Chapter 11 Problem Solutions 11.1 (a)

( )0.7 30.1 23 K

3 1.5 0.05 30 K

EE

CC

RR

RR

− − −= ⇒ =

− = ⇒ =

(b) ( ) ( )2 2 26 2 6 76CE C C E Cv i R R i= − + = −

(c) ( ) 2 2max 0 0.7 Vcm CB CEv v v⇒ = ⇒ =

So ( )2 20.7 6 76 69.74 AC Ci i μ= − ⇒ =

( )( ) ( ) ( ) ( )max 0.7 3

2 0.06974 max 0.908 V23

CMCM

vv

− − −= ⇒ =

( ) ( )min 3 V min 2.3 VCM S CMv V v⇒ = − ⇒ = −

11.2

180, 85 dB

18017,783 0.01012

d dB

dcm

cm cm

A C M RR

AC M RR AA A

= =

= = = ⇒ =

Assume the common-mode gain is negative.

( ) ( )( )

0

0

0

180 0.01012180 2sin mV 0.01012 2sin V0.36sin 0.02024sin

d d cm cm

d cm

v A v A vv v

v t tv t t

ω ωω ω

= += −= −= −

Ideal Output: ( )0 0.360sin Vv tω=

Actual Output: ( )0 0.340sin Vv tω= 11.3 a.

( )

( ) ( )( )

( ) ( )( )( ) ( )

1 1

12 2

4 4

2 2

4 4

10 2 0.71.01 mA

8.51.01 1.01 mA

2 21 11 100 101

100 1.01 0.50 mA101 20 0.7 5 4.3 V

5 0.5 2 0.7 4.7 V

C C

C C

CE CE

CE CE

I I

II I

I I

V V

V V

β β

−= ⇒ =

= = ⇒ ≅+ +

+

⎛ ⎞⎛ ⎞= ⇒ ≅⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

= − − − ⇒ =

= − − − ⇒ =⎡ ⎤⎣ ⎦

b.

( ) ( )( )

( )

4 4

4 4

2 4 2

1 2

1 1

For 2.5 V 0.7 2.5 1.8 V5 1.8 1.6 mA

21 1012 2 1.6 3.23 mA

1003.23 mA

10 2 0.72.66 k

3.23

CE C

C C

C C C

C

V V

I I

I I I

I I

R R

ββ

= ⇒ = − + =−= ⇒ =

⎛ ⎞+ ⎛ ⎞+ = ⇒ =⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

≈ =

−= ⇒ = Ω

11.4 a. Neglecting base currents

1 3 1 1

1 1

30 0.7400 A R 73.25 k0.4

10 V V 9.3 V15 9.3 28.5 k

0.2

CE C

C C

I I R

V

R R

μ −= = ⇒ = ⇒ = Ω

= ⇒ =−= ⇒ = Ω

b. ( )( )

( )

( )( )( )

( )

( )

( )( )( )( )

0 3

0

10

100 0.02613 k

0.250 125 k0.4

We have100 28.5

622 2 13 10

12 1

1

100 28.5 1 0.1132 125 10113 10

113 10

6220log0.113

Cd d

B

Ccm

B

B

cm

dB

r

r Q

RA Ar R

RArr Rr R

A

C M RR C

π

π

π

π

β

ββ

= = Ω

= = Ω

= = ⇒ =+ +

⎧ ⎫⎪ ⎪⎪ ⎪= − ⎨ ⎬++ ⎪ ⎪+⎪ ⎪+⎩ ⎭

⎧ ⎫⎪ ⎪⎪ ⎪= − ⇒ = −⎨ ⎬+ ⎪ ⎪+⎪ ⎪+⎩ ⎭

⎛ ⎞= ⇒⎜ ⎟⎝ ⎠

54.8 dBdBM RR =

c.

( ) ( )

( )

( )( )

0

2 2 13 10 46 k

1 2 121 13 10 2 101 125 12.6 M2

id B id

icm B

icm

R r R R

R r R r

R

π

π β

= + = + ⇒ = Ω

= + + +⎡ ⎤⎣ ⎦

= + + ⇒ = Ω⎡ ⎤⎣ ⎦

11.5

(a) ( )max 0CM CBv V⇒ = so that ( ) ( ) ( ) ( )0.5max 5 5 8

2 2Q

CM C

Iv R= − = −

( )max 3 VCMv =

(b)

( ) ( )2

0.25 0.018 0.08654 mA2 2 0.026 2

0.08654 8 0.692 V

CQd dm

T

C C

IV VI g

VV I R

⎛ ⎞ ⎛ ⎞ ⎛ ⎞Δ = ⋅ = ⋅ = =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠

Δ = Δ ⋅ = =

(c)

( )( )2

0.25 0.010 0.04808 mA0.026 2

0.04808 8 0.385 VC

I

V

⎛ ⎞⎛ ⎞Δ = =⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

Δ = =

11.6

( )( )( )

( )

1 4

1 4 1 1 4

1 1

1 2

so 1.2 2 6 0.1

3 0.7 353

0.13 1For 1 1 40 0.05

C

C C

CM C C C C

P I I V V

I I I I I mA

R R k

v V V V V R R k

+ −= + −

≅ = ⇒ = =

− − −= ⇒ = Ω

−= + ⇒ = = ⇒ = ⇒ = Ω

One-sided output

( )( )

1 0.05 where 1.923 /2 0.026

Then1 1.923 40 38.52

d m C m

d d

A g R g mA V

A A

= = =

= ⇒ =

11.7 a.

( ) ( )

( ) ( )

1 2

1 2

1 2 1

1 2

0 0.7 2 85 52

5 0.7 0.050 mA85 1

100 0.0501 2 101 2

Or 0.0248 mA

5 100 0.7

So 3.22 V

EE

E E

EC C

C C

CE CE C

CE CE

I I

I I

II I

I I

V V I

V V

ββ

= + + −

−= ⇒ =+

⎛ ⎞⎛ ⎞ ⎛ ⎞⎛ ⎞= = =⎜ ⎟ ⎜ ⎟⎜ ⎟⎜ ⎟+ ⎝ ⎠⎝ ⎠⎝ ⎠⎝ ⎠= =

= = − − −⎡ ⎤⎣ ⎦= =

b. ( ) ( )1max for 0 and 5 100 2.52 Vcm CB C Cv V V I= = − =

( )So max 2.52 Vcmv =

( ) 1 2min for and cmv Q Q at the edge of cutoff ( )min 4.3 Vcmv⇒ = −

(c) Differential-mode half circuits

( )

( )( )

( )( )( )

( )( )( )( )

.2

11

Then2

1

1

12 1

100 0.026105 2

0.0248

Then100 1001 16.3

2 105 101 2

dm E

E

d

E

Co m C d

E

TE

CQ

d d

v VV g V R

r

V Rr

vV

Rr

Rv g V R Ar R

Vr k R kI

A A

ππ π

π

ππ

π

π

ππ

π

β

β

ββ

β

⎛ ⎞ ′− = + +⎜ ⎟⎝ ⎠

+⎡ ⎤′= +⎢ ⎥

⎣ ⎦

− /=

+⎡ ⎤′+⎢ ⎥

⎣ ⎦

= − ⇒ = ⋅′+ +

′= = = Ω = Ω

= ⋅ ⇒ =+

11.8 a. For 1 2 0v v= = and neglecting base currents

( )0.7 1062 k

0.15E ER R− − −

= ⇒ = Ω

b.

( )( )( )

( )( )( )

( )

( )( )( )( )

02

10

2

100 0.02634.7 k

0.075

100 5071.0

2 34.7 0.5

12 1

1

100 50 1 0.3982 62 10134.7 0.5

134.7 0.5

71.020log0.398

Cd

d B

T

CQ

d d

Ccm

EB

B

cm

dB dB

v RAv r R

VrI

A A

RARr Rr R

A

C M RR C M RR

π

π

π

π

β

β

ββ

= =+

= = = Ω

= ⇒ =+

⎡ ⎤⎢ ⎥⎢ ⎥= −⎢ ⎥++

+⎢ ⎥+⎢ ⎥⎣ ⎦⎡ ⎤⎢ ⎥

= − ⎢ ⎥ ⇒ = −+ ⎢ ⎥+⎢ ⎥+⎣ ⎦

= ⇒ 45.0 dB=

c. ( )( )

22 34.7 0.5 70.4 k

id B

id id

R r RR R

π= += + ⇒ = Ω

Common-mode input resistance

( )

( )( )

1 2 121 34.7 0.5 2 101 62 6.28 M2

icm B E

icm

R r R R

R

π β= + + +⎡ ⎤⎣ ⎦

= + + ⇒ = Ω⎡ ⎤⎣ ⎦

11.9

(a)

( )( )

1 2

1 2

1 2

1 1.69 1.6 18.97 A

390

9.49 9.39 A2

9.39 0.51 9 4.21 V

E

E

EC C

C C

v v V V

I

I A I I

v v

μ

μ μ

= = ⇒ =−= ⇒

= = =

= = − = −

(b)

( )( )

( )( )

3

6 32

1

9.39 0.361 mA/V0.026

0.361 10 0.005 1.805 A2

1.805 10 510 10 0.921 4.21 0.921 3.29 V

4.21 0.921 5.13 V

m

dm

C C

C

g

VI g

v V v

v

μ−

−

= ⇒

Δ = = × =

Δ = × × = ⇒ = − + ⇒ −

= − − ⇒ −

11.10 (a)

( )( )

( )

1 2

1 2

1 2

1 2

1 2

06 A

605.90 A

5.90 0.360 30.875 V

0.6 0.8751.475 V

E E

C C

C C

EC EC

v vI I

I Iv v

V V

μβ

μ

= == ≅

== == = −= −= = + − −=

(b) (i)

( )( )

5.90 0.227 mA/V0.026

0.227 360 81.70

m

d m C

cm

g

A g RA

= ⇒

= = ==

(ii) ( )( )

( )( )( )( )

60 0.02640.8

2 0.0059264 K

0.227 3600.0442

2 61 40001

264

m Cd

cm

g RA r

A

π= = =

=−

= = −+

11.11

( )( )

( )( )

( )( )

1 2

1 2

1 2

6 3

2 2 2 1

2

For 0.20 V0.1 mA0.1 30 10

7 V0.1 3.846 mA/V

0.026

3.846 0.008 30.77 A2

30.77 10 30 10 0.923 V

7 0.9236.077 V

7 0.9237.923 V

C C

C C

m

dm

C C

C C C

C

v vI Iv v

g

vI g

v I R

v I v v

v

μ

−

= == == = −= −

= =

Δ = = ⇒

Δ = Δ ⋅ = × × =

↑⇒ ↓⇒ ↓⇒ = − += −

= − −= −

11.12

( )

( )( )

1 2

1 2

50 KFor 0

0.7 1075

0.124 mA0.0615 mA

0.0615 2.365 mA/V0.026120 0.026

50.7 K0.0615

C

E

C C

m

Rv v

I

I I

g

rπ

== =

− − −=

== =

= =

= =

Differential Input

1 22 2d dv Vv v= = −

Half-circuit.

( )( )

1

2

1 2

2 2

2

2 22

22

2.365 50 118.25

dm C C

C C

o C C C C

C

dm C

d m C

V RI g v I R

Rv I R

R Rv v v I R I R

IRVg R

A g R

Δ⎛ ⎞Δ = + ⇒ Δ = −Δ +⎜ ⎟⎝ ⎠

Δ⎛ ⎞Δ = +Δ −⎜ ⎟⎝ ⎠Δ Δ⎛ ⎞ ⎛ ⎞= Δ − Δ = −Δ + − Δ −⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠= − Δ

⎛ ⎞= − ⎜ ⎟⎝ ⎠

= − = − = −

Common-mode input.

( )

( )

( ) ( )( )

( )( )

( )( )

[ ]

1 1

2 2

1 2

2

1 1 2

1 211 2

21 2

21 2

2

cm m E

cm

E

m cm cmm

EE

C cm

CE

C cm

CE

C cm C

o C C

Vv V g V RrvV

Rr

g v vI g Vr R

Rr

RR vv IR

r R

RR vv IR

r R

RR v Rv v v

ππ π

π

π

π

ππ

π

π

π

β

βββ

β

β

β

β

β β

⎛ ⎞= + +⎜ ⎟

⎝ ⎠

=⎛ ⎞

+ +⎜ ⎟⎝ ⎠

Δ = = =+ +⎛ ⎞++ ⎜ ⎟

⎝ ⎠Δ⎛ ⎞− + ⋅⎜ ⎟

⎝ ⎠Δ = −Δ =+ +

Δ⎛ ⎞− −⎜ ⎟⎝ ⎠Δ = −Δ =

+ +

Δ⎛ ⎞− + + −⎜ ⎟⎝ ⎠= Δ − Δ = [ ]

( )( )

( )( )( )( )( )( )( )

2

22

1 2

120 0.51 2 50.7 121 2 75

0.0032966118.25 35,870.5

0.003296691.1 dB

cm

cm

E

cmE

dB

R v

R v

r R

RAr R

C M RR

C M R R

π

π

β

β

ββ

Δ⎛ ⎞⎜ ⎟⎝ ⎠

Δ⎛ ⎞− ⎜ ⎟⎝ ⎠=

+ +

−− Δ= =+ + +

= −

= =

=∫

11.13

( )

( ) ( )

1 2

1 2

1

2

00.7 10

750.124 mA

0.0615 mA0.0615 2.365 mA/V0.026

0.01

2.377 mA/V2.353 mA/V120 0.026

50.7 K0.0615

E

C C

m

m

m

m

m

v v

I

I I

g

gggg

rπ

= =− − −

=

== =

= =

Δ=

==

= =

( )

( ) ( )

1 1

2 2

1 2 1 2

1 2

1 2

2

2

2

2 2

250 2.377 2.353 118.25

2 2

dm

dC m c

dC m c

d do C C m C m C

dC m m

Cd m m d

VI g

Vv g R

Vv g R

V Vv v v g R g R

VR g g

RA g g A

Δ =

Δ = −

Δ = +

= Δ − Δ = − −

= − +

−= − + = + ⇒ = −

Common-Mode

( ) ( )

( )

( )

( ) ( )

( ) ( )

1 21 2

1 2

1 11 2 1 2

2.377 2.353 501211 1 2 751 2 50.7

1.2 0.003343358.99

91 dB

m C cm m C cmC C

E E

m m Cocm

cmE

cm

dB

g R v g R vv v

R Rr r

g g RvA

vR

r

A

C M R R

π π

π

β β

β

− −Δ = Δ =

⎛ ⎞ ⎛ ⎞+ ++ +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

− − − −= = =

⎛ ⎞ ⎛ ⎞+ ++ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠−= ⇒ = −

=∫

11.14 (a)

( ) ( )

1 2

1 2

1 2

00.7 V

5 0.7 4.3 mA1

2.132 mA2.132 1 5

2.87 V

E

E

C C

C C

v vv

I

I Iv v

= == +

−= =

= == = −= −

(b) 1 2 2

1

0.5, 0 on off

v v QQ

= =

( ) ( )

1 2

1 2

2

1200, 4.3 mA 4.264 mA121

5 V 4.264 1 5

0.736 V

C C

C C

C

I I

v v

v

⎛ ⎞= = =⎜ ⎟⎝ ⎠

= − = −

= −

(c) 2.1320.7 V 82.0 mA/V0.026E mv g≈ = =

( ) ( )82.01 41.0

2 2 2d d

m C C m C d dv VI g v I R g R V VΔ = Δ = Δ ⋅ = ⋅ = ⋅ =

0.015 0.615 Vd cV v= ⇒ Δ =

2 1C Cv v↓ ↑

1

2

2.87 0.615 2.255 V2.87 0.615 3.485 V

C

C

vv

= − + = −= − − = −

11.15

(a) 1 38.46 mA/V

0.0261 100

0.01

100 38.46 2.6 K

Cm

T

od

d

d m C

C

c

Ig

Vv

Av

A g RR

R

= = =

= = =

==

=

(b) With 1 2 0v v= =

( )( ) ( )1 2 10 1 2.6 7.4 max 7.4 C C cmv v V v V= = − = ⇒ = 11.16 a. i. ( )01 02 0v v− =

ii.

