Microelectronics, Circuit Analysis and Design by Donald A. Neamen, 4th edition-solutions

Power ratings depends on number of pulses per second and duration of pulse.

17

1.42.10 10 250 exp

2 86 10 2508.301 10 exp 32.56

6

18

1.42.10 10 350 exp

2 86 10 3501.375 10 exp 23.26

18

1.15.23 10 100 exp

2 86 10 1005.23 10 exp 63.95

6

19

1.15.23 10 300 exp

2 86 10 3002.718 10 exp 21.32

6

19

1.15.23 10 500 exp

2 86 10 5005.847 10 exp 12.79

1076.510

104

1025.210

105

n

n

1.5

16

2 6 2

1024.310

108

1076.510

104

16

5 10 cm1.5 10

From Problem 1.1(a)(ii) n i =3.97 10 cm× 11 −3

1.7

(a) p-type; p o =5×1016 cm ; −3 ( ) 3

16

2 10 2

105.4105

105

1048.6105

108

10125.1102

105

(c) n o ≅N d = ×5 10 15 cm−3

4.0102.1

=Ε

6.1

67

n

e N N

e

μ

σμ

6.1

11

ρ

n d d

6.1

5

e N N

e

μ

σμ

400106.1

8

1.16

D n =(0.026)(1250)=32.5 cm2/s; D p =(0.026)( )450 =11.7 cm /s 2

001.00

10105.32106.1

12 16

10107.11106.1

16 12

dp

dx

x eD

x J

bi

n

N N V

V

105105ln026

10

15 15

10

15 17

10

18 18

(b) (i) ( ) ( )( )

105105ln026

6

15 15

6

15 17

6

18 18

bi

n

N N V

1076.1026.0

712.0exp10

105.1

i

a

V

V N

V

V C

(1.5 10 ) 0.739

10105ln026

10

17 15

11

60.0

=+

31

60.0

=+

51

60.0

=+

V

V I

I

(a) (i) ( ) 1.03μ

026.0

3.0exp

5.0exp

7.0exp

02.0exp

20.0exp

3.0exp

5.0exp

7.0exp

(iv) ( )13 1 5.37 10 14

026.0

02.0exp

D

I

I V

10

1010ln026

20ln026

4.0exp10

65.0exp10

35.0exp10462

25.0exp10462

8.0exp

0.1exp

2.1exp

02.0exp

8.0exp10

0.1exp10

2.1exp10

02.0exp10

I T

T I

0.6

0.018652.147 10 1.237 109.374 10(100)

2.83 10( 55)

D

V I

By trial and error,

V D =0.282 V, I D =2.52μA

(b)

I D ≅−5×10− 11 A, V D=−2.8 V

1.39

10=I D(2 10× 4)+V D and (0.026 ln) 12

10

D D

80

TH

R V

3 2

10ln026

10ln026

1= = i =

D D

I I

10

105.0ln026

3 2

105

13 14

2 1

D

I

I I I

100909.0ln026

10909.0ln026

635.0exp10

10061.3ln026

3 2

10696.3ln026

3 2

I mA, V O =(0.235)( )20 −5=−0.30 V

(c) (i) ( ) 0.372

25

87.0

3.025 0.652.375 mA

026.

1.54

pn junction diode

105

1072.0ln026

1072.0ln026

3

0 5 10

0 4796exp

S

a T

I

V

V

8 1

12

0 02610

8.6

8

Z

L

I

V R R

Now

T avg

T O

ππυ

υ

2 0 0

6.0sin102

11

9809 0 01911 0

9809 0 01911 0

6.0cos

102

1

x x

π 10 0.9982 0.9982 0.60.9809 0.019112

2.3

10

12

7.0

9869 0 01313 0

7.0sin97.162

9869 0

01313 0

9869 0

01313 0

7.0cos

97.162

1

x x

π 16.97 0.99915 0.99915 0.70.97382

7762 0 2238 0

7.9sin152

1

ππ

π π

π π

5523.07.97628.07628.0152

17

.9cos

152

2238 0

7762 0

2238 0

2.5

2.1

7.915

R R

peak peak

7762 0 2238 0

7.9sin151

Or from Problem 2.4, υR( ) (avg =20.9628)=1.9256V

417.4

9256.1

A (c)

