# 반도체물성과소자-4판-솔루션-semiconductor-physics-and-devices-solution

```lOMoARcPSD|13114859
반도체물성과소자 4판 솔루션 - Semiconductor Physics and
Devices Solution
반도체소자 (한국과학기술원)
StuDocu is not sponsored or endorsed by any college or university
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 1
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Chapter 1
Problem Solutions
1.1
(a) fcc: 8 corner atoms  1 / 8  1 atom
6 face atoms  1 / 2  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  1 / 2  3 atoms
4 enclosed atoms = 4 atoms
Total of 8 atoms per unit cell
_______________________________________
1.2
(a) Simple cubic lattice: a  2r
Unit cell vol  a 3  2r   8r 3
3
 4 r 3
1 atom per cell, so atom vol  1
 3
Then
 4 r 3 


 3 


 100%  52.4%
Ratio 
8r 3
(b) Face-centered cubic lattice
d
d  4r  a 2  a 
 2 2 r
2

Unit cell vol  a 3  2 2  r

3




 4 
Unit cell vol  a 3  
 r 
 3 
Ratio 
3
3

 4r 


 3


(d) Diamond lattice
3




 100%  68%
8
Body diagonal  d  8r  a 3  a 
r
3
 8r 

Unit cell vol  a 3  

 3
3
 4 r 3 

8 atoms per cell, so atom vol  8

 3 
Then
3


8 4 r 
 3 
 100%  34%
Ratio 
3
 8r 


 3


_______________________________________
o
(a) a  5.43 A ; From Problem 1.2d,
a
8
r
3




o
a 3 5.43 3

 1.176 A
8
8
Center of one silicon atom to center of
Then r 
o
nearest neighbor  2r  2.35 A
(b) Number density
8

 5  10 22 cm 3
3
5.43  10 8
(c) Mass density
N  At.Wt. 5  10 22 28.09


NA
6.02  10 23




   2.33 grams/cm 3
_______________________________________
3
 4 r 3
2 atoms per cell, so atom vol  2 
 3

2 4 r
1.3
 16 2  r 3
 4 r 3
4 atoms per cell, so atom vol  4 
 3
Then
3


4 4 r 
 3 
Ratio 
 100%  74%
16 2  r 3
(c) Body-centered cubic lattice
4
d  4r  a 3  a 
r
3
Then




lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 1
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
1.4
(a) 4 Ga atoms per unit cell
4
Number density 
5.65  10 8

(b) a  21.035  2.07 A
o
(c) A-atoms: # of atoms  8 

3
Density 
 Density of Ga atoms  2.22  10 22 cm 3
4 As atoms per unit cell
 Density of As atoms  2.22  10 22 cm 3
(b) 8 Ge atoms per unit cell
8
Number density 
3
5.65  10 8

1.5
From Figure 1.15
 a  3 
 0.4330 a
(a) d   
 2  2 

 3.38  10 cm 3
_______________________________________
# of atoms  8 
o
a
2

2
  2

  54.74
sin   
3
2
2 a 3
2
   109.5
_______________________________________
1.7
(a) Simple cubic: a  2r  3.9 A
o
 5.515 A
 4.503 A
24r 
o
 9.007 A
3
_______________________________________
1.8
(a) 21.035 2  21.035  2rB
o
rB  0.4287 A
4.5 10 
8 3
1.0974 10 12.5
22
6.02  10 23
 0.228 gm/cm 3
(b) a 
4r
o
 5.196 A
3
1
# of atoms 8   1  2
8
2
5.196 10 
8 3
 1.4257  10 22 cm 3
1.4257  10 22 12.5
Mass density   
6.02  10 23
 0.296 gm/cm 3
_______________________________________

o
(d) diamond: a 

Number density 
o
1
 1.097  10 22 cm 3
N  At.Wt.
Mass density   
NA
1.6
3
1
1
8
Number density 
 0.70715.65  d  3.995 A
_______________________________________
(c) bcc: a 

23
o
a
(b) d    2  0.7071a
2
2
4r
8 3
(a) a  2r  4.5 A
 0.43305.65  d  2.447 A
4r
2.07 10 
1.9
o
(b) fcc: a 
1
 1.13  10 23 cm 3
1
B-atoms: # of atoms  6   3
2
3
Density 
3
2.07  10 8

 Density of Ge atoms  4.44  10 22 cm 3
_______________________________________
1
1
8

1.10
From Problem 1.2, percent volume of fcc
atoms is 74%; Therefore after coffee is
ground,
Volume = 0.74 cm 3
_______________________________________
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 1
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
1.11
o
o
(b) a  1.8  1.0  2.8 A
(c) Na: Density 
1 / 2
2.8 10 
8 3
 2.28  10 22 cm 3
Cl: Density  2.28  10 22 cm 3
(d) Na: At. Wt. = 22.99
Cl: At. Wt. = 35.45
So, mass per unit cell
1
1
 22.99    35.45
2
2

 4.85  10  23
6.02  10 23
Then mass density
4.85  10 23

 2.21 grams/cm 3
8 3
2.8  10
_______________________________________


1.12
(a) a 3  22.2   21.8  8 A
o
o
Then a  4.62 A
Density of A:
1

 1.01 10 22 cm 3
3
4.62  10 8
Density of B:
1

 1.01 10 22 cm 3
8 3
4.62  10
(b) Same as (a)
(c) Same material
_______________________________________


1.13
a


22.2   21.8


2
 4.687  10 14 cm 2
o
For 1.12(b), B-atoms: a  4.619 A
1
 4.687  1014 cm 2
a2
For 1.12(a) and (b), Same material
Surface density 
o
For 1.12(b), A-atoms; a  4.619 A
Surface density
1
 3.315  1014 cm 2

2
a 2
B-atoms;
Surface density
1

 3.315  1014 cm 2
2
a 2
For 1.12(a) and (b), Same material
_______________________________________
1.14
(a) Vol. Density 
1
a o3
1
Surface Density 
a
2
o
2
(b) Same as (a)
_______________________________________
1.15
(i) (110) plane
(see Figure 1.10(b))
(ii) (111) plane
(see Figure 1.10(c))
o
 4.619 A
3
(a) For 1.12(a), A-atoms
1
1
Surface density  2 
a
4.619  10 8
(b) For 1.12(a), A-atoms; a  4.619 A
Surface density
1
 3.315  1014 cm 2

2
a 2
B-atoms;
Surface density
1

 3.315  1014 cm 2
2
a 2
1 1 
(iii) (220) plane   , ,    1, 1, 0 
2 2 
Same as (110) plane and [110] direction
 1 1 1
(iv) (321) plane   , ,   2, 3, 6 
 3 2 1
Intercepts of plane at
p  2, q  3, s  6
[321] direction is perpendicular to
(321) plane
_______________________________________
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 1
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
1.16
(a)

1 1 1
 , ,   313
1 3 1
(b)
1 1 1
 , ,   121
4 2 4
_______________________________________
1.17
1 1 1
Intercepts: 2, 4, 3   , ,  
 2 4 3
(634) plane
_______________________________________
1.18
o
(a) d  a  5.28 A
o
a 2
 3.734 A
2
o
a 3
 3.048 A
(c) d 
3
_______________________________________
(b) d 
1.19
(a) Simple cubic
(i) (100) plane:
Surface density 
1
1

a2
4.73  10 8


2
(ii) (110) plane:
1
Surface density 
a
2
2
 3.16  1014 cm 2
(iii) (111) plane:
1
bh
2
o
where b  a 2  6.689 A
Now
 
h  a 2
2
So h 
2
2
 
a 2
 3 a 2

 2 
4


o
6
4.73  5.793 A
2


 6.32  1014 cm 2
(iii) (111) plane:
1
3
6
Surface density 
19.3755 10 16
 2.58  1014 cm 2
(c) fcc
(i) (100) plane:
2
Surface density  2  8.94  1014 cm 2
a
(ii) (110) plane:
2
Surface density 
2
a 2
 6.32  1014 cm 2
(iii) (111) plane:
1
1
3  3
6
2
Surface density 
19.3755  10 16
 1.03  1015 cm 2
_______________________________________
 4.47  1014 cm 2
Area of plane 
Area of plane
1
 6.68923  10 8 5.79304  10 8
2
 19.3755  10 16 cm 2
1
3
6
Surface density 
19.3755  10 16
 2.58  1014 cm 2
(b) bcc
(i) (100) plane:
1
Surface density  2  4.47  1014 cm 2
a
(ii) (110) plane:
2
Surface density 
2
a 2
2
1.20
(a) (100) plane: - similar to a fcc:
2
Surface density 
2
5.43  10 8


 6.78  10 cm 2
14
(b) (110) plane:
Surface density 

4
2 5.43  10 8

2
 9.59  1014 cm 2
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 1
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(c) (111) plane:
Surface density 
2
 3 25.43 10 
8 2
 7.83  1014 cm 2
_______________________________________
1.21
a
4r

42.37 
o
 6.703 A
2
2
1
1
8  6
4
8
2 
(a) #/cm 3 
3
a
6.703  10 8

5  10 17
 100%  10 3 %
5  10 22
2  1015
 100%  4  10  6 %
(b)
5  10 22
_______________________________________

1.25
(a) Fraction by weight
2  10 16 10.82 
 1.542  10  7

5  10 22 28.06 
(b) Fraction by weight
1018 30.98

 2.208  10 5
22
5  10 28.06 
_______________________________________



 3.148  10 cm 2
14
o
a 2 6.703 2

 4.74 A
2
2
1
1
(d) # of atoms  3   3   2
6
2
Area of plane: (see Problem 1.19)
(c) d 


 

Volume density 
o
6a
 8.2099 A
h
2
Area
1
1
 bh  9.4786  10 8 8.2099  10 8
2
2
 3.8909  10 15 cm 2


1.26
o
b  a 2  9.4786 A


(a)
3
 1.328  10 cm
1
1
4  2
2
4
(b) #/cm 2 
a2 2
2

2
6.703  10 8
2


1.24
3
22
1.23
Density of GaAs atoms
8

 4.44  10 22 cm 3
8 3
5.65  10
An average of 4 valence electrons per atom,
So
Density of valence electrons
 1.77  10 23 cm 3
_______________________________________
1
 2  10 16 cm 3
d3
o
So d  3.684  10 6 cm  d  368.4 A

2
3.8909  10 15
= 5.14  1014 cm 2
#/cm 2 
o
a 3 6.703 3

 3.87 A
3
3
_______________________________________
d
1.22
Density of silicon atoms  5 10 22 cm 3 and
4 valence electrons per atom, so
Density of valence electrons  2  10 23 cm 3
_______________________________________
o
We have a o  5.43 A
d
368.4

 67.85
5.43
ao
_______________________________________
Then
1.27
Volume density 
1
 4  10 15 cm 3
d3
o
So d  6.30  10 6 cm  d  630 A
o
We have a o  5.43 A
d
630

 116
a o 5.43
_______________________________________
Then
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 2
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Chapter 2
2.1
2.6
Sketch
_______________________________________
6.625  10 34

550  10 9
 1.205  10 27 kg-m/s
p 1.2045  10 27
 
 1.32  10 3 m/s
m
9.11 10 31
or   1.32  10 5 cm/s
h 6.625  10 34
(b) p  

440  10 9
 1.506  10 27 kg-m/s
p 1.5057  10 27
 
 1.65  10 3 m/s
 31
m
9.11 10
or   1.65  10 5 cm/s
(c) Yes
_______________________________________
2.2
Sketch
_______________________________________
2.3
Sketch
_______________________________________
2.4
From Problem 2.2, phase 
2 x

 t
= constant
Then
2 dx
dx
  
    0, 
  p   

 dt
dt
 2 
2 x
From Problem 2.3, phase 
 t

= constant
Then
2 dx
dx
  
    0, 
  p   

dt
 dt
 2 
_______________________________________
(a) p 
h

2.7
(a) (i)

 
p  2mE  2 9.11 10 31 1.2 1.6 10 19
25
 5.915  10 kg-m/s
h 6.625  10 34
 1.12 10 9 m
 
p 5.915  10  25
o
or   11.2 A

 
(ii) p  2 9.1110 31 12 1.6  10 19
2.5
hc
hc
E  h 
 
E



Gold: E  4.90 eV  4.90 1.6  10 19 J
So,
6.625  10 34 3  1010

 2.54  10 5 cm
4.90 1.6 10 19
or
  0.254  m







Cesium: E  1.90 eV  1.90  1.6  10 19 J
So,
6.625  10 34 3  1010

 6.54 10 5 cm
1.90 1.6 10 19
or
  0.654  m
_______________________________________






 1.87  10 24 kg-m/s
6.625  10 34
 3.54  10 10 m

1.8704  10  24
o
or   3.54 A



(iii) p  2 9.11 10 31 120 1.6  10 19
24
 5.915  10 kg-m/s
6.625 10 34
 1.12 10 10 m

 24
5.915 10
o
or   1.12 A


lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 2
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(b)

p  2 1.67  10
27
2.10
1.21.6 10 
19
 2.532 10 23 kg-m/s
6.625  10 34

 2.62  10 11 m
2.532 10  23
o
or   0.262 A
_______________________________________
2.8
E avg 
3
3
kT   0.0259  0.03885 eV
2
2
Now
p avg  2mE avg


6.625  10 34

85  10 10
 7.794 10 26 kg-m/s
p 7.794 10 26
 
 8.56  10 4 m/s
m
9.11 10 31
or   8.56  10 6 cm/s
1
1
E  m 2  9.11 10 31 8.56  10 4
2
2
 3.33  10 21 J
3.334  10 21
 2.08 10  2 eV
or E 
19
1.6 10
2
1
(b) E  9.11 10 31 8  10 3
2
 2.915 10 23 J
2.915  10 23
or E 
 1.82  10  4 eV
1.6  10 19
p  m  9.11 10 31 8  10 3
(a) p 

 2 9.11 10 31 0.03885 1.6 10 19

or
p avg  1.064  10 25 kg-m/s
h





Now
h 6.625  10 34
 6.225  10 9 m
 
p 1.064  10  25


2



27
 7.288  10 kg-m/s
h 6.625  10 35
 
 9.09  10 8 m
p 7.288  10  27
or
o
  62.25 A
o
_______________________________________
or   909 A
_______________________________________
2.9
E p  h p 
2.11
hc
p
(a) E  h 
Now
p e2
1  h
h

 Ee 
and p e 
e
2m   e
2m
Set E p  E e and  p  10 e
Ee 




1  h


 p 2m   e
hc

1  10h 
 

2m   p 

2
2
which yields
100h
p 
2mc
Ep  E 


hc
p

hc
2mc 2
 2mc 
100h
100

2 9.11 10 31 3  10
100

8 2

 1.99  10
2
Then
hc

6.625 10 3 10 
34
8
1 10 10
15
J
Now
E 1.99  10 15

e
1.6 10 19
V  1.24  10 4 V  12.4 kV
E  e V  V 


(b) p  2mE  2 9.11 10 31 1.99  10 15

23
 6.02  10 kg-m/s
Then
h 6.625  10 34
 1.10  10 11 m
 
p 6.02  10  23
or
o
  0.11 A
_______________________________________
 1.64  10 15 J  10.25 keV
_______________________________________
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 2
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
2.12
 1.054  10 34
p 

x
10  6
 1.054  10 28 kg-m/s
_______________________________________
2.13
(a) (i) px  
1.054  10 34
 8.783  10  26 kg-m/s
10
12  10
dE
d  p2 

  p
 p 
(ii) E 
dp
dp  2m 
pp
2p

 p 
m
2m
p 
Now p  2mE

 
 2 9  10 31 16  1.6  10 19

24
 2.147  10 kg-m/s
2.1466  10 24 8.783  10 26
so E 
9  10 31
 2.095  10 19 J
2.095  10 19
or E 
 1.31 eV
1.6  10 19
(b) (i) p  8.783  10 26 kg-m/s



 
(ii) p  2 5  10 28 16  1.6  10 19


23
 5.06  10 kg-m/s
5.06  10 23 8.783  10 26
E 
5  10  28
 8.888  10 21 J
8.888  10 21
or E 
 5.55  10  2 eV
1.6  10 19
_______________________________________



2.14
 1.054  10 34

 1.054  10 32 kg-m/s
x
10  2
p 1.054  10 32

p  m   
1500
m
p 
  7  10 36 m/s
_______________________________________
2.15
(a) Et  
1.054  10 34
t 
 8.23  10 16 s
0.8 1.6 10 19


 1.054  10 34

x
1.5  10 10
 7.03  10 25 kg-m/s
_______________________________________
(b) p 
2.16
(a) If 1 x, t  and 2 x, t  are solutions to
Schrodinger's wave equation, then
1 x, t 
  2  2 1 x, t 
 V x 1 x, t   j

2
t
2m
x
and
2 x, t 
  2  2  2  x, t 

 V x 2 x, t   j
2
2m
t
x
Adding the two equations, we obtain
 2 2
1 x, t   2 x, t 

2m x 2
 V x 1 x, t   2  x, t 

1 x, t   2 x, t 
t
which is Schrodinger's wave equation. So
1 x, t   2  x, t  is also a solution.
 j
(b) If 1 x, t   2 x, t  were a solution to
Schrodinger's wave equation, then we
could write
2
 2
1  2   V x 1  2 

2m x 2

 j 1  2 
t
which can be written as
 2 1
 2 
  2   2 2
 2
2 1 

1
2
2
x x 
2m 
x
x
1 
 2
 2
 V x 1  2   j 1
t 
t

Dividing by 1  2 , we find
 2
2m
2
 1  2 2
1  1
2 1 2 






2
2

x 


x
1
1  2 x
 2 x
 1 2
1 1 
 V  x   j 





t
1 t 
 2
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 2
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Since 1 is a solution, then
2.19

1  1
1 1
 V  x   j 



2
1 t
2m 1 x
Subtracting these last two equations, we have
2 1 2 
  2  1  2 2




2
1 2 x x 
2m  2 x

2
2
 j 
1 2

2 t
1  2
1 2

 V x   j 



2
2 t
2m 2 x
Subtracting these last two equations, we obtain
 2
 2
2

 1
 V x   0
2m 1 2 x x
This equation is not necessarily valid, which
means that 1 2 is, in general, not a solution
to Schrodinger's wave equation.
_______________________________________
*
0
Function has been normalized.
