Chapter 4 Exercise Solutions EX4.1

( ) ( )( )

( )( )

2

2

2 and

0.75 0.5 0.8 2.025 2 0.5 2.025 0.8 1.22 /

1

1(0.01)(0.75)133 133

m n GS TN D n GS TN

GS GS

m m

oDQ

o

g K V V I K V V

V V Vg g mA V

rI

kr k

λ

= − = −

= − ⇒ == − ⇒ =

=

=

= Ω= Ω

EX4.2

( )( )( )( )

2 2 0.5 0.4 0.8944 mA/V

0.8944 10 8.94

v m D

m n DQ

v

A g R

g K I

A

= −

= = =

= − = −

EX4.3

(a) ( )

( )( )( )

2

1 2

2

320 5 1.905 V520 320

0.20 1.905 0.8 0.244 mA

2 2 0.2 0.244 0.442 mA/V

GS DD

DQ

m n DQ

o

RV VR R

I

g K I

r

⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠

= − =

= = =

= ∞

(b) ( )( )0.422 10 4.22v m DA g R= − = − = − (c) 1 2 520 320 198 KiR R R= = = (d) 10 KO DR R= = EX4.4

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

22

2

At transition point, 1 mA

sat

1 0.2 sat sat 2.236 V

D

D n GSt TN n DS

DS DS

I

I K V V K V

V V

=

= − =

= ⇒ =

( )( )

2

21 2

1 2 1

5 2.236Want 2.236 3.62 V2

5 3.62 2.76 K0.5

0.5 0.2 0.8 2.38 V

1

DSQ

D

GSQ GSQ

GSQ DD DD

V

R

V V

RV V R R VR R R

−= + =

−= =

= − ⇒ =

⎛ ⎞= =⎜ ⎟+⎝ ⎠

( )( ) 1 21

1So 2.38 200 5 420 K and 382 KR RR

= ⇒ = =

( )( )( )( )

2 2 0.2 0.5 0.6325 mA/V

0.6325 2.761.75

v m D

m n DQ

v

A g R

g K I

A

= −

= = =

= −= −

EX4.5 (a)

( ) ( )

( ) ( ) ( )

( )( )

( ) ( )

( )( )

2

1 2

2

2

2

2

2

25010 5 10 5 3 V250 1000

52

3 5 2 0.5 0.62 1.2 0.36

0.2 1.64 0

0.2 0.04 4 1.641.385 V

20.5 1.385 0.6 0.308 mA10 0.3

G

G GSD n GS TN

GS GS

GS GS GS

GS GS

GS

DQ DQ

DSQ

RVR R

V VI K V V

V VV V V

V V

V

I IV

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

= = −

− − + = −− = − +

− − =

± += =

= − ⇒ == − ( )( )08 10 2 6.30 VDSQV+ ⇒ =

(b)

( )( )

( )( )( )( )

2 2 0.5 0.3081

0.7849 mA/V0.7849 10

3.051 0.7849 2

m Dv m n DQ

m S

m

v v

g RA g K I

g Rg

A A

−= = =

+=

−= ⇒ = −

+

EX4.6

( )( )

( )( )

( )( )

2

2

0

5 33 V and 0.5 mA 4 k0.5

0.5 1 1 1.71 V 5 1.71 3.29 V

2 2 1 0.5

1.414 mA/V1.414 4 5.66

0.46sin 5.66 0.0813sin

SDQ DQ D D

DQ P SG TP

SG SG GG GG

v m D

m P DQ

m

v v

sdv i

i i i

V I R R

I K V V

V V V V

A g R

g K I

gA A

v v tA vv v v

ω

−= = ⇒ = ⇒ = Ω

= −

= − ⇒ = ⇒ = − ⇒ =

= −

= =

== − ⇒ = −

−= = = − = − ⇒ = tω

EX4.7 a.

( )( )( )( )

( )

2

2

2

9 ,

9 2 1.2 2

9 2.4 4 4

SG DQ S DQ P SG TP

SG SG

SG SG

V I R I K V V

V V

V V

= − = −

= − −

= − − +

22.4 8.6 0.6 0SG SGV V− + =

( ) ( )( )( )

( )( ) ( )( )

2

2

8.6 8.6 4 2.4 0.62 2.4

3.51 V, 2 3.51 2 4.57 mA

9 9 1.2 1 18 4.57 2.2 7.95 V

SG

SG DQ DQ

SDQ DQ SDQ

V

V I I

V I V

± −=

= = − ⇒ =

= + − + = − ⇒ =

b.

