. Assume the offset voltage polarities are such as to produce the worst case values, but the bias currents are in the same direction. Use superposition:

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Offset voltages ⎛ 100 ⎞ | v01 | = ⎜1 + ⎟ (10) = 110 mV =| v01 | 10 ⎠ ⎝ ⎛ 50 ⎞ | v02 | = (5)(110) + ⎜ 1 + ⎟ (10) ⎝ 10 ⎠ ⇒ | v02 | = 610 mV

Bias Currents: v01 = I B (100 kΩ) = (2 ×10−6 )(100 × 103 ) = 0.2 V Then v02 = (−5)(0.2) + (2 × 10−6 )(50 × 103 ) = −0.9 V Worst case: v01 is positive and v02 is negative, then v01 = 0.31 V and v02 = −1.51 V b.

Compensation network:

If we want ⎛ RB ⎞ + + ⎜ ⎟ V = 20 mV and V = 10 V + R R C ⎠ ⎝ B ⎛ 8.33 ⎞ ⎜ ⎟ (10) = 0.020 ⎝ 8.33 + RC ⎠ or RC ≅ 4.15 MΩ

14.44 Assume bias currents are in same direction, but assume polarity of offset voltages are such as to produce the worst case output. a. Let I B1 = 5.5 μ A, I B 2 = 4.5 μ A Bias Current Effects: v01 = I B1 (50 kΩ) = 0.275 V ⇒ v02 = 0.275 V v03 = I B1 (20 kΩ) − v01 ⇒ v03 = −0.165 V Offset Voltage Effects: ⎛ 50 ⎞ v01 = (5) ⎜1 + ⎟ = 30 mV ⇒ v02 = 30 mV ⎝ 10 ⎠ ⎛ 20 ⎞ v03 = −v01 − 5 ⎜ 1 + ⎟ ⇒ v03 = −40 mV ⎝ 20 ⎠ Total Effect: v01 = 0.305 V and v02 = 0.305 V v03 = −0.205 V 14.45 For circuit (a), effect of bias current: v0 = (50 × 103 )(100 × 10−9 ) ⇒ 5 mV Effect of offset voltage

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⎛ 50 ⎞ v0 = (2) ⎜1 + ⎟ = 4 mV ⎝ 50 ⎠ So net output voltage is v0 = 9 mV

For circuit (b), effect of bias current: Let I B 2 = 550 nA, I B1 = 450 nA, then from Equation (14.79), ⎛ 50 ⎞ v0 = (450 × 10−9 )(50 × 103 ) − (550 × 10−9 )(106 ) ⎜1 + ⎟ ⎝ 50 ⎠ −2 = 2.25 × 10 − 1.1 or v0 = −1.0775 V If the offset voltage is negative, then v0 = (−2)(2) = −4 mV So the net output voltage is v0 = −1.0815 V

14.46 a. At T = 25°C, V0 S = 2 mV so the output voltage for each circuit is v0 = 4 mV b. For T = 50°C, the offset voltage for is V0 S = 2 mV + (0.0067)(25) = 2.1675 mV so the output voltage for each circuit is v0 = 4.335 mV 14.47 a.

At T = 25°C,

V0 S = 1 mV, then

⎛ 50 ⎞ v01 = (1) ⎜ 1 + ⎟ ⇒ v01 = 6 mV ⎝ 10 ⎠ and ⎛ 60 ⎞ ⎛ 60 ⎞ v02 = v01 ⎜ 1 + ⎟ + (1) ⎜ 1 + ⎟ ⎝ 20 ⎠ ⎝ 20 ⎠ = 6(4) + (1)(4) ⇒ v02 = 28 mV

b. At T = 50°C, V0 S = 1 + (0.0033)(25) = 1.0825 mV, then v01 = (1.0825)(6) ⇒ v01 = 6.495 mV and v02 = (6.495)(4) + (1.0825)(4) or v02 = 30.31 mV 14.48 25°C; I B = 500 nA, I 0 S = 200 nA 50°C, I B = 500 nA + (8 nA / °C)(25°C) = 700 nA I 0 S = 200 nA + (2 nA / °C)(25°C) = 250 nA a. Circuit (a): For I B , bias current cancellation, v0 = 0 Circuit (b): For I B , Equation (14.79), ⎛ 50 ⎞ v0 = (500 × 10−9 )(50 × 103 ) − (500 × 10 −9 )(106 ) ⎜ 1 + ⎟ ⎝ 50 ⎠ = 0.025 − 1.00 ⇒ v0 = −0.975 V

b. Due to offset bias currents. Circuit (a):

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v0 = (200 × 10−9 )(50 × 103 ) ⇒ v0 = 0.010 V

Circuit (b): Let I B 2 = 600 nA I B1 = 400 nA Then ⎛ 50 ⎞ v0 = (400 × 10−9 )(50 × 103 ) − (600 × 10−9 )(106 ) ⎜ 1 + ⎟ ⎝ 50 ⎠ = 0.020 − 1.20 ⇒ v0 = −1.18 V

c. Circuit (a): Due to I B , v = 0 Circuit (b): Due to I B , ⎛ 50 ⎞ v0 = (700 × 10−9 )(50 × 103 ) − (700 × 10−9 )(106 ) ⎜1 + ⎟ ⎝ 50 ⎠ = 0.035 − 1.40 ⇒ v0 = −1.365 V

Circuit (a): Due to I 0 S , v0 = (250 × 10−9 )(50 × 103 ) ⇒ v0 = 0.0125 V

Circuit (b): Due to I 0 S , Let I B 2 = 825 nA I B1 = 575 nA Then ⎛ 50 ⎞ v0 = (575 × 10−9 )(50 × 103 ) − (825 × 10−9 )(106 ) ⎜ 1 + ⎟ ⎝ 50 ⎠ = 0.02875 − 1.65 ⇒ v0 = −1.62 V

14.49 25°C; I B = 2 μ A, I 0 S = 0.2 μ A 50°C, I B = 2 μ A + (0.020 μ A / °C)(25°C ) = 2.5 μ A I 0 S = 0.2 μ A + (0.005 μ A / °C)(25°C) = 0.325 μ A a. Due to I B : (Assume bias currents into op-amp). v01 = I B (50 kΩ) = (2 × 10−6 )(50 × 103 ) ⇒ v01 = 0.10 V ⎛ 60 ⎞ ⎛ 60 ⎞ v02 = v01 ⎜ 1 + ⎟ + I B (60 kΩ) − I B (50 kΩ) ⎜ 1 + ⎟ 20 ⎝ ⎠ ⎝ 20 ⎠ 3 −6 −6 = (0.1)(4) + (2 × 10 )(60 × 10 ) − (2 × 10 )(60 × 103 )4

or

v02 = 0.12 V

b. Due to I 0 S : 1st op-amp. Let I B1 = 2.1 μ A 2nd op-amp. Let I B1 = 2.1 μ A I B 2 = 1.9 μ A v01 = I B1 (50 kΩ) = (2.1× 10−6 )(50 × 103 ) ⇒ v01 = 0.105 V ⎛ 60 ⎞ ⎛ 60 ⎞ v02 = v01 ⎜ 1 + ⎟ + I B1 (60 kΩ) − I B 2 (50 kΩ) ⎜1 + ⎟ 20 ⎝ ⎠ ⎝ 20 ⎠ 3 −6 = (0.105)(4) + (2.1× 10 )(60 × 10 ) − (1.9 × 10 −6 )(50 × 103 )(4) or v02 = 0.166 V

c.

Due to I B :

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v01 = (2.5 × 10 −6 )(50 × 103 ) ⇒ v01 = 0.125 V ⎛ 60 ⎞ ⎛ 60 ⎞ v01 = v02 ⎜ 1 + ⎟ + I B (60 kΩ) − I B (50 kΩ) ⎜1 + ⎟ 20 ⎝ ⎠ ⎝ 20 ⎠ −6 3 = (0.125)(4) + (2.5 × 10 )(60 × 10 ) − (2.5 × 106 )(50 × 103 (4) or v02 = 0.15 V

Due to I 0 S : Let I B1 = 2.625 μ A I B 2 = 2.3375 μ A v01 = I B1 (50 kΩ) = (2.6625 × 10−6 )(50 × 103 ) ⇒ v01 = 1.133 V ⎛ 60 ⎞ ⎛ 60 ⎞ v02 = v01 ⎜ 1 + ⎟ + I B1 (60 kΩ) − I B 2 (50 kΩ) ⎜ 1 + ⎟ ⎝ 20 ⎠ ⎝ 20 ⎠ = (0.133)(4) + (2.6625 × 10−6 )(60 × 103 ) − (2.3375 × 10−6 )(50 × 103 )(4) or v02 = 0.224 V

14.50

⎛ R4 ⎞ ⎛ R2 ⎞ vB = ⎜ ⎟ vI 1 and v0 (vI 2 ) = vB ⎜ 1 + ⎟ R1 ⎠ ⎝ ⎝ R3 + R4 ⎠ or ⎛ R4 ⎞ ⎛ R2 ⎞ v0 (vI 2 ) = ⎜ ⎟ ⎜ 1 + ⎟ vI 2 R1 ⎠ ⎝ R3 + R4 ⎠ ⎝

or vI 1 . v0 (vI 1 ) = −

R2 ⋅ vI R1

Then ⎛ R4 ⎞⎛ R2 ⎞ R2 v0 = ⎜ ⎟⎜ 1 + ⎟ vI 2 − ⋅ vI 1 R1 ⎠ R1 ⎝ R3 + R4 ⎠ ⎝ V V we can write vI 2 = vcm + d and vI 1 = Vcm − d ⋅ Then 2 2 ⎛ R4 ⎞⎛ R2 ⎞ ⎛ Vd ⎞ R2 ⎛ Vd ⎞ v0 = ⎜ ⎟⎜ 1 + ⎟ ⎜ Vcm + ⎟ − ⋅ ⎜ Vcm − ⎟ R1 ⎠ ⎝ 2 ⎠ R1 ⎝ 2⎠ ⎝ R3 + R4 ⎠ ⎝ Common-mode gain ⎛ R4 ⎞ ⎛ R2 ⎞ R2 v Acm = 0 = ⎜ ⎟ ⎜1 + ⎟ − Vcm ⎝ R3 + R4 ⎠ ⎝ R1 ⎠ R1

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Differential mode gain v 1 ⎡⎛ R4 ⎞⎛ R2 ⎞ R2 ⎤ Ad = 0 = ⎢⎜ ⎟⎜ 1 + ⎟ + ⎥ Vd 2 ⎣⎝ R3 + R4 ⎠ ⎝ R1 ⎠ R1 ⎦ Then A CM RR = d Acm 1 ⎡⎛ R4 ⎞ ⎛ R2 ⎞ R2 ⎤ ⋅ ⎢⎜ ⎟ ⎜1 + ⎟ + ⎥ R1 ⎠ R1 ⎦ 2 ⎣⎝ R3 + R4 ⎠ ⎝ = ⎡⎛ R4 ⎞ ⎛ R2 ⎞ R2 ⎤ ⎢⎜ ⎟ ⎜1 + ⎟ − ⎥ R1 ⎠ R1 ⎦ ⎣⎝ R3 + R4 ⎠ ⎝ ⎡ ⎢ 1 ⎢ R4 ⋅ 2 ⎢ R3 ⎢ ⎣⎢

⎤ ⎥ ⎛ R2 ⎞ R2 ⎥ 1 ⋅ ⋅ ⎜1 + ⎟ + R1 ⎠ R1 ⎥ ⎛ R4 ⎞ ⎝ + 1 ⎥ ⎜ ⎟ ⎝ R3 ⎠ ⎦⎥ CM RR = ⎛ R2 ⎞ R2 R4 1 ⋅ ⋅ ⎜1 + ⎟ − R3 ⎛ R4 ⎞ ⎝ R1 ⎠ R1 ⎜1 + ⎟ ⎝ R3 ⎠ Minimum CMRR ⇒ Maximum denominator R R ⇒ maximum 4 and minimum 2 . Then R3 R1 R4 (1.02)(50) = = 5.204 R3 (0.98)(10) R2 (0.98)(50) = = 4.804 R1 (1.02)(10) Then 1 ⎡ 5.204 ⎤ ⋅ (5.804) + (4.804) ⎥ 2 ⎢⎣ 6.204 ⎦ CMRR = ⎡ 5.204 ⎤ ⎢⎣ 6.204 ⋅ (5.804) − (4.804) ⎥⎦ 1 ⋅ (9.6725) = 2 (0.06447) CMRR = 75.0 ⇒ CMRRdB = 20 log10 (75.0) ⇒ CMRRdB = 37.5 dB

14.51 Use the results of Problem 14.50: R 1 + x ⎛ 50 ⎞ Let 4 = ⋅ ⎜ ⎟ ≈ (1 + 2 x)(5) R3 1 − x ⎝ 10 ⎠ R 1 − x ⎛ 50 ⎞ Let 2 = ⋅ ⎜ ⎟ ≈ (1 − 2 x)(5) R1 1 + x ⎝ 10 ⎠ Then

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CMRR =

⎤ 1 ⎡ (1 + 2 x ) 5 ⋅ ( 6 − 10 x ) + (1 − 2 x )( 5 ) ⎥ ⎢ 2 ⎣ 6 + 10 x ⎦

⎡ (1 + 2 x ) 5 ⎤ ⋅ ( 6 − 10 x ) − (1 − 2 x )( 5 ) ⎥ ⎢ ⎣ 6 + 10 x ⎦ 1 2 ⎡⎣30 + 10 x − 100 x + 30 − 10 x − 100 x 2 ⎤⎦ = 2 ⎡30 + 10 x − 100 x 2 − ( 30 − 10 x − 100 x 2 ) ⎤ ⎣ ⎦

a. 20 x =

b. 20 x =

1 ⋅ [60 − 200 x 2 ] 30 − 100 x 2 2 = = 20 x 20 x For CMRRdB = 90 dB ⇒ CMRR = 31, 623 x will be small, neglect the x 2 term. Then 30 ⇒ x = 0.0000474 = 0.00474% 31, 623 For CMRRdB = 60 dB ⇒ CMRR = 1000. Then 30 ⇒ x = 0.0015 = 0.15% 1000

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Chapter 15 Exercise Solutions EX15.1 For the circuit shown in Figure 15.7 1 f 3dB = 2π RC or 1 1 RC = = = 3.979 × 10−6 2π f3dB 2π ( 40 × 103 ) For R = 75 K Then C = 5.31×10−11 = 53.1 pF We have C3 = 1.414C = 75.1 pF C4 = 0.707C = 37.5 pF

EX15.2 1 fC = CReq or C=

1 1 = f c Req (105 )( 20 × 106 )

C = 0.5 pF

EX15.3 Low-frequency gain: T = − f 3dB =

C1 30 =− = −6 C2 5

3 −12 fC C2 (100 × 10 )( 5 × 10 ) = ⇒ f 3dB = 6.63 kHz 2π CF 2π (12 × 10−12 )

EX15.4 1

f0 =

2π 3RC 1 1 RC = = = 6.13 × 10−6 3 2π f 0 3 2π (15 × 10 ) 3 Let C = 0.001 μF = 1 nF

Then R = 6.13 kΩ so R2 = 8R = 49 kΩ EX15.5 f0 = C=

1 1 ⇒C = 2π RC 2π f 0 R 1

2π ( 800 ) (104 )

⇒ C ≅ 0.02 μ F

R2 = 2 R1 = 2 (10 ) ⇒ R2 = 20 kΩ

EX15.6

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⎛ R1 ⎞ VTH = ⎜ ⎟ VH ⎝ R1 + R2 ⎠ ⎛ R1 ⎞ 2=⎜ ⎟ (12) ⎝ R1 + 20 ⎠ 2 ( R1 + 20 ) = 12 R1

40 = 10 R1 ⇒ R1 = 4 kΩ

EX15.7 ⎛ R1 ⎞ VTH − VTL = ⎜ ⎟ (VH − VL ) ⎝ R1 + R2 ⎠ ⎛ R1 ⎞ 0.10 = ⎜ ⎟ (10 − [ −10]) ⎝ R1 + R2 ⎠ 1+

R2 R 20 = = 200 ⇒ 2 = 199 R1 0.10 R1

⎛ R2 ⎞ VS = ⎜ ⎟ VREF ⎝ R1 + R2 ⎠ ⎛ R ⎞ 1 ⎞ ⎛ VREF = ⎜1 + 1 ⎟ VS = ⎜ 1 + ⎟ (1) ⇒ VREF = 1.005 V 199 R ⎝ ⎠ 2 ⎠ ⎝ I=

VH − VBE ( on ) − Vγ

R + 0.1 10 − 0.7 − 0.7 R + 0.1 = = 43 kΩ 0.2 R = 42.9 kΩ

EX15.8 At t = 0− , let v0 = −5 so v X = −2.5. For t > 0 ⎛ −t ⎞ v X = 10 + ( −2.5 − 10 ) exp ⎜ ⎟ ⎝ rX ⎠ When v X = 5.0, output switches ⎛ t ⎞ 5.0 = 10 − 12.5 exp ⎜ − 1 ⎟ ⎝ rX ⎠ ⎛ t ⎞ 10 − 5 5.0 = exp ⎜ − 1 ⎟ = ⎝ rX ⎠ 12.5 12.5 ⎛ t ⎞ 12.5 ⎛ 12.5 ⎞ exp ⎜ + 1 ⎟ = ⇒ t1 = rX ⋅ ln ⎜ ⎟ ⇒ t1 = rX ( 0.916 ) ⎝ 5.0 ⎠ ⎝ rX ⎠ 5.0 During the next part of the cycle ⎛ t ⎞ v X = −5 + ( 5 − [ −5]) exp ⎜ − ⎟ ⎝ rX ⎠

When v X = −2.5, output switches

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⎛ t ⎞ −2.5 = −5 + 10 exp ⎜ − 2 ⎟ ⎝ rX ⎠ ⎛ t ⎞ 5 − 2.5 2.5 exp ⎜ − 2 ⎟ = = 10 10 ⎝ rX ⎠ ⎛ t ⎞ 10 ⎛ 10 ⎞ exp ⎜ + 2 ⎟ = ⇒ t2 = rX ⋅ ln ⎜ ⎟ ⇒ t2 = rX (1.39 ) 2.5 r ⎝ 2.5 ⎠ ⎝ X⎠ 1 Period = t1 + t2 = T = ⎡⎣( 0.916 ) + (1.39 ) ⎤⎦ rX = 2.31rX ⇒ Frequency = 2.31rX rX = ( 50 × 103 )( 0.01× 10 −6 ) = 5 × 10 −4 s ⇒ f = 866 Hz Duty cycle =

( 0.916 ) t1 × 100% = × 100% ⇒ Duty cycle = 39.7% t1 + t2 ( 0.916 ) + (1.39 )

EX15.9 a. rX = RX C X ⎛ R1 ⎞ ⎛ 10 ⎞ vY = ⎜ ⎟ v0 = ⎜ ⎟ (12 ) = 1.2 V ⎝ 10 + 90 ⎠ ⎝ R1 + R2 ⎠ R1 = 0.10 β= R1 + R2 0.7 ⎤ ⎡ ⎢ 1 + 12 ⎥ ⎡1 + Vγ /VP ⎤ T = rX ln ⎢ ⎥ ⎥ = rX ln ⎢ ⎣ 1− β ⎦ ⎢1 − (0.10) ⎥ ⎣⎢ ⎦⎥ T = 50 × 10−6 = rX ln [1.18] = (0.162) rX

50 × 10−6 ⇒ RX = 3.09 kΩ (0.1× 10−6 )(0.162) b. Recovery time ⎛ t ⎞ v X = VP + (−1.2 − VP ) exp ⎜ − ⎟ ⎝ rX ⎠ RX =

When v X = Vγ , t = t2

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⎛ t ⎞ 0.7 = 12 + ( −1.2 − 12 ) exp ⎜ − 2 ⎟ ⎝ rX ⎠ ⎛ t ⎞ 12 − 0.7 = 0.856 exp ⎜ − 2 ⎟ = 13.2 ⎝ rX ⎠ ⎛ 1 ⎞ t2 = rX ln ⎜ ⎟ = ( 0.155 ) rX ⎝ 0.856 ⎠

rX = ( 3.09 × 103 )( 0.1× 10−6 ) = 3.09 × 10−4 ⇒ t2 = 48.0 μ s

EX15.10 T = 1.1 RC T = 75 × 10−6 Let C = 10 nF Then 75 × 10−6 R= = 6.82 K (1.1) (10 ×10−9 )

EX15.11 1 1 = ⇒ f = 802 Hz 0.693 ( RA + 2 RB ) C ( 0.693) ⎡⎣ 20 + 2 ( 80 ) ⎤⎦ × 103 × ( 0.01× 10−6 ) R + RB 20 + 80 Duty cycle = A × 100% = × 100% ⇒ Duty cycle = 55.6% 20 + 2 ( 80 ) RA + 2 RB f =

EX15.12 P=

a.

1 VP2 ⋅ 2 RL

VP = 2 RL P = 2 ( 8 )(1) ⇒ VP = 4 V IP =

VP 4 = ⇒ I P = 0.5 A RL 8

VCE = 12 − 4 = 8 V

b.

I C ≈ 0.5 A

So P = I C ⋅ VCE = ( 0.5 )( 8 ) ⇒ P = 4 W EX15.13 VP = 2 RL PL = 2 ( 8 )(10 ) = 12.65 V ⎛ V ⎞ PS = VS ⎜ P ⎟ ⎝ π RL ⎠

VS =

PS π R2 (10 ) π ( 8 ) = VP 12.65

VS = 19.9 V

EX15.14

Line regulation =

dV0 dV dV = 0 ⋅ Z+ + dV dVZ dV

Now

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dV0 ⎛ 10 ⎞ = ⎜1 + ⎟ = 2 dVZ ⎝ 10 ⎠ dVZ ⎛ rZ ⎞ 10 =⎜ = 0.00227 ⎟= + dV ⎝ rZ + R1 ⎠ 10 + 4400 So Line regulation = ( 2 )( 0.00227 ) = 0.00454

0.454% EX15.15

V1 V0 − V1 ⎛1 1⎞ V = ⇒ V1 ⎜ + ⎟ = 0 10 10 ⎝ 10 10 ⎠ 10 V ⎛ 2⎞ V V1 ⎜ ⎟ = 0 ⇒ V0 = 2V1 ⇒ V1 = 0 10 10 2 ⎝ ⎠ V0 − V1 V0 V0 − A0 L (VZ − V1 ) + + =0 RL R0 10 V0 V0 V0 A0 LVZ V1 A0 LV1 + + − = − R0 R0 10 RL R0 10 =

V0 A V − 0L 0 2(10) 2 R0

V0 V 1000 ( 6.3) V0 (1000 ) V0 + I0 + 0 − = − 10 0.5 0.5 20 2 ( 0.5 ) V0 [0.10 + 2.0 − 0.05 + 1000] + I 0 = 12, 600 V0 (1002.05) + I 0 = 12, 600 For I 0 = 1 mA ⇒ V0 = 12.5732

For I 0 = 100 mA ⇒ V0 = 12.4744 Load reg =

V0 ( NL ) − V0 ( FL ) V0 ( NL )

× 100%

12.5732 − 12.4744 × 100% 12.5732 Load reg = 0.786%

=

EX15.16 a.

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IC 3 =

VZ − 3VBE ( on )

IC 3 =

R1 + R2 + R3 5.6 − 3 ( 0.6 )

3.9 + 3.4 + 0.576 ⎛I ⎞ I C 4 R4 = VT ln ⎜ C 3 ⎟ ⎝ IC 4 ⎠

=

3.8 ⇒ I C 3 = 0.482 mA 7.88

⎛ 0.482 ⎞ I C 4 (0.1) = (0.026) ln ⎜ ⎟ ⎝ IC 4 ⎠ By trial and error I C 4 = 0.213 mA VB 7 = 2(0.6) + (0.482)(3.9) ⇒ VB 7 = 3.08 V

b. ⎛ R13 ⎞ ⎜ ⎟ V0 = VB 8 = VB 7 ⎝ R13 + R12 ⎠ ⎛ 2.23 ⎞ ⎜ ⎟ (5) = 3.08 ⎝ 2.23 + R12 ⎠

( 2.23)( 5 ) = ( 3.08 )( 2.23) + ( 3.08) R12

11.15 = 6.868 = 3.08R12 ⇒ R12 = 1.39 kΩ

TYU15.1 1 2π RC 1 1 RC = = = 1.59 × 10−5 2π f3dB 2π (104 ) f 3dB =

Let C = 0.01 μ F ⇒ R = 1.59 kΩ Then C1 = 0.03546 μ F C2 = 0.01392 μ F C3 = 0.002024 μ F 1

T =

6

=

1

⎛ f ⎞ ⎛ 20 ⎞ 1+ ⎜ ⎟ 1+ ⎜ ⎟ ⎝ 10 ⎠ ⎝ f3d ⎠ T = 0.124 or T = −18.1 dB

6

TYU15.2 f 3dB = RC =

1 1 ⇒ RC = 2π RC 2π f3dB 1

2π ( 50 × 103 )

= 3.18 × 10−6

Let C = 0.001 μ F = 1 nF ⇒ R = 3.18 kΩ Then

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R1 = 2.94 kΩ R2 = 3.44 kΩ R3 = 1.22 kΩ R4 = 8.31 kΩ T = 0.01 =

1 ⎛ f ⎞ 1 + ⎜ 3− dB ⎟ ⎝ f ⎠

8

8

2

⎛ f ⎞ ⎛ 1 ⎞ 4 1 + ⎜ 3− dB ⎟ = ⎜ ⎟ = 10 f 0.01 ⎝ ⎠ ⎝ ⎠ 2

⎛ f3− dB ⎞ f 3dB ⇒ f ≅ 15.8 kHz ⎜ ⎟ ≅ 10 ⇒ f = 10 ⎝ f ⎠

TYU15.3 1-pole

2-pole

3-pole

4-pole

T =

T =

T =

T =

1 ⎛ 12 ⎞ 1+ ⎜ ⎟ ⎝ 10 ⎠ 1

2

⎛ 12 ⎞ 1+ ⎜ ⎟ ⎝ 10 ⎠ 1

4

⎛ 12 ⎞ 1+ ⎜ ⎟ ⎝ 10 ⎠

6

1 ⎛ 12 ⎞ 1+ ⎜ ⎟ ⎝ 10 ⎠

8

⇒ −3.87 dB

⇒ −4.88 dB

⇒ −6.0 dB

⇒ −7.24 dB

TYU15.4 1 Req = fC C

or f C C =

1 1 = = 2 × 10−7 Req 5 × 106

If C = 10 pF ⇒ fC = 20 kHz TYU15.5 1 1 f0 = = ⇒ f 0 ≅ 65 kHz 4 2π 6 RC 2π 6 (10 )(100 × 10−12 ) R2 = 29 R = 29 (104 ) ⇒ R2 = 290 kΩ

TYU15.6

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f0 =

1 ⎛ CC ⎞ 2π L ⋅ ⎜ 1 2 ⎟ ⎝ C1 + C2 ⎠

=

1 ⎡ (10−9 ) 2 2π (10 ) ⎢ −9 ⎣ 2 × 10 −6

⎤ ⋅⎥ ⎦

⇒ f 0 = 7.12 MHz

C2 = gm R C1 C2 1 1 ⋅ = ⇒ g m = 0.25 mA / V C1 R 4 × 103 We have ⎛ k ′ ⎞⎛ W ⎞ g m = 2 ⎜ ⎟⎜ ⎟ (VGS − VTh ) ⎝ 2 ⎠⎝ L ⎠ k ′ ≅ 20 μ A / V 2 , VGS − VTh ≅ 1 V gm =

So

W 0.25 × 10−3 = = 12.5 L ( 20 × 10−6 ) (1)

and a value of W / L = 12.5 is certainly reasonable. TYU15.7 ⎛R ⎞ VTH = − ⎜ 1 ⎟ VL ⎝ R2 ⎠ ⎛R ⎞ R 0.10 = − ⎜ 1 ⎟ ( −10 ) ⇒ 1 = 0.010 R2 ⎝ R2 ⎠ Let R1 = 0.10 kΩ then R2 = 10 kΩ

TYU15.8 a. ⎛ R2 ⎞ ⎛ 10 ⎞ VS = ⎜ ⎟ VREF = ⎜ ⎟ ( 2) ⎝ 1 + 10 ⎠ ⎝ R1 + R2 ⎠ VS = 1.82 V ⎛ R1 ⎞ ⎛ 1 ⎞ VTH = VS + ⎜ ⎟ VH = 1.82 + ⎜ ⎟ (10 ) + R R ⎝ 1 + 10 ⎠ ⎝ 1 2 ⎠ VTH = 2.73 V VTL =

⎛ R1 ⎞ ⎛ 1 ⎞ VS + ⎜ ⎟ VL = 1.82 + ⎜ ⎟ ( −10 ) + R R ⎝ 1 + 10 ⎠ ⎝ 1 2 ⎠ VTL = 0.91 V

b.

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TYU15.9 ⎛ R ⎞ VS = ⎜ 1 + 1 ⎟ VREF ⎝ R2 ⎠ ⎛R ⎞ ⎛R ⎞ VTH = VS − ⎜ 1 ⎟ VL and VTL = VS − ⎜ 1 ⎟ VH ⎝ R2 ⎠ ⎝ R2 ⎠ ⎛R ⎞ Hysteresis Width = VTH − VTL = ⎜ 1 ⎟ (VH − VL ) ⎝ R2 ⎠ ⎛R ⎞ ⎛R ⎞ 2.5 = ⎜ 1 ⎟ ( 5 − [ −5]) = 10 ⎜ 1 ⎟ R ⎝ 2⎠ ⎝ R2 ⎠ R So 1 = 0.25 R2

Then ⎛ R ⎞ VS = −1 = ⎜ 1 + 1 ⎟ VREF = (1 + 0.25)VREF ⇒ VREF = −0.8 V R ⎝ 2 ⎠ Then VTH = −1 − ( 0.25 )( −5 ) ⇒ VTH = 0.25 V VTL = −1 − ( 0.25 )( 5 ) ⇒ VTL = −2.25 V

TYU15.10 ⎛ R1 ⎞ 1 ⎛ 10 ⎞ vX = ⎜ ⎟ v0 = ⎜ ⎟ v0 = v0 3 ⎝ 10 + 20 ⎠ ⎝ R1 + R2 ⎠ 10 t = 0, vX = − 3 ⎛ t ⎞ ⎛ 10 ⎞ v X = 10 + ⎜ − − 10 ⎟ exp ⎜ − ⎟ ⎝ 3 ⎠ ⎝ rX ⎠

Output switches when v X =

10 3

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10 = 10 − 13.33 exp 3

⎛ t1 ⎞ ⎜− ⎟ ⎝ rX ⎠

⎛ t ⎞ 10 − 3.33 6.67 = exp ⎜ − 1 ⎟ = 13.33 13.33 ⎝ rX ⎠ ⎛ t ⎞ 13.33 ≅2 exp ⎜ + 1 ⎟ = ⎝ rX ⎠ 6.67 t1 = rX ln (2) = (0.693)rX T = 2(0.693)rX f =

1 2(0.693)rX

rX = RX C X = (104 )( 0.1×10 −6 ) = 1×10 −3 ⇒ f = 722 Hz ⇒ Duty cycle = 50%

TYU15.11 ⎛ R1 ⎞ 20 = 0.333 β =⎜ ⎟= R R + + 40 20 ⎝ 1 2 ⎠

rX = RX C X = (104 )( 0.01× 10−6 ) = 1× 10−4

⎛ 1 + Vγ / VP T = rX ln ⎜ ⎝ 1− β

0.7 ⎤ ⎡ ⎢ 1+ 8 ⎥ ⎞ −4 ⎥ ⇒ T = 48.9 μ s ⎟ = (1× 10 ) ln ⎢ ⎠ ⎢1 − 0.333 ⎥ ⎣⎢ ⎦⎥

Recovery time

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⎛ R1 ⎞ ⎛ 20 ⎞ vY = ⎜ ⎟ v0 = ⎜ ⎟ (8) = 2.667 V R R + ⎝ 20 + 40 ⎠ 2 ⎠ ⎝ 1 ⎛ t ⎞ 0.7 = 8 + ( −2.667 − 8 ) exp ⎜ − 2 ⎟ ⎝ rX ⎠ ⎛ t ⎞ 8 − 0.7 exp ⎜ − 2 ⎟ = = 0.6844 ⎝ rX ⎠ 10.66 ⎛ 1 ⎞ t2 = rX ln ⎜ ⎟ ⇒ t2 = 37.9 μ s ⎝ 0.685 ⎠

TYU15.12 f =

1

( 0.693)( RA + RB ) C

RA + RB =

1 ( 0.693) fC

Let C = 0.01 μ F, RA + RB =

f = 1kHz 1

( 0.693) (103 )( 0.01×10−6 )

Duty cycle = 55 =

= 1.443 × 105

RA + RB × 100% RA + 2 RB

(1.443 ×10 ) (100 ) (1.443 ×10 ) + R (1.443 ×10 ) (100 − 55) ⇒ R = 5

55 =

5

B

5

RB

55

B

= 118 kΩ so RA = 26.2 kΩ

TYU15.13 v01 ⎛ R2 ⎞ ⎛ 30 ⎞ a. = ⎜ 1 + ⎟ = ⎜1 + ⎟ = 2.5 vI ⎝ R1 ⎠ ⎝ 20 ⎠ v02 R 50 =− 4 =− = −2.5 20 vI R3

(b)

P=

1 VL2 1 [12 − (−12)]2 ⋅ = ⋅ = 240 mW 2 RL 2 1.2

Or P = 0.24 W c.

