Chapter 1 Problem Solutions 1.1

/ 23/ 2 gE kTin 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−

⎛ ⎞−= × ⎜ ⎟×⎝ ⎠

34 3/ 2 6.40 101.91 10 expT

T− ⎛ ⎞×

× = −⎜ ⎟⎝ ⎠

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

( )( )9 15 3 / 2

6

1.110 5.23 10 exp2 86 10

TT−

⎛ ⎞−⎜ ⎟= ×⎜ ⎟×⎝ ⎠

37 3 / 2 6.40 101.91 10 expT

T− ⎛ ⎞×

× = −⎜ ⎟⎝ ⎠

By trial and error, 268 KT ≈ ° 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. 15 35 10 cm n typedN −= × ⇒ −

15 30 5 10 cmdn N −= = ×

( )21024 3

0 0150

1.5 104.5 10 cm

5 10inp p

n−

×= = ⇒ = ×

×

b. 15 35 10 cm typedN n−= × ⇒ − 15 35 10 cmo dn N −= = ×

( )( )

( )( ) ( )

3/ 2146

3/ 214 12

6 3

1.42.10 10 300 exp2(86 10 )(300)

2.10 10 300 1.65 10

1.80 10 cm

in −

−

−

⎛ ⎞−= × ⎜ ⎟×⎝ ⎠

= × ×

= ×

( )2624 3

0 0150

1.8 106.48 10 cm

5 10inp p

n− −

×= = ⇒ = ×

×

1.5

(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 −= ×

( )2116 3

16

3.97 103.15 10 cm

5 10op −×

= = ××

1.6 a. 16 310 cm typeaN p−= ⇒ −

16 30 10 cmap N −= =

( )21024 3

0 0160

1.5 102.25 10 cm

10in

n np

−×

= = ⇒ = ×

b. Germanium 16 310 cm typeaN p−= ⇒ −

16 30 10 cmap N −= =

( )( ) ( )( )

( )( ) ( )

3/ 2156

3/ 215 6

13 3

0.661.66 10 300 exp2 86 10 300

1.66 10 300 2.79 10

2.4 10 cm

in−

−

−

⎛ ⎞−⎜ ⎟= ×⎜ ⎟×⎝ ⎠

= × ×

= ×

( )213210 3

0 0160

2.4 105.76 10 cm

10in

n np

−×

= = ⇒ = ×

1.7 (a) p-type (b)

( )

17 3

21023 3

17

2 10 cm

1.5 101.125 10 cm

2 10

o a

io

o

p N

nn

p

−

−

= = ×

×= = = ×

×

(c) 17 32 10 cmop −= ×

From Problem 1.1(a)(i) 8 31.61 10 cmin −= ×

( )283

17

1.61 100.130 cm

2 10on −×

= =×

1.8 (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-typeo on p ⇒

(c) 15 35 10 o dn N cm−≅ = × 1.9 a. Add Donors

15 37 10 cmdN −= ×

b. Want 6 3 210 cm /o i dp n N−= =

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 ≈ ° 1.10

( )( ) ( )42.2 15 10 3.3

I J A EA

I I

σ−

= ⋅ =

= ⇒ = mA

1.11

1

8512

7.08 (ohm cm)

JJ EE

σ σ

σ −

= ⇒ = =

= −

1.12

( )( )( )19

1 1 11.6 10 480 0.80a

p a p

g Ne N e gμ μ −

≈ ⇒ = =×

16 31.63 10 cmaN −= × 1.13

( )( )( )19

15 3

0.51.6 10 1350

2.31 10 cm

n d

dn

d

e N

Ne

N

σ μ

σμ −

−

=

= =×

= ×

1.14 (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 1.15

( )( )15 2

194

2

10 101.6 10 1800.5 10

576 A/cm

n n n

n

dn nJ eD eDdx x

J

−−

Δ= =

Δ⎡ ⎤−

= × ⎢ ⎥×⎣ ⎦=

1.16

( )

( )( )( )

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−= = 1.17 a. 17 3 17 310 cm 10 cma oN p− −= ⇒ =

( )2625 3

17

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

− −

−

= + = × + ⇒ == + = + ⇒ = ×

1.18

(a)

( )( )( )( )

