Micro4ProbSolutionsChapter1

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

Chapter 1 1.1 / 23/ 2 gE kT

in BT e−= (a) Silicon

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

3 / 2156

19

8 3

1.15.23 10 250 exp2 86 10 250

2.067 10 exp 25.581.61 10 cm

i

i

n

n

−

−

⎡ ⎤−⎢ ⎥= ××⎢ ⎥⎣ ⎦

= × −= ×

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

3 / 2156

19

11 3

1.15.23 10 350 exp2 86 10 350

3.425 10 exp 18.273.97 10 cm

i

i

n

n

−

−

⎡ ⎤−⎢ ⎥= ××⎢ ⎥⎣ ⎦

= × −= ×

(b) GaAs

(i) ( )( ) ( )( )

( ) [ ]

3/ 2146

17

3 3

1.42.10 10 250 exp2 86 10 250

8.301 10 exp 32.56

6.02 10 cm

i

i

n

n

−

−

⎡ ⎤−⎢ ⎥= ××⎢ ⎥⎣ ⎦

= × −

= ×

(ii) ( )( ) ( )( )

( ) [ ]

3/ 2146

18

8 3

1.42.10 10 350 exp2 86 10 350

1.375 10 exp 23.26

1.09 10 cm

i

i

n

n

−

−

⎡ ⎤−⎢ ⎥= ××⎢ ⎥⎣ ⎦

= × −

= ×

______________________________________________________________________________________ 1.2

a. 3 / 2 exp2i

Egn BTkT

−⎛ ⎞= ⎜ ⎟⎝ ⎠

12 15 3 / 26

1.110 5.23 10 exp2(86 10 )( )

TT−

⎛ ⎞−= × ⎜ ⎟×⎝ ⎠

3

4 3/ 2 6.40 101.91 10 expTT

− ⎛ ⎞×× = −⎜ ⎟

⎝ ⎠

By trial and error, 368 KT ≈ b. 9 310 cmin −=

( )( )

9 15 3 / 26

1.110 5.23 10 exp2 86 10

TT−

⎛ ⎞−⎜ ⎟= ×⎜ ⎟×⎝ ⎠

3

7 3 / 2 6.40 101.91 10 expTT

− ⎛ ⎞×× = −⎜ ⎟

⎝ ⎠

By trial and error, 268 KT ≈ °______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 1.3 Silicon

(a) ( )( ) ( )( )

( ) [ ]

3/ 2156

18

10 3

1.15.23 10 100 exp2 86 10 100

5.23 10 exp 63.95

8.79 10 cm

i

i

n

n

−

− −

⎡ ⎤−⎢ ⎥= ××⎢ ⎥⎣ ⎦

= × −

= ×

(b) ( )( ) ( )( )

( ) [ ]

3 / 2156

19

10 3

1.15.23 10 300 exp2 86 10 300

2.718 10 exp 21.32

1.5 10 cm

i

i

n

n

−

−

⎡ ⎤−⎢ ⎥= ××⎢ ⎥⎣ ⎦

= × −

= ×

(c) ( )( ) ( )( )

( ) [ ]

3 / 2156

19

14 3

1.15.23 10 500 exp2 86 10 500

5.847 10 exp 12.79

1.63 10 cm

i

i

n

n

−

−

⎡ ⎤−⎢ ⎥= ××⎢ ⎥⎣ ⎦

= × −

= ×

Germanium.

(a) ( ) ( ) ( )( )( ) [ ]3 / 215 18

6

3

0.661.66 10 100 exp 1.66 10 exp 38.372 86 10 100

35.9 cm

i

i

n

n

−

−

⎡ ⎤−⎢ ⎥= × = × −×⎢ ⎥⎣ ⎦

=

(b) ( )( ) ( )( )( ) [ ]3 / 215 18

6

13 3

0.661.66 10 300 exp 8.626 10 exp 12.792 86 10 300

2.40 10 cm

i

i

n

n

−

−

⎡ ⎤−⎢ ⎥= × = × −×⎢ ⎥⎣ ⎦

= ×

(c) ( )( ) ( )( )( ) [ ]3 / 215 19

6

15 3

0.661.66 10 500 exp 1.856 10 exp 7.6742 86 10 500

8.62 10 cm

i

i

n

n

−

−

⎡ ⎤−⎢ ⎥= × = × −×⎢ ⎥⎣ ⎦

= ×

______________________________________________________________________________________ 1.4

(a) n-type; cm ; 1510=on 3− ( ) 1115

2132

1076.510

104.2×=

×==

o

io n

np cm 3−

(b) n-type; cm ; 1510=on 3− ( ) 515

2102

1025.210

105.1×=

×==

o

io n

np cm 3−

______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 1.5

(a) p-type; cm ; 1610=op 3− ( ) 416

262

1024.310

108.1 −×=×

==o

io p

nn cm 3−

(b) p-type; cm ; 1610=op 3− ( ) 1016

2132

1076.510

104.2×=

×==

o

io p

nn cm 3−

______________________________________________________________________________________ 1.6 (a) n-type (b)

( )