12 = V pi = 4.8 V 2.5

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Chapter 15 Problem Solutions 15.1 (a)

For example:

vo ⎛ R2 ⎞ R = ⎜1 + ⎟ = 10 ⇒ 2 = 9 vi ⎝ R1 ⎠ R1 Corner Frequency: 1 f = = 5 × 103 ⇒ RC = 3.18 × 10−5 2π RC (b) For Example:

Low-Frequency:

1 − R2 jω C vo R 1 = =− 2⋅ vi R1 R1 1 + jω R2 C So, set R2 = 15 ⇒ For example, R1 = 10 k Ω, R2 = 150 k Ω R1 R2 C =

1 1 = = 1.06 × 10−5 2π f3− dB 2π (15 × 103 )

Then C = 70.7 pF 15.2 (a)

Av =

1 ⎛ f ⎞ 1+ ⎜ ⎟ ⎝ f3− dB ⎠

2

=

1 1 + (2) 2

= 0.447 ⇒ Av = −7 dB

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(b)

(c)

Av =

Av =

1 ⎛ f ⎞ 1+ ⎜ ⎟ ⎝ f3− dB ⎠ 1

2

⎛ f ⎞ 1+ ⎜ ⎟ ⎝ f3− dB ⎠

2

=

=

1 1 + (2) 4 1 1 + (2)6

= 0.2425 ⇒ Av = −12.3 dB

= 0.1240 ⇒ Av = −18.1 dB

15.3 (a) Figure 15.6 1 f = 2π RC 1 1 RC = = = 7.958 × 10−6 2π f 2π (20 × 103 ) Let R1 = R2 = R = 10 K C = 795.8 pF So C4 = 0.707C = 562.6 pF C3 = 1.414C = 1.125 nF (b) 1 T = = 0.777 (i) 4 ⎛ 18 ⎞ 1+ ⎜ ⎟ ⎝ 20 ⎠ 1 T = = 0.707 (ii) 4 ⎛ 20 ⎞ 1+ ⎜ ⎟ ⎝ 20 ⎠ 1 T = = 0.637 (iii) 4 ⎛ 22 ⎞ 1+ ⎜ ⎟ ⎝ 20 ⎠ 15.4 Use Figure 15.10(b) 1 f3− dB = 2π RC or 1 RC = = 3.18 × 10−6 2π (50 × 103 ) For example, let C = 100 pF Then R = 31.8 kΩ And R1 = 8.97 kΩ R2 = 22.8 kΩ R3 = 157 kΩ From Equation (15.26) 1 T = 6 ⎛ f ⎞ 1 + ⎜ 3dB ⎟ ⎝ f ⎠

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We find f kHz 30 35 40 45

|T| 0.211 0.324 0.456 0.589

15.5 From Equation (15.7). Y1Y2 T (s) = Y1Y2 + Y4 (Y1 + Y2 + Y3 ) For a high-pass filter, let Y1 = Y2 = sC , Y3 =

1 1 , and Y4 = R3 R4

Then s 2C 2 1 ⎛ 1 ⎞ s 2 C 2 + ⎜ sC + sC + ⎟ R4 ⎝ R3 ⎠ 1 = 1 ⎛ 1 ⎞ 1+ ⎜2+ ⎟ sR4 C ⎝ sR3C ⎠

T (s) =

Define r3 = R3C and r4 = R4 C 1 1 ⎛ 1 ⎞ 1+ ⎜2+ ⎟ sr4 ⎝ sr3 ⎠ Set s = jω T (s) =

1 1 ⎛ 1 ⎞ 1+ ⎜2+ ⎟ jω r4 ⎝ jω r3 ⎠ 1 = j ⎛ j ⎞ 1− ⎜2− ⎟ ω r4 ⎝ ω r3 ⎠

T ( jω ) =

=

1 ⎛ 1 ⎞ 2j ⎜1 − 2 ⎟− ω r3 r4 ⎠ ω r4 ⎝ −1/ 2

2 ⎧⎪⎛ 1 ⎞ 4 ⎫⎪ ( ) 1 = − ω T j ⎨⎜ ⎟ + 2 2⎬ 2 ⎪⎩⎝ ω r3 r4 ⎠ ω r4 ⎪⎭ For a maximally flat filter, we want dT =0 dω ω →∞

Taking the derivative, we find d T ( jω ) dω

2 1 ⎧⎪⎛ 1 ⎞ 4 ⎫⎪ = − ⎨⎜ 1 − 2 ⎟ + 2 2⎬ 2 ⎪⎝ ω r3 r4 ⎠ ω r4 ⎪ ⎩ ⎭

−3 / 2

⎡ ⎛ 1 ⎞ ⎛ 2 ⎞ 4(−2) ⎤ × ⎢ 2 ⎜1 − 2 ⎟⎜ 3 ⎟+ 3 2 ⎥ ⎣⎢ ⎝ ω r3 r4 ⎠ ⎝ ω r3 r4 ⎠ ω r4 ⎦⎥

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or d T ( jω ) dω

=0 ω →∞

⎡⎛ 4 ⎞⎛ 1 ⎞ 8 ⎤ = ⎢⎜ 3 ⎟⎜ 1 − 2 ⎟− 3 2 ⎥ ⎣⎢⎝ ω r3 r4 ⎠⎝ ω r3 r4 ⎠ ω r4 ⎦⎥ =

4 ⎡ 1 ⎢ ω 3 ⎣⎢ r3 r4

Then ⎡ 1 ⎛ 1 ⎞ 2⎤ ⎢ ⎜1 − 2 ⎟− 2 ⎥ ⎢⎣ r3 r4 ⎝ ω r3 r4 ⎠ r4 ⎥⎦

⎛ 1 ⎞ 2⎤ ⎜1 − 2 ⎟− 2 ⎥ ⎝ ω r3 r4 ⎠ r4 ⎦⎥ =0

ω →∞

1 2 = ⇒ 2r3 = r4 r3 r4 Then the transfer function can be written as:

So that

2 ⎧⎪ ⎡ ⎤ 1 4 ⎫⎪ T ( jω ) = ⎨ ⎢1 − 2 2 ⎥ + 2 2 ⎬ ω (2r3 ) ⎦ ω (4r3 ) ⎪ ⎩⎪ ⎣ ⎭

−1/ 2

⎧ 1 1 1 ⎫ = ⎨1 − 2 2 + + 2 2⎬ 2 2 2 ω ω ω 4( ) r r r3 ⎭ 3 3 ⎩

−1/ 2

−1/ 2

⎧ ⎫ 1 = ⎨1 + 2 2 2 ⎬ ⎩ 4(ω r3 ) ⎭ 3 − dB frequency 1 2ω 2 r32 = 1 or ω = = 2r3 Define 1 ω= RC So that R R3 = 2

1 2 R3C

We had 2r3 = r4 or 2( R3C ) = R4 C ⇒ R4 = 2 R3 So that R4 = 2 ⋅ R 15.6 From Equation (15.25) 1 T = − 25 dB ⇒ T = 0.0562 2N ⎛ f ⎞ 1+ ⎜ ⎟ ⎝ f 3− dB ⎠ f f 3− dB So

=

20 =2 10

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0.0562 =

1 1 + (2) 2 N

1 + (2) 2 N = 316.6 ⇒ (2)2 N = 315.6 2 N ⋅ ln (2) = ln (315.6) ⇒ N = 4.15 ⇒ N = 5 A 5-pole filter

15.7 T =

1 ⎛ f ⎞ 1+ ⎜ ⎟ ⎝ f 3dB ⎠

2N

At f = 12 kHz, T = 0.9 1

0.9 =

⎛ f ⎞ 1+ ⎜ ⎟ ⎝ f 3dB ⎠

⎛ 12 ⎞ ⎜ ⎟ ⎝ f3dB ⎠ Also

2N

=

2N

=

⎛ 14 ⎞ ⎜ ⎟ ⎝ f3dB ⎠

2N

⎛ 12 ⎞ ⎜ ⎟ ⎝ f3dB ⎠

2N

⎛ 12 ⎞ 1+ ⎜ ⎟ ⎝ f 3dB ⎠

2N

1 − 1 = 0.2346 (0.9) 2 1

0.01 =

⎛ 14 ⎞ ⎜ ⎟ ⎝ f3dB ⎠

1

=

2N

⎛ 14 ⎞ 1+ ⎜ ⎟ ⎝ f 3dB ⎠

2N

1 − 1 = 9999 (0.01) 2

⎛ 14 ⎞ =⎜ ⎟ ⎝ 12 ⎠

2N

=

9999 = 4.262 × 104 0.2346

(1.16667) 2 N = 4.262 × 104 N = 35 Then

0.9 =

⎛ 12 ⎞ ⎜ ⎟ ⎝ f3dB ⎠

1 ⎛ 12 ⎞ 1+ ⎜ ⎟ ⎝ f 3dB ⎠

2N

2N

= 0.2346

⎛ ⎞ ⎜ ⎟ ⎛ 12 ⎞ ⎝ 2N ⎠ = (0.2346)0.014286 ⎜ ⎟ = (0.2346) ⎝ f3dB ⎠ = 0.9795 So f 3dB = 12.25 kHz 1

15.8

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T =

1 2N

⎛ f ⎞ 1+ ⎜ ⎟ ⎝ f 3dB ⎠ (a) For N = 3 1 T = = 0.2841 1 + (1.5)6

For N = 5 1 T = = 0.1306 1 + (1.5)10

(b)

For N = 7 1 T = = 0.05843 1 + (1.5)14

(c)

15.9 Consider

For low-frequency:

vo R2 + R3 = vi R1 + R2 + R3

For high-frequency:

vo R2 = vi R1 + R2

So we need ⎛ R2 ⎞ R2 + R3 = 25 ⎜ ⎟ R1 + R2 + R3 ⎝ R1 + R2 ⎠

Let R1 + R2 = 50 k Ω and R2 = 1.5 k Ω ⇒ R1 = 48.5 k Ω Then

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1.5 + R3 ⎛ 1.5 ⎞ = 25 ⎜ ⎟ ⇒ R3 = 144 k Ω 50 + R3 ⎝ 50 ⎠ Connect the output of this circuit to a non-inverting op-amp circuit.

At low-frequency: vo1 =

R2 + R3 1.5 + 144 ⋅ vi = ⋅ vi = 0.75vi R1 + R2 + R3 48.5 + 1.5 + 144

Need to have vo = 25. ⎛ R ⎞ ⎛ R ⎞ R vo = 25 = ⎜1 + 5 ⎟ ⋅ vo1 = ⎜ 1 + 5 ⎟ (0.75)vi ⇒ 5 = 32.3 R4 ⎝ R4 ⎠ ⎝ R4 ⎠ To check at high-frequency. R2 1.5 vo1 = vi = vi = 0.03vi R1 + R2 1.5 + 48.5 vo = (1 + 32.3)vo1 = (33.3)(0.03)vi = (1.0)vi which meets the design specification Consider the frequency response. 1 R2 + R3 vo1 sC = 1 vi R1 + R2 + R3 sC Now R3 1 = R3 sC 1 + sR3C Then, we find vo1 R3 + R2 (1 + sR3C ) = vi R3 + ( R1 + R2 )(1 + sR3C ) which can be rearranged as ( R2 + R3 ) (1 + s ( R2 || R3 )C ) vo1 = vi ( R1 + R2 + R3 ) 1 + s ( R3 || ( R1 + R2 ) ) C

(

)

So fL ≅

1 1 1 = = 3 2π ( R2 || R3 ) C 2π (1.5 || 144 ) × 10 C ( 9.33 × 103 ) C

fH ≅

1 1 = 2π ( R3 || ( R1 + R2 ) ) C 2π (144 || 50 ) × 103 C

=

1

( 2.33 ×10 ) C 5

Set

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⎤ fL + fH 1 ⎡ 1 1 ⎥ = ⎢ + 3 5 2 2 ⎢ ( 9.33 × 10 ) C ( 2.33 × 10 ) C ⎥ ⎣ ⎦ Which yields C = 2.23 nF 25 kHz =

15.10

Av =

1

⎛ f ⎞ 1+ ⎜ ⎟ ⎝ f3− dB ⎠

− 100 dB ⇒ 10−5

2N

So 10−5 =

1

⎛ 770 ⎞ 1+ ⎜ ⎟ ⎝ 12 ⎠

2N

or 2

⎛ 1 ⎞ 1 + (64.2) 2 N = ⎜ −5 ⎟ = 1010 ⎝ 10 ⎠ or (64.2) 2 N ≅ 1010 Now

N 1 2 3

Left Side 4.112 × 103 1.7 × 107 7 × 1010

So, we need a 3rd order filter. 15.11

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Low-pass: −50 dB ⇒ 3.16 × 10−3 Then 1 1 3.16 × 10−3 = = 4 4 ⎛ f ⎞ ⎛ 60 ⎞ 1+ ⎜ ⎟ 1+ ⎜ ⎟ ⎝ fL ⎠ ⎝ fL ⎠ We find f L = 3.37 Hz High Pass: 1 1 3.16 × 10−3 = = 4 4 ⎛ fH ⎞ ⎛ fH ⎞ 1 + 1+ ⎜ ⎜ 60 ⎟ ⎟ ⎝ ⎠ ⎝ f ⎠ We find f H = 1067 Hz Bandwidth: BW = f H − f L = 1067 − 3.37 ⇒ BW ≅ 1064 Hz 15.12 a. v vI = − 02 − R4 R3

v0 ⎛ 1 ⎞ R1 ⎜ ⎟ ⎝ sC ⎠

(1)

v0 v = − 01 (2) R2 ⎛ 1 ⎞ ⎜ ⎟ ⎝ sC ⎠ v01 v (3) = − 02 ⇒ v01 = −v02 R5 R5 Then ⎛ 1 ⎞ v0 v = + 02 or v02 = v0 ⎜ ⎟ (2) R2 ⎛ 1 ⎞ ⎝ sR2 C ⎠ ⎜ ⎟ ⎝ sC ⎠ And

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⎛ 1 ⎞ v0 ⋅⎜ ⎟− sR C ⎝ 2 ⎠ R ⎛ 1 ⎞ 1 ⎜ ⎟ ⎝ sC ⎠

v vI =− 0 R4 R3

⎡ ⎤ ⎢ ⎥ 1 1 ⎥ = −v0 ⎢ + ⎢ R3 ( sR2 C ) R1 ⋅ (1/ sC ) ⎥ ⎢⎣ R1 + (1/ sC ) ⎥⎦

(1)

⎡ 1 + sR1C ⎤ 1 = −v0 ⎢ + ⎥ R1 ⎦ ⎣ R3 ( sR2 C ) ⎡ R + (1 + sR1C )( sR2 R3C ) ⎤ = −v0 ⎢ 1 ⎥ ( sC ) R1 R2 R3 ⎣ ⎦ Then ⎤ v0 ( sC )( R1 R2 R3 ) 1 ⎡ =− ⎢ ⎥ vI R4 ⎣ R1 + sR2 R3C + s 2 R1 R2 R3C 2 ⎦

or −

Av ( s ) =

1 R4

v0 = 1 1 vI + sC + R1 sCR2 R3 − Av ( jω ) =

b.

1 R4

1 1 + jω C + R1 jω CR2 R3

or − Av ( jω ) =

1 R4

⎡ ⎤ 1 1 + j ⎢ω C − ⎥ ω R1 CR R 2 3 ⎦ ⎣ R 1 =− 1⋅ R4 ⎪⎧ ⎡ R1 ⎤ ⎪⎫ ⎨1 + j ⎢ω R1C − ⎬ ω CR2 R3 ⎥⎦ ⎪⎭ ⎪⎩ ⎣ R 1 Av ( jω ) = 1 ⋅ 2 −1/ 2 R4 ⎧ R1 ⎤ ⎫⎪ ⎪ ⎡ ⎨1 + ⎢ω R1C − ⎬ ω CR2 R3 ⎥⎦ ⎭⎪ ⎩⎪ ⎣ Av

⎡ R1 ⎤ when ⎢ω R1C − ⎥=0 ω CR2 R3 ⎦ ⎣

max

Then Av

max

=

R1 85 = ⇒ Av 3 R4

max

= 28.3

Now ⎡

ω R1C ⎢1 − ⎣

⎤ 1 1 ⎥ = 0 or ω = 2 ω C R2 R3 ⎦ C R2 R3 2

Then

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f =

1 2π C R2 R3

=

1 2π (0.1× 10−6 ) (300) 2

So f = 5.305 kHz To find the two 3 − dB frequencies, ⎡ R1 ⎤ ⎢ω R1C − ⎥ = ±1 ω CR2 R3 ⎦ ⎣

ω 2 R1 R2 R3 C 2 − R1 = ±ω R2 R3 C ω 2 (85 × 103 )(300) 2 (0.1× 10−6 ) 2 − 85 × 103 = ±ω (300) 2 (0.1× 10−6 ) ω 2 (7.65× 10−5 ) − 85 × 103 = ±ω (9 × 10−3 ) ω 2 (7.65× 10−5 ) ± ω (9 × 10−3 ) − 85 × 103 = 0 (9 × 10−3 ) 2 + 4(7.65 × 10−5 )(85 × 10−3 ) ± (9 × 10−3 ) ± 2(7.65 × 10−5 ) 2(7.65 × 10−5 ) We find f = 5.315 kHz and f = 5.296 kHz

ω=

15.13 a. vI − v A v = A (1) R ⎛ 1 ⎞ ⎜ ⎟ ⎝ sC ⎠ vI − vB vB − v0 = (2) R R and v A = vB So vI ⎛1 ⎞ ⎛ 1 + sRC ⎞ = v A ⎜ + sC ⎟ = v A ⎜ ⎟ (1) R R ⎝ ⎠ ⎝ R ⎠ or vI vA = 1 + sRC Then 2vI vI + v0 = 2vB = 2vA = (2) 1 + sRC ⎡ 2 ⎤ ⎡1 − sRC ⎤ v0 = vI ⎢ − 1⎥ = vI ⎢ ⎥ 1 sRC + ⎣ ⎦ ⎣1 + sRC ⎦ Now v0 1 − jω RC = A( jω ) = 1 + jω RC vI A=

1 + ω 2 R2C 2 1 + ω 2 R2C 2

⇒ A =1

Phase: φ = −2 tan −1 (ω RC ) b. f 0

RC = (104 )(15.9 × 10−9 ) = 1.59 × 10−4

φ 0

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−11.4 −53.1 −90° −157 −169

10 2 5 × 103 1/ 2π RC = 103 Hz 5 × 103 10 4

15.14 a. Vi Vi − V0 + =0 R1 R2 || (1/ sC ) Vi Vi − V0 + =0 R1 ⎡ R2 ⎤ ⎢1 + sR C ⎥ 2 ⎣ ⎦ R2 1 (Vi ) + Vi = V0 ⋅ R1 1 + sR2 C V0 R2 + R1 (1 + sR2 C ) ( R2 + R1 ) [1 + s ( R1 || R2 )C ] = = Vi R1 (1 + sR2 C ) R1 (1 + sR2 C ) ⇒

V0 ⎛ R2 ⎞ ⎡1 + s ( R1 || R2 )C ⎤ = ⎜1 + ⎟ ⎢ ⎥ Vi ⎝ R1 ⎠ ⎣ (1 + sR2 C ) ⎦

⇒ f 3dB1 =

1 2π R2 C

⇒ f 3dB2 =

1 2π ( R1 || R2 )C

b. Vi V − V0 + i =0 R1 || (1/ sC ) R2 Vi V V + i = 0 ⎛ R1 ⎞ R2 R2 ⎜ ⎟ ⎝ 1 + sR1C ⎠ ⎡R ⎤ Vi ⎢ 2 ⋅ (1 + sR1C ) + 1⎥ = V0 ⎣ R1 ⎦ Vi ⋅ [ R2 + R1 + sR1 R2 C ] = V0 R1 V0 R2 + R1 V ⎛ R ⎞ 1 = ⋅ [1 + s ( R1 || R2 )C ] ⇒ 0 = ⎜1 + 2 ⎟ [1 + s ( R1 || R2 )C ] ⇒ f3dB = 2π ( R1 || R2 )C Vi R1 Vi ⎝ R1 ⎠

15.15 a.

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−V0 Vi = R1 + (1/ sC1 ) R2 || (1/ sC2 ) ⎛ sC1 ⎞ ⎛ 1 + sR2 C2 ⎞ Vi ⎜ ⎟ = −V0 ⎜ ⎟ ⎝ 1 + sR1C1 ⎠ ⎝ sC2 ⎠ V0 − sR2 C1 − sR2 C1 = = Vi (1 + sR1C1 )(1 + sR2 C2 ) 1 + sR1C1 + sR2 C2 + s 2 R1 R2 C1C2 ⎡ ⎤ ⎢ ⎥ V0 R sC1 ⎥ = − 2 ×⎢ ⎥ Vi R1 ⎢ 1 ⎛ R2 C2 ⎞ 2 ⎢ + sC1 ⎜ 1 + ⋅ ⎟ + s R2 C1C2 ⎥ R1 C1 ⎠ ⎝ ⎣⎢ R1 ⎦⎥ or ⎡ ⎤ ⎢ ⎥ V0 R2 ⎢ 1 ⎥ =− ⋅ T (s) = ⎥ Vi R1 ⎢ 1 ⎛ R2 C2 ⎞ + ⎜ 1 + ⋅ ⎟ + sR2 C2 ⎥ ⎢ R1 C1 ⎠ ⎣⎢ sR1C1 ⎝ ⎦⎥

b. T ( jω ) = −

R2 1 × 2 2 1/ 2 R1 ⎧ 1 ⎞ ⎫⎪ ⎪⎛ R2 C2 ⎞ ⎛ ⎨⎜ 1 + . ⎟ + ⎜ ω R2 C2 − ⎟ ⎬ R1 C1 ⎠ ⎝ ω R1C1 ⎠ ⎪ ⎩⎪⎝ ⎭

⎛ 1 ⎞ when ⎜ ω R2 C2 − ⎟ = 0, we want ω R1C1 ⎠ ⎝ R 1 T ( jω ) = 50 = 2 ⋅ R1 ⎛ R2 C2 ⎞ ⎜1 + ⋅ ⎟ R1 C1 ⎠ ⎝

At the 3 − dB frequencies, we want ⎛ ⎛ R2 C2 ⎞ 1 ⎞ ⎜ ω R2 C2 − ⎟ = ± ⎜1 + ⋅ ⎟ ω R1C1 ⎠ R1 C1 ⎠ ⎝ ⎝ For f = 5 kHz, use + sign and for f = 200 Hz, use − sign.

ω1 = 2π (200) = 1257 ω 2 = 2π (5 × 103 ) = 3.142 × 104 Define r2 = R2 C2 and r1 = R1C1 Then 50 =

R2 ⋅ R1

1 r 1+ 2 r1

⎛ ⎛ r2 ⎞ 1 ⎞ ⎜ ω 2 r2 − ⎟ = + ⎜1 + ⎟ ω 2 r1 ⎠ ⎝ ⎝ r1 ⎠ ⎛ ⎛ r2 ⎞ 1 ⎞ ⎜ ω1 r2 − ⎟ = − ⎜1 + ⎟ r ω ⎝ ⎝ r1 ⎠ 1 1 ⎠ From (2) ω 22 r1r2 − 1 r1 + r2 = ω 2 r1 r1

(1)

(2) (3)

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or

ω 2 r1 r2 −

1

ω2

= r1 + r2

r1 (ω 2 r2 − 1) =

1

ω2

+ r2

So 1

+r

2 ω r1 = 2 ω 2 r2 − 1

Substituting into (3), we find ⎡ ⎤ ⎢ r (ω r − 1) ⎥ 1 ⎥ ω1r2 − = − ⎢1 + 2 2 2 1 ⎡ 1 ⎤ ⎢ ⎥ + r + r 2 2 ⎥ ⎢ω ⎢⎣ ⎥⎦ ω 2 ⎥ ω1 ⎢ 2 ⎢ ω 2 r2 − 1 ⎥ ⎢⎣ ⎥⎦ ⎡⎛ 1 ⎤ ⎡1 ⎤ 1 ⎞ ω1r2 ⎢ + r2 ⎥ − (ω 2 r2 − 1) = − ⎢⎜ + r2 ⎟ + r2 (ω 2 r2 − 1) ⎥ ⎣ω2 ⎦ ω1 ⎠ ⎣⎝ ω 2 ⎦ ω1 ω 1 1 ⋅ r2 + ω1r22 − 2 ⋅ r2 + =− − r2 − ω 2 r22 + r2

ω2

ω1

ω1

ω2

⎛ 1 ⎛ω ω ⎞ 1 ⎞ (ω1 + ω 2 )r22 + ⎜ 1 − 2 ⎟ r2 + ⎜ + ⎟=0 ⎝ ω1 ω 2 ⎠ ⎝ ω 2 ω1 ⎠ (3.2677 × 104 )r22 − 24.96r2 + 8.273 × 10−4 = 0 (24.96) 2 − 4(3.2677 ×10 4 )(8.273 × 10−4 ) 24.96 ± 2(3.2677 ×10 4 ) 2(3.2677 × 104 ) Since ω 2 is large, r2 should be small so use minus sign: r2 =

r2 = 3.47 × 10−5 Then 3.18 × 10−5 + 3.47 × 10−5 r1 = ⇒ r1 = 7.32 × 10−4 9.09 × 10−2 Now R 1 50 = 2 ⋅ R1 3.47 × 10−5 1+ 7.32 × 10−4 Then R2 = 52.37 or R2 = 524 kΩ R1 Also r1 = R1C1 so that C1 = 0.0732 μ F r2 = R2 C2 so that C2 = 66.3 pF

15.16 Gain = 10 dB ⇒ Gain = 3.162 For example, we may have

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Want

R4 = 2.162 R3

For example, let

R3 = 50 kΩ, R4 = 108 kΩ

f1 =

1 = 200 2π R1C1

So R1C1 =

1 = 0.796 × 10−3 2π (200)

For example, let R1 = 200 kΩ ⇒ A large input resistance C1 = 0.00398 μ F 1 1 f2 = = 50 × 103 ⇒ R2 C2 = = 3.18 × 10−6 2π R2 C2 2π (50 × 103 ) For example, let R2 = 10 kΩ and C2 = 318 pF

15.17 f C = 100 kHz 1 Req = fC C a.

For C = 1 pF, Req = 10 MΩ

b.

For C = 10 pF, Req = 1 MΩ

c.

For C = 30 pF, Req = 333 kΩ

15.18 a. From Equation (15.28), V1 − V2 Q= ⋅ TC Req and f C = 100 kHz so that TC =

1 ⇒ 10 μ s 100 × 103

Now Req =

1 1 = ⇒ 1 MΩ 3 f C C (100 × 10 )(10 × 10−12 )

So

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Q=

(2 − 1)(10 × 10−6 ) = 10 × 10−12 C 106

or Q = 10 pC Q 10 × 10−12 or I eq = 1 μ A = TC 10 × 10−6

b.

I eq =

c.

Q = CV so find the time that V0 reaches 99% of its full value.

V0 = V1 (1 − e − t / r ) where r = RC Then 0.99 = 1 − e − t / r or e − t / r = 0.01 or t = r ln (100) r = RC = (103 )(10 × 10−12 ) = 10−8 s Then t = 4.61×10−8 s

15.19 Low frequency gain = −10 ⇒ f 3dB = 10 × 103 Hz =

C1 = 10 C2

fC C2 2π CF

Set f C = 10 f3dB = 100 kHz Then C2 2π (10 × 103 ) = = 0.628 CF 100 × 103 The largest capacitor is C1 , so let C1 = 30 pF Then C2 = 3 pF and CF = 4.78 pF 15.20 a. Req =

Time constant = Req ⋅ CF = r where 1 1 = = 2 × 106 Ω −12 3 f C C1 (100 × 10 )(5 × 10 )

Then r = (2 × 106 )(30 × 10−12 ) or r = 60 μ s b.

v0 = −

1 vI ⋅ dt r∫

or Δv0 =

(1)TC 1 ,TC = r fC

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So Δv0 =

1 (60 × 10 )(100 × 103 ) −6

or Δv0 = 0.167 V c. Now Δv0 = 13 = N (0.167) or N = 78 clock pulses 15.21 1 1 = 2π 3RC 2π 3(104 )(0.10 × 10−12 ) f O = 91.9 MHz fO =

R2 = 8R = 80 kΩ

15.22 a. ⎛ sRCV ⎞ R ⋅ v0 = ⎜ ⎟ ⋅ v0 R + (1/ sCV ) ⎝ 1 + sRCV ⎠ R ⎛ sRC ⎞ ⋅v = ⋅v v2 = 1 1 ⎜⎝ 1 + sRC ⎟⎠ 1 R+ sC R ⎛ sRC ⎞ v3 = ⋅v = ⋅v 1 2 ⎜⎝ 1 + sRC ⎟⎠ 2 R+ sC R2 v0 = − ⋅ v3 R Then 2 R ⎛ sRC ⎞ ⎛ sRCV ⎞ v0 = − 2 ⎜ ⎟ v0 ⎟ ⎜ R ⎝ 1 + sRC ⎠ ⎝ 1 + sRCV ⎠ v1 =

Set s = jω ⎛ ⎞ ⎛ jω RCV ⎞ −ω 2 R 2 C 2 ⎟ ⎜ 2 2 2 ⎟⎜ 1 2 j ω RC ω R C + − ⎝ ⎠ ⎝ 1 + jω RCV ⎠ The real part of the denominator must be zero. 1 − ω 2 R 2 C 2 − 2ω 2 R 2 CCV = 0 so 1 ω0 = R C (C + 2CV ) 1= −

R2 R

b. f 0,max =

1 2π (10 ) (10 4

−11

)(10−11 + 2[10−11 ])

f 0,max = 919 kHz f 0,min =

1 2π (10 ) (10 4

−11

)(10−11 + 2[50 × 10−12 ])

f 0,min = 480 kHz

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15.23 1

fO = RC =

2π 6 RC 1 2π 6 fO

RC = 2.32 × 10

=

1 2π 6(28 × 103 )

−6

2.32 × 10−6 ⇒ R = 18.56 kΩ 125 × 10−12 R2 = 29 R ⇒ R2 = 538.4 kΩ R=

15.24 v0 − v1 v1 v1 − v2 (1) = + 1 1 R sC sC v or (v0 − v1 ) sC = 1 + (v1 − v2 ) sC R v1 − v2 v2 v2 (2) = + 1 1 R +R sC sC v v ( sC ) or (v1 − v2 ) sC = 2 + 2 R 1 + sRC v0 v2 (3) =− 1 R 2 +R sC v v sC or 2 =− 0 1 + sRC R2 so −v0 v2 = (1 + sRC ) sR2 C From (2) 1 sC ⎤ ⎡ v1 ( sC ) = v2 ⎢ sC + + R 1 + sRC ⎥⎦ ⎣ or v (1 + sRC ) ⎡ 1 1 ⎤ ⋅ ⎢1 + + v1 = − 0 ⎥ sR2 C ⎣ sRC 1 + sRC ⎦ From (1) 1 ⎡ ⎤ v0 ( sC ) = v1 ⎢ sC + + sC ⎥ − v2 ( sC ) R ⎣ ⎦ Then v 1 ⎤ ⎡ −v0 (1 + sRC ) ⎤ ⎡1 + sRC 1 ⎤ ⎡ v0 = ⎢ 2 + + + 0 ⋅ (1 + sRC ) ⎢ ⎥×⎢ ⎥ ⎥ sRC ⎦ ⎣ sR2 C 1 + sRC ⎦ sR2 C ⎣ ⎦ ⎣ sRC 2 ⎡1 + 2sRC ⎤ ⎡1 + sRC ⎤ ⎡ (1 + sRC ) + sRC ⎤ 1 + sRC −1 = ⎢ ⎥⎢ ⎥− ⎥⎢ ⎣ sRC ⎦ ⎣ sR2 C ⎦ ⎣ ( sRC )(1 + sRC ) ⎦ sR2 C

−1 =

(1 + 2 sRC )(1 + 2sRC + s 2 R 2 C 2 + sRC ) (1 + sRC )( sRC ) 2 − ( sRC ) 2 ( sR2 C ) ( sRC ) 2 ( sR2 C )

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Set s =

(1 + 2 jω RC ) (1 + 3 jω RC + ω 2 R 2 C 2 ) (1 + jω RC ) ( −ω 2 R 2C 2 ) jω − 1 = − ( −ω 2 R 2C 2 ) ( jω R2C ) ( −ω 2 R 2C 2 ) ( jω R2C )

The real part of the numerator must be zero. 1 − ω 2 R 2 C 2 − 6ω 2 R 2 C 2 + ω 2 R 2 C 2 = 0 6ω 2 R 2 C 2 = 1 so that 1 ω0 = 6 RC Condition for oscillation: 2 jω RC + 3 jω RC − 2 jω 3 R 3 C 3 + jω 3 R 3 C 3 −1 = (−ω 3 R 2 C 2 )( jω R2 C ) 1=

5 − ω 2 R2 C 2 (ω RC )(ω R2 C )

But

ω = ω0 =

1 6RC

Then 1⎞ ⎛ 2 2 1 ⎜ 5 − ⎟ (6 R C ) 6 ⎠ 6 1= =⎝ ( RC )( R2 C ) RR2 C 2 6 R 2C 2 ⎛ 29 ⎞ ⎜ ⎟ (6 R ) R 6 1= ⎝ ⎠ or 2 = 29 R2 R 5−

15.25 Let RF 1 = RF 2 = RF 3 ≡ RF ⎞⎛ 1 ⎞ ⎟ ⎜ 1 + 5R ⎟ vo ⎠⎝ C ⎠ ⎛ R ⎞⎛ 1 ⎞ Vo 2 = ⎜ 1 + F ⎟ ⎜ ⎟ vo1 R ⎠ ⎝ 1 + 5 RC ⎠ ⎝ vo3 + vo 2 v v + o3 + o3 = 0 R 1/ sC R 2 ⎛ ⎞ v vo3 ⎜ + sC ⎟ = o 2 ⎝R ⎠ R ⎛ R Vo1 = ⎜1 + F R ⎝

⎛ 1 ⎞ vo3 = ⎜ ⎟ vo 2 ⎝ 2 + 5RC ⎠ R vo = − F ⋅ vo 3 R R ⎛ 1 ⎞ ⎛ RF ⎞ ⎛ 1 ⎞ ⎛ RF ⎞ ⎛ 1 ⎞ vo = − F ⋅ ⎜ ⎜ ⎟ vo ⎟ 1+ ⎜ ⎟ 1+ R ⎝ 2 + 5 RC ⎠ ⎝⎜ R ⎠⎟ ⎝ 1 + 5 RC ⎠ ⎝⎜ R ⎟⎠ ⎝ 1 + 5 RC ⎠ 1= −

RF R

⎛ RF ⎞ ⎛ 1 ⎞ ⎛ 1 ⎞ ⎛ 1 ⎞ ⎜1 + R ⎟ ⎜ 2 + 5R ⎟ ⎜ 1 + 5R ⎟ ⎜ 1 + 5R ⎟ ⎝ ⎠ ⎝ C ⎠⎝ C ⎠⎝ C ⎠ 2

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Let S = jω 1= −

RF R

⎞⎛ ⎞⎛ ⎞ 1 1 1 ⎛ RF ⎞ ⎛ ⎜ 1 + R ⎟ ⎜ 2 + jω R ⎟ ⎜ 1 + jω R ⎟ ⎜ 1 + jω R ⎟ ⎝ ⎠ ⎝ C ⎠⎝ C ⎠⎝ C ⎠

=−

RF R

⎤ ⎞⎡ 1 1 ⎛ RF ⎞ ⎛ ⎜ 1 + R ⎟ ⎜ 2 + jω R ⎟ ⎢ 2 2 2 ⎥ ⎝ ⎠ ⎝ C ⎠ ⎣1 + 2 jω RC − ω R C ⎦

=−

RF R

⎤ 1 ⎛ RF ⎞ ⎡ ⎜1 + R ⎟ ⎢ 2 2 2 2 2 2 3 3 3 ⎥ ⎝ ⎠ ⎣ 2 + 4 jω RC − 2ω R C + jω RC − 2ω R C − jω R C ⎦

2

2

2

⎤ RF ⎛ RF ⎞ ⎡ 1 1+ ⎢ ⎜ ⎟ 2 2 2 3 3 3⎥ R ⎝ R ⎠ ⎣ 2 − 4ω R C + 5 jω RC − jω R C ⎦ Imaginary Term must be zero 5 jω 0 RC − jω 03 R 3 C 3 = 0 2

1= −

5 − jω 02 R 2 C 2 = 0

ω0 =

5 RC

Then R 1= − F R

1= −

RF R

⎡ ⎤ 2 ⎥ 1 ⎛ RF ⎞ ⎢ ⎢ ⎥ + 1 ⎜ R ⎠⎟ ⎢ 4R2 C 2 − 5 ⎥ ⎝ ⎢⎣ 2 − R 2 C 2 ⎥⎦ 2

⎛ RF ⎞ ⎡ 1 ⎤ 1 RF ⎜ 1 + R ⎟ ⎢ 2 − 20 ⎥ = 18 ⋅ R ⎦ ⎝ ⎠ ⎣

⎛ RF ⎞ ⎜1 + R ⎟ ⎝ ⎠

2

2

R ⎛ R ⎞ R 18 = F ⎜ 1 + F ⎟ ⇒ F = 2 R ⎝ R ⎠ R

15.26 a. v0 − v01 v v −v = 01 + 01 02 R R ⎛ 1 ⎞ ⎜ ⎟ sC ⎝ ⎠ v01 − v02 v02 v −v = + 02 03 R R ⎛ 1 ⎞ ⎜ ⎟ ⎝ sC ⎠ v02 − v03 v v = 03 + 03 R ⎛ 1 ⎞ R ⎜ ⎟ ⎝ sC ⎠ R v0 = − F ⋅ v03 R We can write the equations as v0 − v01 = v01 ( sRC ) + v01 − v02 v01 − v02 = v02 ( sRC ) + v02 − v03 v02 − v03 = v03 ( sRC ) + v03 and R v0 = − F ⋅ v03 R

(1)

(2)

(3)

(4) (1) (2) (3)

(4)

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Combining terms, we find v0 = v01 (2 + sRC ) − v02 (1) v01 = v02 (2 + sRC ) − v03 (2) v02 = v03 (2 + sRC ) (3) and R v0 = − F ⋅ v03 (4) R Combining Equations (3) and (2) v01 = v03 (2 + sRC ) 2 − v03 = v03 ⎡⎣ (2 + sRC ) 2 − 1⎤⎦

(2)

Then Equation (1) is v0 = v03 ⎡⎣(2 + sRC ) 2 − 1⎤⎦ (2 + sRC ) − v03 (2 + sRC ) Using Equation (4), we find R − F ⋅ v03 = V03 ⎡⎣(2 + sRC ) 2 − 1⎤⎦ (2 + sRC ) − (2 + sRC ) R To find the frequency of oscillation, set s = jω and set the imaginary part of the right side of the equation to zero. We will have R − F = (2 + jω RC )[4 + 4 jω RC − ω 2 R 2 C 2 − 1 − 1] R Then jω RC (2 − ω 2 R 2 C 2 ) + 8 jω RC = 0 or jω RC ⎡⎣ 2 − ω 2 R 2 C 2 + 8⎤⎦ = 0

{

}

Then the frequency of oscillation is 1 10 f0 = ⋅ 2π RC The condition to sustain oscillations is determined from R − F = 2[2 − ω 2 R 2 C 2 ] − 4ω 2 R 2 C 2 R or R − F = 4 − 6ω 2 R 2 C 2 R 10 Setting ω 2 = 2 2 , we have RC RF − = 4 − 6(10) R or RF = 56 R b. For R = 5 kΩ and f 0 = 5 kHz, we find 10 ⇒ C = 0.02 μ F 2π (5 × 103 )(5 × 103 ) and RF = 56(5) ⇒ RF = 280 kΩ

C=

15.27 a.