2

16 16

210

ln

10 100.026 ln 0.697 V

1.5 10

a dbi T

i

N NV V

n⎛ ⎞

= ⎜ ⎟⎝ ⎠

⎡ ⎤⎢ ⎥= =⎢ ⎥×⎣ ⎦

(b) ( )( )( )( )

18 16

210

10 100.026 ln 0.817 V

1.5 10biV

⎡ ⎤⎢ ⎥= =⎢ ⎥×⎣ ⎦

(c) ( )( )( )( )

18 18

210

10 100.026 ln 0.937 V

1.5 10biV

⎡ ⎤⎢ ⎥= =⎢ ⎥×⎣ ⎦

1.19

2ln a dbi T

i

N NV V

n⎛ ⎞

= ⎜ ⎟⎝ ⎠

a. ( )( )( )16 16

6 2

10 100.026 ln 1.17 V

(1.8 10 )bi biV V⎡ ⎤⎢ ⎥= ⇒

×⎢ ⎥⎣ ⎦=

b. ( )( )( )18 16

6 2

10 100.026 ln 1.29 V

(1.8 10 )bi biV V⎡ ⎤⎢ ⎥= ⇒

×⎢ ⎥⎣ ⎦=

c. ( )( )( )18 18

6 2

10 100.026 ln 1.41 V

(1.8 10 )bi biV V⎡ ⎤⎢ ⎥= ⇒

×⎢ ⎥⎣ ⎦=

1.20

( )( )16

2 1

10ln 0.026 ln

(1.5 10 )aa d

bi Ti

NN NV V

n

⎡ ⎤⎛ ⎞⎢ ⎥= =⎜ ⎟ ×⎢ ⎥⎝ ⎠ ⎣ ⎦

0 2

For 15 310 , 0.637 a biN cm V−= = VVFor 18 310 , 0.817 a biN cm V−= =

0.817

0.637

1015 1016 1017 1018 Na(cm3)

Vbi(V)

1.21

(0.026)300TkT ⎛ ⎞= ⎜ ⎟

⎝ ⎠

T 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

200

1.45

1.35

1.25

250 300 350 400 450 500

Vbi(V)

T(C) 1.22

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−

⎛ ⎞= + =⎜ ⎟⎝ ⎠

1.23

(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−= ×

1.24

( )( )( )( )

18 15

210

10 100.026 ln 0.757 V

1.5 10biV

⎡ ⎤⎢ ⎥= =⎢ ⎥×⎣ ⎦

a. 1 VRV =1/ 21(0.25) 1 0.164 pF

0.757jC−

⎛ ⎞= + =⎜ ⎟⎝ ⎠

( )( )03 1

1 12 2 2.2 10 0.164 10

fLCπ π − −

= =× × 2

0 8.38 MHzf = b. VR = 10 V

1/ 210(0.25) 1 0.0663 pF0.757jC

−⎛ ⎞= + =⎜ ⎟⎝ ⎠

( )( )03 1

1

2 2.2 10 0.0663 10f

π − −=

× × 2

0 13.2 MHzf = 1.25

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

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

⎠⎞ −− ⎜ ⎟⎢ ⎥⎜ ⎟ ⎝ ⎠⎝ ⎠⎣ ⎦

=−

=

1.26 a.

11

11

11

0.5(10 )exp 2.25 mA0.0260.6(10 )exp 0.105 A

0.0260.7(10 )exp 4.93 A

0.026

I I

I I

I I

−

−

−

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

b. 13

13

13

0.5(10 )exp 22.5 A0.0260.6(10 )exp 1.05 mA

0.0260.7(10 )exp 49.3 mA

0.026

I I

I I

I I

μ−

−

−

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

1.27 (a) ( )/ 1D TV V

SI I e= −

( )/ /6 11 11150 10 10 1 10D T DV V V Ve e− − −× = − ≅ T

Then 6 6

11 11

150 10 150 10ln (0.026) ln10 10D TV V

− −

− −

⎛ ⎞ ⎛ ⎞× ×= =⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

Or 0.430 DV V= (b)

6

13

150 10ln10D TV V

−

−

⎛ ⎞×= ⎜ ⎟

⎝ ⎠

Or 0.549 DV V= 1.28

(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 −× 1.29 (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.− 1.30 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 = ≈ 1.31

(a) (i) ( )6

15

150 10ln 0 026 ln10

= 0.669 V

DD t

S

D

IV V .I

V

−

−

⎛ ⎞ ⎛ ⎞×= =⎜ ⎟ ⎜ ⎟

⎝ ⎠⎝ ⎠

(ii) 6

15

25 100 026)ln10

= 0.622 V

D

D

V .