RDVi

V0

RS CS

gmVSGVSG

( )( )

( )( )0

2 2 2 4.57 6.046 mA/V

6.046 1 6.05

m P DQ

m SG D

v m D v

g K I

V g V RA g R A

= = =

== − = − ⇒ = −

EX4.8

( )( ) ( )( )2 2

2 2

1 2

5 10 1.5 3.33 k

1.5 1 0.8

2.025 V 5 7.025 V 10400

DSQ DD DQ S

S S

DQ n GS TN GS

GS G S G G DD

V V I RR R

I K V V V

R RV V V V V VR R

= −= − ⇒ = Ω

= − ⇒ = −

⎛ ⎞= = − = − ⇒ = = = ⋅⎜ ⎟+⎝ ⎠

So 2 1281 k , 119 kR R= Ω = Ω

( )( )

( )( )

( )( )( )( )

( )( )

0

0

1 10

0

Neglecting , 1

0.015 1.5 44.4 k

3.33 44.4 3.1 k

2 2 1 1.5 2.45 /

2.45 3.10.884

1 2.45 3.1

m SSi v

m S

DQ

S

m n DQ

v v

g R rR A

g R r

r I

R r

g K I mA V

A A

λ− −

=+

⎡ ⎤= = = Ω⎡ ⎤⎣ ⎦⎣ ⎦= = Ω

= = =

= ⇒ =+

EX4.9

( )( )

( )( )( )( )

2

2

1 10

3 2 2 3.22 V5 5 3.223 0.593 k

0.02 3 16.7 k

2 2 2 3 4.9 /

DQ P SG TP

SG SG

SGDQ S

S S

DQ

m P DQ

I K V V

V VVI R

R R

r I

g K I mA V

λ− −

= −

= − ⇒ =− −= ⇒ = ⇒ = Ω

⎡ ⎤= = = Ω⎡ ⎤⎣ ⎦⎣ ⎦

= = =

For ( )

( )0

0

, 1

m SL v

m S

g r RR A

g r R= ∞ =

+

0 16.7 0.593 0.573 kSr R = = Ω

( )( )( )( )4.9 0.573

0.7371 4.9 0.573v vA A= ⇒ =

+

( )( )0

0

If is reduced by 10% 0.737 0.0737 0.663

1

v v

m S Lv

m S L

A A

g r R RA

g r R R

⇒ = − =

=+

Let 0 S Lr R R x=

( )( ) ( )

( ) ( )( )

4.90.663 0.663 4.9 1 0.663

1 4.90.402 0.573

0.573 0.402 0.573 0.402 0.402 0.573 1.35 k0.573

L

LL L

L

xx

xx R

R R RR

= ⇒ = −+

= =

= ⇒ − = ⇒ = Ω+

EX4.10

( )

( ) ( )

2

1 2

2 2

9.3 570.7 9.3

0.581 V

5

G DD

DQ p SG TP P S G TP

S

S

RV VR R

I K V V K V V V

VR

⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟+ +⎝ ⎠⎝ ⎠=

= − = − −

−=

( )( )( )( )( )

2

2

2

2

Then 0.4 5 0.581 0.8 5

2 1.381 5

2 2.762 1.907 5

2 4.52 1.19 0

S S

S S

S S S

S S

V V

V V

V V V

V V

− − = −

− = −

− + = −

− − =

( ) ( )( )( )

( )( )

( )( )( )( )

2

1 2

1 2

4.52 4.52 4 2 1.192 2

5 2.52.50 V 0.5 mA5

2 2 0.4 0.5 0.894 /

10.894 5 70.7 9.3 0.770

1 0.894 5 70.7 9.3 0.5

S

S DQ

m P DQ

m Sv

m S Si

v

V

V I

g K I mA V

g R R RAg R R R R

A

± +=

−= ⇒ = =

= = =

= ⋅+ +

= ⋅ ⇒ =+ +

Neglecting ,SiR Av = 0.817

0 01 15 5 1.12 0.915 k

0.894Sm

R R Rg

= = = ⇒ = Ω

EX4.11

V0

RD RL

gmVsg

Vi

RS Vsg

( )

( )( )( )( )

( )( )( )( )

( )( )

0

2

2

2

2

2

( ) and ( )

5

5 1 4 0.8

5 4 1.6 0.64

4 5.4 2.44 0

5.4 (5.4) 4 4 2.442 4

1.71 V5 1.71 0.822 mA

42 2 1 0.822 1.81 /

m sg D L sg i

v m D L

SGDQ p SG TP

S

SG SG

SG SG SG

SG SG

SG

SG

DQ

m p DQ

v

V g V R R V VA g R R

VI K V VR

V V

V V V

V V

V

V

I

g K I mA V

A

= ==

−= = −

− = −

− = − +

− − =

± +=

=−= =

= = =

( )( ) ( )( )1.81 2 4 1.81 1.33 2.41

1 14 4 0.552 0.485 k1.81

v

in S inm

A

R R Rg

= = ⇒ =

= = = ⇒ = Ω

EX4.12

( )( )