We can write

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⎛ R1 vA = ⎜ ⎝ R1 + R2

⎛ Zp ⎞ ⎟ v0 and vB = ⎜⎜ ⎠ ⎝ Z p + Zs

where Z p = RB and Z s = RA +

⎞ ⎟⎟ v0 ⎠

RB 1 = sCB 1 + sRB CB 1 + sRA C A 1 = sC A sC A

Setting v A = vB , we have RB R1 1 + sRB CB = RB 1 + sRAC A R1 + R2 + sC A 1 + sRB CB R1 sRB C A = (1) R1 + R2 sRB C A + (1 + sRAC A )(1 + sRB CB ) To find the frequency of oscillation, set s = jω and set the real part of the denominator on the right side of Equation (1) equal to zero. The denominator term is jω RB C A + (1 + jω RAC A )(1 + jω RB CB ) or jω RB C A + 1 + jω RAC A + jω RB CB − ω 2 RA RB C ACB (2) Then from (2), we must have 1 − ω 02 RA RB C ACB = 0 or f0 =

1 2π RA RB C ACB

b.

To find the condition for sustained oscillation, combine Equations (1) and (2). Then R1 jω RB C A = R1 + R2 jω RB C A + jω RAC A + jω RB CB )

or R2 R C = 1+ A + B R1 RB C A Then R2 RA CB = + R1 RB C A 1+

15.28 a. We can write ⎛ R1 ⎞ vA = ⎜ ⎟ v0 ⎝ R1 + R2 ⎠ and ⎛ ⎞ R || sL vB = ⎜ ⎟ v0 ⎝ R || sL + R + sL ⎠ Setting v A = vB , we have

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sRL ⎡ ⎤ ⎢ ⎥ R1 + R sL =⎢ ⎥ ⋅ v0 sRL R1 + R2 ⎢ + R + sL ⎥ ⎣⎢ R + sL ⎦⎥ R1 sRL = (1) R1 + R2 sRL + ( R + sL) 2 To find the frequency of oscillation, set s = jω and se the real part of the denominator on the right side of Equation (1) equal to zero. The denominator term is: jω RL + ( R + jω L) 2 or jω RL + R 2 + 2 jω RL − ω 2 L2 (2) Then R 2 − ω 02 L2 = 0 or R 1 f0 = ⋅ L 2π b. To find the condition for sustained oscillations, combine Equations (1) and (2). R1 jω RL 1 = = R1 + R2 jω RL + 2 jω RL 3 Then R 1+ 2 = 3 R1 so that R2 =2 R1

15.29 1 2π RC 1 1 RC = = 2π fO 2π (28 × 103 ) fO =

RC = 5.684 × 10−6 For example, Let R = 20 K Then C = 284.2 pF

Also

R2 =2 R1

15.30 From Equation (15.59) 1 f0 = ⎛ CC ⎞ 2π L ⎜ 1 2 ⎟ ⎝ C1 + C2 ⎠ and from Equation (15.61) C2 = gm R C1

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Now, g m = 2 kn I DQ = 2 (0.5)(1) = 1.414 mA / V We have C1 = 0.01 μ F, R = 4 kΩ, f 0 = 400 kHz So C2 = g m RC1 = (1.414)(4)(0.01) or C2 = 0.0566 μ F and 400 × 103 =

1

⎡ (0.01)(0.0566) ⎤ 2π L ⎢ × 10−6 ⎥ 0.01 0.0566 + ⎣ ⎦ 2

⎡ ⎤ 1 L(8.5 × 10−9 ) = ⎢ = 1.58 × 10−13 3 ⎥ ⎣ 2π (400 × 10 ) ⎦ Then L = 18.6 μ H

15.31 Vπ = −V0 V0

⎛ 1 ⎞ ⎜ ⎟ ⎝ sC2 ⎠

+

V0 V0 − V1 + = g mVπ = − g mV0 RL ⎛ 1 ⎞ ⎜ ⎟ ⎝ sC1 ⎠

⎡ ⎤ 1 V0 ⎢ sC2 + sC1 + + g m ⎥ = V1 ( sC1 ) RL ⎣ ⎦ V1 V0 − V1 + + g mVπ = 0 sL ⎛ 1 ⎞ ⎜ ⎟ ⎝ sC1 ⎠

(1) (2)

⎛ 1 ⎞ V1 ⎜ + sC1 ⎟ = V0 ( sC1 + g m ) ⎝ sL ⎠ V0 ( sC1 + g m ) V1 = ⎛ 1 ⎞ ⎜ + sC1 ⎟ sL ⎝ ⎠ Then ⎡ ⎤ V ( sC1 )( sC1 + g m ) 1 V0 ⎢ s (C1 + C2 ) + + gm ⎥ = 0 RL ⎛ 1 ⎞ ⎣ ⎦ ⎜ + sC1 ⎟ ⎝ sL ⎠ ⎡ ⎤⎛ 1 1 ⎞ + g m ⎥ ⎜ + sC1 ⎟ = sC1 ( sC1 + g m ) ⎢ s (C1 + C2 ) + RL ⎠ ⎣ ⎦ ⎝ sL g C1 + C2 sC 1 + s 2 C1 (C1 + C2 ) + + 1 + sg m C1 + m = s 2 C12 + sg m C1 L sRL L RL sL C1 + C2 sC g 1 + s 2 C1C2 + + 1 + m =0 L sRL L RL sL Set s = jω g C1 + C2 jω C1 1 − ω 2 C1C2 + + + m =0 L jω RL L RL jω L

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Then

ω2 =

C1 + C2 C1 + C2 ⇒ ω0 = C1C2 L C1C2 L

and gm ω C1 1 + = ω L ω RL L RL Then gm (C + C2 )C1 1 + = 1 L RL L C1C2 LRL

gm +

1 C1 + C2 = RL C2 RL

g m RL + 1 =

C1 C + 1 or 1 = g m RL C2 C2

15.32 a. V0 V0 V0 + + g mVπ + =0 (1) 1 sL1 R + sL2 sC ⎛ ⎞ ⎜ sL2 ⎟ Vπ = ⎜ (2) ⎟ V0 ⎜ 1 + sL ⎟ ⎜ 2 ⎟ ⎝ sC ⎠ Then g m ( s 2 L2 C ) ⎪⎫ 1 sC ⎪⎧ 1 V0 ⎨ + + + ⎬=0 2 2 ⎪⎩ sL1 R 1 + s L2 C 1 + s L2 C ⎪⎭ ⎧⎪ R (1 + s 2 L2 C ) + ( sL1 )(1 + s 2 L2 C ) s 2 RL1C + g m ( sRL1 )( s 2 L2 C ) ⎫⎪ + ⎨ ⎬=0 ( sRL1 )(1 + s 2 L2 C ) ( sRL1 )(1 + s 2 L2 C ) ⎩⎪ ⎭⎪ Set s = jω . Both real and imaginary parts of the numerator must be zero.

R(1 − ω 2 L2 C ) + jω L1 (1 − ω 2 L2 C ) − ω 2 RL1C + ( jω g m RL1 )(−ω 2 L2 C ) = 0 Real part: R (1 − ω 2 L2 C ) − ω 2 RL1C = 0 R = ω 2 RC ( L1 + L2 ) or 1 ω0 = C ( L1 + L2 ) b. Imaginary part: jω L1 1 − ω 2 L2C − jω gm RL1 ω 2 L2C = 0

(

)

(

(

L1 = ω L1 L2C + gm RL1 ω L2C 2

2

)

)

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Now ω 2 =

1 ( L1 + L2 )

1=

1 [ L2 C + g m RL2 C ] C ( L1 + L2 )

1=

L2 L (1 + g m R ) ⇒ 1 = (1 + g m R ) L1 + L2 L2

or L1 = gm R L2

15.33

ω 0 = 2π (800 × 103 ) =

1 C ( L1 + L2 )

or C ( L1 + L2 ) = 3.96 × 10−14 Also

L1 = gm R L2

For example, if R = 1 kΩ, then

L1 = (20)(1) = 20 L2

So L1 = 20 L2 Then C (21L2 ) = 3.96 × 10−14 or CL2 = 1.89 × 10−15 C = 0.01 μ F If then

L2 = 0.189 μ H

and

L1 = 3.78 μ H

15.34 v0 − v1 v1 v1 − vB = + R ⎛ 1 ⎞ R ⎜ ⎟ sC ⎝ ⎠ and vB v −v + B 1 =0 R ⎛ 1 ⎞ ⎜ ⎟ ⎝ sC ⎠ or

(1)

(2)

1⎞ v ⎛ vB ⎜ sC + ⎟ = 1 ⇒ v1 = vB (1 + sRC ) R⎠ R ⎝ From (1) 2⎞ v ⎛ v0 ( sC ) = v1 ⎜ sC + ⎟ − B R ⎝ ⎠ R or v0 ( sRC ) = vB (1 + sRC )(2 + sRC ) − vB = vB [ (1 + sRC )(2 + sRC ) − 1]

Now

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⎛ R ⎞⎡ ⎤ ⎛ R2 ⎞ ⎡ sRC sRC ⎤ = ⎜1 + ⎟ ⎢ T ( s ) = ⎜1 + 2 ⎟ ⎢ ⎥ 2 2 2 ⎥ + + − (1 )(2 ) 1 R sRC sRC R + + − 2 3 1 sRC s R C ⎣ ⎦ ⎦ ⎝ ⎝ 1 ⎠⎣ 1 ⎠ or ⎛ R ⎞⎡ sRC ⎤ T ( s ) = ⎜1 + 2 ⎟ ⎢ 2 2 2 ⎥ R1 ⎠ ⎣ s R C + 3sRC + 1 ⎦ ⎝ ⎛ R ⎞⎡ ⎤ jω RC T ( jω ) = ⎜ 1 + 2 ⎟ ⎢ ⎥ 2 2 2 ω ω − + 1 3 R R C j RC ⎦ 1 ⎠⎣ ⎝ Frequency of oscillation: 1 f0 = 2π RC Condition for oscillation: ⎛ R ⎞ ⎡ jω RC ⎤ 1 = ⎜1 + 2 ⎟ ⎢ R1 ⎠ ⎣ 3 jω RC ⎥⎦ ⎝

or R2 =2 R1

15.35

vb − vo vb vb − va + + =0 1 1 R sC sC vb − vo (1) + 2vb ⋅ sC − va ⋅ sC = 0 R Va − Vb Va (2) + =0 1 R sC 1⎞ ⎛ ⎛ 1 + sRC ⎞ Va ⎜ sC + ⎟ = vb ⋅ sC ⇒ vb = va ⎜ ⎟ R⎠ ⎝ ⎝ sRC ⎠ From (1) ⎛1 ⎞ v vb ⎜ + 2 sC ⎟ = o + va ⋅ sC ⎝R ⎠ R Substitute (2) into (1)

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⎛ 1 + sRC ⎞ ⎛ 1 + 2 sRC ⎞ vo va ⎜ ⎟⎜ ⎟ = + va ⋅ sC R ⎝ sRC ⎠ ⎝ ⎠ R ⎡ (1 + sRC )(1 + 2 sRC ) ⎤ v va ⎢ − sC ⎥ = o ( ) sRC R ⋅ ⎣ ⎦ R ⎡ (1 + sRC )(1 + 2 sRC ) ⎤ va ⎢ − sRC ⎥ = vo sRC ⎣ ⎦ 2 2 2 vo (1 + sRC )(1 + 2 sRC ) − s R C = va sRC va sRC = vo 1 + 3sRC + 2( sRC ) 2 − s 2 R 2 C 2 va sRC = vo 1 + 3sRC + ( sRC ) 2 ⎛ R ⎞ sRC T ( s ) = ⎜1 + 2 ⎟ ⋅ R + + ( sRC ) 2 1 3 sRC 1 ⎠ ⎝ ⎛ R ⎞⎡ ⎤ jω RC T ( jω ) = ⎜ 1 + 2 ⎟ ⎢ ⎥ 2 2 2 − + ω ω 1 3 R R C j RC ⎦ 1 ⎠⎣ ⎝

So 1 − ω 02 R 2 C 2 = 0 So f O =

1 2π RC

⎛ R ⎞⎛ 1 ⎞ Also 1 = ⎜ 1 + 2 ⎟ ⎜ ⎟ R1 ⎠ ⎝ 3 ⎠ ⎝ R So 2 = 2 R1

15.36 v0 − v1 v1 v1 − vB = + sL R R ⎛ sL ⎞ vB = ⎜ ⎟ v1 ⎝ R + sL ⎠ or

(1) (2)

⎛ R + sL ⎞ v1 = ⎜ ⎟ vB ⎝ sL ⎠ Then v0 ⎛ 1 2⎞ v = v1 ⎜ + ⎟ − B sL ⎝ sL R ⎠ R or v0 ⎛ R + sL ⎞ ⎛ 1 2 ⎞ vB =⎜ ⎟ ⎜ + ⎟ vB − sL ⎝ sL ⎠ ⎝ sL R ⎠ R ⎧⎛ R + sL ⎞⎛ R + 2 sL ⎞ 1 ⎫ = vB ⎨⎜ ⎟⎜ ⎟− ⎬ ⎩⎝ sL ⎠⎝ sRL ⎠ R ⎭ Then

(1)

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vB =

v0 1 ⋅ sL ⎧ ( R + sL)( R + 2 sL) − ( sL) 2 ⎨ ( sL)( sRL) ⎩

⎫ ⋅⎬ ⎭

Now ⎛ R ⎞⎛ sRL ⎞ T ( s ) = ⎜1 + 2 ⎟ ⎜ 2 2 2 2 2 ⎟ + + − 3 2 R R sRL s L s L ⎠ ⎝ 1 ⎠⎝ or ⎛ R ⎞⎛ sRL ⎞ T ( s ) = ⎜1 + 2 ⎟ ⎜ 2 2 2 ⎟ R + + 3 s L sRL R ⎝ ⎠ ⎝ 1 ⎠ And ⎛ R ⎞⎛ ⎞ jω RL T ( jω ) = ⎜ 1 + 2 ⎟ ⎜ 2 ⎟ 2 2 R − + ω ω 3 R L j RL ⎠ ⎝ 1 ⎠⎝ R Frequency of oscillation: f 0 = 2π L Condition for oscillation: ⎛ R ⎞⎛ 1 ⎞ 1 = ⎜1 + 2 ⎟ ⎜ ⎟ R1 ⎠ ⎝ 3 ⎠ ⎝ or R2 =2 R1

15.37 From Equation (15.65(b)), the crossover voltage is R vI = − 2 ⋅ VREF R1 Let R2 = RVAR + RF where RVAR is the potentiometer and RF is the fixed resistor. Let VREF = −5 V, RF = 10 kΩ, and RVAR = 40 kΩ Then we have R ⎛ 10 ⎞ vI = − F ⋅ VREF = − ⎜ ⎟ (−5) = 1 V R1 ⎝ 50 ⎠ and ⎛ 50 ⎞ vI = − ⎜ ⎟ (−5) = 5 V ⎝ 50 ⎠ 15.38 ⎛ R1 ⎞ VTH − VTL = ⎜ ⎟ (VH − VL ) ⎝ R1 + R2 ⎠ ⎛ R1 ⎞ ⎛ R1 ⎞ 0.2 = ⎜ ⎟ (13 − (−13) ) = ⎜ ⎟ 26 ⎝ R1 + R2 ⎠ ⎝ R1 + R2 ⎠ R1 So = 0.007692 R1 + R2 13 = 0.25 ⇒ R1 + R2 = 52 R1 + R2 Then I=

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R1 = (0.007692)(52) = 0.4 kΩ = R1

So R2 = 51.6 kΩ 15.39 a. ⎛ R1 ⎞ ⎛ 10 ⎞ VTH = ⎜ ⎟ VH = ⎜ ⎟ (10) + R R ⎝ 10 + 40 ⎠ ⎝ 1 2 ⎠ so VTH = 2 V ⎛ R1 ⎞ ⎛ 10 ⎞ VTL = ⎜ ⎟ VL = ⎜ ⎟ (−10) ⎝ 10 + 40 ⎠ ⎝ R1 + R2 ⎠ so VTL = −2 V

b.

vI = 5sin ω t

15.40 a. Upper crossover voltage when v0 = +VP , Now ⎛ R1 ⎞ vB = ⎜ ⎟ (+VP ) ⎝ R1 + R2 ⎠ and ⎛ RA ⎞ ⎛ RB ⎞ vA = ⎜ ⎟ VREF + ⎜ ⎟ VTH + R R B ⎠ ⎝ A ⎝ RA + RB ⎠ v A = vB so that ⎛ R1 ⎞ ⎛ RA ⎜ ⎟ VP = ⎜ ⎝ R1 + R2 ⎠ ⎝ RA + RB or

⎞ ⎛ RB ⎟ VREF + ⎜ ⎠ ⎝ RA + RB

⎞ ⎟ VTH ⎠

⎛ R + RB ⎞ ⎛ R1 ⎞ ⎛ RA ⎞ VTH = ⎜ A ⎟⎜ ⎟ VP − ⎜ ⎟ VREF ⎝ R1 + R2 ⎠ ⎝ RB ⎠ ⎝ RB ⎠ Lower crossover voltage when v0 = −VP So ⎛ R + RB ⎞ ⎛ R1 ⎞ ⎛ RA ⎞ VTL = − ⎜ A ⎟ ⎜ ⎟ VP − ⎜ ⎟ VREF ⎝ R1 + R2 ⎠ ⎝ RB ⎠ ⎝ RB ⎠

b. ⎛ 10 + 20 ⎞ ⎛ 5 ⎞ ⎛ 10 ⎞ VTH = ⎜ ⎟ ⎜ ⎟ (10) − ⎜ ⎟ (2) 5 20 20 + ⎝ ⎠⎝ ⎠ ⎝ 20 ⎠ or VTH = 2 V

and ⎛ 10 + 20 ⎞ ⎛ 5 ⎞ VTL = − ⎜ ⎟ ⎜ ⎟ (10) − 1 ⇒ VTL = −4 V ⎝ 5 + 20 ⎠ ⎝ 20 ⎠

15.41 a.

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vB VREF − vB v0 − vB = + R1 R3 R2 ⎛ 1 v 1 1 ⎞ V + ⎟ = REF + 0 vB ⎜ + R3 R2 ⎝ R1 R2 R3 ⎠ VTH = vB when v0 = +VP and VTL = vB when v0 = −VP So VREF VP + R3 R2 VTH = ⎛ 1 1 1 ⎞ + ⎟ ⎜ + ⎝ R1 R2 R3 ⎠ and VTL =

VREF VP − R3 R2 ⎛ 1 1 1 ⎞ + ⎟ ⎜ + ⎝ R1 R2 R3 ⎠

b. VREF ⎛ 1 1 1 ⎞ + ⎟ R3 ⎜ + ⎝ R1 R2 R3 ⎠ −10 −5 = ⎛ 1 1 1⎞ + ⎟ 10 ⎜ + 10 R R ⎝ 1 ⎠ 2

VS =

1 1 1 1 + = − = 0.10 R1 R2 5 10 ΔVT = VTH − VTL =

0.2 =

2VP R2 ⎛ 1 1 1 ⎞ + ⎟ ⎜ + R R R 2 3 ⎠ ⎝ 1

2(12) R2 (0.10 + 0.10)

So R2 = 600 kΩ Then 1 1 + = 0.10 R1 R2 1 1 + = 0.10 ⇒ R1 = 10.17 kΩ R1 600 c. VTH = −5 + 0.1 = −4.9

VTL = −5 − 0.1 = −5.1 15.42 a.

If the saturated output voltage is VP < 6.2 V, then the circuit behaves as a comparator

where v0 < 6.2 V.

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If the saturated output voltage is VP > 6.2 V, the output will flip to either +VP or −VP and the input has no control. b. Same as part (a) except the curve at vI ≈ 0 will have a finite slope. c. Circuit works as a comparator as long as v01 < 8.7 V and v02 > −3.7 V. Otherwise the input has no control. 15.43 a.

Switching point is when v0 = 0. Then

⎛ R2 ⎞ v+ = vI ≡ VS = ⎜ ⎟ VREF ⎝ R1 + R2 ⎠ VTH occurs when vo = VH , then by superposition ⎛ R1 ⎞ ⎛ R2 ⎞ v+ = VTH = ⎜ ⎟ VH + ⎜ ⎟ VREF ⎝ R1 + R2 ⎠ ⎝ R1 + R2 ⎠ or ⎛ R1 ⎞ VTH = VS + ⎜ ⎟ VH ⎝ R1 + R2 ⎠ VTL occurs when v0 = VL , then by superposition ⎛ R1 ⎞ ⎛ R2 ⎞ v+ = VTL = ⎜ ⎟ VL + ⎜ ⎟ VREF ⎝ R1 + R2 ⎠ ⎝ R1 + R2 ⎠ or ⎛ R1 ⎞ VTL = VS + ⎜ ⎟ VL ⎝ R1 + R2 ⎠

b. Now

For VTH = 2 V and VTL = 1 V, then VS = 1.5 V

⎛ 10 ⎞ 2 = 1.5 + ⎜ ⎟ (10) ⎝ 10 + R2 ⎠ 0.5 10 = ⇒ R2 = 190 kΩ 10 10 + R2 ⎛ 190 ⎞ Now VS = 1.5 = ⎜ ⎟ VREF ⎝ 10 + 190 ⎠ so VREF = 1.579 V

15.44 a. Now

Switching point when v0 = 0.

⎛ R2 ⎞ v+ = VREF = ⎜ ⎟ vI where vI = VS . ⎝ R1 + R2 ⎠ Then ⎛ R + R2 ⎞ ⎛ R1 ⎞ VS = ⎜ 1 ⎟ VREF = ⎜ 1 + ⎟ VREF ⎝ R2 ⎠ ⎝ R2 ⎠

Now upper crossover voltage for v1 occurs when v0 = VL and v+ = VREF . Then

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VTH − VREF VREF − VL = R1 R2 or VTH = −

⎛ R1 R ⎞ ⋅ VL + VREF ⎜1 + 1 ⎟ R2 R 2 ⎠ ⎝

R1 ⋅ VL R2 Lower crossover voltage for vI occurs when v0 = VH and vI = VREF . Then VH − VREF VREF − VTL = R2 R1 or VTH = VS −

or VTL = −

⎛ R1 R ⎞ ⋅ VH + VREF ⎜1 + 1 ⎟ R2 ⎝ R2 ⎠

or VTL = VS −

R1 ⋅ VH R2

For VTH = −1 and VTL = −2, VS = −1.5 V. Then VTL = VS −

b.

R1 R ⋅ VH ⇒ −2 = −1.5 − 1 (12) R2 20

so that R1 = 0.833 kΩ Now ⎛ R ⎞ VS = ⎜ 1 + 1 ⎟ VREF ⎝ R2 ⎠ ⎛ 0.833 ⎞ −1.5 = ⎜ 1 + ⎟ VREF 20 ⎠ ⎝ which gives VREF = −1.44 V

15.45 (a) vo (max) = 4.7 + 0.7 = 5.4 V vo (min) = −5.4 V ⎛ R1 ⎞ VTH − TTL = ⎜ ⎟ ( 5.4 − (−5.4) ) ⎝ R1 + R2 ⎠ ⎛ 2 ⎞ 0.8 = ⎜ ⎟ (10.8) ⎝ 2 + R2 ⎠ 2 + R2 = 27 ⇒ R2 = 25 K

(b) R=

Neglecting current in R1 and R2 , 13.0 − 5.4 ⇒ R = 15.2 K 0.5

15.46 a. v0 = VREF + 2Vγ 5 = VREF + 2(0.7) or VREF = 3.6 V b.

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⎛ R1 ⎞ VTH = ⎜ ⎟ (VREF + 2Vγ ) ⎝ R1 + R2 ⎠ ⎛ R1 ⎞ 0.5 = ⎜ ⎟ (5) ⎝ R1 + R2 ⎠ R2 R = 10 ⇒ 2 = 9 R1 R1 For example, let R2 = 90 kΩ and R1 = 10 kΩ c. For vI = 10 V, and v0 is in its low state. D1 is on and D2 is off. or 1 +

v1 − (v1 + 0.7) VREF − v1 v1 − v0 + = 100 1 1 For v1 = −0.7, then 10 − 0 3.6 − (−0.7) −0.7 − v0 + = 100 1 1 or v0 = −5.1 V

15.47 For v0 = High = (VREF + 2Vγ ). Then switching point is when. ⎛ R1 ⎞ vI = vB = ⎜ ⎟ v0 ⎝ R1 + R2 ⎠ ⎛ R1 ⎞ or VTH = ⎜ ⎟ (VREF + 2Vγ ) ⎝ R1 + R2 ⎠

Lower switching point is when ⎛ R1 ⎞ v1 = vB = ⎜ ⎟ v0 and v0 = −(VREF + 2Vγ ) ⎝ R1 + R2 ⎠ so ⎛ R1 ⎞ VTL = − ⎜ ⎟ (VREF + 2Vγ ) ⎝ R1 + R2 ⎠

15.48 By symmetry, inverting terminal switches about zero. Now, for v0 low, upper diode is on. VREF − v1 = v1 − v0 v0 = 2v1 − VREF where v1 = −Vγ so v0 = −(VREF + 2Vγ ) Similarly, in the high state v0 = (VREF + 2Vγ ) Switching occurs when non-inverting terminal is zero. So for v0 low. VTH − 0 0 − ⎡⎣ − (VREF + 2Vγ ) ⎤⎦ = R1 R2 or VTH =

R1 ⋅ (VREF + 2Vγ R2

)

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By symmetry R VTL = − 1 ⋅ (VREF + 2vγ ) R2 15.49 f =

1 2.2 RX C X

1 1 = 2.2 f (2.2)(12 × 103 ) RX C X = 3, 788 × 10−5

RX C X =

For example, Let RX = 56 K C X = 680 pF Within

1 of 1% of design specification. 2

15.50 (a) ⎛ R1 ⎞ ⎛ 20 ⎞ vx1 = ⎜ ⎟ vo = ⎜ ⎟ vo ⎝ 20 + 5 ⎠ ⎝ R1 + R2 ⎠ So vH = ±9.6 Also vx2 = 12 + (−9.6 − 12)e − t / τ x = 12 − 21.6e − t / τ x Set vx1 = vx2

9.6 = 12 − 21.6e − t / τ x T = τ x ln (9) 2 1 1 f = = T 2τ x ln (9)

e+ t / τ x = 9 ⇒ t1 =

1 2(22 × 10 )(0.2 × 10−6 ) ln (9) f = 51.7 Hz Duty cycle = 50% f =

3

15.51 (a) ⎛ R1 ⎞ ⎛ 20 ⎞ vx+1 = ⎜ ⎟ VH = ⎜ ⎟ (15) = 12 V ⎝ 20 + 5 ⎠ ⎝ R1 + R2 ⎠ ⎛ 20 ⎞ vx−1 = ⎜ ⎟ (−10) = −8 V ⎝ 20 + 5 ⎠ vx+2 = 15 + (−8 − 15)e − t / τ x = 15 − 23e− t / τ x Then 12 = 15 − 23e − t1 / τ x

e − t1 / τ x = 7.667 ⇒ t1 − τ x ln (7.667) ⎛ − t2 − t1 ⎞ ⎜⎜ τ ⎟⎟ x ⎠

vx−2 = −10 + (12 + 10)e⎝ Then

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−8 = −10 + 22e

⎛ − t2 − t1 ⎞ ⎜⎜ τ ⎟⎟ x ⎝ ⎠

− ( t2 − t1 )

e

τx

= 11 ⇒ t2 − t1 = τ x ln (11)

Period = t2 = T = τ x [ ln (7.667) + ln (11) ] = τ x (4.435)

τ x = (22 × 103 )(0.2 × 10−6 ) = 4.4 × 10−3

T = 1.95 × 10−2 1 f = = 51.2 Hz T t1 = τ x ln(7.667) = (4.4 ×10 −3 ) ln (7.667)

t1 = 8.962 × 10−3 t1 8.962 × 10−3 = T 1.951× 10−2 Duty Cycle = 45.9%

Duty cycle =

15.52 t1 = 1.1RX C X = (1.1)(10 4 )(0.1× 10−6 ) ⇒ t1 = 1.1 ms 0 < t < t1 , vY = 10(1 − e− t / rY ) rY = RY CY

Now

= (2 × 103 )(0.02 × 10−6 ) = 4 × 10−5 s

t1 = 2.75 rY

⇒ CY completely charges during each cycle.

15.53 a. Switching voltage ⎛ R1 + R3 ⎞ ⎛ 10 + 10 ⎞ vX = ⎜ ⎟ ⋅ VP = ⎜ ⎟ (±10) ⎝ 10 + 10 + 10 ⎠ ⎝ R1 + R3 + R2 ⎠ So v X = ±6.667 V Using Equation (15.83(b)) 2 ⎛ 2 ⎞ v X = VP + ⎜ − VP − VP ⎟ e − t1 / rX = VP 3 ⎝ 3 ⎠ 5 2 Then 1 − ⋅ e− t1 / rX = 3 3 1 5 − t1 / rX or t1 = rX ln (5) = ⋅e 3 3 T 1 1 t1 = = ⇒ t1 = 0.001 s = 2 2 f 2(500) 10−3 = rX ln (5) ⇒ rX = 6.21× 10−4 = RX (0.01× 10−6 ) So RX = 62.1 kΩ b.

