Solution Manual for Semiconductor Physics and Devices 3ed (Neamen)

Since Ψ1 is a solution, then This equation is not necessarily valid, which means that Ψ Ψ1 2 is, in general, not a solution to Schrodinger’s wave equation... Term with B2 represents inci

Chapter 1

Problem Solutions

1.1

(a) fcc: 8 corner atoms × 1/8 = 1 atom

6 face atoms × ½ = 3 atoms

Total of 4 atoms per unit cell

(b) bcc: 8 corner atoms × 1/8 = 1 atom

1 enclosed atom = 1 atom

Total of 2 atoms per unit cell

(c) Diamond: 8 corner atoms × 1/8 = 1 atom

6 face atoms × ½ = 3 atoms

4 enclosed atoms = 4 atoms

Total of 8 atoms per unit cell

(a) Simple cubic lattice; a=2r

Unit cell vol =a3 =( )r 3= r3

F

HG I KJ

4 43

d =4r=a 3⇒ =a 4 r

3 Unit cell vol = =F

H I K

343

2 atoms per cell, so atom vol = F

HG I KJ

2 43

F

HG I KJ F

H I K

2 4343100%

3 Unit cell vol = =F

H I K

383

8 atoms per cell, so atom vol 8 4

Ratio

r r

F

HG I KJ F

H I K

8 4383100%

From Problem 1.3, percent volume of fcc atoms

is 74%; Therefore after coffee is ground, Volume =0 74 3

cm

8 3

x x

13

11, ,

FH IK ⇒ (313 )(b)

14

12

14, ,

FH IK⇒ 121( )

(i) (100) plane, surface density,

Same as (a),(i); surface density 4.94 1014 2

(iii) (111) plane, surface density,

Same as (a),(iii), surface density 2.85 1014 2

(c)

Face centered cubic

(i) (100) plane, surface density

−

24.50 108 2

atoms x

4

3 4.50 102

(c) (111) plane, surface density,

6 3642

or

valence electrons per atom, so

Density of valence electrons 2 1023 3

An average of 4 valence electrons per atom,

Density of valence electrons 1 77 1023 3

x x

1.21

2 103

5 43

d

a O =146

λ ω

x

t constant Then

x x

6 625 10

5 4 10

34 25

x x

6 625 102.31 10

34 23

or

λ =0 287 A °(c) Tungsten Atom: At Wt = 183.92

x x

6 625 10

313 10

34 22

or

λ =0 0212 A °(d) A 2000 kg traveling at 20 m/s:

x x

6 625 10

4 10

34 4

or

λ =1 66 10− 28 °

6 625 10

7.1 10

34 26

λ E m

h e

e

HG I KJ

12

p

λ 100 2

21002

x x

6 625 10

1822 10

34 26

λ =364 A °(b)

26 31

4

6 625 10 3 10

1 10

10

or

E =1 99 10x − 15 J

.Now

x x

6 625 10

6 02 10

34 23

λ =0 11 A °

8 78 10

5 10

29

8 78 10

5 10

26

32

or ∆t=6 6 10x − 16 s

h aΨ Ψ1 2fwhich can be written as

2 2

2 1 2

2 2 2 1

2 1 2

Since Ψ1 is a solution, then

This equation is not necessarily valid, which

means that Ψ Ψ1 2 is, in general, not a solution to

Schrodinger’s wave equation

HG I KJ z

4

a

a o

2

a

a o

HG I KJ z

a

a o

P=0 865

2 22

1 48 1030

Energy in the (n+1) state is 1 48 1030

x − Joules larger than 10 mJ

42

92

162

2 2

2 2

x

x y z y

x y z z

mE XYZ

2 2

2

2

0h

Dividing by XYZ and letting k2 mE

2

2

=

h , we obtain

2 2

2 2

2 2 2

x y

,h Let ψ( )x y, = X x Y y( ) ( )