16 3

21023 3

16

5 10 cm

1.5 104.5 10 cm

5 10

o d

io

o

n N

np

n

−

−

= = ×

×= = = ×

×

(c) 16 35 10 cmo dn N −= = ×

From Problem 1.1(a)(ii) 11 33.97 10 cmin −= ×

( )211

6 316

3.97 103.15 10 cm

5 10op −×

= = ××

______________________________________________________________________________________ 1.7

(a) p-type; cm ; 16105×=op 3− ( ) 316

2102

105.4105105.1

×=××

==o

io p

nn cm 3−

(b) p-type; cm ; 16105×=op 3− ( ) 516

262

1048.6105108.1 −×=

××

==o

io p

nn cm 3−

______________________________________________________________________________________ 1.8

(a) Add boron atoms (b) cm 17102×== oa pN 3−

(c) ( ) 317

2102

10125.1102105.1

×=××

==o

io p

nn cm 3−

______________________________________________________________________________________ 1.9 (a)

( )

15 3

21024 3

15

5 10

1.5 104.5 10

5 10

o

io o

o

n cm

np p

n

−

cm−

= ×

×= = ⇒ = ×

×

(b) n-type ⇒> oo pn (c) 15 35 10 o dn N cm−≅ = ×______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 1.10 a. Add Donors 15 37 10 cmdN −= ×

6 3 210 cm /o i d

n N−= = b. Want p

So ( )( )2 6 15

2 3

10 7 10 7 10

exp

in

EgB TkT

= × = ×

−⎛ ⎞= ⎜ ⎟⎝ ⎠

21

( ) ( )( )221 15 3

6

1.17 10 5.23 10 exp86 10

TT−

⎛ ⎞−⎜ ⎟× = ×⎜ ⎟×⎝ ⎠

By trial and error, 324 KT ≈ °______________________________________________________________________________________ 1.11

(a) mA ( )( )( ) 15.0105.110 5 =⇒=Ε= − IAI σ

(b) ( )( )( ) 4.2

1024.0102.1

4

3

=×

×==Ε⇒

Ε=

−

−

AIAI ρ

ρV/cm

______________________________________________________________________________________ 1.12

( ) 167.618120 −−Ω==

Ε=⇒Ε= cmJJ σσ

( )( )( )

1619 1033.3

1250106.167.6

×=×

==⇒≅−

nddn e

NNeμσμσ cm 3−

______________________________________________________________________________________ 1.13

(a) ( )( )( )15

19 1069.765.01250106.1

111×=

×==⇒≅

−ρμμρ

nd

dn eN

Necm 3−

(b) ( )( ) 10416065.0 ===Ε⇒Ε

= JJ ρρ

V/cm

______________________________________________________________________________________ 1.14

(a) ( )( )15

19 10375.91000106.1

5.1×=

×==⇒≅

−n

ddn eNNe

μσμσ cm 3−

(b) ( )( )16

19 1025.1400106.1

8.0×=

×== −

pa e

Nμσ cm 3−

______________________________________________________________________________________ 1.15 (a) For n-type, ( )( )191.6 10 8500n d de N Nσ μ −≅ = ×

For ( ) 115 19 3 410 10 1.36 1.36 10dN cm cmσ −−≤ ≤ ⇒ ≤ ≤ × Ω −

(b) ( ) 3 20.1 0.136 1.36 10 /J E Jσ σ= = ⇒ ≤ ≤ × A cm ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 1.16 cm 2 /s; ( )( ) 5.321250026.0 ==nD ( )( ) 7.11450026.0 ==pD cm /s 2

( )( ) 52001.0010105.32106.1

121619 −=⎟⎟

⎠

⎞⎜⎜⎝

⎛−−

×== −

dxdneDJ nn A/cm 2

( )( ) 72.18001.0010107.11106.1

161219 −=⎟⎟

⎠

⎞⎜⎜⎝

⎛−−

×−=−= −

dxdpeDJ pp A/cm 2

Total diffusion current density A/cm 2 7.7072.1852 −=−−=J______________________________________________________________________________________ 1.17

( )

( )( )( )

15

19 15

4

/

110 exp

1.6 10 15 10exp

10 10

2.4 p

p p

pp p

pp

x Lp

dpJ eDdx

xeDL L

xJL

J e

−

−

−

= −

⎛ ⎞ ⎛ ⎞− −= − ⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

× ⎛ ⎞−= ⎜ ⎟⎜ ⎟× ⎝ ⎠

=

(a) x = 0 22.4 A/cmpJ =

(b) 10 mx μ= 1 22.4 0.883 A/cmpJ e−= =

(c) 30 mx μ= 3 22.4 0.119 A/cmpJ e−= =______________________________________________________________________________________ 1.18 a. 17 3 17 310 cm 10 cma oN p− −= ⇒ =

( )262

5 317

1.8 103.24 10 cm

10i

o oo

nn n

p− −

×= = ⇒ = ×

b. 5 15 15 3

17 15 17 3

3.24 10 10 10 cm10 10 1.01 10 cm

o

o

n n n np p p p

δδ

− −

−

= + = × + ⇒ == + = + ⇒ = ×

______________________________________________________________________________________

1.19 ⎟⎟⎠

⎞⎜⎜⎝

⎛= 2ln

i

daTbi n

NNVV

(a) (i) ( ) ( )( )( ) 661.0

105.1

105105ln026.0 210

1515

=⎥⎥⎦

⎤

⎢⎢⎣

⎡

×

××=biV V

(ii) ( ) ( )( )( )

