Chapter 13 Exercise Solutions EX13.1

1 2

1 2 1 2

9.5 A9.5 A 0.0475 A 47.5 nA

200

C C

B B B B

I I

I I I I

μμ μ

= ≅

= = = ⇒ = =

EX13.2

17 13

16

16

5 0.6 0.6 ( 5) 0.22 mA40

0.75 (0.75)(0.22) 0.165 mA0.165 (0.165)(0.1) 0.6200 50

0.000825 0.0123313.2 A

REF

C C B REF

C

C

I

I I I

I

I μ

− − − −= =

= = = =+= +

= +=

EX13.3

3 140.18 10 10 exp D

T

VV

− − ⎛ ⎞× = ⎜ ⎟

⎝ ⎠

3

14

3

14

0.18 10ln10

0.18 10(0.026) ln10

0.6140

D T

D

V V

V

−

−

−

−

⎛ ⎞×= ⎜ ⎟⎝ ⎠

⎛ ⎞×= ⎜ ⎟⎝ ⎠

=

2 1.228 VBB DDV V= ≅

14 20

14

2exp

0.61403 10 exp0.026

BBC C S

T

VI I IV

−

⎛ ⎞/= = ⎜ ⎟⎝ ⎠

⎛ ⎞= × ⎜ ⎟⎝ ⎠

14 20 0.541 mAC CI I= = EX13.4

6100 10.5 M

0.0095or = ⇒ Ω

Then, using results from Example 13.4

[ ] [ ]

( )

1 6 6 2 6

44

1 ( || ) 10.5 1 (0.365)(1 || 547)14.3 M

100 10.5 M0.0095

9.5 10.5 ||14.3 || 4.07 8890.026

act o m

Ao

C

d

R r g R r

VrI

A

π= + = += Ω

= = ⇒ Ω

⎛ ⎞= − = −⎜ ⎟⎝ ⎠

EX13.5

19 13

3 22 19 20

2

17

2

2

100 556 K0.18

(1 )[ || ]7.22 (51)(556 ||111) 4.73 M100 185 K0.54100 185 K0.54

(200)(201)(50)(185 || 4730 ||185)4070[50 [9.63 (201)(0.1)]]

182358786.932450.1

5

o A

i P

act

o

v

v

R r

R r R R

R

R

A

A

π β

= = =

= + += + ⇒ Ω

= =

= =

−=+ +

−=

= − 62

EX13.6

[ ]17 13

17

17

22

19

100 100185 K 185 K0.54 0.54185 1 (20.8)(0.1 || 9.63)566 K7.22 566 ||185 2.88 K

51100 556 K0.18

o o B

C

C

e

C

r r

RR

R

R

= = = =

= +=

+= =

= =

200.65 2.88 || 556 0.0689

5168.9

22 68.9 90.9

e

O

R

R

+= =

= Ω= + = Ω

EX13.7

( )1 2

2

1 1 4

1 | | 30(1 562) 16890 pF4.07 M

|| 14.3 ||10.5 6.05 M

i

i

o act o

C C ARR R r

= + = + == Ω= = = Ω

Then 1 2|| 6.05 || 4.07 2.43 Meq o iR R R= = = Ω Then

6 12

1 12 2 (2.43 10 )(16890 10 )3.88 Hz

PDeq i

fR Cπ π −= =

× ×

=

EX13.8 For 5 ,M 6M

2

2 55

25 5 5

25 5

5

5

40(12.5) 250 A / V2

5 50.25( 0.5)225

56.25( 0.25) 1056.25 55.25 4.0625 0

55.25 3052.5625 914.06252(56.25)

0.902 V10 0.902 40.4 A

225

P

SGSG

SG SG SG

SG SG

SG

SG

set Q

K

VV

V V VV V

V

V

I I

μ

μ

⎛ ⎞= =⎜ ⎟⎝ ⎠

+ −− =

− + = −− + =

± −=

=−= = ⇒

Current in 1 4 20.2 AM M μ− = EX13.9

21 2

40(12.5) 250 A/V2P PK K μ⎛ ⎞= = =⎜ ⎟

⎝ ⎠

From Exercise Ex 13.8, 1 2 20.2 AD DI I μ= =

4 2

1 2 4

27

7 7

7 87

2 7

1 1 2.48 M(0.02)(0.0202)

2 ( || ) 2(0.25)(0.0404)(2480 || 2480)

17680(12.5) 500 A / V2

2 2 (0.50)(0.0404) 0.284 mA/V

1 1 1.24 M(0.02)(0.0404)