360

29.4071.139

2120

From part (a) PIV =2v S(max)−Vγ =2 26.4( )−0.7

or PIV =52.1 V or, from part (b) PIV =2 101.4( )−0.7 or PIV =202.1 V

3.10

=

36.0

122124122

r

M M

D

V

V R

V peak

i

i D(peak)=13.3A

=

36.0

1222

124

1212

36.0212

212

V V

V avg

2120

9219092

r

M M

D

V

V R

V peak

⋅

=

2.0

922

190

99

2.0212

212

V V

V avg

Voltage Divider

0

1

12

4.15min

L

R

( )= =410Ω

03759.0

4.15max

I

R R

15 0.9 10 0.1 20

5000 2504max 1.1875 A

20 10

8.081187.5 50

6.5

6.550

769.5

769.5

6.5

6.550

88.5

88.512

2.25

(a) Set I Z =10mA; 7.5

1

5

(b) 7.5=V ZO+(0.01)( )12 ⇒V ZO =7.38V

For V I =( )( )1.1 12 =13.2V

100012

38.7257

38.7257

450.7556.7

12257

38.712

⇒+

50.7586

Then I Z(max)=0.1 0.02 0.12 + = A and

3.6+

3.65

(a) For − ≤10 v I ≤0, both diodes are conducting ⇒v O =0

For 0≤v I ≤ 3, Zener not in breakdown, so i1=0, v O =0

( )

1

3For 3

(b) For v I <0, both diodes forward biased

1

0

.10

3

3,

20

I I

.7 (b) i D=0 for 0≤v I ≤5

Then for v I >5.7 V

3.422.5

I I

I O

D

v v

I D

L will tend to block the transient signals

D z will limit the voltage to +14 V and −0.7 V.

Power ratings depends on number of pulses per second and duration of pulse

2.39

(a) Square wave between +40V and 0

(b) Square wave between +35V and −5V

(c) Square wave between +5V and −35V

R

I R3=0.5+0.5=1.0mA

( ) 4.4

0.1

56.0

56.0

2 1

V1=10−0.6−( )( )1.11 3 =6.07V

( ) 1.76

5.2

56.0

3 2

2.49

(a) D1 and D2 on

1.1

57.02

7.0

12

1909.315

57.03.2

6.05

.05

55

6.0

+

=

−++

5.0

7.015

107.03

.3

57.015

57.0991.0

107.0991

3.2410

7.015

4 1

+

=+

V A=15−(3.25)(6.15)=−5V

1.4 15

2.72 mA5

2.72 1.071.65 mA

10

+

−+

+

=

−

1010

5

10

1.210

1.2

7.010

10

+

−+

7.04

7.18

7.04

+

−+

V I

Chapter 3

3.1

8.0

102

DS DS

V V

V V

−

−

2.15

(sat) 0.0640(3.5 1.5) (sat) 0.256 mA for 3.5 V

(sat) 0.0640(4.5 1.5) (sat) 0.576 mA for 4.5 V

V V

14 4

t L

10200

1085.89.3600

n

n

t C

W

Then W =(4.026)( )0.8 =3.22μm

1085.89

05.0

05.0

122

63.7 k0.01 1.57

100 V0.01

1054.010200

11

o

I r I

D D

I I

D D

I I

=

3.24

Εox =6×106 V/cm

(a) (i) =Ε =(6×106)(120×10− 8)=7.2V

ox ox

3.26

( )2

3.6 2.046

0.777 mA2

5 0.741

1.42 mA3

1.33 mA1.2

R

V V V

2

12

=

=+

=

R R

3.3

W

Now V G =V GS +I D R S =0.8+( )( )0.2 1 =1.0V

1 1 1 ( )( )200 1.8 1 360

1 1

R R =R =200kΩ ⇒ R =450kΩ

−+

GS S

D GS

L

W k V R

I V

V

( )( )25 0.5( 0.8 0.16)