739.0105.1

10105ln026.0 210

1517

=⎥⎥⎦

⎤

⎢⎢⎣

⎡

×

×=biV V

(iii) ( ) ( )( )( ) 937.0

105.1

1010ln026.0 210

1818

=⎥⎥⎦

⎤

⎢⎢⎣

⎡

×=biV V


(b) (i) ( ) ( )( )( ) 13.1

108.1

105105ln026.0 26

1515

=⎥⎥⎦

⎤

⎢⎢⎣

⎡

×

××=biV V

(ii) ( ) ( )( )( ) 21.1

108.1

10105ln026.0 26

1517

=⎥⎥⎦

⎤

⎢⎢⎣

⎡

×

×=biV V

(iii) ( ) ( )( )( ) 41.1

108.1

1010ln026.0 26

1818

=⎥⎥⎦

⎤

⎢⎢⎣

⎡

×=biV V

______________________________________________________________________________________ 1.20

⎟⎟⎠

⎞⎜⎜⎝

⎛= 2ln

i

daTbi n

NNVV

or

( ) ( ) 16

16

22

1076.1026.0712.0exp

10105.1exp ×=⎟

⎠⎞

⎜⎝⎛×

=⎟⎟⎠

⎞⎜⎜⎝

⎛=

T

bi

d

ia V

VNn

N cm 3−

______________________________________________________________________________________ 1.21

( )( )16

2 1

10ln 0.026 ln

(1.5 10 )aa d

bi Ti

NN NV V

n

⎡ ⎤⎛ ⎞⎢ ⎥= =⎜ ⎟ ×⎢ ⎥⎝ ⎠ ⎣ ⎦

0 2

VV

For 15 310 , 0.637 a biN cm V−= =

For 18 310 , 0.817 a biN cm V−= =

______________________________________________________________________________________ 1.22

(0.026)300TkT ⎛ ⎞= ⎜ ⎟

⎝ ⎠

kT (T)3/2

200 0.01733 2828.4 250 0.02167 3952.8 300 0.026 5196.2 350 0.03033 6547.9 400 0.03467 8000.0 450 0.0390 9545.9 500 0.04333 11,180.3


( )( ) ( )( )14 3/ 2

6

1.42.1 10 exp2 86 10in T

T−

⎛ ⎞−⎜ ⎟= ×⎜ ⎟×⎝ ⎠

2ln a dbi T

i

N NV V

n⎛ ⎞

= ⎜ ⎟⎝ ⎠

T ni Vbi200 1.256 1.405 250 6.02 × 103 1.389 300 1.80 × 106 1.370 350 1.09 × 108 1.349 400 2.44 × 109 1.327 450 2.80 × 1010 1.302 500 2.00 × 1011 1.277

______________________________________________________________________________________ 1.23

1/ 2

1 Rj jo

bi

VC C

V

−⎛ ⎞

= +⎜ ⎟⎝ ⎠

( )( )( )

( )16 15

10 2

1.5 10 4 100.026 ln 0.684 V

1.5 10biV⎡ ⎤× ×⎢ ⎥= =

×⎢ ⎥⎣ ⎦

(a) ( )1/ 210.4 1 0.255 pF

0.684jC−

⎛ ⎞= + =⎜ ⎟⎝ ⎠

(b) ( )1/ 230.4 1 0.172 pF

0.684jC−

⎛ ⎞= + =⎜ ⎟⎝ ⎠

(c) ( )1/ 250.4 1 0.139 pF

0.684jC−

⎛ ⎞= + =⎜ ⎟⎝ ⎠

______________________________________________________________________________________ 1.24

(a) 1 2

1/

Rj jo

bi

VC C

V

−⎛ ⎞

= +⎜ ⎟⎝ ⎠

For VR = 5 V, 1 25(0 02) 1 0 00743

0 8

/

jC . . p.

−⎛ ⎞= + =⎜ ⎟⎝ ⎠

F

For VR = 1.5 V, 1 21 5(0 02) 1 0 0118

0 8

/

j.C . . p.

−⎛ ⎞= + =⎜ ⎟⎝ ⎠

F


0 00743 0 0118( ) 0 00962 2j

. .C avg . pF+= =

( ) ( ) ( ) ( )( ) t /C C C Cv t v final v initial v final e τ−= + −

where 3 1( ) (47 10 )(0 00962 10 )jRC RC avg .τ −= = = × × 2

or 104 52 10 . sτ −= × Then ( ) ( )1 5 0 5 0 it /

Cv t . e τ−= = + −

1 /1

5 5ln1.5 1.5

re tτ τ+ ⎛ ⎞= ⇒ = ⎜ ⎟⎝ ⎠

101 5.44 10 t s−= ×

(b) For VR = 0 V, Cj = Cjo = 0.02 pF

For VR = 3.5 V, ( )1/ 23.50.02 1 0.00863

0.8jC p−

⎛ ⎞= + =⎜ ⎟⎝ ⎠

F

0.02 0.00863( ) 0.0143 2jC avg pF+

= =

( ) 106.72 10jRC avg sτ −= = ×

( ) ( ) ( ) ( )( ) /tC C C Cv t v final v initial v final e τ−= + −

( )2 2/ /3.5 5 (0 5) 5 1t te eτ τ− −= + − = −

so that 102 8.09 10 t s−= ×

______________________________________________________________________________________ 1.25

2/1

1−

⎟⎟⎠

⎞⎜⎜⎝

⎛+=

bi

Rjoj V

VCC ; ( ) ( )( )( )

739.0105.1

10105ln026.0 210

1715

=⎥⎥⎦

⎤

⎢⎢⎣

⎡

×

×=biV V

For V, 1=RV

391.0

739.011

60.0=

+

=jC pF

For V, 3=RV

267.0

739.031

60.0=

+

=jC pF

For V, 5=RV

215.0

739.051

60.0=

+

=jC pF

(a) ( )( )