V

−

−

⎛ ⎞×= ( ⎜ ⎟

⎝ ⎠

(b) (i) 15 120 2(10 )exp 2 19 10 A0 026D

.I ..

− −⎛ ⎞= =⎜ ⎟⎝ ⎠

×

(ii) 0DI =

(iii) 1510 ADI −= −

(iv) 1510 ADI −= − 1.32

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 .

−

−

−

−

⎛ ⎞ ⎛ ⎞×= = =⎜ ⎟ ⎜ ⎟×⎝ ⎠⎝ ⎠

⎛ ⎞×= =⎜ ⎟×⎝ ⎠

≤ ≤

1.33

(a) exp DD S

t

VI IV

⎛ ⎞= ⎜ ⎟

⎝ ⎠

3 21.1012 10 exp 5 07 10 A0.026S SI I .− −⎛ ⎞× = ⇒ = ×⎜ ⎟

⎝ ⎠1

(b) ( )21

4

1.05 07 10 exp0.026

2 56 10 A 0 256 mA

D

D

I .

I . .

−

−

⎛ ⎞= × ⎜ ⎟⎝ ⎠

= × =

1.34

(a) 23 71 010 exp 5 05 10 A0 026D

.I ..

− −⎛ ⎞= =⎜ ⎟⎝ ⎠

×

(b) 23 51 110 exp 2 37 10 A0 026D

.I ..

− −⎛ ⎞= =⎜ ⎟⎝ ⎠

×

(c) 23 31 210 exp 1 11 10 A0 026D

.I ..

− −⎛ ⎞= =⎜ ⎟⎝ ⎠

×

1.35 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≤ ≤

1.36 / 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

⎛ ⎞⎜ ⎟⎝ ⎠= × ×

− ⎛ ⎞⎜ ⎟⎝ ⎠

× ×=

×

= ×−

1.37 3.5 = ID (105) + VD

(a) 995 10 exp 0 026 ln

0 026 5 10D D

D DV II V ..

−−

⎛ ⎞ ⎛= × ⇒ =⎜ ⎟ ⎜ ×⎝ ⎠ ⎝⎞⎟⎠

Trial and error. VD ID VD0.50 53 10−× 0.226 0.40 53 1 10. −× 0.227 0.250 53 25 10. −× 0.228 0.229 53 271 10. −× 0.2284 0.2285 53 2715 10. −× 0.2284 So

5

0 2285 V

3 272 10 AD

D

V .

I . −

≅

≅ × (b)

( )( )

9

9 5 4

5 10 A

5 10 10 5 10 V

3 4995 V

D S

R

D

I I

V

V .

−

− −

= = ×

= × = ×

=

1.38

( )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

=

=

1.39

135 10 AsI −= ×

1.2 V

R1 50 K

R2 30 K

ID

VTH

VD

VD

RTH R1 R2 18.75 K

ID

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= =

1.40

VI

V0

VD VD

IR

I1

I2

1 K

132 10 SI A−= ×

0 0.60 VV =

( )1302

1 2

0.60exp 2 10 exp0.026

2.105 mA0.6 0.60 mA1 K

2.705 mA

ST

R

R

VI I

V

I

I I I

−⎛ ⎞ ⎛ ⎞= = ×⎜ ⎟ ⎜ ⎟⎝ ⎠⎝ ⎠

=

= =

= + =

31

13

0

2.705 10ln (0.026) ln2 10

0.60652 1.81 V

D TS

I D I

IV VI

V V V V

−

−

⎛ ⎞ ⎛ ⎞×= =⎜ ⎟ ⎜ ⎟×⎝ ⎠⎝ ⎠== + ⇒ =

1.41 (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< =

1.42

ID

5 V

VAVB VAVr

3 k

2 k

2 k

2 k

A&VA:

(1) 5

2 2A A

DV V

I−

= +

A& A rV V−

(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− −

2r− −⎡ ⎤+ − =⎢ ⎥⎣ ⎦

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

A

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

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

From (1) 5

2.52 2

A AD A

V VDI V I

−= − = − ⇒ Neg. 0DI⇒ =

Both (a), (b) 0DI =

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

1.43 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 1.44 (a) Vo = 0.7 V

5 0.7 0.215 mA20

I I−= ⇒ =

(b) 10 0.7 0.2325 mA20 20

(20 K) 5 0.35 Vo o

I I

V I V

−= ⇒ =

+= − ⇒ = −

(c) 10 0.7 0.372 mA5 20

0.7 (20) 8 0.14 Vo o

I I

V I V

−= ⇒ =

+= + − ⇒ = +

(d) 0(20) 5 5 Vo o

IV I V

== − ⇒ = −

1.45

(a) ( )95 2 10 DI V= × + ( ) 120.026 ln2 10D

IV −

⎛ ⎞= ⎜ ⎟×⎝ ⎠

VD → ID → VD Vo = VD = 0.482 V 0.6 42.2 10−× 0.481 0.482 42.259 10−× 0.482 0.226 I mA=

(b) ( )410 4 10 DI V= × + ( ) 120.026 ln2 10D

IV −

⎛ ⎞= ⎜ ⎟×⎝ ⎠

Vo → I → VD 0.483 DV V= 0.5 42.375 10−× 0.4834 I = 0.238 mA 0.484 42.379 10−× 0.4834 0.24 oV V= −

(c) ( )410 2.5 10 DI V= × + ( ) 120.026 ln2 10D

IV −

⎛ ⎞= ⎜ ⎟×⎝ ⎠

Vo → I → VD VD = 0.496 V 0.480 43.808 10−× 0.496 I = 0.380 mA 0.496 43.802 10−× 0.496 0.10 oV V= − (d) 122 10 A

5 VS

o

I I IV

−= − ⇒ = ×≅ −

1.46 (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= ⋅ = ⇒ = 1.47

(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

= =

−= ⇒ =

= − = −=

1.48

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= = ⇒ =

1.49

RS

S d

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

⎛ ⎞= = = ⇒ =⎜ ⎟+ +⎝ ⎠

= = ⇒ =+

= = ⇒ =+

1.50

exp , lnaS a T

T S

V II I V VV I

⎛ ⎞⎛ ⎞≅ = ⎜ ⎟⎜ ⎟

⎝ ⎠ ⎝ ⎠

6

14

100 10pn junction, (0 026) ln10aV .

−

−

⎛ ⎞×= ⎜ ⎟

⎝ ⎠

0 599 VaV .=

Schottky diode, 6

9

100 10(0 026) ln 10aV .

−

−

⎛ ⎞×= ⎜ ⎟

⎝ ⎠

0 299 VaV .= 1.51

I

Schottky

pn junction

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 5 100 4796exp exp 0 026

Sa

T

I .I.V.V

−×= =

⎛ ⎞ ⎛ ⎞⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠

124 87 10 ASI . −= × 1.52 (a)

0.5 mA

VD

I1

I2

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)

VD2 VD1

I0.90 V

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

− −

−

⎛ ⎞ ⎛ ⎞−= ×⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠⎛ ⎞ ⎛ −

= × ⎜ ⎟ ⎜⎝ ⎠ ⎝

⎞⎟⎠

81

12

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.

− ⎛ ⎞= ⇒ =⎜ ⎟⎝ ⎠

.

1.53

VPS 10 V

VZ

R 0.5 K

RL

V0

I

IZIL

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

10Zr = Ω

( )( )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 =

1.54

a.

( )( )

9 6 8 11 mA0 2

11 6 8 74 8 mW

Z Z

Z Z

.I I.

P . P .

−= ⇒ =

= ⇒ =

b.

12 6 8 26 mA0 226 11 100 136%

11

Z Z.I I

.

%

−= ⇒ =

−= × ⇒

( )( )26 6.8 176.8 mW176.8 74.8% 100 136%

74.8

ZP = =

−= × ⇒

1.55

( )( )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 .. .−

= ⇒ =+

( )( )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 = − 1.56

For VD = 0, 0.1 ASCI =

For ID = 0 14

0.2ln 15 10D TV V −

⎛ ⎞= +⎜ ⎟×⎝ ⎠

0.754 VD DCV V= =