Chapter 10 Problem Solutions 10.1

a.

( )

( )

1 21 2

2 2 3

23

1 2

2

3 1 2

0 2

2

2 2

1 2 2

BE C

BE C

C BE

V VI I

R RV I R V I R

RV V V V I RR R

RI V V V VR R R

γ

γ

γ γ

γ γ

−

−

−

− −= =

++ = +

+ − − = ++

⎧ ⎫⎛ ⎞⎪ ⎪= − + −⎨ ⎬⎜ ⎟+⎪ ⎪⎝ ⎠⎩ ⎭

b.

( )1 2

3

3

and 1 12 2

2

or 2

BE

C BE

C

V V R R

I V V V VR

VIR

γ

γ γ−

−

= =

⎧ ⎫= − + −⎨ ⎬⎩ ⎭

−=

c. ( )

( ) ( )

33

1 2 1 2 1 21 2

102 mA = 2.5 kΩ

22 0.7 10

2 mA 4.3 kΩ 2.15 kΩ

CI RR

I I R R R RR R

− −= ⇒ =

− − −= = = ⇒ + = ⇒ = =

+

10.2

(a) 11 ln C

BE TS

IV V

I⎛ ⎞

= ⎜ ⎟⎝ ⎠

(i) ( )6

1 1 14

10 1010 A, 0.026 ln 0.5388 V10

10 A

REF C BE

O

I I V

I

μ

μ

−

−

⎛ ⎞×= = = =⎜ ⎟⎝ ⎠

=

(ii) ( )6

1 1 14

100 10100 A, 0.026 ln 0.5987 V10

100 A

REF C BE

O

I I V

I

μ

μ

−

−

⎛ ⎞×= = = =⎜ ⎟⎝ ⎠

=

(iii) ( )3

1 1 14

101 , 0.026 ln 0.6585 V10

1 mA

REF C BE

O

I I mA V

I

−

−

⎛ ⎞= = = =⎜ ⎟

⎝ ⎠=

(b) 21

REFO

II

β

=+

(i)

( )

1 2

6

14

10 9.615 A ln21

509.615 100.026 ln

100.5378 V

OO O BE BE T

S

II I V V V

Iμ

−

−

⎛ ⎞= ⇒ = = = ⎜ ⎟

⎝ ⎠+

⎛ ⎞×= ⎜ ⎟⎝ ⎠

=

(ii) ( )6

1 14

100 96.15 1096.15 A 0.026 ln2 101

500.5977 V

O O BEI I Vμ−

−

⎛ ⎞×= ⇒ = = ⎜ ⎟⎝ ⎠+

=

(iii) ( )3

1 14

1 0.9615 100.9615 mA 0.026 ln2 101

500.6575 V

O O BEI I V−

−

⎛ ⎞×= ⇒ = = ⎜ ⎟⎝ ⎠+

=

10.3

( ) ( )1 1

1

1 2 1 2

1 2

on 3 0.7 30.250

21.2 K

0.250 0.2419 mA2 21 1

604.03 A

BEREF

REFC C C C

B B

V V VI

R RR

II I I I

I Iβ

μ

+ −− − − − −= ⇒ =

=

= = = ⇒ = =+ +

= =

10.4

( ) ( )

( )

1

1 2 1 2

1 2

on 5 0.7 518.3

0.5082 mA0.5082 0.4958 mA

2 21 180

6.198 A

BEREF

REF

REFC C C C

B B

V V VI

RI

II I I I

I Iβ

μ

+ −− − − − −= =

=

= = = ⇒ = =+ +

= =

10.5

(a) ( ) ( )

1 11

15 0.7 15 or 58.6

0.5BE

REF

V V on VI R R k

R

+ −− − − − −= = ⇒ = Ω

(b) ( ) ( )

1 1

0 0.7 1528.6

0.5BE

REF

V V on VR R k

I

+ −− − − − −= = ⇒ = Ω

Advantage: Requires smaller resistance. (c) For part (a):

( ) ( )( )

( ) ( )( )

29.3max 0.526 58.6 0.95

29.3min 0.476 58.6 1.05

0.526 0.476 0.05 5%

O

O

O

I mA

I mA

I mA

= =

= =

Δ = − = ⇒ ±

For part (b):

( ) ( )( )

( ) ( )( )

14.3max 0.526 28.6 0.95

14.3min 0.476 28.6 1.05

0.05 5%

O

O

O

I mA

I mA

I mA

= =

= =

Δ = ⇒ ±

10.6

a. 0

1 1

2 21 2 1 or 2.04 mA100

15 0.7 7.01 kΩ2.04

REF REFI I I

R R

β⎛ ⎞ ⎛ ⎞= + = + =⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠−= ⇒ =

b.

( )

00

00

0

0 0

0 0

80 40 kΩ2

1 1 9.3 0.2325 mA40

0.2325 11.6%2

A

CE

VrI

I IV rI I

I I

= = =

Δ ⎛ ⎞= ⇒ Δ = =⎜ ⎟Δ ⎝ ⎠Δ Δ

= ⇒ =

10.7

0 1

1 01 1 2 1

1 1

00

1 11 1

11 or 11

C

CREF C B B C

REF C C

REF

I nII II I I I I

n nI I I

I nIn In n

β β

β β β

ββ

=

= + + = + +

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

⎝ ⎠

10.8

( )

1 1

20.20 1 0.210 mA2 401

5 0.7 4.3 20.5 K0.21

REFO REF

REF

II I

R RI

β

⎛ ⎞= ⇒ = + =⎜ ⎟⎝ ⎠+

−= = ⇒ =

10.9

a.

0 0

5 0.7 0.239 mA18

0.239 0.230 mA21

50

REFI

I I

−= =

= ⇒ =+

b.

( )

00

0 00

50 218 kΩ0.230

1 1 1.3 0.00597 mA 0.236 mA217

A

EC

VrI

I V Ir

= = =

⎛ ⎞Δ = ⋅ Δ = = ⇒ =⎜ ⎟⎝ ⎠

c. ( )0 01 3.3 0.01516 mA 0.245 mA

217I I⎛ ⎞Δ = = ⇒ =⎜ ⎟

⎝ ⎠

10.10

a. ( )

11

5 0.7 51 9.3 kΩREFI R

R− − −

= = ⇒ =

b. 0 02 2 mAREFI I I= ⇒ =

c. ( )2 2 25 0.7For min 0.7 2.15 kΩ

2EC C CV R R−= ⇒ = ⇒ =

10.11

1 1

0.50 mA 0.25 mA

3 31 0.25 160

0.2625 mA2.5 0.7 6.86 K0.2625

O OA OB

REF OA

REF

I I I

I I

I

R R

β

= ⇒ = =

⎛ ⎞ ⎛ ⎞= + = +⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

=−= ⇒ =

10.12

110 0.7 37.2 K

0.25R −= =

10.13

2 1 3 12 and I 3I I I= = (a) 2 31.0 , 1.5 I mA I mA= = (b) 1 30.25 , 0.75 I mA I mA= = (c) 1 20.167 , 0.333 I mA I mA= = 10.14 a.

( ) ( )

( ) ( )

( )

( )

30 1 1 3 1

13 1 2

2 2

11

2

02

20

and 1

2

21 1

211 1

1

211

EC REF C B C

CBE BEE B B

C BEREF C

BEREF

BEREF

II I I I I I

IV VI I IR R

I VI IR

VI IR

VIR

I

β

β

β β β

β β β

β

β β

= = + = ++

= + + = +

= + ++ +

⎛ ⎞− = +⎜ ⎟⎜ ⎟+ +⎝ ⎠

−+

=⎛ ⎞

+⎜ ⎟⎜ ⎟+⎝ ⎠

b. ( ) ( )( ) ( )( )

( )1

1

2 0.70.70 180 81 81 10

0.700216 0.00086410 2 0.7

0.7011 mA 12.27 kΩ

REF

REF

REF

I

I

I RR

⎛ ⎞= + +⎜ ⎟⎜ ⎟

⎝ ⎠= +

−= = ⇒ =

10.15 a.