57.610391.0105.12

12

1123

=⇒××

==−−

oo fLC

fππ

MHz

(b) ( )( )

MHzff oo 95.710267.0105.12

1123

=⇒××

=−−π


(c) ( )( )

86.810215.0105.12

1123

=⇒××

=−−

oo ffπ

MHz

______________________________________________________________________________________ 1.26

a. exp 1 0.90 exp 1D DS

T T

V VI IV V

⎡ ⎤⎛ ⎞ ⎛ ⎞= − − =⎢ ⎥⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎣ ⎦−

exp 1 0.90 0.10D

T

VV

⎛ ⎞= − =⎜ ⎟

⎝ ⎠

( )ln 0.10 0.0599 VD T DV V V= ⇒ = − b.

0.2exp 1 exp 10.026

0.2exp 1exp 1 0.026

21901

2190

F

TSF

R S R

T

F

R

VVII

I I VV

II

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

⎠⎞ −− ⎜ ⎟⎢ ⎥⎜ ⎟ ⎝ ⎠⎝ ⎠⎣ ⎦

=−

=

______________________________________________________________________________________

1.27 ⎥⎥⎦

⎤

⎢⎢⎣

⎡−⎟⎟

⎠

⎞⎜⎜⎝

⎛= 1exp

T

DSD V

VII

(a) (i) ( ) μ03.1026.0

3.0exp10 11 ⇒⎟⎠⎞

⎜⎝⎛= −

DI A

(ii) ( ) 25.2026.05.0exp10 11 ⇒⎟

⎠⎞

⎜⎝⎛= −

DI mA

(iii) ( ) 93.4026.07.0exp10 11 ⇒⎟

⎠⎞

⎜⎝⎛= −

DI A

(iv) ( ) 1211 1037.51026.0

02.0exp10 −− ×−=⎥⎦

⎤⎢⎣

⎡−⎟

⎠⎞

⎜⎝⎛ −=DI A

(v) ( ) 1111 101026.0

20.0exp10 −− −≅⎥⎦

⎤⎢⎣

⎡−⎟

⎠⎞

⎜⎝⎛ −=DI A

(vi) A ( )1110−−=DI

(b) (i) ( ) μ0103.0026.0

3.0exp10 13 ⇒⎟⎠⎞

⎜⎝⎛= −

DI A

(ii) ( ) μ5.22026.05.0exp10 13 ⇒⎟

⎠⎞

⎜⎝⎛= −

DI A

(iii) ( ) 3.49026.07.0exp10 13 ⇒⎟

⎠⎞

⎜⎝⎛= −

DI mA


(iv) ( ) 1413 1037.51026.0

02.0exp10 −− ×−=⎥⎦

⎤⎢⎣

⎡−⎟

⎠⎞

⎜⎝⎛ −=DI A

(v) A 1310−−≅DI

(vi) A 1310−−≅DI______________________________________________________________________________________

1.28 ⎟⎟⎠

⎞⎜⎜⎝

⎛=

S

DTD I

IVV ln

(a) (i) ( ) 359.010

1010ln026.0 11

6

=⎟⎟⎠

⎞⎜⎜⎝

⎛ ×= −

−

DV V

( ) 419.010

10100ln026.0 11

6

=⎟⎟⎠

⎞⎜⎜⎝

⎛ ×=

−

−

DV V

( ) 479.01010ln026.0 11

3

=⎟⎟⎠

⎞⎜⎜⎝

⎛= −

−

DV V

(ii) 018.01026.0

exp10105 1112 −=⇒⎥⎦

⎤⎢⎣

⎡−⎟

⎠

⎞⎜⎝

⎛=×− −−D

D VV

V

(b) (i) ( ) 479.010

1010ln026.0 13

6

=⎟⎟⎠

⎞⎜⎜⎝

⎛ ×= −

−

DV V

( ) 539.010

10100ln026.0 13

6

=⎟⎟⎠

⎞⎜⎜⎝

⎛ ×=

−

−

DV V

( ) 599.01010ln026.0 13

3

=⎟⎟⎠

⎞⎜⎜⎝

⎛= −

−

DV V

(ii) 00274.01026.0

exp1010 1314 −=⇒⎥⎦

⎤⎢⎣

⎡−⎟

⎠

⎞⎜⎝

⎛=− −−D

D VV

V

______________________________________________________________________________________ 1.29

(a) 3 0.710 exp0.026SI− ⎛ ⎞= ⎜ ⎟

⎝ ⎠

152.03 10 ASI −= × (b)

DV ( ) ( 1)DI A n = ( )( )2DI A n =

0.1 149.50 10 −× 141.39 10 −× 0.2 124.45 10 −× 149.50 10 −× 0.3 102.08 10 −× 136.50 10 −× 0.4 99.75 10 −× 124.45 10 −× 0.5 74.56 10 −× 113.04 10 −× 0.6 52.14 10 −× 102.08 10 −× 0.7 310 − 91.42 10 −× ______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 1.30 (a) 1210SI A−= VD(v) ID(A) log10ID0.10 114 68 10. −× 10 3.− 0.20 92 19 10. −× 8 66.− 0.30 71 03 10. −× 6 99.− 0.40 64 80 10. −× 5 32.− 0.50 42 25 10. −× 3 65.− 0.60 21 05 10. −× 1 98.− 0.70 14 93 10. −× 0 307.− (b) 1410SI A−= VD(v) ID(A) log10ID0.10 134 68 10. −× 12 3.− 0.20 112 19 10. −× 10 66.− 0.30 91 03 10. −× 8 99.− 0.40 84 80 10. −× 7 32.− 0.50 62 25 10. −× 5 65.− 0.60 41 05 10. −× 3 98.− 0.70 34 93 10. −× 2 31.− ______________________________________________________________________________________ 1.31 a.