( ) ( )( )

1 2

01 02 1 1 2 2

2 1 01 02

1 mA

1 7.9 8 0.1 V

C C

C C C C

C C C

I I

v v V I R V I R

I R R v v

+ +

= =

⎡ ⎤ ⎡ ⎤− = − − −⎣ ⎦ ⎣ ⎦= − = − ⇒ − = −

b.

( )

( )( )

( )( )

0 1 2

3

13 13

9

13 91 1 1

13 92 2

exp

2 10So exp10 1.1 109.524 10

exp 10 9.524 10 0.952 mA

1.1 10 9.524 10 1.048 mA

BES S

T

BE

T

BEC S C

T

C C

vI I IV

vV

vI I IV

I I

−

− −

−

−

⎛ ⎞= + ⎜ ⎟

⎝ ⎠⎛ ⎞ ×=⎜ ⎟ + ×⎝ ⎠

= ×

⎛ ⎞= = × ⇒ =⎜ ⎟

⎝ ⎠

= × × ⇒ =

i. ( )( )01 02 2 2 1 1 01 02 01 021.048 0.952 8 0.768 VC C C Cv v I R I R v v v v− = − ⇒ − = − ⇒ − =

ii. ( )( ) ( )( )01 02

01 02 01 02

1.048 7.9 0.952 88.279 7.616 0.663 V

v vv v v v

− = −− = − ⇒ − =

11.17 From Equation (11.12(b))

( ) ( ) ( )

2 /

/

/

110.90

11So 1 0.111

0.90ln 0.111 0.026 ln 0.111 0.0571 V

d T

d T

d T

QC v V

v V

v V

d T d

Ii

e

e

e

v V v

=+

=+

= − =

= = ⇒ = −

11.18 From Example 11.2, we have

( )( ) ( )

( )( )

( )( ) ( )

( ) ( )

max / 0.026

max / 0.026

max / 0.026

max 10.54 0.026 1 0.02

max0.5

4 0.026

max 10.98 0.54 0.026 1

10.490 9.423 max1

d

d

d

dv

d

dv

d v

ve

v

ve

ve

−

−

−

+ −+ =

+

⎡ ⎤+ =⎢ ⎥

+⎢ ⎥⎣ ⎦

+ =+

By trial and error ( )max 23.7 mVdv =

11.19 a.

1 BE3

1 1

12 2

For 1 mA, 0.7 V20 0.7 19.3 k

10.026 1ln ln 0.599 k

0.1 0.1T

Q Q

I V

R R

V IR RI I

= =−= ⇒ = Ω

⎛ ⎞ ⎛ ⎞= ⋅ = ⋅ ⇒ = Ω⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

b. ( )( )

4

04

180 0.02646.8 k

0.10.1 3.846 mA/V

0.026100 1 M0.1

m

r

g

r

π = = Ω

= =

= ⇒ Ω

From Chapter 10 ( )

( ) ( )( )

( ) ( )

( )( ) ( )( )

0 04 4

4

0

01

010

1

0.599 46.8 0.5911 1 3.846 0.591 3.27 M

100 2 M0.05

1 12

181 3.27 181 1

592 181 139 M

m E

E

icm

icm

R r g R r

R rR

r

rR R

R

π

π

β β

= +⎡ ⎤⎣ ⎦= =

= + = Ω⎡ ⎤⎣ ⎦

= ⇒ Ω

⎡ ⎤⎛ ⎞≅ + +⎡ ⎤ ⎢ ⎜ ⎟⎥⎣ ⎦ ⎝ ⎠⎣ ⎦= ⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦= ⇒ = Ω

(c) From Eq. (11.32(b))

( )

( )( )

( )( )( )( )

2 11

0.05 1.923 /0.026180 0.026

93.6 0.05

0Then

1.923 50 0.00760

2 181 32701

93.6

m Ccm

o

B

m

B

cm cm

g RAR

r R

g mA V

r k

R

A A

π

π

β−

=+

++

= =

= = Ω

=

−= ⇒ = −

+

11.19 For 3.5 CMv V= and a maximum peak-to-peak swing in the output voltage of 2 V, we need the quiescent collector voltage to be

3.5 1 4.5 CV V= + = Assume the bias is 10 ,V± and 0.5 .QI mA= Then 0.25 CI mA=

Now 10 4.5 22 0.25C CR R k−= ⇒ = Ω

In this case, ( ) ( )100 0.02610.4

0.25r kπ = = Ω

Then ( )( )( )100 22

1012 10.4 0.5dA = =

+ So gain specification is met.

( ) ( )( )( )( )

4

For 80

1 101 0.51 110 1 1 1.03 2 2 0.026 100

dB

Q o oo

T

CMRR dB

I R RCMRR R M

Vβ

β

= ⇒

⎡ ⎤+⎡ ⎤= = + = + ⇒ = Ω⎢ ⎥⎢ ⎥

⎢ ⎥⎣ ⎦ ⎣ ⎦

Need to use a Modified Widlar current source. ( )

( )( )

( )( )

1

1 1 1 1

2

1

100If 100 , then 200 0.5

100 0.0265.2

0.50.5 19.23 /

0.026Then 1030 200 1 19.23 0.216 5.2 225

Also let 225 and 0.5

o o m E

A o

m

E E E E

E REF

R r g R r

V V r k

r k

g mA V

R r R r k R R

R I mA

π

π

π π

= +⎡ ⎤⎣ ⎦

= = = Ω

= = Ω

= =

= + ⇒ = Ω = ⇒ = Ω⎡ ⎤⎣ ⎦= Ω ≅

11.20

(a) ( )0.7 10

37.2 0.25E ER R k

− − −= ⇒ = Ω

(b)

( ) ( )

( )

( ) ( ) ( )

( )

1 21 2 1 2

1 11 1

2 2

1 2

1 2

1 or 1

Then

11

From this, we find

11

e em m

E E

ee

B B

e

ee e

B E

B

eB

E

V V V Vg V g V V Vr r R r R

V V V rV V V

r R r r RV V V

r VV V V Vr r R R

r RV VrV

r R rR

π ππ π π π

π π π

π ππ

π π π

π

π

π π

π

π

π π

β

β

β

⎛ ⎞++ + + = + =⎜ ⎟⎝ ⎠

⎛ ⎞−= ⇒ = −⎜ ⎟+ +⎝ ⎠= −

⎛ ⎞ ⎡ ⎤+ − + − =⎜ ⎟ ⎢ ⎥+⎝ ⎠ ⎣ ⎦

++ ⋅=

+ + ++

( )

( )( )

2 2

Now

We have120 0.026 0.12525 , 4.81 /

0.125 0.026

B

o m C m C e

m

Rr

V g V R g R V V

r k g mA V

π

π

π

⎡ ⎤+⎢ ⎥⎣ ⎦

= − = − −

= ≅ Ω = =

(i)

( )( )

( )

( )( ) ( )

1 2Set and 2 2

Then

25 0.51 0.022 25 22.02625 0.5 25 0.51

37.2 121 25

So0.00494

Now

4.81 50 0.00494 1192

d d

d d

e

e d

d oo d d

d

V VV V

V V

V

V V

V VV V A

V

= = −

⎛ + ⎞⎛ ⎞− ⎜ ⎟ −⎜ ⎟⎝ ⎠⎝ ⎠= =⎡ ⎤+ ++ +⎢ ⎥⎣ ⎦

= −

⎛ ⎞= − − − − ⇒ = =⎜ ⎟⎝ ⎠

(ii)

( )( )

( )

( )

( )( ) ( )

1 2Set Then

25 0.51 2.02252.0256725 0.5 25 0.51

37.2 121 25

0.9972Then

4.81 50 0.9972

or 0.673

cm

cmcm

e

e cm

o cm cm

ocm

cm

V V V

V VV

V V

V V V

VAV

= =

+⎛ ⎞+⎜ ⎟ −⎝ ⎠= =⎡ ⎤+ ++ +⎢ ⎥⎣ ⎦

=

= − −⎡ ⎤⎣ ⎦

= = −

11.21 From Equation (11.18)

( ) ( )

( ) ( )

( )

0 2 1

For 2 mA, 1 mA1Then 38.46 mA/V

0.026Now 2 38.46 0.015So 3.47 k

Now 10 1 3.476.53 V

For 0 max 6.53 V

C C m C d

CQm

T

Q CQ

m

C

C

C C C

CB cm

v v v g R vI

gV

I I

g

RR

V V I R

V v

+

= − =

=

= =

= =

== Ω

= − = −=

= ⇒ =

11.22 The small-signal equivalent circuit is

1 1 2 2

1 2 1 2

A KVL equation: v V V vv v V V

π π

π π

= − +− = −

A KCL equation

( )

( ) ( )

1 21 2

1 2 1 2

1 2 1 1 1 2 2 1 2

0

1 0

1 1Then 2 and 2 2

m m

m

V Vg V g Vr r

V V g V Vr

v v V V v v V v v

π ππ π

π π

π π π ππ

π π π

+ + + =

⎛ ⎞+ + = ⇒ = −⎜ ⎟

⎝ ⎠

− = ⇒ = − = − −

At the 01v node:

01 01 021 0m

C L

v v v g VR R π

−+ + =

( )01 02 2 11 1 1 1

2 mC L L

v v g v vR R R

⎛ ⎞ ⎛ ⎞+ − = −⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠ (1)

At the 02v node:

02 02 012 0m

C L

v v v g VR R π

−+ + =

( )02 01 1 21 1 1 1

2 mC L L

v v g v vR R R

⎛ ⎞ ⎛ ⎞+ − = −⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠ (2)

From (1):

( )02 01 2 1112

Lm L

C

Rv v g R v vR

⎛ ⎞= + − −⎜ ⎟

⎝ ⎠

Substituting into (2)

( ) ( )

( )

( )

01 2 1/ 01 1 2

01 1 22

011 2

1 1 1 1 1 1 112 2

1 1 1 1 12

122

Lm L m

C C L C L L

L Lm

C C CC

L Lm

C C C

Rv g R v v v g v vR R R R R R

R Rv g v vR R RR

v R Rg v vR R R

⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞+ + − − + − = −⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠⎝ ⎠ ⎝ ⎠⎡ ⎤⎛ ⎞ ⎛ ⎞

+ + = − − +⎢ ⎥⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠ ⎣ ⎦

⎛ ⎞ ⎛ ⎞+ = − −⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

1 2For dv v v− =

011

12

2

m L

vd L

C

g RvA

v RR

−= =

⎛ ⎞+⎜ ⎟

⎝ ⎠

From symmetry: 022

12

2

m L

vd L

C

g RvA

v RR

= =⎛ ⎞

+⎜ ⎟⎝ ⎠

Then 02 01

2

m Lv

d L

C

v v g RAv R

R

−= =

⎛ ⎞+⎜ ⎟

⎝ ⎠

11.23 The small-signal equivalent circuit is

KVL equation: 1 1 2 2 1 2 1 2 or v V V v v v V Vπ π π π= − + − = − KCL equation:

( )

( )( )

( )( )

( )

1 21 2

1 2 1 2

1 2 2 2 1 2

0 2

1 2

01 2

0

1 0

1Then 2 or 2

Now 12

1For 2

m m

m

m C L

m C L

d d m C Ld

V Vg V g V

r r

V V g V Vr

v v V V v v

v g V R R

g R R v v

vv v v A g R Rv

π ππ π

π π

π π π ππ

π π

π

+ + + =

⎛ ⎞+ + = ⇒ = −⎜ ⎟

⎝ ⎠

− = − = − −

= −

= −

− ≡ ⇒ = =

11.23 a.

10 7 6 k0.5D DR R−= ⇒ = Ω

1 2 1 mAQ D D QI I I I= + ⇒ =

b. ( )10 6D DS GSI V V= + −

and DGS TN

n

IV VK

= +

For 0.50.5 mA, 2 3.12 V0.4D GSI V= = + =

and 10.12DSV =

Load line is actually nonlinear. c. Maximum common-mode voltage when 1M and 2M reach the transition point, or

( ) 3.12 2 1.12DS GS TNV sat V V V= − = = = Then

( )02 sat 7 1.12 3.12cm DS GSv v v V= − + = − +

Or ( )max 9 Vcmv =

Minimum common-mode voltage, voltage across QI becomes zero.

( )( )

So min 10 3.12min 6.88 V

cm

cm

vv

= − +⇒ = −

11.24

( )

( )

( ) ( )( )

2 2 2

1 1 1

0 2 1

2 1

1 2

We have and

Then

C m C m b e C

C m C m b e C

C C

m b e C m b e C

m C b b

V g V R g V V R

V g V R g V V R

V V Vg V V R g V V R

g R V V

π

π

= − = − −

= − = − −

= −= − − − − −⎡ ⎤⎣ ⎦= −

Differential gain 0

1 2d m C

b b

VA g RV V

= =−

Common-mode gain 0cmA = 11.25 (a)

1 23 3 10 3Then 70

0.1

cm C C

C C

v V V V V

R R k

= ⇒ = =−= ⇒ = Ω

(b)

( )

( )( )( )( )

75 5623Now

11 12

151 0.215623 1 1.45 2 150 0.026

dB

Q o

T

oo

CMRR dB CMRR

I RCMRR

V

RR M

ββ

= ⇒ =

+⎡ ⎤= +⎢ ⎥

⎣ ⎦⎡ ⎤

= + ⇒ = Ω⎢ ⎥⎢ ⎥⎣ ⎦

Use a Widlar current source. [ ]1o o m ER r g R′= +

Let AV of current source transistor be 100 V.

( )( )

( )

( )( ) ( ) ( )

100 0.2Then 500 , 7.69 /0.2 0.026

150 0.02619.5

0.2So 1450 500 1 7.69 0.247

Now 0.247 19.5 250

Then ln

0.2 0.250 0.026 ln0.2

o m

E E

E E E E

REFQ E T

Q

REFRE

r k g mA V

r k

R R k

R R r R R

II R VI

I I

π

π

= = Ω = =

= = Ω

′ ′= + ⇒ = Ω⎡ ⎤⎣ ⎦′ = ⇒ = ⇒ = Ω

⎛ ⎞= ⎜ ⎟⎜ ⎟

⎝ ⎠⎛ ⎞

= ⇒⎜ ⎟⎜ ⎟⎝ ⎠

( )1 1

1.37

10 0.7 10Then 14.1

1.37

F mA

R R k

=

− − −= ⇒ = Ω

11.26 At terminal A.

( )( )

( )1 15

1 2 2THA A

R R R RR R R kR R

δ δδ δ

+ ⋅ += = = ≅ = Ω

+ + +

Variation in THR is not significant

( )( )( )

( )1 5 5 11 2

ATHA

A

RRV VR R R R

δ δδ δ

+ + +⎛ ⎞= = =⎜ ⎟+ + + +⎝ ⎠

At terminal B.