( )( )

( )

22

2

1 1

2 22

1

1 1

0.015 2 0.030 /2

6 36

36 0.030 1.08 /

1.08 0.015 72

n oxn

n nv

n n

n

C WK mA VL

K KAK K

K mA VW WL L

μ ⎛ ⎞= ⋅ = =⎜ ⎟⎝ ⎠

= − = − ⇒ =

= =

⎛ ⎞ ⎛ ⎞= ⇒ =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

( ) ( )( )The transition point is found from 1 10 1 6 110 1 6 1 2.29 V

1 6

GSt GSt

GSt

v v

v

− = − − −− + += =

+

For Q-point in middle of saturation region 2.29 1 1 1.645 V2GS GSV V−= + ⇒ =

EX4.13 (a) Transition points:

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

2

2 21 1 2 2

22 3

2 3 3 2

For : 5 1.2 3.8

For : 1 1

250 0.01 25 1.2 1 0.01 5 0.01

10 0.01 1.512 0.0144 0.01 0.00144 0.

OtB DD TNL

n OtA OtA n TNL DD OtA

OtA OtA OtA

OtA OtA OtA OtA OtA OtA

M v V V V

M K v v K V V v

v v v

v v v v v v

λ λ

= − = − =

⎡ ⎤ ⎡ ⎤+ = + −⎣ ⎦ ⎣ ⎦⎡ ⎤⎡ ⎤+ = + −⎣ ⎦ ⎣ ⎦

⎡ ⎤+ = − + + −⎣ ⎦ 512 0

which yields 0.388 OtAv V

=

≅

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

( ) ( )

( )

0 1

2 21 1 1 2 2

2 21

21

1

3.8 0.388Then middle of saturation region 0.388 2.094 V2

1 1

250 0.8 1 0.01 2.094 25 1.2 1 0.01 5 2.094

10 0.8 1.0209 1.482

0.8

Q DSQ

n GS TND O n TNL DD O

GS

GS

GS

V V

K V V v K V V v

V

V

V

λ λ

−= + ⇒ =

⎡ ⎤ ⎡ ⎤− + = + −⎣ ⎦ ⎣ ⎦⎡ ⎤ ⎡ ⎤− + = + −⎣ ⎦ ⎣ ⎦

⎡ ⎤− = =⎣ ⎦

− 210.145 1.18 VGSV= ⇒ =

b. ( ) ( )( )( )

( ) ( ) ( )

21 1

2

0.8 1 0.01 2.094

0.25 0.145 1.02094 37.0 A

DQ n GS

DQ DQ

I K V

I I μ

⎡ ⎤= − +⎣ ⎦⎡ ⎤= ⇒ =⎣ ⎦

c. ( ) ( )

( ) ( )( )

( )( )

11 01 02

1 2

1 1 12 2 0.25 1.18 0.8 0.19 mA/V0.19 257

0.037 0.01 0.01

mv m

DQ

m n GS TND

v v

gA g r r

Ig K V V

A A

λ λ−

= = −+

= − = − =−= ⇒ = −

+

EX4.14

( )

( )( )( )

( )2

1

1 666.7 K0.015 0.1

250 666.7 666.7

0.75 mA/V 2 2 0.1

0.0801.406 mA/V2 2

35.2

v m on op

on op

m

m n DQ n

nn

A g r r

r r

g

g K I K

k W WKL L

WL

= −

= = =

− = −

= = =

′ ⎛ ⎞ ⎛ ⎞⎛ ⎞= = =⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠

⎛ ⎞ =⎜ ⎟⎝ ⎠

EX4.15 (a)

( )

2 11 1

10

1

1 1

1 1So 0.5 mA/V2

2

0.5 2 0.2 0.3125 mA

O o om m

m

m n D

D D

R r rg g

gR

g K I

I I

= ≈

= = =

=

= ⇒ =

(b) ( )

( )

( )( )( )

( )( )

1 1 2

1 1 2

1 2

11 320 K

0.01 0.3125

0.5 320 3201 0.5 320 3200.988

m o ov

m o o

o o

v

v

g r rA

g r r

r r

A

A

=+

= = =

=+

=

EX4.16

(a)

( )( )

111

2 1

2 1

1 1

12

1 1

2 0.2 0.01120

0.01 0.01

2.4 0.01 2 0.2

2.39 2 0.2 0.140 mA

2 0.2 0.14 0.335 mA/V

mn D Do

vD D

o o

D D

D D

D D D

D D

m m

gK I IrAI I

r r

I II I

I I I

I I

g g

λλ λ

++

= =++

+=

+

− =

= ⇒ =

= ⇒ =

(b)

( )( )

1 2

1 21 714 K

0.01 0.14357 K

o o o

o o

o

R r r

r r

R

=

= = =

=

EX4.17

0 22

2 2

0 0

1

0.632 mA/V, 8 kΩ18 8 1.58 1.32 k

0.632

Sm

m S

R Rg

g R

R R

=

= =

= = ⇒ = Ω

EX4.18 a.