(

o oD

d P Q o o

d

n

m n Q

o oD

v m

r rI

A K I r r

A

K

g K I

r rI

A g

λ

μ

λ

= = = = Ω

= =

=

⎛ ⎞= =⎜ ⎟⎝ ⎠

= = =

= = = ⇒ Ω

= 7 8

2

|| ) (0.284)(1240 ||1240) 176(176)(176) 30,976

o o

V d V

r rA A A

= == ⋅ = =

EX13.10 (a)

11

1

0.08 0.252 2 (20)2 2 2 2

0.6325 mA/V

Qnm

Ik WgL

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

2

4

1 800 K(0.01)(0.25 2)

1 533.3 K(0.015)(0.25 2)

o

o

r

r

= =/

= =/

1 1 2 4

1

( || ) (0.6325)(800 || 533.3)202.4

d m o o

d

A g r rA

= ==

5 25

0.042 2 (80)(0.25)2 2

1.265 mA/V

nm Q

k Wg IL

′⎛ ⎞⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠=

5

9

1 266.7 K(0.015)(0.25)

1 400 K(0.01)(0.25)

o

o

r

r

= =

= =

2 5 5 9

2

( || ) (1, 265)(266.7 || 400)202.4

m o oA g r rA

= − = −= −

Overall gain (assuming 3 1)A =

1 2 (202.4)( 202.4) 40,966dA A A= ⋅ = − = − EX13.11

1 2

1 8 1 8

6 6

1 6 8 10

44

8

25

402 2 (25)(25) 224 /2 2

802 2 (25)(25) 316 /2 2

1 1 2 (0.02)(25)

1 1 1 (0.02)(50)

D D

pm m DQ m m

nm DQ m

o o o oD

oD

o

I I A

k Wg g I g g A VL

k Wg I g A VL

r r r r MI

r MI

R g

μ

μ

μ

λ

λ

= =

′ ⎛ ⎞ ⎛ ⎞= = = ⇒ = =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

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

= = = = = = Ω

= = = Ω

=( ) ( )

8 8 10

6 6 6 4 1

( ) (224)(2)(2) 896 ( ) || 316(2) 1|| 2 421

m o o

o m o o o

r r MR g r r r M

= = Ω= = = Ω

Then ( ) ( )1 6 8 224 421 896 64,158d m o o dA g R R A= = ⇒ =

EX13.12

1 6 7 1 1

0.6 0.6 0.6 (0.236)(8) 3.69 VEB EB BEV V V V V I R+ −− = + + +

= + + + =

So 1.845 VV V+ −= − =

EX13.13

5 22 ( )

2(1)(0.2)(26)16.4

d P Q i

d

A K I R

A

=

==

EX13.14

1313

2 13 13

01010

01212

1212

1212

1 012

(200)(0.026)0.20

26 k(1 ) 26 201(1)

227 k1 1 500 k

(0.02)(0.1)50 500 k0.1

0.1 3.85 mA / V0.026(200)(0.026) 52 k

0.1[

T

C

i E

D

A

C

Cm

T

T

C

act

VrI

R r R

rI

VrIIgVVr

IR r

π

π

π

β

β

λ

β

= =

= Ω= + + = += Ω

= = = Ω

= = = Ω

= = =

= = = Ω

= 12 12 51 ( || )]500[1 (3.85)(52 || 0.5)] 1453 k

mg r Rπ+= + = Ω

( )5 10 1 122 || ||

2(0.6)(0.2) (500 ||1453 || 227)(0.490)(141) 69.1

d n Q o act

d

A K I r R R

A

= ⋅

= ⋅= ⇒ =

EX13.15 From Example 13.15, 265 HzPDf =

2 (16.4)(1923) 31,537(265)(31,537) 8.36 MHz

v di

T PD v

A A Af f A

= ⋅ = == ⋅ = =

TYU13.1 Computer Analysis TYU13.2 Computer Analysis TYU13.3

i+

i

i

(max) (on) 15 0.6 14.4V(min) 4 (on)