Switching voltage ⎛ ⎞ R1 vX = ⎜ ⎟ (±VP ) ⎝ R1 + R3 + R2 ⎠ 10 1 ⎛ ⎞ =⎜ ⎟ (±VP ) = ⋅ (±VP ) 3 ⎝ 10 + 10 + 10 ⎠

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Using Equation (15.83(b)) 1 ⎛ 1 ⎞ v X = VP + ⎜ − VP − VP ⎟ e − t1 / rX = VP 3 ⎝ 3 ⎠ 4 − t1 / rX 1 = Then 1 − ⋅ e 3 3 2 4 − t1 / rX = ⋅e 3 3 t1 = rX ln (2) = (6.21× 10−4 ) ln (2) = 4.30 × 10−4 s T = 2t1 = 8.6 × 10−4 s 1 f = ⇒ f = 1.16 kHz T 15.54 From Equation (15.92) ⎛ ⎛ Vγ ⎞ ⎞ ⎜1+ ⎜ ⎟ ⎟ V T = rX ln ⎜ ⎝ P ⎠ ⎟ ⎜ 1− β ⎟ ⎜⎜ ⎟⎟ ⎝ ⎠ R1 10 where β = = = 0.2857 R1 + R2 10 + 25 so 0.7 ⎤ ⎡ ⎢ 1+ 5 ⎥ 100 = rX ln ⎢ ⎥ ⎢1 − 0.2857 ⎥ ⎢⎣ ⎥⎦ so τ X = 213.9 μ s = RX C X For example, RX = 10 kΩ, C X = 0.0214 μ F ⎛ R1 ⎞ ⎛ 10 ⎞ vY = ⎜ ⎟ VP = ⎜ ⎟ (5) = 1.43 V ⎝ 10 + 25 ⎠ ⎝ R1 + R2 ⎠ and v X = 0.7 V To trigger the circuit, vY must be brought to a voltage less than v X . Therefore minimum triggering pulse is −0.73 V. Using Equation (15.82) for T < t < T ′ v X = VP + (−0.2857VP − VP )e − t ′ / rX

Recovery period is when v X = Vγ = 0.7 V. 0.7 = 5 + (−6.43)e − t ′ / rX 6.43e− t ′ / rX = 4.3 or t ′ = rX ln (1.495) rX = 213.9 μ s so t ′ = T ′ − T = 86.1 μ s

15.55 (a)

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⎡1 + (Vr / VP ) ⎤ T =τx ⎢ ⎥ ⎣ 1− β ⎦ τ x = Rx Cx β = 1 2 T  0.69τ x = (0.69)(47 × 103 )(0.2 × 10−6 ) T = 6.49 ms Recovery Time  0.4τ x (b) = 3.76 ms

15.56 a. From Equation (15.95) T = 1.1 RC For T = 60 s = 1.1 RC then RC = 54.55 s For example, let C = 50 μ F and R = 1.09 MΩ b.

Recovery time: capacitor is discharged by current through the discharge transistor. 5 − 0.7 If V + = 5 V, then I B ≅ = 0.043 mA 100 If β = 100, I C = 4.3 mA VC =

1 Ic I C dt = ⋅ t ∫ C C

2 + ⋅ V = 3.33 V 3 V ⋅ C (3.33)(50 × 10−6 ) So that t = C = IC ⋅ 4.3 × 10−3

Capacitor has charged to

So recovery time t ≈ 38.7 ms 15.57 T = 1.1 RC 5 × 10−6 = 1.1 RC so RC = 4.545 × 10−6 s For example, let C = 100 pF and R = 45.5 kΩ From Problem (15.53), recovery time V ⋅ C (3.33)(100 × 10−12 ) t≅ C = IC 4.3 × 10−3 or t = 77.4 ns 15.58 From Equation (15.102), 1 f = (0.693)(20 + 2(20)) × 103 × (0.1× 10−6 ) or f = 240.5 Hz Duty cycle =

20 + 20 × 100% = 66.7% 20 + 2(20)

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15.59 f =

1 (0.693)( RA + 2 RB )C

RA = R1 = 10 kΩ, RB = R2 + xR3 So 10 kΩ ≤ RB ≤ 110 kΩ 1 = 627 kHz (0.693)(10 + 2(110)) × 103 × (0.01× 10−6 ) 1 = = 4.81 kHz (0.693)(10 + 2(10)) × 103 × (0.01× 10−6 )

f min = f max

So 627 Hz ≤ f ≤ 4.81 kHz Duty cycle =

RA + RB × 100% RA + 2 RB

Now 10 + 10 × 100% = 66.7% 10 + 2(10) and 10 + 110 × 100% = 52.2% 10 + 2(110) So 52.2 ≤ Duty cycle ≤ 66.7% 15.60 1 kΩ ≤ RA ≤ 51 kΩ 1 kΩ ≤ RB ≤ 51 kΩ 1 = 1.40 Hz (0.693)(1 + 2(51)) × 103 × (0.01× 10−6 ) 1 f max = = 2.72 kHz (0.693)(51 + 2(1)) × 103 × (0.01× 10−6 ) or 1.40 kHz ≤ f ≤ 2.72 kHz f min =

Duty cycle =

RA + RB × 100% RA + 2 RB

1 + 51 × 100% = 50.5% 1 + 2(51) or 51 + 1 × 100% = 98.1% 51 + 2(1) or 50.5% ≤ Duty cycle ≤ 98.1%

15.61 a. I E3 = IE 4 =

V + − 3VEB R1 A + R1B

Assume VEB = 0.7

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I E3 = I E 4 =

22 − 3(0.7) = 0.398 mA 25 + 25

Now ⎛ 20 ⎞ I C 3 = I C 4 = I C 5 = I C 6 = ⎜ ⎟ (0.398) ⎝ 21 ⎠ I C 3 = I C 4 = I C 5 = I C 6 = 0.379 mA 0.398 ⎛ 20 ⎞ ⎜ ⎟ ⇒ I C1 = I C 2 = 0.018 mA 21 ⎝ 21 ⎠ I D = 0.398 mA, current in D1 and D2

I C1 = I C 2 =

b.

⎛I ⎞ ⎛ 0.398 × 10−3 ⎞ VBB = 2VD = 2VT ln ⎜ D ⎟ = 2(0.026) ln ⎜ ⎟ −13 ⎝ 10 ⎠ ⎝ IS ⎠ or VBB = 1.149 V = VBE 7 + VEB 8 Now IC 7 ≈ IC 4 + IC 9 + I E8 I C 4 = 0.379 mA ⎛ 20 ⎞ I B9 = IC 8 = ⎜ ⎟ I E 8 ⎝ 21 ⎠ So ⎛I ⎞ I E 8 = 1.05 I B 9 = 1.05 ⎜ C 9 ⎟ ⎝ 100 ⎠ ⎛ 100 ⎞ ⎛ 21 ⎞ IC 7 = IC 4 + ⎜ ⎟ I E 8 + I E 8 = I C 4 + (96.24) ⎜ ⎟ I C 8 ⎝ 1.05 ⎠ ⎝ 20 ⎠ So I C 7 = 0.379 mA + 101I C 8 and ⎛I ⎞ ⎛I ⎞ VBE 7 = VT ln ⎜ C 7 ⎟ ; VEB 8 = VT ln ⎜ C 8 ⎟ I ⎝ S ⎠ ⎝ IS ⎠

Then ⎡ ⎛I ⎞ ⎛ I ⎞⎤ 1.149 = 0.026 ⎢ ln ⎜ C 7 ⎟ + ln ⎜ C 8 ⎟ ⎥ ⎝ I S ⎠⎦ ⎣ ⎝ IS ⎠ ⎡ I (0.379 × 10−3 ) + 101I C 8 ⎤ 44.19 = ln ⎢ C 8 ⎥ (10−13 ) 2 ⎣ ⎦ (10−13 ) 2 exp (44.19) = 101I C2 8 + 3.79 × 10 −4 I C 8 1.554 × 10−7 = 101I C2 8 + 3.79 × 10−4 I C 8 IC 8 =

(3.79 × 10−4 ) 2 + 4(101)(1.554 × 10−7 ) −3.79 × 10−4 ± 2(101) 2(101) I C 8 = 37.4 μ A

I C 7 = 0.379 + 101(0.0374) ⇒ I C 7 = 4.16 mA ⎛ 21 ⎞ I C 9 = 4.16 − 0.379 − 0.0374 ⎜ ⎟ ⎝ 20 ⎠ I C 9 = 3.74 mA

c.

P = (0.398 + 0.398 + 4.16)(22) ⇒ P = 109 mW

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15.62 a. b.

From Figure 15.47, 3.7 W to the load V + ≈ 19 V

c.

P=

1 VP2 2 RL

or VP = 2 RL P = 2(10)(3.7) ⇒ VP = 8.6 V

15.63 1 VP2 P= 2 RL so VP = 2 RP = 2(10)(20) ⇒ 20 V peak-to-peak output voltage Maximum output voltage of each op-amp = ±10 V. Current is (20 /10) = 2 A. Bias op-amps at ±12 V. For A1 ,

v01 ⎛ R2 ⎞ R = ⎜1 + ⎟ = 15 ⇒ 2 = 14 vI ⎝ R1 ⎠ R1

For A2 ,

v02 R = 4 = 15 vI R3

For example, let R1 = R3 = 10 kΩ, and R2 = 140 kΩ and R4 = 150 kΩ. 15.64 a.

v01 = iR2 + vI where i =

vI R1

Then ⎛ R ⎞ v01 = vI ⎜ 1 + 2 ⎟ R1 ⎠ ⎝ Now ⎛R ⎞ v02 = −iR3 = −vI ⎜ 3 ⎟ ⎝ R1 ⎠ So ⎛ R ⎞ ⎡ ⎛ R ⎞⎤ vL = v01 − v02 = vI ⎜1 + 2 ⎟ − ⎢ −vI ⎜ 3 ⎟ ⎥ R1 ⎠ ⎣ ⎝ ⎝ R1 ⎠ ⎦ Av =

b.

vL R R = 1+ 2 + 3 vI R1 R1

Want Av = 10 ⇒

R2 R3 + =9 R1 R1

⎛ R ⎞ R Also want ⎜1 + 2 ⎟ = 3 R1 ⎠ R1 ⎝ R2 ⎛ R2 ⎞ R + ⎜1 + ⎟ = 9 so 2 = 4 R1 ⎝ R1 ⎠ R1 For R1 = 50 kΩ, R2 = 200 kΩ

Then

and

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R3 = 5 so R3 = 250 kΩ R1 P=

c.

1 VP2 2 RL

or VP = 2 RL P = 2(20)(10) = 20 V

So peak values of output voltages are v01 = v02 = 10 V Peak load current =

20 =1 A 20

15.65 (a) ⎛ R ⎞ vo1 = ⎜1 + 2 ⎟ vI R1 ⎠ ⎝ ⎛ R2 ⎞ ⎜ 1 + ⎟ vI R1 ⎠ ⎝ vL = vo1 − vo 2 vo 2 = −

Av =

R4 R3

vL ⎛ R2 ⎞ ⎛ R4 ⎞ = ⎜1 + ⎟ ⎜1 + ⎟ vI ⎝ R1 ⎠ ⎝ R3 ⎠ For v01 − 12 V, vo 2 = −12 V when R3 = R4

(b) So vL = 24sin ω + (V )

⎛ R ⎞⎛ R ⎞ 10 = ⎜ 1 + 2 ⎟ ⎜1 + 4 ⎟ R1 ⎠ ⎝ R3 ⎠ ⎝ Let R3 = R4

(c)

Then

R2 =4 R1

15.66 (a) From Problem 15.65 vO ⎛ R2 ⎞ ⎛ R4 ⎞ = ⎜1 + ⎟ ⎜1 + ⎟ vI ⎝ R1 ⎠⎝ R3 ⎠ For vo1 = vo 2 ⇒ R3 = R4 Then ⎛ R ⎞ vO = 2 ⎜1 + 2 ⎟ vI R1 ⎠ ⎝ ⎛ R ⎞ ⎛R ⎞ 15 = 2 ⎜1 + 2 ⎟ ⇒ ⎜ 2 ⎟ = 6.5 R1 ⎠ ⎝ R1 ⎠ ⎝ (b)

Peak output voltage

VP = 2 RL PL = 2(16)(20) = 25.3 V

Then Vo1 = Vo 2 = 12.65 V iL (peak) =

25.3 = 1.58 A 16

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15.67 Line regulation =

ΔV0 ΔV +

Now ΔI =

ΔV + and ΔVZ = rZ ⋅ ΔI and ΔV0 = 10ΔVZ R1

So ΔV0 = 10 ⋅ rZ ⋅

ΔV + R1

So Line regulation =

ΔV0 10(15) = ⇒ 1.61% ΔV + 9300

15.68 R0 f = −

ΔV0 ΔI 0

So R0 f =

−(−10 × 10−3 ) 1

or R0 f = 10 mΩ 15.69 For V0 = 8 V V + (min) = V0 + I 0 (max) R11 + VBE11 + VBE10 + VEB 5

This assumes VBC 5 = 0. Then V + (min) = 8 + (0.1)(1.9) + 0.6 + 0.6 + 0.6 V + (min) = 9.99 V 15.70 a. IC 3 = IC 5 =

VZ − 3VBE (npn) R1 + R2 + R3

6.3 − 3(0.6) = 0.571 mA 0.576 + 3.4 + 3.9 1 ⎛ 0.6 ⎞ IC 8 = ⎜ ⎟ = 0.106 mA 2 ⎝ 2.84 ⎠ Neglecting current in Q9 , total collector current and emitter current in Q5 is 0.571 + 0.106 = 0.677 Now I Z 2 R4 + VEB 4 = VEB 5 IC 3 = IC 5 =

⎛I ⎞ VEB 4 = VT ln ⎜ Z 2 ⎟ ⎝ I5 ⎠ ⎛I ⎞ VEB 5 = VT ln ⎜ C 5 ⎟ ⎝ 2I S ⎠

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⎛ I ⎞ Then I Z 2 R4 = VT ln ⎜ C 5 ⎟ ⎝ 2I Z 2 ⎠ ⎛ 0.677 ⎞ 0.026 R4 = ⋅ ln ⎜ ⎟ 0.25 ⎝ 2(0.25) ⎠ or R4 = 31.5 Ω

b.

From Example 15.16, VB 7 = 3.43 V . Then

⎛ R13 ⎞ ⎜ ⎟ V0 = VB 8 = VB 7 ⎝ R12 + R13 ⎠ or ⎛ 2.23 ⎞ ⎜ ⎟ (12) = 3.43 ⎝ 2.23 + R12 ⎠ 3.43(2.23 + R12 ) = (2.23)(12) which yields R12 = 5.57 kΩ

15.71 Line regulation =

ΔV0 ΔV +

Now ΔVB 7 = ΔI C 3 ⋅ R1 ⎛ R13 ⎞ and ⎜ ⎟ (ΔV0 ) = ΔVB 7 = ΔI C 3 R1 ⎝ R12 + R13 ⎠ ΔVZ ΔI Z ⋅ rZ and ΔI C 3 = = R1 + R2 + R3 R1 + R2 + R3

and ΔI Z =

V ΔV + where r0 = A IZ r0

Then ⎛ 0.015 ⎞ (0.4288)( ΔV0 ) = ΔI C 3 (3.9) = (3.9) ΔI Z ⎜ ⎟ ⎝ 7.876 ⎠ 50 = 87.6 kΩ r0 = 0.571 Then ⎛ ΔV + ⎞ (0.4288)(ΔV0 ) = (0.00743) ⎜ ⎟ ⎝ 87.6 ⎠ So ΔV0 = 0.0198% ΔV +

15.72 a.

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IZ =

25 − 5 = 10 R1 + rZ

20 = 2 kΩ = R1 + 0.01 ⇒ R1 = 1.99 kΩ 10 b. In the ideal case; ⎛ R3 + R4 ⎞ ⎜ ⎟ V0 = VZ ⎝ R2 + R3 + R4 ⎠ So R1 + rZ =

⎛ 2 +1 ⎞ ⎜ ⎟ V0 = 5 ⇒ V0 = 6.67 V ⎝ 2 +1+1⎠ and ⎛ ⎞ R4 ⎜ ⎟ V0 = VZ ⎝ R2 + R3 + R4 ⎠ ⎛ 1 ⎞ ⎜ ⎟ V0 = 5 ⇒ V0 = 20 V ⎝ 2 +1+1⎠ So 6.67 ≤ V0 ≤ 20 V

c.

3 3 ⋅ V0 so vd = 5 − ⋅ V0 4 4 and V0 = A0 L vd − VBE

V1 =

⎛I ⎞ and VBE = VT ln ⎜ 0 ⎟ ⎝ IS ⎠ Now 3 ⎛ ⎞ V0 = A0 L ⎜ 5 − ⋅ V0 ⎟ − VBE 4 ⎝ ⎠ ⎛ 3 ⎞ V0 ⎜ 1 + ⋅ A0 L ⎟ = 5 A0 L − VBE 4 ⎝ ⎠ 5 A0 L − VBE V0 = 3 1 + ⋅ A0 L 4

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Load regulation =

V0 (NL) − V0 (FL) V0 (NL)

5 A0 L − VBE (NL) 5 A0 L − VBE (FL) − (1 + 34 A0 L ) (1 + 34 A0 L ) = 5 A0 L − VBE (NL) (1 + 34 A0 L )

⎛ I (FL) ⎞ VT ln ⎜ 0 ⎟ I 0 (NL) ⎠ VBE (FL) − VBE (NL) ⎝ = = 5 A0 L − VBE (NL) 5 A0 L − VBE (NL) V0 6.67 = = 1.67 mA 4 k Ω 4 kΩ 1 ⎛ ⎞ (0.026) ln ⎜ −3 ⎟ ⎝ 1.67 × 10 ⎠ ⇒ 3.33 × 10−4 % Load regulation = 4 5(10 ) − 0.7 I 0 (FL) = 1 A, I 0 (NL) =

15.73 V 5.6 IE = Z = = 1.12 mA 5 R2 I0 =

β 1+ β

⎛ 100 ⎞ ⋅ IE = ⎜ ⎟ (1.12) ⇒ I 0 = 1.109 mA Load current ⎝ 101 ⎠

For VBC = 0 ⇒ V0

= 20 − VZ − 0.6 = 20 − 5.6 − 0.6

or V0 = 13.8 V Then V 13.8 ⇒ RL = 12.4 kΩ RL = 0 = I 0 1.109 So 0 ≤ RL ≤ 12.4 kΩ

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Chapter 16 Exercise Solutions EX16.1 COX =

γ =

( 3.9 ) (8.85 ×10−14 ) 200 × 10−8

= 1.726 ×10−7 F / cm 2

2 (1.6 × 10−19 ) (11.7 ) ( 8.85 ×10 −14 )(1015 ) 1.726 × 10−7

1

= 0.1055 V 2

ΔVTN = r ⎣⎡ 0.576 + VSB − 0.576 ⎦⎤ = ( 0.1055 ) ⎡⎣ 0.576 + 5 + 0.576 ⎤⎦ ΔVTN = 0.169 V

EX16.2 (a) vo = VDD − I D RD ⎛ k′ ⎞⎛ W vo = 3 − ⎜ n ⎟ ⎜ ⎝ 2 ⎠⎝ L vo = 0.1

⎞ 2 ⎟ ⎣⎡ 2 ( 3 − 0.5 ) vo − vo ⎦⎤ RD ⎠

2 ⎛ 0.06 ⎞ ⎡ 0.1 = 3 − ⎜ ⎟ ( 5 ) ⎣( 5 )( 0.1) − ( 0.1) ⎤⎦ RD ⎝ 2 ⎠ 0.1 = 3 − 0.0735RD

RD = 39.5 K (b) 2 ⎛ 0.06 ⎞ ⎜ ⎟ ( 5 )( 39.5 )(VIt − 0.5 ) + (VIt − 0.5 ) − 3 = 0 2 ⎝ ⎠ 5.925 (VIt − 0.5 ) + (VIt − 0.5 ) − 3 = 0 2

(VIt − 0.5 ) = VOt

=

−1 ± 1 + 4 ( 5.925 )( 3) 2 ( 5.925 )

VOt = 0.632 V VIt = 1.132 V

EX16.3 (a) (i) vo = VDD − VTNL = 3 − 0.4 vo = 2.6 V (ii) 2 ⎛W ⎞ ⎛W ⎞ 2 ⎜ ⎟ ⎡⎣ 2 ( vI − 0.4 ) vo − vo ⎤⎦ = ⎜ ⎟ [VDD − vo − 0.4] ⎝ L ⎠D ⎝ L ⎠L 16 ⎡⎣ 2 ( 2.6 − 0.4 ) vo − vo2 ⎤⎦ = 2 [3 − vo − 0.4]

2

35.2vo − 8vo2 = 6.76 − 5.2vo + vo2 9vo2 − 40.4vo + 6.76 = 0 vo =

40.4 ± 1632.16 − 4 ( 9 )( 6.76 ) 2 (9)

vo = 0.174 V

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(b) 2 ⎛ 60 ⎞ iD = ⎜ ⎟ (2) [3 − 0.174 − 0.4] ⎝ 2 ⎠ iD = 353.1 μ A

P = iD ⋅ VDD = 1.06 mW

EX16.4 (a) 2 ⎛W ⎞ ⎛W ⎞ 2 ⎜ ⎟ ⎡⎣ 2 ( vI − 0.4 ) vo − vo ⎤⎦ = ⎜ ⎟ ( − ( −0.8 ) ) L L ⎝ ⎠D ⎝ ⎠L 6 ⎡⎣ 2 ( 3 − 0.4 ) vo − vo2 ⎤⎦ = 2 ( 0.64 ) 6vo2 − 31.2vo + 1.28 = 0 vo =

31.2 ± 973.44 − 4 ( 6 )(1.28 ) 2(6)

vo = 41.4 mV (b) 2 2 ⎛W ⎞ ⎛W ⎞ ⎜ ⎟ ( vIt − 0.4 ) = ⎜ ⎟ ( −(−0.8) ) L L ⎝ ⎠D ⎝ ⎠L 6 ( vIt − 0.4 ) = 2 ( 0.64 ) 2

⇒ vIt = 0.862 V

⎫ ⎬ Driver vOt = 0.862 − 0.4 = 0.462 V ⎭ vIt = 0.862 V ⎫ ⎬ Load vOt = VDD + VTNL = 3 − 0.8 = 2.2 V ⎭ (c) 2 ⎛ 60 ⎞ iD = ⎜ ⎟ (2) ( − ( −0.8 ) ) = 38.4 μ A ⎝ 2 ⎠ P = iD ⋅ VDD = 115.2 μ W

EX16.5 We have

{

} 0.73 ⎤⎦}

VOH = VDD − VTNLO + r ⎡⎣ 2φ fP + VSB − 2φ fP ⎤⎦

{

VOH = 5 − 0.8 + 0.35 ⎡⎣ 0.73 + VOH −

VOH − 4.499 = −0.35 0.73 + VOH Squaring both sides 2 VOH − 8.998VOH + 20.241 = 0.1225(0.73 + VOH ) 2 VOH − 9.1205VOH + 20.15 = 0 9.1205 ± 83.1835 − 4(20.15) 2 = 3.76 V

VOH = VOH

EX16.6 a. i. A = logic 1 = 10 V, B = logic 0 “A” driver in nonsaturation. “B” driver off

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2 ⎛ kn′ ⎞ ⎛ W ⎞ ⎛ k′ ⎞⎛ W ⎞ 2 ⎜ 2 ⎟ ⎜ L ⎟ ( −VTNL ) = ⎜ 2 ⎟ ⎜ L ⎟ ⎡⎣ 2 ( vI − vTND ) VOL − VOL ⎤⎦ ⎝ ⎠ ⎝ ⎠D ⎝ ⎠ ⎝ ⎠L

2 ( 3) = (10 ) ⎡⎣ 2 (10 − 1.5 ) V0 L − V02L ⎤⎦ 2

9 = 5 (17V0 L − V02L )

5V02L − 85V0 L + 9 = 0

(85 )

85 ±

2

− 4 ( 5 )( 9 )

⇒ V0 L = 0.107 V 2(5) ii. A = B = logic 1 2 ⎛ kn′ ⎞ ⎛ W ⎞ ⎛ k′ ⎞⎛ W ⎞ 2 ⎜ 2 ⎟ ⎜ L ⎟ ( −VTNL ) = 2 ⎜ 2 ⎟ ⎜ L ⎟ ⎣⎡ 2 ( vI − vTND )VOL − VOL ⎤⎦ ⎝ ⎠ ⎝ ⎠D ⎝ ⎠ ⎝ ⎠L V0 L =

2 ( 3) = ( 2 )(10 ) ⎡⎣ 2 (10 − 1.5 ) V0 L − V02L ⎤⎦ 2

9 = 10 (17V0 L − V02L )

10V02L − 170V0 L + 9 = 0 V0 L =

170 ±

(170 ) − 4 (10 )( 9 ) ⇒ V0 L = 0.0531 V 2 (10 ) 2

b.

Both cases. 35 2 iD = ⋅ ( 2 )( 3) = 315 μ A ⇒ P = iD ⋅ VDD ⇒ P = 3.15 mW 2

EX16.7 800 = 160 μ A 5 35 ⎛ W ⎞ 2 ⎛W ⎞ ⎛W ⎞ iD = 160 = ⋅ ⎜ ⎟ (1.4 ) = 34.3 ⎜ ⎟ ⇒ ⎜ ⎟ = 4.66 L 2 ⎝ L ⎠L ⎝ ⎠L ⎝ L ⎠L P = iD ⋅ VDD ⇒ iD =

iD = 160 =

35 ⎛ W ⎞ ⎡ 2 ⎛W ⎞ ⋅ ⎜ ⎟ 2 ( 5 − 0.8 )( 0.12 ) − ( 0.12 ) ⎤ ⇒ ⎜ ⎟ = 9.20 ⎦ ⎝ L ⎠D 2 ⎝ L ⎠D ⎣

EX16.8 (a)

VOPt VONt

(b)

VDD 2.1 = = 1.05 V 2 2 = VIt − VTD = 1.05 − (−0.4) = 1.45 V = VIt − VTN = 1.05 − 0.4 = 0.65 V

VIt =

VIt = VOPt

2.1 + (−0.4) + 0.5(0.4)

1 + 0.5 = 1.16 + 0.4 = 1.56 V

= 1.16 V

VONt = 1.16 − 0.4 = 0.76 V

(c)

VIt = VOPt VONt

2.1 + (−0.4) + 2(0.4)

= 0.938 V 1+ 2 = 0.938 + 0.4 = 1.338 V = 0.538 V

EX16.9

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2 P = f ⋅ CL ⋅ VDD

( 0.10 ×10 ) = f ( 0.5 ×10 ) ( 3) −6

−12

2

f = 2.22 × 104 Hz ⇒ f = 22.2 kHz

EX16.10 a. K n / K p = 200 / 80 = 2.5 ⇒ VIt =

10 − 2 + 2.5(2) 1 + 2.5

⇒ VIt = 4.32 V

V0 Pt = 6.32 V V0 Nt = 2.32 V

b. VIL = 2 +

⎤ 10 − 2 − 2 ⎡ 2.5 ⋅ ⎢2 − 1⎥ ⇒ VIL = 3.39 V 2.5 − 1 ⎣ 2.5 + 3 ⎦

1 {(1 + 2.5)(3.39) + 10 − (2.5)(2) + 2} 2 = 9.43 V

V0 HU = V0 HU

VIH = 2 +

⎤ 10 − 2 − 2 ⎡ 2(2.5) ⋅⎢ − 1⎥ ⇒ VIH = 4.86 V 2.5 − 1 ⎣⎢ 3(2.5) + 1 ⎦⎥

(4.86)(1 + 2.5) − 10 − (2.5)(2) + 2 2(2.5) = 0.802 V

V0 LU =

V0 LU c. NM L = VIL − V0 LU = 3.39 − 0.802 ⇒ NM L = 2.59 V NM H = V0 HU − VIH = 9.43 − 4.86 ⇒ NM H = 4.57 V

EX16.11 3 PMOS in series and 3 NMOS in parallel. Worst Case: Only one NMOS is ON in Pull-down mode ⇒ same as the CMOS inverter ⇒ Wn = W . All 3 PMOS are on during pull-up mode ⇒ W p = 3(2W ) = 6W . EX16.12 NMOS: Worst Case, M NA , M NB on, Wn = 2(W ) or M NC , M ND or M NC , M NE on ⇒ Wn = 2(W ). PMOS: M PA and M PC on or M PA and M PB on ⇒ WP = 2(2W ) = 4W If M PD and M PE on, need WP = 2(4W ) = 8W EX16.13 a. vI = φ = 5 V ⇒ v0 = 4 V b.

vI = 3 V, φ = 5 V ⇒ v0 = 3 V

c.

vI = 4.2 V, φ = 5 V ⇒ v0 = 4 V

d.

vI = 5 V, φ = 3 V ⇒ v0 = 2 V

EX16.14 (a) vI = 8V , φ = 10V ⇒ vGSD = 8 V M D in nonsaturation

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K D ⎡⎣ 2(vGSD − VTND )vO − vO2 ⎤⎦ K L [VDD − vO − VTNL ]

2

KD ⎡ K 2 2 2 ( 8 − 2 )( 0.5 ) − ( 0.5 ) ⎤ = [10 − 0.5 − 2] ⇒ D = 9.78 ⎦ KL ⎣ KL (b) vI = φ = 8V ⇒ vGSD = 6 V KD ⎡ K 2 2 2 ( 6 − 2 )( 0.5 ) − ( 0.5 ) ⎤ = [10 − 0.5 − 2] ⇒ D = 15 ⎣ ⎦ KL KL

EX16.15 16 K ⇒ 16384 cells Total Power = 125 mW = (2.5) IT ⇒ IT = 50 mA

50 mA ⇒ I = 3.05 μ A 16384 V 2.5 ⇒ R = 0.82 M Ω or R = DD = 3.05 I

Then, for each cell, I =

Now, I ≅

VDD R

TYU16.1 P = iD ⋅ VDD ⇒ iD =

750 = 150 μ A 5

35 ⎛ W ⎞ 2 ⎜ ⎟ (5 − 0.2 − 0.8) 2 ⎝ L ⎠L ⎛W ⎞ ⎛W ⎞ 150 = 280 ⎜ ⎟ ⇒ ⎜ ⎟ = 0.536 L ⎝ ⎠L ⎝ L ⎠L 150 =

⎛ k′ ⎞⎛ W ⎞ iD = ⎜ n ⎟ ⎜ ⎟ ⎡⎣ 2(vI − VTND )vO − vO2 ⎤⎦ ⎝ 2 ⎠ ⎝ L ⎠D 35 ⎛ W ⎞ 150 = ⎜ ⎟ ⎡⎣ 2(4.2 − 0.8)(0.2) − (0.2) 2 ⎤⎦ 2 ⎝ L ⎠D ⎛W ⎞ ⎛W ⎞ 150 = 23.1 ⋅ ⎜ ⎟ ⇒ ⎜ ⎟ = 6.49 L ⎝ ⎠D ⎝ L ⎠D

TYU16.2 P = iD ⋅ VDD ⇒ I D =

350 = 70 μ A 5

2 ⎛ k′ ⎞⎛ W ⎞ iD = ⎜ n ⎟ ⎜ ⎟ ( −VTNL ) ⎝ 2 ⎠ ⎝ L ⎠L 35 ⎛ W ⎞ 2 ⎛W ⎞ 70 = ⋅ ⎜ ⎟ ( 2 ) ⇒ ⎜ ⎟ = 1 2 ⎝ L ⎠L ⎝ L ⎠L

35 ⎛ W ⎞ ⎡ 2 ⋅ ⎜ ⎟ 2 ( 5 − 0.8 )( 0.05 ) − ( 0.05 ) ⎤ ⎦ 2 ⎝ L ⎠D ⎣ ⎛W ⎞ ⎛W ⎞ 70 = 7.31 ⋅ ⎜ ⎟ ⇒ ⎜ ⎟ = 9.58 ⎝ L ⎠D ⎝ L ⎠D

iD =

TYU16.3

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800 = 160 μ A 5 35 ⎛ W ⎞ 2 ⎛W ⎞ iD = 160 = ⋅ ⎜ ⎟ (1.4 ) ⇒ ⎜ ⎟ = 4.66 2 ⎝ L ⎠L ⎝ L ⎠L P = iD ⋅ VDD ⇒ iD =

iD = 160 μ A =

35 1 ⎛ W ⎞ ⎡ 2 ⎛W ⎞ ⋅ ⋅ ⎜ ⎟ 2 ( 5 − 0.8 )( 0.12 ) − ( 0.12 ) ⎤ ⇒ ⎜ ⎟ = 27.6 ⎦ ⎝ L ⎠D 2 3 ⎝ L ⎠D ⎣

TYU16.4 a. From the load transistor: k 35 2 2 ⎛ ′ ⎞⎛W ⎞ I DL = ⎜ n ⎟ ⎜ ⎟ (VGSL − VTNL ) = ( 0.5 )( 5 − 0.15 − 0.7 ) 2 ⎝ 2 ⎠ ⎝ L ⎠L or I DL = 150.7 μ A Maximum v0 occurs when either A or B is high and C is high. For the two NMOS is series, the effective k N is cut in half, so I DL =

1 ⎡⎛ kn′ ⎞ ⎛ W ⎞ ⎤ 2 ⎢ ⎜ ⎟ ⎥ ⎡ 2 (VGSD − VTND ) VDS − VDS ⎤⎦ 2 ⎣⎜⎝ 2 ⎟⎠ ⎝ L ⎠ D ⎦ ⎣

or 1 ⎡ 35 ⎛ W ⎞ ⎤ ⎡ 2 ⎢ ⋅ ⎜ ⎟ ⎥ 2 ( 5 − 0.7 )( 0.15 ) − ( 0.15 ) ⎤⎦ 2 ⎣ 2 ⎝ L ⎠D ⎦ ⎣ which yields ⎛W ⎞ ⎜ ⎟ = 13.6 ⎝ L ⎠D 150.7 =

b.