2 2 1

1

10 exp

ln (10) 59.9 mV 60 mV

D D D

D T

D T D

I V VI VV V V

⎛ ⎞−= = ⎜ ⎟

⎝ ⎠Δ = ⇒ Δ = ≈

b. ( )ln 100 119.7 mV 120 mVD T DV V VΔ = ⇒ Δ = ≈ ______________________________________________________________________________________ 1.32

(a) (i) ( ) 539.01022ln026.0 9 =⎟

⎠⎞

⎜⎝⎛

×=

−DV V

(ii) ( ) 599.010220ln026.0 9 =⎟

⎠⎞

⎜⎝⎛

×=

−DV V

(b) (i) ( ) 60.9026.04.0exp102 9 ⇒⎟

⎠⎞

⎜⎝⎛×= −

DI mA

(ii) ( ) 144026.065.0exp102 9 ⇒⎟

⎠⎞

⎜⎝⎛×= −

DI A

______________________________________________________________________________________


3

14

3

12

2 10ln (0 026) ln 0 6347 V5 10

2 10(0 026) ln 0 5150 V5 10

0 5150 0 6347 V

DD t

S

D

D

IV V . .I

V . .

. V .

−

−

−

−

⎛ ⎞ ⎛ ⎞×= = =⎜ ⎟ ⎜ ⎟×⎝ ⎠⎝ ⎠

⎛ ⎞×= =⎜ ⎟×⎝ ⎠

≤ ≤

______________________________________________________________________________________ 1.34

(a) AII SS83 1046.1

026.030.0exp105.1 −− ×=⇒⎟

⎠⎞

⎜⎝⎛=×

(b) (i) ( ) 3.10026.035.0exp10462.1 8 =⇒⎟

⎠⎞

⎜⎝⎛×= −

DD II mA

(ii) ( ) 219.0026.025.0exp10462.1 8 =⇒⎟

⎠⎞

⎜⎝⎛×= −

DD II mA

______________________________________________________________________________________ 1.35

(a) ( ) 31.2026.08.0exp10 22 ⇒⎟

⎠⎞

⎜⎝⎛= −

DI nA

( ) μ05.5026.00.1exp10 22 ⇒⎟

⎠⎞

⎜⎝⎛= −

DI A

( ) 1.11026.02.1exp10 22 ⇒⎟

⎠⎞

⎜⎝⎛= −

DI mA

( ) 2322 1037.51026.0

02.0exp10 −− ×−=⎥⎦

⎤⎢⎣

⎡−⎟

⎠⎞

⎜⎝⎛ −=DI A

For V, A 20.0−=DV 2210−−=DI

For V, A 2−=DV 2210−−=DI (b)

( ) 115026.08.0exp105 24 ⇒⎟

⎠⎞

⎜⎝⎛×= −

DI pA

( ) μ253.0026.00.1exp105 24 ⇒⎟

⎠⎞

⎜⎝⎛×= −

DI A

( ) 554.0026.02.1exp105 24 ⇒⎟

⎠⎞

⎜⎝⎛×= −

DI mA

( ) 2424 1068.21026.0

02.0exp105 −− ×−=⎥⎦

⎤⎢⎣

⎡−⎟

⎠⎞

⎜⎝⎛ −×=DI A

For V, A 20.0−=DV 24105 −×−=DI

For V, A 2−=DV 24105 −×−=DI______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 1.36 IS doubles for every 5C increase in temperature. 1210 SI A−= at T = 300K For 120.5 10 T 295 KSI A−= × ⇒ = For 1250 10 , (2) 50 5.64n

SI A n−= × = ⇒ = Where n equals number of 5C increases. Then ( )( )5.64 5 28.2 T KΔ = =

So 295 328.2 T K≤ ≤ ______________________________________________________________________________________ 1.37

/ 5( )2 , 155 C

( 55)TS

S

I TT

IΔ= Δ =

−°

155 / 5 9(100)2 2.147 1

( 55)S

S

II

= = ×−

0

@100 C 373 K 0.03220T TV V° ⇒ ° ⇒ = @ 55 C 216 K 0.01865T TV V− ° ⇒ ° ⇒ =

( )(( )

)

9

9 8

13

3

0.6exp(100) 0.0322(2.147 10 )

0.6( 55) exp0.01865

2.147 10 1.237 10

9.374 10

(100) 2.83 10( 55)

D

D

D

D

II

II

⎛ ⎞⎜ ⎟⎝ ⎠= × ×

− ⎛ ⎞⎜ ⎟⎝ ⎠

× ×=

×

= ×−

______________________________________________________________________________________ 1.38

(a) DDPS VRIV +=

; ( ) DD VI += 6108.2 ( ) ⎟⎠

⎞⎜⎝

⎛×= −

026.0exp105 11 D

DV

I

By trial and error, V, 282.0=DV μ52.2=DI A (b) A, V 11105 −×−≅DI 8.2−=DV

______________________________________________________________________________________


( )410 2 10D DI V= × + and ( ) 120.026 ln10

DD

IV −

⎛ ⎞= ⎜ ⎟⎝ ⎠

Trial and error. VD(v) ID(A) VD(v) 0.50 44.75 10−× 0.5194 0.517 44.7415 10−× 0.5194 0.5194 44.740 10−× 0.5194