4(0.6) 15 12.6 V12.6 (cm) 14.4 V

N EB

N BE

N

V V VV V V

V

+= − = − =≅ += − = −− ≤ ≤

TYU13.4

a. 0 (max) 2 (on) 15 2(0.6)BEV V V+≅ − = −

0

0

0

0

(max) 13.8 V(min) 3 (on) 3(0.6) 15(min) 13.2 V

13.2 13.8 V

BE

VV V VV

V

−

== + = −≅ −− ≤ ≤

b. 0 (max) 5 1.2 3.8 VV = − =

0

0

(min) 3 3(0.6) 5 3.2 V3.2 3.8 V

BEV V VV

−≅ + = − = −− ≤ ≤

TYU13.5

3

14

15 2(0.6) ( 15) 0.72 mA40

0.72 10ln (0.026) ln10

0.650 V

REF

REFBE T

S

I

IV VI

−

−

− − −≅ =

⎛ ⎞ ⎛ ⎞×= =⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

=

So

11

30 2(0.65) 0.718 mA40

0.650 V

REF REF

BE

I I

V

−= ⇒ =

=

10 410

1010

ln

0.718(5) (0.026) ln

REFC T

C

CC

II R VI

II

⎛ ⎞= ⎜ ⎟

⎝ ⎠⎛ ⎞

= ⎜ ⎟⎝ ⎠

By trial and error: 10 18.9 ACI μ=

10 11 10 4 10

106

66

6 614

0.650 (0.0189)(5) 0.556 V

18.9 9.45 A2 2

9.45 10ln (0.026)ln 0.537 V10

BE BE C BE

CC

CBE T BE

S

V V I R V

II

IV V VI

μ

−

−

= − = − ⇒ ≅

≅ = =

⎛ ⎞ ⎛ ⎞×= = ⇒ =⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

TYU13.6

10 410

1010

10 0.6 0.6 ( 10) 0.47 mA40

ln

0.47(5) (0.026) ln

REF REF

REFC T

C

CC

I I

II R VI

II

− − − −= ⇒ =

⎛ ⎞= ⎜ ⎟

⎝ ⎠⎛ ⎞

= ⎜ ⎟⎝ ⎠

By trial and error: 10 17.2 ACI μ⇒ ≅

106 6

13 13

13 13

8.6 A2(0.75) 0.353 mA

(0.25) 0.118 mA

CC C

C B REF C B

C A REF C A

II I

I I I

I I I

μ≅ ⇒ =

= ⇒ =

= ⇒ =

TYU13.7

0 6 14

14 0 01314 1

E

d AE

n

R R Rr R R

R π

β

= ++

=+

The diode resistance can be found as

exp

1 1 exp

DD S

T

D D DS

d D T T T

VI IV

I V IIr V V V V

⎛ ⎞= ⎜ ⎟

⎝ ⎠⎛ ⎞ ⎛ ⎞∂= = ⋅ =⎜ ⎟ ⎜ ⎟∂ ⎝ ⎠ ⎝ ⎠

or

13

017 0132222

013 013

017 017 17 8 17

0.026 1440.18

192.6 k

1 ( ) 283 k

T Td

D C A

BE

P

B B

m

V VrI I

r R RR

R rR r g R r

π

π

β

= = = ⇒ Ω

+=

+= = Ω

= ⎡ + ⎤ = Ω⎣ ⎦

From previous calculations 22

0 22

013 013

1414

14

0 6 14 0

1.51 k2 2(0.144) 1.51 1.80 k

278 k(200)(0.026) 1.04 k

51.04 1.8 || 278 14.1

20127 14.1 41

E

d d E

A A

n T

C

E

E

RR r RR r

VrI

R

R R R R

πβ

= Ω= + = + = Ω

= = Ω

= = = Ω

+= ⇒ Ω

= + = + ⇒ ≅ Ω

TYU13.8 For 6M we have 5 6 1.06 VSG SGV V= = So

6 (sat) 1.06 0.5 0.56 VSDV = − = For 1M and 2M

( )21

21 1

20.0397 0.125( 0.5) 0.898 V

2

QD P SG TP

SG SG

II K V V

V V

= = +

= − ⇒ =

So maximum input voltage 6 1

i

(sat)5 0.56 0.898

(max) 3.54 V

SD SG

N

V V V

V

+= − −= − −⇒ =

For 3,M 2

3

23

3

1

(6.25)(20) 125 /39.7 A

2 239.7 125( )

20.5 0.898 V

(sat) 0.898 0.5 0.398 V

p

QD

GS TN

TN GS

SD

K A VI

I

V V

V V VV

μ

μ

= =

= =

= −

= ⇒ == − =

i 3 1 1(min) (sat)5 0.898 0.398 0.898

N GS SD SGV V V V V−= + + −= − + + −

i

i

(min) 4.60 V

4.60 (cm) 3.54 VN

N

V

V

= −

− ≤ ≤

TYU13.9

0 8

8 5

8

0

0 7

(max) (sat)1.06 V

(sat) 1.06 0.5 0.56 V(max) 5 0.56 4.44 V(min) (sat)

SD

SG SG

SD

DS

V V VV VVVV V V

+

−

= −= =

= − == − == +

7 7

0

0

1.06 (sat) 1.06 0.5 0.56(min) 5 0.56 4.44

4.44 4.44 V

GS DSV VV

V

= ⇒ = − == − + = −

− ≤ ≤

TYU13.10 (a) For 2

5, 5 125 /pM K A Vμ=

2 55 5

2 55

25 5 5

25 5

2

5

5

( )

5 50.125( 0.5)

10012.5( 0.25) 1012.5 11.5 6.875 0

11.5 (11.5) 4(12.5)(6.875)2(12.5)