( )( )

0

1 2

and 1

... 11

ESi CR REF CR BS CR

ES BR B B BN BR

CR

II I I I I I

I I I I I N IN I

β

β

= = + = ++

= + + + + = ++

=

( )( )

( )( )

0

1Then

1

or 1

11

CRREF CR

REFi

N II I

IIN

β β

β β

+= +

+

=⎛ ⎞+

+⎜ ⎟⎜ ⎟+⎝ ⎠

b. ( ) ( )( )( ) ( )

1 1

60.5 1 0.5012 mA50 51

5 2 0.7 517.16 kΩ

0.5012

REFI

R R

⎡ ⎤= + =⎢ ⎥

⎢ ⎥⎣ ⎦− − −

= ⇒ =

10.16

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

0

1 1

2 21 0.5 1 0.5004 mA1 50 51

5 2 0.7 517.19 kΩ

0.5004

REF REFI I I

R R

β β⎛ ⎞ ⎡ ⎤

= + = + ⇒ =⎜ ⎟ ⎢ ⎥⎜ ⎟+ ⎢ ⎥⎝ ⎠ ⎣ ⎦− − −

= ⇒ =

10.17

( )

( ) ( )( )

0

0

1 1

121

2

For 0.8 mA

20.8 1 0.8024 mA25 27

18 2 0.720.69 kΩ

0.8024

REF

REF REF

I I

I

I I

R R

β β

= ⋅⎛ ⎞

+⎜ ⎟⎜ ⎟+⎝ ⎠=

⎛ ⎞= + ⇒ =⎜ ⎟⎜ ⎟

⎝ ⎠−

= ⇒ =

10.18

The analysis is exactly the same as in the text. We have

( )

01

212

REFI I

β β

= ⋅⎛ ⎞

+⎜ ⎟⎜ ⎟+⎝ ⎠

10.19

( )

0 2

1 1

3 1 2

33

1 3

1 1

22 mA, 0.0267 mA7511 mA, 0.0133 mA75

0.0133 0.0267 0.04 mA0.04 0.000526 mA

1 761.000526 1 mA

10 2 0.7 8.6 8.6 kΩ1

B

C B

E B B

EB

REF C B REF

REF

I I

I I

I I III

I I I I

R RI

β

= = =

= = =

= + = + =

= = =+

= + ⇒ = ≈

−= = ⇒ =

10.20 (a)

( )( )

3

3

Assuming 2100 400 K0.25

100 40020 M

2

oO

A AO

O REF

O O

rR

V VrI I

R R

β≈

= = = =

= ⇒ = Ω

(b) 5

20 M 20 M0.25 A

O OO

O

V VR II

I μ

Δ Δ= ⇒ Δ = =Δ Ω Ω

Δ =

10.21

1

1

5 0.79.3

0.4624 mA

0.026 0.4624ln ln1.5

0.46240.01733ln

BEREF

REF

T REFO

E O O

OO

V V VIR

I

V IIR I I

II

+ −− − −= =

=

⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎛ ⎞

= ⎜ ⎟⎝ ⎠

By trial and error

( )( )2

2

2

41.7 A0.70.7 0.0417 1.50.6375 V

O

BE O E

BE

BE

IV I RVV

μ= −= −=

10.22 (a)

( )

( )

1

1

3

5 0.7 593

100

93 10 ln 10 0.026ln

BEREF REF

REFO E T O

O O

V V VI I AR

I mAI R V II I

μ+ −

−

− − −− −= = ⇒ =

⎛ ⎞ ⎛ ⎞×= ⇒ =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

By trial and error, 6.8OI Aμ≅

( )2 21o o m ER r g R′= + Now

( )( )

2

2

2

30 4.41 6.80.0068 0.2615 /0.026100 0.026

382.4 0.0068π

= = Ω

= =

= = Ω

o

m

r M

g mA V

r k

So 2 || 382 ||10 9.74 E ER r R kπ′ = = = Ω

Then ( )( )4.41 1 0.262 9.74 15.6 = + ⇒ = Ω⎡ ⎤⎣ ⎦o oR R M

(b) ( )( )1 2 1 20.0068 10 0.068 − = = ⇒ − =BE BE o E BE BEV V I R V V V 10.23

( )

( )( )

( ) ( )( )

00

0 02 2

020

02

2

2

0

180 11.76 M6.80.0068 0.2615 mA/V0.026

80 0.026306 k

0.006810 306 9.68 K

11.76 1 0.2615 9.68 41.54 M

C

m E

A

mT

E E

II VR

R r g RVrII

gV

r

R R rR

π

π

Δ = ⋅ Δ

′= +

= = = Ω

= = =

= = Ω

′ = = == + = Ω⎡ ⎤⎣ ⎦

Now

( )0 01 5 0.120 A

41.54I I μ⎛ ⎞Δ = ⇒ Δ =⎜ ⎟

⎝ ⎠

10.24

(a) ( )

1

1

5 0.7 50.50

18.6 K

ln

0.026 0.50ln0.050 0.0501.20 K

REF

REFO E T

O

E

E

IR

R

II R VI

R

R

− − −= =

=

⎛ ⎞= ⎜ ⎟

⎝ ⎠⎛ ⎞= ⎜ ⎟⎝ ⎠

=

(b) [ ]

( )( )

( )( ) ( )

2 2

2

2 2

2

1

75 0.026 0.05039 K 1.923 mA/V0.050 0.026

100 2 M 1.20 39 1.164 K0.05

2 1 1.164 1.923 6.477 M

O c E m

E E

m

Ao E

O

O O

R r R gR R r

r g

Vr RI

R R

π

π

′= +′ =

= = = =

′= = ⇒ Ω = =

= + ⇒ = Ω⎡ ⎤⎣ ⎦

(c) 5 0.772 A6.477

0.772100% 100 1.54%50

OO

O

O

VIR

II

μΔΔ = = =

Δ× = × =

10.25 Let 1 5 ,= ΩR k Then

( )12 0.7 124.66

5− − −

= ⇒ =REF REFI I mA

Now 0.026 4.66ln ln 1 0.10 0.10

REFO E T E E

O

II R V R R kI

⎛ ⎞ ⎛ ⎞= ⇒ = ⇒ ≅ Ω⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

10.26

( )

( )

315

3

15

1 1

0 0

ln

100.7 0.026 ln 2.03 10 A

2 10At 2 mA, 0.026 ln2.03 10

0.718 V15 0.718 7.14 kΩ

20.026 2ln ln 1.92 k0.050 0.050

REFBE T

S

SS

BE

T REFE E

IV VI

II

V

R R

V IR RI I

−−

−

−

⎛ ⎞= ⎜ ⎟

⎝ ⎠⎛ ⎞

= ⇒ = ×⎜ ⎟⎝ ⎠

⎛ ⎞×= ⎜ ⎟×⎝ ⎠=

−= ⇒ =

⎛ ⎞ ⎛ ⎞= = ⋅ ⇒ = Ω⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

10.27 a.

10 0.7 0.465 mA20

Let 0

REFI

V −

−≈ =

=

( )3

15

ln

100.7 0.026 ln 2.03 10 A

REFBE T

S

SS

IV VI

II

−−

⎛ ⎞≅ ⎜ ⎟

⎝ ⎠⎛ ⎞

= ⇒ = ×⎜ ⎟⎝ ⎠

Then

( )3

15

0.465 100.026 ln 0.680 V2.03 10

−

−

⎛ ⎞×≅ =⎜ ⎟×⎝ ⎠BEV

Then

10 0.680 0.466 mA20

−≅ ⇒ =REF REFI I

b. 0 0

0.026 0.466ln 4000.10 0.10

lnT REFE E

V IR RI I

⎛ ⎞ ⎛ ⎞= = ⇒ = Ω⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

⋅

10.28

( )

( )3

15

10 0.7 100.4825 mA

40

ln

100.7 0.026 ln 2.03 10 A

REF

REFBE T

S

SS

I

IV VI

II

−−

− − −≈ =

⎛ ⎞≅ ⎜ ⎟

⎝ ⎠⎛ ⎞

= ⇒ = ×⎜ ⎟⎝ ⎠

( )3

15

1

Now

0.4825 100.026 ln 0.681 V2.03 10

0.681 V

BE

BE

V

V

−

−

⎛ ⎞×= =⎜ ⎟×⎝ ⎠=

( )

( ) ( )

00

00

So10 0.681 10

0.483 mA40

ln

0.48312 0.026 ln

REF REF

REFE T

I I

II R VI

II

− − −≅ ⇒ =

⎛ ⎞= ⎜ ⎟

⎝ ⎠⎛ ⎞

= ⎜ ⎟⎝ ⎠

( )( )0

2 1 0 2

By trial and error.8.7 A

0.681 0.0087 12 0.5766 VBE BE E BE

I

V V I R V

μ⇒ ≅

= − = − ⇒ =

10.29

1 1 2 0 2

1 2 0 2 1

BE REF E BE E

BE BE E REF E

V I R V I RV V I R I R

+ = +− = −

For matched transistors

1

02

0 2 10

ln

ln

Then ln

REFBE T

S

BE TS

REFT E REF E

IV VI

IV VI

IV I R I RI

⎛ ⎞= ⎜ ⎟

⎝ ⎠⎛ ⎞

= ⎜ ⎟⎝ ⎠⎛ ⎞

= −⎜ ⎟⎝ ⎠

Output resistance looking into the collector of Q2 is increased. 10.30

(a) ( )1

1 1

5 0.7 50.3174 mA

27.3 20.3174 mA

BEREF

E

O REF

V V VIR R

I I

+ − − − −− −= = =+ +

= =

(b) Using the same relation as for the widlar current source.