5 2

2.5

THB

THB

RR R R k

RV V VR R

+

= = = Ω

⎛ ⎞= =⎜ ⎟+⎝ ⎠

From Eq. (11.27) ( )

( )( )( )

( )( )( )( ) ( )

( )

( ) ( )

2 12 1

2 12 1

2 1

where and 2

120 0.0265 , 12.5

0.25120 3

So 10.32 12.5 5

We can find

5 12.5

2

2.5 2 5 1 2.5 52 2

2.5 1.2

CO THB THA

B

B

O

THB THA

THB THA

R V VV V V V V

r R

R k r k

V VV V V

V V V V

V V

π

π

β

δδ

δ δ δ δδ δ

δ

− −= = =

+

= Ω = = Ω

− −= = − −

+− = −

+⎡ ⎤− = − ⎢ ⎥+⎣ ⎦

+ − + −= =+ +

−≅ = −

( )( )25

Then 10.3 1.25 12.9OV

δ

δ δ= − − =

So for 0.01 0.01δ− ≤ ≤ We have 20.129 0.129 OV V− ≤ ≤ 11.27 a.

( )( )2180 0.026

23.4 k0.2

So 46.8 k

id

id

R r

r

R

π

π

=

= = Ω

= Ω

b. Assuming ,rμ → ∞ then

( )( )( )

01

181 1

181 181 M

icm

icm

icm

R R

R

R

β≅ +⎡ ⎤⎣ ⎦= ⎡ ⎤⎣ ⎦= ⇒ = Ω

11.28 (a)

( )1 1

1

2 2

10 0.7 100.5 38.6 K

0.026 0.5ln 236 0.14 0.14

I RR

R R

− − −= = ⇒ =

⎛ ⎞= ⇒ = Ω⎜ ⎟⎝ ⎠

(b)

( )

( )

( )( )

( ) ( )

( )( )

4 4 4

4

4

10.141 5.385 mA/V

0.026180 0.026

33.4 K0.14

33.4 0.236 0.234 K100 714 K0.14

714 1 5.385 0.234

1614 K181 1614 292 MΩ

icm o

o o m E m

E

o

o

icm

R R

R r g R g

r

R

r

R

R

π

β≈ +

′= + = =

= =

′ = =

= =

= +⎡ ⎤⎣ ⎦== ≈

(c)

( )

( )( )

( )( )( )( )

11

1

1

0.07 2.692 mA/V2 1 0.026

1

180 0.02666.86 K

0.072.692 40

2 181 16141

66.860.0123

m Ccm m

o

cm

cm

g RA gR

r

r

A

A

π

π

β−

= = =+

+

= =

−=

+

= −

11.29

( )

( )( )

( )

1 1 1 3

11 1

32 2 2

23 22 3 2

22 2 2

/ 219.23

2 100 0.026 5.2/ 2

/ 2, 19.23

219.23

Then 30 3.12 2

d m

Qm Q

T

T

Q Q Q

Qmd m Q

T

QQ

A g R r

Ig I

V

VrI I I

Ig RA g IV

IR I R V

π

πβ

=

= =

= = =

= = =

= ⋅ ⇒ =

Maximum 2 1 18 o ov v mV− = ± for linearity ( ) ( )( )3

2 2

max 18 30 0.54 so 3.12 is OK.

o

Q

v mV VI R V

= ± ⇒ ±=

From 1dA :

( )

( )

( )( )

12

1 1 3 1

12

1 1

2 1

11 1 1

2 12 1

1 1 11

21

5.2

20 19.23 19.235.2

19.23 5.220

5.2

Let 5 10 2

19.23 10 5.2Then 20 44.8

5.2

10Now 10

10So

QQ Q

Q

Q

Q

QQ

QQ

QQ

QQ

RI

I R r IR

I

I RI R

IR V I R V

I R VI R

I R RI

II

π

⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠= = ⎜ ⎟⎛ ⎞⎜ ⎟+ ⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

=+

⋅ = ⇒ =

= ⇒ =+

= ⇒ =

⎛ ⎞⎜⎜⎝

2

1

1 2

2 2 2

1 1 1

44.8 4.48

Let 100 , 448

Then3.12 6.96

10 100

Q

Q

Q Q

Q

Q

II

I A I A

I R R k

I R R k

μ μ

⎛ ⎞= ⇒ =⎟ ⎜ ⎟⎟ ⎜ ⎟

⎠ ⎝ ⎠= =

= ⇒ = Ω

= ⇒ = Ω

11.30 a.

( )

( )

( ) ( )( )( )

( )( )

231 3

23 3 3

23 3

2

3 3

1 1

1 1

21 1

4

200.25 2

5020 12.5 4 4

12.5 49 30 0

49 49 4 12.5 303.16 V

2 12.520 3.16 0.337 mA

50

0.168 mA2

0.168 0.25 2 2.82 V2.82 10

GSGS

GS GS GS

GS GS

GS GS

Q

QD D

GS GS

DS DS

VI V

V V V

V V

V V

I I I

II I

V VV V

−= = −

− = − +

− + =

± −= ⇒ =

−= ⇒ = =

= ⇒ =

= − ⇒ == − − − ⇒

( )( )( )

4

1

1 1

7.18 V

10 0.168 24 5.97 V5.97 2.82 8.79 V

D

DS DS

VV V

=

= − == − − ⇒ =

(b)

(c)

( )

( )( )( ) ( )( )

( ) ( )( )( ) ( )( )

1 2 1

1

1

1

4 1

4 4

Max 2.82 2 0.82

Now 10 0.168 24 5.97 max 5.97 5.97 0.82max 5.15

max max 5.15 2.82max 7.97

min3.16 2 1.16

CM DS DS DS GS TN

D

S DS

S

CM S GS

CM

CM DS GS

DS GS TN

v V V V sat V VV

V VV V satV Vv V Vv V

v V V sat VV sat V V

−

⇒ = = = −− =

= − == − = −=

= + = +=

= + += − = − =( ) ( )

Then min 10 1.16 2.82 min 6.02 CM CM

Vv v V= − + + ⇒ = −

11.31 a.

( )

( )

21 2 1 1 2

1 2 1 2

1 2 1

21 3 3

1 1

120 A 100 V 1.2 2.30 VFor 5.4 V and 12 V 5.4 2.30 12 4.3 V

10 4.3 47.5 k0.12

240 A

240 200 1.2 2.30 V20 2.3 73.7

0.24

D D GS GS GS

DS DS D

D D

Q D D Q

GS GS

I I V Vv v V V V

R R

I I I I I

I V V

R R

μ

μ

= = = − ⇒ = == = − = = ⇒ − − + = =

−= ⇒ = Ω

= + ⇒ = =

= = − ⇒ =−= ⇒ = 5 kΩ

b.

( )( )04

04

1 1 416.7 k0.01 0.24

1 5.4 13 A416.7

Q

Q DS Q

rI

I V Ir

λ

μ

= = = Ω

Δ = ⋅ Δ = ⇒ Δ ≅

11.32 (a)

( )

( )( )

( )

( )

2

2

25

160 A

28080 4 0.52

80 160 0.5

80 0.5 1.207 V160

5 2 37.5 K 2 1.207 3.21 V0.08

Q

nD GS TN

OS

o

GS

D DS

Ik WI V V

L

V

V

V

R V

μ=′ ⎛ ⎞= −⎜ ⎟⎝ ⎠

= −

= −

= + =

−= = = − − =

(c) ( )

( )2

1 2

1.207 0.5 0.707 VThen 2 0.707 1.29 VAnd 1.207 1.29

2.50 V

DS GS TN

S O DS

cm GS S

cm

V sat V VV V V sat

v v v V Vv

= − = − == − = − = +

= = = + = +=

(b)

11.33

( )( )

( )

( )

5 0.2 8 3.4 V

0.2 0.8 1.694 V0.25

1.694 0.80.894 V3.4 0.8942.506

2.506 1.694 4.2 V

D

DGS TN

n

GS TNDS

S D DS

CM S GS CM

v

IV VK

V VV sat

V V V sat

v V V v

= − =

= +

= + =

= − = −== −= −=

= + = + ⇒ =

(b)

( )( )( )( )( )( )6 3

2

2 2

22

2 0.25 0.2 0.4472 mA/V

0.4472 0.05 22.36 A

22.36 10 8 10 0.179 V

3.43.4 0.179 3.58 V

dD D D D m m n D

D

D

D D

D D

Vv I R I g g K I

I

v

v vv v

μ−

Δ = Δ ⋅ Δ = ⋅ =

= =

Δ = ⇒

Δ = × × =

= + Δ= + ⇒ =

(c)

( )( )( )( )6 3

2 2

50 mV0.4472 0.025 11.18 A

11.18 10 8 10 0.0894 V

3.4 0.0894 3.31 V

d

D

D

D D

vI

v

v v

μ−

= −Δ = − ⇒ −

Δ = − × × = −

= − ⇒ =

11.34 a.

( )

1 2

01 02 1 1 2 2

01 02 2 2 1 1 2 1

0.5 mAD D

D D D D

D D D D D D D

I I

v v V I R V I R

v v I R I R I R R

+ +

= =

⎡ ⎤ ⎡ ⎤− = − − −⎣ ⎦ ⎣ ⎦− = − = −

i. 1 2 01 026 k , 0D DR R v v− = Ω − =

ii. 1 26 k , 5.9 kD DR R= Ω = Ω

( )( )01 02 01 020.5 5.9 6 0.05 Vv v v v− = − ⇒ − = − b.

( )( )( )( ) ( )

( )( )( )( )

2 21 2

1 22

1 22 2

1

2

0.4 / , 0.44 /

1 0.4 0.44 1.190.4 1.19 0.476 mA0.44 1.19 0.524 mA

n n

GS GS

Q n n GS TN

GS TN GS TN

D

D

K mA V K mA VV V

I K K V V

V V V VII

= ==

= + −

= + − ⇒ − == == =

i.

( )( )1 2

01 02 01 02

6 k0.524 0.476 6 0.288 V

D DR Rv v v v

= = Ω− = − ⇒ − =

ii.

( )( ) ( )( )1 2

01 02

01 02

6 k , 5.9 k0.524 5.9 0.476 6

3.0916 2.856 0.236 V

D DR Rv v

v v

= Ω = Ω− = −

= − ⇒ − =

11.35 (a) From Equation (11.69)

( ) ( )( ) ( )

( )

22

2

2

2

1 12 2 2

0.1 0.10.90 0.50 12 0.25 2 0.25

0.40 0.4472 1 0.2

0.8945 1 0.2

n nDd d

Q Q Q

d d

d d

d d

K Ki v vI I I

v v

v v

v v

⎛ ⎞= − ⋅ − ⎜ ⎟⎜ ⎟

⎝ ⎠

⎡ ⎤= − ⋅ − ⎢ ⎥

⎢ ⎥⎣ ⎦

+ = − −

= − −

Square both sides [ ]( )

( )( )( )( )( )

2 2

22 2

2 2 2

0.80 1 0.2

0.2 0.80 0

1 1 4 0.2 0.804 or 1

2 0.2

d d

d d

d

v v

v v

v V V

= −

− + =

± −= =

Then 2 V or 1 Vdv = ± ±

max

0.25But 1.580.1

So 1V, 1V

Qd

n

d d

Iv

kv v

= = =

= ± ⇒ = −

b. From part (a), ,max 1.58 Vdv = 11.36

( )

( ) ( ) ( )

( )

( )

0

1

2

1

2max max max

max

max

12 2

2

1So linear 2 2

1 1 12 2 2 2 2

Then 0.0212 2

1 10.982 2 2 2

d

D

Q n nd d v

d Q Q

n

Q

nDd

Q Q

n n nd d d

Q Q n

nd

Q

n nd

Q Q

idI K K v v

dv I I

KI

Ki vI I

K K Kv v vI I I

K vI

K Kv vI I

=

⎛ ⎞⎜ ⎟⎜ ⎟ ⎛ ⎞⎝ ⎠ = ⋅ − +⎜ ⎟⎜ ⎟

⎝ ⎠

=

= + ⋅

⎡ ⎤⎛ ⎞⎢ ⎥+ ⋅ − + ⋅ ⋅ − ⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦ =

+ ⋅

⎡ ⎤+ ⋅ = + ⋅⎢ ⎥

⎢ ⎥⎣ ⎦( ) ( )

( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

2max max

2max max max

2max max

2max max

12

0.15 1 0.15 0.150.49 0.98 12 0.2 2 2 0.2 2 0.2

0.49 0.600 max 0.50 0.6124 1 0.6124

0.600 max 0.010 0.6124 1 0.6124

nd d

Q

d d d

d d d

d d d

K vI

v v v

v v v

v v v

⎡ ⎤⎛ ⎞⎢ ⎥⋅ − ⎜ ⎟⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

⎡ ⎤⎛ ⎞⎢ ⎥+ ⋅ = + ⋅ ⋅ − ⎜ ⎟⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

+ = + ⋅ −

= + ⋅ −

By trial and error ( )max 0.429 Vdv ≈ 11.37 (b)

( )( )

( )( )( )( )

2 2

1

2 2 0.05 0.008696

0.0417 mA/V

0.0417 0.05 0.002085 mA2

0.002085 510 1.0631.063 4.565 3.502 V

1.063 4.565 5.628 V

m p D

dm

D

D D

D

g K I

VI g

vv vv

= =

=

Δ = = =

Δ = =↑⇒ = − = −= − − = −

( )( )( )( )( )

( )

( )( )( )

( )( )

2

2

2

2

1 2

1 2

9 12

8 2

8 2 0.05 390 0.8

8 39 1.6 0.64

39 61.4 16.96 0

61.4 3769.96 4 39 16.962 39

1.217 V 2.2170.01739 mA 8.696 A

8.696 0.510 9

S S SG

S D

P S SG TP SG

SG SG

SG SG SG

SG SG

SG

S

S D D

D D

I R VI I

K R V V V

V V

V V V

V V

V

VI I Iv v

μ

= + +=

= + +

= − +

= − + +

− + =

± −=

= == = ⇒

= = − = −4.565 V

(b)

( )( )

( )( )( )( )

2 2 0.05 0.008696 0.0417 mA/V

0.0417 0.05 0.002085 mA2

0.002085 510 1.063 V

m P DQ

dD D D D m

D

g K I

Vv I R I g

v

= = =

Δ = Δ ⋅ Δ = ⋅ = =

Δ = =

1 1 1, ,D Dv I v↑ ↓ ↓

1

2

4.565 1.063 5.628V4.565 1.063 3.502V

D

D

vv

= − − = −= − + = −

11.38 (a)

( )

( )( )( )

1 22

0

6 A

6 0.430

0.847 V0.847 V

36 0.36 3 0.84 V

0.847 0.841.69 V

D n SG TP

D

SG

SG

S

D D D

SD S D

SD

v v

I K V VI

V

VVv I R

V V vv

μ

= =

= +=

+ =

== += −= − = −

= − = − −=

(b) (i)

( )( )( )( )

2

2 30 6 26.83 A/V

26.83 0.36 9.660

d m D m n D

d d

cm

A g R g K I

A AA

μ

= =

= =

= ⇒ ==

(ii) ( )( )

( )( )( )( )

26.83 0.364.83

2 226.83 0.36

0.04481 2 1 2 26.83 4

m Dd d

m Dcm

m O

g RA A

g RAg R

= = ⇒ =

−−= = = −

+ +

11.39

( )( )

( )( )

( )( )

( ) ( )( )

1 2

1 2

1 2

2 2 2

1 1

For 0.30 V0.1 mA

0.1 1 2 V0.1

0.1 30 107 V

2 2 0.1 0.1 0.2 mA/V

0.2 0.1 0.02 mA2

0.02 30 0.6 V7 0.6 6.4 V

7 0.6 7.6 V

D D

DSG TP

P

D D

m p D

dD m

D D D

D D D

D D

v vI I

IV VK

v v

g K I

VI g

v I Rv v vv v

= = −= =

= −

= + =

= = −= −

= = =

⎛ ⎞Δ = = =⎜ ⎟⎝ ⎠

Δ = Δ = =↑⇒ = − + ⇒ = −= − − ⇒ = −

11.40

( )( )( )( )

( )( )( )( )

1 2

2

2

2

2

For 00 2 10

10 2

2 0.15 75 1

22.5 44 12.5 0

So 1.61 and 0.15 1.61 1 55.9 A

2 2 0.15 0.0559

0.1831 mA/V

GS D S

GS n S GS TN

GS GS

GS GS

GS D

m n D

m

v vV I R

V K R V V

V V

V V

V V I

g K I

g

μ

= == + −

= + −

= + −

− + =

= = − ⇒

= =

=

Use Half-circuits – Differential gain

1

2

1 2

2 2

2 2

dD m D

do m D

o D D m d D

od m D

d

V Rv g R

V Rv g R

v v v g V Rv

A g RV

Δ⎛ ⎞⎛ ⎞= − +⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠Δ⎛ ⎞⎛ ⎞= −⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

= − = −

= = −

Now – Common-Mode Gain

( )

( )

( )

( )

( )( )

( )( )

( )( )( )( )

( )( )( )

1

2

1 2

2

1 2

21 2

21 2

So 1 2

1 2Then

0.1831 50 9.16

0.1831 0.50.0

1 0.1831 2 75

i gs m gs S cm

cmgs

m S

Dm D cm

Dm S

DD cm

Dm S

O D D

m D cmo

m D

m Docm

cm m S

d

cm

V V g V R VV

Vg R

Rg R Vv

g R

Rgm R Vv

g Rv v v

g R Vv

g R

g RvA

V g R

A

A

= + =

=+

Δ⎛ ⎞− +⎜ ⎟⎝ ⎠=+

Δ⎛ ⎞− −⎜ ⎟⎝ ⎠=+

= −− Δ

=+

− Δ= =

+

= − = −

−= = −

+03216

69.1 dBbB

C M R R =∫

11.41 a.