1.33 V

SGp SG TP

set

SGSG

SG SG SG

SG SG

SG

SG

V V VK V V

RV

V

V V VV V

V

V

+ −− −+ =

+ −− =

− + = −− − =

± +=

=

Then

8 7

1 2 3 4

10 1.33 86.7 A100

43.35 A2

REF Q D D

QD D D D

I I I I

II I I I

μ

μ

−= = = = ⇒

= = = = =

(b) 21 2 125 /p pK K A Vμ= =

2 41 1153

(0.02)(0.04335)o oD

Ir r kIλ

= = = = Ω

Input stage gain

1 2 42 ( )

2(0.125)(0.0867) (1153 1153) 84.9

d p Q o o

d

A K I r r

A

= ⋅

= ⋅ ⇒ =

Transconductance of 7M

7 7 72 2 (0.250)(0.0867)0.294 /

m n Dg K ImA V

= ==

7 87

1 1 577(0.02)(0.0867)o o

D

r r kIλ

= = = = Ω

Second stage gain 2 7 7 8 2

2

( ) (0.294)(577 577) 84.8

Overall gain (84.9)(84.8) 7, 200v m o o v

d v

A g r r A

A A A

= = ⇒ =

= ⋅ = ⇒ =

TYU13.11 (a) ( )1 6 8d m o oA Bg r r=

1 1

1

802 2 (20)(50)2 2

400 /

nm D

m

k Wg IL

g A Vμ

′⎛ ⎞⎛ ⎞ ⎛ ⎞= =⎜ ⎟ ⎜ ⎟⎜ ⎟⎝ ⎠ ⎝ ⎠⎝ ⎠=

( )

66

88

1 1 0.333 (0.02)(150)

1 1 0.333 (0.02)(150)

3(400) 0.333 || 0.333 200

oP D

on D

d d

r MI

r MI

A A

λ

λ

= = = Ω

= = = Ω

= ⇒ =

(b) 12 ( )PD

o L P

fR C Cπ

=+

where 6 8|| 0.333 || 0.333 o o oR r r M= = Ω

( ) 6 12

3

1 477 2 0.333 || 0.333 10 2 10

(477 10 )(200) 95.5

PD PD

PD d

f f kHz

f A MHzπ −= ⇒ =

× × ×⋅ = × ⇒

TYU13.12 (a) From Exercise TYU 13.11, 1 400 /mg A Vμ=

6 8 10 12 0.333 o o o or r r r M= = = = Ω

10 10 10

12 12 12

402 2 (20)(150) 490 /2 2

802 2 (20)(150) 693 /2 2

Pm D m

nm D m

k Wg I g A VL

k Wg I g A VL

μ

μ

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

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

( )

10 10 10 6

12 12 12 8

1 10 12

( ) (490)(0.333)(0.333) 54.4 ( ) (693)(0.333)(0.333) 77.0

|| 3(400)(54.4 || 77.0) 38, 254

o m o o

o m o o

d m o o d

R g r r MR g r r MA Bg R R A

= = = Ω= = = Ω

= = ⇒ =

(b) 10 12|| 54.4 || 77.0 31.9 o o oR R R M= = = Ω

6 12

3

1 2.50 2 (31.9 10 )(2 10 )

(2.5 10 )(38, 254) 95.6

PD

PD d

f kHz

f A MHzπ −= =

× ×⋅ = × ⇒

TYU13.13 (a) ( )1 6 8||d m o oA g R R= From Example 13.10,

1 316 / ,mg A Vμ= 8 316 oR M= Ω Now

6 6 6 4 1

1 4

66

66

6

( )( || )1 , 0.5

50 1.923 /0.026

80 1.6 50

o m o o o

o o

Cm

T

Ao

C

R g r r rr M r M

Ig mA VV

Vr M

I

== Ω = Ω

= = ⇒

= = = Ω

Then 6 (1.923)(1600)(0.5 ||1) 1026

(316)(1026 || 316) 76,343o

d d

R MA A

= = Ω= ⇒ =

(b) 6 12

1 329 2 (316 ||1026) 10 2 10

(329)(76,343) 25.1

PD PD

PD d

f f Hz

f A MHzπ −= ⇒ =

× × ×⋅ = ⇒

TYU13.14 For 7Q and 1R

7 1 1 10.6 (5)SG BEV V I R I= + = + For 8 :M

22

22

( )0.3( 1.4)

p SG TP

SG

I K V VI V

= += −

By trial and error:

1 2

2.54 V

0.388 mASGV

I I

=

= =

TYU13.15 For 6J biased in the saturation region

3 300 AC DSSI I μ⇒ = =

1,Q 2 ,Q 3Q are matched

1 2 3 300 AC C CI I I μ⇒ = = =