Then substituting,

Y X

Y y

mE XY

We can also define 1

2 2 2

Y y( =b)= ⇒0 k b y =n yπ

so that

k n

b y y

Now − −k x2 k y2+ mE =

2

20hwhich yields

E E

m

n a

n b

π b g

For n2 =2,n1=1Then

h π

Term with B2 represents incident wave, and

term with A2 represents the reflected wave

Term involving B1 represents the transmitted

wave, and the term involving A1 represents the

reflected wave; but if a particle is transmitted

into region I, it will not be reflected so that

x x

x x

F

HG I KJThe reflection coefficient is

T|

U V| W|

2 2

2.35

Region I, V =0(x<0 ; Region II, )

V =V O (0< <x a); Region III, V =0(x>a) (a) Region I;

ψ1( )x = A1expa fjK x1 +B1expa−jK x1 f (incident) (reflected) Region II;

ψ2( )x = A2expa fK x2 +B2expa f−K x2

Region III;

ψ3( )x = A3expa fjK x1 +B3expa−jK x1 f(b)

In region III, the B3 term represents a reflected wave However, once a particle is transmitted into region III, there will not be a reflected wave

which means that B3 =0 (c)

so from the boundary conditions, we want to

solve for A3 in terms of A1 Solving for A1 in

4

= + l b − gexpa f a f−exp −

−2jK K1 2 expa f a f s a fK a2 +exp −K a2 exp jK a2

We then find that

24

2 24

1 2 2 2 2

22

1

=FH IK FHG IKJ− ( )

hThen

The transmission coefficient is defined as

From boundary conditions, solve for A3 in terms

of A1 The boundary conditions are:

The general solutions can be written, keeping in

mind that ψ1 must remain finite for x<0, as

= −F

HG I KJ tanwhich gives

h

or E

mE a

/expand

a

r a

ψ100 is independent of θ and φ, so the wave

equation in spherical coordinates reduces to

/exp

d

r a

r d

r a

2 100

5 2 2

so that d

d dr

r a

r a

r a

r a

r a

m E

r a o

/exp

/exp

which gives 0 = 0, and shows that ψ100 is indeed

a solution of the wave equation

2.41

All elements from Group I column of the periodic table All have one valence electron in the outer shell

Chapter 3

Problem Solutions

3.1 If a o were to increase, the bandgap energy

would decrease and the material would

begin to behave less like a semiconductor

and more like a metal If a o were to

decrease, the bandgap energy would

increase and the material would begin to

behave more like an insulator

Region I, V x( )=0 , so substituting the proposed

solution into the wave equation, we obtain

exp

h = FH IK− ⋅ ( ) L FH −FH IK IK

exp

h +∂

exp

h = + ( ) L FH −FH IK IK

hThis equation can then be written as

In region II, V x( )=V O Assume the same form

exp

h +∂

( ) L F H F H I K I K

NM O QP UV W

2 2

u x

E t

exp

h + ( ) L FH −FH IK IK

h = ( ) L FH −FH IK IK

hThis equation can be written as

2

2 22

2 2 2

22

For the differential equation in u x2( ) and the

proposed solution, the procedure is exactly the

dx

du dx

du

dx

du dx

−(α+k B) exp −j(α+k a)

−(β−k C) exp −j(β−k b)

+(β+k D) exp j(β+k b) =0

3.5 Computer plot 3.6 Computer plot 3.7

2 2h

E1=1 0163 eV

E2 =15041 eV

so ∆E=0 488 eV

E1=0 8665 eV

E2 =1504 eV

so ∆E=0 638 eV

(c)

2π <ka<3π

1st point: αa =2.42 π

2nd point: αa=3 π

2

2 1h

2

2 2 1

2 22h

219

0 7 1 6 10 10 1 054 10

219

1 42 1 6 10

6 625 10

19 34

(c) Curve A: Effective mass is a constant

Curve B: Effective mass is positive around

k =0, and is negative around k = ±π

m

d E dk

2 2

2 2

x

x y z y

x y z z

2 2

2 2

+2mE2 ⋅XYZ =0

hDividing by XYZ, we obtain

2 2

2 2

Also, ψ(x y z, , )=0 at x =a , then X a( )=0 so

we must have k a x =n xπ, where

n x =1 2 3, , , Similarly, we have 1

2 2 2

The total number of quantum states in the

3-dimensional potential well is given (in k-space)