( )2 2 2

2 2

1

80 0.3174252 K 12.21 mA/V0.3174 0.026

O o m E

Ao m

O

R r g R r

Vr gI

π⎡ ⎤= +⎣ ⎦

= = = = =

( )( )

( )( )2 2

100 0.0268.192 K || 2 || 8.192 1.608 K

0.3174252 1 12.21 1.608 5.2 M

E

O O

r R r

R R

π π= = = =

= + ⇒ = Ω⎡ ⎤⎣ ⎦

(c) ( )

2

5 0.7 50.3407 mA

27.380 235 K

0.3407

O REF

AO o O

O

I I

VR r RI

− − −= = =

= = = ⇒ =

10.31 Assume all transistors are matched. a.

1 3 0

1

03

2

ln

ln

BE BE E

REFBE T

S

BE TS

V V I R

IV VI

IV VI

= +

⎛ ⎞= ⎜ ⎟

⎝ ⎠⎛ ⎞

= ⎜ ⎟⎝ ⎠

00

2

00

2

00

2 ln ln

ln ln

ln

REFT T E

S S

REFT E

S S

REFT E

S

IIV V I RI I

IIV I RI I

IV I RI I

⎛ ⎞ ⎛ ⎞− =⎜ ⎟ ⎜ ⎟

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

=⎜ ⎟⎝ ⎠

b.

( )

3 15

3

15

0 0

0.70.7 V at 1 mA 10 exp or 2.03 10 A0.026

0.1 10 at 0.1 mA 0.026 ln 0.640 V2.03 10

0.640Since , then or 6.4 k0.1

BE S S

BE BE

REF BE E E E

V I I

V V

I I V I R R R

− −

−

−

⎛ ⎞= ⇒ = = ×⎜ ⎟⎝ ⎠

⎛ ⎞×⇒ = =⎜ ⎟×⎝ ⎠

= = ⇒ = = Ω

10.32 (a)

( )1

1

2 2

3 2

5 0.7 50.80 mA

11.6 K

0.026 0.80ln 1.44 K0.050 0.0500.026 0.80ln 4.80 K0.020 0.020

REF

E E

E E

IR

R

R R

R R

− − −= =

=

⎛ ⎞= ⇒ =⎜ ⎟⎝ ⎠⎛ ⎞= ⇒ =⎜ ⎟⎝ ⎠

(b) ( )( )( )( )

2 2

3 3

0.7 0.05 1.44 0.628 V0.7 0.02 4.80 0.604 V

BE BE

BE BE

V VV V

= − ⇒ == − ⇒ =

10.33 (a)

1 2

1

1 2

1 2 3

1 3 2

1 3

2

Now2or

2

We have

ln and ln

BE BE

BEREF

BE REF BE O E

O E BE BE REF

OREFBE T BE T

S S

V VV V VI

R R

V I R V I R

I R V V I R

IIV V V VI I

+ −

=− −=

+

+ = +

= − +

⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

(b) Let 1 2R R= and 1 3O REF BE BE BEI I V V V= ⇒ = ≡

( )

( )

2 2

2

2

2 2

Then

so22

2Then

2

BE O E REF O E

O EREF O

EO O

O

V I R I R I R R

V V I R RI I

R

RV V I IR R

V VIRε

+ −

+ −

+ −

= − = −

− − −= =

⎛ ⎞−= − +⎜ ⎟⎝ ⎠

−=

(c) Want 0.5 OI mA=

( )( )

( ) ( )2

5 5So 10

2 0.5

5 2 0.7 52 17.2

0.5

E ER R k

R k

− −= ⇒ = Ω

− − −= = Ω

1 2Then 8.6 R R k= = Ω 10.34 a.

01

02

03

20 0.7 0.7 1.55 mA12

2 3.1 mA1.55 mA

3 4.65 mA

REF

REF

REF

REF

I

I II II I

− −= =

= == == =

b. ( ) ( )( )

( )( )( )( )

1 01 1 1

2 02 2 2

3 03 3 3

10 3.1 2 10 3.8 V

10 10 1.55 3 5.35 V

10 10 4.65 1 5.35 V

CE C CE

EC C EC

EC C EC

V I R V

V I R V

V I R V

= − − − = − + ⇒ =

= − = − ⇒ =

= − = − ⇒ =

10.35 a. Ist approximation

( )

20 1.4 2.325 mA8

2.32Now 0.7 0.026 ln 0.722 V1

REF

BE BE EB

I

V V V

−≅ =

⎛ ⎞− = ⇒ = =⎜ ⎟⎝ ⎠

Then 2nd approximation ( )

01

02

03

20 2 0.7222.32 mA

82 4.64 mA

2.32 mA3 6.96 mA

REF

REF

REF

REF

I

I II II I

−≅ =

= == == =

b.

( )1 1

2 2

3 3

At the edge of saturation,0.722 V

0 0.722 102.0 k

4.6410 0.722 4.0 k

2.3210 0.722 1.33 k

6.96

CE BE

C C

C C

C C

V V

R R

R R

R R

= =− − −

= ⇒ = Ω

−= ⇒ = Ω

−= ⇒ = Ω

10.36

( )

( )

1 2

3 4

55

10 0.7 0.7 101.86 mA

101.86 mA

1.860.5 0.026ln

C C

C C

CC

I I

I I

II

− − − −= = =

= =

⎛ ⎞= ⎜ ⎟

⎝ ⎠

By Trial and error. 5 6 70.136 mAC C CI I I⇒ = = =

( ) ( )( )3 3 3

3

2 0.8 10 10 2 1.86 0.87.02 V

C CE CE

CE

I V VV

+ = ⇒ = −=

( )( )( )

6 5 5

5

5

5 0.5 105 10 0.7 0.136 0.514.2 V

EB CE C

CE

CE

V V IVV

= + + −= + − −=

( )( )( )

7 7

7

7

5 0.85 0.136 0.84.89 V

EC C

EC

EC

V IVV

= += −=

10.37

( )

( )

1 2 1 2

4 5

13 1 3

3 3

10 0.7 0.7 101.86 mA

101.86 mA

1.86ln 0.3 0.026ln

C C C C

C C

CC E T C

C C

I I I I

I I

II R V I

I I

− − − −= = ⇒ = =

= =

⎛ ⎞ ⎛ ⎞= ⇒ =⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

By trial and error 3 0.195 mACI =

( )56 2 6

6 6

1.86ln 0.5 0.026lnCC E T C

C C

II R V I

I I⎛ ⎞ ⎛ ⎞

= ⇒ =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

By trial and error 6 0.136 mACI =

10.38

( )( )( )

10 0.7 1 mA6.3 3

0.7 V as assumed1 3 3 V

REF

BE R

RER REF ER

I

V QV I R

−= =+

== ⋅ = =

11 1 1

01

22 2 2

02

33 3 3

03

33 V 3 k133 V 1.5 k233 V 0.75 k4

= ⇒ = = ⇒ = Ω

= ⇒ = = ⇒ = Ω

= ⇒ = = ⇒ = Ω

RERE E E

RERE E E

RERE E E

VV R RI

VV R RI

VV R RI

01

02

03

1 mA2 mA4 mA

===

III

10.38

( )