( )21 1 1 1DQ n GS TNI K V V= −

( )( )( )

21 1 2

1

1 1.2 2 2.91 V10 k 10 1 10 10 0

GS GS GS

S S DQ S

V V VR V I R

= − ⇒ = == Ω ⇒ = − = − =

( )

( )

31

1 2 3

33

2.91 10

10 145.5 k500

GR

VR R R

R R

⎛ ⎞= = ⎜ ⎟+ +⎝ ⎠

⎛ ⎞= ⇒ = Ω⎜ ⎟⎝ ⎠

1 2 2 23.5 3.5 V 3.5 2.91 6.41DSQ S G GV V V V= ⇒ = ⇒ = + ⇒ =

( )

( )

2 32

1 2 3

2 3

10 6.41

10500

GR R

VR R R

R R

⎛ ⎞+= =⎜ ⎟+ +⎝ ⎠

+⎛ ⎞= ⎜ ⎟⎝ ⎠

2 3 2 2320.5 145.5 175 kΩR R R R+ = = + ⇒ =

Then 1 2 3 1 1500 175 145.5 179.5 kR R R R R+ + = = + + ⇒ = Ω

Now 2 2 2 23.5 3.5 3.5 7 VS D S SDQV V V V= ⇒ = + = + =

So 10 7 3 k1D DR R−= ⇒ = Ω

b. From Example 6-18:

( )( )( )( )

1

1 12 2 1.2 1 2.19 /

2.19 3 6.57

v m D

m n DQ

v v

A g R

g K I mA V

A A

= −

= = =

= − ⇒ = −

EX4.19

( )( )

( )( )

( )

2

2

2 2

1 2

1.2 2.7 3.24 V3.24 12 15.24

20 15.24 3.97 k1.2

1

1.2 4 1 0.4523

0.4523 3 1.3573.24 1.357 1.883 V

20500

S DQ S

D S DSQ

D D

GSD DSS

P

GS GS

P P

GS

G S GS

G

V I RV V V

R R

VI I

V

V VV V

VV V V

R RVR R

= = == + = + =

−= ⇒ = Ω

⎛ ⎞= −⎜ ⎟

⎝ ⎠

⎛ ⎞= − ⇒ =⎜ ⎟

⎝ ⎠= − = −= + = − =

⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜+ ⎝ ⎠⎝ ⎠( )

( )( )

( )( )

( ) ( )( )

2 1

0

0

20 1.88 47 k . 453 k

1 1 167 k0.005 1.2

2 42 1.3571 1 1.46 mA/V3 3

1.46 167 3.97 4 2.87

DQ

DSS GSm

P P

v m D L v

R R

rI

I VgV V

A g r R R A

λ

= ⇒ = Ω = Ω⎟

= = = Ω

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

EX4.20 a.

( )( )

2 2

1 2 8 1 0.5

0.5 3.5 1.75

GS GS GSDQ DSS

P P P

GS GS

V V VI I

V V VV V

⎛ ⎞ ⎛ ⎞= − = − ⇒ =⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠= − ⇒ = −

( )10 1.75 10Also 2 5.88 kGSDQ S

S S

VI R

R R− − − += = ⇒ = Ω

b.

gmVgs

V0

Vgs

Vi

RS

( )

( )( )

( )[ ]( )[ ]

0

0 0

1 0

2 82 1.751 1 2.29 mA/V3.5 3.5

1 50 k0.01 2

12.29 5.88 50

0.92341 1 2.29 5.88 50

DSS GSm

P P

ii gs m S gs gs

m S

m Sv v

m S

I Vg

V V

r

VV V g R V Vg R

V g R rA A

V g R r

⎛ ⎞ ⎛ ⎞= − = − =⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

= = Ω

= + ⇒ =+

= = = ⇒ =+ +

c. ( )

( ) ( )( )

( )( )( )( )