( )

( )( )

02 04

202

2

404

4

2

150 375 kΩ0.4100 250 kΩ0.40.4 15.38 mA/V

0.02615.38 375 250 2307

d m

A

C

A

C

Cm

T

d d

A g r r

VrIVrII

gV

A A

=

= = =

= = =

= = =

= ⇒ =

b.

02 04 375 250 150 kΩL LR r r R= = ⇒ =

11.41 From 11.40

1 2 55.9 A0.183 mA/V

D D

m

I Ig

μ= ==

( )

( ) ( )

( )

1 1 2 2

1 2 1 2

2 1

1 2

: 2 2

2 2

2 2 2 20.183 50 9.15

2 2:

1 2

d dd D m D D m D

d dO D D m D m D

d d m mO D m m D m m

d m D

M Mm D cm m

CM O D Dm S

V VA v g R v g R

V Vv v v g R g R

V V g gv R g g R g g

A g R

g gg R v g

A v v vg R

+⎛ ⎞Δ = − ⋅ Δ = + ⎜ ⎟⎝ ⎠

= Δ − Δ = − −

− − ⎛ Δ Δ ⎞⎛ ⎞= ⋅ + = ⋅ − + −⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

= − = − = −

Δ Δ⎛ ⎞ ⎛− + −⎜ ⎟⎝ ⎠ ⎝= Δ − Δ = +

+ ( )

( ) ( ) ( )

( ) ( )( ) ( ) ( )

1 2

0.01 0.183 0.001831 2

0.00183 500.003216

1 0.183 2 75

69.1 dB

D CM

m S

O m Dcm m

cm m S

cm

dB

R v

g Rv g R

A gv g R

A

C M R R

⎞⎜ ⎟

⎠+

−Δ= = Δ = =

+

−= = −

+

=∫

11.42 (a)

( )( )( )( )

( )( )( )

( )( )

1 2

2

2

2

2

1 2

1 2 1

05 2

5 2

5 2 0.5 2 1.6 0.64

5 2 2.2 1.282 2.2 3.72 0

2.2 4.84 4 2 3.722 2

2.02 V5 2.022.02 V, 1.49 mA

20.745 mA

0.745 1 5

D S SG

p S SG TP SG

SG SG SG

SG SG

SG SG

SG

SG

S S

D D

D D D D

v vI R V

K R V V V

V V V

V VV V

V

V

v I

I I

v v v v

= == +

= + +

= − + +

= − +− − =

± +=

=−= = =

= =

= = − ⇒ = 2 4.26 V= −

(b)

( )( ) ( )

( ) ( ) ( ) ( )

2

1 2 2

2 21 2 2

1 2

2 22 2 2

2 22 2 2 2 2

22 2

22 2

55

5

1

5 0.5 2 1.8 0.8

5 3.6 3.24 1.6 0.64

5 2 4.2 3.882 4.2 1.12 0

S S SG

D D S SG

p SG TP p SG TP S SG

SG SG

SG SG SG

SG SG SG SG SG

SG SG

SG SG

S

I R VI I R V

K V V K V V R V

V V

V V V

V V V V V

V VV V

V

= += + +

⎡ ⎤= + + + +⎣ ⎦= −

⎡ ⎤= − + − +⎣ ⎦⎡ ⎤= − + + − + +⎣ ⎦

= − +− − =

( ) ( )( )2

4.2 17.64 4 2 1.12

2 2G

± +=

( ) ( )

( ) ( ) ( ) ( )

2 1

2 21 2

1 2

1 2

1 2

2.339 V 1.339 V2.339 V

0.5 1.339 0.8 0.5 2.339 0.80.1453 mA 1.184 mA0.1453 1 5 1.184 1 54.855 V 3.816 V

SG SG

S

D D

D D

D D

D D

V Vv

I II Iv vv v

= ==

= − = −= == − = −= − = −

(c)

( )( )

( )( )( ) ( )( )

2 1

1 2

1 2

22

2.02 V 2 0.5 0.745

1.22 mA/V1.22 0.1 0.122 mA

0.122 1 0.122 V

4.26 0.122 4.26 0.1224.138 V 4.382 V

dm m p D

S

m

D D

D D

D D

D D

VI g g K I

v

gIv I R

v vv vv v

Δ = =

≈ =

=Δ = =Δ = Δ = =

↓ ↑= − + = − −= − = −

11.43

a. ( ) ( )( )( )4 8 4 0.0264

Qf Q f T

T

Ig I g V

V= ⇒ = =

0.832 mAQI⇒ =

Neglecting base currents.

1 130 0.7 35.2 k0.832

R R−= ⇒ = Ω

b. 04 02100 240 k

0.416A

CQ

Vr rI

= = = = Ω

( ) ( )02 04

0.416 16 mA / V0.026|| 16 240 || 240

1920

CQm

T

d m

d

Ig

VA g r r

A

= = =

= =⇒ =

( )( )

0 02 04 0

180 0.0262 , 11.25 k

0.41622.5 k

|| 120 k

id

id

R r r

R

R r r R

π π= = = Ω

⇒ = Ω

= ⇒ = Ω

c. Max. common-mode voltage when 0CBV = for 1Q and 2Q .

Therefore ( ) ( )( )

3max 15 0.7max 14.3 V

cm EB

cm

v V V Qv

+= − = −=

Min. common-mode voltage when 0CBV = for 5.Q

Therefore ( ) ( )min 0.7 0.7 15 13.6 V

So 13.6 14.3 Vcm

cm

vv

= + + − = −− ≤ ≤

( )( )

( )( )

0

0

1 1 22

100 120 k0.832

181 120 21.7 M

icm

A

Q

icm icm

R R

VRI

R R

β≅ +

= = = Ω

= ⇒ = Ω

11.43 (a)

( )( )

( )

2

2 0.4 1

1.265 mA/V1 10

0.1

10 1.2657.91 K

m n D

m

od

d

d m D

D

D

g K I

gvAv

A g RR

R

=

=

=

= = =

===

(b)

( )( )

( )( )

1 2

1 2

Quiescent 010 1 7.91 2.09 V

1 0.8 2.38 V0.4

2.38 0.8 1.58

So 2.09 1.58 2.38

2.89 V

D D

DGS TN

n

DS

cm D DS GS

cm

v vv v

IV VK

V sat

v v V sat V

v

= == = − =

= + = + =

= − =

= − += − +

=

11.44

( )

( )

1 2

1 2

1 2

2For 2.5

0.25 2

10 3Let 3 , then 28 0.25

28Then 100 7.14 /

2

And 22

0.0807.14 2 0.252

1274 (Extremely la

m Dd

CM

QD D

D D D D

mm

nm D

g RA

v VI

I I mA

V V V R R k

gg mA V

k Wg IL

WL

W WL L

=

=

= = =

−= = = ⇒ = Ω

= ⇒ =

′ ⎛ ⎞= ⎜ ⎟⎝ ⎠

⎛ ⎞⎛ ⎞= ⇒⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

rge transistors to meet the gain requirement.)

Need 0.10CMA =

From Eq. (11.64(b))

( )( )( )

1 27.14 28

So 0.10 140 1 2 7.14

m DCM

m o

oo

g RA

g R

R kR

=+

= ⇒ = Ω+

For the basic 2-transistor current source

( ) ( )1 1 200

0.01 0.5o oQ

R r kIλ

= = = = Ω

This current source is adequate to meet common-mode gain requirement. 11.45 Not in detail, Approximation looks good. a.

( ) ( )

( )( )

( )

( ) ( )( )( )

( )( )

211

211

21 1 1

21 1

2

1 1

1 2 1 2

02 02

5 and 2 2

5 2 0.050 120

5 2 2 1

2 3 3 0

3 3 4 2 32.186 V

2 25 2.186 0.141 mA

20

0.0704 mA2

5 0.0704 25 3.24

GSS S D n GS TN

S

GSGS

GS GS GS

GS GS

GS GS

S S

SD D D D

VI I I K V V

RV V

V V V

V V

V V

I I

II I I I

v v

− − −= = = −

−= −

− = − +

− − =

± += ⇒ =

−= ⇒ =

= = ⇒ = =

= − ⇒ = V

b. ( ) ( ) ( )

( ) ( )0

2 2 0.05 2.186 10.119 mA/V1 1 710 k

0.02 0.0704

m n GS TN

m

DQ

g K V Vg

rIλ

= − = −=

= = = Ω

1 1 2 2

01 011

0

,

0

gs S gs S

Sm gs

D

V v V V v Vv v V

g VR r

= − = −−

+ + =

( )01 10 0

1 1 0Sm S

D

Vv g v V

R r r⎛ ⎞

+ + − − =⎜ ⎟⎝ ⎠

(1)

02 022

0

0Sm gs

D

v v Vg VR r

−+ + =

( )02 20 0

1 1 0Sm S

D

Vv g v V

R r r⎛ ⎞

+ + − − =⎜ ⎟⎝ ⎠

(2)

( ) ( )

01 021 2

0 0

01 021 2

0 0 0

2

S S Sm gs m gs

S

S Sm S m S

S

v V v V Vg V g Vr r R

v v V Vg v V g v Vr r r R

− −+ + + =

− + + − + − =

( ) 01 021 2

0 0 0

2 12m S mS

v vg v v V g

r r r R⎧ ⎫

+ + + = + +⎨ ⎬⎩ ⎭

(3)

From (1)

10

01

0

1

1 1

S m m

D

V g g vr

v

R r

⎛ ⎞+ −⎜ ⎟

⎝ ⎠=⎛ ⎞

+⎜ ⎟⎝ ⎠

Then

( )1

0 021 2

0 00

0

12 12

1 1

S m m

m S mS

D

V g g vr v

g v v V gr r R

rR r

⎛ ⎞+ −⎜ ⎟ ⎧ ⎫⎝ ⎠+ + + = + +⎨ ⎬

⎛ ⎞ ⎩ ⎭+⎜ ⎟⎝ ⎠

(3)

( )

( )

1 2 0 1 02 00 0 0 0 0

0 01 2 1 02

0 0 0

01

1 1 1 1 1 2 1 1 12

1 1 2 1 11 2 1

m S m m S mD D S D

m m S m mD D S D

mD

g v v r V g g v v V g rR r r R r r R R r

r rg v v g v v V g g

R R r r R R r

rg vR

⎛ ⎞ ⎛ ⎞ ⎛ ⎞ ⎧ ⎫ ⎛ ⎞+ + + + − + + = + + ⋅ +⎨ ⎬⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎩ ⎭ ⎝ ⎠⎧ ⎫⎛ ⎞ ⎛ ⎞ ⎛ ⎞⎛ ⎞ ⎛ ⎞⎪ ⎪+ + − + + = + + + − +⎨ ⎬⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎪ ⎪⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎩ ⎭

⋅ 0 0 02 2 02

0 0 0

1 1 2 1 2 12 2S m m mD D S D D S D

r r rv v v V g g gR R r r R R R R R r

⎛ ⎞ ⎧ ⎫⎛ ⎞+ + ⋅ + + = + + + ⋅ + + − −⎨ ⎬⎜ ⎟⎜ ⎟

⎝ ⎠ ⎝ ⎠ ⎩ ⎭

( )0 0 01 2 2 02 0

0 0

1 1 1 1 22 1 1m S m mD D D S D D

r r rg v v v v V g g r

R R R r r R R R⎧ ⎫⎛ ⎞⎛ ⎞ ⎛ ⎞⎪ ⎪⋅ + + ⋅ + + = + + + + +⎨ ⎬⎜ ⎟⎜ ⎟ ⎜ ⎟⎪ ⎪⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎩ ⎭

(4)

Then substituting into (2), 02 20 0

1 1 1m S m

D

v g v V gR r r

⎛ ⎞ ⎛ ⎞+ + = +⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

Substitute numbers: ( )

( )( )

1 2 2 02710 710 1 10.11925 25 25 710

1 1 710 20.119 1 1 0.119 710710 20 25 25S

v v v v

V

⎡ ⎤ ⎡ ⎤+ + + +⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦⎧ ⎫⎛ ⎞= + + + + +⎡ ⎤⎨ ⎬⎜ ⎟ ⎣ ⎦⎝ ⎠⎩ ⎭

(4)

( )[ ] ( ) ( )

1 2 02

1 2 02

0.119 28.4 29.4 0.0414 0.1204 1.470 6.83928.4296

or 0.4010 0.4150 0.00491

S

S

S

v v v VV

V v v v

+ + = + +=

= + +

( )02 21 1 1Then 0.119 0.11925 710 710Sv v V⎛ ⎞ ⎛ ⎞+ + = +⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠ (2)

( ) ( ) ( )[ ]( ) ( ) ( )

( ) ( )

( ) ( )

02 2 1 2 02

02 1 2

02 1 2

1

2

02

02

0.0414 0.119 0.1204 0.401 0.4150 0.004910.0408 0.04828 0.0690

1.183 1.691

Now 2

2

So 1.183 1.6912 2

Or 1.437 0.508 1.437,

dcm

dcm

d dcm cm

d cm d

v v v v vv v vv v v

vv v

vv v

v vv v v

v v v A A

+ = + += −

= −

= +

= −

⎛ ⎞ ⎛ ⎞= + − −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= − ⇒ =

10

0.508

1.43720log 9.03 dB0.508

cm

dB dBC M R R C M R R

= −

⎛ ⎞= ⇒ =⎜ ⎟⎝ ⎠

11.46

KVL:

1 1 2 2

1 2 1 2So gs gs

gs gs

v V V vv v V V

= − +− = −

KCL:

( ) ( )1 2 1 2

1 1 2 2 1 2

01 1So ,2 2

m gs m gs gs gs

gs gs

g V g V V V

V v v V v v

+ = ⇒ = −

= − = − −

Now 02 02 01

2

0102

1 1

m gsD L

D L L

v v v g VR R

vvR R R

−+ = −

⎛ ⎞= + −⎜ ⎟

⎝ ⎠

(1)

01 01 021

0201

1 1

m gsD L

D L L

v v v g VR R

vvR R R

−+ = −

⎛ ⎞= + −⎜ ⎟

⎝ ⎠

(2)

From (1): 01 02 21 Lm L gs

D

Rv v g R VR

⎛ ⎞= + +⎜ ⎟

⎝ ⎠

Substitute into (2):

( ) ( )

( )

021 02 2

1 2 1 2 02 2

02 021 2 2

1 2

1 1 1 11

1 1 112

11 222

2

Lm gs m L gs

D D L D L L

L Lm m

D D DD

m LL L

m dD D D L

D

vRg V v g R VR R R R R R

R Rg v v g v v vR R RR

g Rv vR Rg v v AR R R v v R

R

⎛ ⎞⎛ ⎞ ⎛ ⎞− = + + + + −⎜ ⎟⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠ ⎝ ⎠⎛ ⎞⎛ ⎞⎛ ⎞− ⋅ − + + − = + +⎜ ⎟⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠ ⎝ ⎠

⋅⎛ ⎞ ⎛ ⎞− = + ⇒ = =⎜ ⎟ ⎜ ⎟ − ⎛ ⎞⎝ ⎠ ⎝ ⎠ +⎜ ⎟

⎝ ⎠

From symmetry 011

1 2

12

2

m L

dL

D

g RvAv v R

R

− ⋅= =

− ⎛ ⎞+⎜ ⎟

⎝ ⎠

Then 02 01

1 2 2

m Lv

L

D

v v g RAv v R

R

−= =

− ⎛ ⎞+⎜ ⎟

⎝ ⎠

11.47

( )

( ) ( )( )

( )

1 2 1 2 1 2 1 2

1 2 2

2 1 2

0 2 1 2

and 0Then 2

1Or 2

2

Or 2

gs gs m gs m gs gs gs

gs

gs

mm gs D L D L

md D L

v v V V g V g V V Vv v V

V v v

gv g V R R R R v v

gA R R

− = − + = ⇒ = −− = −

= − −

= − = −

=

11.48

From Equation (11.64(a)), 2n Q

d D

K IA R= ⋅

We need 2 100.2dA = =

( )0.5Then 10 or 20

2n

D n D

KR K R= ⋅ ⋅ =

If we set 20 ,DR k= Ω then 21 /nK mA V=

For this case ( )( )10 0.25 20 5 VDV = − =

( )( ) ( )

( )

0.25 1 1.5 V1

1.5 1 0.5

Then max sat5 0.5 1.5

Or max 6 V

GS

DS GS TN

cm D DS GS

cm

V

V sat V V V

v V V V

v

= + =

= − = − =

= − += − +

=

11.49

( )( )

( ) ( )( )( )

1 1 1

2 2 2

2 1 2 1

1 2

1 2

Now

Define

Then and 0

d m gs D m D s

d m gs D m D s

o d d m D s m D s

o m D

d

od m D cm

d

V g V R g R V VV g V R g R V V

V V V g R V V g R V V

V g R V VV V V

VA g R AV

= − = − −= − = − −

= − = − − − − −

= −− ≡

= = =

11.49

( )( )( )

( )( )

( )( )( )( )

02 04

02

04

2 2 0.12 0.075

0.1897 mA/V1 1 889 kΩ

0.015 0.0751 1 667 kΩ

0.02 0.075

0.1897 889 667 72.3

d m

m n DQ

n DQ

p DQ

d d

A g r r

g k I

rI

rI

A A

λ

λ

=

= =

=

= = =

= = =

= ⇒ =

11.50 (a)

( )

( )( )

21 2

1 2

1

1 2 1 1

1 2

0.080 10 0.40 /2 2

0.1 1 1.5 0.4

1.5 1 0.5 For 3

3 1.5 0.5 2 10 2 80

0.1

nn n

DGS GS TN

n

DS

CM D D CM GS DS

D D

D D

K WK K mA VL

IV V V VK

V sat Vv V V V v V V sat

V V V

R R k

′⎛ ⎞⎛ ⎞ ⎛ ⎞= = = =⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠

= = + = + =

= − == + ⇒ = = − +

= − + ⇒ = =−= ⇒ = Ω

(b)

( )( )

( )( )

( )( )( )

1 and 2 0.4 0.1 0.4 /2

1Then 0.4 80 162

1645 177.8

So 0.090

1 20.4 80

0.090 443 1 2 0.4

d m D m

d

dBcm

cm

m Dcm

m o

oo

A g R g mA V

A

C M R R C M R RA

Ag RA

g R

R kR

= = =

= =

= ⇒ = =

=

=+

= ⇒ = Ω+

If we assume 10.01 Vλ −= for the current source transistor, then

( )( )1 1 500

0.01 0.2oQ

r kIλ

= = = Ω

So the CMRR specification can be met by a 2-transistor current source.

( )

( ) ( ) ( )

3 4

23 4 3

3

3 1 3 3

Let 1

0.080 0.2Then 1 0.040 / and 1 3.24 2 0.04

For 3 , 3 3 1.5 4.5 min 4.5 10 5.5

Qn n GS TN

n

CM D GS DS DS

W WL L

IK K mA V V V V

Kv V V V V V V V sat

⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ ⎞= = = = + = + =⎜ ⎟⎝ ⎠

= − = − − = − − = − ⇒ = − − − = > So design is OK.

On reference side: For ( )1, max 3.24GSW V VL

⎛ ⎞ ≥ =⎜ ⎟⎝ ⎠

320 20 3.24 16.76 GSV V− = − =

Then 16.67 5.173.24

= ⇒ We need six transistors in series.

( )

( )

2

2

20 3.24 2.793 6

20.0800.2 2.793 1 1.56 for each of the 6 transistors.

2

GS

nREF GS TN

V V

K WI V VLW WL L

−= =

′⎛ ⎞⎛ ⎞= −⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠⎛ ⎞⎛ ⎞ ⎛ ⎞= − ⇒ =⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠

11.51

( )( )

( )( )

122 2 0.25 0.25 0.50 /

1 0.50 3 0.752

d m D

m n D

d

A g R

g K I mA V

A

=

= = =

= =

From Problem 11.26

( )

( ) ( )( )

1 2 1 2

2 1 2

5 1, 2.5 and 1.25

2Then

0.75 1.25 0.9375

A B

o d

V V V V V V V

V A V V

δδ

δ

δ δ

+= = = = − =

+

= ⋅ − = =

So for 0.01 0.01δ− ≤ ≤ 29.375 9.375 oV mV− ≤ ≤

11.52 From previous results

( )

2 11 1 1 1 1 1

1 2

32 3 2 3 2 2

2 1

1 1 2 2

1 2

1 22

1

2 20

1 1and 2 302 2

Set 5 and 2.5 2 2

Let 0.1 Then 100 , 50

0.06 20Then 2 0.12 100

o od m n Q

od m n Q

o o

Q Q

Q Q

v vA g R K I Rv v

vA g R K I R

v vI R I R

V V

I I mAR k R k

W WL L

−= = = ⋅ =

−

= = = ⋅ =−

= =

= == Ω = Ω

⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞= ⇒⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎝ ⎠

( ) ( )1 2

2

3 3 4

6.67

2 300.060and 2 0.1 2402 50

WL

W W WL L L

⎛ ⎞= =⎟ ⎜ ⎟⎝ ⎠

⎛ ⎞⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞= ⇒ = =⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎝ ⎠⎝ ⎠

11.53

a. 2

11 1 GS

D DSSP

vi IV

⎛ ⎞= −⎜ ⎟

⎝ ⎠

2

22 1 GS

D DSSP

vi IV

⎛ ⎞= −⎜ ⎟

⎝ ⎠

( )

( )

( ) ( )

( ) ( )( )

( )( )

1 21 2

2 1

1 2 2 1

2 21 1 2

21 1 1 1 2

21 1 2

1 1

2

1Then 2

GS GSD D DSS DSS

P P

DSSGS GS

P

DSS DSSd d

P P

D D Q D Q D

DSSD Q D d

P

DSSD D Q D Q D d

P

DSSD Q D Q d

P

v vi i I I

V V

Iv v

V

I Iv v

V Vi i I i I i

Ii I i v

VI

i i I i I i vV

Ii I i I vV

⎛ ⎞ ⎛ ⎞− = − − −⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

= −

= − ⋅ = ⋅−

+ = ⇒ = −

− − = ⋅−

− − + − = ⋅−

⎡ ⎤− = − ⋅⎢

−⎢⎣ ⎦⎥⎥

Square both sides

( )

( )

( ) ( )

2

2 21 1 2

2

2 22

1

22 2

2 21 2 2

1 04

144

2

212 2

DSSD D Q Q d

P

DSSQ Q Q d

PD

Q Q DSS d DSS dD Q Q

P P

Ii i I I v

V

II I I vV

i

I I I v I vi I IV V

⎡ ⎤− + − ⋅ =⎢ ⎥

−⎢ ⎥⎣ ⎦

⎡ ⎤⎛ ⎞± − − ⋅⎢ ⎥⎜ ⎟⎝ ⎠ −⎢ ⎥⎣ ⎦=

⎡ ⎤⎛ ⎞⎢ ⎥⎜ ⎟= ± − − +

⎜ ⎟⎢ ⎥− −⎝ ⎠⎣ ⎦

Use + sign

( ) ( )

( )

2

2 21 2 2

2 2

1

2 2

1

2 1

2

212 2

212 2

Or

21 12 2

We had

Then

1 12 2

Q Q DSS DSSD d d

P P

Q Q DSS DSS dD d

P Q Q P

DSS DSS dDd

Q P Q Q P

D Q D

D

Q P

I I I Ii v v

V V

I I I I vi vV I I V

I I vi vI V I I V

i I i

iI V

⎛ ⎞⎜ ⎟= + ⋅ − ⋅⎜ ⎟− −⎝ ⎠

⎛ ⎞ ⎛ ⎞= + − ⎜ ⎟ ⎜ ⎟⎜ ⎟− ⎝ ⎠⎝ ⎠

⎛ ⎞⎛ ⎞ ⎛ ⎞= + ⋅ − ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟−⎝ ⎠ ⎝ ⎠⎝ ⎠

= −

⎛ ⎞= − ⎜ ⎟−⎝ ⎠

2 22 DSS DSS d

dQ Q P

I I vv

I I V⎛ ⎞ ⎛ ⎞

⋅ − ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

b. 1

2 2

2 2

If , then

21 112 2

2

D Q

DSS DSS dd

P Q Q P

DSS DSS dP d

Q Q P

i I

I I vvV I I V

I I vV vI I V

=

⎛ ⎞⎛ ⎞ ⎛ ⎞= + ⋅ − ⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟−⎝ ⎠ ⎝ ⎠⎝ ⎠

⎛ ⎞ ⎛ ⎞= − ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

Square both sides

( )

( )

( )

2 22 2

2 22 22 2

2 2 22

22 2

22

2

21 0

2 2 14

2 12

DSS DSS dP d

Q Q P

DSS DSSd d P

Q P Q

DSS DSS DSSP

Q Q Q Pd

DSS

Q P

Qd P

DSS

I I vV vI I V

I Iv v V

I V I

I I I VI I I V

vII V

Iv V

I

⎡ ⎤⎛ ⎞ ⎛ ⎞⎢ ⎥= − ⎜ ⎟ ⎜ ⎟⎜ ⎟⎢ ⎥⎝ ⎠⎝ ⎠⎣ ⎦

⎛ ⎞ ⎛ ⎞− ⋅ + =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

⎛ ⎞ ⎛ ⎞ ⎛ ⎞± −⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎝ ⎠⎝ ⎠ ⎝ ⎠=

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

⎛ ⎞= ⎜ ⎟

⎝ ⎠1/ 2

Or Qd P

DSS

Iv V

I⎛ ⎞

= ⎜ ⎟⎝ ⎠

c. For vd small,

( )

( )

1

10

212 2

212d

Q Q DSSD d

P Q

Q DSSDf v

vd P Q

I I Ii v

V I

I Idigd V I→

≈ + ⋅ ⋅−

= = ⋅ ⋅−

( ) 1Or max2

Q DSSf

P

I Ig

V⎛ ⎞

⇒ = ⎜ ⎟−⎝ ⎠

11.53

( )2d m o oA g r R=

Want 400dA = From Example 11.15, 2 1 or M= Ω Assuming that 0.283 /mg mA V= for the PMOS from Example 11.15, then 285 .oR M= Ω

( )

( )1

1 1 2

So 400 1000 285000 0.4014 / 22

0.0800.04028 0.1 10.12

nm m DQ

k Wg g mA V IL

W W WL L L

′⎛ ⎞⎛ ⎞= ⇒ = = ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

⎛ ⎞⎛ ⎞ ⎛ ⎞ ⎛ ⎞= ⇒ = =⎜ ⎟⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠ ⎝ ⎠

11.54 a.

( )1 2

0

1 mA

7 10 0.5 6 kΩQ D D Q

D D

I I I I

v R R

= + ⇒ =

= = − ⇒ =

b.

( )

( ) ( )( ) ( )

1max2

1 21max max 0.25 mA/V4 2

Q DSSf

P

f f

I Ig

V

g g

⋅⎛ ⎞= ⎜ ⎟−⎝ ⎠

⎛ ⎞= ⇒ =⎜ ⎟⎝ ⎠

c.

( )( )( )

max2

0.25 6 1.5

m Dd f D

d d

g RA g R

A A

= = ⋅

= ⇒ =

11.55 a.

( ) ( )

( )( )( ) ( )

( )

( )( )( )

( )

2

2

2

2

1 2

2

52 1

5 2 0.8 20 12

15 2 16 14

8 33 27 0

33 1089 4 8 272 8

1.125 V5 1.125

200.306 mA

0.153 mA1.17 V

GS GSS DSS

S P

GSGS

GS GS GS

GS GS

GS

S

D D

o

V VI IR V

VV

V V V

V V

V

I

I Iv

− − − ⎛ ⎞= = −⎜ ⎟

⎝ ⎠

⎛ ⎞− = −⎜ ⎟⎜ ⎟−⎝ ⎠

⎛ ⎞− = + +⎜ ⎟⎝ ⎠

+ + =

− ± −=

= −− −

=

== ==

(b) 11.56 Equivalent circuit and analysis is identical to that in problem 11.36.