by

g k dk T( ) = πk dk⋅a

π

2 3 3where

Substituting these expressions into the density of

states function, we obtain

31 3 2

34 3

π

πb g* /Now

31 3 2

34 3

π

m m C V n p

As a 1st approximation for T>0, assume the

probability of n=5 state being occupied is the

same as the probability of n=4 state being

empty Then

1

11

n n n x y z =221 122 contains both an electron =

and an empty state, so

1 1

11

111

a f=

=+

E kT E kT

a f

Hence, we have that

f E1a f1 = −1 f E2a f2 Q.E.D

1

1 0 011

1

1

11

Then

1exp

exp(a) T=0, For

dE F

dE F

At E E df

dE F

11

11

Using results of Problem 3.35, the answers to

part (b) are exactly the same as those given in

kT F

d g f

x kT

C F

a f ∝1 − FH IK−

2

1 2 /exp

/expThis yields

Let E V − ≡E x

NM O QP

exp

expwhere

kT i

( ) ( ) =47.5

1 04 10

19 19

0which becomes

kT O

F

E C

11

kT O

NM O QP1

r O

1 =∈ FH IK*For Germanium, ∈ =r 16,m*=0 55 m O

E m

O S

r O

p O =4.5 10x 15cm− 3

, p O >n O ⇒ p-type (b)

n

O i

HG I KJln

x x

.and

n

C O

HG I KJln

x x

b gexp .

or p O =9.67 10x − 3cm− 3

HG I KJln

HG I KJln

n n

p

x x O

i O

HG I KJln

x x

i O

Then

p x( )′ = ′K x′exp( )− ′x

To find the maximum of p E( )→ p x( )′ , set

dp x dx

b gexp .

= 2.8 1019 (−4.737)

= 4.7 1017 (−3 224)

b gexp .

= 7 1018 (−4.332)

i O

13

2.4 102.95 10

n O =1 95 10x 13 cm− 3

.(b)

i

2 2Then

HG I KJ

5 102

p n

n

x x O

i O

2exp =

d d

=8 85 10− 4.And

E−E F =aE E− Cf a+ E C −E Ff

or

E−E F =kT+0 245.Then

Also

p n

n

x x O

i O

i O

p O =1 62 10x − 3cm− 3

.(b)

p O = N a =1016cm− 3

n n

p

x O

i O

16

18 1010

n O =3 24 10x − 4cm− 3

.(c)

n O = p O =n i = x cm−

18 106 3

p O = N a = cm−

1014 3and

n n

p

x O

i O

14

3 28 1010

n O =1 08 10x 5 cm− 3

.(e)

p n

n

x O

N a > N d ⇒ p-type (b)

i

2 2

n n

p

x x O

i O

n O = x cm−

4.23 1011 3(c)

Total ionized impurity concentration

i O

n O = x cm−

1125 1015 3

n O > p O ⇒ n-type

p O = n i ⇒

.And

p n

n

x O

i O

16

15 1010

p O =2.25 10x 4 cm− 3(b)

N a > N d ⇒ p-type

p O = N a−N d =3 10x 16−2 10x 15

or

p O =2.8 10x 16 cm− 3Then

n n

p

x x O

i O

16

15 102.8 10

p n

n

x x O

i O

4

15 104.5 10

N d =5 10x 15cm− 3

Donor impurity concentration

HG I KJlnFor Germanium:

p O N a N a n

i

= 2 + FH IK2 +

2 2

HG I KJlnFor Germanium,

so

N d = x cm−

1 2 1016 3

x x

HG I KJln

p O < N a, Donors must be added

HG I KJln

x x

HG I KJln

HG I KJln

HG I KJln

x x

E F −E Fi =0 0024 eV

HG I KJln

x x

HG I KJln

HG I KJln

HG I KJln

x x

E Fi−E F =0 3294 eV

(page left blank)

(ii) For GaAs doped at N a = cm−

σ ≈eµp p O ⇒

0 01 1 6 1019 480 =b x − g ( )p O

5.3

A

L A

v d =(1100 500)( )⇒ v d =5 5 10x 5cm s

or

µn =3333cm V2 −s

/(b)

1 2 10

4 6 t t =8 33 10x − 11s

−107.5 10

4

6 t t =1 33 10x − 11 s

.(b)