( )

( ) ( )

( )

( ) ( )

2 2 2 2 2

22 2

2

2

2 2

2 11

2

2 1

3

2 1.5 3.5 12

250 20 3.5 1.5 3.125

12

100 20 3.5 1.5 1.25

Now 10

DS GS TN GS GS

O n ox GS TN

REF n ox GS TN

GS

V sat V V V V V VWI C V VL

W WL L

WI C V VL

W WL L

V

μ

μ

= = − = − ⇒ =

⎛ ⎞⎛ ⎞= −⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

⎛ ⎞ ⎛ ⎞= − ⇒ =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ ⎞⎛ ⎞= −⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

⎛ ⎞ ⎛ ⎞= − ⇒ =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

=

( ) ( )2

2

3 3

10 3.5 6.5

So 100 20 6.5 1.5 0.2

GSV VW WL L

− = − =

⎛ ⎞ ⎛ ⎞= − ⇒ =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

10.39

( )( )

( )

2

2

2

2.5 0.08 6 0.515 2

2.5 3.6 0.25

3.6 2.6 1.6 0

GSREF GS

GS GS GS

GS GS

VI V

V V V

V V

− ⎛ ⎞= = −⎜ ⎟⎝ ⎠

− = − +

− − =

( )( )

( )

2.6 6.76 23.042 3.6

1.12 V 1.11932.5 1.1193 92.0 A 92.05

1592.0 A

GS

GS

REF REF

o

V

V

I I

I

μ

μ

± +=

=−= ⇒ =

=

( )( )

2

2

sat 1.1193 0.5sat 0.619 V

DS GS TN

DS

V V VV

= − = −=

10.39 a. From Equation (10.50),

( ) ( )

( ) ( )

( ) ( )( )( )

1 2

1 22 2

1 1

5 5125 255 0.5

5 51 125 25

0.447 1 0.4475 0.51 0.447 1 0.447

1.74 V

18 25 1.74 0.5 0.692

GS GS

GS GS

REF n GS TN REF

V V

V V

I K V V I mA

⎛ ⎞ ⎛ ⎞−⎜ ⎟ ⎜ ⎟

⎜ ⎟ ⎜ ⎟= = +⎜ ⎟ ⎜ ⎟

+ +⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

−⎛ ⎞ ⎛ ⎞= +⎜ ⎟ ⎜ ⎟+ +⎝ ⎠ ⎝ ⎠= =

≅ − = − ⇒ =

b. ( ) ( )

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

22 2

22

0

0

1 12

18 15 1.74 0.5 1 0.02 2

415 104 0.432 mA

O n ox GS TN DSWI C V V VL

I

I

μ λ⎛ ⎞⎛ ⎞= − +⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

= − +⎡ ⎤⎣ ⎦= ⇒ =

c. ( ) ( )( )0 0415 1 0.02 4 0.448 mAI I= + ⇒ =⎡ ⎤⎣ ⎦ 10.40 (a)

( )2

1

8050 0.52

2.00.050

REF GS

GSREF

WI VLV

IR

⎛ ⎞⎛ ⎞= = −⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

−= =

Design such that ( )2 sat 0.25 0.5DS GSV V= = − 0.75 VGSV =

So 2 0.750.050 25 KRR

−= ⇒ =

( )2

1 1

1

2

2 2

8050 0.75 0.5 202

20 50 40100

REF

O

W WL L

WIL W

W WI LL L

⎛ ⎞⎛ ⎞ ⎛ ⎞= − ⇒ =⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠

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

(b) ( )( )

1 1 667 K0.015 0.1O O

O

R RIλ

= = ⇒ =

(c) 1 1.5 A666

1.5100% 100% 1.5%100

OO

O

O

VIR

II

μΔΔ = = ⇒

Δ ⎛ ⎞× = × ⇒⎜ ⎟⎝ ⎠

10.41

(a) ( )( )

( )( )

2

2

80250 3 12

2.44 V250 A at 2.44 V

1 1 200 K0.02 0.25

REF GS

GS

O DS GS

OO

I V

VI V V

RI

μ

λ

⎛ ⎞= = −⎜ ⎟⎝ ⎠

== = =

= = =

(i) 3 2.44 2.8 A200

252.8 A

OO

O

VIR

I

μ

μ

Δ −Δ = = ⇒

=

(ii) 4.5 2.44 10.3 A200

260.3 A

OO

O

VIR

I

μ

μ

Δ −Δ = = ⇒

=

(iii) 6 2.44 17.8 A200

267.8 A

OO

O

VIR

I

μ

μ

Δ −Δ = = ⇒

=

(b) ( )

( )( )

4.5 250 375 A at 2.44 V31 1 133.3 K

0.02 0.375

O DS

OO

I V

RI

μ

λ

= = =

= = =

(i) 3 2.44 4.20 A133.3

379.2 A

OO

O

VIR

I

μ

μ

Δ −Δ = = ⇒

=

(ii) 4.5 2.44 15.5 A133.3

390.5 A

OO

O

VIR

I

μ

μ

Δ −Δ = = ⇒

=

(iii) 6 2.44 26.7 A133.3

401.7 A

OO

O

VIR

I

μ

μ

Δ −Δ = = ⇒

=

10.41

( )

( )

( )

2 2

22

2

2

2 2

0.25 0.4 0.65

24025 0.65 0.4 202

SD SG TP SG SG

pO SG TP

V sat V V V V Vk WI V V

LW WL L

= = + = − ⇒ =′ ⎛ ⎞= +⎜ ⎟⎝ ⎠

⎛ ⎞ ⎛ ⎞= − ⇒ =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

( )( )

( )

( )

1

12

23

3

3

2

3 3

75 60

23 0.65 2.35

40Then 75 2.35 0.4 0.9862

REF O

pREF SG TP

SG

W L WI A IW L L

k WI V VL

V VW WL L

μ/ ⎛ ⎞= = ⋅ ⇒ =⎜ ⎟/ ⎝ ⎠

′ ⎛ ⎞= +⎜ ⎟⎝ ⎠

= − =

⎛ ⎞ ⎛ ⎞= − ⇒ =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

10.42 (a)

11

2

2 21 1

0.51 2 0.5

REFGS TN

n

REF REFO n n

n n

IV V VK

I II K KK K

= + = + =

⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

( ) ( )( ) ( )

( ) ( )( ) ( )

0 0

0 0

0.5max 0.5 1.05 max 0.525 mA0.50.5min 0.5 0.95 min 0.475 mA0.5

I I

I I

⎛ ⎞= ⇒ =⎜ ⎟⎝ ⎠⎛ ⎞= ⇒ =⎜ ⎟⎝ ⎠

0So 0.475 0.525 mAI≤ ≤ (b)

( ) ( )

( ) ( )

2

2 1 21

2

0 0

2

0 0

0.5min 0.5 1 1.05 (min) 0.451 mA0.5

0.5max 0.5 1 0.95 (max) 0.551 mA0.5

REFO n TN TN

n

II K V VK

I I

I I

⎡ ⎤= + −⎢ ⎥

⎢ ⎥⎣ ⎦

⎡ ⎤= + − ⇒ =⎢ ⎥

⎣ ⎦

⎡ ⎤= + − ⇒ =⎢ ⎥

⎣ ⎦

0So 0.451 0.551 mAI≤ ≤ 10.43

(1) 2x A

x m gso

V VI g V

r−

= +

(2) 1

1 2,

Ax m gs

o

gs x gs A

VI g Vr

V V V V

= +

= = −

So

(1) 1xx A m

o o

VI V g

r r⎛ ⎞

= − +⎜ ⎟⎝ ⎠

(2) [ ]Ax m x A o x m x

o

VI g V V r I g Vr

= + ⇒ = −

( )

[ ]

2

2

2

Then

1

12

xx o x m x m

o o

x x mx o m x x m x

o o o

xx x m o x m x m o x

o

x m o x m m oo

VI r I g V gr r

V I gI r g I V g V

r r rVI I g r I g V g r Vr

I g r V g g rr

⎛ ⎞= − − +⎜ ⎟

⎝ ⎠⎡ ⎤

= − + − ⋅ −⎢ ⎥⎣ ⎦

= − − + +

⎡ ⎤+ = + +⎢ ⎥

⎣ ⎦

Since 1m

o

gr

>>

[ ] ( )( )2 1x m o x m m oI g r V g g r+ ≅ +

( )2

Then 1

x m oo

x m m o

V g rR

I g g r+

= =+

Usually, 2,m og r >> so that 1o

m

Rg

≅

10.44

( )