(a) Now
ao
P
2
 2
  x 
 dx
exp


 a o 
 a o
4

0
ao
Since 2 is also a solution, we have
2

ao
2
2
    dx  1
Note that

2
ao
2
1
0
  ao

 2
 x 
cos 2 
dx  1
 2 
ao
2
4
ao
2

ao
 3   1 
A      1
 2  2 
2
or A 
2

ao
1
2
1
2
  2x 
 dx
a o 
 exp
ao
4
  ao

 2
  2 x  ao


 exp


 a o  ao
2
4
or
2
_______________________________________
2.18

  1 
P   1exp 1  exp 
 2 

which yields
P  0.239
(c)
1 / 2

 
 1 
  1  1  exp 

 2 
 
 2
  x 
 dx
exp


 a o 
 a o
2

P
 x sin x    3
A2  
1
2  1
2
so A 2 
  2 x  ao 4


 exp


 ao  0
   2a o
P   1exp
  4a o
which yields
P  0.393
(b)
ao
A
 exp
or
2.17
3
  2x 
 dx
a o 
4
ao
A 2 cos 2 nx dx  1
P
1 / 2
 x sin 2nx   1 / 2
A  
1
4n  1 / 2
2
2
 1  1 
1
A 2       1  A 2  
4
4
2




or A  2
_______________________________________

0

2
 2
  x 
 dx
exp


 a o 
 a o
2
ao
ao
  2x 
 dx
a o 
 exp
0
  2 x  ao
  ao 

 exp


 2 
 ao  0
  1exp 2   1
which yields
P  0.865
_______________________________________

2
ao
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 2
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
2.20
2.21
P   x  dx

a/4
(a)

0
2
a/4
(a) P 
2
2 x 
dx
  cos 
a
 
 2 
 a 
 
   sin   
4
2
   
 2 

  
 4  
 a  2
 


 a  

 2   a 1a  
   
4 
 a  8
or P  0.409
a/2
2
  a  cos
a/4
2




 2   a sin   
  
 a   8  8  
 

 a 

or P  0.25
(b) P 
2
  a  sin
a/4
a / 2
(c) P 

 2x    a / 2
sin 


2 x
 a 
   
 4  
 a  2
  a / 2


 a  





 2  a sin     a  sin    

   

 a   4  4   4   4  
 




 a  
 a 

or P  1
_______________________________________
 2  2  2x 
dx
  sin 
 a 
a
a / 2


 4x    a / 2
sin


2 x
 a 
   
 2  
 a  2
4
  a / 2

 a  

a / 2
2
2  x 
  cos  dx
 a 
a
a / 2
 2x 
dx

 a 










sin
sin
2
a
a
2
 
 

   
 
 a   4  8   8   8  
 
 

 a 
 a 

or P  0.25
1 1 
1
 2  0  
8 4 
4
or P  0.0908

2

 4x   a / 2
sin 


2 x
 a 
   
 2  
 a  2
4
  a/4

 a  

 x 
 dx
 a 

 
sin   

 2  a sin   a
 2 
 
   

 a  4  4  8  4  
 




 a  
 a 

 2x 
dx

 a 

 4x   a / 4
sin 


2
x
 
 a 
   
 2  
 a  2
4
  0

 a  

a/2

 2x   a / 2
sin 


 2  x
 a 
  
  
 a  2
4   a / 4

a 

(c) P 
2
0

 2x   a / 4
sin 


2 x
 a 
   
  
 a  2
4   0

a 

(b) P 
2
  a  sin




 2  a sin 2    a  sin  2  

   

 8  
 a   4  8   4 
  
 

 a  
 a 

or P  1
_______________________________________
2.22
or

8  1012
 10 4 m/s
k 8  10 8
 p  10 6 cm/s
(a) (i)  p 


2
2

 7.854  10 9 m
k
8  10 8
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 2
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
2
 4.32  10 8 m 1
 1.454  10 8
  k  4.32  10 8 5  10 4
o
  78.54 A
or

 
(ii) p  m  9.11 10 31 10 4
 9.11 10 kg-m/s
2
1
1
E  m 2  9.11 10 31 10 4
2
2
 4.555  10 23 J
4.555  10 23
or E 
 2.85  10  4 eV
1.6  10 19
 1.5  1013
 10 4 m/s
(b) (i)  p  
k  1.5  10 9
or  p  10 6 cm/s

 
2
2

 4.19  10 9 m
9
k
1.5  10

 9.11 10 kg-m/s
6.625  10 34
 7.27  10 10 m

 25
9.11 10
2
 8.64  10 9 m 1
k
7.272  10 10
  8.64  10 9 10 6  8.64  10 15 rad/s
_______________________________________

p  9.11 10 27 kg-m/s



For electron traveling in  x direction,
  9.37  10 6 cm/s
p  m  9.11 10 31  9.37  10 4 
 8.537  10 26 kg-m/s
h 6.625  10 34

 7.76  10 9 m

 26
p 8.537  10
2
 8.097  10 8 m 1
 7.76  10 9
  k    8.097  10 8 9.37 10 4
k
2




or   7.586  10 rad/s
_______________________________________
13
2.24
(a) p  m  9.11 10 31 5  10 4


26


or

 4.555  10 kg-m/s
h 6.625  10 34
 
 1.454  10 8 m
p 4.555  10  26

2



2


n 2 1.0698  10 21
1.6  10 19
or E n  n 2 6.686  10 3 eV
Then
E1  6.69  10 3 eV

1
 m 2
2
1
9.11 10 31  2
2
so   9.37  10 4 m/s  9.37  10 6 cm/s


E n  n 2 1.0698  10 21 J
En 
2.23
(a)  x, t   Ae  j kx  t 

 
n 2 1.054  10 34  2
 2 n 2 2

2
2ma
2 9.11 10 31 75  10 10
En 
(b) E  0.025 1.6  10
 
2.25
E  2.85  10 4 eV
_______________________________________

25
or   41.9 A
19

(b) p  9.11 10 31 10 6
o
(ii)


27

2
k

E 2  2.67  10 2 eV
E 3  6.02  10 2 eV
_______________________________________
2.26


2
n 2 1.054  10 34  2
 2 n 2 2
(a) E n 

2
2ma
2 9.11 10 31 10  10 10




n 6.018  10 
 n 0.3761 eV

 n 2 6.018  10 20 J
20
2
or
En
2
1.6  10 19
Then
E1  0.376 eV
E 2  1.504 eV
E 3  3.385 eV
hc
E
E  3.385  1.504  1.6  10 19
(b)  

 3.01 10
19
J


2
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 2
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________

6.625 10 3 10 
34
8
19
3.01 10
 6.604  10 7 m
or
  660.4 nm
_______________________________________
2.27
(a) E n 
 2 n 2 2
2ma 2
15  10 3 


2
n 2 1.054  10 34  2



2
2 15  10 3 1.2  10  2

15  10 3  n 2 2.538  10 62

or n  7.688  10
(b) E n 1  15 mJ
(c) No
_______________________________________
29
2.28
For a neutron and n  1 :
E1 



2
1.054  10 34  2
 2 2

2ma 2 2 1.66  10  27 10 14


2
 3.3025  10 13 J
or
3.3025  10 13
 2.06  10 6 eV
1.6  10 19
For an electron in the same potential well:
E1 
1.054 10  

29.11 10 10 
34 2
E1
 31
2
14 2
 6.0177  10 10 J
or
6.0177  10 10
 3.76  10 9 eV
1.6  10 19
_______________________________________
E1 
2.29
Schrodinger's time-independent wave
equation
 2 x  2m
 2 E  V x  x   0

x 2
We know that
a
a
 x   0 for x  and x 
2
2
We have
a
a
x
V x   0 for
2
2
so in this region
 2 x  2mE
 2  x   0

x 2
The solution is of the form
 x   A cos kx  B sin kx
where
2mE
k
2
Boundary conditions:
a
a
 x   0 at x 
, x
2
2
First mode solution:
 1 x   A1 cos k1 x
where

 2 2
k1   E1 
a
2ma 2
Second mode solution:
 2 x   B 2 sin k 2 x
where
2
4 2  2
 E2 
k2 
a
2ma 2
Third mode solution:
 3 x   A3 cos k 3 x
where
3
9 2  2
k3 
 E3 
a
2ma 2
Fourth mode solution:
 4 x   B 4 sin k 4 x
where
4
16 2  2
k4 
 E4 
a
2ma 2
_______________________________________
2.30
The 3-D time-independent wave equation in
cartesian coordinates for V x, y, z   0 is:
 2 x, y, z 
x 2

 2 x, y, z 
y 2

 2 x, y, z 
z 2
2mE
  x, y , z   0
2
Use separation of variables, so let
 x, y, z   X x Y  y Z z 
Substituting into the wave equation, we
obtain

lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 2
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
YZ
  x, y 
 Ak y sin k x x  cos k y y
y
2 X
 2Y
2Z


XZ
XY
x 2
y 2
z 2
2mE
XYZ  0
2
2mE
Dividing by XYZ and letting k 2  2 , we

find
1  2 X 1  2Y 1  2 Z
(1)
k2  0

 
 
X x 2 Y y 2 Z z 2
We may set
1 2 X
2 X
 2   k x2 
 k x2 X  0
X x
x 2
Solution is of the form
X x   A sin k x x   B cosk x x 
Boundary conditions: X 0  0  B  0

n 
and X x  a   0  k x  x
a
where n x  1, 2, 3....
Similarly, let
1  2Y
1 2Z
  k z2

 2   k y2 and
Z z 2
Y y
Applying the boundary conditions, we find
n y
, n y  1, 2, 3....
ky 
a
n
k z  z , n z  1, 2, 3...
a
From Equation (1) above, we have
 k x2  k y2  k z2  k 2  0
or
k x2  k y2  k z2  k 2 
2mE
2
so that
 2 2 2
n x  n 2y  n z2
2
2ma
_______________________________________
E  E nx n y nz 


2.31
 2 x, y   2 x, y  2mE
(a)

 2  x, y   0
x 2
y 2

Solution is of the form:
 x, y   A sin k x x  sin k y y
 2 x, y 
  Ak y2 sin k x x  sin k y y
y 2
Substituting into the original equation, we
find:
2mE
(1)
 k x2  k y2  2  0

From the boundary conditions,
o
A sin k x a  0 , where a  40 A
So k x 
n x
, n x  1, 2, 3, ...
a
o
Also A sin k y b  0 , where b  20 A
So k y 
n y
, n y  1, 2, 3, ...
b
Substituting into Eq. (1) above
2 2
n y2  2 
 2  n x 

E nx n y 

2m  a 2
b 2 
(b)Energy is quantized - similar to 1-D result.
There can be more than one quantum state
per given energy - different than 1-D result.
_______________________________________
2.32
(a) Derivation of energy levels exactly the
same as in the text
 2 2 2
n 2  n12
(b) E 
2
2ma
For n 2  2, n1  1
Then
3 2  2
E 
2ma 2


o
(i) For a  4 A
E 



2
3 1.054  10 34  2

2 1.67  10  27 4  10 10

2
 6.155  10 22 J
6.155  10 22
or E 
 3.85  10 3 eV
1.6  10 19
We find
 x, y 
 Ak x cos k x x  sin k y y
x
 2 x, y 
  Ak x2 sin k x x  sin k y y
x 2
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 2
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(ii) For a  0.5 cm
E 


3 1.054  10


 34 2
2
2 1.67  10  27 0.5  10  2

2
Combining these two equations, we find
 k  k1 
  B 2
A2   2
 k 2  k1 
 2k 2 
  B 2
B1  
 k 2  k1 
The reflection coefficient is
 3.939  10 36 J
or
3.939  10 36
 2.46  10 17 eV
19
1.6  10
_______________________________________
E 
2.33
(a) For region II, x  0
 2 2 x  2m
 2 E  VO  2 x   0

x 2
General form of the solution is
 2 x   A2 exp jk 2 x   B 2 exp jk 2 x 
where
2m
k2 
E  V O 
2
Term with B 2 represents incident wave and
term with A2 represents reflected wave.
Region I, x  0
 2 1 x  2mE
 2  1 x   0

x 2
General form of the solution is
 1 x   A1 exp jk1 x   B1 exp jk1 x 
where
2mE
k1 
2
Term involving B1 represents the
transmitted wave and the term involving A1
represents reflected wave: but if a particle is
transmitted into region I, it will not be
reflected so that A1  0 .
Then
 1 x   B1 exp jk 1 x 
 2 x   A2 exp jk 2 x   B 2 exp jk 2 x 
(b)
Boundary conditions:
(1)  1 x  0    2 x  0 
 1
 2
(2)

x x  0
x x  0
Applying the boundary conditions to the
solutions, we find
B1  A2  B 2
k 2 A2  k 2 B 2  k 1 B1
2
 k  k1 

  2
*
B 2 B 2  k 2  k 1 
The transmission coefficient is
4k 1 k 2
T  1 R  T 
k1  k 2 2
_______________________________________
R
2.34
A2 A2*
 2 x   A2 exp k 2 x 
P
 x 
2
2mVo  E 
where k 2 

 exp 2k 2 x 
A2 A2*
2



2 9.11 10 31 3.5  2.8 1.6  10 19
1.054  10
k 2  4.286  10 9 m 1

 34
o
(a) For x  5 A  5  10 10 m
P  exp 2k 2 x 
 

 exp  2 4.2859  10 9 5  10 10
 0.0138

o
(b) For x  15 A  15  10 10 m
 

P  exp  2 4.2859  10 9 15  10 10
 2.61 10

6
o
(c) For x  40 A  40  10 10 m
 

P  exp  2 4.2859  10 9 40  10 10

15
 1.29  10
_______________________________________
2.35
 E
T  16
 Vo
where k 2 



E
1 
 V
o


 exp 2k 2 a 


2mV o  E 
2


2 9.11 10 31 1.0  0.1 1.6  10 19
1.054  10
 34

lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 2
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
or
k 2  4.860  10 9 m 1
k2 =
(a) For a  4  10 10 m
 0.1  0.1 
10
9
T  16
 exp  2 4.85976  10 4  10
1 
 1.0  1.0 
 0.0295
(b) For a  12  10 10 m
 0.1  0.1 
9
10
T  16
 exp  2 4.85976  10 12  10
1 
 1.0  1.0 
 


 


 1.24  10 5
(c) J  N t e , where N t is the density of
transmitted electrons.
E  0.1 eV  1.6  10 20 J
1
1
 m 2  9.11 10 31  2
2
2
   1.874  10 5 m/s  1.874  10 7 cm/s
1.2  10 3  N t 1.6  10 19 1.874  10 7




3
2.36
 E 
E
1 
T  16
 V
V
O
 O 
(a) For m  0.067 m o


 20.067  9.11 10 31 0.8  0.2  1.6  10 19

2

1.054  10 34




 

T  1.27  10 5
_______________________________________
2.37
2mV o  E 
where k 2 


 exp 2k 2 a 



E
1 
 V
o

 E
T  16
 Vo
2



2 1.67  10 27 12  1 10 6  1.6  10 19
1.054  10

 34
 7.274  10 m 1
(a)
1
 1 
T  16 1   exp  2 7.274  10 14 10 14
12
12

 
14


1/ 2
 5.875  10 7
(b)
T  10  5.875  10 7


 1.222 exp 27.274  10 a 

 1.222
27.274  10 a  ln

14
14
6

or a  0.842  10 14 m
_______________________________________
k 2  1.027  10 9 m 1
Then
 0.2  0.2 
T  16
1 

 0.8  0.8 
2.38

 exp  2 1.027  10 15  10
T  0.138
(b) For m  1.08m o


 5.875  10
 
1/ 2
or

or
or


 1.222 exp 14.548
2


 

 exp 2k 2 a 


2mVO  E 
k2 

 exp  2 4.124  10 9 15  10 10

N t  4.002  10 electrons/cm
Density of incident electrons,
4.002  10 8
Ni 
 1.357  10 10 cm 3
0.0295
_______________________________________
8

 21.08 9.11 10 31 0.8  0.2 1.6  10 19

2

1.054  10 34
or
k 2  4.124  10 9 m 1
Then
 0.2  0.2 
T  16
1 

 0.8  0.8 
9
10

Region I x  0  , V  0 ;
Region II 0  x  a  , V  VO
Region III x  a  , V  0
(a) Region I:
 1 x   A1 exp jk1 x   B1 exp jk1 x 
(incident)
(reflected)
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 2
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
where
A1 A1* 
2mE
2
k1 
2mVO  E 

d 1 d 2


dx
dx
jk 1 A1  jk 1 B1  k 2 A2  k 2 B 2
At x  a :  2   3 
A2 expk 2 a   B 2 exp k 2 a 
 A3 exp jk 1 a 
d 2 d 3


dx
dx
k 2 A2 expk 2 a   k 2 B 2 exp k 2 a 
 jk 1 A3 exp jk 1 a 
The transmission coefficient is defined as
A A*
T  3 3*
A1 A1
so from the boundary conditions, we want
to solve for A3 in terms of A1 . Solving
for A1 in terms of A3 , we find
 jA3
k 22  k 12 expk 2 a   exp k 2 a 
4k 1 k 2

k
2
2

 k12 expk 2 a 
 exp k 2 a 
 4k 12 k 22 expk 2 a   exp k 2 a 
2
We have
k2 
2
Region III:
 3 x   A3 exp jk1 x   B3 exp jk1 x 
(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 so that B3  0 .
(c) Boundary conditions:
At x  0 :  1   2 
A1  B1  A2  B 2
A1 
4k1 k 2 
2
2
Region II:
 2 x   A2 expk 2 x   B 2 exp k 2 x 
where
k2 
A3 A3*

 2 jk 1 k 2 expk 2 a   exp k 2 a  
 exp jk 1 a 
We then find

2mVO  E 
2
If we assume that VO  E , then k 2 a will
be large so that
expk 2 a   exp k 2 a 
We can then write
A3 A3*
2
A1 A1* 
k 2  k 2 expk 2 a 
4k1 k 2 2 2 1


 4k12 k 22 expk 2 a 
2

which becomes
A3 A3*
A1 A1* 
k 2  k 2 exp2k 2 a 
4k1 k 2 2 2 1
Substituting the expressions for k1 and
k 2 , we find

k12  k 22 

2mVO
2
and
 2mVO  E    2mE 
k12 k 22  
 2 
2
  

2
 2m 
  2  VO  E E 
 

E
 2m 
  2  VO 1 
 
 VO
2

E 


Then
2
 2mVO 
 exp2k 2 a 
A3 A3* 
2
  
A1 A1* 
 2m  2 
E  
E 
16  2  VO 1 

  
 VO  

A3 A3*
 E
16
 VO

E
1 
 V
O


 exp 2k 2 a 


Finally,
 E 
A A*
E 
1 
 exp 2k 2 a 
T  3 3*  16



A1 A1
 VO  VO 
_____________________________________
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 2
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
2.39
Region I: V  0
 2 1 x  2mE
 2  1 x   0 

x 2
 1 x   A1 exp jk1 x   B1 exp jk1 x 
incident
reflected
where
2mE
k1 
2
Region II: V  V1
 2 2  x 
2mE  V1 

 2 x   0 
2
x 2
 2 x   A2 exp jk 2 x   B 2 exp jk 2 x 
transmitted
reflected
where
k2 
2mE  V1 
_______________________________________
2.40
(a) Region I: Since VO  E , we can write
 2 1 x 
2
 2 3  x 
2 m E  V 2 

 3 x   0 
2
x 2
 3 x   A3 exp jk 3 x 
transmitted
where
2 m E  V 2 
2
There is no reflected wave in Region III.
The transmission coefficient is defined as:
 A A* k A A*
T  3  3 3*  3  3 3*
1 A1 A1 k 1 A1 A1
From the boundary conditions, solve for A3
in terms of A1 . The boundary conditions are:
At x  0 :  1   2 
A1  B1  A2  B 2
 1  2


x
x
k1 A1  k1 B1  k 2 A2  k 2 B 2
At x  a :  2   3 
A2 exp jk 2 a   B 2 exp jk 2 a 
 A3 exp jk 3 a 
2mVO  E 
 1 x   0
2
x
Region II: V  0 , so
 2 2 x  2mE
 2  2 x   0

x 2
Region III: V     3  0
The general solutions can be written,
keeping in mind that  1 must remain
finite for x  0 , as
 1 x   B1 expk1 x 
2

Region III: V  V 2
k3 
 2  3


x
x
k 2 A2 exp jk 2 a   k 2 B 2 exp jk 2 a 
 k 3 A3 exp jk 3 a 
But k 2 a  2n 
exp jk 2 a   exp jk 2 a   1
Then, eliminating B1 , A2 , B 2 from the
boundary condition equations, we find
4k 1 k 3
k
4k 12
T 3

2
k1 k 1  k 3 
k1  k 3 2

 2 x   A2 sin k 2 x   B 2 cosk 2 x 
 3 x   0
where
2mVO  E 
and k 2 
2
(b) Boundary conditions
At x  0 :  1   2  B1  B 2
k1 
2mE
2
 1  2
 k1 B1  k 2 A2

x
x
At x  a :  2   3 
A2 sin k 2 a   B 2 cosk 2 a   0
or
B 2   A2 tan k 2 a 
(c)
k 
k1 B1  k 2 A2  A2   1  B1
 k2 
and since B1  B 2 , then
k 
A2   1  B 2
 k2 
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 2
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
From B 2   A2 tan k 2 a  , we can write
k
B 2   1
 k2
  2r 

 r 2 exp

a o 
 ao 
To find the maximum probability
dPr 
0
dr
  2r 
4   2  2

 r exp


3 

 a 
a o   a o 
 o 
P

 B 2 tan k 2 a 

or
k 
1   1  tan k 2 a 
 k2 
This equation can be written as
 2mE 
V E
 tan 
 a
1  O
2
E

 
or
 2mE 
E
  tan 
 a
2
VO  E

 
This last equation is valid only for specific
values of the total energy E . The energy
levels are quantized.