0.80 0.9234 0.73871

2.290.7387 1.235 k

1 2.295.261 k

5.261 1.235 k 1.61 k5.261

m S L ov

m S L o

S L oS L o

S L o

S o

LL

L

g R R rA

g R R r

R R rR R r

R R rR r

R RR

= = =+

= ⇒ = Ω+

= Ω

= Ω ⇒ = Ω+

TYU4.1

( ) ( )

( )( )

( )

( )

( )( )

2

22

1 12

1

2 and

and 2 2

3.41.445 /

4 4 2

2

1.445 0.018 80.3

0.015 2 33.3

m n GS TN D n GS TN GS TN

DQm n n DQ

n

mn

DQ

ox

o n GS TN DQ

o o

CnKn

IDQg K V V I K V V V VKn

Ig K K I

K

gK mA VI

WL

W WL L

r K V V I

r r k

μ

λ λ− −

−

= ⋅

= − = − ⇒ − =

= =

= = =

⎛ ⎞= ⇒ =⎜ ⎟⎝ ⎠

⎡ ⎤ ⎡ ⎤= − = ⎣ ⎦⎣ ⎦

= ⇒ = Ω⎡ ⎤⎣ ⎦

TYU4.2 a. ( )

( )( )( )

2

20.4 0.5 2 2.894 V

10 0.4 10 6 V

DQ n GS TN

GS GS

DSQ DD DQ D DSQ

I K V V

V V

V V I R V

= −

= − ⇒ =

= − = − ⇒ =

b. ( ) ( )( )

( )( )

1

0 0

0

2 2 0.5 2.894 2 0.894 mA/V

, 0

0.894 10 8.94

m n GS TN m

DQ

v m D vi

g K V V g

r I r

vA g R A

v

λ λ−

= − = − ⇒ =

⎡ ⎤= = ⇒ = ∞⎣ ⎦

= = − = − ⇒ = −

c. ( )( )0.4sin 8.94 0.4 sin3.58sin

i ds

ds

v t v tv t

ω ωω

= ⇒ = −= −

( )

1

1

1

1 1

At 6 3.58 2.422.89 0.4 3.29

3.29 2 1.29 satSo Biased in saturation region

DS

GS

GS TN DS

DS GS TN

VVV V V

V V V

= − == + =− = − = =

> − ⇒

TYU4.3 a. ( ) ( ) ( )

( )( )

2 20.25 2 0.8 0.36 mA

5 0.36 5 3.2 VDQ n GS TN DQ

DSQ DD DQ D DSQ

I K V V I

V V I R V

= − = − ⇒ =

= − = − ⇒ =

b. ( ) ( )( ) 02 2 0.25 2 0.8 0.6 mA/V, m n GS TN mg K V V g r= − = − ⇒ = = ∞

c. ( )( )0 0.6 5 3.0v m D vi

vA g R Av

= = − = − ⇒ = −

TYU4.4

( )( )

( )( ) ( )

0.1sin0.6 0.1 sin

0.06sin mA3 0.1 sin 0.3sin V

Then 0.36 0.06sinmA

3.2 0.3sin

i gs

d m gs

d

ds

D DQ d

D

DS DSQ ds DS

v v ti g v ti tv t t

i I i ti

v V v t v

ωω

ωω ω

ω

ω

= == === − = −

= + = +

== + = − =

TYU4.5 a.

( )( )( )

2

2

7 12 6 0.833 mA

0.833 2 1 1.65 V

SDQ DD DQ D

DQ DQ

DQ P SG TP

SG SG

V V I RI I

I K V V

V V

= −= − ⇒ =

= −

= − ⇒ =

b. ( ) ( )( )

( )( )

0

0

2 2 2 1.65 1 2.58 mA/V,

2.58 6 15.5

m P SG TP m

v m D vi

g K V V g r

vA g R A

v

= − = − ⇒ = = ∞

= = − = − ⇒ = −

V0

gmVsg RD

VSGVi

TYU4.6

( )

( )

2

2 2

So 2

DQDQ n GS TN GS TN

n

DQm n GS TN n

n

m n DQ

II K V V V V

K

Ig K V V K

K

g K I

= − ⇒ − =

= − =

=

TYU4.7

2 2 f SBvγη

φ=

+

(a) ( )0.40 0.153

2 2 0.35 1η η= ⇒ =

+

(b) ( )

( )( )

0.40 0.1042 2 0.35 3

2 2 0.5 0.75 1.22 /m n DQg K I mA V

η η= ⇒ =+

= = =

For (a), ( )( )1.22 0.153 0.187 /mb m mbg g g mA Vη= = ⇒ =

For (b), ( )( )1 22 0 104 0 127 mb mbg . . g . mA / V= ⇒ = TYU4.8

( )( )

( )( )

( )( ) ( )

