6.4 (a)

Q(x) = WCox (VGS − V (x) − VTH)

= WCox (VGS − VTH) − WCoxV (x)

Increasing VDS

L

WCox (VGS − VTH)

x

Q(x

)

The curve that intersects the axis at x = L (i.e., the curve for which the channel begins to pinchoff) corresponds to VDS = VGS − VTH .

(b)

RLocal(x) ∝1

µQ(x)

L x

RL

ocal(x

)

Increasing VDS

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

6.15

VGS

ID

VTH

Increasing VDS

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

6.17

ID =1

2µnCox

W

L(VGS − VTH)α

, α < 2

gm ,∂ID

∂VGS

=α

2µnCox

W

L(VGS − VTH)

α−1

=αID

VGS − VTH

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 casewhere 1.01ID2 = ID1 (which implies ID2 < ID1) will produce an invalid value for λ (you can check thisyourself). Thus, we need only consider the case when 0.99ID2 = ID1.

0.99ID2 = 0.991

2µnCox

W

L(VB − VTH)

2(1 + λVDS2)

= ID1

=1

2µnCox

W

L(VB − VTH)

2(1 + λVDS1)

0.99 (1 + λVDS2) = 1 + λVDS1

λ = 0.02 V−1

5.27

VDD − IDRD = VGS = VTH +

√

2ID

µnCoxWL

2ID

µnCoxWL

= (VDD − VTH − IDRD)2

ID =1

2µnCox

W

L

[

(VDD − VTH)2− 2IDRD (VDD − VTH) + I2

DR2

D

]

We can rearrange this to the standard quadratic form as follows:

(

1

2µnCox

W

LR2

D

)

I2

D −

(

µnCox

W

LRD (VDD − VTH) + 1

)

ID +1

2µnCox

W

L(VDD − VTH)

2= 0

Applying the quadratic formula, we have:

ID =

(

µnCoxWL

RD (VDD − VTH) + 1)

±

√

(

µnCoxWL

RD (VDD − VTH) + 1)2

− 4(

1

2µnCox

WL

RD (VDD − VTH))2

2(

1

2µnCox

WL

R2

D

)

=µnCox

WL

RD (VDD − VTH) + 1 ±

√

(

µnCoxWL

RD (VDD − VTH) + 1)2

−(

µnCoxWL

RD (VDD − VTH))2

µnCoxWL

R2

D

=µnCox

WL

RD (VDD − VTH) + 1 ±√

1 + 2µnCoxWL

RD (VDD − VTH)

µnCoxWL

R2

D

Note that mathematically, there are two possible solutions for ID. However, since M1 is diode-connected, 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 orsaturation, so it would be rejected in favor of a positive value).

6.33 (a) Assume M1 is operating in saturation.

VGS = 1 V

VDS = VDD − IDRD = VDD −1

2µnCox

W

L(VGS − VTH)

2(1 + λVDS)RD

VDS = 1.35 V > VGS − VTH , which verifies our assumption

ID = 4.54 mA

gm = µnCox

W

L(VGS − VTH) = 13.333 mS

ro =1

λID

= 2.203 kΩ

+

vgs

−

gmvgs ro RD

(b) Since M1 is diode-connected, we know it is operating in saturation.

VGS = VDS = VDD − IDRD = VDD −1

2µnCox

W

L(VGS − VTH)

2(1 + λVGS)RD

VGS = VDS = 0.546 V

ID = 251 µA

gm = µnCox

W

L(VGS − VTH) = 3.251 mS

ro =1

λID

= 39.881 kΩ

+

vgs

−

gmvgs ro RD

(c) Since M1 is diode-connected, we know it is operating in saturation.

ID = 1 mA

gm =

√

2µnCox

W

LID = 6.667 mS

ro =1

λID

= 10 kΩ

+

vgs

−

gmvgs ro

(d) Since M1 is diode-connected, we know it is operating in saturation.

VGS = VDS

VDD − VGS = ID(2 kΩ) =1

2µnCox

W

L(VGS − VTH)2 (1 + λVGS) (2 kΩ)

VGS = VDS = 0.623 V

ID = 588 µA

gm = µnCox

W

L(VGS − VTH) = 4.961 mS

ro =1

λID

= 16.996 kΩ

+

vgs

−

gmvgs ro 2 kΩ

(e) Since M1 is diode-connected, we know it is operating in saturation.

ID = 0.5 mA

gm =

√

2µnCox

W

LID = 4.714 mS

ro =1

λID

= 20 kΩ

+

vgs

−

gmvgs ro

6.38 (a)

vin

+

vgs1

−

gm1vgs1 ro1

−

vgs2

+

gm2vgs2 ro2

vout

RD

(b)

vin

+

vgs1

−

gm1vgs1

gm2vgs2 ro2

−

vgs2

+

RD

vout

ro1

(c)

vin

+

vgs1

−

gm1vgs1 ro1

vout

RD

+

vgs2

−

gm2vgs2 ro2

(d)

vin

+

vgs1

−

gm1vgs1 ro1

+

vgs2

−

vout

gm2vgs2 ro2

(e)

+

vgs1

−

gm1vgs1

vin

gm2vgs2 ro2

−

vgs2

+

RD

vout

ro1

6.43 (a) Assume M1 is operating in triode (since |VGS | = 1.8 V is large).

|VGS | = 1.8 V

VDD − |VDS | = |ID| (500 Ω) =1

2µpCox

W

L

[

2 (|VGS | − |VTH |) |VDS | − |VDS |2

]

(500 Ω)

|VDS | = 0.418 V < |VGS | − |VTH | , which verifies our assumption

|ID| = 2.764 mA

(b) Since M1 is diode-connected, we know it is operating in saturation.

|VGS | = |VDS |

VDD − |VGS | = |ID|(1 kΩ) =1

2µpCox

W

L(|VGS | − |VTH |)

2(1 kΩ)

|VGS | = |VDS | = 0.952 V

|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) =1

2µpCox

W

L(|VGS | − |VTH |)

2(1 kΩ)

|VGS | = |VGS | = 0.952 V

|ID| = 848 µA

6.44 (a)

VX

IX

VDD − VTH VDD

Saturation Cutoff

M1 goes from saturation to cutoff when VX = VDD − VTH = 1.4 V.

(b)

VX

IX

1 + VTH VDD

Saturation Triode

M1 goes from saturation to triode when VX = 1 + VTH = 1.4 V.

(c)

VX

IX

VDD − VTH VDD

Saturation Cutoff

M1 goes from saturation to cutoff when VX = VDD − VTH = 1.4 V.

(d)

VX

IX

VTH VDD

Cutoff Saturation

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