_______________________________________
4
3
 
  2r 

 2r exp

 a o 
which gives
r
0
 1  r  ao
ao
or r  a o is the radius that gives the greatest
probability.
_______________________________________
2.43
2.41
En 


 mo e 4
(J)
4 o 2 2 2 n 2
 mo e 3
4 o 2 2 2 n 2

(eV)


 21.054 10  n
 9.11 10 31 1.6  10 19
4 8.85 10
12
2
3
 34 2
2
 100 is independent of  and  , so the wave
equation in spherical coordinates reduces to
1   2   2m o
 r
E  V r   0

r   2
r 2 r 
where
 e2
 2

V r  
4 o r m o a o r
For
or
 100 
13.58
(eV)
n2
n  1  E1  13.58 eV
n  2  E 2  3.395 eV
n  3  E 3  1.51 eV
En 
 100
1

r

2.42
We have
 100
 1

 
  ao




3/ 2
r 

exp

 ao 
and
*
P  4 r 2 100 100
 4 r 2 
1  1

  a o
3




3/ 2
r 

exp

 ao 
Then
n  4  E 4  0.849 eV
_______________________________________
1
 1
 
  ao
1

  2r 
 exp


 a 

 o 
 1
 
 ao




3/ 2
 1 
r 
  exp

a 
a 
 o
 o 
so
 100
1  1


r
  a o
We then obtain
r2
  2  100
r
r 
r




5/2
r 

r 2 exp

 ao 
 1  1 
 
 
  a o 

5/ 2

  r 
 r   r2 

    exp
 2r exp
 a 



 o 
 ao   ao 

Substituting into the wave equation, we have
or
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 2
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
 1
 
  ao
1
r2





5/ 2

 r  r2
  r 


exp
2r exp



 ao  ao
 a o 
2m o 
2 
E



mo a o r 
2 
 1   1
 
 
 
    ao




3/ 2
r 
0
exp

 ao 
where
 mo e 4
 2
4 o  2 2 2mo a o2
Then the above equation becomes
E  E1 
 1
 
  ao
1
2





3/ 2

   r    1 
r2 
   2  2r  
exp

ao 
  a o   r a o 
2m o    2
 2 

  0

2 
  2m o a o m o a o r 
or
 1
 
  ao
1




3/ 2
   r 

exp

  a o 
  2 1   1
2 

 2   2 
0
 a o r a o  a o a o r 
which gives 0 = 0 and shows that  100 is
indeed a solution to the wave equation.
_______________________________________
2.44
All elements are from the Group I column of
the periodic table. All have one valence
electron in the outer shell.
_______________________________________
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 3
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Chapter 3
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.
_______________________________________
3.2
Schrodinger's wave equation is:
  2  2   x, t 
 V  x     x, t 
2m
x 2
 x, t 
t
Assume the solution is of the form:
 
 E  
 x, t   u x  exp  j  kx    t 
   
 
 j
Region I: V x   0 . Substituting the
assumed solution into the wave equation, we
obtain:
 
 2  
 E  
 jku x  exp  j  kx   t 
2m x 
   
 

 
u x 
 E   
exp  j kx   t  
x
    
 
 
  jE 
 E  
 j 
  u  x  exp  j  kx   t 

   


 
which becomes
 
 2 
 E  
2
  jk  u x  exp  j  kx   t 
2m 
   
 
 2 jk

 
u x 
 E  
exp  j  kx   t 
x
   
 
 
 2 u x 
 E   
exp  j  kx   t  
2
x
    
 
 
 E  
  Eu x  exp  j  kx   t 
   
 
This equation may be written as
u x   2 u  x  2mE
 k 2 u x   2 jk

 2 u x   0
x

x 2
Setting u x   u1 x  for region I, the equation
becomes:
d 2 u1 x 
du x 
 2 jk 1  k 2   2 u1 x   0
dx
dx 2
where
2mE
2  2
Q.E.D.

In Region II, V x   VO . Assume the same
form of the solution:
 
 E  
 x, t   u x  exp  j kx    t 
   
 
Substituting into Schrodinger's wave
equation, we find:
 
 2 
 E  
2
  jk  u  x  exp  j  kx   t 
2m 
   
 

 
u x 
 E  
exp  j kx   t 
x
   
 
 2 jk


 
 2 u x 
 E   
exp  j  kx   t  
2
x
    
 
 
 E  
 VO u  x  exp  j  kx   t 
   
 
 
 E  
 Eu  x  exp  j  kx   t 
   
 
This equation can be written as:
u x   2 u x 
 k 2 u x   2 jk

x
x 2
2mVO
2mE

u x   2 u x   0
2

Setting u x   u 2 x  for region II, this
equation becomes
du x 
d 2 u 2 x 
 2 jk 2
2
dx
dx
2mVO 

  k 2   2 
u 2 x   0
2 

where again
2mE
2  2
Q.E.D.

_______________________________________
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 3
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
3.3
We have
d 2 u1 x 
du1 x 
 k 2   2 u1 x   0
dx
dx
Assume the solution is of the form:
u1 x   A exp j   k x 
2

 2 jk

 B exp j   k x 
The first derivative is
du1 x 
 j   k A exp j   k x 
dx
 j   k B exp j   k x
and the second derivative becomes
d 2 u1 x 
2
  j   k  A exp j   k x 
dx 2
2
  j   k  B exp j   k x 
Substituting these equations into the
differential equation, we find
2
   k  A exp j   k x 
   k  B exp j   k x 
2
 2 jk  j   k A exp j   k x
 j   k B exp j   k x


 k 2   2 A exp j   k x 
 B exp j   k x   0
Combining terms, we obtain
  2  2k  k 2  2k   k   k 2   2
 A exp j   k x 








   2  2k  k 2  2k   k   k 2   2
 B exp j   k x   0
We find that
Q.E.D.
00
For the differential equation in u 2 x  and the
proposed solution, the procedure is exactly
the same as above.
_______________________________________
3.4
We have the solutions
u1 x   A exp j   k x 
 B exp j   k x 
for 0  x  a and
u 2 x   C exp j   k x 
 D exp j   k x 
for b  x  0 .
The first boundary condition is
u1 0  u 2 0
which yields
A B C  D  0
The second boundary condition is
du1
du
 2
dx x  0 dx x  0
which yields
  k A    k B    k C
   k D  0
The third boundary condition is
u1 a   u 2  b 
which yields
A exp j   k a   B exp j   k a 
 C exp j   k  b 
 D exp j   k  b 
and can be written as
A exp j   k a   B exp j   k a 
 C exp j   k b
 D exp j   k b  0
The fourth boundary condition is
du1
du
 2
dx x  a dx x  b
which yields
j   k A exp j   k a 
 j   k B exp j   k a 
 j   k C exp j   k  b 
 j   k D exp j   k  b 
and can be written as
  k A exp j   k a 
   k B exp j   k a 
   k C exp j   k b
   k D exp j   k b  0
_______________________________________
3.5
(b) (i) First point: a  
Second point: By trial and error,
a  1.729
(ii) First point: a  2
Second point: By trial and error,
a  2.617
_______________________________________
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 3
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
3.6
(b) (i) First point: a  
Second point: By trial and error,
a  1.515
(ii) First point: a  2
Second point: By trial and error,
a  2.375
_______________________________________
3.7
sin a
 cos a  cos ka
a
Let ka  y , a  x
Then
sin x
P
 cos x  cos y
x
d
of this function.
Consider
dy
P



d
1
P   x  sin x  cos x   sin y
dy
We find

dx
dx 
2
1
P  1x  sin x    x  cos x  
dy
dy


dx
 sin x
  sin y
dy
Then

dx    1
cos x 
sin x 
 sin x    sin y
P 
dy   x 2
x 

For y  ka  n , n  0, 1, 2, ...  sin y  0
So that, in general,
dx
d a  d
0

dy
d ka  dk
And
2mE

2
So
1 / 2
d 1  2mE 
 2m  dE
  2 
 2 
dk 2   
   dk
This implies that
d
dE
n
0
for k 
dk
dk
a
_______________________________________
3.8
(a)  1 a  
2 m o E1
2
a  
 2 2
E1 
2m o a 2


 2 1.054 10 34 2

2 9.11 10 31 4.2  10 10

2
 3.4114  10 19 J
From Problem 3.5
 2 a  1.729
2m o E 2
 a  1.729
2
E2 
1.729 2 1.054 10 34 2


2 9.11 10 31 4.2  10 10

2
 1.0198  10 18 J
E  E 2  E1
 1.0198  10 18  3.4114  10 19
 6.7868  10 19 J
6.7868  10 19
or E 
 4.24 eV
1.6  10 19
(b)  3 a  2
2m o E 3
2
E3 

 a  2
2 2 1.054 10 34 2

2 9.11 10 31 4.2  10 10

2
 1.3646  10 18 J
From Problem 3.5,
 4 a  2.617
2m o E 4
2
E4 
 a  2.617
2.617 2 1.054 10 34 2


2 9.11 10 31 4.2  10 10

2
 2.3364  10 18 J
E  E 4  E 3
 2.3364  10 18  1.3646  10 18
 9.718  10 19 J
9.718  10 19
or E 
 6.07 eV
1.6  10 19
_______________________________________
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 3
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
3.9
(a) At ka   ,  1 a  
2 m o E1
E1 
2m o E1
a  
2

3.10
(a)  1 a  
2
 2 1.054 10 34 2


2 9.11 10 31 4.2  10 10
E1 

2
 3.4114  10 19 J
At ka  0 , By trial and error,
 o a  0.859


2 9.11 10 31 4.2  10 10
2

 3.4114  10  2.5172  10
 8.942  10 20 J
8.942  10 20
or E 
 0.559 eV
1.6  10 19
(b) At ka  2 ,  3 a  2
E2 
E3 
19
 a  2
1.054 10 
29.11 10 4.2  10 
2 
34 2
2
 31
2
E2 
 a  1.729
1.054 10 
29.11 10 4.2  10 
1.729 
 31
2
10 2
 1.0198  10 18 J
E  E 3  E 2
 1.3646  10 18  1.0198  10 18
 3.4474  10 19 J
3.4474  10 19
or E 
 2.15 eV
1.6  10 19
_______________________________________
 a  1.515



2
 7.830  10 19 J
E  E 2  E1
 7.830  10 19  3.4114  10 19
 4.4186  10 19 J
4.4186  10 19
or E 
 2.76 eV
1.6  10 19
(b)  3 a  2
2m o E 3
2
E3 

 a  2
2 2 1.054 10 34 2

2 9.11 10 31 4.2  10 10

2
 1.3646  10 18 J
From Problem 3.6,  4 a  2.375
2m o E 4
34 2
2

2 9.11 10 31 4.2  10 10
10 2
 1.3646  10 18 J
At ka   . From Problem 3.5,
 2 a  1.729
2m o E 2

1.515 2 1.054 10 34 2
2
19
2

2 9.11 10 31 4.2  10 10
2m o E 2
 2.5172  10 19 J
E  E1  E o
2m o E 3
 2 1.054 10 34 2
 3.4114  10 19 J
From Problem 3.6,  2 a  1.515
0.859 2 1.054 10 34 2
Eo 
a  
2
E4 
 a  2.375
2.375 2 1.054 10 34 2


2 9.11 10 31 4.2  10 10

2
 1.9242  10 18 J
E  E 4  E 3
 1.9242  10 18  1.3646  10 18
 5.597  10 19 J
5.597  10 19
or E 
 3.50 eV
1.6  10 19
_____________________________________
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 3
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
3.11
(a) At ka   ,  1 a  
2m o E1
E g  1.170 
a  
2
E1 
3.12
For T  100 K,

2 9.11 10 31 4.2  10 10
2
Eo 
2
T  400 K, E g  1.097 eV
T  500 K, E g  1.066 eV
T  600 K, E g  1.032 eV
_______________________________________
0.727 2 1.054 10 34 2

2 9.11 10
 1.8030  10
E  E1  E o
4.2 10 
19
10 2
J
 3.4114  10 19  1.8030  10 19
 1.6084  10 19 J
1.6084  10 19
or E 
 1.005 eV
1.6  10 19
(b) At ka  2 ,  3 a  2
2m o E 3
2
E3 

E2 


2
 a  1.515

2 9.11 10 34 4.2  10 10
 7.830  10
E  E 3  E 2
19
 1 d 2E 

m *p   2 

dk 2 

We have that
2
d 2E
curve A  d E2 curve B 
2
dk
dk
so that m *p curve A  m *p curve B 
_______________________________________
1.515 2 1.054 10 34 2

1
 1 d 2E 
m *   2  2 
  dk 
We have
d 2E
d 2E
curve
A



curve B 
dk 2
dk 2
so that m * curve A  m * curve B 
_______________________________________
1
2 2 1.054 10 34 2
 1.3646  10 18 J
At ka   , From Problem 3.6,
 2 a  1.515
2
3.13
The effective mass is given by
3.14
The effective mass for a hole is given by
 a  2
2 9.11 10 31 4.2  10 10
2m o E 2

T  300 K, E g  1.125 eV
 a  0.727
 31
2
T  200 K, E g  1.147 eV

 3.4114  10 19 J
At ka  0 , By trial and error,
 o a  0.727
2m o E o
4
636  100
E g  1.164 eV
 2 1.054 10 34 2

4.73 10 100

2
J
 1.3646  10 18  7.830  10 19
 5.816  10 19 J
5.816  10 19
or E 
 3.635 eV
1.6  10 19
_______________________________________
3.15
dE
 0  velocity in -x direction
dk
dE
Points C,D:
 0  velocity in +x direction
dk
d 2E
Points A,D:
0
dk 2
negative effective mass
2
d E
Points B,C:
0
dk 2
positive effective mass
_______________________________________
Points A,B:
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 3
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
3.16
For A: E  C i k 2
At k  0.08  10 10 m 1 , E  0.05 eV
Or E  0.05 1.6  10 19  8  10 21 J
So 8  10 21
Now m  
10 2
1



1.054  10 34
2

2C1
2 1.25  10 38

2
31
 4.44  10 kg
4.4437  10 31
or m  
 mo
9.11 10 31
m   0.488 m o
At k  0.08  10 10 m 1 , E  0.5 eV
Or E  0.5 1.6  10 19  8  10 20 J

So 8  10


 C1 0.08  10



1.054  10 34
2
Now m 

2C1
2 1.25  10 37

2
32
 4.44  10 kg
4.4437  10 32
or m  
 mo
9.11 10 31
m   0.0488 m o
_______________________________________
3.17
For A: E  E  C 2 k 2


 0.025 1.6  10 19  C 2 0.08  1010
  2  1.054  10 34
m 

2C 2
2 6.25  10 39



2
2

 0.3 1.6  10 19  C 2 0.08  1010

.
2
_______________________________________
E  E O  E1 cos k  k O 
Then
dE
  E1    sin k  k O 
dk
  E1 sin k  k O 
and
d 2E
 E1 2 cos k  k O 
dk 2


 8.8873 10 31 kg
 8.8873 10 31
or m  
 mo
9.11 10 31
m     0.976 m o
For B: E  E  C 2 k 2


3.19
(c) Curve A: Effective mass is a constant
Curve B: Effective mass is positive
around k  0 , and is negative
3.20
 C 2  6.25  10 39


around k  

2
3.18
(a) (i) E  h
E 1.42  1.6  10 19
or   
h
6.625  10 34
 3.429  1014 Hz
hc c
3  1010
(ii)  
 
E  3.429  10 14
 8.75  10 5 cm  875 nm
E 1.12 1.6 10 19
(b) (i)   
h
6.625  10 34
 2.705  1014 Hz
3  1010
c
(ii)   
 2.705  1014
 1.109  10 4 cm  1109 nm
_______________________________________


10 2
 C1  1.25  10 37



For B: E  C i k 2
20


 7.406  10 32 kg
 7.406 10 32
 mo
or m  
9.11 10 31
m   0.0813 m o
_______________________________________


 C 0.08  10 
 C1  1.25  10 38

  2  1.054  10 34
m 

2C 2
2 7.5  10 38


2
 C 2  7.5  10 38
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 3
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Then
2
1
1 d E
 2 2
*
m
 dk

E1
YZ
2
2
k  ko
2
E1 2
_______________________________________
m* 


0.082m  1.64m 

 4 2 / 3 mt  ml
(a) m dn
2
 42/ 3
1/ 3
2
1/ 3
o
o

m dn
 0.56m o
(b)
3
2
1
2
1





m cn mt ml 0.082m o 1.64m o

24.39 0.6098

mo
mo

m cn
 0.12m o
_______________________________________
3.22

 m  
 0.45m   0.082m  

(a) m dp
 m hh 
3/ 2 2/3
3/ 2
lh
3/ 2 2/3
3/ 2
o
 0.30187  0.02348
2/3
o
 mo

m dp
 0.473m o
m hh 3 / 2  mlh 3 / 2
m hh 1 / 2  mlh 1 / 2
0.453 / 2  0.0823 / 2  m

0.451 / 2  0.0821 / 2 o


(b) m cp

m cp
 0.34m o
_______________________________________
3.23
For the 3-dimensional infinite potential well,
V x   0 when 0  x  a , 0  y  a , and
0  z  a . In this region, the wave equation
is:
 2 x, y, z   2 x, y, z   2 x, y, z 


z 2
y 2
x 2
2mE
  x, y , z   0
2
Use separation of variables technique, so let
 x, y, z   X x Y  y Z z 
Substituting into the wave equation, we have

2mE
 XYZ  0
2
Dividing by XYZ , we obtain
1  2 X 1  2 Y 1  2 Z 2mE
 

 
 2 0
X x 2 Y y 2 Z z 2

Let
1 2 X
2 X
 2   k x2 
 k x2 X  0
2
X x
x
The solution is of the form:
X x   A sin k x x  B cos k x x
Since  x, y, z   0 at x  0 , then X 0  0
so that B  0 .
Also,  x, y, z   0 at x  a , so that

or
3.21
2 X
 2Y
2Z


XY
XZ
x 2
y 2
z 2
X a   0 . Then k x a  n x  where
n x  1, 2, 3, ...
Similarly, we have
1 2Z
1  2Y

 k z2
 2  k y2 and
Z z 2
Y y
From the boundary conditions, we find
k y a  n y  and k z a  n z 
where
n y  1, 2, 3, ... and n z  1, 2, 3, ...