2

0.5 mA

Let 25

20 25 500 /

0.5 1.5 2.5 V 2.5 V0.5

2 0.5 2.5 1.5 1 mA/VFor 4.0 4 k

5 0.5 4 3 V 3 2.5 5.5 V

DQ Q

n

GS S

v m D

m

v D

D DSQ

I IWL

K A V

V V

A g Rg

A RV V

μ

= =

=

= =

= + = ⇒ = −

= −= − =

= − ⇒ = Ω= − = ⇒ = − − =

TYU4.9 a. With G GS DSR V V⇒ = ⇒ transistor biased in sat. region

( ) ( )( )

( )( )( )( )

( ) ( )( )( )( )

( )( )

2 2

2

2

2

2

2

2

15 0.15 10 1.8

15 1.5 3.6 3.24

1.5 4.4 10.14 0

4.4 4.4 4 1.5 10.144.45 V

2 1.5

0.15 4.45 1.8 1.05 mA

D n GS TN n DS TN

DS DD D D DD n D DS TN

DS DS

DS DS

DS DS

DS DSQ

DQ DQ

I K V V K V V

V V I R V K R V V

V V

V V

V V

V V

I I

= − = −

= − = − −

= − −

= − − +

− − =

± += ⇒ =

= − ⇒ =

b. Neglecting effect of RG: ( )( ) ( )( )

( )( )2 2 0.15 4.45 1.8 0.795 mA/V

0.795 10 5 2.65

v m D L

m n GS TN m

v v

A g R Rg K V V gA A

= −= − = − ⇒ == − ⇒ = −

c. GR ⇒ establishes GS DSV V= ⇒ essentially no effect on small-signal voltage gain. TYU4.10 a.

( )

( )

( )

2

2

2

2

0.8 24

3.2 4 4

3.2 13.8 12.8 0

DQ n GS TN

SG SGDQ SG

S

SG SG SG

SG SG

I K V VV V

I VR

V V V

V V

= −

= − = =

− + =

− + =

( ) ( )( )( )

( )

2

2

13.8 13.8 4 3.2 12.82 3.2

1.35 V 0.8 2 1.35 0.338 mA

SG

SG DQ DQ

V

V I I

± −=

= ⇒ = − ⇒ =

b. ( )

( )( )( )( )

6 10 0.338 410 0.338 4 6

7.83 k0.338

DSQ DD DQ D S

D

D D

V V I R RR

R R

= − += − +

− −= ⇒ = Ω

c.

Vi RD

V0

RS

gmVgsVgs

( ) ( )( )( )( )

( )( )

0

0

1

2 2 0.8 1.35 2 1.04 mA/V1.04 7.83

1.581 1 1.04 4

ii gs m gs S gs

m S

m gs D

m n GS TN

m Dv v

i m S

VV V g V R Vg R

V g V Rg K V V

V g RA A

V g R

= + ⇒ =+

= −= − = − + =

−−= = = ⇒ = −

+ +

TYU4.11 a.

( )( )

( )( )( )( )

2

2

5 and ( )0.8 0.5( 0.8) 0.465 V5 0.8 0.465 5.67 k

103 10 0.8 5.67

10 0.8 5.67 33.08 k

0.8

DQ S SG DQ p SG TP

SG SG

S S

SDQ DQ S D

D

D D

I R V I K V VV VR R

V I R RR

R R

= + = += + ⇒ =

= + ⇒ = Ω

= − += − +

− −= ⇒ = Ω

b.

r0 RD

V0

VSG

Vi gmVsg

( ) ( )

( )

( )( )( )( )

0 0 0

00

00

2 ( ) 2(0.5)(0.465 0.8) 1.265 mA/V1 1 62.5 k

0.02 0.81.265 3.08 62.5 3.71

m sg D m i D

v m Di

m p SG TP

v v

V g V R r g V R rV

A g R rV

g K V V

rI

A Aλ

= = −

= = −

= + = + =

= = = Ω

= − ⇒ = −

TYU4.12 (a)

( )( )( )

( )( )

0 0

0 0

0 0 0

0 0

0

So 4 50

0.9951 1 4 50

m gs

i gs gs i

m i

mv v

i m

V g V rV V V V V V

V g r V V

V g rA AV g r

== + ⇒ = −

= −

= = = ⇒ =+ +

r0

Ix

gmVgs

Vgs

Vx

0

0 0 00

and

1 150 0.25 k4

xx m gs gs x

xx m x

m

VI g V V Vr

VI g V R r Rr g

+ = = −

= + ⇒ = = ⇒ ≅ Ω

(b) With ( )

( )0

0

4 k1

m SS v

m S

g r RR A

g r R= Ω ⇒ =

+

( )( )( )( )0

4 3.7|| 50 || 4 3.7 k 0.937

1 4 3.7s v vr R A A= = Ω ⇒ = ⇒ =+

TYU4.13 (a)

( )

( )

( ) ( )

2

2 2

2 2 2 0.8 1.25 /

1.25 0.0202

So 62.5

0.8 1.25 2 2.8 V

m n DQ n n

n oxn

DQ n GS TN GS GS

g K I K K mA V

C W WKL L

WL

I K V V V V

μ

= ⇒ = ⇒ =

⎛ ⎞= ⋅ ⇒ = ⎜ ⎟⎝ ⎠

=

= − ⇒ = − ⇒ =

b.

