Chapter 1 Exercise Problems EX1.1

3 / 2 exp2

gi

En BT

kT−⎛ ⎞

= ⎜ ⎟⎝ ⎠

GaAs: ( ) ( ) ( )( )3/ 214

6

1.42.1 10 300 exp2 86 10 300in

−

⎛ ⎞−⎜ ⎟= ×⎜ ⎟×⎝ ⎠

or 6 31.8 10 in cm−= ×

Ge: ( )( ) ( )( )3/ 213

6

0.661.66 10 300 exp2 86 10 300in

−

⎛ ⎞−⎜ ⎟= ×⎜ ⎟×⎝ ⎠

or 13 32.40 10 in cm−= ×

EX1.2 (a) majority carrier: holes, 17 310 op cm−= minority carrier: electrons,

( )21023 3

17

1.5 102.25 10

10i

oo

nn c

p−

×= = = × m

cm

(b) majority carrier: electrons, 15 35 10 on −= × minority carrier: holes,

( )21024 3

15

1.5 104.5 10

5 10i

oo

np c

n−

×= = = ×

×m

EX1.3 For n-type, drift current density EnJ e nμ≅ or ( )( )( )19 16200 1.6 10 7000 10 E−= × which yields

E 17.9 /V cm= EX1.4 Diffusion current density due to holes:

( )16 110 exp

p p

pp p

dpJ eDdx

xeDL L

= −

⎛ ⎞ ⎛− −= − ⎜ ⎟ ⎜⎜ ⎟ ⎜

⎝ ⎠ ⎝

⎞⎟⎟⎠

(a) At 0x =

( )( )( )19 162

3

1.6 10 10 1016 /

10pJ A cm−

−

×= =

(b) At 310 x cm−= 3

23

1016exp 5.89 /10pJ A cm

−

−

⎛ ⎞−= =⎜ ⎟

⎝ ⎠

EX1.5

( )( )( )( )

16 17

2 26

10 10ln 0.026 ln

1.8 10a d

bi Ti

N NV V

n

⎡ ⎤⎡ ⎤ ⎢ ⎥= =⎢ ⎥ ⎢ ⎥⎣ ⎦ ×⎣ ⎦

or 1.23 biV V=

EX1.6

1/ 2

1 Rj jo

bi

VC C

V

−⎛ ⎞

= +⎜ ⎟⎝ ⎠

and

( )( )( )( )

2

17 16

210

ln

10 100.026 ln 0.757

1.5 10

a dbi T

i

N NV V

n

V

⎡ ⎤= ⎢ ⎥

⎣ ⎦⎡ ⎤⎢ ⎥= =⎢ ⎥×⎣ ⎦

Then ( )1/ 2

1/ 250.8 1 7.610.757jo joC C

−−⎛ ⎞= + =⎜ ⎟

⎝ ⎠

or 2.21 joC p= F

EX1.7

exp 1DD S

T

vi IV

⎡ ⎤⎛ ⎞= −⎢ ⎥⎜ ⎟

⎢ ⎥⎝ ⎠⎣ ⎦

so ( )3 1310 10 exp 10.026

Dv− − ⎡ ⎤⎛ ⎞= −⎢ ⎥⎜ ⎟⎝ ⎠⎣ ⎦

Solving for the diode voltage, we find ( )3

13

100.026 ln 110Dv

−

−

⎡ ⎤= +⎢ ⎥

⎣ ⎦

or ( ) ( )100.026 ln 10Dv ≅

which yields 0.599 Dv V=

EX1.8

and exp DPS D D D S

T

VV I R V I I

V⎛ ⎞

= + ≅ ⎜ ⎟⎝ ⎠

so ( ) ( )33

44 4 10

4 10D

D D D

VI V I

−= × + ⇒ =

×

and

( )1210 exp0.026

DD

VI − ⎛ ⎞= ⎜ ⎟

⎝ ⎠

By trial and error, we find 0.864 and 0.535 D DI mA V V≅ ≅ EX1.9

(a) 5 0.7 1.08 4

PSD D

V VI I m

Rγ− −

= = ⇒ = A

(b) PS PSD

D

V V V VI R

R Iγ γ− −

= ⇒ =

Then 8 0.7 6.79 Ω1.075

R k−= =

(c)

1.25

1.08

0.7 2 4 6 80

Diode curve

Load lines

VD(v)

(a)

(b)

ID(mA)

EX1.10 PSpice analysis EX1.11

Quiescent diode current 10 0.7 0.465 20

PSDQ

V VI m

Rγ− −

= = = A

Time-varying diode current:

We find that 0.026 0.0559 Ω0.465

Td

DQ

Vr k

I= = =

Then ( )( )