From the wave equation, we can write
2mE
 k x2  k y2  k z2  2  0

The energy can be written as
2 2
 
n x  n y2  n z2  
2m
a
_______________________________________
E  E nx n y nz 


2
3.24
The total number of quantum states in the
3-dimensional potential well is given
(in k-space) by
 k 2 dk 3
g T k dk 
a
3

where
2mE
2
We can then write
k2 
2mE

Taking the differential, we obtain
k
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 3
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
m
1
1 1
1
 2m  
 dE  
 dE

2 E
 2E
Substituting these expressions into the density
of states function, we have
 a 3  2mE  1
m
 dE
g T E dE  3  2   
     2E
Noting that
h

2
this density of states function can be
simplified and written as
4 a 3
2m 3 / 2  E  dE
g T E dE 
h3
Dividing by a 3 will yield the density of
states so that
3/ 2
4 2m 
g E  
 E
h3
_______________________________________
dk 
3.25
For a one-dimensional infinite potential well,
2m n E n 2  2

 k2
2
a2
Distance between quantum states
  
 
k n 1  k n  n  1   n   
a
a a
 
Now
2  dk
g T k dk 
 
 
a
Now
1
k   2m n E

dk 

1 1 2m n
 
 dE
 2
E
2m n
1


E

20.067  9.11 10 31
1

1.054  10 34  
E


g E  
1.055  10 18
E
_______________________________________
3.26
(a) Silicon, m n  1.08m o
g c E  
gc 

4 2m n

4 2m n



h3

3/ 2
E  Ec

3 / 2 Ec  2 kT

h3

4 2m n

h
h


3/ 2
3
2
3/ 2
  E  E c 
3

 3/ 2
n
3
4 2m
E  E c  dE
Ec

Ec  2 kT
Ec
2
3/ 2
  2kT 
3
4 21.08 9.11 10 31
6.625 10 
 7.953  10 2kT 

3/ 2
2
3/ 2
  2kT 
3
 34 3
3/ 2
55
(i) At T  300 K, kT  0.0259 eV
 0.0259  1.6  10 19

 4.144 10

Then g c  7.953  10
55
21

J
24.144 10 
3/ 2
21
 6.0  10 25 m 3
g c  6.0  10 19 cm 3
or
 400 
(ii) At T  400 K, kT  0.0259

 300 
 0.034533 eV
 5.5253 10

n

m 3 J 1

2m
2a 1
 
 dE
E
 2
Divide by the &quot;volume&quot; a, so
g E  
g E  
 0.034533 1.6  10 19
Then
g T E dE 
So
Then
 

21
J
g c  7.953  10 55 2 5.5253 10 21
or

3/ 2
 9.239 10 25 m 3
g c  9.24  1019 cm 3
(b) GaAs, m n  0.067 m o
gc 



4 20.067  9.11 10 31
6.625 10 
 1.2288  10 2kT 
 34 3
54
3/ 2

3/ 2
2
3/ 2
  2kT 
3
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 3
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(i) At T  300 K, kT  4.144  10 21 J
 

g c  1.2288 10 54 2 4.144 10 21

(i)At T  300 K, kT  4.144  10 21 J
g   2.3564  10 55 3 4.144  10  21

3/2
 3.266  10 25 m 3
or g   3.27  1019 cm 3
 9.272  10 23 m 3
or g c  9.27  1017 cm 3
(ii) At T  400 K, kT  5.5253  10 21 J
 

 

3/ 2
g c  1.2288  10 54 2 5.5253  10 21

(ii)At T  400 K, kT  5.5253  10 21 J
 

3/ 2
g   2.3564  10 55 3 5.5253  10 21

3/ 2
 1.427  10 24 m 3
g c  1.43  1018 cm 3
_______________________________________
 5.029  10 25 m 3
or g   5.03  10 19 cm 3
_______________________________________
3.27
(a) Silicon, m p  0.56m o
3.28
g  E  
g 




4 2m p

3/ 2

4 2m p

3/ 2

4 2m p

h3
4 2m p
h3

3/ 2

3/ 2

2
3/ 2
E  E 

3



2
3/ 2
  3kT 

 3 
4 20.56  9.11 10 31
6.625 10 
 2.969  10 3kT 

3/ 2
 34 3
E
(i)At T  300 K, kT  4.144  10
 

g   2.969  10 55 3 4.144  10 21
 2.614  10 46 m 3 J 1
E  E c  0.4 eV;
 3.018  10 46 m 3 J 1
(b) g  
(b) GaAs, m p  0.48m o

4 20.48 9.11 10 31
6.625 10 
 2.3564  10 3kT 
h3


3/ 2
E  E

4 20.56  9.11 10 31
6.625 10 

3/ 2
 34 3
E  E
 4.4541 10 55 E  E
g  0
E  E  0.1 eV; g   5.634  10 45 m 3 J 1

3/ 2
E  E  0.2 eV;
 7.968  10 45 m 3 J 1
E  E  0.3 eV;
 9.758  10 45 m 3 J 1
 1.127  10 46 m 3 J 1
E  E  0.4 eV;
_______________________________________

 34 3
55

4 2m p
3/ 2
 6.337  10 25 m 3
or g   6.34  1019 cm 3

gc  0
E  E c  0.3 eV;

g   2.969  10 55 3 5.5253  10 21
E  Ec
 34 3
For E  E ;
 

6.625 10 
 2.134  10 46 m 3 J 1
3
(ii)At T  400 K, kT  5.5253  10 21 J
3/ 2
E  E c  0.2 eV;

 4.116  10 m
or g   4.12  10 19 cm 3
25

E  E c  0.1 eV; g c  1.509  10 46 m 3 J 1

J
E  Ec
4 21.08 9.11 10 31
For E  E c ;
E  3 kT
2
3/ 2
 3kT 
3
21
3/ 2
 1.1929  10 56 E  E c
3/ 2
55
g 

E  E  dE
E  3 kT




h3
E

h3
4 2m n
E  E
h3

(a) g c E  
3/ 2
2
3/ 2
 3kT 
3
3.29


m

m
(a)
gc
m
 n
g
m p
(b)
gc
g
3/ 2




3/ 2
3/ 2
 3/ 2
n
 3/ 2
p
 1.08 


 0.56 
3/ 2
 0.067 


 0.48 
 2.68
3/ 2
 0.0521
_______________________________________
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 3
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
3.30
Plot
_______________________________________
3.31
(a) Wi 
gi!
10!

N i ! g i  N i ! 7!10  7 !
10987!  1098  120
7!3!
321
121110!
12!
(i) Wi 

10!12  10! 10!21

(b)
 66
12111098!
12!
(ii) Wi 

8!12  8! 8!4321
 495
_______________________________________
3.32
f E  
1
 E  EF
1  exp
 kT



(a) E  E F  kT , f E  
f E   0.269
1

1  exp1
(b) E  E F  5kT , f E  
f E   6.69  10
1

1  exp5
3
(c) E  E F  10kT , f E  
1

1  exp10 
f E   4.54  10 5
_______________________________________
3.34
  E  E F  
f F  exp 

kT


  0.30 
6
E  E c ; f F  exp 
  9.32  10
0
.
0259


(a)
Ec 
kT
  0.30  0.0259 2  
; f F  exp 

2
0.0259


 5.66  10 6
  0.30  0.0259  
E c  kT ; f F  exp 

0.0259


 3.43  10 6
3kT
  0.30  30.0259 2  
; f F  exp 
Ec 

0.0259
2


 2.08  10 6
  0.30  20.0259  
E c  2kT ; f F  exp 

0.0259


 1.26  10 6
1
(b) 1  f F  1 
 E  EF 
1  exp 

 kT 
  E F  E  
 exp 

kT


  0.25 
5
E  E ; 1  f F  exp 
  6.43  10
 0.0259 
E 
kT
  0.25  0.0259 2  
; 1  f F  exp 

2
0.0259


 3.90  10 5
  0.25  0.0259  
E  kT ; 1  f F  exp 

0.0259


3.33
1
1  f E   1 
 E  EF
1  exp
 kT
or
1
1  f E  
 EF  E 
1  exp

 kT 
 2.36  10 5



(a) E F  E  kT , 1  f E   0.269
(b) E F  E  5kT , 1  f E   6.69  10 3
(c) E F  E  10kT , 1  f E   4.54  10 5
_______________________________________
E 
3kT
;
2
  0.25  30.0259 2  
1  f F  exp 

0.0259


 1.43  10 5
E  2kT ;
  0.25  20.0259  
1  f F  exp 

0.0259


 8.70  10 6
_______________________________________
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 3
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
1.054 10    2  2  1 

29.11 10 12  10 
3.35
  E c  kT  E F  
  E  E F  
 exp 
f F  exp 


kT
kT




and
  E F  E  
1  f F  exp 

kT


  E F  E  kT  
 exp 

kT


  E c  kT  E F  
So exp 

kT


  E F  E  kT  
 exp 

kT


Then E c  kT  E F  E F  E  kT
E c  E
 E midgap
2
_______________________________________
Or E F 
3.36
 2 n 2 2
2ma 2
For n  6 , Filled state
En 
E6 
1.054 10  6  
29.11 10 12  10 
34 2
2
2
10 2
 31
 1.5044  10 18 J
1.5044  10 18
or E 6 
 9.40 eV
1.6  10 19
For n  7 , Empty state
1.054 10  7  

29.11 10 12  10 
34 2
E7
2
2
10 2
 31
 2.048  10 18 J
2.048  10 18
or E 7 
 12.8 eV
1.6  10 19
Therefore 9.40  E F  12.8 eV
_______________________________________
34 2

E5 



2
2
2
10 2
 3.761 10 19 J
3.761 10 19
 2.35 eV
or E 5 
1.6  10 19
For the next quantum state, which is empty,
the quantum state is n x  1, n y  2, n z  2 .
This quantum state is at the same energy, so
E F  2.35 eV
(b) For 13 electrons, the 13th electron
occupies the quantum state
n x  3, n y  2, n z  3 ; so
E13 
1.054 10    3  2  3 
29.11 10 12  10 
34 2
2
2
2
2
10 2
 31
 9.194  10 19 J
9.194  10 19
 5.746 eV
or E13 
1.6  10 19
The 14th electron would occupy the quantum
state n x  2, n y  3, n z  3 . This state is at
the same energy, so
E F  5.746 eV
_______________________________________
3.38
The probability of a state at
being occupied is
1
f 1 E1  
 E  EF
1  exp 1
 kT
E1  E F  E

1
 E 
1  exp

 kT 



The probability of a state at E 2  E F  E
being empty is
1
1  f 2 E 2   1 
 E2  EF 
1  exp

 kT 
  E 
exp

1
 kT 

 1
  E 
  E 
1  exp

 1  exp
kT
 kT 


2
2 2
 
n x  n 2y  n z2  
2m
a
2
 31
3.37
(a) For a 3-D infinite potential well
2mE
 
 n x2  n 2y  n z2  
2

a
For 5 electrons, the 5th electron occupies
the quantum state n x  2, n y  2, n z  1 ; so
2
or
1  f 2 E 2  
1
 E 
1  exp

 kT 
so f 1 E1   1  f 2 E 2 
Q.E.D.
_______________________________________
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 3
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
3.39
(a) At energy E1 , we want
1
1

 E1  E F
 E1  E F 
exp
 1  exp
 kT
 kT 
1
 E  EF 
1  exp 1

 kT 
This expression can be written as
 E  EF 
1  exp 1

 kT 
 1  0.01
 E1  E F 
exp

 kT 
or
 E  EF 
1  0.01 exp 1

 kT 
Then
E1  E F  kT ln 100 
or
E1  E F  4.6kT
(b)
At E  E F  4.6kT ,



 0.01
f  E1  
1
1

 E  E F  1  exp4.6 
1  exp 1

 kT 
which yields
f E1   0.00990  0.01
_______________________________________
3.40
(a)
  E  E F  
  5.80  5.50  
f F  exp 
  exp 

kT
0.0259




 9.32  10 6
 700 
(b) kT  0.0259 
  0.060433 eV
 300 
  0.30 
3
f F  exp 
  6.98  10
 0.060433 
  E F  E  
(c) 1  f F  exp 

kT


  0.25 
0.02  exp 

 kT 
1
  0.25 
or exp 
 50

 kT  0.02
0.25
 ln 50 
kT
or
0.25
 T 
 0.063906  0.0259 
kT 

ln 50 
 300 
which yields T  740 K
_______________________________________
3.41
(a)
f E  
1
 0.00304
 7.15  7.0 
1  exp

 0.0259 
or 0.304%
(b) At T  1000 K, kT  0.08633 eV
Then
1
 0.1496
f E  
 7.15  7.0 
1  exp

 0.08633 
or 14.96%
1
(c) f E  
 0.997
 6.85  7.0 
1  exp

 0.0259 
or 99.7%
(d)
1
At E  E F , f E   for all temperatures
2
_______________________________________
3.42
(a) For E  E1
f E  
1
 E  EF
1  exp 1
 kT
   E1  E F  
 exp 

kT





Then
  0.30 
6
f E1   exp
  9.32  10
 0.0259 
For E  E 2 , E F  E 2  1.12  0.30  0.82 eV
Then
1
1  f E   1 
  0.82 
1  exp

 0.0259 
or
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 3
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
  E F  E 2  
1  f E   exp 

kT



  0.82 
1  f E   1  1  exp

 0.0259 

  0.82 
14
 exp
  1.78  10
 0.0259 
(b) For E F  E 2  0.4 eV,
E1  E F  0.72 eV
At E  E1 ,
   E1  E F  
  0.72 
 exp
f E   exp 


kT
 0.0259 


or
f E   8.45  10 13
  0.4 
 exp

 0.0259 
or 1  f E   1.96  10 7
_______________________________________
3.44

 E  EF
f E   1  exp
 kT


df E 
 E  EF
  11  exp
dE
 kT

  0.4 
 exp

 0.0259 
or
1  f E   1.96  10 7
_______________________________________
3.43
(a) At E  E1
  E1  E F  
  0.30 
f E   exp 
 exp


kT
 0.0259 


f E   9.32  10 6
At E  E 2 , E F  E 2  1.42  0.3  1.12 eV
So
  E F  E 2  
1  f E   exp 

kT


  1.12 
 exp

 0.0259 
1  f E   1.66  10 19
(b) For E F  E 2  0.4 ,
E1  E F  1.02 eV
2



 E  EF 
 1 

 exp

kT


 kT 
  E F  E 2  
1  f E   exp 

kT


or
1
so
At E  E 2 ,
or



or
 E  EF
 1 
 exp

 kT 
 kT
df E 

dE

 E  EF
1  exp
 kT

(a) At T  0 K, For






2
E  E F  exp    0 
df
0
dE
df
0
E  E F  exp     
dE
df
At E  E F 
 
dE
(b) At T  300 K, kT  0.0259 eV
df
For E  E F ,
0
dE
df
For E  E F ,
0
dE
At E  E F ,
 1 
1

df
 0.0259 

 9.65 (eV) 1
dE
1  12
At E  E1 ,
   E1  E F  
  1.02 
 exp
f E   exp 


kT
 0.0259 


or
f E   7.88  10 18
At E  E 2 ,
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 3
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(c) At T  500 K, kT  0.04317 eV
df
For E  E F ,
0
dE
df
For E  E F ,
0
dE
At E  E F ,
3.46
(a)
 1 

1
df
 0.04317 
 5.79 (eV) 1

dE
1  12
_______________________________________
3.45
(a) At E  E midgap ,
f E  
1
 E  EF
1  exp
 kT




1
 Eg
1  exp
 2kT




Si: E g  1.12 eV,
f E  
or
1
 1.12 
1  exp 

 20.0259  
f E   4.07  10 10
or
1
 0.66 
1  exp 

 20.0259  
f E   2.93  10 6
GaAs: E g  1.42 eV
f E  
or


0.60
 ln 10 8
kT
0.60
kT 
 0.032572 eV
ln 10 8
 
 T 
0.032572  0.0259

 300 
so T  377 K
  0.60 
(b) 10  6  exp 

 kT 


0.60
 ln 10  6
kT
0.60
kT 
 0.043429
ln 10 6
 
 T 
0.043429  0.0259 

 300 
or T  503 K
_______________________________________
3.47
(a) At T  200 K,
Ge: E g  0.66 eV
f E  
or
  E  E F  
f F  exp 

kT


  0.60 
10 8  exp 

 kT 
1
 1.42 
1  exp 

 20.0259  
f E   1.24  10 12
(b) Using the results of Problem 3.38, the
answers to part (b) are exactly the same as
those given in part (a).
_______________________________________
 200 
kT  0.0259
  0.017267 eV
 300 
1
f F  0.05 
 E  EF 
1  exp

 kT 
1
 E  EF 
exp
 1  19


 kT  0.05
E  E F  kT ln 19   0.017267  ln 19 
 0.05084 eV
By symmetry, for f F  0.95 ,
E  E F  0.05084 eV
Then E  20.05084   0.1017 eV
(b) T  400 K, kT  0.034533 eV
For f F  0.05 , from part (a),
E  E F  kT ln 19   0.034533 ln 19 
 0.10168 eV
Then E  20.10168  0.2034 eV
_______________________________________
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 4
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Chapter 4
(b)
4.1
  Eg
n i2  N c N  exp
 kT




2
25
3
  Eg
 T 
 N cO N O 
 exp
 300 
 kT
3
38








By trial and error, T  417.5 K
_______________________________________
where N cO and N O are the values at 300 K.
4.4
T (K)
kT (eV)
(a) Silicon
n i (cm 3 )
200
0.01727
7.68  10 4
400
0.03453
2.38  10 12
600
0.0518
9.74  1014
 200 
At T  200 K, kT  0.0259

 300 
 0.017267 eV
 400 
At T  400 K, kT  0.0259

 300 
 0.034533 eV
200
2.16  1010
1.38
400
600
8.60 10
3.82  1016
3.28 10
5.72  1012
7.70 10 
200 1.40 10 
n i2 400 
(b) Germanium (c) GaAs
n i (cm 3 )
n i (cm 3 )
T (K)

  2.5 10
  1.12300 
 T 
 2.912  10 
 exp 
0.0259T  
300
n i2  5  1012
n i2
2 2
_______________________________________
4.2
Plot
_______________________________________
 3.025  10 17
  Eg 
exp 

 0.034533 


3
  Eg 
 200 
exp 



 300 
 0.017267 
Eg
 Eg

 8 exp 


 0.017267 0.034533 
 400 


 300 
9
14
10 2

3


3.025  10 17  8 exp E g 57.9139  28.9578
or
4.3
  Eg
(a) n i2  N c N  exp
 kT




T 
5 10   2.8 10 1.04 10  300

11 2
19

  38.1714


19




 1.12
 exp 

 0.0259T 300 

3
 3.025 1017
E g 28.9561  ln
8

or E g  1.318 eV

Now
7.70 10 
10 2
 400 
 N co N o 

 300 
  1.318 
 exp

 0.034533 
3
 T 
2.5  10 23  2.912  10 38 

 300 
  1.12 300 
 exp 

 0.0259T  
By trial and error, T  367.5 K
3

5.929  10 21  N co N o 2.370  2.658  10 17
6

so N co N o  9.41 10 cm
_______________________________________
37
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 4
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(b)
4.5
  1.10 
exp

n i B 
  0.20 
 kT 

 exp

n i  A
  0.90 
 kT 
exp

 kT 
For T  200 K, kT  0.017267 eV
For T  300 K, kT  0.0259 eV
For T  400 K, kT  0.034533 eV
(a) For T  200 K,
n i B 
  0.20 
6
 exp
  9.325  10
0
.
017267
n i  A


(b) For T  300 K,
n i B 
  0.20 
4
 exp
  4.43  10
n i  A
 0.0259 
(c) For T  400 K,
n i B 
  0.20 
3
 exp
  3.05  10
0
.