P = iD ⋅ VDD = (150.7 )( 5 ) ⇒ P = 753 μ W

TYU16.5 a. v0 (max) occurs when A = B = 1 and C = D = 0 or A = B = 0 and C = D = 1 1 ⎛W ⎞ ⎛W ⎞ 2 2 ⎜ ⎟ (−VTNL ) = ⋅ ⎜ ⎟ ⎡⎣ 2(vI − VTND )vO − vO ⎤⎦ 2 ⎝ L ⎠D ⎝ L ⎠L 1 ⎛W ⎞ (0.5)(1.2)2 = ⋅ ⎜ ⎟ ⎣⎡ 2(5 − 0.7)(0.15) − (0.15) 2 ⎦⎤ 2 ⎝ L ⎠D ⎛W ⎞ ⎛W ⎞ 0.72 = (0.634) ⎜ ⎟ ⇒ ⎜ ⎟ = 1.14 ⎝ L ⎠D ⎝ L ⎠D

b. 2 ⎛ k′ ⎞⎛ W ⎞ ⎛ 35 ⎞ iD = ⎜ n ⎟ ⎜ ⎟ (−VTNL ) 2 = ⎜ ⎟ (0.5) [ −(−1.2) ] ⎝ 2⎠ ⎝ 2 ⎠ ⎝ L ⎠L iD = 12.6 μ A

P = iD ⋅ VDD = (12.6)(5) ⇒ P = 63 μ W

TYU16.6 a. K n = K p = 50 μ A / V 2 VIt = 2.5 V iD (max) = K n (VIt − VTN ) 2 = 50(2.5 − 0.8) 2 ⇒ iD (max) = 145 μ A

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b. K n = K p = 200 μ A / V 2 VIt = 2.5 V iD (max) = (200)(2.5 − 0.8) 2 ⇒ iD (max) = 578 μ A

TYU16.7 a. 5 − 2 + (1)(0.8) VIt = 1+1 VIt = 1.9 V V0 Pt = 3.9 V V0 Nt = 1.1 V b. 3 VIL = 0.8 + ⋅ [5 − 2 − 0.8] ⇒ VIL = 1.63 V 8 1 V0 HU = {2(1.63) + 5 − 0.8 + 2} 2 V0 HU = 4.73 V 5 VIH = 0.8 + (5 − 2 − 0.8) ⇒ VIH = 2.18 V 8 1 V0 LU = {2(2.18) − 5 − 0.8 + 2} 2 V0 LU = 0.275 V c. NM L = VIL − V0 LU = 1.63 − 0.275 ⇒ NM L = 1.35 V NM H = V0 HU − VIH = 4.73 − 2.18 ⇒ NM H = 2.55 V

TYU16.8

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TYU16.9 NMOS − 2 transistors in series Wn = 2 (W ) = 2W PMOS − 2 transistors in series W p = 2 ( 2W ) = 4W

TYU16.10 The NMOS part of the circuit is:

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TYU16.11 The NMOS part of the circuit is:

TYU16.12 Exclusive-OR

A 0 1 0 1

B 0 0 1 1

f 0 1 1 0

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TYU16.13

NMOS conducting for 0 ≤ vI ≤ 4.2 V ⇒ NMOS Conducting: 0 ≤ t ≤ 8.4 s NMOS Cutoff: 8.4 ≤ t ≤ 10 s PMOS cutoff for 0 ≤ vI ≤ 1.2 V ⇒ PMOS Cutoff: 0 ≤ t ≤ 2.4 s PMOS Conducting: 2.4 ≤ t ≤ 10 s

TYU16.14 (a) 1 K ⇒ 32 × 32 array Each row and column requires a 5-bit word ⇒ 6 transistors per row and column, ⇒ 32 × 6 + 32 × 6 = 384 transistors plus buffer transistors. (b) 4 K ⇒ 64 × 64 array Each row and column requires a 6-bit word ⇒ 7 transistors per row and column ⇒ 64 × 7 + 64 × 7 = 896 transistors plus buffer transistors. (c) 16 K ⇒ 128 × 128 array Each row and column requires a 7-bit word ⇒ 8 transistors per row and column ⇒ 128 × 8 + 128 × 8 = 2048 transistors plus buffer transistors. TYU16.15 From Equation (16.84) (W / L )nA 2 (VDDVTN ) − 3VTN2 2(2.5)(0.4) − 3(0.4) 2 = = = 0.526 2 2 (W / L )n1 (VDD − 2VTN ) ( 2.5 − 2(0.4) ) From Equation (16.86)

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(W / L ) p (W / L )nB

=

2 ⎡ 2(2.5)(0.4) − 3(0.4) 2 ⎤ kn′ 2 (VDDVTN ) − 3VTN ⋅ = (2.5) ⎢ ⎥ = 0.862 2 k ′p (2.5 − 0.4) 2 (VDD + VTP ) ⎣ ⎦

⎛W ⎞ ⎛W ⎞ So ⎜ ⎟ of transmission gate device must be < 0.526 times the ⎜ ⎟ of the NMOS transistors in the L ⎝ ⎠ ⎝L⎠ ⎛W ⎞ ⎛W ⎞ inverter cell. The ⎜ ⎟ of the PMOS transistors must be < 0.862 times the ⎜ ⎟ of the transmission gate ⎝L⎠ ⎝L⎠ W W ⎛ ⎞ ⎛ ⎞ devices. Then the ⎜ ⎟ of the PMOS devices must be < 0.453 times ⎜ ⎟ of NMOS devices in cell. ⎝L⎠ ⎝L⎠

TYU16.16 Initial voltage across the storage capacitor = VDD − VTN = 3 − 0.5 = 2.5 V . Now dV I −I = C or V = − ⋅ t + K dt C 2.5 = 1.25 V , and C = 0.05 pF . Then where K = 2.5 V , t = 1.5 ms, V = 2 I (1.5 × 10−3 ) 1.25 = 2.5 − ⇒ (0.05 × 10−12 ) I = 4.17 × 10−11 A ⇒ I = 41.7 pA

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Chapter 16 Problem Solutions 16.1 (a) ΔVTN = Cax =

2e ∈s N a Cax

⎡ 2φ fp + VSB − 2φ fp ⎤ ⎣ ⎦

∈ax (3.9)(8.85 × 10 −14 ) = = 7.67 × 10−8 450 × 10−8 tax

2e ∈s N a = ⎡⎣ 2 (1.6 × 10 −19 ) (11.7 ) ( 8.85 × 10−14 )( 8 × 1015 ) ⎤⎦ Then 5.15 × 10−8 ⎡ ΔVTN = ⋅ 2(0.343) + VSB − 2(0.343) ⎤⎦ 7.67 × 10−8 ⎣ For VSB = 1 V :

1/ 2

= 5.15 × 10 −8

ΔVTN = 0.671 ⎡⎣ 1.686 − 0.686 ⎤⎦ ⇒ ΔVTN = 0.316 V For VSB = 1 V : ΔVTN = 0.671 ⎣⎡ 2.686 − 0.686 ⎦⎤ ⇒ ΔVTN = 0.544 V (b) For VGS = 2.5 V, VDS = 5 V, transistor biased in the saturation region. ID For VSB ID For VSB

= K n (VGS − VTN ) 2 = 0, = 0.2(2.5 − 0.8) 2 = 0.578 mA = 1,

I D = 0.2 ( 2.5 − [ 0.8 + 0.316]) = 0.383 mA 2

For VSB = 2, I D = 0.2 ( 2.5 − [ 0.8 + 0.544]) = 0.267 mA 2

16.2 (a) ID =

VDD − vO = K n ⎡⎣ 2(VGS − VTN )vO − vO2 ⎤⎦ RD

5 − (0.1) 2 = K n ⎡ 2 ( 5 − 0.8 )( 0.1) − ( 0.1) ⎤ 3 ⎣ ⎦ 40 × 10 −5 8 × 10 ⎛ W ⎞ or K n = 1.476 × 10−4 A / V 2 = ⎜ ⎟ 2 ⎝L⎠ ⎛W ⎞ So that ⎜ ⎟ = 3.69 ⎝L⎠ b. From Equation (16.10).

K n RD [VIt − VTN ] + [VIt − VTN ] − VDD = 0 2

(0.1476)(40) [VIt − 0.8] + [VIt − 0.8] − 5 = 0 2

−1 ± (1) 2 + 4(0.1476)(40)(5) 2(0.1476)(40) or [VIt − 0.8] = 0.839 or [VIt − 0.8] =

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So that VIt = 1.64 V P = I D (max) ⋅ VDD

5 − (0.1) = 0.1225 mA 40 or P = 0.6125 mW

and I D (max) =

16.3 a. From Equation (16.10), the transistor point is found from K n RD (VIt − VTN ) 2 + (VIt − VTN ) − VDD = 0 K n = 50 μ A / V 2 , RD = 20 k Ω, VTN = 0.8 V (0.05)(20)(VIt − VTN ) 2 + (VIt − VTN ) − 5 = 0 −1 ± 1 + 4(0.05)(20)(5) = 1.79 V So VIt = 2.59 V 2(0.05)(20) V0t = 1.79 V Output voltage for vI = 5 V is determined from Equation (16.12): VIt − VTN =

v0 = 5 − (0.05)(20) ⎡⎣ 2(5 − 0.8)v0 − v02 ⎤⎦ v02 − 9.4v0 + 5 = 0 So v0 =

b.

9.4 ± (9.4) 2 − 4(1)(5) = 0.566 V 2(1) For RD = 200 kΩ,

−1 ± 1 + 4(0.05)(200)(5) = 0.659 V So VIt = 1.46 V 2(0.05)(200) V0t = 0.659 V

(VIt − VTN ) =

v0 = 5 − (0.05)(200) ⎡⎣ 2 ( 5 − 0.8 ) v0 − v02 ⎤⎦ or 10v02 − 85v0 + 5 = 0 v0 =

16.4 (a)

85 ±

(85 ) − 4 (10 )( 5 ) = 0.0592 V 2 (10 ) 2

P = IV

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0.25 = I (33) ⇒ I = 75.76 μA 3.3 − 0.15 R= ⇒ R = 41.6 K 0.07576 2 ⎛ k ′ ⎞⎛ W ⎞ I = ⎜ n ⎟ ⎜ ⎟ (VGS − VTN ) 2 L ⎝ ⎠⎝ ⎠ 2 ⎛ 80 ⎞ ⎛ W ⎞ ⎛W ⎞ 75.76 = ⎜ ⎟ ⎜ ⎟ ( 3.3 − 0.8 ) ⇒ ⎜ ⎟ = 0.303 ⎝ 2 ⎠⎝ L ⎠ ⎝L⎠ (b) VDS (sat) = VGS − VTN V − VDS (sat) 2 I D = K n (VGS − VTN ) = DD R 3.3 − (VGS − 0.8 ) ⎛ 0.08 ⎞ 2 ⎜ ⎟ ( 0.303) (VGS − 1.6VGS + 0.64 ) = 41.6 ⎝ 2 ⎠ 0.504 (VGS2 − 1.6VGS + 0.64 ) = 4.1 − VGS 0.504VGS2 + 0.1936VGS − 3.777 = 0 −0.1936 ± 0.03748 + 4(0.504)(3.777) 2(0.504) = 2.55 V

VGS = VGS

For 0.8 ≤ VGS ≤ 2.55 V Transistor biased in saturation region 16.5 (a) P = I ⋅ VDD 0.25 = I (3.3) ⇒ I = 75.76 μA For Sat Load 2 ⎛ 80 ⎞ ⎛ W ⎞ ⎛W ⎞ I = 75.76 = ⎜ ⎟ ⎜ ⎟ ( 3.3 − 0.15 − 0.8 ) ⇒ ⎜ ⎟ = 0.343 2 L ⎝ ⎠ ⎝ ⎠L ⎝ L ⎠L 2 2 ⎛ 80 ⎞ ⎛ 80 ⎞ ⎛ W ⎞ ⎜ ⎟ ( 0.343)( 3.3 − 0.15 − 0.8 ) = I = 75.76 = ⎜ ⎟ ⎜ ⎟ ⎡⎣ 2 ( 2.5 − 0.8 )( 0.15 ) − ( 0.15 ) ⎤⎦ 2 2 L ⎝ ⎠ ⎝ ⎠ ⎝ ⎠D W ⎛ ⎞ ⎜ ⎟ = 3.89 ⎝ L ⎠D

Eq 16.21 ⎛ 3.89 ⎞ 3.3 − 0.8 + 0.8 ⎜⎜1 + ⎟⎟ 0.343 ⎝ ⎠ = 5.994 VIt = 4.3677 3.89 1+ 0.343 0.8 ≤ VGS ≤ 1.372 V

16.6 (a) From Equation (16.23) KD ⎡ K 2 2 2 ( 3 − 0.5 )( 0.25 ) − ( 0.25 ) ⎤ = ( 3 − 0.25 − 0.5 ) ⇒ D = 4.26 ⎣ ⎦ KL KL

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KD ⎡ K 2 2 2 ( 2.5 − 0.5 )( 0.25 ) − ( 0.25 ) ⎤ = ( 3 − 0.25 − 0.5 ) ⇒ D = 5.4 ⎣ ⎦ KL KL

(b)

iD = K L (VGSL − VTNL ) = K L (VDD − vO − VTNL ) 2 2

(c)

⎛ 0.080 ⎞ 2 =⎜ ⎟ (1)(3 − 0.25 − 0.5) ⇒ iD = 0.203 mA ⎝ 2 ⎠ P = iD ⋅ VDD = (0.203)(3) ⇒ P = 0.608 mW

for both parts (a) and (b). 16.7 P = 0.4 mW = iD ⋅ VDD = iD (3) ⇒ iD = 0.1333 mA iD = K L (VDD − vO − VTNL ) 2 2 ⎛ 0.080 ⎞ ⎛ W ⎞ ⎛W ⎞ 0.1333 = ⎜ ⎟ ⎜ ⎟ ( 3 − 0.1 − 0.5 ) = ( 0.2304 ) ⎜ ⎟ 2 L ⎝ ⎠ ⎝ ⎠L ⎝ L ⎠L ⎛W ⎞ So ⎜ ⎟ = 0.579 ⎝ L ⎠L

KD ⎡ 2 2 2 ( 2.5 − 0.5 )( 0.1) − ( 0.1) ⎤ = ( 3 − 0.1 − 0.5 ) ⎣ ⎦ KL ⇒

KD ⎛W ⎞ = 14.8 so that ⎜ ⎟ = 8.55 KL ⎝ L ⎠D

VIt =

(

3 − 0.5 + 0.5 1 + 14.8

)

1 + 14.8 or VIt = 1.02 V , VOt = 0.52 V

16.8 We have KD 2 ⎡⎣ 2 ( vI − VTND ) vO − vO2 ⎤⎦ = (VDD − vO − VTNL ) KL

(W / L )D ⎡ 2 2 2 (VDD − VTN − VTN )( 0.08VDD ) − ( 0.08VDD ) ⎤ = (VDD − 0.08VDD − VTN ) ⎣ ⎦ (W / L )L (W / L )D ⎡ 2 2 2 ⎤⎦ = ⎡⎣( 0.92 − 0.2 )VDD ⎤⎦ = 0.5184VDD 2 (VDD − 2 ( 0.2 )VDD ) ( 0.08VDD ) − 0.0064VDD ⎣ (W / L )L (W / L )D (W / L )D = 5.4 [0.096] = 0.5184 ⇒ (W / L )L (W / L )L 16.9 VOH = VB − VTN = Logic 1 So (a) VB = 4 V ⇒ VOH (b) VB = 5 V ⇒ VOH (c) VB = 6 V ⇒ VOH (d) VB = 7 V ⇒ VOH For vI = VOH

= 3V = 4V = 5V = 5 V ,since VDS = 0

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K D ⎡⎣ 2 ( vI − VT ) vO − vO2 ⎤⎦ = K L [VB − vO − VT ]

2

Then (a)

(1) ⎡⎣ 2 ( 3 − 1)VOL − VOL2 ⎤⎦ = ( 0.4 ) [ 4 − VOL − 1]

(b)

(1) ⎣⎡ 2 ( 4 − 1)VOL − V

(c)

(1) ⎡⎣ 2 ( 5 − 1)VOL − VOL2 ⎤⎦ = ( 0.4 ) [6 − VOL − 1]

2

2 OL

⇒ VOL = 0.657 V

⎦⎤ = ( 0.4 ) [5 − VOL − 1] ⇒ VOL = 0.791 V 2

2

⇒ VOL = 0.935 V

(d) Load in non-sat region iDD = iOL 2 2 ⎤⎦ = ( 0.4 ) ⎡ 2 ( 7 − VOL − 1)( 5 − VOL ) − ( 5 − VOL ) ⎤ (1) ⎡⎣ 2 ( 5 − 1) VOL − VOL ⎣ ⎦ 2 2 8VOL − VOL = ( 0.4 ) ⎡⎣ 2 ( 6 − VOL )( 5 − VOL ) − ( 25 − 10VOL + VOL )⎤⎦ 2 = ( 0.4 ) ⎡⎣ 2 ( 30 − 11VOL + VOL ) − 25 + 10VOL − VOL2 ⎤⎦ 2 2 ⎤⎦ = ( 0.4 ) ⎡⎣60 − 22VOL + 2VOL − 25 + 10VOL − VOL 2 2 8VOL − VOL = 14 − 4.8VOL + 0.4VOL 2 1.4VOL − 12.8VOL + 14 = 0

VOL =

12.8 ± 163.84 − 4 (1.4 )(14 ) 2 (1.4 )

VOL = 1.27V For load VDS ( sat ) = 7 − 1.27 − 1 = 4.73V VDS = 5 − 1.27 = 3.73 non-sat

16.10 a.

For load VOt = VDD + VTNL = 5 − 2 = 3 V

KD ⋅ (VIt − VTND ) = −VTNL KL 500 (VIt − 0.8 ) = − ( −2 ) 100 ⇒ VIt = 1.69 V ⎫ ⎬ Load VOt = 3 V ⎭

Driver: VOt = VIt − VTND = 1.69 − 0.8 = 0.89 V VIt = 1.69 V ⎫ ⎬ Driver V0t = 0.89 V ⎭ From Equation (16.29(b)):

b. 2 500 ⋅ ⎣⎡ 2(5 − 0.8)v0 − v02 ⎦⎤ = ⎣⎡ − ( −2 ) ⎦⎤ 100 5v02 − 42v0 + 4 = 0 v0 =

c.

42 ±

( 42 ) − 4 ( 5)( 4 ) ⇒ v0 = 0.0963 V 2 (5) 2

2

iD = K L ( −VTNL ) = 100 ⎡⎣ − ( −2 ) ⎤⎦ ⇒ iD = 400 μ A 2

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16.11 2 2 ⎛ 500 ⎞ ⎡ ⎜ ⎟ ⎣ 2 ( 3 − 0.5 )( 0.1) − ( 0.1) ⎤⎦ = ( −VTNL ) ⎝ 50 ⎠ So

( −VTNL )

2

= 4.9 ⇒ VTNL = −2.21 V

16.12 (a) P = iD ⋅ VDD 150 = iD ⋅ ( 3) ⇒ iD = 50 μ A iD = K L (−VTNL ) 2 2 ⎛ 80 ⎞ ⎛ W ⎞ ⎛W ⎞ 50 = ⎜ ⎟ ⎜ ⎟ ⎡⎣ − ( −1) ⎦⎤ ⇒ ⎜ ⎟ = 1.25 2 L ⎝ ⎠ ⎝ ⎠L ⎝ L ⎠L 2 KD ⎡ 2 2 ( 3 − 0.5 )( 0.1) − ( 0.1) ⎤ = ⎡⎣ − ( −1) ⎤⎦ ⎦ KL ⎣

K D (W / L ) D ⎛W ⎞ = = 2.04 ⇒ ⎜ ⎟ = 2.55 K L (W / L ) L ⎝ L ⎠D For the Load: VOt = VDD + VTNL = 3 − 1 ⇒ VOt = 2 V

2.04 (VIt − 0.5 ) = ⎡⎣ − ( −1) ⎤⎦ ⇒ VIt = 1.20 V

For the Driver: VOt = VIt − VTND = 1.20 − 0.5 ⇒ VOt = 0.70 V VIt = 1.20 V

(b)

NM L = VIL − VOLU NM H = VOHU − VIH

VIL = 0.5 +

⎡⎣ − ( −1) ⎤⎦ = 0.902 V ( 2.04 )(1 + 2.04 )

2 ⎡ − ( −1) ⎤⎦ = 1.31 V VIH = 0.5 + ⎣ 3 ( 2.04 ) Then VOHU = ( 3 − 1) + ( 2.04 )( 0.902 − 0.5 ) = 2.82 V

VOLU =

(1.31 − 0.5)

= 0.405 V 2 NM L = 0.902 − 0.405 ⇒ NM L = 0.497 V NM H = 2.82 − 1.31 ⇒ NM H = 1.51 V

16.13 a. From Equation (16.29(b)): 2 ⎛W ⎞ ⎡ ⎛W ⎞ 2 ⎜ ⎟ ⎣ 2 ( 2.5 − 0.5 )( 0.05 ) − ( 0.05 ) ⎤⎦ = ⎜ ⎟ [− ( −1)] L L ⎝ ⎠D ⎝ ⎠L W ⎛ ⎞ ⎜ ⎟ =1 ⎝ L ⎠L

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⎛W ⎞ ⎜ ⎟ = 5.06 ⎝ L ⎠D

Then

2 ⎛ 80 ⎞ iD = ⎜ ⎟ (1) ⎡⎣ − ( −1) ⎤⎦ ⎝ 2⎠ or iD = 40 μ A

b.

P = iD ⋅ VDD = ( 40 )( 2.5 ) ⇒ P = 100 μ W

16.14 a.

vI = 0.5 V ⇒ iD = 0 ⇒ P = 0 vI = 5 V, From Equation (16.12),

i. ii.

v0 = 5 − ( 0.1)( 20 ) ⎡⎣ 2 ( 5 − 1.5 ) v0 − v02 ⎤⎦ 2v02 − 15v0 + 5 = 0 v0 =

15 ±

(15 ) − 4 ( 2 )( 5 ) ⇒ v0 = 0.35 V 2 ( 2) 2

5 − 0.35 = 0.2325 mA 20 P = iD ⋅ VDD = ( 0.2325 )( 5 ) ⇒ P = 1.16 mW

iD =

b. ii.

vI = 0.25 V ⇒ iD = 0 ⇒ P = 0 i. vI = 4.3 V, From Equation (16.23),

100 ⎡⎣ 2 ( 4.3 − 0.7 ) v0 − v02 ⎤⎦ = 10 [5 − v0 − 0.7 ]

2

10 ⎡⎣7.2v0 − v02 ⎤⎦ = 18.49 − 8.6v0 + v02

Then 11v02 − 80.6v0 + 18.49 = 0 v0 =

80.6 ±

(80.6 ) − 4 (11)(18.49 ) ⇒ v0 = 0.237 V 2 (11) 2

Then iD = 10 [5 − 0.237 − 0.7 ] = 165 μ A 2

P = iD ⋅ VDD = (165 )( 5 ) ⇒ P = 825 μ W

c.

i. ii.

vI = 0.03 V ⇒ iD = 0 ⇒ P = 0 vI = 5 V 2

iD = K L ( −VTNL ) = (10 ) ⎣⎡ − ( −2 ) ⎦⎤ = 40 μ A 2

P = iD ⋅ VDD = ( 40 )( 5 ) ⇒ P = 200 μ W

16.15 vo1 = 3.8V Load & Driver in Sat, region, ML1, MD1

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iDL = iDD 2 2 ⎛W ⎞ ⎛W ⎞ ⎜ ⎟ ( vGSL − VTNL ) = ⎜ ⎟ ( vGSD − VTND ) ⎝ L ⎠L ⎝ L ⎠D

(1) ( 5 − vo1 − 0.8 ) = (10 )( vI − 0.8 ) 2

0.16 = 10 ( vI − 0.8 )

2

2

vI = 0.9265V

Now MD2: Non Sat and ML2: Sat iDL = iDD 2 ⎛W ⎞ ⎛W ⎞ 2 ⎜ ⎟ ( vGSL − VTNL ) = ⎜ ⎟ ⎣⎡ 2 ( vo1 − VTND ) vo 2 − vo 2 ⎦⎤ ⎝ L ⎠L ⎝ L ⎠D

(1)( 5 − vo 2 − 0.8 ) ( 4.2 − vo 2 )

2

2

= (10 ) ⎡⎣ 2 ( 3.8 − 0.8 ) vo 2 − vo22 ⎤⎦

= 10 ⎡⎣ 6vo 2 − vo22 ⎤⎦

17.64 − 8.4vo 2 + vo22 = 60vo 2 − 10vo22 11vo22 − 68.4vo 2 + 17.64 = 0 vo 2 =

68.4 ± 4678.56 − 4 (11)(17.64 ) 2 (11)

vo 2 = 0.270 V

16.16 a. VIH

From Equation (16.41), 2 ⎡ − ( −2 ) ⎤⎦ = 0.8 + ⎣ ⇒ VIH = 1.95 V = v01 3( 4)

M D 2 in non-saturation and M L 2 in saturation. 2 ⎤⎦ = K L ( −VTNL ) K D ⎡⎣ 2 ( v01 − VTND ) v02 − v02 2 4 ⎣⎡ 2 (1.95 − 0.8 ) v02 − v02 ⎦⎤ = (1) ⎡⎣ − ( −2 ) ⎤⎦

2

2

2 4v02 − 9.2v02 + 4 = 0

v02 =

9.2 ±

( 9.2 ) − 4 ( 4 )( 4 ) ⇒ v02 = 0.582 V 2 ( 4) 2

Both M D1 and M L1 in saturation region. From Equation (16.28(b)). 4 ⋅ ( vI − 0.8 ) = − ( −2 )

or vI = 1.8 V b.

VIL = 0.8 +

( +2 ) = 1.25 V = v01 4 (1 + 4 )

M D 2 in saturation, M L 2 in non-saturation

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2 2 K D [ vO1 − VTND ] = K L ⎡ 2 ( −VTNL )( 5 − vO 2 ) − ( 5 − vO 2 ) ⎤ ⎣ ⎦

4 (1.25 − 0.8 ) = 2 ( 2 )( 5 − v02 ) − ( 5 − v02 ) 2

( 5 − v02 ) 5 − v02 =

2

2

− 4 ( 5 − v02 ) + 0.81 = 0

( 4)

4±

2

− 4 (1)( 0.81)

2 (1)

= 0.214 V

so v02 = 4.786 V To find vI : 4 ( v01 − 0.8 ) = (1) ( − ( −2 ) ) 2

2

v01 − 0.8 = 1 v01 − 1.8 = V VIH = 1.95 V, VIL = 1.25 V c.

16.17 a. i. Neglecting the body effect, v0 = VDD − VTN Assume VDD = 5 V, then v0 = 4.2 V ii.

Taking the body effect into account: From Problem 16.1. VTN = VTN 0 + 0.671 ⎡⎣ 0.686 + VSB − 0.686 ⎤⎦ and VSB = v0 Then v0 = 5 − ⎡ 0.8 + 0.671 0.686 + v0 − 0.686 ⎤ ⎣ ⎦

(

)

v0 = 4.756 − 0.671 0.686 + v0 0.671 0.686 + v0 = 4.756 − v0 0.450 ( 0.686 + v0 ) = 22.62 − 9.51v0 + v02 v02 − 9.96v0 + 22.3 = 0 v0 =

b.

9.96 ±

( 9.96 )

2

− 4 ( 22.3)

⇒ v0 = 3.40 V 2 PSpice results similar to Figure 16.13(a).

16.18 Results similar to Figure 16.13(b). 16.19 a. M X on, M Y cutoff. From Equation (16.29(b)): 2 KD ⎡ 2 2 ( 5 − 0.8 )( 0.2 ) − ( 0.2 ) ⎤ = ⎣⎡ − ( −2 ) ⎦⎤ ⎣ ⎦ KL or b.

KD = 2.44 KL

For v X = vY = .5 V

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2 ( 2.44 ) ⎡⎣ 2 ( 5 − 0.8 ) v0 − v02 ⎤⎦ = ⎡⎣ − ( −2 ) ⎤⎦

2

4.88v02 − 41.0v0 + 4 = 0 v0 =

41 ±

( 41) − 4 ( 4.88 )( 4 ) 2 ( 4.88 ) 2

or v0 = 0.0987 V c. 2 ⎛ 80 ⎞ iD = ⎜ ⎟ (1) ⎡⎣ − ( −2 ) ⎤⎦ = 160 μ A 2 ⎝ ⎠ P = (160 )( 5 ) ⇒ P = 800 μ W

for both parts (a) and (b). 16.20 (a)

Maximum value of vO in low state- when only one input is high, then,

2 KD ⎡ 2 2 ( 3 − 0.5 )( 0.1) − ( 0.1) ⎤ = ⎡⎣ − ( −1) ⎤⎦ ⎦ KL ⎣

KD = 2.04 KL

(b) P = iD ⋅ VDD 0.1 = iD (3) ⇒ iD = 33.3 μ A 2 ⎛ k′ ⎞⎛ W ⎞ iD = ⎜ n ⎟ ⎜ ⎟ ( −VTNL ) L 2 ⎝ ⎠ ⎝ ⎠L 2 ⎛ 80 ⎞⎛ W ⎞ ⎛W ⎞ 33.3 = ⎜ ⎟⎜ ⎟ ⎡⎣ − ( −1) ⎤⎦ ⇒ ⎜ ⎟ = 0.8325 ⎝ 2 ⎠⎝ L ⎠ L ⎝ L ⎠L

⎛W ⎞ Then ⎜ ⎟ = 1.70 ⎝ L ⎠D

(c)

2

3 ( 2.04 ) ⎡⎣ 2 ( 3 − 0.5 ) vO − vO2 ⎤⎦ = ⎡⎣ − ( −1) ⎤⎦ ⇒ vO = 0.0329 V

16.21 (a)

One driver in non-sat, 2 2 ⎤⎦ I D = K L ( −VTNL ) = K D ⎡⎣ 2 (VGS − VTND ) VDSD − VDSD

2 KD ⎡ 2 ⎤ ⎣⎡ − ( −1) ⎦⎤ = K ⎣ 2 ( 3.3 − 0.5 )( 0.1) − ( 0.1) ⎦ L

KD = 1.82 KL

(b)

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P = ± VDD 0.1 = ± ( 3.3) ⇒ I = 30.3 μ A 2 ⎛ 80 ⎞ ⎛ W ⎞ ⎛W ⎞ 30.3 = I = ⎜ ⎟ ⎜ ⎟ ⎡⎣ − ( −1) ⎤⎦ ⇒ ⎜ ⎟ = 0.7575 ⎝ 2 ⎠ ⎝ L ⎠L ⎝ L ⎠L

⎛W ⎞ ⎜ ⎟ = 1.38 ⎝ L ⎠D

(c) 0.1 = 0.05 V 2 0.1 = 0.0333 V Three inputs High, vo ≈ 3 0.1 = 0.025 V Four inputs High, vo ≈ 4

Two inputs High, vo ≈

(i) (ii)

(iii)

16.22 a. P = iD ⋅ VDD 250 = iD ( 5 ) ⇒ iD = 50 μΑ ⎛ k' ⎞⎛W ⎞ 2 iD = ⎜ n ⎟ ⎜ ⎟ [ −VTNL1 ] 2 L ⎝ ⎠ ⎝ ⎠ ML1 2 ⎛ 60 ⎞ ⎛ W ⎞ 50 = ⎜ ⎟ ⎜ ⎟ ⎡⎣ − ( −2 ) ⎤⎦ ⎝ 2 ⎠ ⎝ L ⎠ ML1 ⎛W ⎞ So that ⎜ ⎟ = 0.417 ⎝ L ⎠ ML1

KD 2 ⎡⎣ 2 ( vI − VTND ) vO − vO2 ⎤⎦ = [ −VTNL ] KL 2 KD ⎡ 2 2 ( 5 − 0.8 )( 0.15 ) − ( 0.15 ) ⎤ = ⎡⎣ − ( −2 ) ⎤⎦ ⎣ ⎦ KL

or

KD ⎛W ⎞ = 3.23 ⇒ ⎜ ⎟ = 1.35 KL ⎝ L ⎠ MD1

b. For v X = vY = 0 ⇒ v01 = 5 and v03 = 4.2 Then 2 K D 2 ⎡⎣ 2 ( vO1 − VTND ) vO 2 − vO2 2 ⎤⎦ + K D 3 ⎡⎣ 2 ( vO 3 − VTND ) vO 2 − vO2 2 ⎤⎦ = K L 2 [ −VTNL 2 ] K D 2 ∝ 8, K D 3 ∝ 8, K L 2 ∝ 1 2 2 ⎤⎦ + 8 ⎡⎣ 2 ( 4.2 − 0.8 ) v02 − v02 ⎤⎦ = (1) ⎡⎣ − ( −2 ) ⎤⎦ 8 ⎡⎣ 2 ( 5 − 0.8 ) v02 − v02

2

2 2 + 54.4v02 − 8v02 =4 67.2v02 − 8v02 Then 16v02 − 121.6v0 + 4 = 0

v02 =

121.6 ±

(121.6 ) − 4 (16 )( 4 ) 2 (16 ) 2

So v02 = 0.0330 V 16.23

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a.

We can write

2 2 ⎤⎦ = K y ⎡⎣ 2 ( vY − vDSX − VTN ) vDSY − vDSY ⎤⎦ = K L [VDD − vO − VTN ] K x ⎡⎣ 2 ( v X − VTN ) vDSX − vDSX

2

where v0 = vDSX + vDSY We have v X = vY = 9.2 V , VDD = 10 V , VTN = 0.8 V 2 2 and vDSY terms. Let v0 ≈ 2vDSX . Then from the first and As a good first approximation, neglect the vDSX third terms in the above equation.