0.5194 V

0.4740 mAD

D

V

I

=

=

______________________________________________________________________________________ 1.40 135 10 AsI −= ×

2

1 2

30(1.2) (1.2) 0.45 V80TH

RV

R R⎛ ⎞ ⎛ ⎞= = =⎜ ⎟ ⎜ ⎟+ ⎝ ⎠⎝ ⎠

0.45 , ln DD TH D D T

S

II R V V V

I⎛ ⎞

= + = ⎜ ⎟⎝ ⎠

By trial and error: 2.56 A, 0.402 VD DI Vμ= = ______________________________________________________________________________________


(a) mA 121 == DD II

(i) ( ) 599.01010ln026.0 13

3

21 =⎟⎟⎠

⎞⎜⎜⎝

⎛==

−

−

DD VV V

(ii) ( ) 617.0105

10ln026.0 14

3

1 =⎟⎟⎠

⎞⎜⎜⎝

⎛

×=

−

−

DV V

( ) 557.0105

10ln026.0 13

3

2 =⎟⎟⎠

⎞⎜⎜⎝

⎛

×=

−

−

DV V

(b) 21 DD VV =

(i) 5.0221 === i

DDI

II mA

( ) 581.010

105.0ln026.0 13

3

21 =⎟⎟⎠

⎞⎜⎜⎝

⎛ ×== −

−

DD VV V

(ii) 10.0105105

13

14

2

1

2

1 =××

==−

−

S

S

D

D

II

II

So 21 10.0 DD II = mA 11.1 221 ==+ DDD III So mA, mA 909.02 =DI 0909.01 =DI Now

( ) 554.0105

100909.0ln026.0 14

3

1 =⎟⎟⎠

⎞⎜⎜⎝

⎛

××

=−

−

DV V

( ) 554.0105

10909.0ln026.0 13

3

2 =⎟⎟⎠

⎞⎜⎜⎝

⎛

××

= −

−

DV V

______________________________________________________________________________________ 1.42

(a) ( ) 426.2026.0635.0exp106 14

3 ⇒⎟⎠⎞

⎜⎝⎛×= −

DI mA

635.01635.0

==RI mA

mA 061.3635.0426.221 =+== DD II

( ) 641.0106

10061.3ln026.0 14

3

21 =⎟⎟⎠

⎞⎜⎜⎝

⎛

××

==−

−

DD VV V

V ( ) 917.1635.0641.02 =+=IV(b) mA 426.23 =DI

27.15.0

635.0==RI mA

mA 696.327.1426.221 =+== DD II

( ) 6459.0106

10696.3ln026.0 14

3

21 =⎟⎟⎠

⎞⎜⎜⎝

⎛

××

==−

−

DD VV V

V ( ) 927.1635.06459.02 =+=IV

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ ______________________________________________________________________________________ 1.43 (a) Assume diode is conducting. Then, 0.7 DV V Vγ= =

So that 20 7 23 3 30R.I . Aμ= ⇒

11 2 0 7 50

10R. .I Aμ−

= ⇒

Then 1 2 50 23 3D R RI I I .= − = − Or 26 7 DI . Aμ=

(b) Let Diode is cutoff. 1 50 R k= Ω

30 (1 2) 0 45 30 50DV .= ⋅ =

+. V

Since , 0D DV V Iγ< =

______________________________________________________________________________________ 1.44

At node VA:

(1) 52 2

A AD

V VI

−= +

At node γVVV AB −=

(2) ( ) ( )52 2A r A r

D

V V V VI

− − −+ =

So ( )5 5

3 2 2A r A A AV V V V V V− − − −⎡ ⎤+ − =⎢ ⎥⎣ ⎦ 2

r

)A

Multiply by 6: ( ) (10 2 15 6 3A r A A rV V V V V− − + − = −

25 2 3 11r rV V V+ + = (a) 0.6 VrV =

( )11 25 5 0.6 28 2.545 VA AV V= + = ⇒ =


From (1) 52.5

2 2A A

D AV V

DI V I−

= − = − ⇒ Neg. 0DI⇒ =

Both (a), (b) 0DI =

VA = 2.5, 2 5 2 V 0.50 V5B DV V= ⋅ = ⇒ =

______________________________________________________________________________________ 1.45

(a) ; ; for mA ( )1iO IV = 0=DI 7.00 ≤≤ iI V; mA; for mA 7.0=OV ( 7.0−= iD II ) 7.0≥iI (b) ; ; for ( )1iO IV = 0=DI 7.10 ≤≤ iI mA V; mA; for mA 7.1=OV ( 7.1−= iD II ) 7.1≥iI (c) V; ; ; for 7.0=OV iD II =1 02 =DI 20 ≤≤ iI mA

______________________________________________________________________________________ 1.46 Minimum diode current for VPS (min) (min) 2 , 0.7 D DI mA V V= =

2 12 1

0 7 5 0 7 4 3, . .I IR R

−= = =

1

.R

We have 1 2 DI I I= +

so (1) 1 2

4 3 0 7 2. .R R

= +

Maximum diode current for VPS (max) ( )10 0 7 14 3 D D D DP I V I . I . mA= = ⇒ = 1 2 DI I I= + or

