Fundamental Of MIcroelectronics Bahzad Razavi Chapter 4 Solution Manual

4.4 According to Equation (4.8), we have  AE qDn n2i  VBE /VT −1 e NB WB 1 ∝ WB

IC =

We can see that if WB increases by a factor of two, then IC decreases by a factor of two .

4.11 VBE = 1.5 V − IE (1 kΩ) ≈ 1.5 V − IC (1 kΩ) (assuming β ≫ 1)   IC = VT ln IS IC = 775 µA VX ≈ IC (1 kΩ) = 775 mV

4.12 Since we have only integer multiples of a unit transistor, we need to find the largest number that divides both I1 and I2 evenly (i.e., we need to find the largest x such that I1 /x and I2 /x are integers). This will ensure that we use the fewest transistors possible. In this case, it’s easy to see that we should pick x = 0.5 mA, meaning each transistor should have 0.5 mA flowing through it. Therefore, I1 should be made up of 1 mA/0.5 mA = 2 parallel transistors, and I2 should be made up of 1.5 mA/0.5 mA = 3 parallel transistors. This is shown in the following circuit diagram. I1

VB

I2

+ −

Now we have to pick VB so that IC = 0.5 mA for each transistor.   IC VB = VT ln IS   5 × 10−4 A = (26 mV) ln 3 × 10−16 A = 732 mV

4.15 VB − VBE = IB R1 IC = β β [VB − VT ln(IC /IS )] IC = R1 IC = 786 µA

4.17 First, note that VBE1 = VBE2 = VBE . VB = (IB1 + IB2 )R1 + VBE R1 = (IX + IY ) + VT ln(IX /IS1 ) β 5 IS2 = IS1 3 5 ⇒ IY = IX 3 8R1 VB = IX + VT ln(IX /IS1 ) 3β IX = 509 µA IY = 848 µA

4.21 (a) VBE = 0.8 V IC = IS eVBE /VT = 18.5 mA VCE = VCC − IC RC = 1.58 V Q1 is operating in forward active. Its small-signal parameters are gm = IC /VT = 710 mS rπ = β/gm = 141 Ω ro = ∞ The small-signal model is shown below. B

C + rπ

gm vπ

vπ −

E

(b) IB = 10 µA IC = βIB = 1 mA VBE = VT ln(IC /IS ) = 724 mV VCE = VCC − IC RC = 1.5 V Q1 is operating in forward active. Its small-signal parameters are gm = IC /VT = 38.5 mS rπ = β/gm = 2.6 kΩ ro = ∞ The small-signal model is shown below. B

C + rπ

gm vπ

vπ −

E

(c) VCC − VBE 1+β = IC RC β β VCC − VT ln(IC /IS ) IC = 1+β RC

IE =

IC = 1.74 mA VBE = VT ln(IC /IS ) = 739 mV VCE = VBE = 739 mV Q1 is operating in forward active. Its small-signal parameters are gm = IC /VT = 38.5 mS rπ = β/gm = 2.6 kΩ ro = ∞ The small-signal model is shown below. B

C + rπ

gm vπ

vπ −

E

4.22 (a) IB = 10 µA IC = βIB = 1 mA VBE = VT ln(IC /IS ) = 739 mV VCE = VCC − IE (1 kΩ) 1+β (1 kΩ) = VCC − β = 0.99 V Q1 is operating in forward active. Its small-signal parameters are gm = IC /VT = 38.5 mS rπ = β/gm = 2.6 kΩ ro = ∞ The small-signal model is shown below. B

C + rπ

gm vπ

vπ −

E

(b) 1+β VCC − VBE = IC 1 kΩ β β VCC − VT ln(IC /IS ) IC = 1+β 1 kΩ

IE =

IC = 1.26 mA VBE = VT ln(IC /IS ) = 730 mV VCE = VBE = 730 mV Q1 is operating in forward active. Its small-signal parameters are gm = IC /VT = 48.3 mS rπ = β/gm = 2.07 kΩ ro = ∞ The small-signal model is shown below.

B

C + rπ

gm vπ

vπ −

E

(c) IE = 1 mA β IC = IE = 0.99 mA 1+β VBE = VT ln(IC /IS ) = 724 mV VCE = VBE = 724 mV Q1 is operating in forward active. Its small-signal parameters are gm = IC /VT = 38.1 mS rπ = β/gm = 2.63 kΩ ro = ∞ The small-signal model is shown below. B

C + rπ

gm vπ

vπ −

E

(d) IE = 1 mA β IC = IE = 0.99 mA 1+β VBE = VT ln(IC /IS ) = 724 mV VCE = VBE = 724 mV Q1 is operating in forward active. Its small-signal parameters are gm = IC /VT = 38.1 mS rπ = β/gm = 2.63 kΩ ro = ∞ The small-signal model is shown below.

