Fundamental Of MIcroelectronics Bahzad Razavi Chapter 6 Solution Manual

6.4 (a)

Q(x)

Q(x) = W Cox (VGS − V (x) − VT H ) = W Cox (VGS − VT H ) − W Cox V (x)

W Cox (VGS − VT H )

Increasing VDS

L

x

The curve that intersects the axis at x = L (i.e., the curve for which the channel begins to pinch off) corresponds to VDS = VGS − VT H . (b) 1 µQ(x)

RLocal (x)

RLocal (x) ∝

Increasing VDS

L

x

Note that RLocal diverges at x = L when VDS = VGS − VT H .

6.15

ID

Increasing VDS

VT H

VGS

Initially, when VGS is small, the transistor is in cutoff and no current flows. Once VGS increases beyond VT H , the curves start following the square-law characteristic as the transistor enters saturation. However, once VGS increases past VDS + VT H (i.e., when VDS < VGS − VT H ), the transistor goes into triode and the curves become linear. As we increase VDS , the transistor stays in saturation up to larger values of VGS , as expected.

6.17 1 W µn Cox (VGS − VT H )α , α < 2 2 L ∂ID , ∂VGS α W α−1 = µn Cox (VGS − VT H ) 2 L αID = VGS − VT H

ID = gm

6.21 Since they’re being used as current sources, assume M1 and M2 are in saturation for this problem. To find the maximum allowable value of λ, we should evaluate λ when 0.99ID2 = ID1 and 1.01ID2 = ID1 , i.e., at the limits of the allowable values for the currents. However, note that for any valid λ (remember, λ should be non-negative), we know that ID2 > ID1 (since VDS2 > VDS1 ), so the case where 1.01ID2 = ID1 (which implies ID2 < ID1 ) will produce an invalid value for λ (you can check this yourself). Thus, we need only consider the case when 0.99ID2 = ID1 . W 1 2 (VB − VT H ) (1 + λVDS2 ) 0.99ID2 = 0.99 µn Cox 2 L = ID1 1 W 2 = µn Cox (VB − VT H ) (1 + λVDS1 ) 2 L 0.99 (1 + λVDS2 ) = 1 + λVDS1 λ = 0.02 V−1

5.27 VDD − ID RD = VGS = VT H +

s

2ID µn Cox W L

2ID 2 = (VDD − VT H − ID RD ) W µn Cox L i W h 1 2 2 2 (VDD − VT H ) − 2ID RD (VDD − VT H ) + ID RD ID = µn Cox 2 L

We can rearrange this to the standard quadratic form as follows:     1 W 2 W W 1 2 2 µn Cox RD ID − µn Cox RD (VDD − VT H ) + 1 ID + µn Cox (VDD − VT H ) = 0 2 L L 2 L Applying the quadratic formula, we have:
q
2
2 R (V − V ) + 1 ± µn Cox W − 4 21 µn Cox W µn Cox W DD TH L D L RD (VDD − VT H ) + 1 L RD (VDD − VT H )
ID = 2 2 21 µn Cox W L RD q
2
2 W µn Cox W − µn Cox W µn Cox L RD (VDD − VT H ) + 1 ± L RD (VDD − VT H ) + 1 L RD (VDD − VT H ) = 2 µn Cox W L RD q µn Cox W 1 + 2µn Cox W L RD (VDD − VT H ) + 1 ± L RD (VDD − VT H ) = 2 µn Cox W L RD Note that mathematically, there are two possible solutions for ID . However, since M1 is diodeconnected, we know it will either be in saturation or cutoff. Thus, we must reject the value of ID that does not match these conditions (for example, a negative value of ID would not match cutoff or saturation, so it would be rejected in favor of a positive value).