( )( )( )

( )

( )( )( )( )

1 10

0

0

0

0 0 0

0.01 0.8 125 k

1125 4 3.88

2 3.880.886

1 2 3.88

1 1 125 0.5 k2

DQ

m Lv

m L

L

v v

m

r I

g r RA

g r Rr R

A A

R r Rg

λ− −

⎡ ⎤= = = Ω⎡ ⎤⎣ ⎦⎣ ⎦

=+= =

= ⇒ =+

= = ⇒ ≈ Ω

TYU4.14

( ) ( )( )

( )( )( ) ( )( )

( )

0

2

2

2

1 0.35 k 2.86 mA/V

42.4 44

4 2.4 2.4 4 6 k

2

2.86 2 0.5 4.09 /

0.5 4.09 1 1.35 V 1.35 V, 5 0.5 6 2 V2 1.35 3.35 V

in mm

DD L D

i D

D D

m n DQ

n n

DQ n GS TN

GS GS S D

DS D S

R gg

V RR R RI R

R R

g K I

K K mA V

I K V V

V V V VV V V

= = Ω ⇒ =

= = = =+

− = ⇒ = Ω

=

= ⇒ =

= −

= − ⇒ = ⇒ = − = − == − = − − =

We have 3.35 1.35 1 0.35 Biased in the saturation regionDS GS TNV V V V= > − = − = ⇒ TYU4.15

( )( )

( )( )

21

1

22

2

1

2

0.020 80 1.6 /2

0.020 1 0.020 /2

1.6 8.940.020

n oxn

n oxn

nv v

n

C WK mA VL

C WK mA VL

KA A

K

μ

μ

⎛ ⎞= ⋅ = =⎜ ⎟⎝ ⎠

⎛ ⎞= ⋅ = =⎜ ⎟⎝ ⎠

= − = − ⇒ = −

( )

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

1

2

The transition point is determined from

0.8 5 0.8 8.94 0.85 0.8 8.94 0.8 0.8

1 8.941.22 V

nGSt TND DD TNL GSt TND

n

GSt GSt

GSt

GSt

Kv V V V v VK

v v

v

v

− = − − −

− = − − −− + +

=+

=

For Q-point in middle of saturation region 1.22 0.8 0.8 1.01 V2GS GSV V−= + ⇒ =

TYU4.16 a.

( )( )

2 2

2 2 2 2

22 2 2 2

22 2 2

2 , 10 V10 2 5 k

2 1 2 3.41 V 3.41 V

DQ DSQ

DQ S S S

DQ n GS TN

GS SG G

I mA VI R R R

I K V V

V V V

= =⋅ = = ⇒ = Ω

= −

= − ⇒ = ⇒ =

( )

1 1

1 1

1 1

10 3.41Then 3.29 k2

For 10 V 3.41 10 6.59 V6.59 10

Then 1.71 k2

D D

DSQ S

S S

R R

V V

R R

−= ⇒ = Ω

= ⇒ = − = −− − −

= ⇒ = Ω

( )( )

( )

( )( ) ( )( )

21 1 1 1

21 1

2 21 1 1

1 2 1 2 1

11

2 1 2 3.41 V

120

13.41 200 20 2 1.71 586 k

D n GS TN

GS GS

GS DQ S in

I K V V

V V

R RV I R RR R R R R

RR

= −

= − ⇒ =

⎛ ⎞= − = ⋅⎜ ⎟+ +⎝ ⎠

= − ⇒ = Ω

( ) ( )( )22 2

2

586 200 586 200 200 586 304 k586

R R RR

= ⇒ − = ⇒ = Ω+

b. ( )( )1 1 1 1 22 2 1 2 2.828 mA/Vm n DQ m mg K I g g= = ⇒ = = From Example 6.17

( )( )