B

C + rπ

gm vπ

vπ −

E

4.31 IC = IS eVBE /VT

  VCE 1+ VA

IC,T otal = nIC = nIS eVBE /VT



1+

VCE VA



∂IC ∂VBE IS = n eVBE /VT VT IC ≈n VT = ngm

gm,T otal =

IB,T otal rπ,T otal

ro,T otal

= n × 0.4435 S 1 = IC,T otal β −1  ∂IB,T otal = ∂VBE  −1 IC,T otal ≈ βVT −1  nIC = βVT rπ = n 225.5 Ω = (assuming β = 100) n  −1 ∂IC,T otal = ∂VCE −1  IC,T otal ≈ VA VA = nIC ro = n 693.8 Ω = n

The small-signal model is shown below. B

C +

rπ,T otal

gm,T otal vπ

vπ −

E

ro,T otal

4.32 (a) VBE = VCE (for Q1 to operate at the edge of saturation) VT ln(IC /IS ) = VCC − IC RC IC = 885.7 µA VB = VBE = 728.5 mV ′ ′ (b) Let IC′ , VB′ , VBE , and VCE correspond to the values where the collector-base junction is forward biased by 200 mV. ′ ′ VBE = VCE + 200 mV

VT ln(IC′ /IS ) = VCC − IC′ RC + 200 mV IC′ = 984.4 µA VB′ = 731.3 mV Thus, VB can increase by VB′ − VB = 2.8 mV if we allow soft saturation.

4.34 VBE = VCC − IB RB VT ln(IC /IS ) = VCC − IC RB /β IC = 1.67 mA VBC = VCC − IB RB − (VCC − IC RC ) < 200 mV IC RC − IB RB < 200 mV 200 mV + IB RB RC < IC 200 mV + IC RB /β = IC RC < 1.12 kΩ

4.41

VCC

VEB = VEC (for Q1 to operate at the edge of saturation) − IB RB = VCC − IC RC IC RB /β = IC RC RB /β = RC β = RB /RC = 100

4.44 (a) IB = 2 µA IC = βIB = 200 µA VEB = VT ln(IC /IS ) = 768 mV VEC = VCC − IE (2 kΩ) 1+β = VCC − IC (2 kΩ) β = 2.1 V Q1 is operating in forward active. Its small-signal parameters are gm = IC /VT = 7.69 mS rπ = β/gm = 13 kΩ ro = ∞ The small-signal model is shown below. B

C + rπ

gm vπ

vπ −

E

(b) IE =

VCC − VEB 5 kΩ

VCC − VT ln(IC /IS ) 1+β IC = β 5 kΩ IC = 340 µA VEB = 782 mV VEC = VEB = 782 mV Q1 is operating in forward active. Its small-signal parameters are gm = IC /VT = 13.1 mS rπ = β/gm = 7.64 kΩ ro = ∞ The small-signal model is shown below.

B

C + rπ

gm vπ

vπ −

E

(c) IE =

1+β IC = 0.5 mA β

IC = 495 µA VEB = 971 mV VEC = VEB = 971 mV Q1 is operating in forward active. Its small-signal parameters are gm = IC /VT = 19.0 mS rπ = β/gm = 5.25 kΩ ro = ∞ The small-signal model is shown below. B

C + rπ

gm vπ

vπ −

E

4.49 The direction of current flow in the large-signal model (Fig. 4.40) indicates the direction of positive current flow when the transistor is properly biased. The direction of current flow in the small-signal model (Fig. 4.43) indicates the direction of positive change in current flow when the base-emitter voltage vbe increases. For example, when vbe increases, the current flowing into the collector increases, which is why ic is shown flowing into the collector in Fig. 4.43. Similar reasoning can be applied to the direction of flow of ib and ie in Fig. 4.43.

4.53 (a) VCB2 < 200 mV IC2 RC < 200 mV IC2 < 400 µA VEB2 = VE2 = VT ln(IC2 /IS2 ) < 741 mV β2 IE2 RC 1 + β2 β2 1 + β1 IC1 RC 1 + β2 β1 IC1 VBE1 Vin

< 200 mV < 200 mV < 396 µA = VT ln(IC1 /IS1 ) < 712 mV = VBE1 + VEB2 < 1.453 V

(b) IC1 = 396 µA IC2 = 400 µA gm1 = 15.2 mS rπ1 = 6.56 kΩ ro1 = ∞ gm2 = 15.4 mS rπ2 = 3.25 kΩ ro2 = ∞ The small-signal model is shown below. + vin

rπ1 −

B1 + vπ1

C1 gm1 vπ1

− E1 /E2

rπ2

− vπ2 + B2

gm2 vπ2 vout

C2 RC

4.55 (a)

VBE2 − (VCC

VBC2 < 200 mV − IC2 RC ) < 200 mV

VT ln(IC2 /IS2 ) + IC2 RC − VCC < 200 mV IC2 < 3.80 mA VBE2 < 799.7 mV 1 + β1 IC1 = IB2 = IC2 /β2 IE1 = β1 IC1 < 75.3 µA VBE1 < 669.2 mV Vin = VBE1 + VBE2 < 1.469 V (b) IC1 = 75.3 µA IC2 = 3.80 mA gm1 = 2.90 mS rπ1 = 34.5 kΩ ro1 = ∞ gm2 = 146.2 mS rπ2 = 342 Ω ro2 = ∞ The small-signal model is shown below. B1 +

+ vin

rπ1 −

C1 gm1 vπ1

vπ1 − E1 /B2

C2

vout

+ rπ2

gm2 vπ2

vπ2 −

E2

RC