2

1

02 01

12

2

12

2

2

m L

dL

D

m L

dL

D

m Lv

d L

D

g RA

RR

g RA

RR

v v g RA

v RR

⋅=⎛ ⎞

+⎜ ⎟⎝ ⎠

− ⋅=⎛ ⎞

+⎜ ⎟⎝ ⎠

−= =

⎛ ⎞+⎜ ⎟

⎝ ⎠

11.57 (a)

( )

( )( )

2 4

2

4

0.1 3.846 mA/V0.026120 1200 K0.180 800 K0.13.846 1200 800

1846

d m o o

m

o

o

d

d

A g r r

g

r

r

A

A

=

= =

= =

= =

=

=

(b)

( )( )For 923 3.846 1200 800

480240 480 480 K480

d L

LL L

L

A R

RR RR

= =

= = ⇒ =+

11.58 (a)

( ) ( )1 1

2250 A 1

2250 1 252.8 A180

5 0.7 536.8 K

0.2528

Q REF QI I I

R R

μβ

μ

⎛ ⎞= = +⎜ ⎟

⎝ ⎠⎛ ⎞= + =⎜ ⎟⎝ ⎠

− − −= ⇒ =

(b)

( )

( )( )

2 4

2

4

0.125 4.808 mA/V0.026150 1200 K

0.1251004.808 1200 800 800 K

0.1252308

d m o o m

o

d o

d

A g r r g

r

A r

A

= = =

= =

= = =

=

(c) ( )( )

2 4

2 180 0.0262 74.9 K

0.1251200 800 480 K

id id

o o o o

R r R

R r r R

π= = ⇒ =

= = = =

(d) ( )( )max 5 0.7 4.3 Vmin 0.7 0.7 5 3.6 V

cm

cm

vv

= − == + − = −

11.59 a.

0 3 4

0 0

122

0.2 2 A100

QB B

Q

II I I

II I

β

μβ

⎛ ⎞⎛ ⎞= + ≈ ⎜ ⎟⎜ ⎟

⎝ ⎠⎝ ⎠

= = ⇒ =

b.

( ) ( )( )

02 04

02 04

100 1000 k0.1

0.1 3.846 mA/V0.026

3.846 1000 1000 1923

A

CQ

CQm

T

d m d

Vr rI

Ig

VA g r r A

= = = = Ω

= = =

= = ⇒ =

c. ( )

( )( )02 04

3.846 1000 1000 250 641d m L

d d

A g r r R

A A

=

= ⇒ =

11.60 a.

( )02 04

202

404

02 04

2125

80

If 2 mA, then 38.46 mA/V125 kΩ, 80 kΩ

So 38.46 125 80 200

Or 1508

d m L

CQ Qm

T T

A

CQ CQ

A

CQ CQ

Q m

d

d

A g r r R

I Ig

V VVrI IVrI I

I gr r

A

A

=

= =

= =

= =

= == =

= ⎡ ⎤⎣ ⎦=

For each gain of 1000. lower the current level

02

04

For 0.60 mA, 0.30 mA0.3 11.54 mA/V

0.026125 417 kΩ0.380 267 kΩ0.311.54 417 267 200 1036

So 0.60 mA is adequate

Q CQ

m

d

Q

I I

g

r

r

A

I

= =

= =

= =

= =

= ⎡ ⎤ =⎣ ⎦=

b.

( ) ( )( )

For 10 V, 0.6 VFor 0, max 2 10 2 0.6Or max 8.8 V

BE EB

CB cm EB

cm

V V VV v V V

v

+

+

= = == = − = −

=

11.61 a. From symmetry.

( )( )

3 4 3 4

3 4

1 2

1 2 1 3

1 2

0.1 10.1

Or 2 V

0.1 1 2 V0.1

102 2 10

Or 10 V

GS GS DS DS

DS DS

SG SG

SD SD SG DS

SD SD

V V V V

V V

V V

V V V V

V V

= = = = +

= =

= = + =

= = − −= − −

= =

b.

( )( )

( )( )( )

( )( )( ) ( )( )

0

0

1 1 1 MΩ0.01 0.1

1 1 0.667 MΩ0.015 0.1

22 0.1 2 1 0.2 /

0.2 1000 667 80

nn DQ

pP DQ

m p SG TP

d m on op d

rI

rI

g K V VmA V

A g r r A

λ

λ

= = ⇒

= = ⇒

= += − =

= = ⇒ =

(c)

( )( )

( )( )

2 1

44

22

2 4

0.1 2

1 1 1000 0.01 0.1

1 1 667 0.015 0.1

667 1000 400

QD D

on D

oP D

o o o

II I mA

r kI

r kI

R r r k

λ

λ

= = =

= = = Ω

= = = Ω

= = = Ω

11.62

( )

( )( )

( )( )

( )( )( )( )

4 2

4

2

0.082 2.5 0.052

0.1414 mA/V1 1000 K

0.02 0.051 1333 K

0.015 0.050.1414 1000 1333

80.8

d m o o

m

o

o

d

d

A g r r

g

r

r

A

A

=

⎛ ⎞= ⎜ ⎟⎝ ⎠

=

= =

= =

=

=

11.63

( )

( )( )

( )( )

( )( )

04 04 4 4

04

4

4

4

3

3

04

0

0

1

80 800 K0.1

0.1 3.8460.026100 0.026

0.126 K

1 26 0.963 K

Assume 100100 0.026

26 kΩ0.1

0.1 3.846 mA/V0.026800 1 3.846 0.963 3.763 MΩ

3.763MThen

m

m

m

v m

R r g R r

r

g

r

R r

r

g

R

R

A g r

π

π

π

π

β

⎡ ⎤= +⎣ ⎦

= =

= =

=

=

= =

=

= =

= =

= + ⇒⎡ ⎤⎣ ⎦⇒ = Ω

= − ( )

( )

2 0

02120 1200 kΩ0.10.1 3.846 mA/V

0.0263.846 1200 3763 3499

m

v v

R

r

g

A A

= =

= =

= − ⎡ ⎤ ⇒ = −⎣ ⎦

b. For

( )( )

04

02 04

800, 800 kΩ0.1

3.846 1200 800 1846v m

v

R r

A g r r

A

= = =

= −

= − ⎡ ⎤ ⇒ = −⎣ ⎦

(c) For part (a), ( )3.763 1.2 0.910 oR M= = Ω

For part (b), ( )1.2 0.8 0.48 oR M= = Ω 11.64

( )

( )

( )

5 3 4 3 45

3 4

5

166

1 1 1Now

So 1

1 1

E B B C CB

C C Q

QB

QEB

I I I I II

I I II

I

III

β β β β

β β

β β β

+ += = =

+ + ++ ≈

≈+

= =+ +

6 5For balance, we want B BI I=

1So that Q QI I=

11.65 Resistance looking into drain of M4.

4 1

44 4

04

14 1

04 04

10 04 4 1

04

1

Or 1

sg X

X sgX m sg

XX m

m

V I RV V

I g Vr

R VI g Rr r

RR r g Rr

≅−

± =

⎡ ⎤+ + =⎢ ⎥

⎣ ⎦⎡ ⎤

= + +⎢ ⎥⎣ ⎦

a.

( )( )( )

( )( )( )( )

( )( )

( )( )

( )

2 2

2

2

4

4

0

2 2 0.080 0.1

0.179 /1 1 667

0.015 0.1

2 2 0.080 0.1

0.179 /1 1 500

0.02 0.1

1500 1 0.179 1 590.5 kΩ500

0.179 667 590.5 56.06

d m o o

m n DQ

on DQ

m P DQ

op DQ

d d

A g r R

g K I

mA V

r kI

g K I

mA V

r kI

R

A A

λ

λ

=

= =

=

= = = Ω

= =

=

= = = Ω

⎡ ⎤= + + =⎢ ⎥⎣ ⎦= ⎡ ⎤ ⇒ =⎣ ⎦

b.

( )1 0 04When 0, 500 kΩ

0.179 667 500 51.15d d

R R rA A

= = == ⎡ ⎤ ⇒ =⎣ ⎦

(c) For part (a), 2 667 590.5 313 o o o oR r R R k= = ⇒ = Ω

For part (b), 2 4 667 500 286 Ωo o o oR r r R k= = ⇒ = 11.66 Let 100, 100 AV Vβ = =

[ ]

( )( )

( )( )( ) ( )

2

4 4

4

2 4

100 1000 0.1

1 where Now

100 0.02626

0.10.1 3.846 /

0.02626 1 0.963

Then 1000 1 3.846 0.963 4704

3.846 1000 4704 3172

Ao

CQ

o o m E E E

m

E

o

d m o o d

Vr kI

R r g R R r R

r k

g mA V

R kR k

A g r R A

π

π

= = = Ω

′ ′= + =

= = Ω

= =

′ = = Ω= + = Ω⎡ ⎤⎣ ⎦

= = ⇒ =

11.67 (a) For Q2, Q4

(1) 42 2 4 4

2 4

x xx m m

o o

V V VI g V g Vr r

ππ π

−= + + +

(2) 4 42 2

2 4 2

xm

o

V V Vg V

r r rπ π

ππ π

−+ =

(3) 4 2V Vπ π= −

From (2) 4 22 4 2 2

1 1xm

o o

V V gr r r rπ

π π

⎡ ⎤= + +⎢ ⎥

⎢ ⎥⎣ ⎦

( )

( )( )

( )( )

( )

4

2 22

2

2

2

4

Now

120 0.5 0.496 1 2 121

1 1200.5 0.0041 2 1 1 121

So120 0.026

761 0.0041

0.0041 0.158 0.026100 24.4

0.0041120 0.

QC

QC C

m

o

II mA

II I mA

r k

g mA V

r M

r

π

π

ββ

ββ β

⎛ ⎞⎛ ⎞ ⎛ ⎞= = =⎜ ⎟⎜ ⎟ ⎜ ⎟+ ⎝ ⎠⎝ ⎠⎝ ⎠⎛ ⎞⎛ ⎞⎛ ⎞⎛ ⎞⎜ ⎟= = ⇒ =⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠⎝ ⎠ ⎝ ⎠

= = Ω

= = /

= ⇒ Ω

=( )

( )

4

4

4 42 2

0266.29

0.4960.496 19.08 /0.026100 202

0.496Now

1 1 0.158 which yields 6.29 761 24400 0.318

m

o

x x

o o

k

g mA V

r k

V VV V

r rπ π

= Ω

= =

= = Ω

⎡ ⎤= + + ⇒ =⎢ ⎥

⎢ ⎥⎣ ⎦

From (1),

( )( )

4 4 22 4 2

2

6

2 6

1

119.08 0.1581 1 24400 which yields 135

24400 202 0.318 24400

80Now 160 0.5

Then 135 160 73.2

x xx m m

o o o

x xo

x x

o

o o o o

V VI V g gr r r

I VR kV I

r k

R R r R k

π⎛ ⎞

= + + − −⎜ ⎟⎝ ⎠

⎡ ⎤⎛ ⎞− −⎜ ⎟⎢ ⎥⎝ ⎠⎢ ⎥= + + = = Ω⎢ ⎥⎢ ⎥⎣ ⎦

= = Ω

= = ⇒ = Ω

(b)

where / 2

c cd m o m

d

iA g R gv

Δ= =

( )

( ) ( )

1 1 3 3 1 3

11 1 3 3

1

1 3 31

1 3 1

1 1

1 3 1

and 2

Also

1So

121Or 6.29761

Then 22 4

So 0.158 19.08 9.624 2

So

dm m

m

d d

d dm m

cm

vi g V g V V V

Vg V r V

r

V r Vr

V V V

v vV V

v vi g g V

igv

π π π π

ππ π π

π

π π ππ

π π π

π π

π

β

Δ = + + =

⎛ ⎞+ =⎜ ⎟

⎝ ⎠⎛ ⎞+ =⎜ ⎟⎝ ⎠⎛ ⎞ = ≅⎜ ⎟⎝ ⎠

= ⇒ =

⎛ ⎞ ⎛ ⎞Δ = + = + =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

Δ= ( )( )

( )( )( )

1 3

9.62 9.62 73.2 704/ 2

Now 2 where 1761 121 6.29 1522

Then 3.044

d dd

id i i

i

id

A A

R R R r rR k

R M

π πβ

= ⇒ = ⇒ =

= = + += + = Ω

= Ω

11.69 (a)

( )

( )( )

( )( )( )

( )

2 4

2

4

100

Let 0.5 1 1 200

0.02 0.251 1 160

0.025 0.25Then 100 200 160 1.125 /

22

0.0801.125 2 0.25 31.62

d m o o

Q

on D

oP D

m m

nm D

n

A g r r

I mA

r kI

r kI

g g mA V

K Wg IL

W WL L

λ

λ

= =

=

= = = Ω

= = = Ω

= ⇒ =

′⎛ ⎞⎛ ⎞= ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

⎛ ⎞⎛ ⎞ ⎛ ⎞= ⇒ =⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠

Now P

WL

⎛ ⎞⎜ ⎟⎝ ⎠

somewhat arbitrary. Let 31.6P

WL

⎛ ⎞ =⎜ ⎟⎝ ⎠

11.70

( )( )( )

( )( )

( )( )

( )( )

( )

2 4

2

4

Let Then 0.5 2 3 3 0.0417

1 1 3205 0.015 0.0208

1 1 2404 0.02 0.0208

Then

80 3205 2404 0.0582

22

0.00.0582 2

d m o o

Q REF

Q REF

Q Q REF

on D

oP D

d m m

nm D

n

A g r r

P I I V V

I II I I mA

r kI

r kI

A g g mA V

k Wg IL

λ

λ

+ −

=

= + −

== − − ⇒ = =

= = = Ω

= = = Ω

= = ⇒ = /

′⎛ ⎞⎛ ⎞= ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

= ( )80 0.0208 1.022 n n

W WL L

⎛ ⎞⎛ ⎞ ⎛ ⎞⇒ =⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠

11.71

( )

( )( )( )

( )

2

2

2

1 2

1

1 666.7 K0.015 0.1

400 666.70.60 mA/V

22

0.080.60 2 0.12

0.090 0.004

22.5

d m o o

m o

on D

d m

m

nD

A g r R

g r

rI

A gg

k W IL

WL

WL

W WL L

λ

=

≈

=

= =

= ==

′⎛ ⎞⎛ ⎞= ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

⎛ ⎞⎛ ⎞= ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

⎛ ⎞= ⎜ ⎟⎝ ⎠

⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

11.72

( )

[ ][ ]

( )( )

( )( )

( )( )

( )( )

4 6

4 4 2 4 4

6 6 8 6 6

2 4

6 8

4

6

4

where11

We have1 1667

0.015 0.0401 1250

0.02 0.040

0.0602 15 0.040 0.268 2

0.0252 10 0.040 0.141 2

Then

1667 166

d m o o

o o o m o

o o o m o

o o

o o

m

m

o

A g R R

R r r g rR r r g r

r r k

r r k

g mA V

g mA V

R

=

= + += + +

= = = Ω

= = = Ω

⎛ ⎞= = /⎜ ⎟⎝ ⎠

⎛ ⎞= = /⎜ ⎟⎝ ⎠

= + ( )( )( )( )6

7 1 0.268 1667 748

1250 1250 1 0.141 1250 222.8 o

M

R M

+ ⇒ Ω⎡ ⎤⎣ ⎦= + + ⇒ Ω⎡ ⎤⎣ ⎦

(a)

4 6 748 222.8 172 o o o oR R R R M= = ⇒ = Ω

(b) ( ) ( )( )4 4 6 0.268 172000 46096d m o o dA g R R A= = ⇒ =

11.73

( )

( )( )( )( )

( )( )

2 4

2 4

2 4

1

1 500 K0.02 0.1

2 2 0.5 0.1

0.4472 mA/V0.4472 500 500 112

500 500 250 K

d m o o

o oD

m n D

d d

o o o o

A g r r

r rI

g K I

A A

R r r R

λ

=

= =

= =

= =

== ⇒ =

= = ⇒ =

11.74 (a)

( )

( )

( )( )