(2) 1 2

9 3 0 7 14 3. . .R R

= +

Using Eq. (1), 11 1

9 3 4 3 2 14 3 0 41 Ω. . . R . kR R

= − + ⇒ =

Then 2 82 5 82 5R . .= Ω Ω ______________________________________________________________________________________ 1.47

(a) (i) 215.020

7.05=

−=I mA, 7.0=OV V

(ii) 220.020

6.05=

−=I mA, 6.0=OV V

(b) (i) ( ) 2325.040

57.05=

−−−=I mA, ( )( ) 35.05202325.0 −=−=OV V

(iii) ( ) 235.040

56.05=

−−−=I mA, ( )( ) 30.0520235.0 −=−=OV V


(c) (i) ( ) 372.025

87.02=

−−−=I mA, ( )( ) 14.05372.02 =−=OV V

(ii) ( ) 376.025

86.02=

−−−=I mA, ( )( ) 12.05376.02 =−=OV V

(d) (i) , V 0=I 5−=OV(ii) , V 0=I 5−=OV

______________________________________________________________________________________ 1.48

(a) 20

5 OVI

−= , ( ) ⎟

⎠

⎞⎜⎝

⎛×= −

026.0exp105 14 DV

I

By trial and error, =DV 5775.0=OV V, 221.0=I mA

(b) 40

10 DVI −= , ( ) DO VIV −−= 205

mA, V, 2355.0=I 579.0=DV 289.0−=OV

(c) 25

10 DVI −= , ( )52 IVO −=

mA, V, 3763.0=I 5913.0=DV 1185.0=OV

(d) A, V 14105 −×−=I 5−≅OV______________________________________________________________________________________ 1.49 (a) Diode forward biased VD = 0.7 V 5 (0.4)(4.7) 0.7 2.42 VV V= + + ⇒ = (b) (0.4)(0.7) 0.28 mωDP I V P= ⋅ = ⇒ = ______________________________________________________________________________________ 1.50

(a) 2 1 1

2

02 1

1 1

0.65 0.65 mA1

2(0.65) 1.30 mA2 5 3(0.65) 1.30 2.35 K

R D D

D

I rD

I I I

IV V VI R

R R

= = = =

= =− − −

= = = ⇒ =

(b) 2

2 2

1 2 2

1

0.65 0.65 mA1

8 3(0.65) 3.025 mA2

3.025 0.652.375 mA

R

D D

D D R

D

I

I I

I I II

= =

−= ⇒ =

= − = −=

______________________________________________________________________________________ 1.51


a. (0 026) 0 026 k 26

1

0 05 50 A peak-to-peak

(26)(50) A 1 30 mV peak-to-peak

Td

DQ

d DQ

d d d d

V . .I

i . I

v i v .

τ

μ

τ μ

= = = Ω = Ω

= =

= = ⇒ =

b. For (0 026)0 1 mA 2600 1DQ d.I .

.τ= ⇒ = = Ω

0 05 5 A peak-to-peakd DQi . I μ= = (260)(5) V 1 30 mV peak-to-peakd d d dv i v .τ μ= = ⇒ = ______________________________________________________________________________________ 1.52

(a) 1026.0026.0

===DQ

Td I

Vr k Ω

(b) Ω⇒= 10026.0

026.0dr

(c) Ω⇒= 106.2

026.0dr

______________________________________________________________________________________ 1.53

a. diode resistance d Tr V I= /

d Td S

Td SS

Td s o

T S

r V Iv vVr R RI

Vv v vV IR

⎛ ⎞⎜ ⎟⎛ ⎞ /

= = ⎜ ⎟⎜ ⎟+⎝ ⎠ ⎜ ⎟+⎜ ⎟⎝ ⎠

⎛ ⎞= =⎜ ⎟+⎝ ⎠

Sv

b. 260SR = Ω

( ) ( )

0 0

0 0

0 0

0 0261 mA, 0 09090 026 (1)(0 26)

0 0260 1 mA, 0 500 026 0 1 0 26

0 0260 01 mA. 0 9090 026 (0 01)(0 26)

T

S T S S

s S

S S

v vV .I .v V IR . . v

v v.I . .v . . . v

v v.I . .v . . . v

⎛ ⎞= = = ⇒ =⎜ ⎟+ +⎝ ⎠

= = ⇒ =+

= = ⇒ =+

______________________________________________________________________________________

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 1.54 pn junction diode

( ) 548.0105

1072.0ln026.0 13

3

=⎟⎟⎠

⎞⎜⎜⎝

⎛

××

=−

−

DV V

Schottky diode

( ) 249.0105

1072.0ln026.0 8

3

=⎟⎟⎠

⎞⎜⎜⎝

⎛

××

=−

−

DV V

______________________________________________________________________________________ 1.55

Schottky: exp aS

T

VI I

V⎛ ⎞

≅ ⎜ ⎟⎝ ⎠

3

7

0.5 10ln (0.026) ln5 10

0.1796

a TS

IV VI

V

−

−

⎛ ⎞ ⎛ ⎞×= =⎜ ⎟ ⎜ ⎟×⎝ ⎠⎝ ⎠

=

Then

of pn junction 0.1796 0.30

0.4796aV = +

=

30 5 10

0 4796exp exp 0 026

Sa

T

I .I.V.V

−×= =

⎛ ⎞ ⎛ ⎞⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

124 87 10 ASI . −= × ______________________________________________________________________________________ 1.56 (a)

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 3

1 2 0 5 10I I . −+ = ×

8 125 10 exp 10 exp 0 5 10D D

T T

V V.

