Microelectronics circuit analysis and design donald a neamen 4th edition solutions

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D... Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D... Microelectronics: Circuit Analys

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Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1

By D A Neamen Problem Solutions

6 19

1.15.23 10 250 exp

2 86 10 2502.067 10 exp 25.58

6 19

11 3

1.15.23 10 350 exp

2 86 10 3503.425 10 exp 18.27

6 17

6 18

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6 18

6 19

0.66

2 86 10 10035.9 cm

1076.510

104

1025.210

105

n

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1.5

(a) p-type; p o =1016 cm ; −3 ( ) 4

16

2 6 2

1024.310

108

1076.510

104

16

5 10 cm1.5 10

105.4105

105

1048.6105

108

10125.1102

105

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102

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

(b) ( 19) ( ) 1.25 1016

400106.1

8

Trang 7

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

(a) (i) ( ) ( )( )

(1.5 10 ) 0.661

105105ln026

10

15 15

10

15 17

10

18 18

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

(1.8 10 ) 1.13

105105ln026

6

15 15

6

15 17

6

18 18

bi

n

N N V

1076.1026.0

712.0exp10

105.1

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Trang 10

j

V

V C

(1.5 10 ) 0.739

10105ln026

10

17 15

11

60.0

=+

31

60.0

=+

51

60.0

=+

12

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(c)

(1.5 10 )(0.215 10 ) 8.862

D

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

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026.0

02.0exp

D

I

I V

10

1010ln026

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2ln026

20ln026

4.0exp10

65.0exp10

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1.33

3 14 3

30.0exp10

35.0exp10462

25.0exp10462

8.0exp

0.1exp

2.1exp

02.0exp

8.0exp10

0.1exp10

2.1exp10

02.0exp10

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S

I T

T I

0.6exp(100) (2.147 10 ) 0.0322

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

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1.39

10=I D(2 10× 4)+V D and (0.026 ln) 12

10

D D

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3 2

10ln026

I I

10

105.0ln026

3 2

105

13 14 2

1 2

D

I

I I I

100909.0ln026

10909.0ln026

635.0exp10

10061.3ln026

3 2

10696.3ln026

3 2

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5− − − =

=

I mA, V O =(0.235)( )20 −5=−0.30 V

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(c) (i) ( ) 0.372

25

87.0

3.025 0.652.375 mA

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026

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1.54

pn junction diode

105

1072.0ln026

0.5 10

ln (0.026) ln

5 100.1796

3

0 5 10

0 4796exp

V

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8 1

12

0 02610

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0

8.6

8

Z

L

I

V R R

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Now

( ) ( )t dt [ x ]dx

T avg

T

O

ππυ

υ

2

0 0

6.0sin102

11

9809 0 01911 0

6.0cos

102

1

x x

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2.3

10

12

7.0

9869 0

01313 0

7.0sin97.162

9869 0 01313 0

7.0cos

97.162

1

x x

7762 0 2238 0

7.9sin152

1

ππ

π π

5523.07.97628.07628.0152

17

.9cos

152

2238 0

7762 0 2238 0

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2.5

(a) ( )= ( )⇒ = − =4.417Ω

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

( ) ( ) 0.436

417.4

9256.1

i D υR

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

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By D A Neamen Problem Solutions (b)

3.10

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2120

=

36.0

122124122

r

M M

D

V

V R

V peak

i

i D(peak)=13.3A

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⋅

=

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

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Voltage Divider

0

1

12

Trang 33

4.1560

4.15min

L

R

( )= =410Ω

03759.0

4.15max

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6.5

6.550

769.5

6.5

6.550

88.5

88.512

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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

450.7556.7

38.712

⇒+

50.7586

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

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35

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

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

1

0

.10

0 1 0

0 0

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I I

D

v v

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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

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07.6 0.589 7.01 mA

R

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I R3=0.5+0.5=1.0mA

( ) 4.4

0.1

56.0

2 1

V1=10−0.6−( )( )1.11 3 =6.07V

( ) 1.76

5.2

56.0

3 2

Trang 45

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

Trang 46

57.02

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

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2.72 mA5

2.72 1.071.65 mA

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7.010

10

+

−+

+

=

−

1010

5

10

1.210

1.2

7.010

10

+

−+

Trang 49

7.05

7.04

7.18

7.04

+

−+

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I

V I

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Microelectronics: Circuit Analysis and Design, 4 edition Chapter 3

Chapter 3

3.1

8.0

102

DS DS

V V

−

2.15

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