034533
n i  A


_______________________________________
4.6
  E  E F  
(a) g c f F  E  E c exp 

kT


  E  E c  
 E  E c exp 

kT


  E F  E  
 exp 

kT


Let E  E  x
x
Then g  1  f F   x exp

 kT 
To find the maximum value
d g  1  f F  d 
  x 

  0
 x exp
dx
dx 
 kT 
Same as part (a). Maximum occurs at
kT
x
2
or
kT
E  E 
2
_______________________________________
4.7
n  E1 

nE 2 
  E c  E F  
 exp 

kT


E1  E c  4kT and E 2  E c 
x
Then g c f F  x exp

 kT 
To find the maximum value:
d g c f F  1 1 / 2
x
exp
 x

2
dx
 kT 
Then
n  E1 

nE 2 
kT
2
  E1  E 2  
exp 

kT
kT


2
4kT
 
1 
 2 2 exp   4    2 2 exp 3.5
2 
 
1 1/ 2
x
 x exp
0
kT
 kT 
which yields
x1 / 2
kT
1

x
1/ 2
kT
2
2x
The maximum value occurs at
kT
E  Ec 
2
   E1  E c  
E1  E c exp 

kT


  E 2  E c  
E 2  E c exp 

kT


where
Let E  E c  x

  E F  E  
g  1  f F   E  E exp 

kT


   E  E  
 E  E exp 

kT


or
n  E1 
 0.0854
nE 2 
_______________________________________
4.8
Plot
_______________________________________
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 4
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
4.9
Plot
_______________________________________
4.13
Let g c E   K  constant
Then

no 
4.10
E Fi  E midgap
 m *p
3
 kT ln *
 mn
4





E Fi  E midgap  0.0128 eV
m n*  0.067 m o
E Fi  E midgap  0.0382 eV
_______________________________________
4.11
 1.04  1019
1
 kT  ln
19
2
 2.8 10
T (K)
200
400
600
kT (eV)
0.01727
0.03453
0.0518





  0.4952kT 


( E Fi  E midgap )(eV)
 0.0086
 0.0171
 0.0257
_______________________________________
(a) E Fi  E midgap
 m *p
3
 kT ln *
 mn
4





3
 0.70 
 0.0259 ln

4
 1.21 
 10.63 meV
3
 0.75 
(b) E Fi  E midgap  0.0259  ln

4
 0.080 
 43.47 meV
_______________________________________
dE

Let
E  Ec
so that dE  kT  d
kT
We can write
E  E F  E c  E F   E  E c 
so that
  E c  E F  
  E  E F  
 exp 
exp 
  exp  

kT
kT




The integral can then be written as

  E c  E F  
n o  K  kT  exp 
 exp  d
kT
0

which becomes
  E c  E F  
n o  K  kT  exp 

kT


_______________________________________


4.14
Let g c E   C1 E  E c  for E  E c
Then

4.12



  E  E F  
 K exp 
dE
kT


Ec
E Fi  E midgap  0.0077 eV
Gallium Arsenide: m *p  0.48m o ,
1
 E  EF
Ec 1  exp
 kT



Germanium: m *p  0.37m o , m n*  0.55m o
N
1
kT  ln 
2
 Nc
E  f F E dE
c

K
Silicon: m *p  0.56m o , m n*  1.08m o
E Fi  E midgap 
g
Ec
no 
g
E  f F E dE
c
Ec
E  E c 

 C1

Ec
 E  EF
1  exp
 kT

 C1
 E  E
Ec
C



dE
 E  E F  
 dE
kT


 exp
Let
E  Ec
so that dE  kT  d
kT
We can write
E  E F   E  E c   E c  E F 

lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 4
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
The ionization energy is
Then
  E c  E F  
n o  C1 exp 

kT




 E  E
c
Ec
or
 E  E c  
 dE
kT


 exp 
  E c  E F  
n o  C1 exp 

kT


 kT  exp kT d
0
We find that


 o

 
 s
2

0.067
 13.6  
13.6

13.12

or
E  0.0053 eV
_______________________________________
4.17


 m*
E  
 mo

 exp  d  exp     1  1
0
0
N
(a) E c  E F  kT ln c
 no
 2.8  1019
 0.0259  ln
15
 7 10
 0.2148 eV
(b) E F  E  E g  E c  E F 
So
  E c  E F  
2
n o  C1 kT  exp 

kT


_______________________________________
4.15
r
m 
We have 1 r  o* 
ao
m 
For germanium, r  16 , m *  0.55m o
Then
 1 
r1  16 
a o  29 0.53
 0.55 
or
o
r1  15.4 A
The ionization energy can be written as




(c)




 1.12  0.2148  0.90518 eV
   E F  E  
p o  N  exp 

kT




  0.90518 
 1.04  1019 exp 

 0.0259 
 6.90 10 3 cm 3
(d) Holes
n 
(e) E F  E Fi  kT ln o 
 ni 
 7  1015 

 0.0259  ln
10 
 1.5  10 
 0.338 eV
_______________________________________
4.18




 m  o 
  13.6 eV
E  
 
 m o  s 
0.55
13.6  E  0.029 eV

162
_______________________________________
N
(a) E F  E  kT ln 
 po
4.16
 1.12  0.162  0.958 eV
  0.958 
(c) n o  2.8  1019 exp

 0.0259 
*
2
r
m 
We have 1 r  o* 
ao
m 
For gallium arsenide, r  13.1 ,
m  0.067 m o
 1.04  1019
 0.0259  ln
16
 2 10
 0.162 eV
(b) E c  E F  E g  E F  E 


 2.4110 3 cm 3
*
Then
o
 1 
r1  13.1
0.53  104 A
 0.067 




lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 4
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
p 
(d) E Fi  E F  kT ln o 
 ni 
4.21
 2  1016 

 0.0259  ln
10 
 1.5  10 
 0.365 eV
_______________________________________
4.19
N
(a) E c  E F  kT ln c
 no





 2.414  1014 cm 3
(c) p-type
_______________________________________
4.20
 375 
(a) kT  0.0259
  0.032375 eV
 300 


3/ 2
  0.28 
exp 

 0.032375 
 1.15  1014 cm 3
E F  E  E g  E c  E F   1.42  0.28
 1.14 eV


 375 
p o  7  1018 

 300 
3/ 2
  1.14 
exp 

 0.032375 
 4.99  10 3 cm 3
 4.7 1017 

(b) E c  E F  0.0259 ln
14 
 1.15 10 
 0.2154 eV
E F  E  E g  E c  E F   1.42  0.2154
 1.2046 eV

p o  7  10
18


3/ 2
  0.28 
exp 

 0.032375 
 6.86  1015 cm 3
E F  E  E g  E c  E F   1.12  0.28

  0.27637 
(b) p o  1.04  1019 exp

 0.0259 
 375 
n o  4.7  1017 

 300 

 375 
n o  2.8  1019 

 300 
 0.840 eV




 2.8  1019
 0.0259  ln
5
 2  10
 0.8436 eV
E F  E  E g  E c  E F 
 1.12  0.8436
E F  E  0.2764 eV

 375 
(a) kT  0.0259
  0.032375 eV
 300 
  1.2046 
exp 

 0.0259 

 375 
p o  1.04  1019 

 300 
 7.84  10 7 cm 3
N
(b) E c  E F  kT ln c
 no
3/ 2
  0.840 
exp 

 0.032375 




 2.8  1019 

 0.0259  ln
15 
 6.862  10 
 0.2153 eV
E F  E  1.12  0.2153  0.9047 eV


  0.904668 
p o  1.04  1019 exp 

 0.0259 
 7.04 10 3 cm 3
_______________________________________
4.22
(a) p-type
Eg
1.12
 0.28 eV
4
4
  E F  E  
p o  N  exp 

kT


(b) E F  E 



  0.28 
 1.04  1019 exp 

 0.0259 
 2.10  1014 cm 3
E c  E F  E g   E F  E 
 1.12  0.28  0.84 eV
  E c  E F  
n o  N c exp 

kT




  0.84 
 2.8 1019 exp 

 0.0259 
 2.30  10 5 cm 3
_______________________________________
 4.42  10 2 cm 3
_______________________________________
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 4
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
4.25
4.23
 E  E Fi 
(a) n o  n i exp  F


 kT
 0.22 
 1.5  1010 exp 

 0.0259 

 400 
kT  0.0259
  0.034533 eV
 300 


 7.33  1013 cm 3
 E  EF 
p o  n i exp  Fi

 kT

  0.22 
 1.5  10 10 exp 

 0.0259 




 5.6702  10 24
 n i  2.381 10 12 cm 3

N
(a) E F  E  kT ln 
 po

 3.68  10 2 cm 3
_______________________________________
4.24




 1.04 1019
 0.0259 ln
15
 5  10
 0.1979 eV
(b) E c  E F  E g  E F  E 




 1.12  0.19788  0.92212 eV
  0.92212 
(c) n o  2.8 10 19 exp 

 0.0259 

 9.66  10 3 cm 3
(d) Holes
p
(e) E Fi  E F  kT ln o
 ni

  1.12 
 exp 

 0.034533 
 8.80  10 9 cm 3
 E  EF 
p o  n i exp  Fi


 kT
  0.22 
 1.8 10 6 exp 

 0.0259 

3/ 2
 4.3109  1019 cm 3
n  4.3109  1019 1.601 10 19
2
i
 0.22 
 1.8  10 6 exp 

 0.0259 
N
(a) E F  E  kT ln 
 po

 400 
N c  2.8  1019 

 300 
 3.07 10 6 cm 3
 E  E Fi 
(b) n o  n i exp  F

 kT


3/ 2
 1.6011019 cm 3



 400 
N   1.04  1019 

 300 




 1.601 1019 

 0.034533 ln
15


 5  10
 0.2787 eV
(b) E c  E F  1.12  0.27873  0.84127 eV


  0.84127 
(c) n o  4.3109  10 19 exp 

 0.034533 
 1.134 10 9 cm 3
(d) Holes
p 
(e) E Fi  E F  kT ln o 
 ni 
 5  1015 

 0.034533 ln
12 
 2.38110 
 0.2642 eV
_______________________________________
4.26


  0.25 
(a) p o  7 1018 exp 

 0.0259 
 4.50  1014 cm 3
E c  E F  1.42  0.25  1.17 eV




 5  1015 

 0.0259  ln
10 
 1.5  10 
 0.3294 eV
_______________________________________


  1.17 
n o  4.7  1017 exp 

 0.0259 
 1.13  10 2 cm 3
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 4
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(b) kT  0.034533 eV

N   7  10
18

 400 


 300 

 8.49  10 9 cm 3
_______________________________________
 1.078 1019 cm 3


 400 
N c  4.7  1017 

 300 
3/ 2
4.28
(a) n o 
 7.236  1017 cm 3
N 
E F  E  kT ln  
 po 
 1.078 1019
 0.034533 ln
14
 4.50  10
 0.3482 eV
E c  E F  1.42  0.3482  1.072 eV





no 
4.27

po 

n o  7.23  10 cm
(b) kT  0.034533 eV

N   1.04  10
19
2.8 10 1.0
19
4.7 10 1.0
2
17

2

N  F1 / 2  F 
2
5  1019 
3

1.04 10 F   
19
1/ 2
F
So F1 / 2  F   4.26

 400 


 300 
E  E F
kT
E  E F  3.00.0259   0.0777 eV
_______________________________________
We find  F  3.0 
3/ 2
 1.6011019 cm 3


4.29
  0.870 
n o  2.8  10 19 exp 

 0.0259 
4
2
 5.30  1017 cm 3
_______________________________________
  0.25 
exp 

 0.0259 
 6.68  1014 cm 3
E c  E F  1.12  0.25  0.870 eV

N c F1 / 2  F 

 2.40  10 4 cm 3
_____________________________________


 3.16  1019 cm 3
2
(b) n o 
N c F1 / 2  F 


2
For E F  E c  kT 2 ,
E  E c kT 2
F  F

 0.5
kT
kT
Then F1 / 2  F   1.0
  1.07177 
n o  7.236  10 17 exp 

 0.034533 
(a) p o  1.04 10 19

  0.77175 
n o  4.311 10 19 exp 

 0.034533 
3/ 2

 400 
N c  2.8  1019 

 300 
3/ 2
4.30
E F  E c 4kT

4
kT
kT
Then F1 / 2  F   6.0
(a)  F 
 4.311 1019 cm 3
N 
E F  E  kT ln  
 po 
 1.601 10
 0.034533 ln
14
 6.68  10
 0.3482 eV
E c  E F  1.12  0.3482  0.7718 eV
19




no 
2

2


N c F1 / 2  F 
2.8 10 6.0
19
 1.90  10 20 cm 3
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 4
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(b) n o 
4.7 10 6.0
2
p E  
17

 3.18  1018 cm 3
_______________________________________
4.31
For the electron concentration
nE   g c E  f F E 
The Boltzmann approximation applies, so
nE  
or
nE  

4 2m
h
E  Ec
  E  E F  
 exp 

kT



4 2m n*
h

* 3/ 2
n
3

3/ 2
3
 kT
  E c  E F  
exp 

kT


E  Ec
  E  E c  
exp 

kT
kT


Define
E  Ec
x
kT
Then
nE   nx   K x exp x 
To find maximum nE   nx  , set
dnx 
1
 0  K  x 1 / 2 exp x 
dx
2
 x 1 / 2  1 exp x 

or

1
0  Kx 1 / 2 exp x   x 
2


which yields
1 E  Ec
1
 E  E c  kT
x 
2
2
kT
For the hole concentration
pE   g  E 1  f F E 
Using the Boltzmann approximation
p E  
or

4 2m *p
h3

3/ 2
E  E
  E F  E  
 exp 

kT



4 2m *p
h

3
 kT
3/ 2
   E F  E  
exp 

kT


E  E
   E  E  
exp 

kT
kT


Define
x 
E  E
kT
Then
px   K  x  exp x 
To find maximum value of pE   px  , set
dpx 
 0 Using the results from above,
dx 
we find the maximum at
1
E  E  kT
2
_______________________________________
4.32
(a) Silicon: We have
  E c  E F  
n o  N c exp 

kT


We can write
E c  E F  E c  E d   E d  E F 
For
E c  E d  0.045 eV and E d  E F  3kT eV
we can write
  0.045 
n o  2.8  10 19 exp 
 3

 0.0259


 2.8  10 exp 4.737 
19
or
n o  2.45  1017 cm 3
We also have
  E F  E  
p o  N  exp 

kT


Again, we can write
E F  E  E F  E a   E a  E 
For
E F  E a  3kT and E a  E  0.045 eV
Then
0.045 

p o  1.04 1019 exp  3 
0.0259 



 1.04  10  exp 4.737 
19
or
p o  9.12  10 16 cm 3
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 4
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(b) GaAs: assume E c  E d  0.0058 eV
Then
  0.0058 
n o  4.7  1017 exp 
 3

 0.0259


 4.7  10  exp 3.224 
17
Assume E a  E  0.0345 eV
Then
  0.0345 
p o  7 1018 exp 
 3

 0.0259
p o  9.20  10 16 cm 3
_______________________________________
4.33
Plot
_______________________________________
4.34
(a) p o  4  15  10 15  3  1015 cm 3
 7.5  10 4 cm 3
3  1015
(b) n o  N d  3 10 16 cm 3
 7.5  10 cm
3
3  10
(c) n o  p o  n i  1.5  10 10 cm 3

3
p o  N a  4 10 15 cm 3
7.334 10 
11 2
4  10

15


 450 
(e) n i2  2.8  1019 1.04  1019 

 300 
3
  1.12300  
 exp 

 0.0259 450 
 n i  1.722  10 13 cm 3
2
 1.08  10 3 cm 3
3  10
(c) n o  p o  n i  1.8  10 6 cm 3



 375 
(d) n i2  4.7  1017 7.0  1018 

 300 
3
  1.42 300  
 exp 

 0.0259 375 
 n i  7.580  10 8 cm 3
p o  N a  4 10 15 cm 3
7.580 10 
8 2
4  1015
 1.44  10 2 cm 3


 450 
(e) n i2  4.7  1017 7.0  1018 

 300 
3
  1.42 300  
 exp 

 0.0259 450  
 n i  3.853  1010 cm 3
n o  N d  10 14 cm 3
3.853 10 
10 2
 1.48  10 7 cm 3
10 14
_______________________________________
po 
 1.34  10 8 cm 3

 1.08  10  4 cm 3
16

  1.12300 
 exp 

 0.0259 375 
 n i  7.334  1011 cm 3
no 
po
no 
3
 375 
(d) ni2  2.8  1019 1.04  1019 

 300 
1.8 10 

6 2
10 2


n i2
1.8  10 6

po
3  1015
(b) n o  N d  3 10 16 cm 3
1.5 10 

13 2
no 
or
16
2
1.722 10 
 3 10 15 cm 3
18
po

4.35
(a) p o  N a  N d  4  10 15  10 15


 7  10  exp 4.332 
10 2

 2.88  1012 cm 3
1.029  1014
_______________________________________
n o  1.87  1016 cm 3
1.5 10 

2

  1.722  10 13


 1.029  1014 cm 3
po 
or
no 
 1014
10 14
no 
 
2
 2
4.36
(a) Ge: n i  2.4  1013 cm 3
2
(i) n o 

Nd
N 
  d   n i2
2
 2 
 2  1015
2  1015
 
2
 2
2

  2.4  10 13




2
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 4
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Then
or
n o  N d  2 1015 cm 3
po 

f F E  

13 2
n i2
2.4  10

no
2  10 15
 2.88  1011 cm 3
(ii) p o  N a  N d  10 16  7  10 15
 3 10 15 cm 3
no 

n i2
2.4  10 13

po
3  1015

2
 1.92  10 11 cm 3
(b) GaAs: n i  1.8  10 6 cm 3
(i) n o  N d  2 1015 cm
1.8 10 
6 2
 1.62  10 3 cm 3
2  10 15
(ii) p o  N a  N d  3 10 15 cm 3
po 
1.8 10 

6 2
no
 1.08  10 3 cm 3
3  10
(c) The result implies that there is only one
minority carrier in a volume of 10 3 cm 3 .
_______________________________________
15
4.37
(a) For the donor level
nd
1

Nd
 E  EF 
1
1  exp d

2
 kT 
1

1
 0.20 
1  exp

2
 0.0259 
or
nd
 8.85  10  4
Nd
(b) We have
1
f F E  
 E  EF 
1  exp

 kT 
Now
E  E F   E  E c   E c  E F 
or
E  E F  kT  0.245
1
0.245 

1  exp1 

 0.0259 
or
f F E   2.87  10 5
_______________________________________
4.38
(a) N a  N d  p-type
(b) Silicon:
p o  N a  N d  2.5  1013  1 10 13
or
p o  1.5  10 13 cm 3
Then


2
n i2
1.5  1010

po
1.5  10 13
Germanium:
no 
po 
 1.5  10 7 cm 3
Na  Nd
 N  Nd
  a
2
2

 1.5  10 13 
 1.5  10 13
 
 


2
2



or
p o  3.26  10 13 cm 3
Then

2
2

  n i2


  2.4  10 13




2

2
n i2
2.4  1013
 1.76  1013 cm 3

p o 3.264  10 13
Gallium Arsenide:
p o  N a  N d  1.5  1013 cm 3
and
no 


2
n i2
1.8  10 6
 0.216 cm 3

13
po
1.5  10
_______________________________________
no 
4.39
(a) N d  N a  n-type
(b) n o  N d  N a  2  10 15  1.2  1015
 8 1014 cm 3
po 

n i2
1.5  1010

no
8  10 14

2
 2.81 10 5 cm 3
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 4
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(c) p o  N a  N a   N d
4  10
15
 N a  1.2  10  2  10
15
4.43
Plot
_______________________________________
15
 N a  4.8  1015 cm 3
1.5 10 

10 2
3
 5.625  10 cm
4  1015
_______________________________________
no
4
4.44
Plot
_______________________________________
4.45
4.40


no 
10 2
n
1.5  10
 1.125  10 15 cm 3

po
2  10 5
n o  p o  n-type
_______________________________________
no 
2
i
Nd  Na
 N  Na
  d
2
2

1.1 1014 
4.41



 250 
n i2  1.04  1019 6.0  10 18 

 300 
3
1.110


 0.66
 exp 




0
.