( )

( )

2 2 2

2

2 2

1 1

1

2

3

2

3 3

(sat) 2 0.8 2.8 V60200 2.8 0.8 1.672

0.4 3.330.2 1.67

6 2.8 3.2 V60400 3.2 0.8 2.312

DS GS GS

O

REF

O

GS

REF

V V VW WIL L

W WI L L W

WI LL

VW WIL L

= = − ⇒ =

⎛ ⎞ ⎛ ⎞= = − ⇒ =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

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

= − =

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

10.45 (a)

( )( ) ( )( )

( )

( )( )

( )( )

2 21 3

1 3

1 1

1 1 2

22

2

60 6020 0.7 3 0.72 2

5

20 0.7 5 0.73

3.582 6.107 1.705 V60 12 1.705 0.7 363.6 A at 1.705 V260 20 1.705 0.7 606 A2

REF GS GS

GS GS

GS GS

GS GS GS

O DS

REF

I V V

V V

V V

V V V

I V

I

μ

μ

⎛ ⎞ ⎛ ⎞= − = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

+ =

− = − −

= ⇒ = =

⎛ ⎞= − = =⎜ ⎟⎝ ⎠⎛ ⎞= − =⎜ ⎟⎝ ⎠

(b) ( )( )

1 1 183.4 K0.015 0.3636

1.5 1.705 1.12 A183.4

362.5 A

OO

OO

O

RIVI

RI

λ

μ

μ

= = =

Δ −Δ = = ⇒ −

=

(c) 3 1.705 7.06 A183.4

370.7 A

OO

O

VIR

I

μ

μ

Δ −Δ = = ⇒

=

10.46

( )( ) ( )( )

( )

( )( )

2 21 3

1 3 3 1

1 1

1 1

2

2 2

50 5015 0.5 3 0.52 2

10 10

15 0.5 10 0.53

3.236 10.618 3.28 V50 15 3.28 0.5 2.90 mA2

2.90 mA(sat) 3.28 0.5

REF SG SG

SG SG SG SG

SG SG

SG SG

REF REF

O REF

SD SG TP

I V V

V V V V

V V

V V

I I

I IV V V

⎛ ⎞ ⎛ ⎞= − = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

+ = ⇒ = −

− = − −

= ⇒ =

⎛ ⎞= − ⇒ =⎜ ⎟⎝ ⎠

= == + = − 2 (sat) 2.78 VSDV⇒ =

10.47

( )

( )( )

( )

( )

2 2 2

2

2 2

1 1

12

1 3 3

2

3 3

(sat) 1.2 0.35 1.55 V50100 1.55 0.35 2.782

200 5.56100 2.78

4 2.45 V50200 2.45 0.35 12

SD SG SG

O

REF

O

SG SG SG

REF

V V VW WIL L

W WI WL L

WI LL

V V VW WIL L

= = − ⇒ =

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

⎛ ⎞= ⇒ = ⇒ =⎜ ⎟⎝ ⎠

+ = ⇒ =

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

.81

10.48

( )( ) ( )( )

( )

( )( )

( )

2 21 3

11 3 3

11

1 1

2

2 2 2

80 8025 1.2 4 1.22 2

102 102

1025Then 1.2 1.24 2

3 6.8 2.27 V80 25 2.267 1.2 1.14 mA2

(sat) 2.27 1.2 sat

REF SG SG

SGSG SG SG

SGSG

SG SG

REF REF O

SD SG TP SD

I V V

VV V V

VV

V V

I I I

V V V V

⎛ ⎞ ⎛ ⎞= − = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

−+ = ⇒ =

−− = −

= ⇒ =

⎛ ⎞= − ⇒ = =⎜ ⎟⎝ ⎠

= + = − ⇒ = 1.07 V

10.49

( )

( )( )( )

2 2 2

2

2 2

1

2

1

1

(sat) 1.8 1.4 3.2 V80 3.2 1.4 100 0.7722

200 1.54100 0.772

SD SG SG

O

REF

O

V V VW WIL L

WI L

WIL

WWLL

= = − ⇒ =

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

=

⎛ ⎞= ⇒ =⎜ ⎟⎝ ⎠

Assume M3 and M4 are matched.

( )

3 1 3

2

3,4

3,4

10 3.22 10 3.4 V2

80200 3.4 1.42

1.25

SG SG SG

REF

V V V

WIL

WL

−+ = ⇒ = =

⎛ ⎞⎛ ⎞= = −⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

⎛ ⎞ =⎜ ⎟⎝ ⎠

10.50 (a)

( )

( )

21

1

23

3

2

2

pREF SG TP

pSG TP

k WI V VL

k W V VL

′⎛ ⎞⎛ ⎞= +⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

′⎛ ⎞⎛ ⎞= +⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

( ) ( )3 1

2 21 1

1

3

But 3

So 25 0.4 5 3 0.4which yields 1.08

and 1.92

SG SG

SG SG

SG

SG

V V

V VV V

V V

= −

− = − −=

=

( )( )( )( )

( )( )

2

2

1

20 25 1.08 0.4 231

/ 15 0.6/ 25

Then 0.6 231 139

REF REF

O

REF

O

I I A

W LII W L

I A

μ

μ

= − ⇒ =

= = =

= =

(b) ( )2 1.08 0.4 0.68 3 0.68 2.32

then2.32 16.7

0.139

DS

R O

V sat VV I R

R R k

= − == − = =

= ⇒ = Ω

10.51

( ) ( )

( )( )

( ) ( )

( ) ( )

2 2 2

2

2 2

1 1

1

2

3

2

3 3

(sat) 0.35 0.4 0.75 V

80 20 0.75 0.4 32.7

50 20.480 32.7

3 0.75 2.25

50 20 2.25 0.4 0.730

SD SG SG

O

REF

O

SG

REF

V V VW WIL L

W WI L L W

LWIL

VW WIL L

= = − ⇒ =

⎛ ⎞ ⎛ ⎞= = − ⇒ =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

= ⇒ = ⇒ =

= − =

⎛ ⎞ ⎛ ⎞= = − ⇒ =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

10.52 a. ( )

( )

2

2

4 0

100 100 2 3 VFor 3 V, 100 A

REF n GS TN

GS GS

D

I K V V

V VV I μ

= −

= − ⇒ == − =

b. ( )

( )( )( ) ( )( )

( )( )

0 04 02 04

02 040

0

0

0 4 00

11 1 500 k

0.02 0.12 2 0.1 3 2 0.2 /500 500 1 0.2 500

51 M1 6 0.118 A

51

m

m n GS TN

D

R r r g r

r rI

g K V V mA VR

R

I V IR

λ

μ

= + +

= = = = Ω

= − = − == + +⎡ ⎤⎣ ⎦= Ω

Δ = ⋅ Δ = ⇒ Δ =

10.53

( )( )

( )

4 02

6 4 04 02

02 04 02

6 02 02 04 61

= −

= − +

= + += + + = −⎡ ⎤⎣ ⎦

gs X

S X m gs X

X m X X

S X m gs

V I r

V I g V r I r

I g I r r I rV I r g r r V

( )

66 6

06 06 06

02 02 0406 06

1

1 1

⎛ ⎞−= + = − +⎜ ⎟

⎝ ⎠⎛ ⎞

= − + + +⎡ ⎤⎜ ⎟ ⎣ ⎦⎝ ⎠

X S XX m gs S m

XX X m m

V V VI g V V gr r r

VI I g r g r rr r

( )

( ) ( )

( )

( ) ( )( )

( )( )

02 02 0406 06

0 06 06 02 02 04

20

02 04 060

11 1

1 1

0.2 mA 0.2 12 V2 2 0.2 2 1 0.4 /

1 1 250 k0.02 0.2λ

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

= = + + + +⎡ ⎤⎣ ⎦

≈ = = −== − = − =

= = = = = Ω

XX m m

Xm m

X

REF GS

GS

m n GS TN

VI g r g r rr r

V R r g r r g r rI

I I VVg K V V mA V

r r rI

( )( ) ( )( ) ( ) 0

90 0

250 1 0.4 250 250 1 0.4 250 250

2575750 k 2.58 10

= + + × + +⎡ ⎤ ⎡ ⎤⎣ ⎦ ⎣ ⎦= Ω ⇒ = × Ω

R

R R

10.54

( ) ( )