9 ⎡⎣ 2 ( 9.2 − 0.8 ) vDSX ⎤⎦ ≅ (1)(10 − 2vDSX − 0.8 )

(151.2 ) vDSX

2

≅ 84.64 − 36.8vDSX

So that vDSX = 0.450 V From the first and second terms of the above equation. 9 ⎡⎣ 2 ( 9.2 − 0.8 ) vDSX ⎤⎦ ≅ 9 ⎡⎣ 2 ( 9.2 − vDSX − 0.8 ) vDSY ⎤⎦

or (16.8)( 0.45) = 2 ( 9.2 − 0.45 − 0.8) vDSY which yields vDSY = 0.475 V Then v0 = vDSX + vDSY = 0.450 + 0.475 or v0 = 0.925 V We have vGSX = 9.2 V and vGSY = 9.2 − vDSX = 9.2 − 0.45 or vGSY = 8.75 V b. Since v0 is close to ground potential, the body effect will have minimal effect on the results. From a PSpice analysis: For part (a): vDSX = 0.462 V, vDSY = 0.491 V, v0 = 0.9536 V, vGSX = 9.2 V, and vGSY = 8.738 V For part (b): vDSX = 0.441 V, vDSY = 0.475 V, v0 = 0.9154 V, vGSX = 9.2 V, and vGSY = 8.759 V 16.24 a. We can write 2 2 2 ⎤⎦ = K y ⎡⎣ 2 ( vY − vDSX − VTNY ) vDSY − vDSY ⎤⎦ = K L [ −VTNL ] K x ⎡⎣ 2 ( v X − VTNX ) vDSX − vDSX 2 From the first and third terms, (neglect vDSX ),

4 ⎡⎣ 2 ( 5 − 0.8 ) vDSX ⎤⎦ = (1) ⎡⎣ − ( −1.5 ) ⎤⎦

2

or vDSX = 0.067 V 2 From the second and third terms, (neglect vDSY ),

4 ⎡⎣ 2 ( 5 − 0.067 − 0.8 ) vDSY ⎤⎦ = (1) ⎡⎣ − ( −1.5 ) ⎤⎦

2

or vDSY = 0.068 V Now vGSX = 5, vGSY = 5 − 0.067 ⇒ vGSY = 4.933 V and v0 = vDSX + vDSY ⇒ v0 = 0.135 V Since v0 is close to ground potential, the body-effect has little effect on the results. 16.25

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(a)

0.2 = 0.05 V 4 2 ⎤⎦ − VDSD

We have VDS of each driver ≈

K L [ −VTNL ] = K D ⎡⎣ 2 (VGSD − VTN ) VDSD 2

2 KD ⎡ 2 ⎤ ⎣⎡ − ( −1) ⎦⎤ = K ⎣ 2 ( 3.3 − 0.4 )( 0.05 ) − ( 0.05 ) ⎦ L

KD = 3.478 KL

(b) P = I VDD 0.15 = I ( 3.3) ⇒ I = 45.45 μ A 2 ⎛ 80 ⎞ ⎛ W ⎞ ⎛W ⎞ 45.45 = ⎜ ⎟ ⎜ ⎟ ⎡⎣ − ( −1) ⎤⎦ ⇒ ⎜ ⎟ = 1.14 ⎝ 2 ⎠ ⎝ L ⎠L ⎝ L ⎠L

⎛W ⎞ ⎜ ⎟ = 3.95 ⎝ L ⎠D

16.26 Complement of (B AND C) OR A ⇒ ( B ⋅ C ) + A 16.27 Considering a truth table, we find A B Y 0 0 0 0 1 1 1 0 1 1 1 0 which shows that the circuit performs the exclusive-OR function. 16.28 ( A + B )(C + D) 16.29 (a)

Carry-out = A • ( B + C ) + B • C

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(b)

For vO1 = Low = 0.2 V

2 KD ⎡ 2 2 ( 5 − 0.8 )( 0.2 ) − ( 0.2 ) ⎤ = ⎣⎡ − ( −1.5 ) ⎦⎤ ⇒ ⎦ KL ⎣

⎛W ⎞ ⎛W ⎞ For ⎜ ⎟ = 1, then ⎜ ⎟ = 1.37 ⎝ L ⎠L ⎝ L ⎠D ⎛W ⎞ So, for M 6 : ⎜ ⎟ = 1.37 ⎝ L ⎠6

To achieve the required composite conduction parameter, ⎛W ⎞ For M 1 − M 5 : ⎜ ⎟ = 2.74 ⎝ L ⎠1− 5 16.30

AE: VDS ≈ 0.075 2 2 ⎛W ⎞ (1) ⎡⎣ − ( −1) ⎤⎦ ≅ ⎜ ⎟ ⎡ 2 ( 3.3 − 0.4 )( 0.075 ) − ( 0.075 ) ⎤ ⎣ ⎦ L ⎝ ⎠D ⎛W ⎞ ⎛W ⎞ ⎛W ⎞ ⎛W ⎞ ⎛W ⎞ ⎜ ⎟ = ⎜ ⎟ = 2.33 ⇒ ⎜ ⎟ = ⎜ ⎟ = ⎜ ⎟ = 4.66 ⎝ L ⎠ A ⎝ L ⎠E ⎝ L ⎠ B ⎝ L ⎠C ⎝ L ⎠ D

16.31 Not given

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16.32 a.

From Equation (16.43), 5 − 0.8 + 0.8 vI = VIt = = VIt = 2.5 V 1+1 p − channel, V0 Pt = 2.5 − ( −0.8 ) ⇒ V0 Pt = 3.3 V n − channel, V0 Nt = 2.5 − 0.8 ⇒ V0 Nt = 1.7 V

c

For vI = 2 V, NMOS in saturation and PMOS in nonsaturation. From Equation (16.49),

2 = ⎡ 2 ( 5 − 2 − 0.8 )( 5 − v0 ) − ( 5 − v0 ) ⎤ ⎣ ⎦ 1.44 = 4.4(5 − v0 ) − (5 − v0 ) 2

( 2 − 0.8)

2

So ( 5 − v0 ) − 4.4 ( 5 − v0 ) + 1.44 = 0 2

( 5 − v0 ) =

4.4 ±

( 4.4 )

2

− 4 (1)(1.44 )

2 or 5 − v0 = 0.356 ⇒ v0 = 4.64 V

By symmetry, for vI = 3 V, v0 = 0.356 V 16.33 (a)

⎛ 80 ⎞ K n = ⎜ ⎟ ( 2 ) = 80 μ A / V 2 ⎝ 2⎠ ⎛ 40 ⎞ K p = ⎜ ⎟ ( 4 ) = 80 μ A / V 2 ⎝ 2 ⎠

VDD + VTP +

(i)

VIt =

Kn ⋅ VTN Kp Kn Kp

1+

=

3.3 − 0.4 + (1)(0.4) 1+1

VIt = 1.65 V

PMOS: VOt = VIt − VTP = 1.65 − ( −0.4 ) ⇒ VOt = 2.05 V NMOS: VOt = VIt − VTN = 1.65 − ( 0.4 ) ⇒ VOt = 1.25 V (iii) For vO = 0.4 V : NMOS: Non-sat: PMOS:Sat 2 ⎤⎦ = K p [VSGP + VTP ] K n ⎡⎣ 2 (VGSN − VTN )VDS − VDS

2

2 ( vI − 0.4 )( 0.4 ) − ( 0.4 ) = ( 3.3 − vI − 0.4 ) ⇒ vI = 1.89 V 2

2

For vO = 2.9 V , By symmetry vI = 1.65 − (1.89 − 1.65 ) ⇒ vI = 1.41 V

(b)

⎛ 80 ⎞ K n = ⎜ ⎟ ( 2 ) = 80 μ A/V 2 ⎝ 2⎠ ⎛ 40 ⎞ K p = ⎜ ⎟ ( 2 ) = 40 μ A/V 2 ⎝ 2 ⎠

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80 ⋅ ( 0.4 ) 40 ⇒ VIt = 1.44 V 80 1+ 40

3.3 − 0.4 +

VIt =

(i)

PMOS: VOt = 1.44 − ( −0.4 ) ⇒ VOt = 1.84 V NMOS: VOt = 1.44 − 0.4 ⇒ VOt = 1.04 V For vO = 0.4 V

(iii)

⎤ = 40 3.3 − vI − 0.4]2 ⇒ vI = 1.62 V ⎦ ( )[ For vO = 2.9 V : NMOS: Sat, PMOS: Non-sat

(80 ) ⎣⎡ 2 ( vI − 0.4 )( 0.4 ) − ( 0.4 ) (80 ) [vI − 0.4]

2

16.34 (a) (i)

2 = ( 40 ) ⎡ 2 ( 3.3 − vI − 0.4 )( 0.4 ) − ( 0.4 ) ⎤ ⇒ vI = 1.18 V ⎣ ⎦

From Eq. (16.43), switching voltage VDD + VTP +

vI t =

(ii)

2

Kn ⋅ VTN Kp

Kn 1+ Kp

=

2 ( 4) ( 0.4 ) 3.2266 12 = ⇒ vIt = 1.776 V 1.8165 2 ( 4) 1+ 12

3.3 + ( −0.4 ) +

v0 = 3.1 V , PMOS, non-sat; NMOS, sat

⎛ k p′ ⎞ ⎛ W ⎞ 2 ⎛ kn′ ⎞ ⎛ W ⎞ 2 ⎜ ⎟ ⎜ ⎟ ⎡⎣ 2 (VSG + VTP ) VSD − VSD ⎤⎦ = ⎜ ⎟ ⎜ ⎟ (VGS − VTN ) L L 2 2 ⎝ ⎠ ⎝ ⎠n ⎝ ⎠⎝ ⎠p 2 2 ⎛ 40 ⎞ ⎛ 80 ⎞ ⎜ ⎟ (12 ) ⎡⎣ 2 ( 3.3 − vI − 0.4 )( 3.3 − 3.1) − ( 3.3 − 3.1) ⎤⎦ = ⎜ ⎟ ( 4 ) [ vI − 0.4] ⎝ 2 ⎠ ⎝ 2⎠ 12 [1.16 − 0.4vI − 0.04] = 8 ⎡⎣ vI2 − 0.8vI + 0.16 ⎤⎦ 8vI2 − 1.6vI − 12.16 = 0 vI =

(iii)

1.6 ± 2.56 + 4 ( 8 ) (12.16 ) 2 (8)

⇒ vI = 1.337 V

v0 = 0.2 V PMOS: sat, NMOS, non-sat.

2 2 ⎛ 40 ⎞ ⎛ 80 ⎞ ⎜ ⎟ (12 ) [3.3 − vI − 0.4] = ⎜ ⎟ ( 4 ) ⎡⎣ 2 ( vI − 0.4 )( 0.2 ) − ( 0.2 ) ⎤⎦ ⎝ 2 ⎠ ⎝ 2⎠

12 ⎡⎣8.41 − 5.8vI + vI2 ⎤⎦ = 8 [ 0.4vI − 0.2] 12vI2 − 72.8vI + 102.52 = 0 vI =

72.8 ± 5299.84 − 4 (12 )(102.52 ) 2 (12 )

vI = 2.222 V (b)

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vIt =

(i)

2 ( 6) ( 0.4 ) 3.5928 4 = 2.732 2 (6) 1+ 4

3.3 + ( −0.4 ) +

vIt = 1.315 V

(ii) From (a), (ii) 4 [1.16 − 0.4vI − 0.04] = 12 ⎡⎣vI2 − 0.8vI + 0.16 ⎤⎦ 12vI2 − 8vI − 2.56 = 0 vI =

8 ± 64 + 4 (12 )( 2.56 ) 2 (12 )

⇒ vI = 0.903 V

(iii) From (a), (iii) ⎡ 4 ⎣8.41 − 5.8vI + vI2 ⎤⎦ = 12 [ 0.4vI − 0.2] 4vI2 − 28vI + 36.04 = 0 vI =

28 ± 784 − 4 ( 4 )( 36.04 ) 2 ( 4)

16.35 a.

⇒ vI = 1.70 V

For vO1 = 0.6 < VTN ⇒ vO 2 = 5 V

N1 in nonsaturation and P1 in saturation. From Equation (16.45), ⎡ 2 ( vI − 0.8 )( 0.6 ) − ( 0.6 )2 ⎤ = [5 − vI − 0.8]2 ⎣ ⎦ 1.2vI − 1.32 = 17.64 − 8.4vI + vI2 or vI2 − 9.6vI + 18.96 = 0

vI =

9.6 ±

( 9.6 )

2

− 4 (1)(18.96 ) 2

or vI = 2.78 V V0 Nt ≤ v02 ≤ V0 Pt b. From symmetry, VIt = 2.5 V V0 Pt = 2.5 + 0.8 = 3.3 V and V0 Nt = 2.5 − 0.8 = 1.7 V So 1.7 ≤ v02 ≤ 3.3 V

16.36 a. V0 Nt ≤ v01 ≤ V0 Pt By symmetry, VIt = 2.5 V V0 Pt = 2.5 + 0.8 = 3.3 V and V0 Nt = 2.5 − 0.8 = 1.7 V So 1.7 ≤ v01 ≤ 3.3 V b.

For vO 2 = 0.6 < VTN ⇒ vO 3 = 5 V

N 2 in nonsaturation and P2 in saturation. From Equation (16.57),

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⎡ 2 ( vI 2 − 0.8 )( 0.6 ) − ( 0.6 )2 ⎤ = [5 − vI 2 − 0.8]2 ⎣ ⎦ 1.2vI 2 − 1.32 = 17.64 − 8.4vI 2 + vI22 or vI22 − 9.6vI 2 + 18.96 = 0 So vI 2 = v01 = 2.78 V

For v01 = 2.78, both N1 and P1 in saturation. Then vI = 2.5 V 16.37 a. iPeak = K n ( vI − VTN ) iPeak = 0.1 ⋅ ( 2.5 − 0.8 ) = 0.538 ( mA )

1/ 2

b.

iPeak = 0.1 ⋅ (1.65 − 0.8 ) = 0.269 ( mA )

1/ 2

16.38 (a)

⎛ 50 ⎞ K n = ⎜ ⎟ ( 2 ) = 50 μ A / V 2 ⎝ 2⎠ ⎛ 25 ⎞ K p = ⎜ ⎟ ( 4 ) = 50 μ A / V 2 ⎝ 2 ⎠

I D , peak = K n ( v1 − VTN ) = 50 ( 2.5 − 0.8 ) 2

2

or I D , peak = 144.5 μ A (b)

K n = 50 μ A / V 2 , K p = 25 μ A/V 2

From Equation (16.55), 50 5 − 0.8 + (0.8) 25 vIt = = 2.21 V 50 1+ 25 Then I D , peak = K n (VIt − VTN ) 2 = 50 ( 2.21 − 0.8 )

2

or I D , peak = 99.4 μ A 16.39

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(a)

Switching Voltage, Eq. (16.43)

2 ( 4) ( 0.4 ) 8 vIt = = 1.65 V = vIt 2 ( 4) 1+ 8 80 2 ⎛ ⎞ iD, peak = ⎜ ⎟ ( 4 )(1.65 − 0.4 ) ⇒ iD , peak = 250 μA ⎝ 2⎠ (b) 3.3 − 0.4 +

2 ( 4) ( 0.4 ) 4 = 1.436 V = vIt vIt = 2 ( 4) 1+ 4 2 ⎛ 80 ⎞ iD , peak = ⎜ ⎟ ( 4 )(1.436 − 0.4 ) ⇒ iD , peak = 172 μA ⎝ 2⎠ (c) 3.3 − 0.4 +

2 ( 4) ( 0.4 ) 12 vIt = ⇒ vIt = 1.776 V 2 ( 4) 1+ 12 2 ⎛ 80 ⎞ iD , peak = ⎜ ⎟ ( 4 )(1.776 − 0.4 ) ⇒ iD , peak = 303 μA ⎝ 2⎠ 3.3 − 0.4 +

16.40 2 a. P = fCLVDD For VDD = 5 V, P = (10 × 106 )(0.2 ×10−12 )(5) 2 or P = 50 μ W For VDD = 15 V, P = (10 × 106 )(0.2 × 10−12 )(15) 2 or P = 450 μ W b. For VDD = 5 V, P = (10 × 106 )(0.2 ×10−12 )(5) 2 or P = 50 μ W 16.41 (a)

2 P = fCLVDD = (150 × 106 )( 0.4 ×10−12 ) ( 5 ) = 1.5 × 10 −3 W / inverter 2

Total power: PT = ( 2 × 106 )(1.5 × 10−3 ) ⇒ PT = 3000 W !!!! (b)

For f = 300 MHz

2 1.5 ×10 = ( 300 ×106 )( 0.4 × 10−12 ) VDD ⇒ VDD = 3.54 V

−3

16.42 3 = 3 × 10−7 W 7 10

(a)

P=

(b)

2 P = fCLVDD ⇒ CL =

P 2 fVDD

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(i)

CL =

(ii)

CL =

(iii)

CL =

3 × 10−7

( 5 ×10 ) ( 5) 6

⇒ CL = 0.0024 pF

2

3 × 10−7

2

⇒ CL = 0.00551pF

2

⇒ CL = 0.0267 pF

( 5 ×10 ) ( 3.3) 6

3 × 10−7

( 5 ×10 ) (1.5) 6

16.43 10 = 2 × 10−6 W 5 × 106 P CL = 2 fVDD P=

(a) (b)

2 × 10−6

⇒ CL = 0.01 pF

(i)

CL =

(ii)

CL =

(iii)

CL =

16.44 (a)

For vI ≅ VDD , NMOS in nonsaturation

(8 ×10 ) ( 5) 6

2

2 × 10−6

2

⇒ CL = 0.023 pF

2

⇒ CL = 0.111 pF

(8 ×10 ) ( 3.3) 6

2 × 10−6

(8 ×10 ) (1.5) 6

2 ⎤⎦ and vDS ≅ 0 iD = K n ⎡⎣ 2 ( vI − VTN ) vDS − vDS

So

di 1 = D ≅ K n ⎡⎣ 2 (VDD − VTN ) ⎤⎦ rds dvDS

Or rds =

1 ⎛ kn′ ⎞ ⎛ W ⎞ ⎜ 2 ⎟ ⎜ L ⎟ ⋅ 2 (VDD − VTN ) ⎝ ⎠ ⎝ ⎠n

or 1

rds =

⎛W kn′ ⎜ ⎝L For vI ≅ 0,

⎞ ⎟ ⋅ (VDD − VTN ) ⎠n PMOS in nonsaturation

2 ⎤⎦ iD = K p ⎡⎣ 2 (VDD − vI + VTP ) vSD − vSD

and vSD ≅ 0 for vI ≅ 0. So ⎛ k ′p ⎞ ⎛ W ⎞ di 1 = D ≅ ⎜ ⎟ ⎜ ⎟ ⋅ 2 (VDD + VTP ) rsd dvSD ⎝ 2 ⎠ ⎝ L ⎠ p or rsd =

1 1 k ′p

⎛W ⎜⎜ ⎝ Lp

⎞ ⎟⎟ ⋅ (VDD + VTP ) ⎠

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⎛W ⎞ ⎛W ⎞ (b) For ⎜ ⎟ = 2, ⎜ ⎟ = 4 ⎝ L ⎠n ⎝ L ⎠p rds = rsd =

1 ⇒ r = 2.38 k Ω 50 2 ( )( )( 5 − 0.8) ds 1

( 25 )( 4 )( 5 − 0.8 )

⇒ rsd = 2.38 k Ω

⎛W ⎞ For ⎜ ⎟ = 2,. ⎝ L ⎠p rsd =

1 ⇒ r = 4.76 k Ω 25 2 ( )( )( 5 − 0.8 ) sd

Now, for NMOS: vds = id rds or id =

vds 0.5 = ⇒ id = 0.21 mA rds 2.38

For PMOS: For rsd = 2.38 k Ω, id =

vsd 0.5 = ⇒ id = 0.21 mA rsd 2.38

For rsd = 4.76 k Ω , id =

vsd 0.5 = ⇒ id = 0.105 mA rsd 4.76

16.45 From Equation (16.63) 3 VIL = 1.5 + ⋅ (10 − 1.5 − 1.5 ) ⇒ VIL = 4.125 V 8 and Equation (16.62) 1 V0 HU = ⋅ ⎡⎣ 2 ( 4.125 ) + 10 − 1.5 + 1.5⎤⎦ 2 or V0 HU = 9.125 V From Equation (16.69) 5 VIH = 1.5 + ⋅ (10 − 1.5 − 1.5 ) ⇒ VIH = 5.875 V 8 and Equation (16.68) 1 V0 LU = ⋅ ⎡⎣ 2 ( 5.875 ) − 10 − 1.5 + 1.5⎤⎦ 2 or V0 LU = 0.875 V Now NM L = VIL − V0 LU = 4.125 − 0.875 ⇒ NM L = 3.25 V NM H = V0 HU − VTH = 9.125 − 5.875 ⇒ NM H = 3.25 V

16.46 From Equation (16.71) ⎡

VIL = 1.5 +

⎤ 100 ⎥ 50 − 1 ⎥ = 1.5 + 7 ⎡ 2 ( 0.632 ) − 1⎤ 2 ⎣ ⎦ ⎢ 100 + 3 ⎥ ⎢⎣ ⎥ 50 ⎦

(10 − 1.5 − 1.5 ) ⎢⎢ ⎛ 100 ⎞ − 1⎟ ⎜ ⎝ 50 ⎠

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or VIL = 3.348 V From Equation (16.70) 1 ⎧⎛ 100 ⎞ ⎫ ⎛ 100 ⎞ V0 HU = ⋅ ⎨⎜1 + ⎟ ( 3.348 ) + 10 − ⎜ ⎟ (1.5 ) + 1.5⎬ 2 ⎩⎝ 50 ⎠ ⎝ 50 ⎠ ⎭ or V0 HU = 9.272 V From Equation (16.77)

VIH

⎡ ⎤ ⎛ 100 ⎞ ⎢ 2⎜ ⎥ ⎟ (10 − 1.5 − 1.5 ) ⎢ ⎝ 50 ⎠ = 1.5 + − 1⎥ = 1.5 + 7 [1.51 − 1] ⎥ ⎛ 100 ⎞ ⎢ ⎛ 100 ⎞ − 1⎟ ⎢ 3 ⎜ +1 ⎥ ⎜ ⎟ ⎝ 50 ⎠ ⎣⎢ ⎝ 50 ⎠ ⎦⎥

or VIH = 5.07 V From Equation (16.76) 100 100 ( 5.07 ) ⎛⎜1 + ⎞⎟ − 10 − ⎛⎜ ⎞⎟ (1.5 ) + 1.5 50 ⎝ ⎠ ⎝ 50 ⎠ V0 LU = ⎛ 100 ⎞ 2⎜ ⎟ ⎝ 50 ⎠ or V0 LU = 0.9275 V Now NM L = VIL − V0 LU = 3.348 − 0.9275 or NM L = 2.42 V NM H = V0 HU − VIH = 9.272 − 5.07 or NM H = 4.20 V

16.47 (a) Kn = KP 3 (VDD + VTP − VTN ) 8 3 = 0.4 + ( 3.3 − 0.4 − 0.4 ) ⇒ VIL = 1.3375 V 8 1 VOHu = {2 (1.3375 ) + 3.3 − 0.4 + 0.4} 2 VOHu = 2.9875 V

VIL = VTN +

VIH = 0.4 +

5 ( 3.3 − 0.4 − 0.4 ) ⇒ VIH = 1.9625 V 8

1 {2 (1.9625) − 3.3 − 0.4 + 0.4} 2 = 0.3125 V

VOLu = VOLu

NM H = VOHu − VIH = 2.9875 − 1.9625 ⇒ NM H = 1.025 V NM L = VIL − VOLu = 1.3375 − 0.3125 ⇒ NM L = 1.025 V

(b)

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⎡ ⎤ 2 ( 4) ⎢ ⎥ ( 3.3 − 0.4 − 0.4 ) ⎢ 2.5 12 VIL = 0.4 + 2 − 1⎥ = 0.4 + ⎡( −0.147 ) ⎦⎤ 2 4 ⎢ ⎥ ( ) +3 ⎛ ( 2 )( 4 ) ⎞ ( −0.333) ⎣ 1 − ⎥ ⎜ ⎟ ⎢ 12 ⎦ ⎝ 12 ⎠ ⎣ VIL = 1.505 V VOHu =

⎫⎪ 1 ⎛ 2 ( 4) ⎞ 1 ⎪⎧⎛ ( 2 )( 4 ) ⎞ ⎟ (1.505 ) + 3.3 − ⎜ ⎟ ( 0.4 ) + 0.4 ⎬ = {2.5083 + 3.3 − 0.2667 + 0.4} ⎨⎜1 + 2 ⎪⎩⎝ 12 ⎠ ⎪⎭ 2 ⎝ 12 ⎠

VOHu = 2.9708 V ⎡ ⎤ ⎛ 2 ( 4) ⎞ ⎢ 2⎜ ⎥ ⎟ ( 3.3 − 0.4 − 0.4 ) ⎢ ⎝ 12 ⎠ 2.5 − 1⎥ = 0.4 + VIH = 0.4 + [ −0.2302] ⇒ VIH = 2.1282 V ⎢ ⎥ −0.333) ⎛ 2 ( 4) ⎞ ( 2 )( 4 ) ( − 1⎟ ⎢ 3 +1 ⎥ ⎜ ⎥⎦ 12 ⎝ 12 ⎠ ⎢⎣ ⎛ 2 ( 4) ⎞ ⎛ 2 ( 4) ⎞ ( 2.1282 ) ⎜1 + ⎟ − 3.3 − ⎜ ⎟ ( 0.4 ) + 0.4 12 ⎠ 12 ⎠ 3.547 − 3.3 − 0.2667 + 0.4 ⎝ ⎝ VOLu = = ⇒ VOLu = 0.2853 V 1.333 ⎛ 2 ( 4) ⎞ 2⎜ ⎟ ⎝ 12 ⎠ NM H = VOHu − VIH = 2.9708 − 2.1282 ⇒ NM H = 0.8426 V NM L = VIL − VOLu = 1.505 − 0.2853 ⇒ NM L = 1.22 V

16.48 a. v A = vB = 5 V N1 and N 2 on, so vDS1 ≈ vDS 2 ≈ 0 V P1 and P2 off So we have a P3 − N 3 CMOS inverter. By symmetry, vC = 2.5 V (Transition Point). b. For v A = vB = vC ≡ vI Want K n ,eff = K p ,eff kn′ ⎛ W ⎞ k ′ ⎛ 3W ⎞ ⋅⎜ ⎟ = P ⋅⎜ ⎟ 2 ⎝ 3L ⎠ n 2 ⎝ L ⎠ P With kn′ = 2k P′ , then 2 1 ⎛W ⎞ 1 ⎛W ⎞ ⋅ ⋅⎜ ⎟ = ⋅3⋅⎜ ⎟ 2 3 ⎝ L ⎠n 2 ⎝ L ⎠ P 9 ⎛W ⎞ ⎛W ⎞ Or ⎜ ⎟ = ⋅ ⎜ ⎟ L ⎝ ⎠n 2 ⎝ L ⎠ P

c.

We have ⎛ k ′ ⎞ ⎛ W ⎞ ⎛ 2k ′p ⎞ ⎛ 9 ⎞ ⎛ W ⎞ Kn = ⎜ n ⎟ ⎜ ⎟ = ⎜ ⎟⎜ ⎟⎜ ⎟ ⎝ 2 ⎠ ⎝ L ⎠n ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ L ⎠ p ⎛ k ′p ⎞ ⎛ W ⎞ Kp = ⎜ ⎟⎜ ⎟ ⎝ 2 ⎠⎝ L ⎠p Then from Equation (16.55)

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Kn ⋅ ( 0.8 ) Kp

5 + ( −0.8 ) + VIt = 1+

Kn Kp

Now Kn ⎛9⎞ = ( 2) ⎜ ⎟ = 9 Kp ⎝ 2⎠ Then VIt =

5 + ( −0.8 ) + 3 ( 0.8 ) 1+ 3

⇒ VIt = 1.65 V

16.49 By definition, NMOS is on if gate voltage is 5 V and is off if gate voltage is 0 V. State N1 N2 N3 N4 N5

v0

1 2 3 4

0 0 5 0

off off on on

on off on on

off on off off

on on off on

Logic function ( v X OR vY ) ⊗ ( v X AND vZ ) Exclusive OR of ( vX OR vY ) with ( vX AND vZ ) 16.50 ⎛W ⎞ NMOS in Parallel ⇒ ⎜ ⎟ = 2 ⎝ L ⎠n ⎛W ⎞ 4-PMOS in series ⇒ ⎜ ⎟ = 4 ( 4 ) = 16 ⎝ L ⎠p

(b)

CL doubles ⇒ current must double to maintain switching speed.

⎛W ⎞ ⇒⎜ ⎟ =4 ⎝ L ⎠n ⎛W ⎞ ⎜ ⎟ = 32 ⎝ L ⎠p

16.51 ⎛W ⎞ 4-NMOS in series ⎜ ⎟ = 4 ( 2 ) = 8 ⎝ L ⎠n ⎛W ⎞ 4-PMOS in parallel ⎜ ⎟ = 4 ⎝ L ⎠p

(b)

⎛W ⎜ ⎝L ⎛W ⎜ ⎝L

⎞ ⎟ = 16 ⎠n ⎞ ⎟ =8 ⎠p

16.52

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on off on on

⎛W ⎞ NMOS in parallel ⇒ ⎜ ⎟ = 2 ⎝ L ⎠n ⎛W ⎞ 3-PMOS in series ⇒ ⎜ ⎟ = 3 ( 4 ) = 12 ⎝ L ⎠P

(a)

(b)

⎛W ⎜ ⎝L ⎛W ⎜ ⎝L

⎞ ⎟ =4 ⎠n ⎞ ⎟ = 24 ⎠p

16.53 ⎛W ⎞ 3-NMOS in series ⎜ ⎟ = 3 ( 2 ) = 6 ⎝ L ⎠n ⎛W ⎞ 3-PMOS in parallel ⎜ ⎟ = 4 ⎝ L ⎠p

(a)

(b)

16.54 (a) (b)

⎛W ⎜ ⎝L ⎛W ⎜ ⎝L

⎞ ⎟ = 12 ⎠n ⎞ ⎟ =8 ⎠p

Y = A( B + C )( D + E )

⎛W ⎞ For NMOS in pull down mode, 3 in series ⇒ ⎜ ⎟ = 3 ( 2 ) = 6 ⎝ L ⎠n For PMOS ⎛W ⎞ ⎜ ⎟ =4 ⎝ L ⎠P, A

(c)

⎛W ⎞ = 2 ( 4) = 8 ⎜ ⎟ ⎝ L ⎠ P , B ,C , D , E

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16.55 (a) (b)

Y = A( BD + CE )

(c)

NMOS: 3 transistors in series for pull down mode. ⎛W ⎞ For twice the speed: ⎜ ⎟ = 2 ( 3)( 2 ) = 12 ⎝ L ⎠n ⎛W ⎞ PMOS: ⎜ ⎟ = 2 ( 4 ) = 8 ⎝ L ⎠P, A

⎛W ⎞ = 2 ( 2 )( 4 ) = 16 ⎜ ⎟ ⎝ L ⎠ P , B ,C , D , E 16.56 (a) (b)

Y = A + BC + DE

⎛W ⎞ ⎛W ⎞ NMOS: ⎜ ⎟ = 2 =4 ⎜ ⎟ ⎝ L ⎠n, A ⎝ L ⎠ n , B ,C , D , E PMOS: 3 transistors in series for the pull-up mode

(c)

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⎛W ⎞ ⎜ ⎟ = 3 ( 4 ) = 12 ⎝ L ⎠p

16.57 (a) (b)

Y = A + ( B + D)(C + E )

(c)

⎛W ⎞ NMOS: ⎜ ⎟ = 2 ( 2 ) = 4 ⎝ L ⎠n, A

⎛W ⎞ = ( 2 )( 4 ) = 8 ⎜ ⎟ ⎝ L ⎠ n , B ,C , D , E

PMOS: ⎛W ⎞ ⎜ ⎟ = ( 2 ) 3 ( 4 ) = 24 ⎝ L ⎠p 16.58 (a) A classic design is shown:

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A, B, C signals supplied through inverters. ⎛W ⎞ ⎛W ⎞ (b) For Inverters, ⎜ ⎟ = 1 and ⎜ ⎟ = 2 ⎝ L ⎠p ⎝ L ⎠n ⎛W ⎞ ⎛W ⎞ For PMOS in Logic function, let ⎜ ⎟ = 1 , then for NMOS in Logic function, ⎜ ⎟ = 2.25 ⎝ L ⎠p ⎝ L ⎠n

16.59 (a)

A classic design is shown:

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(b)

⎛W ⎞ ⎛W ⎞ =2 ⎜ ⎟ = 1, ⎜ ⎟ L ⎝ ⎠ ND ⎝ L ⎠ NA, NB , NC ⎛W ⎞ ⎛W ⎞ = 8, ⎜ ⎟ =4 ⎜ ⎟ ⎝ L ⎠ PA, PB ⎝ L ⎠ PC , PD

16.60

( A OR B )

AND C

16.61 ⎛W ⎞ 5-NMOS in series ⇒ ⎜ ⎟ = 5 ( 2 ) = 10 ⎝ L ⎠n ⎛W ⎞ 5-PMOS in parallel ⇒ ⎜ ⎟ = 4 ⎝ L ⎠p

16.62 By definition: NMOS off if gate voltage = 0 NMOS on if gate voltage = 5 V PMOS off if gate voltage = 5 V PMOS on if gate voltage = 0 State 1 2

N1 off on

P1 on off

NA off on

NB off off

NC off off

v01 5 5

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N2 on on

P2 off off

v02 0 0

3 4 5 6

off on off on

on off on off

off off off off

off off off on

off on off on

5 5 5 0

on on on off

off off off on

Logic function is v02 = ( v A OR vB ) AND vC 16.63 State 1 2 3 4 5 6 Logic function: v03 = ( vX OR vZ ) AND vY

v01

v02

v03

5 0 5 5 5 0

5 0 5 0 5 5

0 5 0 5 0 0

16.64

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0 0 0 5

16.65

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16.66

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16.67

2 I = −C

dVC dt

So 1 ( 2I ) ⋅ t C For ΔVC = −0.5 V ΔVC = −

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−0.5 = −

16.68 (a) (i) (ii) (iii) (b) (i) (ii) (iii) 16.69 (a) (i) (ii) (iii) (b) (i) (ii) (iii)