0259
250
300


So
2
N
N 
p o  a   a   n i2
2
 2 
  4 10 
13 2
so n i  5.74  1013 cm 3
N a  N d  p-type
Majority carriers are holes
p o  N a  N d  3  10 16  1.5  1016


2
n i2
1.5  1010
 1.5  10 4 cm 3

po
1.5  1016
(b) Boron atoms must be added
p o  N a  N a  N d
5 1016  N a  3  1016  1.5 1016
2

  1.8936  10 24


N
7.5735  10 24  2.752  1012 N a   a
 2
N
  a
 2
 ni2
 4  1013
no 
2

2
14
 1.5 1016 cm 3
Minority carriers are electrons
Then p o  2.75  10 12 cm 3
N
  a
 2

  n i2


4.9  10 27  1.6  10 27  n i2
4.46
(a)
1
 n o  ni
2
n o  6.88  1011 cm 3 ,



2
 2  1014  1.2  1014
 
2

po 
n i2
n2
1
 i  n o2  n i2
p o 4n o
4
N

 2.752  1012  a
2

2  1014  1.2  1014
2
n i2 3.3  10 27

 3  1013 cm 3
n o 1.1 10 14
_______________________________________
 1.8936  10 24
 n i  1.376  10 12 cm 3
no 
2

  n i2

So N a  3.5  1016 cm 3



2

  1.8936  10 24

1.5  10 
10 2
2
 4.5  10 3 cm 3
5  10 16
_______________________________________
no 
so that N a  2.064  10 12 cm 3
_______________________________________
4.42
Plot
_______________________________________
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 4
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________




 Nd
 0.0259 ln
10
 1.5  10
N
(b) E F  E Fi  kT ln d
 ni
4.47
(a) p o  ni  n-type
(b) p o 
ni2
n2
 no  i
no
po
no 
1.5 10 
10 2
 1.125  1016 cm 3
2  10
 electrons are majority carriers
4
p o  2 10 4 cm 3
 holes are minority carriers
(c) n o  N d  N a
For 1014 cm 3 , E F  E Fi  0.2280 eV
1015 cm 3 , E F  E Fi  0.2877 eV
1016 cm 3 , E F  E Fi  0.3473 eV
1017 cm 3 , E F  E Fi  0.4070 eV
_______________________________________
1.125 1016  N d  7  1015
so N d  1.825  10 16 cm 3
_______________________________________
4.50
(a) n o 
p 
E Fi  E F  kT ln o 
 ni 
For Germanium
T (K)
kT (eV)
1.05 10
ni (cm 3 )
2.16  1010
8.60 1014
3.82  1016
0.01727
0.03453
0.0518
15
 0.5 1015


2
N a  1015 cm 3
T (K)
p o (cm 3 )
E Fi  E F  (eV)
200
400
600
1.0 1015
1.49 1015
3.87 1016
0.1855
0.01898
0.000674
_______________________________________
4.49




 2.8  1019
 0.0259 ln
 Nd

2
 ni2
so n i2  5.25  10 28
Now



 T 
ni2  2.8  1019 1.04 1019 

 300 
3


 1.12
 exp 

 0.0259T 300 
Na
N 
  a   n i2 and
2
 2 
N
(a) E c  E F  kT ln c
 Nd

  n i2

 0.5  1015
2
po 
2
Nd
N
  d
2
 2
n o  1.05 N d  1.05  1015 cm 3
4.48
200
400
600







For 1014 cm 3 , E c  E F  0.3249 eV
1015 cm 3 , E c  E F  0.2652 eV
1016 cm 3 , E c  E F  0.2056 eV


 T 
5.25  10 28  2.912  10 38 

 300 
3
  12972.973 
 exp 

T


By trial and error, T  536.5 K
(b) At T  300 K,
N 
E c  E F  kT ln c 
 no 
 2.8 1019 

E c  E F  0.0259 ln
15


 10
 0.2652 eV
At T  536.5 K,
 536.5 
kT  0.0259 
  0.046318 eV
 300 


 536.5 
N c  2.8 1019 

 300 
 6.696  1019 cm 3
1017 cm 3 , E c  E F  0.1459 eV
3/ 2
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 4
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
N
E c  E F  kT ln c
 no




 6.696  1019 

E c  E F  0.046318 ln
15 
 1.05  10 
 0.5124 eV
then E c  E F   0.2472 eV
(c) Closer to the intrinsic energy level.
_______________________________________
Then, T  200 K, E Fi  E F  0.4212 eV
T  400 K, E Fi  E F  0.2465 eV
T  600 K, E Fi  E F  0.0630 eV
_______________________________________
4.52
(a)
N
E Fi  E F  kT ln a
 ni

 Na
  0.0259 ln
6

 1.8  10




For N a  1014 cm 3 , E Fi  E F  0.4619 eV
4.51
p 
E Fi  E F  kT ln o 
 ni 
At T  200 K, kT  0.017267 eV
T  400 K, kT  0.034533 eV
T  600 K, kT  0.0518 eV
N a  10 15 cm 3 , E Fi  E F  0.5215 eV
16
N a  10 cm 3 , E Fi  E F  0.5811 eV
N a  10 17 cm 3 , E Fi  E F  0.6408 eV
(b)
N
E F  E  kT ln 
 Na
At T  200 K,



 200 
n i2  2.8  1019 1.04  1019 

 300 
3
N a  10 15 cm 3 , E F  E  0.2293 eV
16
N a  10 cm 3 , E F  E  0.1697 eV
N a  10 17 cm 3 , E F  E  0.1100 eV
_______________________________________
 n i  7.638 10 4 cm 3
At T  400 K,



3
4.53
(a) E Fi  E midgap 
  1.12 
 exp 

 0.034533 
 n i  2.381 10 cm
At T  600 K,
12

n  2.8 10
2
i
19

3
1.04 10 
19
 600 


 300 
3
0.0259 ln10
4
(b) Impurity atoms to be added so
E midgap  E F  0.45 eV
(i) p-type, so add acceptor atoms
(ii) E Fi  E F  0.0447  0.45  0.4947 eV
Then
 E  EF 
p o  ni exp Fi

 kT

p o  N a  31015 cm 3
At T  600 K,
 
 0.4947 
 10 5 exp

 0.0259 
2

  n i2

 3 1015
3  1015
 
2
 2




E Fi  E midgap  0.0447 eV
3
 n i  9.740  1014 cm 3
At T  200 K and T  400 K,

 m *p
3
kT ln *
 mn
4

or
  1.12 
 exp 

 0.0518 
N
N
p o  a   a
2
 2




For N a  1014 cm 3 , E F  E  0.2889 eV
  1.12 
 exp 

 0.017267 
 400 
n i2  2.8  1019 1.04  1019 

 300 

 7.0  1018
  0.0259 ln

 N
a


or
2

  9.740  1014




2
p o  N a  1.97  1013 cm 3
_______________________________________
 3.288  1015 cm 3
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 4
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
4.54
  E c  E F  
n o  N d  N a  N c exp 

kT


so


  0.215 
N d  5  1015  2.8 1019 exp

 0.0259 
 5  1015  6.95  1015
N
(b) E F  E Fi  kT ln c
 Nd




 2.8  1019
 0.0259 ln
16
 2  10
(c) For part (a);
p o  2  10 16 cm 3

n2
1.5  1010
no  i 
po
2 1016
or
N d  1.2  10 16 cm 3
_______________________________________
For part (b):
n o  2 1016 cm 3




po 
 2.8  1019 
  0.2188 eV
 0.0259 ln
15 
 6  10 
(ii) E c  E F  0.2188  0.0259  0.1929 eV
  E c  E F  
N d  N c exp 

kT


0
.
1929



 2.8  1019 exp 

 0.0259 

 N d  1.031 1016 cm 3 Additional
donor atoms
(b) GaAs
 4.7  1017 

(i) E c  E F  0.0259 ln
15


 10
 0.15936 eV
(ii) E c  E F  0.15936  0.0259  0.13346 eV

N d  4.7  10
17

  0.13346 
exp 

 0.0259 
 2.718  1015 cm 3  N d  10 15
 N d  1.718  1015 cm 3 Additional
donor atoms
_______________________________________
4.56
N 
(a) E Fi  E F  kT ln  
 Na 
 1.04  1019 
  0.1620 eV
 0.0259 ln
16 
 2  10 
2

ni2
1.5  1010

no
2 1016

2
 1.125  10 4 cm 3
_______________________________________
4.57
 E  E Fi 
n o  ni exp  F

 kT

 0.55 
 1.8  10 6 exp 

 0.0259 


N d  1.631 1016 cm 3  N d  6 1015

 1.125  10 4 cm 3
4.55
(a) Silicon
N
(i) E c  E F  kT ln c
 Nd

  0.1876 eV



 3.0  1015 cm 3
no  N d  N a
3  1015  7  10 15  N a
 N a  4 10 15 cm 3
_______________________________________
4.58
p 
(a) E Fi  E F  kT ln o 
 ni 
 3  1015
 0.0259 ln
10
 1.5  10
n
(b) E F  E Fi  kT ln o
 ni




 3  1016
 0.0259 ln
10
 1.5  10
(c) E F  E Fi

  0.3161 eV



  0.3758 eV


lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 4
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
p 
(d) E Fi  E F  kT ln o 
 ni 
4.60
n-type
15

 375   4  10

 0.0259
 ln
11 
300

7
.
334
10
 


 0.2786 eV
n 
(e) E F  E Fi  kT ln o 
 ni 
14
 450   1.029 10 
 0.0259
 ln
13
 300   1.722  10 
 0.06945 eV
_______________________________________
n
E F  E Fi  kT ln o
 ni
 1.125  10 16 
  0.3504 eV
 0.0259  ln


10
 1.5  10

______________________________________
4.61
po 
Na
N
  a
2
 2
5.08  1015 
4.59
N
(a) E F  E  kT ln 
 po

  0.2009 eV


 7.0  10 18 

(b) E F  E  0.0259  ln
4 
 1.08  10 
 1.360 eV
 7.0  1018 

(c) E F  E  0.0259 ln
6 
 1.8  10 
 0.7508 eV
 375 
(d) E F  E  0.0259

 300 


 7.0 1018 375 3003 / 2 
 ln 

4  1015


 0.2526 eV
 450 
(e) E F  E  0.0259

 300 

2

  n i2

5 1015
2




 7.0  1018
 0.0259 ln
15
 3  10

 7.0 10 450 300 
 ln 

1.48  10 7


 1.068 eV
_______________________________________
18




5.08 10
 5  1015
 
 2
15
2

  n i2



 2.5 10 
2
 2.5 1015
15 2
 n i2
6.6564  10 30  6.25  10 30  n i2
 n i2  4.064  10 29
 Eg 
ni2  N c N  exp 

 kT 
 350 
kT  0.0259
  0.030217 eV
 300 


2


2
 350 
3
19
N c  1.2  1019 
  1.633  10 cm
300


 350 
3
19
N   1.8  1019 
  2.45 10 cm
300


Now
4.064  10 29  1.633  1019 2.45  10 19



  Eg 
 exp 

 0.030217 
3/ 2
So



 1.633  1019 2.45  1019 
E g  0.030217  ln 

4.064  10 29


 E g  0.6257 eV
_______________________________________
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 4
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
4.62
(a) Replace Ga atoms  Silicon acts as a
donor
N d  0.05 7  1015  3.5  10 14 cm 3
Replace As atoms  Silicon acts as an
acceptor
N a  0.95 7  1015  6.65  1015 cm 3




(b) N a  N d  p-type
(c) p o  N a  N d  6.65  1015  3.5  1014
 6.3 1015 cm 3


2
n2
1.8 10 6
no  i 
 5.14  10  4 cm 3
po
6.3  1015
p 
(d) E Fi  E F  kT ln o 
 ni 
 6.3 1015 
  0.5692 eV
 0.0259 ln
6 
 1.8  10 
_______________________________________
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 5
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Chapter 5
5.1

 
1
 0.208 (  -cm) 1
 4.8077
_______________________________________
(b)  
5.2
1

1
1.6  10 220 8  1016
 0.355  -cm
(b)   e n N d

1
1
(a)  

e n N d
1.6  10 19 1300 1015
 4.808  -cm

 2.96 1016 cm 3
_______________________________________
5.3
(a)   e n N d
then  n  3800 cm 2 /V-s which gives
5.5
R
L
A
or  n 


From Figure 5.3, for N d  6 10 cm
16
3
find  n  1050 cm /V-s which gives
19
10506 10 
16
 10.08 (  -cm) 1
1
(b)  
e p N a
19
L
A

L
e n N d A
L
eN d RA
2.5
2  1015 700.1
1.6 10 
19

 1116 cm /V-s
_______________________________________
2

p Na
From Figure 5.3, for N a  10 17 cm 3 we
find  p  320 cm 2 /V-s which gives
1
1.6 10 32010 
19
17
 0.195  -cm
_______________________________________
5.4
1
e p N a
1
1.6 10 
19
p
Na
From Figure 5.3, for N a  8 10 16 cm 3 we
find  p  220 cm /V-s which gives
2

5.6
(a) n o  N d  10 16 cm 3
and
1
1.6 10 

we
2
0.35 

From Figure 5.3, for N d  2  1017 cm 3 ,
10  1.6  10 19  n N d
(a)  

 121.6 (  -cm) 1
_______________________________________
1.80

or N a 

e p 1.6  10 19 380


  1.6  10 19 3800 2  10 17 
  e p N a
0.20 

120  1.6  10 19  n N d

  1.6  10

19
n2
1.8  10 6
po  i 
no
10 16
(b)
J  e n n o 

2
 3.24  10  4 cm 3
For GaAs doped at N d  10 16 cm 3 ,
 n  7500 cm 2 /V-s
Then
J  1.6  10 19 7500  10 16 10 
or
J  120 A/cm 2
(b) (i) p o  N a  1016 cm 3


no 
 
n i2
 3.24  10  4 cm 3
po
(ii) For GaAs doped at N a  10 16 cm 3 ,
 p  310 cm 2 /V-s
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 5
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
J  e p p o 

 1.6  10
19
(b) R  L  R  68.933  206.79 
31010 10
16
or
J  4.96 A/cm 2
_______________________________________
29.0  10 3
 34.12 A/cm 2
8.5  10  4
J
34.12
Then  d 

ep o
1.6  10 19 2 1016
For (a), J 
5.7
(a) V  IR  10  0.1R
or
R  100 
(b)
L
L
R
 
RA
A
or
10 3
 0.01 (  -cm) 1

100 10 3
(c)   e n N d
or
0.01  1.6  10 19 1350 N d
Then
N d  4.63  10 13 cm 3
(d)   e p p o

or


5.9
(a) For N d  2 1015 cm 3 , then
 n  8000 cm 2 /V-s

e
L
p Na A





25  10 3
1.6  10 19 2  1015
5 10 
5
 1.56  10 cm/s
(c) I  en o d A

0.075
1.6 10 4002 10 8.5 10 
16
 68.93 



4




 1.6  10 19 2  10 15 5  10 6 5  10 5
 0.080 A
or I  80 mA
_______________________________________
V
2

 0.0290 A
R 68.93
I  29.0 mA
I
or

 1.6  10 19 8000  2  10 15 200  5  10 5
 0.0256 cm
I
(b) J   en o d
A
I
or  d 
Aen o 
6
 p  400 cm 2 /V-s
19
or L  e n N d RA

For N a  2  10 16 cm 3 , then
R
5
V

 200 
I 25  10 3
L
R
e n N d A
R
5.8
L

 3.55  10 cm/s
_______________________________________

A

3
0.01  1.6  10 19 480  p o
Then
p o  1.30  1014 cm 3  N a  N d
So
N a  1.30  1014  1015  1.13  10 15 cm 3
Note: For the doping concentrations
obtained, the assumed mobility values are
valid.
_______________________________________
(a) R 

 1.066 10 cm/s
9.67  10 3
For (b), J 
 11.38 A/cm 2
8.5  10  4
11.38
d 
1.6  10 19 2  1016



4


2
V

 0.00967 A
R 206.79
or I  9.67 mA
(c) J  ep o d
I
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 5
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(b) N a  N d  10 16 cm 3
5.10
V 3
(a)     3 V/cm
L 1
 d  n    n 
d

  n  1250 cm 2 /V-s

 p  410 cm 2 /V-s
10 4
3

or
 n  3333 cm 2 /V-s

 2.51 10  -cm
(c) N a  N d  10 18 cm 3
 d   n   8003
  n  290 cm 2 /V-s
or
 p  130 cm 2 /V-s
 d  2.4  10 3 cm/s
_______________________________________
5.11
(a) Silicon: For   1 kV/cm,
 d  1.2  10 6 cm/s
Then
10 4
d

 8.33  10 11 s
tt 
 d 1.2  10 6
1
1

e n n o  e p p o e  n   p n i



5.13
(a) GaAs:
  e p p o  5  1.6  10 19  p p o

  n  1350 cm 2 /V-s
 p  480 cm 2 /V-s
1

From Figure 5.3, and using trial and error, we
find
p o  1.3  10 17 cm 3 and
 p  240 cm 2 /V-s
Then


2
n i2
1.8  10 6
 2.49  10 5 cm 3

po
1.3  1017
(b) Silicon:
1
   e n n o
no 

or
1
1

e n 8 1.6 10 19 1350
which gives
n o  5.79  10 14 cm 3
and
no 




2
n i2
1.5  1010
 3.89  10 5 cm 3

no
5.79  1014
Note: For the doping concentrations obtained
in part (b), the assumed mobility values are
valid.
_______________________________________
po 
(a) N a  N d  1014 cm 3
1.6 10 1350  4801.5 10 
19
1
290  130 1.5 1010
1.6 10 
19
 9.92  10  -cm
_______________________________________
5.12


5
For GaAs:  d  7.5  10 6 cm/s
Then
d
10 4
tt 

 1.33  10 11 s
 d 7.5  10 6
(b) Silicon: For   50 kV/cm,
 d  9.5  10 6 cm/s
Then
10 4
tt 
 1.05  10 11 s
9.5  10 6
For GaAs:  d  7  10 6 cm/s
Then
10 4
tt 
 1.43  10 11 s
7  10 6
_______________________________________


5
(b)

1
1250  410 1.5 1010
1.6 10 
19
10
 2.28  10 5  -cm
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 5
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
5.14
 i  eni  n   p 
Then
10 6  1.6  10 19 1000  600n i
or
n i (300 K)  3.91 10 9 cm 3
Now
  Eg 

n i2  N c N  exp

 kT 
or
N N 
E g  kT ln c 2  
 ni 


 
 1019 2
 0.0259  ln 
 3.91 10 9
which gives
E g  1.122 eV


2



 
2
(i) Si:
R
200  10 4
 1.06  10 6 
2.23  10  2 85  10 8
(iii) GaAs:
200  10 4
R
 9.19  1012 
2.56  10 9 85  10 8
_______________________________________
R







1.5 10 1350  480


 0.2496 (  -cm)
(b) Using Figure 5.2,
(i) For T  250 K ( 23 C),
  n  1800 cm 2 /V-s
  1.6  10 19 1800 1.2  10 15 
 0.346 (  -cm) 1
(ii) For T  400 K ( 127 C),
  n  670 cm 2 /V-s
  1.6  10 19 670 1.2  1015 
 0.129 (  -cm) 1
_______________________________________
5.17
t
 i  4.39  10 6 (  -cm) 1
 avg
(ii) Ge:
 i  1.6  10 19 2.4  1013 3900  1900 
or
 i  2.23  10 2 (  -cm) 1
(iii) GaAs:
 i  1.6  10 19 1.8  10 6 8500  400 
or
 i  2.56  10 9 (  -cm) 1


1
10
or


So   1.6  10 19 1300  1.2  10 15

5.15
(a) (i) Silicon:  i  eni  n   p
 i  1.6  10

then  n  1300 cm 2 /V-s
n i (500 K)  2.27  1013 cm 3
Then
 i  1.6  10 19 2.27  1013 1000  600 
which gives
 i (500 K)  5.81 10 3 (  -cm) 1
_______________________________________


From Figure 5.3, for N d  1.2  10 15 cm 3 ,
or
19




 1.122
exp 




0
.