Silicon: ForΕ =50 kV cm/ ,

v d =9.5 10x 6 cm s

/ Then

t d

t d

−109.5 10

t d

t d

−10

n n

p

x

x O

Note: For the doping concentrations obtained in

part (b), the assumed mobility values are valid

N d =9.26 10x 14cm− 3(b)

which yields 2.625 10

300

11012

HG I KJ.

exp

exp

E kT E kT

1500 =0 00050 0 000667 0 0020 + +

0 01019

16 15

5.24

J eD dn

n x

4

x n

or

J x n( =0)=2 A cm2

/Then

p dif, = − pb g1015 FH IK FH IK−1 exp −

or

x

x d

exp

x

x d

5.31

kT i

2

.(i) At x=0, J n = −2.95 10x 3 A cm2

/(ii) At x=5µm , J n = −23 7 A cm2

x x

= −( ) FH IK− − −

3 3

dN x dx

(b)

E V

W

x H

n I B

edV

x z H

µn

x x

I L enV Wd

= − ⇒ n=8 68 10x 14cm− 3

(c) µn

x x

I L enV Wd

R n

x n

6 625 10 3 10

6300 10

10

6.5

We have ∂

We can write ∇ •( )pΕ = • ∇ + ∇ •Ε p p Ε

and ∇ • ∇ = ∇p 2p

so

τδ

Multiply Equation (1) by µn n and Equation (2)

by µp p , and then add the two equations

In steady-state, δp= ′gτ

So that ∆σ =ebµn+µpgb gg′τpO

6.11

n-type, so that minority carriers are holes Uniform generation throughout the sample means we have

δτ

( )

t nO

δτ

δ

The solution is of the form

δn= ′gτnO 1 exp− a−t τnOfNow

14 20

or

τnO = − s

106(c)

Determine t at which

0 75 1014 3

We have

0 75 1014 10 114 x = −expa−t τnOfwhich yields

R p p

R

x pO

O

pO

pO O pO

τ τ

2.25 1010

4 11

pO

δ

τ

102.25 10

14 7

From part (a), τpO =2.25 10x − 7 s

nO

2

δ δτ

For x→ ∞, δn remains finite, so that B=0

Then the solution is

x x n

n

10

115

δ

expSubstituting into the differential equation

pO

2

0exp( )−µ exp( )− exp( )=

In order that δp=0 for x>0, use the minus

sign for x>0 and the plus sign for x<0

Then the solution is

n L n

n O

µ =FH IK so we can define

µn n O O

L L n

.Then

α =5 75 102 − 1

(c)

Force on the electrons due to the electric field is

in the negative x-direction Therefore, the

effective diffusion of the electrons is reduced and

the concentration drops off faster with the

applied electric field

continuous at x= −L; The flux must be continuous so that

δp G

O p

= ′ (3 − ) for L< <x 3L

δp G L

O p

= ′ (3 + ) for −3L< < −x L

6.26

µp O

18756

2

x Dt

x Dt

/

expAlso

into the differential equation, we find 0 0= ,

HG I KJln

x x

HG I KJln

x x

x x

n

O i

HG I KJlnand

n

O i

i

O i

HG I KJ ( )

0 01Then

p p

p O O

+

δ

exp 0 01 1 010

x x

g O = R O since these are the thermal equilibrium

generation and recombination rates If g′ =0 ,

δ =τ =

+ −1

107 1

(b)

Intrinsic, n O = p O =n i

Then R

δ =τ +τ = − + − ⇒

107 5 107 R

δ =

+ −

1 67 106 1

2 2

2

δ δτ

Applying the boundary conditions, we find

δn δn x

W O

D O n

6.42

Computer Plot

(page left blank)

HG I KJln

x x

HG I KJln

x x

n

a d i

HG I KJ

n d a

HG n N I KJ

a i

x x

x x

We can write

N N C V = N N CO VOF H T I K

3003

g

2

HG I KJexpNow

a d i

a d i

2 1

2 2

2 1

x

l −ln 2.8 1019 1 04 10 19 + UV W

2

kT g

1

kT g

bi bi

g

g

2 1

E kT

so that E

HG I KJln

x D n

=0 1320

x x

x x

NM O QP F

a d i

2

22

a d i

or ∆V bi =17.95mV

18 15

18 16

16 15

1 2

( ) ( )