( )

2 21 3

1 3

24

4

2 2

2

n nGS TN GS TN

pGS TP

k kW WV V V VL L

k W V VL

′ ′⎛ ⎞ ⎛ ⎞− = −⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

′ ⎛ ⎞= +⎜ ⎟⎝ ⎠

(1) ( )( ) ( )( )2 21 350 20 0.5 50 5 0.5GS GSV V− = −

(2) ( )( ) ( )( )2 21 450 20 0.5 20 10 0.5GS GSV V− = −

(3) 4 3 1 6SG GS GSV V V+ + =

From (1) ( ) ( ) ( )2 21 3 3 14 0.5 0.5 2 0.5 0.5GS GS GS GSV V V V− = − ⇒ = − +

From (2) ( ) ( ) ( )2 21 4 4 15 0.5 0.5 5 0.5 0.5− = − ⇒ = − +GS GS SG GSV V V V

Then (3) becomes ( ) ( )1 1 1

1 3 4

5 0.5 0.5 2 0.5 0.5 6which yields 1.36 and 2.22 , 2.42

GS GS GS

GS GS SG

V V VV V V V V V

− + + − + + == = =

( ) ( )( )

( ) ( )

2 21

1

1 2

2 2 2

Then 50 20 1.36 0.5 or 0.740 2

1.36 1.36 0.5 0.86

nREF GS TN REF O

GS GS

DS GS TN DS

k WI V V I I mAL

V V VV sat V V V sat V

′ ⎛ ⎞= − = − = =⎜ ⎟⎝ ⎠

= == − = − ⇒ =

10.55

( )

( )

( )

2 2 2 2

22

2

2

2 2

0.5 0.5 1

50 2

50 1 0.5 4

DS GS TN GS GS

nO GS TN

V sat V V V V V V

k WI A V VL

W WL L

μ

= = − = − ⇒ =′ ⎛ ⎞= = −⎜ ⎟⎝ ⎠

⎛ ⎞ ⎛ ⎞= − ⇒ =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

( ) ( )2 21 2 1

1 1 1

1 150 50 1 0.5 122′ ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= = ⇒ = = − = − ⇒ =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

nGS GS REF GS TN

k W W WV V V I V VL L L

3 4 1

3 3

62 6 1 5 2.5

GS SG GS

GS GS

V V VV V V V

+ + == − = ⇒ =

( )

( )

( )

2

3 3

24

4

2

4 4

150 50 2.5 0.5 0.75

2

150 20 2.5 0.5 1.88

REF

pREF SG TP

W WIL L

k WI V VL

W WL L

⎛ ⎞ ⎛ ⎞= = − ⇒ =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

′ ⎛ ⎞= +⎜ ⎟⎝ ⎠

⎛ ⎞ ⎛ ⎞= − ⇒⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

10.56 a. As a first approximation

( )( )

21 1

1

80 80 1 2 VThen 2 2 4 V

REF GS GS

DS

I V VV

= = − ⇒ =≅ =

The second approximation ( ) ( )( )

( )

21

21 1

80 80 1 1 0.02 4

80Or 1 1.96286.4

GS

GS GS

V

V V

= − +⎡ ⎤⎣ ⎦

= − ⇒ =

( ) ( )( ) ( )( )

21 1

2

0

Then

1

80 1.962 1 1 0.02 1.962

Or 76.94 A

O n GS TN n GSI K V V V

I

λ

μ

= − +

= − +⎡ ⎤⎣ ⎦=

b. From a PSpice analysis, 0 77.09 Aμ=I for 3 1 VDV = − and 0 77.14 Aμ=I for 3 3 V.DV = The change is 0 0.05 A or 0.065%.I μΔ ≈ 10.57 a. For a first approximation,

( )24 480 80 1 2 VREF GS GSI V V= = − ⇒ =

As a second approximation ( ) ( )( )

( ) ( )

24

4 12

2 2

80 80 1 1 0.02 2

Or 1.98 V

1

REF GS

GS GS

O n GS TN GS

I V

V V

I K V V Vλ

= = − +⎡ ⎤⎣ ⎦= =

= − +

To a very good approximation 0 80 Aμ=I b. From a PSpice analysis, 0 80.00 Aμ=I for 3 1 VDV = − and the output resistance is

0 76.9 M .= ΩR Then

0 30

0

For 3 V1 4 0.052 A

76.980.05 A

D

D

V

I VR

I

μ

μ

= +

Δ = ⋅ = =

=

10.58 (a) ( ) ( )

( )

( )

3 3 3 3

23

2

3

or 0.2 0.8 1.0

2

50 48 0.2 26

DS GS TN GS DS TN

nD GS TN

V sat V V V V sat Vk WI V V

LW WL L

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

(b) ( )( )

5 3

5 5

20.8 2 0.2 1.2

GS TN GS TN

GS GS

V V V VV V V

− = −= + ⇒ =

(c) ( ) ( ) ( ) ( )1 1min 2 2 0.2 min 0.4 D DS DV V sat V V= = ⇒ =

10.59 (a)

( )

( )( )

( )( )( )( )

21

1

1

1 1 2

50 5 250 /2

/1 1/

1 51 8.944 0.6838500.25 0.05

6.12

nn

n D

k WK A VL

W LR

W LK I

R k

μ′ ⎛ ⎞= = =⎜ ⎟⎝ ⎠

⎛ ⎞⎜ ⎟= −⎜ ⎟⎝ ⎠

⎛ ⎞= − =⎜ ⎟⎜ ⎟

⎝ ⎠= Ω

(b)

( )( )

( )( )( )

( )( )( )

3 1

3 32

1 3 3

32

1 1 1

min

50 20 5 0.5 1.207 Then 1.21 0.5 0.707

Also 50 50 5 0.5 0.9472

Then 0.71 0.947 1.66

SD GS

SD SG TP

D SG SG

SD

D GS GS

V V V sat VV sat V V

I V V VV sat V

I V V V

V V V

+ −

+ −

− = += +

= = − ⇒ == − =

= = − ⇒ =

− = + =

(c)

( )

( )

21

5 5

22

6 6

25 50 0.947 0.5 2.5

75 20 1.207 0.5 7.5

O

O

W WIL L

W WIL L

⎛ ⎞ ⎛ ⎞= = − ⇒ =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞= = − ⇒ =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

10.60

( )

( )

( ) ( )

( )

3

23

3

2

3 3 4 5

21 3

1

3

01

1

01

1

1 5 1.667 V3

12

100 20 1.667 1 11.25

12

Or

GS

REF n ox GS TN

O n ox GS TN

REF

REF

V

WI C V VL

W W W WL L L L

WI C V VL

WI L

WIL

IW WL I L

μ

μ

= =

⎛ ⎞⎛ ⎞= −⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

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

⎛ ⎞⎛ ⎞= −⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

⎛ ⎞⎜ ⎟⎝ ⎠=⎛ ⎞⎜ ⎟⎝ ⎠

⎛ ⎞⎛ ⎞ = ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠

( )

( )

3 1

02

2 3 2

0.2 11.25 22.50.1

0.3And 11.25 33.750.1REF

WL

IW W WL I L L

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

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

10.61

( )( )( )( )

( ) ( )

( )

2

2

24

Also 40 1 1.2

18 1 1.2

Then 40 1.2 18 1.26.325which yields 1.2 1.24.243

SGP GSNREF

REF GSN

REF SGP

GSN SGP

SGP GSN

V VIR

I V

I V

V V

V V

− −=

= −

= −

− = −

= − +

( ) ( )

( )( )( )( )

( )

2

1

2

3

4

Then 0.040 1.2 24 1.49 1.2 1.2

which yields 2.69 and 3.43

24 3.42 2.69Now 89.4 200

89.4 17.9 5

1.25 89.4 112 0.8 89.4 71.5

4 89.4 358

GSN GSN GSN

GSN

SGP

REF REF

V R V V

V VV V

I I A

I A

I AI AI A

μ

μ

μμ

μ

⎡ ⎤− ⋅ = − − − −⎣ ⎦=

=− −= ⇒ =

= =

= == == =

10.61 a. ( )