0.2sin0.0559 20 Ω

Id

d

Vv tir R k

ω= = ⋅

+ + or ( )9.97sindi t Aω μ=

EX1.12

For the pn junction diode, ( )3

15

1.2 10ln 0.026 ln4 10

DD T

S

IV V

I

−

−

⎛ ⎞ ⎛ ⎞×≅ =⎜ ⎟ ⎜ ⎟×⎝ ⎠⎝ ⎠

or 0.6871 DV V=

The Schottky diode voltage will be smaller, so 0.6871 0.265 0.4221 DV V= − =

Now exp DD S

T

VI I

V⎛ ⎞

≅ ⎜ ⎟⎝ ⎠

or 3

101.2 10 1.07 10 0.4221exp0.026

S SI I−

−×= ⇒ = ×

⎛ ⎞⎜ ⎟⎝ ⎠

A

EX1.13

( )10 5.6 1.79 ZP I V I I mA= ⋅ ⇒ = ⇒ =

Also 10 5.6 1.79 2.46 ΩI R kR−

= = ⇒ =

Test Your Understanding Exercises TYU1.1 (a) T = 400K

Si: 3/ 2 exp2

gi

En BT

kT−⎛ ⎞

= ⎜ ⎟⎝ ⎠

( )( ) ( )( )3 / 215

6

1.15.23 10 400 exp2 86 10 400in

−

⎡ ⎤−⎢ ⎥= ××⎢ ⎥⎣ ⎦

or 12 34.76 10 in c −= × m

( )( ) ( )( )Ge: 3/ 215

6

0.661.66 10 400 exp2 86 10 400in

−

⎡ ⎤−⎢ ⎥= ××⎢ ⎥⎣ ⎦

m

or 14 39.06 10 in c −= ×

GaAs:

( )( ) ( )( )3 / 214

6

1.42.1 10 400 exp2 86 10 400in

−

⎡ ⎤−⎢ ⎥= ××⎢ ⎥⎣ ⎦

m

or 9 32.44 10 in c −= ×

(b) T = 250 K

( )( ) ( )( )Si: 3/ 215

6

1.15.23 10 250 exp2 86 10 250in

−

⎡ ⎤−⎢ ⎥= ××⎢ ⎥⎣ ⎦

m

or 8 31.61 10 in c −= ×

( )( ) ( )( )Ge: 3/ 215

6

0.661.66 10 250 exp2 86 10 250in

−

⎡ ⎤−⎢ ⎥= ××⎢ ⎥⎣ ⎦

m

or 12 31.42 10 in c −= ×

( )( ) ( )( )GaAs: 3/ 214

6

1.42.10 10 250 exp2 86 10 250i −

⎡ ⎤−⎢ ⎥= ××⎢ ⎥⎣ ⎦

m

−= × <<<

n

or 3 36.02 10 in c −= ×

TYU1.2

( ) (( )(a) so 16 35 10 , ,n cm p n )19 161.6 10 1350 5 10ne nσ μ −≅ = × × or

( ) 110.8 cmσ −= Ω −

(b) 16 35 10 , ,p cm n p−= × <<< so ( )( )( )19 161.6 10 480 5 10pe pσ μ −≅ = × × or

( ) 13.84 cmσ −= Ω − TYU1.3

( )( )10 15J Eσ= = or 2150 /J A cm= TYU1.4

(a) n n ndn nJ eD eDdx x

Δ= =

Δ so ( )( )

15 1619

4

10 101.6 10 350 2.5 10nJ −

−

⎛ ⎞−= × ⎜ ⎟− ×⎝ ⎠

or 2202 /nJ A cm=

(b) p p pdp pJ eD eDdx x

Δ= − = −

Δ so ( )( )

14 1519

4

10 5 101.6 10 12.50 4 10pJ −

−

⎛ ⎞− ×= − × ⎜ ⎟− ×⎝ ⎠

or 224.5 /pJ A cm= −

TYU1.5

(a)

( )

15 3

21024 3

15

8 10

1.5 102.81 10

8 10

o d

io

o

n N cm

np cn

−

m−

= = ×

×= = = ×

×

(b) 15 158 10 0.1 10on n nδ= + = × + ×or

15 38.1 10 n c −= × m4

4 12.81 10 10op p pδ= + = × +

or 14 310 p cm−≅

TYU1.6

(a) 2ln a dbi T

i

N NV V

n⎡ ⎤

= ⎢ ⎥⎣ ⎦

so ( )( )( )( )

15 17

210

10 100.026 ln 0.697

1.5 10biV V

⎡ ⎤⎢ ⎥= =⎢ ⎥×⎣ ⎦

(b) ( )( )( )( )

17 17

210

10 100.026 ln 0.817

1.5 10biV V

⎡ ⎤⎢ ⎥= =⎢ ⎥×⎣ ⎦

TYU1.7

(a) exp 1DD S

T

VI I

V⎡ ⎤⎛ ⎞

= −⎢ ⎥⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

14 0.510 exp0.026DI − ⎛ ⎞≅ ⎜ ⎟

⎝ ⎠

Then, for VD = 0.5 V, 2.25 DI Aμ= VD = 0.6 V, 0.105 DI mA= VD = 0.7 V, 4.93 DI mA= (b) 1410 D SI I A−≅ − = − for both cases. TYU1.8

100T CΔ = so 2 100 200 DV mΔ ≅ × = VVThen 0 650 0 20 0 450 DV . . .= − =

TYU1.9

1 2 3 40

Diode0.450.500.55

0.0330.2251.54

Load line

VD(v)

ID(mA)

1.00.87

0.54v

VD ID

TYU1.10

( )1.05 0.7D D DP I V I= ⇒ = so 1 5 DI . mA=

Now 10 0.7 6.2 1.5

PS

D

V VR R

Iγ− −

= = ⇒ = kΩ

TYU1.11

0.8 30.8 0.026

Dd

T

Ig mS

V= = =

TYU1.12

0.026 0.0265050

Td D

D D

Vr I

I I= ⇒ = ⇒ =

or 0.52 DI mA=

TYU1.13 For the pn junction diode,

4 0.7 0.825 4DI mA−

= =

For the Schottky diode, 4 0.3 0.925 4DI mA−

= =

TYU1.14

z zo z z zo z z zV V I r V V I r= + ⇒ = − so ( )( )35.20 10 20 5.18 zoV V−= − =

Then ( )( )35.18 10 10 20 5.38 z zV V−= + × ⇒ = V