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

1 2 1 2

2 2

2

0 2 02

15 4 2.222 k

2.828 2.828 3.29 2.2228.03

1 2.828 2.222

1 1 5 0.3536 5 0.330 k2.828

m m D S Lv

m S L

S L

v v

Sm

g g R R RA

g R RR R

A A

R R Rg

−=

+= = Ω

−= ⇒ = −

+

= = = ⇒ = Ω

TYU4.17 From Example 6.19

0

1 2

3.0 mA/V, 41.7 k420 180 126 k

mg rR R

= = Ω= = Ω

1 2

1 2

126 0.863126 20

gs ii

i i

R RV VR R R

V V

⎛ ⎞= ⎜ ⎟+⎝ ⎠⎛ ⎞= =⎜ ⎟+⎝ ⎠

( )

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

0

3.0 0.863 41.7 2.7 42.589 41.7 1.61 2.589 1.55 4.01

m gs D Lv

i

v

g V r R RA

V

A

−=

= −= − = − ⇒ = −

TYU4.18 a.

( )

( )

21

1 2

1430 20 17.2 V

430 70

G DD

G

RV VR R

V

⎛ ⎞= ⎜ ⎟+⎝ ⎠⎛ ⎞= =⎜ ⎟+⎝ ⎠

2 21 1

1

2 21 1 1

1

21 1

1 6 12

2017.26 1 6 7.6 and 2 2 2 4

20Then 6 7.6

4 2

GS G SDQ DSS

P

S S SDQ

S S

V V VI IV

V V VI

V V

⎛ ⎞ −⎛ ⎞= − = −⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

−⎛ ⎞ ⎛ ⎞= − + = − =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

− ⎛ ⎞= −⎜ ⎟⎝ ⎠

21

1 1

21 1

21 1

20 24 7.6 57.764

6 182.4 1386.24

6 181.4 1366.24 0

SS S

S S

S S

VV V

V V

V V

⎛ ⎞− = − +⎜ ⎟

⎝ ⎠= − +

− + =

( ) ( )( )( )

2

1

1 1

181.4 181.4 4 6 1366.242 6

14.2 V 17.2 14.2 3 V

S

S GS P

V

V V V

± −=

= ⇒ = − = >

1 1

1 1

So 16.0 17.2 16 1.220 16on 1 mA

4

S GS P

DQ DQ

V V V Q

I I

= ⇒ = − = < −−= ⇒ =

( )( )( )

1 1 1 1

1

2020 1 8 12 V

SDQ DQ S D

SDQ

V I R RV

= − += − ⇒ =

( )( )1 1 1 21 4 4 VD DQ D GV I R V= = = =

( )

22

2 22

2 22 2 2 2

22

1 6 12

46 1 6 3 and 2 2 2 4

GS G SDQ DSS

P

S S S SDQ

S

V V VI IV

V V V VI

R

⎛ ⎞⎛ ⎞ −= − = −⎜ ⎟⎜ ⎟ ⎜ ⎟−⎝ ⎠ ⎝ ⎠

⎛ ⎞ ⎛ ⎞= + − = − = =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

( ) ( )( )( )

22 2

22

2 2

22 2

2

2 2

Then 6 34 2

24 9 34

6 73 216 0

73 73 4 6 2167.09 V or 5.08 V

2 6

S S

SS S

S S

S S

V V

VV V

V V

V V

⎛ ⎞= −⎜ ⎟⎝ ⎠

⎛ ⎞= − +⎜ ⎟

⎝ ⎠− + =

± −= ⇒ = =

2 2

2 2

2 2 2

For 5.08 V 4 5.08 1.08 transistor biased ON5.08 1.27 mA

420 20 5.08 14.9 V

S GS

DQ DQ

DS S DS

V V

I I

V V V

= ⇒ = − = −

= ⇒ =

= − = − ⇒ =

b.

RL

gm2Vgs2

V0

gm1Vsg1

Vgs2

RS2

RD1

Vg2

Vi Vsg1

( )

( )

( )

2 1 1 1 1 1 1

0 2 2 2

22 2 0 2

2 2

0 1 1

2 2

1

1

g m sg D m D

m gs S L

gg gs gs

m S L

m Dv

i m S L

V g V R g V RV g V R R

VV V V V

g R RV g R

AV g R R

= = −=

= + ⇒ =+

−= =

+

( )

( )

1

2

2 1

2 6 1.21 2.4 mA/V2 2

2 6 1.081 2.76 mA/V2 2

DSS GSm

P P

m

I VgV V

g

⎛ ⎞= −⎜ ⎟

⎝ ⎠

⎛ ⎞= − =⎜ ⎟⎝ ⎠

⎛ ⎞= − =⎜ ⎟⎝ ⎠

( )( )( )( )( )

2.4 4Then 2.05

1 2.76 4 2v vA A−

= = − =+