0259
500
300




(ii) Ge:
0.25  1.6  10 19  n N d
 5.15  10 26

200  10 4
 5.36  10 9 
4.39  10  6 85  10 8
5.16
(a)   e n N d
Now
n i2 (500K)  1019
L
A
(b) R 




t
1
1
x

 x dx   o exp
dx
t 0
t 0
 d 




o
t

 d  exp  x 
 od
t
t
 d 0
  t  
exp   1
  d  
200.3 1  exp  1.5 


1.5 
 0.3 
 3.97 (  -cm) 1
_______________________________________
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 5
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
5.18
2
V
 133.3 V/cm
(a)   
L 150  10  4
(b)  x   e n N d  x 
 avg  e n 
1
T

T
x 

 2 10 1  1.111T dx
16
0

T
e 2  1016 
x2
 n

x


21.111T   0
T






e n 2  1016 
T2
T



21.111T  
T



 1.6  10
(c) I 
7502 10 0.55
V 
L
150  10  4
 1.32  10 5 A
or I  13.2  A
(d) Top surface;
  1.6  10 19 750  2  1016







5.19
Plot
_______________________________________
5.20
(a)   10 V/cm
so
 d   n   1350 10   1.35  10 4 cm/s
or
 d  1.35  10 2 m/s
Then
1
T  m n* d2
2
2
1
 1.08 9.11 10 31 1.35  10 2
2
or
T  8.97  10 27 J  5.60  10 8 eV


 7.18  1019
n i  8.47  10 9 cm 3
 0.24 (  -cm) 1
J    0.24133.3  32 A/cm 2
_______________________________________






or
 2.4 (  -cm)
J    2.4 133.3  320 A/cm 2
Bottom surface:
  1.6  10 19 750  2  1015


  1.10 
 2  10 19 1 10 19 exp

 0.0259 
2
1


  Eg
(a) n i2  N c N  exp
 kT
16
 avg  1.32 (  -cm) 1
 avg A
1.327.5 10 4 10 4 

5.21
 e n 2  10 16 0.55
19
(b)   1 kV/cm
 d  1350 1000   1.35  10 6 cm/s
or
 d  1.35  10 4 m/s
Then
2
1
T  1.08 9.11 10 31 1.35  10 4
2
or
T  8.97  10 23 J  5.60  10 4 eV
_______________________________________

For N d  1014 cm 3 &gt;&gt; n i  n o  1014 cm 3
Then
J    e n n o 


 
 1.6  10 19 1000  1014 100 
or
J  1.60 A/cm 2
(b) A 5% increase is due to a 5% increase in
electron concentration, so
2
n o  1.05  1014 
Nd
N 
  d   ni2
2
 2 
which becomes
1.05 10
14
 5  1013
  5 10 
2
13 2
 n i2
and yields
n i2  5.25  10 26
3
  Eg
 T 
 2 1019 1 1019 
 exp
 300 
 kT







or


 1.10
 T 
2.625  10 12  
 exp 

 300 
 0.0259T 300 
By trial and error, we find
T  456 K
_______________________________________
3
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 5
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(b) From Figure 5.3,
n-type:  n  1100 cm 2 /V-s
5.22
n2
(a)   e n n o  e p p o and n o  i
po
Then
e n 2
  n i  e p p o
po
To find the minimum conductivity, set
 1e n ni2
d
0
 e p
dp o
p o2
p-type:  p  400 cm 2 /V-s
compensated:  n  1000 cm 2 /V-s
(c) n-type:   e n n o

 8.8 (  -cm)
p-type:   e p p o

 
p o  ni  n  (Answer to part (b))
p 


Substituting into the conductivity expression
e n n i2
   min 
1/ 2
ni  n  p
 
 e p n i  n  p
 
 min  2en i  n  p
The intrinsic conductivity is defined as
i
n   p
The minimum conductivity can then be
written as
 min 



2 i  n  p

1


 1.6  10 19 1000  3  10 16
 4.8 (  -cm)
J
(d) J     

1

1/ 2
which simplifies to
 i  eni  n   p   eni 

 1.28 (  -cm)
compensated:   e n n o
1/ 2
 

1
 1.6  10 19 400 2  1016
which yields
 

 1.6  10 19 1100 5  1016
120
n-type:  
 13.6 V/cm
8.8
120
p-type:  
 93.75 V/cm
1.28
120
compensated:  
 25 V/cm
4.8
_______________________________________
5.24
1
n   p

_______________________________________

1
1

1
2

1
3
1
1
1


2000 1500 500
 0.00050  0.000667  0.0020

5.23
(a) n-type: n o  N d  5 10 16 cm 3

n2
1.5  1010
po  i 
no
5  1016

2
 4.5  10 3 cm 3
p-type: p o  N a  2  1016 cm 3
1.5 10 
10 2
 1.125  10 4 cm 3
2  10 16
compensated: n o  N d  N a
no 
 5  1016  2  1016
 3  1016 cm 3
po
1.5 10 
10 2
3  1016
 7.5  10 3 cm 3
or
1

 0.003167
Then
  316 cm 2 /V-s
_______________________________________
5.25
 T 

 300 
(a) At T  200 K,
 n  1300
 300 

 200 
 n  1300
3 / 2
 300 
 1300

 T 
3 / 2
3/ 2
 2388 cm 2 /V-s
(b) At T  400 K,  n  844 cm 2 /V-s
_______________________________________
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 5
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
5.26
1
1
1
1
1




 0.006
  1  2 250 500
Then
  167 cm 2 /V-s
_______________________________________
5.27
Plot
_______________________________________
5.28
Plot
_______________________________________
5.29
 5  1014  n0 
dn

J n  eD n
 eD n 

dx
 0.01  0 
 5  1014  n0 

0.19  1.6  10 19 25

0.010


Then
0.190.010  5 1014  n0
1.6  10 19 25
which yields
n0   0.25  10 14 cm 3
_______________________________________




5.30
dn
n
 eDn
dx
x
 2  10 16  5  1015 
J n  1.6  10 19 27 

0  0.012


J n  eDn


J n  5.4 A/cm 2
_______________________________________
5.31
n
dn
 eD n
x
dx
 1015  nx1  
 2  1.6  10 19 30
4 
 0  20  10 
(a) J n  eD n


4  10 3  4.8  10 3  4.8  10 18 nx1 
which yields
nx1   1.67  10 14 cm 3
 1015  n x1  
(b)  2  1.6  10 19 230
4 
 0  20  10 
4  10 3  3.68  10 2  3.68  10 17 nx1 


nx1   8.91 1014 cm 3
_______________________________________
5.32
2
dp
d  16 
x 
10
1
 eD p


 

dx
dx 
 L  
J p  eD p
 eD p 
x
1016 
 21  
L
 L
(a) For x  0 ,
 1.6  10 19 10 1016 2
Jp 
12  10  4
 26.7 A/cm 2
(b) For x  6  m,

  

  
6

 1.6  10 19 10  1016 2 1  
12


Jp 
12  10  4
 13.3 A/cm 2
(c) For x  12  m,
Jp 0
_______________________________________
5.33
For electrons:
dn
d
 eD n
10 15 e  x / Ln
J n  eD n
dx
dx


 
 eD n 10 15 e  x / Ln
Ln
At x  0 ,
 1.6  10 19 25 10 15
Jn 
 2 A/cm 2
2  10 3
For holes:
dp
d
x / Lp
J p  eD p
 eD p
5  10 15 e
dx
dx


  




 eD p 5  10 15 e

 x / Lp
Lp
For x  0 ,
 1.6  10 19 10  5  10 15
Jp 
 16 A/cm 2
5  10  4
J Total  J n x  0   J p x  0 

 

 2   16   18 A/cm 2
_______________________________________
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 5
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
5.34

dp
d
x / Lp
J p  eD p
 eD p
5  10 15 e
dx
dx

(a) (i) J p 


eD p 5  1015 e

5.36
x / Lp
15

 

 

 



dn
dx

x / L

 5.76e
 10 A/cm
(c) We have J p    e p p o 
5.76e

x / L
2


 10  1.6  10

19

42010 
16
So   8.57e  x / L  14.88 V/cm
_______________________________________
5.37
(a) J  e n nx   eD n
dnx 
dx
We have  n  8000 cm 2 /V-s, so that
  
 1.6  10 19 25 10 16
 1 
x

 exp

4
 18  10 
 18 
Then
   x 
x
 40  1.536 exp

   22.22 exp
 18 
  18 
We find
22.22 exp  x   40
 18 

1.536 exp  x 
 18 
or
x
  14.5  26.0  exp

 18 
_______________________________________



 1.6  10 19 207 


  x 
 40  1.6  10 19 96010 16 exp
 
 18 



(b) J p  J Total  J n  10   5.76e  x / L

or


D n  0.0259 8000  207 cm 2 /s
Then
100  1.6  10 19 8000 12 nx 
5.35
J n  e n n  eD n
 




1.6 10 105 10 
50  10  4
 1.6 A/cm 2
1.6  10 19 48 5  1015
(ii) J p 
22.5  10  4
 17.07 A/cm 2
1.6  10 19 10  5  10 15 e 1
(b) (i) J p 
50  10  4
 0.589 A/cm 2
1.6  10 19 48 5  10 15 e 1
(ii) J p 
22.5  10  4
 6.28 A/cm 2
_______________________________________

 eD n 2  1015 e  x / L
L
 1.6  10 19 27  2  1015 e  x / L

15  10  4
x / L
 5.76e

Lp
19

dn
d
 eD n
2  10 15 e  x / L
dx
dx
(a) J n  eD n
dnx 
dx
which yields



100  1.536  10 14 n x   3.312  10 17
 dndxx 
Solution is of the form
x
nx   A  B exp

 d 
so that
dnx   B
x

exp

dx
d
 d 
Substituting into the differential equation, we
have

  x 
100  1.536  10 14  A  B exp

 d 




3.312 10  B exp  x 
17
d
This equation is valid for all x, so
100  1.536  10 14 A
or
A  6.51 1015


 d 
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 5
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Also
x
1.536  10 14 B exp

 d 

3.312 10  B exp  x   0
17
 d 
d
which yields
d  2.156  10 3 cm
At x  0 , e n n0   50
so that
50  1.6  10 19 8000 12  A  B 
which yields
B  3.255  1015
Then
x
3
nx   6.51 1015  3.255  1015 exp
 cm
 d 
(b)
At x  0 , n0   6.51 1015  3.255  10 15
Or
n0   3.26  10 15 cm 3
At x  50  m,


  50 
n50   6.51 10 15  3.255  10 15 exp

 21.56 
or
n50   6.19  10 15 cm 3
(c)
At x  50  m,
J drf  e n n50




 1.6  10 19 8000  6.19  10 15 12 
or

 
J diff  x  50   4.92 A/cm 2
_______________________________________
5.38
 E  E Fi 
n  ni exp F


 kT
(a) E F  E Fi  ax  b , b  0.4


0.15  a 10 3  0.4
which yields
a  2.5  10 2


 0.4  2.5  10 2 x 

 exp

0.0259


or
 0.4  2.5  10 2 x 

J n  5.79  10  4 exp

0.0259


(i) At x  0 , J n  2.95  10 3 A/cm 2
(ii) At x  5  m, J n  23.7 A/cm 2
_______________________________________
5.39
(a) J n  e n n  eD n


dn
dx
 
x

 80  1.6  10 19 1000  10 16 1  
L


J drf x  50  95.08 A/cm 2
Then
J diff x  50   100  95.08
or
Then
E F  E Fi  0.4  2.5  10 2 x
so
 0.4  2.5  10 2 x 

n  ni exp

kT


dn
(b) J n  eDn
dx
  2.5  10 2 
 0.4  2.5  10 2 x 
 exp

 eDn n i 



kT
kT




Assume T  300 K, so kT  0.0259 eV and
n i  1.5  1010 cm 3
Then
 1.6  10 19 25 1.5  1010 2.5  10 2
Jn 
0.0259
  1016
 1.6  10 19 25.9
 L


where L  10  10 4  10 3 cm
We find
 x 
 80  1.6  1.6  3   41.44
 10 
or
x 
80  1.6  1  41.44
L 
Solving for the electric field, we find
24.1

V/cm
x 
  1
L 




lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 5
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(b) For J n  20 A/cm 2
5.42
x 
20  1.6  1  41.44
L 
Then
13.3
V/cm

x

1  
 L
_______________________________________

or

 0.0259    1 
   N do e  x / L
N do e  x / L  L 

0.0259
0.0259

L
10  10  4
 X  25.9 V/cm


 25.9  10  10 4  0.0259 V
or   25.9 mV
_______________________________________
5.41
From Example 5.6
0.0259 1019  0.0259 10 3
x 
1016  10 19 x
1  10 3 x
 



  dx
x





or
0
10 4
 

 0.0259  10
3
0
 
dx
1  10 3 x


 1 
 0.0259  10 3  3  ln 1  10 3 x
 10 
 0.0259 ln 1  0.1  ln 1
or

   x 
 1.6  10 19 6000  5  10 16 exp
 
  L 
10 4
V 


Now
J drf  e n n
 


dN d x 
dn
 eD n
dx
dx
eD n
x

 N exp

 L  do  L 
We have
 kT 
Dn   n 
  6000 0.0259 
 e 
or
D n  155.4 cm 2 /s
Then
 1.6  10 19 155.4  5  1016
x
J diff 
exp

0.1 10  4
 L 
or
x
2
J diff  1.243  10 5 exp
 A/cm
 L 
(b)
0  J drf  J diff

L
(b)     X dx  25.9 L  0 

0.0259
 500 V/cm
L
Which yields L  5.18  10 5 cm
_______________________________________
So  X 
J diff  eD n

0
For N d x   N do e  x / L
5.43
(a) We have
5.40
dN d x 
1
 kT 
(a)  X  


dx
 e  N d x 
 0.0259  d
N do e  x / L


N do e  x / L dx
dN d x 
1
 kT 

 x  

dx
 e  N d x 

10 4
0
V  2.73 mV
_______________________________________
   x 
J drf  48exp
 
  L 
We have
J drf   J diff
so


48exp  x   1.243 10 5 exp  x 
 L 
  L 
which yields
  2.59  10 3 V/cm
_______________________________________
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 5
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
5.44
Plot
_______________________________________
5.48
(a) V H  0  n-type
(b) n 
5.45
(a) (i) D n  0.0259 1150   29.8 cm 2 /s
8
 308.9 cm 2 /V-s
0.0259
35
(ii)  p 
 1351 cm 2 /V-s
0.0259
_______________________________________
(b) (i)  p 
5.46
L  10 1 cm, W  10 2 cm, d  10 3 cm


 I X BZ
 1.2  10 3 5  10 2
VH 

ned
2  10 22 1.6  10 19 10 5




 1.875  10 3 V
or V H  1.875 mV
(b)
V
 1.875  10 3
H  H 
 0.1875 V/cm
W
10  2
_______________________________________







(c) n 

6
1.6 10
I x Bz
edV H
5





 0.5  10 3 6.5  10 2
1.6  10 19 5  10 5  0.825  10 3






0.5 10 0.5 10 
4.924 10 1.255 10 5 10 
3
1.6 10
19
21
2
4
5
or
 n  0.1015 m 2 /V-s  1015 cm 2 /V-s
_______________________________________
250 10 10 
5 10 0.12 10 5 10 
21
4
21
n  4.924  10 21 m 3  4.924  10 15 cm 3
(d)
IxL
n 
enV xWd
V H  0.3125 mV
(b)
V
 0.3125  10 3
H  H 
W
2  10  2
or
 H  1.56  10 2 V/cm
(c)
IxL
n 
enV xWd
19
19
3
or
or

0.5 10 10 
6.0110 1510 10 
3
1.6 10

3
5.49
(a) V H   H W   16.5  10 3 5  10 2
or
V H  0.825 mV
(b) V H  negative  n-type
(a) V H 


 0.03466 m 2 /V-s
or  n  346.6 cm 2 /V-s
_______________________________________
5.47
I x Bz
ned
 250  10 6 5  10 2

5  10 21 1.6  10 19 5  10 5

 6.01 10 m
or n  6.01 1015 cm 3
IX L
(c)  n 
enV X Wd
(ii) D n  0.0259 6200   160.6 cm /s


21
2
(a)

 I X BZ
 0.50  10 3 0.10 

edV H
1.6  10 19 10 5  5.2  10 3
3
4
5
or
 n  0.3125 m 2 /V-s  3125 cm 2 /V-s
_______________________________________

lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 5
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
5.50
(a) V H  negative  n-type
I x Bz
edV H
(b) n 




 2.5  10 3 2.5  10 2
1.6  10 19 0.01 10  2  4.5  10 3




or
n  8.68  10 20 m 3  8.68  10 14 cm 3
IxL
(c)  n 
enV x Wd





2.5  10 3 0.5  10 2


19
20
1
.
6
10
8
.
68
10
2
.
2








1

2
2 
0
.
05
10
0
.
01
10










or
 n  0.8182 m 2 /V-s  8182 cm 2 /V-s
(d)  
1

 e n n



 1.6  10  19 8182  8.68  10 14

or
  0.88 (  -cm)
_______________________________________
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 6
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Chapter 6
6.1
n o  N d  5 10 15 cm 3


2
n i2
1.5  10 10
 4.5  10 4 cm 3

Nd
5  10 15
(a) Minority carrier hole lifetime is a
constant.
 pt   p 0  2 10 7 s
po 
R po 
po

 p0
(b) R po 
4.5  10 4
 2.25  10 11 cm 3 s 1
2  10  7
p o  p
 p0

4.5  10 4  1014
2  10  7
3
(b) Generation rate = recombination rate
Then
2.25  10 4
G
 1.125  10 9 cm 3 s 1
20  10  6
(c)
R  G  1.125  10 9 cm 3 s 1
_______________________________________
6.4
(a) E  h 
p o  N a  2  10 16 cm 3

n i2
1.8  10 6

po
2  1016
(a) R  
2
 1.62  10  4 cm 3
po

 pt
no
 nt


n  p  g  3.17  1019 10  10 6 
or
n  p  3.17  10 14 cm 3
 

p
2  10 16
 5  10  7
 o  n 0 
4
no
1.62  10

6300  10

_______________________________________
6.5
 6.17  10 13 s
_______________________________________
We have
p
p
   F p  g p 
p
t
6.3
(a) Recombination rates are equal
no
p
 o
and
J p  e p p  eD p p
 nO
 pO
n o  N d  10 16 cm 3
po 
Then
1016

n i2
1.5  1010

no
10 16
2.25  10 4
 nO
20  10  6
which yields
 nO  8.89  10 6 s


8
10
(b)
no
 n0
34
 3.17  1019 e-h pairs/cm 3 -s
n 5  1014

 10 21 cm 3 s 1
 n 0 5  10  7
(b) R p 
 pt

6.625 10 3 10 
E  3.15  10 19 J; energy of one photon
Now
1 W = 1 J/s  3.17  1018 photons/s
Volume = (1)(0.1) = 0.1 cm 3
Then
3.17  1018
g
0. 1
1
6.2
no 


or
 5 10 cm s
_______________________________________
20
hc
2
 2.25  10 4 cm 3
The hole particle current density is
Jp
F p 
  p  D p p
 e  p
Now
  F p   p    p   D p   p
We can write
   p     p  p  
and
  p   2 p
so
  F p   p   p  p     D p  2 p
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 6
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
By charge neutrality,
n  p  n  n   p 
and
 n   p 

 2 n    2 p  and
t
t
Also
p
n

R
gn  g p  g ,
Then
p
   p   p  p   
t
 Dp2 p  g p 
p
p
We can then write
D p  2 p   p   p  p   
gp 
p
p

p
t
_______________________________________
p
n
Then we have
(1) D p  2 n    p   n   p  
gR 
6.6
From Equation (6.18),
p
p
   F p  g p 
t
p
and
(2) D n  2 n    n   n   n  
 n 
t
Multiply Equation (1) by  n n and Equation
(2) by  p p , and add the two equations.
gR 
p
0
t
Then
0    F p  g p  R p
For a one-dimensional case,
dF p
 g p  R p  10 20  2  1019
dx
or
dF p
 8  1019 cm 3 s 1
dx
_______________________________________
6.7
From Equation (6.18),
dF p
0
 0  2  1019
dx
or
dF p
 2  1019 cm 3 s 1
dx
_______________________________________
6.8
We have the continuity equations
(1) D p  2 p    p   p   p  
gp 
p
p
 p 

t
and
(2) D n  2 n    n   n   n  
 gn 
n
n
 n 
t

 n 
t
We find
 n nD p   p pD n  2 n 


  n  p  p  n   n 


  n n   p p g  R 

 nn   p p


 tn
Divide by  n n   p p , then
  n nD p   p pD n  2
 n 

 nn   p p 


  n  p  p  n
 
   n 
  n n   p p 
 g  R  
Define
D 
and
 
 n nD p   p pD n
nn   p p

 n 
t
D n D p n  p 
Dn n  D p p
 n  p  p  n
nn   p p
Then we have
 n 
t
Q.E.D.