V V− −⎛ ⎞ ⎛ ⎞

× + =⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

3−×

8 35 0001 10 exp 0 5 10D

T

V. .

V− −⎛ ⎞

× =⎜ ⎟⎝ ⎠

×

3

8

0 5 10(0 026) ln 0 23955 0001 10D D

.V . V ..

−

−

⎛ ⎞×= ⇒⎜ ⎟×⎝ ⎠

=

Schottky diode, 2 0 49999 mAI .=

pn junction, 1 0 00001 mAI .= (b)

12 81 210 exp 5 10 expD D

T T

V VI

V V− −⎛ ⎞ ⎛

= = ×⎜ ⎟ ⎜⎝ ⎠ ⎝

⎞⎟⎠

1 2 0.9D DV V+ =

12 81 1

8 1

0.910 exp 5 10 exp

0.95 10 exp exp

D D

T T

D

T T

V VV V

VV V

− −

−

⎛ ⎞ ⎛ ⎞−= ×⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎛ ⎞ ⎛ −

= × ⎜ ⎟ ⎜⎝ ⎠ ⎝

⎞⎟⎠

8

112

2 5 10 0 9exp exp0 02610

D

T

V .V .

−

−

⎛ ⎞ ⎛ ⎞× ⎛ ⎞=⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠⎝ ⎠

8

1 12

5 102 ln 0 9 1 181310D TV V . .

−

−

⎛ ⎞×= + =⎜ ⎟

⎝ ⎠

1 0 5907 pn junctionDV .=

2 0 3093 Schottky diodeDV .=

12 0 590710 exp 7 35 mA0 026.I I.

− ⎛ ⎞= ⇒ =⎜ ⎟⎝ ⎠

.

______________________________________________________________________________________ 1.57

0 5 6 V at 0 1 mAZ Z ZV V . I .= = = 10Zr = Ω

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ ( )( )0 1 10 1 mVZ ZI r .= = VZ0 = 5.599 a. LR →∞⇒

10 5 599 4 401 8 63 mA0 50 0 01Z

Z

. .I .R r . .−

= = =+ +

( )( )0 5.599 0.00863 10Z Z Z ZV V I r= + = + 0 5 685 VZV V .= =

b. 11 5 59911 V 10 59 mA0 51PS Z

.V I ..

−= ⇒ = =

( )( )0 5.599 0.01059 10 5.7049 VZV V= = + =

9 5 5999 V 6 669 mA0 51PS Z

.V I ..

−= ⇒ = =

( )( )0 5.599 0.006669 10 5.66569 VZV V= = + =

0 05 7049 5 66569 0 0392 VV . . V .Δ = − ⇒ Δ = c. I = IZ + IL

0 0 0, , 0PS ZL Z

L Z

V V V V VI I IR R r

− −= = =

0 010 5 5990 50 0 010 2

V V . V. .− −

= + 0

010 5 599 1 1 1

0 50 0 010 0 50 0 010 2. V

. . . .⎡ ⎤+ = + +⎢ ⎥⎣ ⎦

20.0 + 559.9 = V0 (102.5) 0 5.658 VV = ______________________________________________________________________________________ 1.58

(a) 4.65.0

8.610=

−=ZI mA

mW ( )( ) 5.438.64.6 === ZZVIP(b) mA ( )( ) 64.04.61.0 ==ZI 76.564.04.6 =−=LI mA

18.176.58.6===⇒=

Z

ZL

L

ZL I

VRRVI kΩ

______________________________________________________________________________________ 1.59 ( )( )0.1 20 2 mVZ ZI r = =

0 6 8 0 002 6 798 VZV . . .= − = a. LR = ∞

10 6 798 6 158 mA0 5 0 02Z Z

.I I .. .−

= ⇒ =+

Microelectronics: Circuit Analysis and Design, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ ( )( )0 0 6.798 0.006158 20Z Z Z ZV V V I r= = + = + 0 6 921 VV .= b. Z LI I I= +

0 010 6 7980 50 0 020 1

V V . V. .− −

= + 0

010 6 798 1 1 1

0 30 0 020 0 50 0 020 1. V

. . . .⎡ ⎤+ = + +⎢ ⎥⎣ ⎦

359.9 = V0 (53) V0 = 6.791 V 0 6 791 6 921V . .Δ = − 0 0.13 VVΔ = − ______________________________________________________________________________________ 1.60 For VD = 0, 0.1 ASCI =

For ID = 0 14

0.2ln 15 10D TV V −

⎛ ⎞= +⎜ ⎟×⎝ ⎠

0.754 VD DCV V= = ______________________________________________________________________________________ 1.61 A 2.0,0 == DD IV V, A 60.0=DV 1995.0=DI V, A 65.0=DV 1964.0=DI V, A 70.0=DV 1754.0=DI V, A 72.0=DV 1468.0=DI V, A 74.0=DV 0853.0=DI V, 7545.0=DV 0=DI______________________________________________________________________________________ 1.62

(a) ( ) 7126.01026.0

exp10520.016.0 14 =⇒⎥⎦

⎤⎢⎣

⎡−⎟

⎠

⎞⎜⎝

⎛×−= −D

D VV

V

(b) W ( )( ) 114.07126.016.0 ==P______________________________________________________________________________________

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