( ) ( )( ) ( )

( )( )

( )( )

0

0 0

0 0

0 0

2

2 0.25 0.2 0.447 mA/V

1 1 250 k0.02 0.2

1 1 167 k0.03 0.2

m n REF

m m

n nn REF

p pp REF

g M K I

g M g M

r rI

r rI

λ

λ

=

= ⇒ =

= = ⇒ = Ω

= = ⇒ = Ω

b. ( ) ( )( )0 0|| 0.447 250 ||167 44.8v m n p vA g r r A= − = − ⇒ = −

c. 0 0205 ||167 || or 100 kL n p LR r r R= = = Ω 10.62

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

( )

1

2

3

4

We have 2.69 and 3.43 10 2.69 3.43 3.88So 19.4

200Then 0.2 19.4 3.88

1.25 19.4 24.3 0.8 19.4 15.5

4 19.4 77.6

GSN SGP

REF REF

V V V V

I I AR

I AI AI AI A

μ

μμ

μμ

= =− −= = ⇒ =

= == == == =

10.63

( )( ) ( )

( )( ) ( )

( )( )

( )

22 2

1

42

3

24

4

4 4

9 200 120 A15

20 120 267 A9

40266.7 20 0.62

1.416 V(sat) 1.416 0.6 sat 0.816 V

D REF D

O D O

O SG

SG

SD SD

WLI I I

WL

WLI I I

WL

I V

VV V

μ

μ

= ⋅ = ⇒ =

⎛ ⎞= ⋅ = ⇒ =⎜ ⎟⎝ ⎠

⎛ ⎞= = −⎜ ⎟⎝ ⎠

== − ⇒ =

10.64

( )21

1

1

4050 0.62

1.7550

REF SG

SGREF

WI VL

VI

R

⎛ ⎞⎛ ⎞= = −⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠−

= =

( )

2 2 2

2

1 1

(sat) 0.35 0.6 0.95 V1.75 0.95 16 K

0.054050 0.95 0.6 20.42

SD SG SGV V V

R R

W WL L

= = − ⇒ =−= ⇒ =

⎛ ⎞⎛ ⎞ ⎛ ⎞= − ⇒ =⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠ ⎝ ⎠

( )( )

( )( )

1 2

2

3 3

3

120 4950 20.4

25 10.250 20.4

O

REF

D

REF

WI WLI L

WI WLI L

⎛ ⎞= = ⇒ =⎜ ⎟⎝ ⎠

⎛ ⎞= = ⇒ =⎜ ⎟⎝ ⎠

( )

( )

5 5 5

22

5 5

34 4

42 2

(sat) 0.35 0.4 0.75 V100 0.75 0.4 150 24.5

2

25 4.08150 24.5

DS GS GS

O

DD

O O

V V VW WIL L

WII WL

I I L

= = − ⇒ =

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

⎛ ⎞= = = ⇒ =⎜ ⎟⎝ ⎠

10.65 For ( )0, 1GS D DSS DSv i I vλ= = + a.

( ) ( )( )5 V, 52 1 0.05 5 2.5 mA

D DS

D D

V vi i

= − == + ⇒ =⎡ ⎤⎣ ⎦

b. ( ) ( )( )0, 102 1 0.05 10 3 mA

D DS

D D

V vi i

= == + ⇒ =⎡ ⎤⎣ ⎦

c. ( ) ( )( )5 V, 15 V2 1 0.05 15 3.5 mA

D DS

D D

V vi i

= == + ⇒ =⎡ ⎤⎣ ⎦

10.66

( )( )

( )

2

0

2

0

0

1

2 4 1

21 0.2934

So 0.293 4 1.17 V

Then and

1.170.586 k

2

⎛ ⎞= −⎜ ⎟

⎝ ⎠

⎛ ⎞= −⎜ ⎟

⎝ ⎠

= − =

= − = −

= = −

−−= = − ⇒ = Ω

GSDSS

P

GS

P

GS

P

GS

SS GS

GS

VI IV

VV

VV

VVI V VR

VR R

I

Finish solution: See solution 10.66 Completion of solution

( )( )

( )

Need sat1.17 4

2.83 VSo sat 2.83 1.17 4 V

DS DS GS P

DS

D DS S D

v v v V

vV v V V

≥ = −= − − −

≥≥ + = + ⇒ ≥

10.67

a.

( )

11

3

1 1131

exp

1 10or ln 0.026 ln 0.55685 10

EBREF S

T

REFEB T EB

S

VI IV

IV V VI

−

−

⎛ ⎞= ⎜ ⎟

⎝ ⎠⎛ ⎞ ⎛ ⎞×= = ⇒ =⎜ ⎟ ⎜ ⎟×⎝ ⎠⎝ ⎠

b. 1 15 0.5568 4.44 k

1R R−= ⇒ = Ω

c. From Equations (10.79) and (10.80) and letting 0 2 2.5 V= =CE ECV V

( )

( )

12 3

12 3

9

2.512.5 8010 exp 1 100.5568120 1

801.031251.0208333 10 exp 101.00696

Then 0.026ln 1.003222 10

So 0.5389

I

T

I

T

I

I

VV

VV

V

V V

− −

− −

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

⎛ ⎞ ⎛ ⎞× =⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

= ×

=

d. ( )

( ) ( )1/

1/ 1/1

38.460.0261 1 0.00833 0.0125

120 801846

Tv

AN AP

v

v

VA

V V

A

A

−=

+

− −= =++

= −

10.68

a. ( )3

121

0.5 10ln 0.026 ln 0.520810

REFBE T BE

S

IV V V

I

−

−

⎛ ⎞ ⎛ ⎞×= = ⇒ =⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

b. 1 15 0.5208 8.96 k

0.5R R−= ⇒ = Ω

c. Modify Eqs. 10.79 and 10.80 to apply to pnp and npn, and set the two equation equal to each other.

22 2

13 12

13 12

exp 1 exp 1

2.5 2.55 10 exp 1 10 exp 180 120

5.15625 10 exp 1.020833 10 exp

EBO ECO CEBECO SO C S

T AP T AN

EBO BE

T T

EBO BE

T T

V V VVI I I IV V V V

V VV V

V VV V

− −

− −

⎛ ⎞⎛ ⎞⎛ ⎞ ⎛ ⎞= + = = +⎜ ⎟⎜ ⎟⎜ ⎟ ⎜ ⎟

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

⎝ ⎠ ⎝ ⎠⎝ ⎠ ⎝ ⎠⎛ ⎞ ⎛ ⎞

× = ×⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

( ) ( ) ( )

exp1.9798 exp

exp

ln 1.9798 0.5208 0.026 ln 1.97980.5386 5 0.5386 4.461 V

EBO

T EBO BE

TBE

T

EBO BE T

EBO I I

VV V V

VVV

V V VV V V

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

d. ( )

( ) ( )1/

1/ 1/1

38.460.0261 1 0.00833 0.0125

120 801846

Tv

AN AP

v

v

VA

V V

A

A

−=

+

− −= =++

= −

10.69 a. M1 and M2 matched. For 0 ,=REFI I we have 2 3 0 2.5 V= = = =SD SG SG DSV V V V For M1 and M3:

( ) ( )

( ) ( )( )

2

1

2

1 1 3 2

1 12

100 10 2.5 1 1 0.02 2.5 4.23

REF p ox SG TP P SDWI C V V VL

W W W WL L L L

μ λ⎛ ⎞⎛ ⎞= + +⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

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

For M0:

( ) ( )

( ) ( )( )

2

0

2

0 0

1 12

100 20 2 1 1 0.02 2.5 4.76

O n ox GS TN n DSWI C V V VL

W WL L

μ λ⎛ ⎞⎛ ⎞= − +⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

⎛ ⎞ ⎛ ⎞= − + ⇒ =⎡ ⎤⎜ ⎟ ⎜ ⎟⎣ ⎦⎝ ⎠ ⎝ ⎠

b.