2

12.02

3232.1

=

=+

=

R R

30036536

2 1

=

R R

4.38 K0.5

2

5 2.18

11.3 K0.25

2.18 2 4.18 V

D

DQ p

2 2

R

R R

6.002124.06

.05.1

2

4

O O

W

2 1

n TN GS n

V V L

W k V

V L

2

126.06.059

.12

126.06.031

1 0.6 0.23247 0.6523

.23138.913167

.010.06.052

12.009733

W

08.008.04.0322

12.009733

W

12.02

W L

W

(b) M1 nonsaturation, M2 cutoff

6.052623.02

12.050

5

O O O

2 2

3 2

(b) V GS1=0.9V⇒V D1=V DS1−V GS1=2−0.9=1.1V

125.0

1.15

L

W L

W I

REF

L

W L

W I

(b) V SGA =1.2V

V DA =V SGA−V SDA=1.2−4=−2.8V

20.0

58

0

2 2

0

3 2

0

4 2

V V

V

V V

5.85 mA0.2

3.64

( )2

V

V V

2

2 2

1 2 1 V1

R

R R

GS

V

V V

R

R R

V V

1.025

W

(ii) (8.33) ( 0.4) 1.0

2

1.015

2

04.02

W

2

04.050

DS D

11

DS D

o

V I

102%

500

DS D

o D

D

V I

r I

I

μΔ

2

1.02

W

2

08.022

5.0015.0

11

Ratio of signal at 2 to that at :

27.002.0

11

−

=

Si D

o m

R R R

R R R r g

A

2 1

2 1 υ

We find r o R D =1858=7.668 kΩ

R1 R2 =60240=48 kΩ

248

48668.7078

4.15

2251751752

D m

R g

R g

456

m

L D m

R R R

R R

g

R R g

156.11

56.1

+

−

=+

103464.3

+

−

=+

−

=

S m

L D m

R g

R R g

g R

A

g R

R R

m S

g R A

g R g

D

g g

R

+

=mA/V

(b)

( )( ) ( )

1 1610

0.6 K

v

S S

A

R R

L W k I

5.002.0

+

−

=+

−

=

S m

L D m

R g

R R g

(b) g m=2 K n I DQ =2 ( )(2 1.244)=3.155 mA/V

009.0155.31

22155.3

+

−

=+

−

=

S m

L D m

R g

R R g

Aυ

4.24

1.059 1.5 0.441 V0.441 3

10.2 K0.25

DQ p SGQ TP

SGQ SGQ

1 2.2 1.33

2.2

1 2.93 2.2

13.93 2.2

11

0.614 3 2.386 V2.386 10

15.2 K0.5

1005

+

=+

=

o m

o m

r g

r g

Aυ

= = 100=0.2100⇒ ≅200Ω

5

11

o o

m

g

R

(b) ( )

(o S)

m

S o m

R r g

R r g

762.4

0.49 K

m o

v

m o v

4.33

15.010

5.10

15.002.0

11

708.9549.1

+

=+

=

S o m

S o m

R r g

R r g

Aυ

549.1

11

25.15.2

DQ

R

V V

(b)

