半导体物理与器件第三版(尼曼)09章答案

半导体物理与器件第三版(尼曼)09章答案,本文库还有其它章节答案

Chapter 9

Problem Solutions

9.1

(a) We have eφFCn=eVtln

GHNINd

19 =(0.0259)lnFGH2.8x101016

=0.206eV

(c)

φBO=φm χ=4.28 4.01 or

φBO=0.27V and

Vbi=φBO φn=0.27 0.206 or

Vbi=0.064V Also

x=LM2∈V1/2bi

d

d

141/2

=LNOeNPQ

M

x10 19

1016

or

N2(11.7)b8.85x10g(0.064)O1.6PQ

x 6

d=9.1x10cm Then ΕeNdxd

max= =b∈

1.6x10 19

gb1016

gb9.1x10 6

g(11.7)8.85x10 14

or

Ε4

max=1.41x10V/cm (d)

Using the figure, φBn=0.55V So

Vbi=φBn φn=0.55 0.206 or

Vbi=0.344V We then find

x 5

d=2.11x10cm and Ε4

max=3.26x10V/cm

9.2

(a) φBO=φm χ=51. 4.01 or

φBO=1.09V (b)

φFn=Vtln

GHNCNd

19 =(0.0259)lnFGH2.8x101015

=0.265V

Then

Vbi=φBO φn=1.09 0.265 or

Vbi=0.825V

(c)

1/2 W=x=LMN2∈Vbi

Od

eNd

PQ

141/2

=LM

19

15

or

N2(117.)b8.85x10g(0.825)O1.6x1010

W=1.03x10 4

cm (d) ΕeNdxd

max=

∈

=

b

1.6x10 19

gb1015

gb1.03x10 4

g(11.7)8.85x10 14

or

9.3

(a) Gold on n-type GaAs

χ=4.07V and φm=51.V φBO=φm χ=51

. 4.07 and

φBO=1.03V (b) φn=Vtln

FGHNCN d

半导体物理与器件第三版(尼曼)09章答案,本文库还有其它章节答案

17 =(0.0259)lnFGH4.7x105x1016

or

φn=0.0580V (c)

Vbi=φBO φn=1.03 0.058 or

Vbi=0.972V (d)

1/2

x=LMN2∈aVbi

+VR

fOd

eNd

PQ

2(131. 141/2

=LMN)b8.85x10g(0.972+5)O1.6x10 19

5x1016

PQ

or

xd=0.416µm

(e) ΕeNdxd

max=

∈

b1.6x10 19

gb5x1016

gb0.416x 4

=

10g(131.)8.85x10 14

or

Ε5

max=2.87x10V/cm

9.4

φBn=0.86Vandφn=0.058V (Problem 9.3) Then

Vbi=φBn φn=0.86 0.058 or

Vbi=0.802V and

L2∈V+1/2

x=MNbi

VR

Od

eNd

PQ

141/2

=LM2(131

.)b8.85x10g(0.802+5)O1.6x105xor]

N 19

1016

xd=0.410µm

Also ΕeNdxd

max=

∈

19

16

4

=

b1.6x10gb5x10gb0.410x10g(131.)8.85x10 14

or

Ε5

max=2.83x10V/cm

9.5

Gold, n-type silicon junction. From the figure, φBn=0.81V

For Nd=5x1015cm 3

, we have

φFn=Vtln

GHNCNd

19 =(0.0259)lnFGH2.8x105x1015

= φ

n

=0.224V

Then

Vbi=0.81 0.224=0.586V (a) Now

1/2

C′=LMe∈Nd

Obi

R

P

LN2V+VbQ1. 19

14

15

1/2

=MM6x10

or

N

g(117.)b8.85x10gb5x10gO2(0.586+4) C′=9.50x10

9

F/cm2

For A=5x10 4

cm2

, C=C′A So

C=4.75pF (b)

For Nd=5x1016

cm 3

, we find

19

φn

=(0.0259)lnFGH2.8x105x1016

=0.164V

Then

Vbi=0.81 0.164=0.646V Now

C′=LMNb1.6x10 19

g(11.7)b8.85x10 14

gb5x1016

gO1/2

2(0.646+4)Por

Q

C′=2.99x10 8

F/cm2

and

C=C′A so

C=15pF

半导体物理与器件第三版(尼曼)09章答案,本文库还有其它章节答案

9.6 (a) From the figure, Vbi=0.90V (b) We Ffind H12

C′3x1015

V= 0x1015

R

2 ( 0.9)=1.03and

1.03x1015

=

2

e∈N

dThen we can write Nd=2

1.6x10 19

(131

.)8.85x10 14

1.03x1015

or

N16

3

d=1.05x10cm

(c)

φln

FNC

n=VtGHNd

17 =(0.0259)lnFGH4.7x101.05x1016

or

φn=0.0985V (d)

φBn=Vbi+φn=0.90+0.0985 or

φBn=0.9985V 9.7

From the figure, φBn=0.55V (a)

φ=VlnFGHNCn

t

Nd

19 =(0.0259)lnFGH2.8x101016

=0.206V

Then

Vbi=φBn φn=0.55 0.206 or

Vbi=0.344V We find 1/2

xN2∈VbiOd=

LMeNP

d

Q

=LMN2(11.7)b8.85x10 14

g

(0.344)OP1/21.6x10 19

1016Q or

xd=0.211µm

Also ΕeNdxd

max=

∈

=

b1.6x10 19

gb1016

gb0.211x10 4

g(11.7)8.85x10 14

or

(b) φ=

eΕ

Lb1.6x10 19

gb3.26x104g1/2

4π∈=Mor

NO4π(11.7)8.85x10 14

PQ

φ=20.0mV Also xm=

e16π∈Ε

19O1/2

=M

or

Nb1.6x10g16π(11.7)8.85x10 14

3.26x104

x0.307x10 6

m=cm (c)

For VR=4V 141/2

x=LMN2(11.7)b8.85x10g(0.344+4)Od

1.6x10 19

1016

Por

Q

xd=0.75µm and 19

16

4

Εmax=

b1.6x10gb10gb0.75x10g(11.7)8.85x10 14

or

φ=

eΕ4π∈

φ=37.8mV

and

半导体物理与器件第三版(尼曼)09章答案,本文库还有其它章节答案

xem=16π∈Ε

xm=0.163x10 6

cm

9.8

We have φ(x)= e16π∈x

Εx

or eφ(x)=e

2

16π∈x+Εex

Now

d(eφ(x))

e

2

dx

=0=

16π∈x

2

+Εe

Solving for x2

, we find

x2

=

e16π∈Ε

or

x=xm=

e16π∈Ε

Substituting this value of xm=x into the

equation for the potential, we find φ=

e

16π∈

e+Ε

e16π∈Ε

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