( )( )

( )( )( )

( ) ( )( )

0 00

0 0

1 1 500 k0.02 0.1

12 22

2 0.02 4.76 0.1

0.195 mA/V

|| 0.195 500 || 500 48.8

n p

m n O n ox Oo

m

v m n p v

r rI

Wg K I C IL

g

A g r r A

λ

μ

= = = = Ω

⎛ ⎞⎛ ⎞= = ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

=

=

= − = − ⇒ = −

10.70 a. ( )

( )

21

2100 100 1 2 VREF p SG TP

SG SG

I K V V

V V

= +

= − ⇒ =

b. From Eq. 10.89 ( ) ( )

( )( )( ) ( )

( )( )

( )

2

2

2

2

1

1 0.02 10 2 100 15

0.02 0.02 100 0.02 0.02

15 29

0.041 0.961.98 V

p SG n I TNO

n p REF n p

I

I

I

I

V V K V VV

I

V

V

VV

λ

λ λ λ λ

+⎡ ⎤+ − −⎣ ⎦= −+ +

+ −⎡ ⎤ −⎣ ⎦= −+ +

−= −

− ==

c. ( )

( )( )( )( )

( )( )

0 0

0 0

||

1 1 500 k0.02 0.1

2 2 0.1 0.1 0.2 /

0.2 500 || 500 50

v m n p

n pREF

m n REF

v v

A g r r

r rI

g K I mA V

A A

λ

= −

= = = = Ω

= = =

= − ⇒ = −

10.71

1

5 0.6 5 0.6 0.22 mA20REFI

R− −= = =

From Eq. 10.96

0.220.026

0.22 1 0.221140 90

8.4615 8.46151 10.0015714 0.002444 0.004016

C

Tv

C C

LAN L AP

L L

IV

AI I

RV R V

R R

⎛ ⎞−⎜ ⎟ −⎝ ⎠= =

⎛ ⎞ + ++ +⎜ ⎟⎝ ⎠

− −= =+ + +

(a) , 2107L vR A= ∞ = − (b) 250 K, 1056L vR A= = − (c) 100 K, 604L vR A= = − 10.72

5 0.6 0.1257 mA35

Then2 0.2514 mA

REF

CO REF

I

I I

−= =

= =

From Eq. 10.96 0.2514

9.66920.0260.2514 0.2514 1 10.002095 0.0031425

120 809.6692

10.0052375

v

L L

v

L

A

R R

A

R

−−= =

+ + + +

−=+

(a) 1846= ∞ = −L vR A (b) 250 K, 1047L vR A= = − 10.73 (a) To a good approximation, output resistance is the same as the widlar current source.

( )0 02 2 21 ||m ER r g r Rπ= +⎡ ⎤⎣ ⎦

(b) ( )0 0 0|| ||v m LA g r R R= − 10.74 Output resistance of Wilson source

( )

( )( ) [ ]

030

0 0

03

0

2Then ||

80 400 k0.2

120 600 k0.2

0.2 7.692 mA/V0.026

80 4007.69 600 7.69 600 ||16,000 4448

2

v m

AP

REF

AN

REF

REFm

T

v v

rR

A g r RVrI

VrIIgV

A A

β≅

= −

= = = Ω

= = = Ω

= = =

⎡ ⎤= − = − ⇒ = −⎢ ⎥

⎢ ⎥⎣ ⎦

10.75 (a) 2 0 200 AD D REFI I I μ= = =

( )( )

( )( )

( )( )

( )( )

2 22

2 2

2

0

1 1For ; 250 K0.02 0.2

0.042 2 35 0.22

0.748 mA/V1 1For ; 333 K

0.015 0.2

0.082 20 0.2 0.80 mA/V2

oP D

m P D

m

n Do

mo mo

M rI

g K I

g

M rI

g g

λ

λ∞

= = =

⎛ ⎞= = ⎜ ⎟⎝ ⎠

=

= = =

⎛ ⎞= ⇒ =⎜ ⎟⎝ ⎠

(b) ( ) ( )( )2 || 0.80 250 || 333114.3

v mo o oo

v

A g r rA

= − = −= −

(c) ( )Want 57.15 0.80 142.8 ||

142.8142.8 || 71.375 143 K142.8

v L

LL L

L

A RRR RR

= − = −

= = ⇒ =+

10.76 Assume M1, M2 matched

( )( )

( )( )( )( )

( )

2

22

0

2

0

0

200 A1 1 250 K

0.02 0.21 1 333 K

0.015 0.2

100 250 333 0.70 mA/V

0.082 0.2 0.702

15.3

REF D Do

op D

oon D

v mo o oo

mo mo

mo

I I I

rI

rI

A g r r

g g

WgL

WL

μ

λ

λ

= = =

= = =

= = =

= −

− = − ⇒ =

⎛ ⎞⎛ ⎞= =⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

⎛ ⎞ =⎜ ⎟⎝ ⎠

( )

0 2

2

2 1

Now 2 2

80 4015.32 2

30.6

pn kk W WL L

WL

W WL L

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

⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

10.77

Since 3 0,sgV = the circuit becomes

( )

( )

22 2 3

02

33

2 2

32 3

2

2 3 2

1 1

and

Then

1

so that

1

or1

x sgx m sg sg x o

o xx m o

o o

x oo o m o

x o

o o o m o

ov m o o

i

V VI g V V I r

r

r VI g r

r r

V rR r g rI r

R r r g rvA g r Rv

−′= − + =

⎛ ⎞′+ + =⎜ ⎟⎝ ⎠

⎛ ⎞′= = + +⎜ ⎟⎝ ⎠

′= + +

= = −

( )( )( )

( )( )

1

1

Now

2 0.050 20 0.10 0.632 /

1 1 500 0.02 0.10

m

on DQ

g mA V

r kIλ

= =

= = = Ω

( )( )( )

( )( )2 3

2 2 0.020 80 0.1 0.80 /

1 1 500 0.020 0.1

m p DQ

o op DQ

g K I mA V

r r kIλ

′ = = =

= = = = Ω

( )( )( )( )

Then

500 500 1 0.8 500 201

0.632 500 201000 315o

v v

R M

A A

= + + ⇒ Ω⎡ ⎤⎣ ⎦= − ⇒ = −

10.78 From Eq. 10.105

( ) ( ) ( )

2

3 4 1 2

1

1 1

22

2 0.050 20 0.08

0.5657 /

mv

o o o o

nm D

m

gA

r r r r

k Wg IL

g mA V

−=

+

′⎛ ⎞ ⎛ ⎞= ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

=

=

( )( )( )

( ) ( )( )

2

2 2

1 1 625 K0.02 0.08

0.5657 0.32001 1 2 0.00000256

625 62562,500

oD

v

v

rI

A

A

λ= = =

− −= =+

= −

10.79

(1) ( )22 2

1 2 22 1 2

Om i m

o o

V VV Vg V g Vr r r

ππ ππ

π

− −= + + +

(2) ( )2

2 23 2

0OOm

O o

V VV g VR r

ππ

− −+ + =

(1) 1 2 22 1 2 2

1 1 1 Om i m

o o o

Vg V V g

r r r rππ

⎛ ⎞= + + + +⎜ ⎟

⎝ ⎠

(2) 2 23 2 2

1 1 1 0O mO o o

V V gR r rπ

⎛ ⎞ ⎛ ⎞+ + + =⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

1m

o

gr

>>

(1) 1 22 2

1 Om i

o

Vg V V

r rππ

β⎛ ⎞+= +⎜ ⎟⎝ ⎠

(2) 2 23 2

1 1 0O mO o

V V gR r π

⎛ ⎞+ + ⋅ =⎜ ⎟

⎝ ⎠

(3) 22 3 2

1 1O

m O o

VV

g R rπ⎛ ⎞

= − +⎜ ⎟⎝ ⎠

Then

12 3 2 2 2

3 2 2

3

1 3

1 1 1(1)

1 1 1

1

1

O Om i

m O o o

OO

O o o

O

O

Om O

i

V Vg Vg R r r r

VVR r r

VR

V g RV

π

β

ββ

ββ

ββ

⎛ ⎞⎛ ⎞+= − + +⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

⎛ ⎞⎛ ⎞+= − + +⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

⎛ ⎞+≈ − ⎜ ⎟⎝ ⎠⎛ ⎞

= − ⎜ ⎟+⎝ ⎠

From Equation (10.20) 3 3O OR rβ≈ So

( )( )( )

21 3

3

2

0.25 9.615 mA/V1 0.026

80 320 K0.25

9.615 320 120366,165

121

O m ov m

i

o

v

V g rA gV

r

A

ββ

−= = = =

+

= =

−= = −