2

3

2

1

1 1

3 1 1

0.4 1 1.894 V0.5

0.4 1 1.894 V0.5

1.894 1 0.894 V

1.894 0.894 1.894 4 4.89 V

DP p SG TP

SG

DN n GS TN

GS

DS GS TN

SG DS GS CM

I K V V

V

I K V V

V

V sat V VV V V sat V vV V V

+

+ + −

= +

+ = =

= −

+ = =

= − = − == + − += + − + ⇒ = = −

(b)

( )

( )( )( )( )

( )( )

2 4

2 41 1 166.7 K

0.015 0.4

2 2 0.5 0.4 0.8944 mA/V

0.8944 166.7 166.7 74.5

d m o o

o oD

m n D

d d

A g r r

r rI

g K I

A A

λ

=

= = = =

= = =

= ⇒ =

11.75 (a) For 2V 2.7 Vcmv V += + ⇒ = If is a 2-transistor current source,

0.7 0.73.4 V 3.4 V

Q

cm

I

V vV V V

−

− + −

= − −= − ⇒ = − =

(b)

( )

( )( )

2 4 2

4

100 1000 K0.160 600 K0.1

0.1 3.846 mA/V0.026

3.846 1000 600 1442

d m o o o

o

m

d d

A g r r r

r

g

A A

= = =

= =

= =

= ⇒ =

11.76 (a) 3.4 VV V+ −= − = (b)

( )( )

2

4

75 1250 K0.0640 666.7 K

0.060.06 2.308 mA/V0.0262.308 1250 666.7

1004

o

o

m

d

d

r

r

g

A

A

= =

= =

= =

=

=

11.77

( )( )

( )( )

1 1

2

2

2 2 0.2 0.25 0.447 mA/V

0.75 28.85 mA/V0.026120 0.026

4.16 k0.75

m n Bias

CQm

T

T

CQ

g K I

Ig

V

VrIπβ

= = =

= = =

= = = Ω

( )

( )

( ) ( )( )( )( )

1 1 2 20

2 1 1 2 1 2

1 1 2 1 20

1 1 1 2 11 2

10

1 2

10

1 2

and

and 1

11

1 0.447 1211 1 0.447 4.16

18.9 mA/V

m gs m

m gs i gs

gs m m m

ii gs m gs gs

m

mi

m

mCm

i mCm

i g V g VV g V r v V Vi V g g g r

vv V g V r Vg r

gi v

g rgi

gv g r

g

π

π π π

π

ππ

π

π

β

β

= += = +

= + ⋅

= + =+

+= ⋅

++

= = =+ +

⇒ =

11.78

( ) ( )( )

( )

( ) ( )( )

( ) ( ) ( )

0 2

0 2

2

2 0 2 0 2

1 1 500 kΩ0.01 0.2

80 400 kΩ0.2

2 2 0.2 0.2

0.4 mA/V

0.4 500 400 88.9

n DQ

A

CQ

m n DQ

d m

d

r MI

Vr QI

g M K I

A g M r M r Q

A

λ= = =

= = =

= =

=

⎡ ⎤= ⎣ ⎦= ⎡ ⎤ ⇒ =⎣ ⎦

If the IQ current source is ideal, 0cmA = and dBC M RR = ∞ 11.79 a.

b. Assume RL is capacitively coupled. Then

( )( )

( )( )

1

1 1

2 2

2 2

0.7 0.0875 mA8

0.9 0.0875 0.8125 mA

2 2 1 0.0875 0.592 mA/V

0.8125 31.25 mA/V0.026100 0.026

3.2 k0.8125

CQ DQ Q

BEDQ

CQ

m P DQ m

CQm m

T

T

CQ

I I IVIR

I

g K I g

Ig g

V

Vr rIπ πβ

+ =

= = =

= − =

= = ⇒ =

= = ⇒ =

= = ⇒ = Ω

c. ( )

( )( )( )

( ) ( )( )

( )

( )

0 1 2 2

0 0

2 1 1 2

0 1 2 1 1 2

0 0 1 2 1 1 2

1 2 1 1 20

1 2 1 1 2

1 2 1 1 2

1

We find

0.592 31.2

m sg m L

i sg sg i

m sg

m sg m m sg L

i m m m L

m m m Lv

i m m m L

m m m

V g V g V R

V V V V V V

V g V R r

V g V g g V R r R

V V V g g g R r R

g g g R r RVAV g g g R r R

g g g R r

π

π π

π

π

π

π

π

= − −

+ = ⇒ = −

=

⎡ ⎤= − +⎣ ⎦⎡ ⎤= − − +⎣ ⎦

⎡ ⎤+⎣ ⎦= =⎡ ⎤+ +⎣ ⎦

+ = + ( )( )( )

( )( )( )( )

5 0.592 8 3.2

42.8842.88

Then 1 42.88

Lv

L

RA

R

=

=+

11.80 a. Assume RL is capacitively coupled.

( )( )

( )( )

1 1

2 2

2 2

0.7 0.0875 mA8

1.2 0.0875 1.11 mA

2 2 1 0.0875 0.592 mA/V

1.11 42.7 mA/V0.026100 0.026

2.34 kΩ1.11

DQ

CQ

m p DQ m

CQm m

T

T

CQ

I

I

g K I g

Ig g

V

Vr rIπ πβ

= =

= − =

= = ⇒ =

= = ⇒ =

= = ⇒ =

b.

( )( )( )

( )

( )( )( )

2 2 1

1 1 2 2

1 2 1 1 2

01 2 1 1 2

0

1

1 21.6 Ω0.592 0.592 42.7 8 2.34

sg X

X m m sg

m sg

X X m m m

X

X m m m

V VI g V g V

g V R r V

I V g g g R r

VRI g g g R r

R

π

π π

π

π

== +

=

⎡ ⎤= +⎣ ⎦

= =+

= ⇒ =+

11.81 (a)

(1) ( )

22

0om

o

V Vg V

rπ

π− −

+ =

(2) ( )

2 1 12 1 1

1 1 or 0om m i m i

o o o

V V V Vg V g V g V V

r r r r rπ π π

π ππ π

− − ⎛ ⎞− −+ = + + = − +⎜ ⎟

⎝ ⎠

Then 1

1

1 1m i

o

g VV

r r

π

π

=⎛ ⎞

+⎜ ⎟⎝ ⎠

From (1)

( )( )

22 2

22

2 2 2 12

1

1 2 22

1

1

2

1

1 0

11

1 1

1

1 1

Now

2 2 0.25 0.025 0.158 /

0.025 0.9615 /0.026

1

om

o o

mo

o o m o m io

o

m o moo

vi

o

m n Q

Qm

T

o

Vg Vr r

gr

V r g V r g Vr

r r

g r grV

AV

r r

g K I mA V

Ig mA V

V

rI

π

π

π

π

λ

⎛ ⎞+ + =⎜ ⎟

⎝ ⎠⎛ ⎞

+⎜ ⎟⎛ ⎞ ⎝ ⎠= − + = −⎜ ⎟⎛ ⎞⎝ ⎠ +⎜ ⎟⎝ ⎠

⎛ ⎞− +⎜ ⎟

⎝ ⎠= =⎛ ⎞

+⎜ ⎟⎝ ⎠

= = =

= = =

=( )( )

( )( )

( )( )

2

1 2000 0.02 0.025

50 2000 0.025

100 0.026104

0.025Then

10.158 2000 0.96152000 30039

1 12000 104

Q

Ao

Q

T

Q

v v

k

Vr kI

Vr kI

A A

πβ

= = Ω

= = = Ω

= = = Ω

⎛ ⎞− +⎜ ⎟⎝ ⎠= ⇒ = −

⎛ ⎞+⎜ ⎟⎝ ⎠

To find Ro; set 10 0i m iV g V= ⇒ =

( )

( )

( )( )

( )

( )

22

1

2 12 2

2 1 22

Then

1

Combining terms,

1 1

12000 1 2000 104 0.9615 192.2 2000

xx m

o

x o

xx m x o

o o

xo o o m

x o

o

V VI g V

rV I r r

VI g I r rr r

VR r r r gI r

R M

ππ

π π

π

π

− −= +

= −

⎛ ⎞= + − +⎜ ⎟⎝ ⎠

⎡ ⎤⎛ ⎞= = + +⎢ ⎥⎜ ⎟

⎝ ⎠⎣ ⎦⎡ ⎤⎛ ⎞= + + ⇒ = Ω⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

(b)

(1) ( )3

3 33

0o gsm gs

o

V Vg V

r

− −+ =

(2) ( ) ( )3 3 2 3

3 3 2 2 2 23 2 2 2

1 or 0o gs gs gsm gs m m

o o o o

V V V V Vg V g V V g

r r r rπ

π π

− − − − − ⎛ ⎞+ = + = + −⎜ ⎟

⎝ ⎠

(3) ( ) ( )3 2 22

2 2 12 2 1

gsm m i

o o

V V VV g V g Vr r r

π πππ

π

− − − −+ + = +

From (2), 32

2 22

1gs

o mo

VV

r gr

π =⎛ ⎞

+⎜ ⎟⎝ ⎠

Then

(3) 32 2 1

2 2 1 2

1 1 1 gsm m i

o o o

VV g g V

r r r rππ

⎛ ⎞+ + + = +⎜ ⎟

⎝ ⎠

or

3 32 1

2 2 1 22 2

2

1 1 11

gs gsm m i

o o oo m

o

V Vg g V

r r r rr g

rπ

⎡ ⎤+ + + = +⎢ ⎥⎛ ⎞ ⎣ ⎦+⎜ ⎟

⎝ ⎠

3 31 1 10.9615 0.96151 104 2000 2000 20002000 0.9615

2000

gs gsi

V VV⎡ ⎤+ + + = +⎢ ⎥⎛ ⎞ ⎣ ⎦+⎜ ⎟

⎝ ⎠

Then 53 1.83 10gs iV V= ×

From (1), ( )53 3

3 3

1 1 or 2000 0.158 1.83 102000

om gs o i

o o

Vg V V V

r r⎛ ⎞ − ⎛ ⎞+ = = − + ×⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠

75.80 10ov

i

VAV

= = − ×

To find Ro

(1) ( )3

3 33

x gsx m gs

o

V VI g V

r

− −= +

(2) ( ) ( )3 3 2

3 3 2 23 2

x gs gsm gs m

o o

V V V Vg V g V

r rπ

π

− − − − −+ = +

(3) ( )2 1 2x oV I r rπ π= −

3 33 3

3

3

1From (1)

10.1582000 2000

2000So 0.1585

xx gs m

o o

xx gs

xx

gs

VI V gr r

VI V

VIV

⎛ ⎞= + +⎜ ⎟

⎝ ⎠⎛ ⎞= + +⎜ ⎟⎝ ⎠

−=

( ) ( )( )

3 3 2 23 2 3 2

3 2

10

From (2),

1 1 1

1 1 10.158 0.96152000 2000 2000 2000

/ 2000Then 0.159 2000 104 0.9620.1585 2000

We find 6.09 10

xgs m m

o o o o

xgs

x x xx

xo

x

VV g V g

r r r r

VV V

I V V I

VRI

π

π

⎡ ⎤ ⎛ ⎞+ + + = +⎜ ⎟⎢ ⎥

⎣ ⎦ ⎝ ⎠⎡ ⎤ ⎛ ⎞+ + + = +⎜ ⎟⎢ ⎥⎣ ⎦ ⎝ ⎠

−⎡ ⎤ + = −⎢ ⎥⎣ ⎦

= = × Ω

11.82 Assume emitter of Q1 is capacitively coupled to signal ground.

( )( )

( )

( ) ( )( )( )

1

1

1

800.2 0.1975 81

0.2 0.00247 81

80 0.02610.5

0.19750.1975 7.60 /0.0262 2 0.2 0.00247

0.0445 /

CQ

DQ

m

m n D

m

I mA

I mA

r k

g Q mA V

g M K I

g M mA V

π

⎛ ⎞= =⎜ ⎟⎝ ⎠

= =

= = Ω

= =

= =

=

( ) ( )

( )( )

( ) ( )

( )( )( )

( )( )

11

1

1

11

1

and or

1Then 1 or 11

11

7.60 20Then 48.4

110.0445 10.5

i gs m gs gsm

ii

m

m

m Coo m C v

i

m

v v

VV V V V g M V r V

g M r

VV V Vg M r

g M r

g Q RVV g Q V R AV

g M r

A A

ππ π π

π

π ππ

π

π

π

= + = =

⎛ ⎞= + =⎜ ⎟⎜ ⎟ ⎛ ⎞⎝ ⎠ +⎜ ⎟⎜ ⎟

⎝ ⎠−

= − ⇒ = =⎛ ⎞

+⎜ ⎟⎜ ⎟⎝ ⎠

−= ⇒ = −⎛ ⎞

+⎜ ⎟⎜ ⎟⎝ ⎠

11.83 Using the results from Chapter 4 for the emitter-follower:

( )( )

( )( )

[ ]( )

9 07 0118

0 4

88

89

9

07

011

011 011

11

1||1

100 0.0262.6 k

11 0.01 mA

100100 0.026

260 k0.01100 500 k0.2

100 500 k0.2

0.21 , 7.690.026

100 0.0

T

C

CC

A

Q

A

Q

m E m

r r Rr

R R

VrIII

r

VrIVrI

R r g R g

r

ππ

π

π

π

ββ

β

β

⎡ + ⎤+⎢ ⎥+⎢ ⎥=

⎢ ⎥+⎢ ⎥⎣ ⎦

= = = Ω

≈ = =

= = Ω

= = = Ω

= = = Ω

′= + = =

=( )

( )( )011

0

0

2613 k

0.20.2 13 0.197 k500 1 7.69 0.197 1257 k

Then

260 500 12572.6

1015 ||101

5 0.0863 0.0848 K 84.8

ERR

R

R

= Ω

′ = = Ω= + = Ω⎡ ⎤⎣ ⎦

⎡ + ⎤+⎢ ⎥

⎢ ⎥=⎢ ⎥⎢ ⎥⎣ ⎦

= ⇒ = ⇒ Ω

11.84

( )( )( )

( )( )( )

( ) ( )

( )( )( )( )

1 2

2

2

1

30 3

0 0

1100 0.026

5.2 k0.5

100 0.026 100 0.026520 k

0.5 /100 0.5520 101 5.2 1.05 M

100 0.026505 , 2.6 k101 1

2.6 505 5 0.521 0.472 k101

i

i i

R r r

r

r

R R

rR r

R R

π π

π

π

ππ

β= + +

= = Ω

= = = Ω

= + ⇒ ≅ Ω

+= = = Ω

+= = ⇒ = Ω

( )

( )

30 3 3

3

0 33

5

1 5

mVV g Vr

V Vr

ππ

π

ππ

β

⎛ ⎞= − +⎜ ⎟

⎝ ⎠⎛ ⎞+= − ⎜ ⎟⎝ ⎠

(1)

( )0 332 2

3

02 2 3

3

50

1 150 50

m

m

V VVg V

r

Vg V V

r

πππ

π

π ππ

−= +

⎛ ⎞= + −⎜ ⎟

⎝ ⎠ (2)

12 1 1 2

1

1 21

1

mVV g V rr

V rr

ππ π π

π

π ππ

β

⎛ ⎞= +⎜ ⎟⎝ ⎠

⎛ ⎞+= ⎜ ⎟⎝ ⎠

(3)

and 1 2inV V Vπ π= + (4)

20.5 19.23 mA/V

0.026mg = =

Then

( ) ( )0 3 3 0101 5 0.0051492.6

V V V Vπ π⎛ ⎞= − ⇒ = −⎜ ⎟⎝ ⎠

(1)

( )

( )( )

02 0

0

2 0

And1 119.23 0.005149

2.6 50 500.02208

Or 0.001148

VV V

V

V V

π

π

⎛ ⎞= − + −⎜ ⎟⎝ ⎠

= −

= −

(2)

And ( )1 2 0 0.001148in inV V V V Vπ π= − = + (4)

So

( ) ( ) ( )0 01010.001148 0.001148 5.2520inV V V ⎛ ⎞− = +⎡ ⎤ ⎜ ⎟⎣ ⎦ ⎝ ⎠

(3)

( ) ( ) ( ) 00 00.001148 0.001159 1.01 438in v

in

VV V V AV

− − = ⇒ = = −

11.85

( )

2

2

1

5 1 mA5

1 0.8 2.21 V0.5

2.21 50.206 mA

35

GS

I

V

I

= =

= + =

− −= =

( )( )( )( )

( )( )

( ) ( )( )( ) ( )

( )( )( )

( )( )

0 2 2 2 02

2 1 1 01 1 0 1

2 1 01 1 0

0 2 2 02 1 01 1 0

2 2 02 1 01 10

2 2 02

2 2 2

1 1 1

and

So

Then

1

2 2 0.5 1 1.414 /

2 2 0.2 0.206 0.406

m gs

gs m sg sg in

gs m in

m m in

m mv

in m

m n D

m p D

V g V R r

V g V r R V V V

V g V r R V

V g R r g V r R V

g R r g r RVAV g R r

g K I mA V

g K I mA

=

= − = −

= − −

⎡ ⎤= − −⎣ ⎦−

= =+

= = =

= = =

( )( )

( )( )

( )( )( )( )( )( )

011 1

022 2

2 02

1 01

/

1 1 485 k0.01 0.206

1 1 100 k0.01 1

5 100 4.76 k

35 485 32.6 k

1.414 4.76 0.406 32.6Then

1 1.414 4.76So 11.5

D

D

v

v

V

rI

rI

R r

R r

A

A

λ

λ

= = = Ω

= = = Ω

= = Ω

= = Ω

−=

+⇒ = −

Output Resistance—From the results for a source follower in Chapter 6.