2 ( 2 x 10−12 ) ⋅ t 25 x 10−15

⇒ t = 3.125 ms

vO = 0 vO = 4.2 V vO = 2.5 V vO = 0 vO = 3.2 V vO = 2.5 V

vo = 0 vo = 2.9 V vo = 2.4 V vo = 0 vo = 2.0 V vo = 2.0 V

16.70 Neglect the body effect. v01 (logic 1) = 4.2 V , v02 (logic 1) = 5 V a. vI = 5 V ⇒ vGS 1 = 4.2 V b. M 1 in nonsaturation and M 2 in saturation. From Equation (16.23) ⎛W ⎜ ⎝L ⎛W ⎜ ⎝L Or ⎛W ⎜ ⎝L

2 ⎞ ⎛W ⎞ 2 ⎟ ⎡⎣ 2 ( vGS 1 − VTND ) vO1 − vO1 ⎤⎦ = ⎜ ⎟ (VDD − vO1 − VTNL ) L ⎠D ⎝ ⎠L 2 2 ⎞ ⎡ ⎟ ⎣ 2 ( 4.2 − 0.8 )( 0.1) − ( 0.1) ⎤⎦ = (1) [5 − 0.1 − 0.8] ⎠D

⎞ ⎛W ⎞ ⎟ ( 0.67 ) = 16.81 ⇒ ⎜ ⎟ = 25.1 ⎠D ⎝ L ⎠D

Now v01 = 4.2 V ⇒ vGS 3 = 4.2 V M 3 in nonsaturation and M 4 in saturation. From Equation (16.29(b)). ⎛W ⎜ ⎝L ⎛W ⎜ ⎝L

2 ⎞ ⎛W ⎞ 2 ⎟ ⎡⎣ 2 ( vGS 3 − VTND ) vO 2 − vO 2 ⎤⎦ = ⎜ ⎟ [ −VTNL ] ⎠D ⎝ L ⎠L 2 2 ⎞ ⎟ ⎡⎣ 2 ( 4.2 − 0.8 )( 0.1) − ( 0.1) ⎤⎦ = ( 2 ) ⎡⎣ − ( −1.5 ) ⎤⎦ ⎠D

⎛W ⎞ ⎜ ⎟ (0.67) = 2.25 ⎝ L ⎠D

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⎛W ⎞ Or ⎜ ⎟ = 3.36 ⎝ L ⎠D

16.71 A B Y 0 0 1 0 1 0 1 0 0 1 1 ⇒ indeterminate 0.1 Without the top transistor, the circuit performs the exclusive-NOR function. 16.72 B A A 0 1 0 0 1 1 1 0 0 1 0 1 Y = A + AB = A + B Z = Y or Z = AB

B 1 0 1 0

Y 0 1 1 1

Z 1 0 0 0

16.73

16.74 For φ = 1, φ = 0, then Y = B. And for φ = 0, φ = 1, then Y = A . A multiplexer. 16.75 Y = AC + BC 16.76 Y = AB + AB = A ⊗ B 16.77

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A 0 1 0 1

B 0 0 1 1

Y 0 1 1 0

Exclusive-OR function. 16.78 This circuit is referred to as a two-phase ratioed circuit. The same width-to-length ratios between the driver and load transistors must be maintained as discussed previously with the enhancement load inverter. When φ1 is high, v01 becomes the complement of vI . When φ2 goes high, then v0 becomes the complement of v01 or is the same as vI . The circuit is a shift register. 16.79 Let Q = 0 and Q = 1 ; as S increases, Q decreases. When Q reaches the transition point of the M 5 − M 6 inverter, the flip-flop with change state. From Equation (16.28(b)), KL VIt = ⋅ ( −VTNL ) + VTND KD where K L = K 6 and K D = K 5 . Then 30 VIt = ⋅ ⎡ − ( −2 ) ⎤⎦ + 1 ⇒ VIt = Q = 2.095 V 100 ⎣ This is the region where both M 1 and M 3 are biased in the saturation region. Then S=

K3 ⋅ ( −VTNL ) + VTND = K1

30 ⋅ ⎡ − ( −2 ) ⎤⎦ + 1 200 ⎣

or S = 1.77 V This analysis neglects the effect of M 2 starting to turn on at the same time. 16.80 Let vY = R, v X = S , v02 = Q, and v01 = Q. Assume VThN = 0.5 V and VThP = −0.5 V. For S = 0, we have the following:

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If we want the switching to occur for R = 2.5 V, then because of the nonsymmetry between the two circuits, we cannot have Q and Q both equal to 2.5 V. Set R = Q = 2.5 V and assume Q goes low. For the M 1 − M 5 inverter, M 1 in nonsaturation and M 5 in saturation. Then 2 2 K n ⎡⎢ 2 ( 2.5 − 0.5 ) Q − Q ⎤⎥ = K p [ 2.5 − 0.5] ⎣ ⎦ Or 2 ⎛ Kp ⎞ 4Q − Q = 4 ⎜ ⎟ ⎝ Kn ⎠

For the other circuit, M 2 − M 4 in saturation and M 6 in nonsaturation. Then

(

)

2 K n ( 2.5 − 0.5 ) + K n (Q − 0.5) 2 = K p ⎡ 2 5 − Q − 0.5 ( 2.5 ) − ( 2.52 ) ⎤ ⎣ ⎦

Combining these equations and neglecting the Q 3 term, we find Kp

Q = 1.4 V and

kn

= 0.9

16.81 3.3 + ( −0.4 ) + 0.5 vIt = = 1.7 V 1+1 vI = 1.5 V NMOS Sat; PMOS Non Sat 2 = ⎡ 2 ( 3.3 − vI − 0.4 )( 3.3 − vo1 ) − ( 3.3 − vo1 ) ⎤ ⇒ vo1 = 2.88 V ⎣ ⎦ vI = 1.6 V vo1 = 2.693 V vI = 1.7 V vo1 = variable (switching region)

( vI − 0.5)

2

vI = 1.8 V

NMOS Non Sat; PMOS Sat

( 3.3 − VI − 0.4 )

2

= ⎡⎣ 2 ( vI − 0.5 ) vo1 − vo21 ⎤⎦ ⇒ vo1 = 0.607 V

Now vI = 1.5 V, vo1 = 2.88 V ⇒ vo ≈ 0V

vI = 1.6 V, vo1 = 2.693 V NMOS Non Sat; PMOS Sat 2 ( 3.3 − vo1 − 0.4 ) = ⎡⎣ 2 ( vo1 − 0.5 ) vo − vo2 ⎤⎦ vo = 0.00979 V vI = 1.7 V, v o1 = Switching Mode ⇒ v0 = Switching Mode. vI = 1.8 V, vo1 = 0.607 V NMOS Sat; PMOS Non Sat

( v01 − 0.5)

2

2 = ⎡ 2 ( 3.3 − v01 − 0.4 )( 3.3 − v0 ) − ( 3.3 − v0 ) ⎤ ⇒ v0 = 3.298 V ⎣ ⎦

16.82 For R = φ = VDD and S = 0 ⇒ Q = 0, Q = 1 For S = φ = VDD and R = 0 ⇒ Q = 1, Q = 1 The signal φ is a clock signal. For φ = 0, The output signals will remain in their previous state. 16.83 a.

Positive edge triggered flip-flop when CLK = 1, output of first inverter is D and then Q = D = D .

www.elsolucionario.net

b.

For example, put a CMOS transmission gate between the output and the gate of M 1 driven by a

CLK pulse.

16.84 For J = 1, K = 0, and CLK = 1; this makes Q = 1 and Q = 0 .

For J = 0, K = 1, and CLK = 1 , and if Q = 1, then the circuit is driven so that Q = 0 and Q = 1. If initially, Q = 0, then the circuit is driven so that there is no change and Q = 0 and Q = 1. J = 1, K = 1, and CLK = 1, and if Q = 1, then the circuit is driven so that Q = 0. If initially, Q = 0 , then the circuit is driven so that Q = 1. So if J = K = 1, the output changes state. 16.85 For J = v X = 1, K = vY = 0, and CLK = vZ = 1, then v0 = 0. For J = v X = 0, K = vY = 1, and CLK = vZ = 1, then v0 = 1. Now consider J = K = CLK = 1. With v X = vZ = 1, the output is always v0 = 0, So the output does not change state when J = K = CLK = 1. This is not actually a J − K flip-flop. 16.86 64 K ⇒ 65,536 transistors arranged in a 256 × 256 array. (a) Each column and row decoder required 8 inputs. (b) (i) Address = 01011110 so input = a7 a6 a5 a4 a3 a2 a1a0 (ii) Address = 11101111 so input = a7 a6 a5 a4 a3 a2 a1a0 (c) (i) (ii)

Address = 00100111 so input = a7 a6 a5 a4 a3 a2 a1a0 Address = 01111011 so input = a7 a6 a5 a4 a3 a2 a1a0

16.87 (a)

1-Megabit memory ⇒

= 1, 048,576 ⇒ 1024 × 1024 Nuclear & input row and column decodes lines necessary = 10

(b) 250K × 4 bits ⇒ 262,144 × 4 bits ⇒ 512 × 512 For 512 lines ⇒ 9 row and column decoder lines necessary. 16.88 Put 128 words in a 8 × 16 array, which means 8 row (or column) address lines and 16 column (or row) address lines. 16.89 Assume the address line is initially uncharged, then dV 1 I I = C C or VC = ∫ Idt = ⋅ t C C dt −12 VC ⋅ C ( 2.7 ) ( 5.8 × 10 ) ⇒ = I 250 × 10−6 t = 6.26 × 10 −8 s ⇒ 62.6 ns

Then t =

16.90

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(a) ⎛W or ⎜ ⎝L (b)

5 − 0.1 ⎛ 35 ⎞⎛ W = ⎜ ⎟⎜ 1 ⎝ 2 ⎠⎝ L

2 ⎞⎡ ⎟ ⎣ 2 ( 5 − 0.7 )( 0.1) − ( 0.1) ⎤⎦ ⎠

⎞ ⎟ = 0.329 ⎠ 16 K ⇒ 16,384 cells

2 = 2 μA 1 Power per cell = (2 μ A)(2 V ) = 4 μW Total Power = PT = (4 μW )(16,384) ⇒ PT = 65.5 mW iD ≅

Standby current = (2 μ A)(16,384) ⇒ IT = 32.8 mA 16.91 16 K ⇒ 16,384 cells 200 ⇒ 12.2 μW 16,384 V P 12.2 2.5 iD = = = 4.88 μ A ≅ DD = ⇒ R = 0.512 M Ω VDD R R 2.5 PT = 200 mW ⇒ Power per cell =

If we want vO = 0.1 V for a logic 0, then ⎛ k′ ⎞⎛ W ⎞ iD = ⎜ n ⎟ ⎜ ⎟ ⎡⎣ 2 (VDD − VTN ) vO − vO2 ⎤⎦ ⎝ 2 ⎠⎝ L ⎠ 2 ⎛ 35 ⎞ ⎛ W ⎞ 4.88 = ⎜ ⎟ ⎜ ⎟ ⎡ 2 ( 2.5 − 0.7 )( 0.1) − ( 0.1) ⎤ ⎣ ⎦ ⎝ 2 ⎠⎝ L ⎠ ⎛W ⎞ So ⎜ ⎟ = 0.797 ⎝L⎠

16.92 Q = 0, Q = 1 So D = Logic 1 = 5 V A very short time after the row has been addressed, D remains charged at VDD = 5 V . Then M p 3, M A, and M N 1 begin to conduct and D decreases. In steady-state, all three transistors are biased in the nonsaturation region. Then 2 2 2 ⎤ ⎡ ⎤ ⎡ ⎤ K p 3 ⎡⎣ 2 (VSG 3 + VTP 3 ) VSD 3 − VSD 3 ⎦ = K nA ⎣ 2 (VGSA − VTNA ) VDSA − VDSA ⎦ = K n1 ⎣ 2 (VGS 1 − VTN 1 ) VDS 1 − VDS 1 ⎦

Or 2 2 K p 3 ⎡ 2 (VDD + VTP 3 )(VDD − D ) − (VDD − D ) ⎤ = K nA ⎡ 2 (VDD − Q − VTNA )( D − Q ) − ( D − Q ) ⎤ ⎣ ⎦ ⎣ ⎦

= K n1 ⎡⎣ 2 (VDD − VTN 1 ) Q − Q 2 ⎤⎦ (1)

Equating the first and third terms: 2 ⎛ 20 ⎞ ⎡ ⎛ 40 ⎞ 2 ⎜ ⎟ (1) ⎣ 2 ( 5 − 0.8 )( 5 − D ) − ( 5 − D ) ⎦⎤ = ⎜ ⎟ ( 2 ) ⎣⎡ 2 ( 5 − 0.8 ) Q − Q ⎦⎤ (2) ⎝ 2 ⎠ ⎝ 2 ⎠ As a first approximation, neglect the ( 5 − D ) and Q 2 terms. We find 2

Q = 1.25 − 0.25 D (3) Then, equating the first and second terms of Equation (1): 2 2 ⎛ 20 ⎞ ⎡ ⎛ 40 ⎞ ⎜ ⎟ (1) ⎣ 2 ( 5 − 0.8 )( 5 − D ) − ( 5 − D ) ⎤⎦ = ⎜ ⎟ (1) ⎡⎣ 2 ( 5 − Q − 0.8 )( D − Q ) − ( D − Q ) ⎤⎦ 2 2 ⎝ ⎠ ⎝ ⎠

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Substituting Equation (3), we find as a first approximation: D = 2.14 V Substituting this value of D into equation (2), we find 2 8.4 ( 5 − 2.14 ) − ( 5 − 2.14 ) = 4 ⎡⎣8.4Q − Q 2 ⎤⎦ We find Q = 0.50 V Using this value of Q, we can find a second approximation for D by equating the second and third terms of equation (1). We have 2 20 ⎡ 2 ( 4.2 − Q )( D − Q ) − ( D − Q ) ⎤ = 40 ⎡⎣ 2 ( 4.2Q ) − Q 2 ⎤⎦ ⎣ ⎦ Using Q = 0.50 V , we find D = 1.79 V 16.93 Initially M N 1 and M A turn on. M N 1, Nonsat; M A , sat. K nA [VDD − Q − VTN ] = K n1 ⎡⎣ 2 (VDD − VTN 1 ) Q − Q 2 ⎤⎦ 2

2 ⎛ 40 ⎞ ⎛ 40 ⎞ 2 ⎜ ⎟ (1) [5 − Q − 0.8] = ⎜ ⎟ ( 2 ) ⎣⎡ 2 ( 5 − 0.8 ) Q − Q ⎦⎤ 2 2 ⎝ ⎠ ⎝ ⎠ which yields Q = 0.771 V

Initially M P 2 and M B turn on Both biased in nonsaturation reagion 2 2 K P 2 ⎡ 2 (VDD + VTP 3 ) VDD − Q − VDD − Q ⎤ = K nB ⎡⎢ 2 (VDD − VTNB ) Q − Q ⎤⎥ ⎥⎦ ⎣ ⎦ ⎣⎢ 2 2 ⎛ 20 ⎞ ⎡ ⎤ ⎛ 40 ⎞ ⎡ ⎤ ⎜ ⎟ ( 4 ) ⎣⎢ 2 ( 5 − 0.8 ) 5 − Q − 5 − Q ⎦⎥ = ⎜ ⎟ (1) ⎢⎣ 2 ( 5 − 0.8 ) Q − Q ⎥⎦ ⎝ 2 ⎠ ⎝ 2 ⎠

(

) (

(

) (

)

)

which yields Q = 3.78 V Note: (W / L) ratios do not satisfy Equation (16.86) 16.94 For Logic 1, v1:

( 5 )( 0.05 ) + ( 4 )(1) = (1 + 0.05 ) v1 ⇒ v1 = 4.0476 V v2 : (5)(0.025) + (4)(1) = (1 + 1.025)v2 ⇒ v2 = 4.0244 V

For Logic 0, v1: (0)(0.05) + (4)(1) = (1 + 0.05)v1 ⇒ v1 = 3.8095 V v2 : (0)(0.025) + (4)(1) = (1 + 0.025)v2 ⇒ v2 = 3.9024 V

16.95 Not given 16.96 Not given 16.97 Not given

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16.98 Quantization error =

(a)

Or LSB ≤ 0.10 V For a 6-bit word , LSB = 1 − LSB =

(b)

1 LSB ≤ 1% ≤ 0.05 V 2

5 = 0.078125 V 64

5 = 0.078125 V 64

3.5424 × 64 = 45.34 ⇒ n = 45 5 Digital Output = 101101 45 × 5 = 3.515625 64 . 1 Δ = 3.5424 − 3.515625 = 0.026775 < LSB. 2

(c)

16.99 Quantization error =

(a)

1 − LSB = 0.10 V

1 LSB ≤ 0.5% ≤ 0.05 V 2

10 = 0.078125 V 128 1 − LSB = 0.078125 V

For a 7-beit word, LSB = (b)

3.5424 ×128 = 45.34272 ⇒ n = 45 10 Digital output = 0101101

(c)

Now

45 × 10 = 3.515625 128

Δ = 3.5424 − 3.515625 = 0.026775 V <

1 LSB 2

16.100 (a)

(b)

⎛0 1 0 1 ⎞ vo = ⎜ + + + ⎟ ( 5 ) ⎝ 2 4 8 16 ⎠ vo = 1.5625 V ⎛1 0 1 0 ⎞ vo = ⎜ + + + ⎟ ( 5 ) ⎝ 2 4 8 16 ⎠ vo = 3.125 V

16.101 ⎛1⎞ LSB = ⎜ ⎟ ( 5 ) = 0.3125 V ⎝ 16 ⎠ 1 LSB = 0.15625 V 2 ⎛ 10 ⎞ Now vo = ⎜ ⎟ (5) ⎝ 20 + ΔR1 ⎠

(a)

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For vo = 2.5 + 0.15625 = 2.65625 V

(10 )( 5)

⇒ ΔR1 = −1.176 K 2.65625 For vo = 2.5 − 0.15625 = 2.34375 V 20 + ΔR1 =

20 + ΔR1 =

(10 )( 5 )

2.34375 V ΔR1 = +1.333 K

For ΔR1 = 1.176 K ⇒ ΔR1 = 5.88% ⎛ ⎞ 10 For R4 : vo = ⎜ ⎟ ( 5) ⎝ 160 + ΔR4 ⎠ vo = 0.3125 + 0.15625 = 0.46875 V

(b)

(10 )( 5 )

⇒ ΔR4 = −53.33K 0.46875 Or vo = 0.3125 − 0.15625 = 0.15625 V

160 + ΔR4 =

(10 )( 5 )

⇒ ΔR4 = 160 K 0.15625 For ΔR4 = 53.33K ⇒ ΔR4 = 33.33% 160 + ΔR4 =

16.102 (a)

R5 = 320 kΩ R6 = 640 kΩ R7 = 1280 kΩ R8 = 2560 kΩ

(b)

⎛ 10 ⎞ vo = ⎜ ⎟ ( 5 ) = 0.01953125 V ⎝ 2560 ⎠

16.103 (a) V −5 ⇒ I1 = −0.50 mA I1 = REF = 2 R 10 I I 2 = 1 = −0.25 mA 2 I2 I 3 = = −0.125 mA 2 I3 I 4 = = −0.0625 mA 2 I4 I 5 = = −0.03125 mA 2 I I 6 = 5 = −0.015625 mA 2 Δvo = I 6 RF = ( 0.015625 )( 5 ) (b) Δvo = 0.078125 V

(c)

vo = − [ I 2 + I 5 + I 6 ] RF = [ 0.25 + 0.03125 + 0.015625] ( 5 ) vo = 1.484375 V

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(d)

For 101010; vo = ( 0.50 + 0.125 + 0.03125 )( 5 ) = 3.28125 V

For 010101; vo = ( 0.25 + 0.0625 + 0.015625 )( 5 ) = 1.640625 V Δvo = 1.640625 V

16.104 1 5 ⎛V ⎞⎛ R ⎞ V = 0.3125 V LSB = ⎜ REF ⎟ ⎜ ⎟ = REF = 2 8 2 16 16 R ⎝ ⎠⎝ ⎠ Ideal 3 ⎛ 3V ⎞ v A for 011 ⇒ ⎜ REF ⎟ ( R ) = VREF = 1.875 V 8 R 8 ⎝ ⎠ 1 Range of v A = 1.875 ± LSB 2 or 1.5625 ≤ vA ≤ 2.1875 V 16.105 6-bits ⇒ 26 = 64 resistors 26 − 1 = 63 comparators 16.106 (a) 10- bit output ⇒ 1024 clock periods 1 1 1 clock period = = 6 = 1 μS f 10 May conversion time = 1024 μS = 1.024 mS (b) 1 1⎛ 5 ⎞ LSB = ⎜ ⎟ = 0.002441406 V 2 2 ⎝ 1024 ⎠ ⎛ 5 ⎞ v′A = (128 + 16 + 2 ) ⎜ ⎟ = 0.712890625 V ⎝ 1024 ⎠ 1 So range of v A = v′A ± LSB 2 0.710449219 ≤ v A ≤ 0.715332031 V (c) 0100100100 ⇒ 256 + 32 + 4 = 292 clock pulses 16.107 N ×5 ⇒ N = 640 ⇒ 512 + 128 1024 Output = 1010000000 (b) N ×5 ⇒ N = 381.19 ⇒ N = 381 ⇒ 256 + 64 + 32 + 16 + 8 + 4 + 1 1.8613 = 1024 Output = 0101111101 (a)

3.125 =

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Chapter 17 Exercise Solutions EX17.1 a. −0.7 − (−5) iE = = 1 mA ⇒ RE = 4.3 kΩ RE iC1 = iC 2 = 0.5 mA =

5 − 3.5 ⇒ RC1 = RC 2 = 3 kΩ RC1

v1 = 1 V i. (1 − 0.7) − (−5) iE = ⇒ iE = 1.23 mA 4.3 iC1 = iE = v01 = 5 − (1.23)(3) ⇒ v01 = 1.31 V

b.

v02 = 5 V v1 = −1 V

ii.

iE = 1 mA ⇒ v02 = 5 − (1)(3) ⇒ v02 = 2 V v01 = 5 V

EX17.2 P (iCXY + iCR + i3 + i4 )(5.2) a. v X = vY = logic 1 ⇒ iCXY = 3.22 mA iCR = 0 −0.7 + 5.2 i3 = = 3 mA 1.5 −1.4 + 5.2 i4 = = 2.53 mA 1.5 P = (3.22 + 0 + 3 + 2.53)(5.2) ⇒ P = 45.5 mW b. v X = vY = logic 0 ⇒ iCXY = 0 iCR = 2.92 mA i3 = 2.53 mA i4 = 3 mA P = (0 + 2.92 + 2.53 + 3)(5.2) ⇒ P = 43.9 mW EX17.3 3.5 − (2.45 + 0.7) = 1.4 K 0.25 3.5 − 2(0.7) R1 + R2 = = 8.4 k ⇒ R2 = 7 K 0.25 2.45 R5 = = 9.8 K 0.25 R1 =

EX17.4

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I MAX = 1 =

−0.7 − (−5.2) ⇒ R3 = R4 = 4.5 K R3

−0.7 − 0.7 − (−5.2) = 3.22 mA 1.18 i5 = i1 = 1.40 mA i3 = 1.0 mA −1.4 − (−5.2) i4 = = 0.844 mA 4.5 P = (3.22 + 1.4 + 1.4 + 1.0 + 0.844)(5.2) = 40.8 mW iCXY =

EX17.5 ⎛ v − 0.7 − (−5.2) ⎞ iL = ⎜ oR ⎟ (10) (1.18)(51) ⎝ ⎠ voR − (−5.2) i3 = 1.5 v + 5.2 ⎛ voR + 4.5 ⎞ ⎡ 0 − (VoR + 0.7) ⎤ +⎜ (51) = oR ⎟ (10) ⎢ ⎥ 0.24 1.5 ⎣ ⎦ ⎝ (1.18)(51) ⎠ ⎡⎛ 51 ⎞ 1 1 − (6) ⎤ (0.7)(51) 5.2 4.5(10) −voR ⎢⎜ + = + + ⎟+ ⎥ 0.24 1.5 (1.18)(51) ⎣⎝ 0.24 ⎠ 1.5 (1.18)(51) ⎦ −voR [212.50 + 0.6667 + 0.166168] = 148.75 + 3.4666 + 0.747757 −voR [213.3328] = 152.9644 voR = −0.7170

EX17.6 (100)(0.026) rπ 3 = = 2.6 K 1 1 g m3 = = 38.46 mA / V 0.026 Vn Vn = Ib3 = RC 2 + rπ 3 + (1 + β ) R3 0.24 + 2.6 + (101)(4.5) Ib3 =

Vn 457.34

⎛ Vn ⎞ VO = − I b 3 ( RC 2 + rπ 3 ) = − ⎜ ⎟ (0.24 + 2.6) ⎝ 457.34 ⎠ VO = −0.00621 Vn ⎛ Vn ⎞ VO′ = (1 + β ) I b 3 R3 = (101) ⎜ ⎟ (4.5) ⎝ 457.34 ⎠ VO′ = 0.9938 Vn

EX17.7 P = I Q ⋅ VCC ⇒ 0.2 = I Q (1.7) ⇒ I Q = 117.6 μ A QR on ⇒ v0 = 1.7 − I Q RC = 1.7 − 0.4 ⇒ RC = VR =

0.4 ⇒ RC = 3.40 kΩ 0.1176

1.7 + 1.3 ⇒ VR = 1.5 V 2

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EX17.8 (a) v X = vY = 5 V v1 = VBE ( sat ) + 2VY = 0.8 + 2(0.7) = 2.2 V 5 − 2.2 i1 = = 0.70 mA 4 V − VCE ( sat ) 5 − 0.1 iRC = CC = = 1.225 mA RC 4 P = (i1 + iRC )VCC = (0.70 + 1.225)(5)

or P = 9.625 mW (b) v X = vY = 0 ⇒ v1 = 0.70 V V − v 5 − 0.70 i1 = CC 1 = = 1.075 mA 4 R1 P = i1 ⋅ VCC = (1.075)(5) ⇒ P = 5.375 mW

EX17.9 vB1 = 0.1 + 0.8 = 0.9 V (a) 5 − 0.9 i1 = = 0.683 mA 6 All other currents ≈ 0 vB1 = 0.8 + 0.8 + 0.7 = 2.3 V (b) 5 − 2.3 = 0.45 mA i1 = 6 iB 2 = (1 + 0.2)(0.45) = 0.54 mA 5 − (0.8 + 0.1) i2 = = 2.733 mA 1.5 iE 2 = 2.733 + 0.54 = 3.273 mA 0.8 i4 = = 0.533 mA 1.5 iB 0 = 3.273 − 0.533 = 2.74 mA 5 − 0.1 i3 = = 2.23 mA 2.2 i 2.733 For Q2 : 2 = = 5.06 < β F 0.54 iB 2 For Q0 :

i3 2.23 = = 0.81 < β F iB 0 2.74

EX17.10 v X = vY = 3.6 V, Output low: 5 − 2.3 = 0.45 mA 6 = (1 + 2 β R )i1

i1 = iB 2

iB 2 = 0.54 mA vB 3 = 0.8 + 0.1 = 0.9

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5 − 0.9 = 2.05 mA 2 0.8 ⇒ iB 0 = 2.06 mA iB 0 = 2.05 + 0.54 − 1.5 5 − 0.9 iL′ = = 0.683 mA 6 β F ⋅ iB 0 = N ⋅ iL′ ⇒ (20)(2.06) = N (0.683) i2 =

⇒ N = 60

EX17.11 iRC =

a.

5 − 0.4 = 2.04 mA 2.25

2 + 2.04 ⇒ iC′ = 3.67 mA 1 1+ 10 3.67 iB′ = = 0.367 mA 10 iD = 2 − 0.367 ⇒ iD = 1.63 mA iC′ =

iD = 0 ⇒ iB′ = iB = 2 mA b. iC′ = β iB′ = (10)(2) = 20 mA = iRC + iL iL = 20 − 2.04 ⇒ iL (max) ≈ 18 mA

EX17.12 v1 = 0.4 + 0.3 = 0.7, i1 =

(a)

5 − 0.7 = 0.1075 mA 40

All transistor currents are zero. P = (0.1075)(5 − 0.4) ⇒ 495 μ W 5 − 1.4 = 0.090 mA 40 5 − 1.1 vC 2 = 0.7 + 0.4 = 1.1 V, i2 = iC 2 = = 0.325 mA 12 iB 0 ≈ iB 2 + iC 2 = 0.09 + 0.325 = 0.415 mA v1 = 1.4 V, i1 = iB 2 =

(b)

iC 0 ≈ 0 P = (i1 + i2 )(5) = (0.09 + 0.325)(5) = 2.08 mW

TYU17.1 QR on a. iE =

1.5 − 0.7 − ( −3.5) = 2 mA ⇒ RE = 2.15 kΩ RE

iCR ≈ iE = 2 mA =

v X = vY = 2 V ⇒ Q1 and Q2 on

b. iE =

3.5 − 2 ⇒ RC 2 = 0.75 kΩ RC 2

2 − 0.7 − (−3.5) 4.8 = ⇒ iE = 2.233 mA 2.15 RE

RC1 =

3.5 − 2 1.5 = ⇒ RC1 = 0.672 kΩ 2.233 iCXY

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TYU17.2 logic 1 = −0.7 V Q1 and Q2 on when v X = vY = −0.7 V −0.7 − 0.7 − (−5.2) = 2.5 ⇒ RE = 1.52 kΩ RE

iE =

vNOR = −1.5 ⇒ RC1 =

0 − (−1.5 + 0.7) ⇒ RC1 = 320 Ω 2.5

−1.5 − 0.7 ⇒ VR = −1.1 V 2 −1.1 − 0.7 − (−5.2) = 2.237 mA QR on ⇒ iE = 1.52 0 − (−1.5 + 0.7) ⇒ RC 2 = 358 Ω RC 2 = 2.237 −0.7 − (−5.2) ⇒ R3 = R4 = 1.8 kΩ R3 = R4 = 2.5

VR =

TYU17.3

N M H = −0.70 − (−0.93) ⇒ N M H = 0.23 V N M L = −1.17 − ( −1.40) ⇒ N M L = 0.23 V

TYU17.4 State 1 2 3 4 5 6 7 8

A 0 1 0 0 1 1 0 1

B 0 0 1 0 1 0 1 1

C 0 0 0 1 0 1 1 1

Q01 off “on” off off on on off on

Q02 off off on off on off on on

Q03 off off off on off on on on

( A AND C ) OR ( B AND C )  

 

true for

true for

states 6 and 8 states 3 and 5  

Output goes high for these 4 states

TYU17.5 A 0

B 0

C 0

v0 0

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Q1 off off off on off on on on

Q2 on on on off on off off off

QR on on “off” on off “off” on off

v0 0 0 1 0 1 1 0 1

1 0 0 1 1 0 1

0 1 0 1 0 1 1

0 0 1 0 1 1 1

1 1 1 0 0 0 1

⇒ ( A ⊕ B) ⊕ C

TYU17.6 v X = vY = 0.1 ⇒ v1 = 0.8 a. 5 − 0.8 i1 = ⇒ i1 = 0.7 mA 6 i2 = iR = iB = iRC = 0 v0 = 5 V

b. c.