6.33 (a) Assume M1 is operating in saturation. VGS = 1 V 1 W 2 VDS = VDD − ID RD = VDD − µn Cox (VGS − VT H ) (1 + λVDS ) RD 2 L VDS = 1.35 V > VGS − VT H , which verifies our assumption ID = 4.54 mA W (VGS − VT H ) = 13.333 mS gm = µn Cox L 1 = 2.203 kΩ ro = λID

+ vgs

gm vgs

ro

RD

−

(b) Since M1 is diode-connected, we know it is operating in saturation. W 1 2 (VGS − VT H ) (1 + λVGS ) RD VGS = VDS = VDD − ID RD = VDD − µn Cox 2 L VGS = VDS = 0.546 V ID = 251 µA W (VGS − VT H ) = 3.251 mS gm = µn Cox L 1 = 39.881 kΩ ro = λID

+ vgs

gm vgs

ro

RD

−

(c) Since M1 is diode-connected, we know it is operating in saturation. ID = 1 mA r

gm = ro =

2µn Cox

W ID = 6.667 mS L

1 = 10 kΩ λID

+ vgs

gm vgs

ro

−

(d) Since M1 is diode-connected, we know it is operating in saturation. VGS = VDS 1 W µn Cox (VGS − VT H )2 (1 + λVGS ) (2 kΩ) 2 L = 0.623 V

VDD − VGS = ID (2 kΩ) = VGS = VDS

ID = 588 µA W gm = µn Cox (VGS − VT H ) = 4.961 mS L 1 ro = = 16.996 kΩ λID

+ gm vgs

vgs

2 kΩ

ro

−

(e) Since M1 is diode-connected, we know it is operating in saturation. ID = 0.5 mA r

gm = ro =

2µn Cox

W ID = 4.714 mS L

1 = 20 kΩ λID

+ vgs −

gm vgs

ro

6.38 (a) vout

+ vgs2

gm2 vgs2

ro2

gm1 vgs1

ro1

RD

−

vin

+ vgs1 −

(b) vin

vout

+ vgs1

gm1 vgs1

ro1

RD

−

+ gm2 vgs2

ro2

vgs2 −

(c) vin

vout

+ vgs1

gm1 vgs1

ro1

gm2 vgs2

ro2

−

+ vgs2 −

(d)

RD

vin

+ vgs1

gm1 vgs1

ro1

− vout + vgs2

gm2 vgs2

ro2

−

(e) vout

+ vgs1

gm1 vgs1

ro1

RD

− vin + gm2 vgs2

ro2

vgs2 −

6.43 (a) Assume M1 is operating in triode (since |VGS | = 1.8 V is large). |VGS | = 1.8 V i W h 1 2 2 (|VGS | − |VT H |) |VDS | − |VDS | (500 Ω) µp Cox 2 L |VDS | = 0.418 V < |VGS | − |VT H | , which verifies our assumption

VDD − |VDS | = |ID | (500 Ω) =

|ID | = 2.764 mA (b) Since M1 is diode-connected, we know it is operating in saturation. |VGS | = |VDS | 1 W 2 µp Cox (|VGS | − |VT H |) (1 kΩ) 2 L |VGS | = |VDS | = 0.952 V

VDD − |VGS | = |ID |(1 kΩ) =

|ID | = 848 µA (c) Since M1 is diode-connected, we know it is operating in saturation. |VGS | = |VDS | |VGS | = VDD − |ID |(1 kΩ) = VDD − |ID |(1 kΩ) = |VGS | = |VGS | = 0.952 V |ID | = 848 µA

1 W 2 µp Cox (|VGS | − |VT H |) (1 kΩ) 2 L

6.44 (a)

IX

Saturation

Cutoff

VDD − VT H

VDD

VX

VDD

VX

M1 goes from saturation to cutoff when VX = VDD − VT H = 1.4 V. (b)

IX 1 + VT H

Saturation M1 goes from saturation to triode when VX = 1 + VT H = 1.4 V. (c)

Triode

IX VDD − VT H

Saturation

VDD

VX

VDD

VX

Cutoff

M1 goes from saturation to cutoff when VX = VDD − VT H = 1.4 V. (d)

IX

Saturation

Cutoff

VT H M1 goes from cutoff to saturation when VX = VT H = 0.4 V.