0 2 022

0

1 1 5 1001.4140.707 4.76

So 0.616 k

m

R R rg

R

= =

== Ω

11.86 a.

( )

2 2

22 2

2

1 1

5 10 k0.5

0.5 1 2.41 0.25

5 2.4174.1 k

0.1

DSG TP

p

R R

IV V VK

R R

= ⇒ = Ω

= − = + =

− −= ⇒ = Ω

b.

( )( )( )( )

( )( )( )( )( )( )( )( )( )( )

( )( )

( )( )

0 2 2 02 2

2 0 1 1 01 1 1

2 02 2 1 01 10

2 02 2

1 1 1

2 2 2

011 1

022 2

and

1

2 2 0.1 0.1 0.2 /

2 2 0.25 0.5 0.707 /

1 1 1000 k0.01 0.1

1 1 200 0.01 0.5

m sg

sg m gs gs in

m mv

in m

m n D

m p D

D

D

V g V r R

V V g V r R V V

g r R g r RVAV g r R

g K I mA V

g K I mA V

rI

rI

λ

λ

= −

⎡ ⎤= − − =⎣ ⎦−

= =+

= = =

= = =

= = = Ω

= = =

( )( )( )( )( )( )

02 2

01 1

0 2 022

0

k

200 10 9.52 k

1000 74.1 69.0 k

0.707 9.52 0.2 69Then

1 0.707 9.52So 12.0

1 1 10 2000.707

1.414 9.52Or 1.23 k

v

v

m

r R

r R

A

A

R R rg

R

Ω

= = Ω

= = Ω

−=

+⇒ = −

= =

== Ω

11.87 a.

( )

( ) ( )

2

023 1 1

1

03

3

034

2

2 2

0.25 mA5 2 12 k0.25

on 2 0.7 2.6 k0.5

5 5 3 4 k0.5

on 5

3 0.7 5 2.43 k3

C

BEC E E

E

C CC

BEC

E

E E

I

R R

v VI R R

Rv

R RI

v VI

R

R R

=−= ⇒ = Ω

− −= ⇒ = ⇒ = Ω

− −= = ⇒ = Ω

− − −⎡ ⎤⎣ ⎦=

− += ⇒ = Ω

b. Input resistance to base of Q3, ( )

( )( )

( )( )

( )

( )( )( )( )( )

( )( )

3 3 1

3

3

021 2 3

2

1 1

403

02 3 1

4 4 2

20

03 4

1100 0.026

5.2 k0.5

5.2 101 2.6 267.8 k12

0.25 9.62 mA/V0.0261 9.62 12 267.8 55.22

Now 1

where 1

1and

1

i E

i

d m id

m

d d

C i

E

i E

E

E

R r R

r

Rv

A g R Rv

g

A A

R Rvv r R

R r R

Rvv r R

π

π

π

π

π

β

βββ

ββ

= + +

= = Ω

= + = Ω

= =

= =

= ⇒ =

−=

+ +

= + +

+=

+ +

( )( )

( )( )( )( )( )( )

( )( )( )( )

( )( )( )

2

4

0

03

4

3

03

02

0

100 0.0260.867 k

3101 2.43

0.99650.867 101 2.430.867 101 2.43 246.3 k5.2 k

100 4 246.3So 1.47

5.2 101 2.6

So 55.2 0.9965 1.47 80.9

i

d dd

r

vvRr

vv

vA Av

π

π

= = Ω

= =+

= + = Ω= Ω

−= = −

+

= = − ⇒ = −

c. Using Equation (11.32b) ( )

( )

( )( )

( )( )( )( )

( )( )( )

2 31

0

2

2

1 1

0 031

03 02

10

2 11

100 0.02610.4 k

0.259.62 12 267.8

0.05692 101 100

110.4

Then

0.9965 1.47 0.0569 0.08335

80.920log 590.08335

m icm

cm cm

cm cm

cm

dB dB

g R RA

Rr

r

A A

v vA A

v vA

C M RR C M RR

π

π

β−

=+

+

= = Ω

−= = − =

+

⎛ ⎞⎛ ⎞= ⋅⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

= − − ⇒ =

⎛ ⎞= ⇒ =⎜ ⎟⎝ ⎠

.7 dB

11.88 a.

011 1

1

042 2

4

10 10 2 80 k0.1

10 10 6 20 k0.2

C CC

C CC

vR RI

vR RI

− −= = ⇒ = Ω

− −= = ⇒ = Ω

b.

( )

( )( )

( ) ( )

( )( )( )( )

01 021 1 1 3

1

3

1 1

042 4 2

01 02

4

2

0.1 3.846 mA/V0.026180 0.026

23.4 k0.2

3.846 80 23.4 69.6

12

0.2 7.692 mA/V0.0261 7.692 20 76.92

Then 76.9 69.6 5352

d m Cd

m

d d

d m C

m

d

d d

v vA g R rv

g

r

A A

vA g Rv v

g

A

A A

π

π

−= = −

= =

= = Ω

= − ⇒ = −

= =−

= =

= =

= − ⇒ = −

11.89 a. Neglect the effect of r0 in determining the differential-mode gain.

( ) ( )

( )( )

( )

( )( )

( )( )

( )( )

021 2 3 3 3

22

3

1 51

2

33

02

3

3

1 where 12

1

12 0.7 12 23.3 1.94 mA12

1 1.942 37.3 mA/V

0.026200 0.026

112 1.94 8 4.24 V2

4.24 0.7 1.07 mA3.3

200 0.0264.86 k

1.07

d m C i i Ed

C

E

C

m

C

C

vA g R R R r Rv

RA

r R

I IR

g

rI

v

I

r

R

π

π

π

π

β

ββ

= = = + +

−=

+ +

− − −= = = ≈

⋅= =

=

= − =

−= =

= = Ω

( )( )

( )3

1

4.86 201 3.3 668 k1 37.3 8 668 147.42

i

dA

= + = Ω

= ⎡ ⎤ =⎣ ⎦

( )( )

( )( )

( )( )( )

1 2

0 055

2 31

0

2

2

Then 147.4 1.197 176

80 41.2 k1.94

2 11

200 0.0265.36 k

1 1.942

d d d

A

C

m C icm

A A A A

VR rI

g R RA

Rr

r

π

π

β

= ⋅ = − ⇒ = −

= = = = Ω

−=

++

= = Ω⋅

( )( )( )( )

( )( )

1

2

37.3 8 6680.09539

2 201 41.21

5.361.197

0.09539 1.197 0.114

cm

cm cm

A

AA A

−= = −

+

= −= − − ⇒ =

b.

( )

( )( ) ( )( )

1 2

1 2

03

2.015sin 1.985sin0.03sin V

2.0sin2

176 0.03 0.114 2

d

d

cm

d d cm cm

v v v t tv t

v vv t

v A v A v

ω ωω

ω

= − = −=

+= =

= += − +

Or 03 5.052sinv tω= −

Ideal, 0cmA =

( )( )03

03

So176 0.03

5.28sind dv A v

v tω= = −= −

c. ( )

( ) ( )

( )

( )( ) ( )( )

2

0 0

02

2 2 5.36 10.72 kΩ

2 2 1 1

80 82.5 k1 1.942

2 2 201 41.2 201 82.5

16.6 M 16.6 MSo 4.15 M

id id

icm

A

C

icm

icm

R r R

R R r

VrI

R

R

π

β β

= = ⇒ =

≅ + +

= = = Ω⋅

= ⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦= Ω Ω

⇒ = Ω

11.90 a.

( )

( ) ( ) ( )( )

( ) ( ) ( )( )

( )

( )

241 4

12

4 4

24 4 4

24 4

2

4

1

02

202 33 3

5

24

24 55 0.2 2

24 11 4 4

11 43 20 0

43 43 4 11 203.37 V

2 1124 3.37 0.375 mA

550.37512 40 4.5 V

2

GSn GS Th

GS GS

GS GS GS

GS GS

GS

Q

GSD n GS Th

VI k V V

R

V V

V V V

V V

V

I I

v

v VI k V V

R

−= = −

− = −

− = − +

− + =

± −= =

−= = =

⎛ ⎞= − =⎜ ⎟⎝ ⎠

−= = −

( ) ( ) ( )

( ) ( ) ( )( )

23 3 3

23 3

2

3

3

4.5 0.2 6 4 4

1.2 3.8 0.3 0

3.8 3.8 4 1.2 0.33.09 V

2 1.24.5 3.09 0.235 mA

6

GS GS GS

GS GS

GS

D

V V V

V V

V

I

− = − +

− + =

± −= =

−= =

( )

( ) ( )

( ) ( )

( ) ( )( ) ( )

( ) ( )

( ) ( )

2 2

1 2 1

3 22

3 5

3 3

2

1 2

0 05

0.3752 2 0.22

0.387 mA/V1 1 0.387 40 7.742 2

1

2 2 0.2 0.235

0.434 mA/V0.434 4

0.4821 0.434 6

So 7.74 0.482 3.73

1 1 133 0.02 0.375

m n D

d m D d

m D

m

m n D

d d d

Q

g K I

A g R A

g RA

g R

g K I

A

A A A A

R rIλ

⎛ ⎞= = ⎜ ⎟⎝ ⎠

=

= = ⇒ =

−=

+

= =

=−

= = −+

= ⋅ = − ⇒ = −

= = = =

( ) ( )( ) ( )

( ) ( )

21

2 0

k

0.387 401 2 1 2 0.387 133

0.1490.149 0.482 0.0718

m Dcm

m

cm cm

g RA

g R

A A

Ω

−−= =

+ += −= − − ⇒ =

b.

( )( ) ( )( ) ( )

1 2

1 2

03

03

0.3sin

2sin2

3.73 0.3 0.0718 2 0.975sin V

d

cm

d d cm cm

v v v tv vv t

v A v A vv t

ω

ω

ω

= − =+= =

= += − + ⇒ = −

Ideal, 0cmA =

( )( )03 3.73 0.3d dv A v= = − Or ( )03 1.12sin Vv tω⇒ = −

11.91 The low-frequency, one-sided differential gain is

( )( )

( )( )

( )

( )( )

( )

( )

022

2 2

3 12

12

12

100 0.0265.2 k

0.51 100 102 87.7

5.2 0.51

0.5 19.23 mA/V0.0262 1 19.23 10 387 pF

12

1 2 5.2 0.5 10 8 387 10

v m Cd B

C

B

v v

M m C

m

M M

HB M

v rA g R

v r R

R

r R

r

A A

C C g R

g

C C

fr R C C

π

π

π

π

μ

π π

β

π

π −

⎛ ⎞= = ⎜ ⎟+⎝ ⎠

⋅=

+

= = Ω

⋅= ⇒ =

+= +

= =

= + ⇒ =⎡ ⎤⎣ ⎦

=⎡ ⎤ +⎣ ⎦

=⎡ ⎤× × + ×⎣ ⎦

So 883 kHzHf⇒ =

11.92

a. From Equation (11.117), ( ) ( )6 120 0

1 12 2 5 10 0.8 10Zf R Cπ π −

= =× ×

Or 39.8 kHzZf =

b. From Problem 11.69, 883 .Hf kHz= From Equation (11.116(b)), the low-frequency common- mode gain is

( )

( )( )( )( )

0

6

3

4

10 4

2 11

5.2 k , 19.23 mA/VSo

19.23 10

2 101 5 100.515.2 5.2 10

9.9 1087.720log 98.9 dB

9.9 10

m Ccm

B

m

cm

dB

g RARR

r r

r g

A

C M RR

π π

π

β

−

−

−=⎡ ⎤+⎛ ⎞

+ +⎢ ⎥⎜ ⎟⎝ ⎠⎣ ⎦

= Ω =

−=⎡ ⎤×⎛ ⎞⎢ ⎥+ +⎜ ⎟ ×⎝ ⎠⎢ ⎥⎣ ⎦

= − ×

⎛ ⎞= =⎜ ⎟×⎝ ⎠

11.93

a. From Equation (7.72), ( )2m

Tgf

C Cπ μπ=

+

( )

( ) ( )( )

( )( )( )

( )

36

12

3 12

1 38.46 mA/V0.026

38.46 10Then 800 102

Or 7.65 10 F 7.65 pFAnd 6.65 pF

1 1 1 38.46 10

386 pF1

2

120 0.0263.12 k

11

2 3.12 1 10 6.65 386 10

O

m

M m C

HB M

H

g

C C

C CC

C C g R

fr R C C

r

f

π μ

π μ

π

μ

π π

π

π

π

π

−

−

−

= =

×× =+

+ = × ==

= + = +⎡ ⎤⎣ ⎦=

=⎡ ⎤ +⎣ ⎦

= = Ω

=⎡ ⎤× × + ×⎣ ⎦

r 535 kHzHf =

b. From Equation (11.140), ( )( )6 120 0

1 12 2 10 10 10Zf R Cπ π −

= =×

Or 15.9 kHzZf = 11.94 The differential-mode half circuit is:

( )

( )( )

( )( )

( ) ( )

02

12 2 or

111

100 0.0265.2 k

0.51 100 10

50025.2 101 5.2 101

dm C C

vE

E

vE E

vg R Rv A

r RR

r

r

AR R

π

π

π

β

ββ

⎛ ⎞ ⎛ ⎞⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠= =

+ +⎛ ⎞++ ⎜ ⎟⎝ ⎠

= = Ω

⎛ ⎞⎜ ⎟⎝ ⎠= =

+ +

a. For 0.1 k : 32.7E vR A= Ω =

b. For 0.25 k : 16.4E vR A= Ω =