Same as part (a). v X = vY = 5 V ⇒ v1 = 0.8 + 0.7 + 0.7 = 2.2 V

i1 = i2 =

5 − 2.2 ⇒ i1 = i2 = 0.467 mA 6

0.8 ⇒ iR = 0.053 mA 15 iB = i2 − iR ⇒ iB = 0.414 mA iR =

iRC =

5 − 0.1 ⇒ iRC = 0.98 mA 5 vC = 0.1 V

TYU17.7 (a) From Tyu17.6, iB = 0.414 mA, iRC = 0.98 mA v1′ = 0.1 + 0.7 = 0.8, iL′ =

5 − 0.8 = 0.70 mA 6

iC ,max = β iB = iRC + NiL′ (25)(0.414) = 0.98 + N (0.70) N = 13.4 → N = 13

(b)

iC , max = β iB = (25)(0.414) = 10.35 mA < 15 mA

So from part (a), N = 13 TYU17.8 From EX17.9, iB 0 = 2.74 mA, i3 = 2.23 mA vo = 0.1 V, so vB′ 1 = 0.1 + 0.8 = 0.9 V 5 − 0.9 = 0.6833 mA 6 iC 0, max = β iB 0 = i3 + Ni1′ (20)(2.74) = 2.23 + N (0.6833) N = 76.9 ⇒ N = 76

i1′ =

TYU17.9

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Q1 in saturation 5 − 0.9 iB1 = ⇒ iB1 = 0.683 mA 6 iC1 = iB 2 = iC 2 = 0 iB 0 = iC 0 = 0 vB 4 = 0.1 + 0.7 = 0.8 V 0.1 iB 4 = ⇒ iB 4 = 1.19 μ A (21)(4) iC 4 = 23.8 μ A iB 3 = iC 3 = 0

TYU17.10 (a) v X = vY = 0.4, vB1 = 0.4 + 0.7 = 1.1 V 5 − 1.1 = 1.393 mA 2.8 P = i1 (5 − 0.4) = (1.393)(5 − 0.4) = 6.41 mW

i1 =

(b)

v X = vY = 3.6 V

5 − 2.1 = 1.036 mA 2.8 5 − 1.1 = 5.132 mA vC 2 = 0.7 + 0.4 = 1.1 V, i2 = 0.76 0.4 vE 4 = 1.1 − 0.7 = 0.4 V, iR 4 = = 0.1143 mA 3.5 ⎛ β ⎞ ⎛ 25 ⎞ iR 3 = ⎜ ⎟ iR 4 = ⎜ ⎟ (0.1143) = 0.1099 ≈ 0.11 mA ⎝ 26 ⎠ ⎝1+ β ⎠ vB1 = 2.1, i1 =

P = (i1 + i2 + iR 3 )(5) = (1.036 + 5.132 + 0.11)(5) or P = 31.4 mW

TYU17.11 (a) v X = 0.4 V, vC1 = 0.4 + 0.7 = 1.1 V 5 − 1.1 ⇒ 97.5 μ A 40 v X = 3.6 V, vC1 = 3(0.7) = 2.1 V (b) i1 =

i1 =

5 − 2.1 ⇒ 72.5 μ A 40

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Chapter 17 Problem Solutions 17.1 a.

vI = −1.5 Vi Q1 off , Q2 on −0.7 − (−3.5) ⇒ iE = 0.56 mA iE = 5 iC1 = 0 ⇒ v01 = 3.5 V

iC 2 = iE ⇒ v02 = 3.5 − iE RC 2 = 3.5 − ( 0.56 )( 2 ) or v02 = 2.38 V vI = 1.0 Vi Q1 on, Q2 off

b. iE = iC 2

(1 − 0.7 ) − ( −3.5)

5 = 0 ⇒ v02 = 3.5 V

⇒ iE = 0.76 mA

c. logic 0 at v02 (low level) = 2.38 V Then v01 = 2.38 = 3.5 − (0.76) RC1 or RC1 = 1.47 kΩ 17.2 (a)

iC 2 = I Q = 0.5 =

3−0 ⇒ RC 2 = 6 K RC 2

(b)

iC1 = I Q = 0.5 =

3 −1 ⇒ RC1 = 4 K RC1

(c)

⎛V ⎞ I S exp ⎜ BE1 ⎟ iC1 ⎝ VT ⎠ = IQ ⎡ ⎛V ⎞ ⎛ V ⎞⎤ I S ⎢ exp ⎜ BE1 ⎟ + exp ⎜ BE 2 ⎟ ⎥ V ⎝ T ⎠ ⎝ VT ⎠ ⎦ ⎣ 1 ⎛ VBE 2 − VBE1 ⎞ 1 + exp ⎜ ⎟ VT ⎝ ⎠ vI = VBE1 − VBE 2 =

So iC1 = IQ

1 ⎛ −v ⎞ 1 + exp ⎜ I ⎟ ⎝ VT ⎠ 0.1 1 = = 0.2 0.5 ⎛ −v ⎞ 1 + exp ⎜ I ⎟ ⎝ VT ⎠

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⎛ −v ⎞ 1 exp ⎜ I ⎟ = −1 = 4 ⎝ VT ⎠ 0.2 ( −vI ) = ( 0.026 ) ln (4) vI = −0.0360 V

17.3 (a) vI = 0.5 V, Q1 on, Q2 off = v02 = 3 V v01 = 3 − (1)(0.5) = 2.5 V (b) vI = −0.5 V Q1 off, Q2 on ⇒ v01 = 3 V v02 = 3 − (1)(0.5) = 2.5 V 17.4 (a)

Q2 on, vE = −1.2 − 0.7 = −1.9V

iE = iC 2 =

−1.9 − ( −5.2 )

= 1.32 mA 2.5 v2 = −1V = −iC 2 RC 2 = − (1.32 )( RC 2 ) RC 2 = 0.758 k Ω Q1 on, vE = −0.7 − 0.7 = −1.40 V

(b) iE = iC1 =

−1.4 − ( −5.2 )

= 1.52 mA 2.5 v1 = −1V = −iC1 RC1 = − (1.52 )( RC1 ) RC1 = 0.658 k Ω

(c) For vin = −0.7 V , Q1 on, Q2 off ⇒ vO1 = −0.70V vO 2 = −1 − 0.7 ⇒ vO 2 = −1.7 V

For vin = −1.7 V , Q1 off , Q2 on ⇒ vO 2 = −0.7 V vO1 = −1 − 0.7 ⇒ vO1 = −1.7 V (d) (i) For vin = −0.7V , iE = 1.52 mA iC 4 = iC 3 =

−1.7 − ( −5.2 ) 3 −0.7 − ( −5.2 )

= 1.17 mA

= 1.5 mA 3 P = ( iE + iC 4 + iC 3 )( 5.2 ) = (1.52 + 1.17 + 1.5 )( 5.2 )

or P = 21.8 mW (ii) For vin = −1.7V , iE = 1.32 mA iC 4 = iC 3 =

−0.7 − ( −5.2 ) 3 −1.7 − ( −5.2 )

= 1.5 mA

= 1.17 mA 3 P = (1.32 + 1.5 + 1.17 )( 5.2 )

or P = 20.7 mW

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17.5

3.7 − 0.7 = 1.5 mA 0.67 + 1.33 VR = I 3 R4 + Vγ = (1.5 )(1.33) + 0.7 I3 =

a.

or VR = 2.70 V logic 1 level = 3.7 − 0.7 ⇒ 3.0 V

b.

For v X = vY = logic 1. 3 − 0.7 iE = = 2.875 mA = iRC1 0.8 vB 3 = 3.7 − ( 2.875 )( 0.21) = 3.10 V ⇒ v01 ( logic 0 ) = 2.4 V For v X = vY = logic 0, QR on 2.7 − 0.7 iE = = 2.5 mA = iRC 2 0.8 vB 4 = 3.7 − ( 2.5 )( 0.24 ) = 3.1 V ⇒ v02 ( logic 0 ) = 2.4 V 17.6 I REF =

(a)

1.1 − 0.7 − ( −1.9 )

I Q = I REF

20 = 0.115 mA

= 0.115 mA

vo1 (max) = 1.1 − 0.7 = 0.4 V

(b)

i5 (max) =

vo1 (max) − ( −1.9 ) R5

0.4 + 1.9 = 11.5 K 0.2 vo1 = −0.4 V ⇒ vB 5 = +0.3 V 1.1 − 0.3 RC1 = = 6.96 K 0.115 vo 2 = −0.4 V ⇒ vB 6 = +0.3V 1.1 − 0.3 RC 2 = = 6.96 K 0.115 R5 = R6 =

(c)

(d)

17.7 logic 1 + logic 0 1 + 0 = = 0.5 V 2 2 For i2 = 1 mA VR =

R5 =

0.5 − ( −2.3)

1 For QR on, iE =

⇒ R5 = 2.8 kΩ

VR − VBE − ( −2.3) RE

or

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0.5 − 0.7 + 2.3 ⇒ RE = 2.1 kΩ 1 = VR + VBE = 0.5 + 0.7 = 1.2 V

RE = VB 2

i1 =

1.2 − 1.4 − ( −2.3) R2

or 1.2 − 1.4 + 2.3 ⇒ R2 = 2.1 kΩ 1 1.7 − 1.2 R1 = ⇒ R1 = 0.5 kΩ 1 1 − (−2.3) i3 = = 3 ⇒ R3 = 1.1 kΩ R3 R2 =

i4 =

0 − (−2.3) = 3 ⇒ R4 = 0.767 kΩ R4

For QR on, v0 R = logic 0 = 0 V ⇒ vB 3 = 0.7 V iE = iCR = 1 mA So 1.7 − 0.7 RC 2 = ⇒ RC 2 = 1 kΩ 1 For vI = logic 1 = 1 V, iE =

1 − 0.7 − ( −2.3)

For vNOR

= 1.238 mA 2.1 = logic 0 = 0 V,

vB 4 = 0.7 V Then 1.7 − 0.7 RC1 = ⇒ RC1 = 0.808 kΩ 1.238

17.8 Maximum iE for vI = logic 1 = 3.3 V Then iE = 5 mA =

3.3 − 0.7 ⇒ RE = 0.52 kΩ RE

For v02 = logic 1 = 3.3 V iE 3 =

3.3 − 0 = 5 mA R3

or R3 = 0.66 kΩ By symmetry, R2 = 0.66 kΩ For Q1 on, iE = iRC1 = 5 mA =

4 − ( 2.7 + 0.7 ) RC1

So RC1 = 0.12 kΩ

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For QR on, 3 − 0.7 = 4.423 mA = iRC 2 0.52 4 − ( 2.7 + 0.7 ) and iRC 2 = 4.423 = RC 2 iE =

⇒ RC 2 = 0.136 kΩ

17.9 Neglecting base currents: (a) I E1 = 0, I E 3 = 0 5 − 0.7 ⇒ I E 5 = 1.72 mA 2.5 Y = 0.7 V I E5 =

(b)

5 − 0.7 ⇒ I E1 = 0.239 mA 18 =0

I E1 = I E3

5 − 0.7 ⇒ I E 5 = 1.72 mA 2.5 Y = 0.7 V I E5 =

(c)

I E1 = I E 3 =

5 − 0.7 ⇒ I E1 = I E 3 = 0.239 mA 18

I E 5 = 0, Y = 5 V

(d)

Same as (c).

17.10 (a) (b)

VR = −(1)(1) − 0.7 ⇒ VR = −1.7 V QR off , then vO1 = Logic 1 = −0.7 V

QR on, then vO1 = −(1)(2) − 0.7 ⇒ vO1 = Logic 0 = −2.7 V

QA / QB − off , then vO 2 = Logic 1 = −0.7 V QA / QB − on, then vO 2 = −(1)(2) − 0.7 ⇒ vO 2 = Logic 0 = −2.7 V

(c)

A = B = Logic 0 = −2.7 V , QR on, VE = −1.7 − 0.7 ⇒ VE = −2.4 V

A = B = Logic 1 = −0.7 V , QA / QB on, VE = −0.7 − 0.7 ⇒ VE = −1.4 V

(d)

A = B = Logic 1 = −0.7 V , QA / QB on,

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−2.7 − (−5.2) = 1.67 mA 1.5 −0.7 − (−5.2) iC 2 = = 3 mA 1.5 P = (1.67 + 1 + 1 + 1 + 3)(5.2) ⇒ P = 39.9 mW iC 3 =

A = B = Logic 0 = −2.7 V iC 3 = 3 mA, iC 2 = 1.67 mA P = 39.9 mW

17.11 a. b.

AND logic function logic 0 = 0 V

5 − (1.6 + 0.7) = 2.25 mA 1.2 V2 = (2.25)(0.8) ⇒ logic 1 = 1.8 V Q3 on, i =

c.

5 − 0.7 ⇒ iE1 = 1.65 mA 2.6 5 − (0.7 + 0.7) = ⇒ iE 2 = 3 mA 1.2 iC 3 = 0, iC 2 = iE 2 = 3 mA

iE1 = iE 2

V2 = 0

d.

5 − (1.8 + 0.7) ⇒ iE1 = 0.962 mA 2.6 5 − (1.6 + 0.7) = ⇒ iE 2 = 2.25 mA 1.2 iC 2 = 0, iC 3 = iE 2 = 2.25 mA

iE1 = iE 2

V2 = 1.8 V

17.12 a. b.

3.5 + 3.1 − 0.7 ⇒ VR = 2.6 V 2 For Q1 on, v X = vY = logic 1 = 3.5 V VR =

3.5 − (0.7 + 0.7) = 0.175 mA 12 0.175 0.4 Want iRC1 = = ⇒ RC1 = 4.57 kΩ 2 RC1 iE =

c.

For Q2 on, iE =

Want iRC 2 =

2.6 − 0.7 = 0.158 mA 12

0.158 0.4 = ⇒ RC 2 = 5.06 kΩ 2 RC 2

For vY = logic 1 = 3.5 V 3.5 − 0.7 = 0.35 mA, i E = 0.175 mA iR1 = 8 P = ( iR1 + iE )(VCC ) = ( 0.35 + 0.175 )( 3.5 ) ⇒ P = 1.84 mW

d.

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17.13 a. logic 1 = 0.2 V logic 0 = −0.2 V iE =

b.

( 0 − 0.7 ) − ( −3.1) RE

= 0.8

So RE = 3 kΩ 0.8 0.4 = ⇒ R1 = 1 kΩ 2 R1

c.

Want iR1 =

d.

For v X = vY = logic 1 = 0.2 V

iE =

( 0.2 − 0.7 ) − ( −3.1)

iR 2 =

e.

3

= 0.867 mA

0.4 ⇒ iR 2 = 0.4 mA 1 iD 2 = 0.467 mA iE = 0.867 mA

0.2 − (−3.1) i3 = = 1 mA 3.3 −0.2 − (−3.1) i4 = = 0.879 mA 3.3 P = (0.867 + 1 + 0.879)(0.9 − (−3.1))

or P = 11.0 mW 17.14 a. ( −0.9 − 0.7 ) − ( −3) i1 = ⇒ i1 = 1.4 mA 1 ( −0.2 − 0.7 ) − ( −3) ⇒ i3 = 0.14 mA i3 = 15 ( −0.2 − 0.7 ) − ( −3) ⇒ i4 = 0.14 mA i4 = 15 i2 + iD = i1 + i3 = 1.4 + 0.14 = 1.54 mA 0.4 i2 = ⇒ i2 = 0.8 mA 0.5 iD = 0.74 mA v0 = −0.4 V

b.

i1 = 1.4 mA

(0 − 0.7) − (−3) ⇒ i3 = 0.153 mA 15 i4 = i3 ⇒ i4 = 0.153 mA i3 =

i2 + iD = i4 ⇒ i2 = 0.153 mA iD = 0 v0 = −(0.153)(0.5) ⇒ v0 = −0.0765 V

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c.

i1 =

( 0 − 0.7 − 0.7 ) − ( −3) 1

⇒ i1 = 1.6 mA

(−0.2 − 0.7) − (−3) ⇒ i3 = 0.14 mA 15 i4 = i3 ⇒ i4 = 0.14 mA i3 =

i2 + iD = i3 ⇒ i2 = 0.14 mA iD = 0.0 v0 = −(0.14)(0.5) ⇒ v0 = −0.07 V (0 − 0.7 − 0.7) − (−3) ⇒ i1 = 1.6 mA 1 (0 − 0.7) − (−3) ⇒ i3 = 0.153 mA i3 = 15 i4 = i3 ⇒ i4 = 0.153 mA

d.

i1 =

i2 + iD = i1 + i4 = 1.6 + 0.153 = 1.753 mA 0.4 i2 = ⇒ i2 = 0.8 mA 0.5 iD = 0.953 mA v0 = −0.40 V

17.15 a. i. A = B = C = D = 0 ⇒ Q1 − Q4 cutoff So VDD = 2 I E R 1 + VEB + I B R2 and I I IB = E = E 1 + β P 51 Then I 2.5 = 2 I E (2) + 0.7 + E ⋅ (15) 51 15 ⎛ ⎞ 2.5 = 0.7 = I E ⋅ ⎜ 4 + ⎟ 51 ⎠ ⎝ So I E = 0.419 mA and Y = 2.5 − 2(0.4192)(2) ⇒ Y = 0.823 V ii. A = C = 0, B = D = 2.5 V Now vB 5 = vB 6 = 2.5 − 0.7 = 1.8 V and Y = vB 5 + 0.7 ⇒ Y = 2.5 V

b.

Y = ( A OR B ) AND (C OR D)

17.16 a. logic 1 = 0 V logic 0 = −0.4 V

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v01 = A OR B

b.

v02 = C OR D v03 = v01 OR v02 or v03 = ( A OR B ) AND (C OR D)

17.17 a.

For CLOCK = high, I DC flows through the left side of the circuit.. If D is high, I DC flows through

the left R resistor pulling Q low. If D is low. I DC flows through the right R resistor pulling Q low. For CLOCK = low, I DC flows through the right side of the circuit maintaining Q and Q in their previous state. P = ( I DC + 0.5 I DC + 0.1I DC + 0.1 I DC )( 3) b. P = 1.7 I DC ( 3) = (1.7 )( 50 )( 3) ⇒ P = 255 μ W

17.18 (a) vI = 0 ⇒ V1 = 0.7 V 3.3 − 0.7 = 0.433 mA 6 iB = iC = 0

i1 =

vo = 3.3 V (b) vI = 3.3 V v1 = 0.7 + 0.8 = 1.5 V 3.3 − 1.5 i1 = = 0.3 mA 6 0.8 iR = = 0.04 mA 20 iB = 0.3 − 0.04 = 0.26 mA 3.3 − 0.1 iC = = 0.8 mA 4 vo = 0.1 V

17.19 i. i1 =

For v X = vY = 0.1 V ⇒ v ′ = 0.8 V

5 − 0.8 ⇒ i1 = 0.525 mA 8 i3 = i4 = 0

ii. For v X = vY = 5 V, v′ = 0.8 + 0.7 + 0.7 ⇒⇒ v′ = 2.2 V 5 − 2.2 ⇒ i1 = 0.35 mA 8 0.8 ⇒ i4 = 0.297 mA i4 = i1 − 15 5 − 0.1 i3 = ⇒ i3 = 2.04 mA 2.4

i1 =

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17.20 a.

For v X = vY = 5 V , both Q1 and Q2 driven into saturation.

v1 = 0.8 + 0.7 + 0.8 ⇒ v1 = 2.3 V 5 − 2.3 ⇒ i1 = iB1 = 0.675 mA 4 5 − (0.8 + 0.7 + 0.1) i2 = ⇒ i2 = 1.7 mA 2 i4 = iB1 + i2 ⇒ i4 = 2.375 mA

i1 =

0.8 ⇒ i5 = 0.08 mA 10 = i4 − i5 ⇒ iB 2 = 2.295 mA

i5 = iB 2

i3 =

5 − 0.1 ⇒ i3 = 1.225 mA 4 v0 = 0.1V

b.

iL′

5 − (0.1 + 0.7) = 1.05 mA 4 iC (max) = β iB 2 = NiL′ + i3

=

(20)(2.295) = N (1.05) + 1.225 So N = 42

17.21 DX and DY off, Q1 forward active mode v1 = 0.8 + 0.7 + 0.7 = 2.2 V 5 = i1 R1 + i2 R2 + v1 and i1 = (1 + β )i2

So 5 − 2.2 = i2 [ (1 + β ) R1 + R2 ]

Assume β = 25 5 − 2.2 i2 = ⇒ i2 = 0.0589 mA (26)(1.75) + 2 i1 = (1 + β )i2 = (26)(0.05895) ⇒ i1 = 1.53 mA i3 = β i2 ⇒ i3 = 1.47 mA 0.8 = 0.0589 + 1.47 − 0.16 ⇒ 5 iBo = 1.37 mA Qo in saturation

iBo = i2 + i3 −

iCo =

5 − 0.1 ⇒ iCo = 0.817 mA 6

17.22 vI = 0 V, Q1 forward actions (a) 5 − 0.7 iB = = 0.717 mA 6 iC = (25)(0.71667) = 17.9 mA iE = (26)(0.71667) = 18.6 mA

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VI = 0.8 V

(b)

5 − (0.8 + 0.7) = 0.583 mA 6 Because of the relative doping levels of the Emitter and collector, and because of the difference in B-C and B-E areas, we have −iC ≈ iB = 0.583 mA and iE = small value. (c) vI = 3.6 Q1 inverse active. 5 − (0.8 + 0.7) iB = = 0.583 mA 6 iE = − β R iE = −(0.5)(0.583) = −0.292 mA iC = −iB − iE = −0.583 − 0.292 ⇒ iC = −0.875 mA iB =

17.23 a. i1 =

i.

v X = vY = 0.1 V, so Q1 in saturation.

5 − (0.1 + 0.8) ⇒ i1 = 0.683 mA 6 ⇒ iB 2 = i2 = i4 = iB 3 = i3 = 0 v X = vY = 5 V, so Q1 in inverse active mode.

ii.

Assume Q2 and Q3 in saturation. 5 − (0.8 + 0.8 + 0.7) ⇒ i1 = iB 2 = 0.45 mA 6 5 − (0.8 + 0.1) ⇒ i2 = 2.05 mA i2 = 2 0.8 ⇒ i4 = 0.533 mA i4 = 1.5 i1 =

iB 3 = ( iB 2 + i2 ) − i4 = 0.45 + 2.05 − 0.533

or iB 3 = 1.97 mA i3 =

b.

5 − 0.1 ⇒ i3 = 2.23 mA 2.2 For Q3 :

i3 2.23 = = 1.13 < β iB 3 1.97

For Q2 : i2 2.05 = = 4.56 < β iB 2 0.45

Since ( I C / I B ) < β , then each transistor is in saturation. 17.24 (a)

v X = vY = Logic 1

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v ′ = 0.8 + 2 ( 0.7 ) = 2.2 V 5 − 2.2 i1 = = 0.35 mA 8 0.8 i4 = i1 − = 0.35 − 0.0533 = 0.2967 mA 15 5 − 0.1 i3 = = 2.04 mA 2.4 5 − ( 0.1 + 0.7 ) iL′ = = 0.525 mA 8

Assume β = 25 Then ( 25 )( 0.2967 ) = 2.04 + N ( 0.525 ) So N = 10.2 ⇒ N = 10 (b) Now 5 = 2.04 + N ( 0.525 ) So N = 5.64 ⇒ N = 5 17.25 a. iB1 = iB 2

For v X = vY = 5 V, Q, in inverse active mode.

5 − ( 0.8 + 0.8 + 0.7 )

= 0.45 mA 6 = iB1 + 2 β R iB1 = 0.45(1 + 2 [ 0.1]) = 0.54 mA

iC 2 =

5 − ( 0.8 + 0.1) 2

iB 3 = ( iB 2 + iC 2 ) −

= 2.05 mA

0.8 = 0.54 + 2.05 − 0.533 1.5

or iB 3 = 2.06 mA Now 5 − (0.1 + 0.8) = 0.683 mA 6 Then iC 3 (max) = β F iB 3 = NiL′ or (20)(2.06) = N (0.683) ⇒ N = 60 iL′ =

b.

From above, for v0 high, I L′ = (0.1)(0.45) = 0.045 mA. Now

⎛ 5 − 4.9 ⎞ (21)(0.1) I L′ (max) = (1 + β F ) ⎜ ⎟ = R 2 2 ⎝ ⎠ = 1.05 mA So I L (max) = NI L′ or 1.05 = N (0.045) ⇒ N = 23

17.26

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Vin = 0.1V : QI , Sat : Qs , Qo , Cutoff

(a)

5 − (0.1 + 0.8) = 1.025 mA 4 P = iI (5 − 0.1) = (1.025)(4.9) ⇒

iI =

P = 5.02 mW Vin = 5V , QI , Inverse Active; QS , Qo , Saturation

(b)

vBI = 0.7 + 0.8 + 0.7 = 2.2V 5 − 2.2 = 0.7 mA 4 iEI = β R ⋅ iI = (0.1)(0.7) = 0.07 mA iI =

Vout = 0.7 + 0.1 = 0.8 V 5 − 0.8 i2 = = 4.2 mA 1 P = (iI + iEI + i2 )(5) = (0.7 + 0.07 + 4.2)(5) ⇒ P = 24.9 mW

17.27 a. iB1 =

v X = vY = vZ = 0.1 V

5 − (0.1 + 0.8) ⇒ iB1 = 1.05 mA 3.9

Then iC1 = iB 2 = iC 2 = iB 3 = iC 3 = 0 v X = vY = vZ = 5 V

b. iB1 =

5 − (0.8 + 0.8 + 0.7) ⇒ iB1 = 0.692 mA 3.9

Then iC1 = iB 2 = iB1 (1 + 3β R ) = (0.692)(1 + 3[0.5]) ⇒ iC1 = iB 2 = 1.73 mA 5 − (0.1 + 0.8) ⇒ iC 2 = 2.05 mA 2 0.8 = iB 2 + iC 2 − = 1.73 + 2.05 − 1.0 0.8 ⇒ iB 3 = 2.78 mA

iC 2 = iB 3

5 − 0.1 = 2.04 mA 2.4 5 − (0.1 + 0.8) iL′ = = 1.05 mA 3.9 iC 3 = iR 3 + 5iL′ = 2.04 + (5)(1.05) ⇒ iC 3 = 7.29 mA iR 3 =

17.28 a.

v X = vY = vZ = 2.8 V, Q1 biased in the inverse active mode.

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2.8 − (0.8 + 0.8 + 0.7) ⇒ iB1 = 0.25 mA 2 = iB1 (1 + 3β R ) = 0.25(1 + 3 [0.3]) ⇒ iB 2 = 0.475 mA

iB1 = iB 2

vC 2 = 0.8 + 0.1 = 0.9 V 0.9 − (0.7 + 0.1) 0.1 = (1 + β F )(0.5) (101)(0.5)

iB 4

=

iR 2

= 0.00198 mA (Negligible) 5 − 0.9 = = 4.56 mA 0.9 ⇒ iC 2 = 4.56 mA 0.8 = 0.475 + 4.56 − 0.8 1 = 4.235 mA

iB 3 = iB 2 + iC 2 − ⇒ iB 3

v X = vY = vZ = 0.1 V

b.

5 − (0.1 + 0.8) ⇒ iB1 = 2.05 mA 2 From part (a), iL′ = β R ⋅ iB1 = (0.3)(0.25) = 0.075 mA Then 5iL′ 5(0.075) iB 4 = = ⇒ iB 4 = 0.00371 mA 1+ βF 101 iB1 =

17.29 a.

v X = vY = vZ = 0.1 V

2 − (0.1 + 0.8) + iB 3 RB1 where (2 − 0.7) − (0.9) 0.4 iB 3 = = RB 2 1 iB1 =

⇒ iB 3 = 0.4 mA

Then iB1 =

1.1 + 0.4 ⇒ iB1 = 1.5 mA 1 iB 2 = 0 = iC 2

Q3 in saturation iC 3 = 5iL′ For v0 high, vB′ 1 = 0.8 + 0.7 = 1.5 V ⇒ Q3′ off

2 − 1.5 = 0.5 mA 1 iL′ = β R iB′ 1 = (0.2)(0.5) = 0.1 mA iB′ 1 =

Then

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iC 3 = 0.5 mA v X = vY = vZ = 2 V b. From part (a), ⇒ iB1 = 0.5 mA iB 3 = 0 = iC 3 iB 2 = iB1 (1 + 3β R ) = (0.5)(1 + 3 [0.2]) iB 2 = 0.8 mA iC 2 = 5iL′ , and from part (a), iL′ = 1.5 mA

So iC 2 = 7.5 mA 17.30 5.8 − 0.7 = 0.51 mA 10 5 − (0.7 − 0.3) = 4.6 mA IC − I D = 1 Now I I I D = 0.51 − I B = 0.51 − C = 0.51 − C 50 β Then I ⎞ 1 ⎞ ⎛ ⎛ I C − I D = I C − ⎜ 0.51 − C ⎟ = I C ⎜ 1 + ⎟ − 0.51 = 4.6 50 ⎠ ⎝ 50 ⎠ ⎝ So I C = 5.01 mA

IB + ID =

(a)

IC

5.01 ⇒ I B = 0.1002 mA 50 I D = 0.51 − 0.1002 ⇒ I D = 0.4098 mA

IB =

β

=

VCE = 0.4 V I D = 0, VCE = VCE ( sat ) = 0.1 V

(b)

5.8 − 0.8 ⇒ I B = 0.5 mA 10 5 − 0.1 ⇒ I C = 4.9 mA IC = 1 IB =

17.31 a.

v X = vY = 0.4 V

vB1 = 0.4 + 0.7 ⇒ vB1 = 1.1 V 5 − 1.1 ⇒ iB1 = 1.39 mA 2.8 = 0.4 + 0.4 ⇒ vB 2 = 0.8 V

iB1 = vB 2

iB 2 = iC 2 = iB 0 = iC 0 = iB 5 = iC 5 = iB 3 = iC 3 = 0 ( No load )

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5 = iB 4 R2 + VBE + (1 + β )iB 4 R4 5 − 0.7 ⇒ iB 4 = 0.0394 mA 0.76 + (31)(3.5) = β F iB 4 ⇒ iC 4 = 1.18 mA

iB 4 = iC 4

vB 4 = 5 − (0.0394)(0.76) ⇒ vB 4 = 4.97 V v X = vY = 3.6 V

b.

vB1 = 0.7 + 0.7 + 0.3 ⇒ vB1 = 1.7 V vB 2 = 1.4 V vB 0 = 0.7 V vC 2 = 1.1 V 5 − 1.7 ⇒ iB1 = 1.1786 mA 2.8 = iB1 (1 + 2 β R ) = 1.18(1 + 2 [0.1])

iB1 = iB 2

iB 2 = 1.41 mA 1.1 − 0.7 ⇒ iB 4 = 0.00369 mA (31)(3.5) 5 − 1.1 = = 5.13 mA ⇒ iC 2 ≈ 5.13 mA 0.76 ≈ iB 2 + iC 2

iB 4 = iR 2 iB 0

iB 0 = 6.54 mA

17.32 (a)

vI = 0, v1 = 0.3 V

1.5 − 0.3 = 1.2 mA 1 iB = iC = 0, vO = 1.5 V i1 =

(b)

vI = 1.5 V v1 = 0.7 + 0.3 = 1 V

1.5 − 1 = 0.5 mA 1 0.7 iR = = 0.035 mA 20 iB = 0.5 − 0.035 = 0.465 mA i1 =

1.5 − 0.4 = 0.917 mA 1.2 vO = 0.4 V

iC =

17.33 a.

Assuming the output transistor Q2 is a Schottky transistor, then

v0 = 0.4 V, iL′ =

2.5 − (0.4 + 0.3) = 0.5 RB1

Then

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RB1 = 3.6 kΩ

Then iB1 =

2.5 − (0.7 + 0.8) 1.0 = = 0.278 mA 3.6 RB1

0.7 = 1.50 mA 0.7 iE1 = iB1 + iC1 ⇒ iC1 = 1.50 − 0.278 = 1.222 mA iB 2 = 0.5 mA, iE1 = 0.5 +

and iC1 =

2.5 − (0.7 + 0.1) = 1.222 mA RC1

⇒ RC1 = 1.39 kΩ

b.

v X = vY = 0.4 V, vB1 = 0.7 V

vC 2 = 2.5 − 0.7 ⇒ vC 2 = 1.8 V

All transistor currents are zero. c. vB1 = 1.5 V, vC1 = 0.8 V Currents calculated in part (a). iB 2 = 0.5 mA, iL′ = 0.5 mA d. iC 2 (max) = β iB 2 = NiL′ or (50)(0.5) = N (0.5) So N = 50 17.34 a.

For v X = vY = 3.6 V

5 − 2.1 = 0.29 mA 10 5 − 1.8 vC1 = 0.7 + 0.7 + 0.4 = 1.8 V ⇒ iC1 = = 0.32 mA 10 1.4 iB 2 = iB1 + iC1 − = 0.29 + 0.32 − 0.0933 15 So iB 2 = 0.517 mA vC 2 = 0.7 + 0.4 = 1.1 V 5 − 1.1 iC 2 = = 0.951 mA 4.1 vB1 = 3(0.7) = 2.1 ⇒ iB1 =

0.7 = 0.517 + 0.951 − 0.175 4 or iB 5 = 1.293 mA For v0 = 0.4 V, vB′ 1 = 0.4 + 0.7 = 1.1 V Then 1.1 − 0.7 iB′ 1 = = 0.00086 mA (31)(15) 5 − 1.1 iL′ = − 0.00086 or iL′ ≈ 0.39 mA 10 So iC 5 (max) = β iΒ 5 = NiL′ iB 5 = iB 2 + iC 2 −

(30)(1.293) = N (0.39) ⇒ N = 99

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b. P = (0.29 + 0.32 + 0.951)(5) + (99)(0.39)(0.4) P = 7.805 + 15.444 or P = 23.2 mW (Assumming 99 load circuits which is unreasonably large.) 17.35 a.

Assume no load. For v X = logic 0 = 0.4 V

5 − (0.4 + 0.7) iE1 = = 0.0975 mA 40 Essentially all of this current goes to ground from VCC . P = iE1 ⋅ VCC = (0.0975)(5) ⇒ P = 0.4875 mW

5 − (3)(0.7) = 0.0725 mA 40 5 − (0.7 + 0.7 + 0.4) = 0.064 mA iR 2 = 50 5 − (0.7 + 0.4) = 0.26 mA iR 3 = 15 P = (0.0725 + 0.064 + 0.26)(5) P = 1.98 mW

b.

iR1 =

c.

For v0 = 0, vC 7 = 0.7 + 0.4 = 1.1 V

iR 7 =

17.36 (a)

5 − 1.1 ⇒ iR 7 = 78 mA ≈ iSC 0.050 vl = vO = 25 V ; A transient situation

vDS ( M N ) = 2.5 − 0.7 = 1.8 V vGS ( M N ) = 2.5 − 0.7 = 1.8 V ⇒ M N in saturation vSD ( M P ) = 5 − (2.5 + 0.7) = 1.8 V vSG ( M P )5 − 2.5 = 2.5 V ⇒ M P in saturation iDN = K n (vGSN − VTN ) 2 = (0.1)(1.8 − 0.8) 2 ⇒ iDN = 0.1 mA iDP = K P (vSGP + VTP ) 2 = (0.1)(2.5 − 0.8) 2 ⇒ iDP = 0.289 mA iC1 = β iDP = (50)(0.289) ⇒ iC1 = 14.45 mA iC 2 = β iDN = (50)(0.1) ⇒ iC 2 = 5 mA

Difference between iE1 and iDN + iC 2 is a load current. (b) Assume iC1 = 14.45 mA is a constant i ⋅t (V )(C ) 1 VC = ∫ iC1dt = C1 ⇒ t = C C C iC1 t=

(c)

(5)(15 × 10−12 ) ⇒ t = 5.19 ns 14.45 × 10−3 t=

(5)(15 × 10−12 ) ⇒ t = 260 ns 0.289 × 10−3

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17.37

(a)

Assume R1 = R2 = 10 kΩ; β = 50

0.7 = 0.07 mA 10 NMOS in saturation region; vGSN = 2.5 − 0.7 = 1.8 V

Then iR1 = iR 2 =

iDN = K n ( vGSN − VTN ) = ( 0.1)(1.8 − 0.8 ) iDN = 0.10 mA 2

2

Then iB 2 = 0.03 ⇒ iC 2 = (50)(0.03) = 1.5 mA iE1 = 1.53 mA ⇒ iB1 = 0.03 mA ⇒ iC1 = 1.5 mA So iDP = 0.10 mA Now, M P biased in non-saturation region vSGP = 2.5 V 2 ⎤⎦ iDP = 0.10 = 0.10 ⎡⎣ 2(2.5 − 0.8)vSD − vSD 2 0.10 vSD − 0.34 vSD + 0.10 = 0

vSD =

0.34 ± (0.34) 2 − 4(0.10)(0.10) 2(0.10)

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Or vSD = 0.325 V Then vo = 5 − 0.325 − 0.7 vo = 3.975 V 1 i ⋅t idt = ∫ C C Cv (15 × 10−12 )(5) t= = i 1.53 × 10−3 t = 49 ns (c) Cv (15 × 10−12 )(5) t= = i 0.1× 10−3 t = 0.75 μ s

(b)

v=

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