=

+

++

18 16

18 15

1 2

C

V j

1 21

+

−L

N

eV

x x d

1 2

x x

x x

14 19

A one-sided junction and assume V R >>V bi, then

eN p

R a

NM2 O QP

1 2 /

so

14 16

y=mx b+

V

x R

HG I KJ

(c) p-region d

Ε = −

eN

x x aO

a f

We also have

At x= +x O and x= −x O, Ε =0

So 0

2Set φ=0 at x= −x O, then

0

3 3

3

1 2

HG I KJln(a)

D

a n

N

D N D a

n nO

a n

p pO

which yields N

N a d

=14.24 Now

N a = x cm−

1 01 1016 3

eD n

L

eD p L

n

n pO n

n pO n

p nO p

i a

p nO

n pO

a d

ττ

nO pO

n N e

L

n N

e L

n N

n

n n i a n

n

i a p p i d

D t

eD n L

eD p L

D

L N D

p nO p

n

n a n

D L

N N

n n n n p p a d

0 95

25

15 825

15 8

1010

HG I KJln

x x

HG I KJln

110

10 910

HG I KJexp

I

x x

−

−

3 76 104.48 10

16 15

a t

p nO

p

p

pO i d

10

2.4 1010

6

13 2 16

or

I S = x − A

2.91 109(a)

or I = − = −I S 2.91nA

n

a d i

The total current is

I I eV

kT

x x L

32.83

16

10 32.4 10

= L FH IK−

Then we find the total number of excess

electrons in the p-region to be:

eV kT i

HG I KJ

exp

exp

or I

2

x x

a n

or

I CT E

kT S

g

HG I KJ

3exp

eV kT f

0 510

b g b g

or kT

kT

C V a n

110

1010

x

E kT

g

exp

or exp +

1104.66 10

HG I KJ

L

NMexp 1 O QP = Aexpb+x L n pg+Bexpb−x L n pg

and the boundary condition at x=x n+W n gives

nO

a t

n p

a t

p

n p

HG I KJexp

sinh

sinh1

HG I KJ

V W L

p nO

a t

n p

coshThen

J eD p

L

W L

V V p

p nO

p

n p

a t

HG I KJ

2expFor the temperature range 300≤ ≤T 320K,

neglect the change in N C and N V

So

kT

eV kT D

1

1 1 2 2

−

−10

ω

ωω

For a forward-bias voltage, the junction capacitance is

HG21 I KJ b τ τ gwhere

I Aen

N

V pO

i d p

pO

a t

HG I KJ2

τ exp

a

n nO

a t

or

V pO

a t

or

V nO

a t

or

V d

a t

pO pO

HG21 I KJ b τ gNow

τpO t

n pO

n

a t

HG I KJexp

L

A e N n

2 5

HG I KJln

3 10

3 10

or V =1 98 V

(b) For R=0

V =0 477 V

a t

D S

D

a

S t

a t

x x

+ − O

QP

110

a t gen

a t

x

a t gen S

2

2.73 104.16 10 6 56 10

7 11

D

a n

10

77.710

−

O QP

x x

14 19

17 34

V rec

i O

a t

HG I KJ

2τ exp 2 =

14 19

17 34

Reverse-bias; ratio of generation to ideal diffusion current:

I

I

x x gen

Ratio =1 29 106

x For V a =0 5 V

kT i

O

a a

+

η η

R n eV

kT i

x x

3 10

1810

610

HG I KJexpAlso

which yields exp V

D t

22

I J A V

V S

a t

B d

19 18 14

From Figure 8.25, the breakdown voltage is

approximately 300 V So, in each case,

breakdown is reached first

τ =0 2

Then

erf t

I I

erf S

F

τ = + =

11

0 2

where

erf 0 2 =erf(0 447 )=0 473

We obtain I

I R F

0 473 1

I I R F

pO

pO

R F

HG I KJ − FH IK

τ ln 1 10 ln 1 2

17

b g

or

t S =11 10x − 7 s

.Also

C avg =18 4.2+ = pF

The time constant is

x x

(page left blank)

HG I KJln

bi d

N

HG I KJln

C= ′C A

so

C=15pF