S m

S m

R g

R g

5.085

2

1.02

1.02

W

5098.11

5098.1

+

=+

=

o m

o m

r g

r g

Aυ

2

1.0

11

o o

11

778.1828.21

=+

=+

=

L S o m

L S o m

R R r g

R R r g

Aυ

4

11

o

g

R

i i

A

υ

υυ

υυ

(d) 416 0.35363.2

828.2

11

1

=

DQ o

I r

R r g

R r g A

857.2472

11

R o= 219Ω

(b) (i) g m =2 ( )( )1 2 =2.828mA/V

( )( ) 25

202.0

448.3828

11

40.95

12

2

m

m L v

20 k0.01 5

W

2

1.0

9524.05

+

=+

=

S o m

S o m

R r g

R r g

Aυ

5

11

o S

o m

6452.05

+

=+

=

L S o m

L S o m

R R r g

R R r g

5 4.09 2

0.870

m m

1014.14

+

=+

=

o m

o m

r g

r g

Aυ

(b) = = 10=0.0707210⇒ =70.2Ω

14.14

11

o o

R r g

R r g

R r

R r

14.141

14.1490

V V

15

.01

2

1

3.04.02.12.25

.0

3.02

=+

D

DS DD

DQ

I DQ =2.2mA

(c) ( )2

TN GSQ n

11

i m

1101

i i

L

R R

15.08

2

1.02

4.51

TN GSQ n Q

11

i m

493.13.3

I

V V

D

L W

L W K

1

516.06.03.3

=+

++

1.0

GSDQ D

V V L

W V

So ( )0.8 0.6 0.9266

2.1

2

For point A: V OtA =0.3266V, V GSDQ =0.9266V

For point B: V OtB =4.2V, V GSDQ =0.9266V

(b) V GSDQ =0.9266V,

2

3266.02

11

oD

I r

1

r V

V K r

TN DS n

(b) I D =(0.6173)(3−0.6)2 =3.56mA

= 2(0.6173)(3−0.6)⇒ =337Ω

1

r r

2 0

4.0830

0.40 k

LD m

1

=

DQ D oD

I r

5.004.0

11

=

=

=

DQ L

oL

I r

2

1.02

=

DQ D oD

I r

25.002.0

11

=

=

=

DQ L oL

I r

2

04.02375

11

=

=

=

DQ D oD

I r

1.002.0

11

=

=

=

DQ L oL

I r

2

04.0215

W

50 2.125 mA/V2

133.3 K0.075 0.1

2 1

200 K0.05 0.1

25

V V

2

04.02

121.0025.0

oD

I r

3.1656957.0

+

=+

=

oL oD mD

oL oD mD

r r g

r r g

o

K

K g

g V

V

5

200 K0.05 0.1

133.3 K0.075 0.1

1.085

0.956

0.956 0.922 200 133.370.5

(2)

2

2 2 2

4.68

1 1 1

6.25

2 1

2 1

R

R R R

907.35

907

2

=11

3.13

3

D

V S2 =V D1+V SGQ2 =1.3+1.1=2.4V

25.0

4.23

0

6.0

.0

8.12

.0

8.14739.0

2 2 1 1

1 11

S m S m

D m

R g

R g R g

R g

42.4191.264.01

204

+

⋅+

−

=

υ

A

(c) = = 4.42=0.45644.42⇒ =414Ω

191.2

11

2 2

o S

m m v

10 6.12

38.8 k0.1

VGS GS GS

V V V

10 V2

5 mA2

5 k265

265

R

R R

//

90 110 49.5 k49.5

Chapter 5

5.1

8.2

3251

80.1

9918.0

980074.0

βαβ

5.4

25.1

=+

V

V I

1080.0ln026

9910.0

6160

0.983661

76.1

E

V

V I

10785.1ln026

615.0exp105

T

BE S

C

V I

15 1

1.69 10

6.94 10

Eo Eo

I I

5.12

For transistor A:

108

10275ln026.0

I

I V

10275ln026

C

V I

C B

C E

C E

BV BV

C B

C E

BV BV

βββ

3

3 3

BV

ββ

75

760.7 4 3.3 V

1.93 mA

10 3.3

3.47 K1.93

C

V

V I

105

1035.0ln026

(b) 0.7

2

1.15

E

R

V V

106923.0ln026

2.07.0

1

2.0

5.22

120005.0

7.03

R

on V V

I CQ =βI BQ =( )(100 0.005)=0.5mA

5.0

5.1

1 0.7 1.7 V

0.2708 mA4.8

0.2708

0.0020.99261

E

E

E

E E

V

V

I

R I

91

90

1 ββ

9010

242

67.80

3

E

V V

10

8371.03

73

C

I

V V

80

25

5.33

a

( )0

0

0

10Cutoff 5

10 53.33 V

BB

L CC

(iii) 2 mA Transistor is in saturation( ) ( ) 0.7 0.2 0.5 V

5.38

2

4 0.16 mA

25

0.057 mA

0.057 0.16 0.217 mA0.217 15 0.7 3.96 V

0.00417 mA120

5 0.7

1032 K0.00417

R R

2 2

−

= O E

V I

( )1.25 1.45

4

51

2.0

So 1.6, 4.8

O C

I

V I

5.44

(a) For V I ≥4.3, Q is off and V O =0

When transistor enters saturation, 5 101 ( )1 0.2 ( )4 0.958 mA

2.0

C

R

sat V V

I

mA

2.1

7.0

R

on V V

−

=++

−

E B

BE I

R R

on V V

8

I B

R

V on V V

8

I

mA

704.0

57.099

R

V on V

7.0

B

I

on V V

(b) ΔV O =3.3−0.2=2.8V, peak-to-peak

0.17.0

6.1

8.2max

V TH =I BQ R TH +V BE( )on +I EQ R E =(0.00125)( )200 +0.7+(0.15125)( )2 =1.2525V

R

V R R

152

R

EQ EB

E EQ

+++

=

β1

101

57.87

.01

085.1

01.5101

57.81

29.47.0

⋅+

7.008

I V

242

+

−

=+

+

−

E TH

BE TH

R R

on V V

I CQ =βI BQ =( )(80 0.01098)=0.8782mA, I EQ =( )(81 0.01098)=0.8892mA

V CEQ =9−(0.8782)(5.25) (− 0.8892)( )1 =3.50V

7.080

8782.09415

500.3108

From this quadratic, we find R2 =48 64.5kΩ ⇒ R1 = kΩ

(b) Standard resistor values:

R

V+ =I R +V ( )on +I R +V

So ( )

25.17.05.2

TH EB

R R

V on V V

5914.06563

135.19874

5.58

(a)

R TH =3668=23.5kΩ; ( )10 5 1.54

6836

ECQ

I V

76

For β =75, 0.0103I BQ =5.051.145 0.7( )( )76 0.5− = mA

+ Then I CQ =( )(75 0.0103)=0.775 m A

For β =150, 0.00552I BQ =5.051.145 0.7( )( )151 0.5− = mA

+ Then I CQ =0.829 mA

I I

I I

=

ββ

774

So 2.34 ( )( )121 0.2 9.08μ

7.0059.2

066

+

E C

TH EB

R R

V on V V

R

R

R R

R R

6.15

=

−

=+

CQ

CEQ CC E C

I

V V R

−

=++

−

E TH

BE TH

R R

on V V

For β =120,

5.05 ( )( )121 0.5 1.695μ

7.08111

1

on1

R R

R R

BQ

C CQ

C

I

I I

I

−

+Δ

Design criterion is satisfied

which gives 97.3 , and 48.4

2

2

1.67 mA3

1

0.0165 mA1

E E

Ω

2

.3⇒R =37.8 kΩ

5.70

a R TH =R1 R2=1020=6.67kΩ

1020

205

102 1

10 0.7 1.67 7.63

0.0593 mA 6.67 61 2 128.7

3.56 mA, =3.62 mA

2.76 V

10 3.56 2.2 10 2.17 V

R TH =( )(0.1 1+β)R E =( )( )( )0.1 61 10

Or R TH =61 kΩ

( )2

−

−

E C

80

8115

.05

−

I EQ =0.1519mA, I BQ =1.875μA

R =( )( )( )0.1 81 2 =16.2kΩ

V+ =I EQ R E+V EB( )on +I BQ R TH+V TH

2.5=(0.1519)( )2 +0.7+(0.001875)( )16.2 +V TH ⇒V TH =1.466V

1 ( )5 2.5 1.466 1 ( )( )16.2 5 2.5

1 1

EB TH

R R

on V V V

145.01531

777.2655

3500

=

−+

−V TH V TH V TH

70

1500

1500

170

5500

5.74

8.0

5

11

(0.00667)(22.6) 0.7 (0.807)(1.87) 1 (22.6)(10

1

R

=+

R

R R

R

R V

R R

which yields R1=68.7kΩ and R2 =22.4kΩ

(b) For standard resistor values:

Let R E =1.5kΩ , R C =5.6kΩ, R1=68kΩ , R2=22kΩ

R TH =R1 R2 =6822=16.62 kΩ

6822

223

62 1

3

=

⇒+

−

−

=+

+

−+

E TH

BE TH

R R

on V V

7

3

6− =

=+ E

TH EB

R R

V on V V

I CQ =0.1027mA, I EQ =0.1034mA

V ECQ =6−(0.1027)( ) (23 − 0.1034)( )7 =2.914V

5

689

182 1

TH EB

R R

V on V V

4010

2 1

−

=+

+

−

=

E TH

BE TH

B

R R

on V V

0.7 0.7 1.4

3.6 mA1

0.0444 mA3.56 mA

2.1

2.5

2.0

5

2.14.22

1 2

R

B B

I

V V

020.0

4.25

V C2=2V CE+V RE =2( )1.2 +0.5=2.9V

20.0

9.252

I

V V

5.82

R TH =4080=26.67kΩ

8040

+

−

E n TH

BE TH

R R

on V V

I I

V

=+

−

3.81.01

7.091

0

2 1

p

C B

C EB

E

V V

I V on V

=+

−

−

=

⇒++

=

β

1.8

3.82

9

1 1

−

C C

214.73

1005

102 1

I

I

R R

5.21 k0.792

Chapter 6

6.1

026.0

5

026.0

25

026.0

08

026.0

026.0

2

( )( ) 5.85

8.0

026.0

026.0

=π

12.012

=+

346

R

on V V

026.0

−

=

11015.7

15.7183478.16

B o

C m

R r

r r R g

A

π

π υ

Aυ =−4.0

t t

120 0.026

5.40 k0.578

0.578 22.2 mA/V0.026

100 173 k0.578

6.7

( )( ) 6.24

5.0

026.0

100 0.026

1.73 K1.5

t A

0.50.005100

100 0.026

5.2 k0.5

20.60sin mA

100

2 K50

ω μβ

2 2

026.0

+

−

=++

r

R A

E C

502

V ECQ =V CC−I CQ R C −I EQ R E =3.3−(0.506)( ) (2 − 0.511)( )1

V ECQ =1.78V

506.0

026.0

+

−

=++

R A

β

βπ

5

908.17.03.3

026.0

5

066.27.03.3

−

=+

r

R A

E C

I

R R

100 11.3

55.110.4 101 0.1

1

50 10.4 101 0.1

C v

6.16

22.2

20.5

+

−

=+

−

≅

E C

CEQ CC

CQ

R R

V V

026.0

0

026.0

8873

21512.24

7.0707.2

6.17

(a) (i) 0.009877

81

8

2

79

−

=

448.2

48.2

s m

448.2

48.238

i G

(b) (i) 0.00661

121

8

2

7934

459.252

25

026.0

=π

5.248.12

48.1259.10615

−

=

S L

C m

R r

r R R g A

π

π υ

(iii) υo =−(27.49) (5×10− 3sinωt)=−0.137sinωt (V)

6.19

(a) (i) 2.5 ( )( )81 10 0.005292

7.0

4234

026.0

=π

5.291.4

91.45528

−

=

S L

C m

R r

r R R g A

π

π υ

L s L

o o

R

A R

i G

7.05

4256

026.0

=π

5.233.7

33.75537

1.35 9.64 1.184 K1.184

S S

77.2

(c)

S B

B

R r R

r R g

−

=

π

π υ

026.0

28.43

B

R r R

r R g

−

=

π

π υ

R B rπ =107.49=4.28kΩ

5.028.4

28.43

026.0

r R R g

2 1

2 1

R1 R2 rπ =61.51.67=0.698kΩ

( ) ( )1.2 45.8

2.0698.0

698.0

=

υ

A