_______________________________________
D  2 n      n    g  R  
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 6
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
6.9
 
p-type material;
minority carriers are electrons
(a)     n
 1340 cm 2 /V-s
(b) For holes,  pt   p 0  2  10 6 s
 kT 
(b) D   D n  
   n  0.0259 1300
 e 
For electrons,
p
n

 33.67 cm 2 /s
  n 0  10 7 s
(c)  nt
 nt
p o  N a  7  10 15 cm 3

n2
1.5  1010
no  i 
Na
7  1015

3.214  10
7  10

7
 pt
10
  nt  9.12  10 6 s
_______________________________________
6.11
_______________________________________
6.10
For Ge: n i  2.4  1013 cm 3

  n i2

 4  1013
4  1013

 
2
 2
 5.124  10 13 cm

2

  2.4  10 13




2
3

2
 p  1900 cm /V-s, D p  49.2 cm /s
2
2
For very, very low injection,
D n D p n  p 
D 
Dn n  D p p
10149.25.124 1013  1.124 1013 

1015.124 1013   49.21.124 1013 
 54.2 cm 2 /s
  e n n o  e p p o  e n   p p 
so
n i2
2.4  10 13
 1.124  10 13 cm 3

n o 5.124  10 13
(a) We have:
 n  3900 cm 2 /V-s, D n  101 cm 2 /s
po 
With excess carriers
n  n o  n and p  p o  p
For an n-type semiconductor, we can write
n  p  p
Then
  e n n o  p   e p  p o  p 
or
2
Nd
N
  d
2
 2
  e n n  e p p
15
so  pt  2.18  10 4 s
no 
1.124  10 13
2  10  6

 nt
 pt
4
 p0
5.124  1013
2
 3.21 10 4 cm 3
no
p
 o
 nt
nn   p p
390019001.124 1013  5.124 1013 
39005.124 1013   19001.124 1013 

From Figure 5.3,  n  1300 cm 2 /V-s
 n  p  p  n


  e  n   p p 
In steady-state, p  g  pO
So that
  e  n   p g  pO



_______________________________________
6.12
(a) p o  N a  10 16 cm 3


2
n i2
1.5  10 10

 2.25  10 4 cm 3
po
1016
  e n n o  n   e p  p o  p 
no 


 e p p o  e  n   p n

Now n  p  g  n 0 1  e t /  n 0

 8  10
 4  10
and
14

5 10 1  e
1  e  cm
20
7
t /  n 0
t /  n 0
3

lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 6
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
38010 
 1.6  10 900  380 

 4  10 1  e
 (  -cm)
  0.608  0.08191  e
Then
  1.6  10
19
16
where p  g  p 0 e

t /  n 0
t /  n 0
1

 

 1.6  10 1300  400 
 4  10 e
19

  1.248  0.109e



 cm
 2  10 1  e
 4  10 21 5  10 8 1  e
t /  p 0
t /  p 0
14
6
At t  10 s,

p 10 6   2  1014 1  e 10
6

8


  n 0  2.5  10 7 s

(b) n  p  g  n 0 1  e  t /  n 0




 2  1014 1  e

 6.0  0.250 1  e
t /  p 0
For t  10 6 s,

  6.0  0.250e

 5  10 1  e
14
R 

 1.6  10 19 7500  310 

 2.18  10 4 e
t /  p 0
A
 1.4  1016 cm 3
(a) n  p  g  n 0
cm 3
For 0  t  10 s,
  1.6  10 19 7500  5  10 15

0.05
3
p o  N a  N d  2  10 16  6  10 15
6

5
6.15
(b) n o  5  10 15 cm 3
  e n n o  e  n   p p

10 10
5  10 14  2  10 21 n 0
 t 10 6 /  p 0

t /  p 0
or I  2.496  0.218e
mA
_______________________________________

3
/ 510
t /  p 0
t /  p 0
 2 10 14 cm 3
Then for t  10 6 s,
p  2  1014 e
1.248  0.109e
 2.496  10

t /  p 0
14
I
n  p  g  p 0 1  e
t /  p 0
t / 
(ii)     0.690 (  -cm) 1
_______________________________________
t /  p 0

 4  10 14 e p 0 cm 3
  1.6  10 19 1300  8  1015  2  10 15
(b) (i)  0   0.608 (  -cm) 1
6.13
(a) For 0  t  10 6 s,

 8  10 20 5  10 7 e
19
14
t /  p 0
t /  p 0

 t 10 6 /  p 0


5  10
n

1  e t /  n 0
 n 0 2.5  10  7
14



 2  10 21 1  e  t /  no cm 3 s 1

 (  -cm)

t /  n0
(c)
1
1
(  -cm)
_______________________________________



1
(i)   5  10 14  5  10 14 1  e t /  n 0
4

t   n 0 ln 1.3333  7.19  10 8 s




1
(ii)   5  10 14  5  1014 1  e t /  n 0
2
t   n 0 ln 2   1.73  10 7 s
6.14
L
V
; R
A
R
A
I
V
L
For N I  N d  N a  8  1015  2  1015
I
 1016 cm 3
Then,  n  1300 cm 2 /V-s



3
(iii)   5  10 14  5  1014 1  e t /  n 0
4
t   n 0 ln 4   3.47  10 7 s




(iv) 0.95 5  10 14  5  10 14 1  e  t /  n 0

t   n 0 ln 20   7.49  10 s
_______________________________________
 p  400 cm 2 /V-s
  e n n o  e n   p p
7
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 6
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
6.16
At t  2 10 6 s,
15
15
6
7
n o  N d  N a  8  10  2  10
n  5  10 14 e  210  / 510 
1
 6  10 15 cm 3
po 

(a) R o 
po
 p0

6 2
n
1.8  10

no
6  10 15
2
i
 5.4  10  4 cm 3
so  p 0  1.35  10 s

 p0
 4.908  10 14 1  e t /  n 0
8
(b) (i) n0   5  10 cm
14


(b) p  g  p 0  2  10 21 1.35  10 8


_______________________________________
6.17
(a) (i)For 0  t  5  10 7 s
t /  p 0



 5  10 20 5  10 7 1  e

 2.5  1014 1  e
t /  p 0
At t  5 10 7 s,
p  2.5  10 14 1  e 1 / 1

 1.58 10 cm
For t  5  10 7 s
14
pt   1.58  1014 e

t /  p 0
 cm




 t  510 7 /  pO
6
t /  p 0


 2.454  10 cm
For t  2 10 6 s,
14


3
p t   2.454  10 14 e


 t  210 6 /  pO








 1.6 10 19 31.08 2  1014  x / Ln
e
5.575  10 3

_______________________________________
cm 3
(ii) p 2  10 6  2.454  1014 cm 3
_______________________________________
6.18
(a) For 0  t  2  10 6 s
nt   g  n 0 e t /  n 0


J n  0.1784e  x / Ln A/cm 2
Holes diffuse at same rate as minority carrier
electrons, so
J p  0.1784e  x / Ln A/cm 2
3
6
7
p  2.5  1014 1  e  210  / 510 
1/ 2


cm 3
 cm

d n 
d
 eDn
2  1014 e  x / Ln
dx
dx
eDn

2  1014 e  x / Ln
Ln
(b) J n  eDn
(b) (i) For 0  t  2  10 s
At t  2  10 6 s,

L n  D n n 0  31.08 10 6
 5.575  10 3 cm
(a) n x   p x   2  10 14 e  x / Ln cm 3




 31.08 cm 2 /s
3
3
pt   2.5  10 1  e
3
6.19
p-type; minority carriers - electrons
 kT 
Dn  
  n  0.02591200
 e 
(ii) p 5  10 7  1.58  1014 cm 3
14
 9.16  1012
 9.16  1012 cm 3
(iii) n   5 1014 cm 3
_______________________________________
 2.7  10 cm
(c)    p 0  1.35  10 8 s
pt   g  p 0 1  e


(ii) n 2 10 6  9.16  1012 cm 3
3
13



5.4  10 4
 4  10 4 

 9.16  1012 cm 3
For t  2 10 6 s
n  5  10 14  9.16  10 12 1  e t /  n 0
6.20
(a) p-type; p pO  1014 cm 3
and

 10 21 5  10 7 e t /  n 0
 5  1014 e t /  n 0 cm 3


2
ni2
1.5  1010

 2.25  10 6 cm 3
p pO
10 14
(b) Excess minority carrier concentration
n  n p  n pO
n pO 
At x  0 , n p  0 so that
n0  0  n pO  2.25  10 6 cm 3
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 6
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(c) For the one-dimensional case,
d 2 n  n
Dn

0
 nO
dx 2
or
d 2 n  n
 2  0 where L2n  D n nO
dx 2
Ln
The general solution is of the form
x
x

  B exp
n  A exp

L 
 n 
 Ln 
For x   , n remains finite, so B  0 .
Then the solution is
x

n  n pO exp

 Ln 
_______________________________________
6.21
6.22
n-type, so we have
d 2 p 
d p  p
Dp
  po

0
dx
 pO
dx
Assume the solution is of the form
p  A expsx 
Then
d p 
d 2 p 
 As 2 expsx 
 As expsx  ,
dx
dx 2
Substituting into the differential equation
D p As 2 expsx    p  o As expsx 

 
where Ln  D n n 0  25 10
6

s2 
3
 5  10 cm
d n 
d
 eDn
5 1014 e  x / Ln
J n  eDn
dx
dx
eD
  n 5  1014 e  x / Ln
Ln





1.6 10 255 10  e
5 10 
19
14
3
 x / Ln
J n  0.4e  x / Ln A/cm 2
(a) For x  0 ,
Define
J n 0   0.4 A/cm 2

J p 0   0.4 A/cm 2
nL n   5  10 e
J n Ln   0.4e
1
 pO
0
 po
Dp
s
1
0
L2p
 p Lpo
2D p
Then
(b) For x  L n  5  10 3 cm,
1
1
The solution for s is
2



 p
1 p
4
s  
o  
 o   2 

 Dp
2 Dp
Lp




which can be rewritten as
2


  p Lpo 
1   p Lpo


 1 
s

 2D p 
L p  2D p




n0   5  10 14 cm 3
14
0
Dividing by D p , we have
1/ 2

 pO
or
Dp s 2   pos 
nx   5  10 14 e  x / Ln cm 3
A expsx 
 1.84  10 cm
14
3
 0.147 A/cm 2
J p L n   0.4e 1  0.147 A/cm 2
(c) For x  15  10 3 cm  3Ln
n3L n   5  10 14 e 3  2.49  1013 cm 3
J n 3L n   0.4e 3  0.020 A/cm 2
J p 3Ln   0.4e 3  0.020 A/cm 2
_______________________________________
s
1
Lp
  1   2 


In order that p  0 as x   , use the
minus sign for x  0 and the plus sign for
x  0 . Then the solution is
p  A exps  x  for x  0
p  A exps  x  for x  0
where
1 
  1   2 
s 

L p 
_______________________________________
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 6
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
6.23
Plot
_______________________________________
6.24
(a) From Equation (6.55)
d 2 n 
d n  n
 no

0
Dn
2
 nO
dx
dx
or
d n  n
d 2 n   n

o
 2 0
Dn
dx
Ln
dx 2
We have that
D n  kT 

 so we can define
n  e 
n
o
1
o 

Dn
kT e L 
Then we can write
d 2 n  1 d n  n
 
 2 0
L  dx
Ln
dx 2
The solution is of the form
n  n0 exp x  where   0
Then
d n 
d 2 n 
  n  and
  2 n 
dx
dx 2
Substituting into the differential equation, we
find
1
n
 2 n     n   2  0
L
Ln
or
1

2   2 0
L  Ln
which yields
2


 L 
1  Ln

  n   1 

Ln  2 L 
 2L  


We may note that if  o  0 , then L   
1
and  
Ln
(b)
 kT 
Ln  D n nO where D n   n 

 e 
so
D n  12000.0259   31.1 cm 2 /s
and
Ln 
31.15 10 7   39.4 10 4 cm
or
Ln  39.4  m
For  o  12 V/cm, then
L 
kT e  0.0259  21.6 10  4 cm
o
12
and
  5.75  10 2 cm 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.
_______________________________________
6.25
p-type so the minority carriers are electrons
and
n n 

D n  2 n    n   n   g  
 nO
t
Uniform illumination means that
n    2 n   0 . For  nO   , we are
left with
d n 
 g  which gives n  g t  C1
dt
For t  0 , n  0  C1  0
Then
n  G o t for 0  t  T
For t  T , g   0 so that
d n 
0
dt
And
n  G o T (no recombination)
_______________________________________
6.26
n-type, so minority carriers are holes and
p  p 
D p  2 p    p   p   g  

 pO
t
We have  pO   ,   0 , and
p 
 0 (steady-state). Then we have
t
d 2 p 
d 2 p 
g
Dp

 g   0 or
2
Dp
dx
dx 2
For  L  x   L , g   G o = constant. Then
G
d p 
  o x  C1
dx
Dp
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 6
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
and
We find
D p 10.42

 0.02668 V
 p 390.6
G
p   o x 2  C1 x  C 2
2D p
For L  x  3L , g   0 so we have
d 2 p 
d p 
 C 3 and
 0 so that
2
dx
dx
p  C 3 x  C 4
For 3L  x   L , g   0 so that
d 2 p 
2
d p 
 C 5 and
dx
 0 so that
dx
p  C 5 x  C 6
The boundary conditions are:
(1) p  0 at x  3L
(2) p  0 at x  3L
(3) p continuous at x  L
(4) p continuous at x   L
d p 
continuous at x  L
dx
d p 
(6)
continuous at x   L
dx
Applying the boundary conditions, we find
G
p  o 5L2  x 2 for  L  x   L
2D p
(5)

G o L
3L  x  for 3L  x   L
Dp
_______________________________________
6.27
8
V

 20 V/cm
L 0.4
d
0.25

p 
 0 t 0 20  32  10  6
0 

 390.6 cm /V-s
 p  0 2 t 2
Dp 
16t 0

2

6.28
(a)
  x2
exp
 4 Dt
is the solution to the differential equation
  2 f  f
D 2  
 x  t
Assume that f  x, t   4 Dt 

2 f
x
2
 4 Dt 

16 32  10
D p  10.42 cm /s
2





2
1 / 2
  x2
  2 x 
 exp

 4 Dt 
 4 Dt




  x2
 2 

 exp
 4 Dt 
 4 Dt
Also
2
f
1 / 2   x
 4 Dt  
t
 4D
  x2
  1 
 2  exp
 4 Dt
 t 










  x2 
  1  3 / 2

exp
 t

 2 
 4 Dt 
2 f
and
Substituting the expressions for
x 2
f
into the differential equation, we find
t
0 = 0.
Q.E.D.
(b)
Consider
 4 D 
  x2
exp
 4 Dt


6




and

2
2


390.6 20  9.35  10 6 

1 / 2
To prove: we can write
  x2
f
1 / 2   2 x 
 4 Dt  
 exp
x
 4 Dt 
 4 Dt

G L
p  o 3L  x  for L  x  3L
Dp
p 
This value is very close to 0.0259 for
T  300 K.
_______________________________________
1 / 2

dx


Let u  x 2 , then du  2 x  dx or
du
du

dx 
2x 2 u
Let a 
1
4 Dt
Now
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 6
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________

  x2
exp
 4 Dt



2
2
0


a
1
u

  x2

dx  2 exp
 4 Dt



0

exp au du 


0
1
u
6.31
(a) p-type

dx


exp au du
 4Dt
4D t
  x2 
dx 
1
exp

4D t
4D t
 4 Dt 

_______________________________________

 5  1015
 0.0259 ln
10
 1.5  10
1
6.29
Plot
_______________________________________
E Fi  E F  0.3294 eV
(b)
n  p  5  10 14 cm 3
and
no 
n
(a) E F  E Fi  kT ln o
 ni

 10 cm

2
 4.5  10 4 cm 3
 4.5  10 4  5  1014
 0.0259 ln
1.5  1010





or
E Fn  E Fi  0.2697 eV
and
 p  p 

E Fi  E Fp  kT ln o

 ni 

3
 5  1015  5  1014
 0.0259 ln
1.5  1010

 n  n 

E Fn  E Fi  kT ln o

 ni 
E Fi  E Fp

 n  n 

E Fn  E Fi  kT ln o

 ni 




 4  1016 

 0.0259  ln
10 
 1.5  10 
 0.383225 eV
(b) n  p  g  p 0  2 10 21 5  10 7

n i2
1.5  1010

po
5 1015
Then
6.30
15




or
Then

p 
E Fi  E F  kT ln o 
 ni 
 4 1016  1015
 0.0259 ln
10
 1.5 10
 0.383865 eV
 p  p 

 kT ln o

 ni 




 1015 

 0.0259 ln
10 
 1.5 10 
 0.28768 eV
(c) E Fn  E F  0.383865  0.383225
 0.000640 eV
or
 0.640 meV
_______________________________________




or
E Fi  E Fp  0.3318 eV
_______________________________________
6.32
(a) For n-type,
E Fn  E F  E Fn  E Fi   E F  E Fi 
n
 n  n 
  kT ln o
 kT ln o

n
 i
 ni 
 n  n 

 kT ln o

 no 
 5 1015  n 

So 0.00102  0.0259 ln
15


 5 10
 0.00102 
5  1015  n  5 1015 exp

 0.0259 
Which yields n  2  1014 cm 3




lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 6
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
 n  n 

(b) E Fn  E Fi  kT ln o

 ni 
 5  1015  2  1014
 0.0259 ln
1.5 1010

 0.33038 eV
 p 
(c) E Fi  E Fp  kT ln 
 ni 
6.34
n n

(a) (i) E Fn  E Fi  kT ln o

 ni 




  
 2  1014 

 0.0259 ln
10 
 1.5  10 
 0.2460 eV
_______________________________________
6.33
 n 
(a) E Fn  E Fi  kT ln 
 ni 
 E  E Fi 
or n  n i exp  Fn

kT




 0.270 
 1.5  1010 exp 

 0.0259 
 5.05  1014 cm 3
 p  p 

(b) E Fi  E Fp  kT ln o

 ni 
 6 1015  5.05  1014 

 0.0259 ln

1.5  1010


 0.33618 eV
(c) (i) E F  E Fp  E Fi  E Fp  E Fi  E F 




 0.02  1016 

 0.0259  ln
6 
 1.8  10 
 0.47982 eV
 1.11016 

(b) (i) E Fn  E Fi  0.0259 ln
6 
 1.8  10 
 0.58361 eV
 0.1 1016 

(ii) E Fi  E Fp  0.0259  ln
6 
 1.8  10 
 0.52151 eV
_______________________________________
6.35
Quasi-Fermi level for minority carrier
electrons:
 n  n 

E Fn  E Fi  kT ln o

 ni 


2
n2
1.8 10 6
 3.24  10  4 cm 3
no  i 
16
po
10
We have
 x 
n  1014  
 50 
Then
 3.24  10 4  10 14 x 50 
E Fn  E Fi  kT ln 

1.8  10 6


We find
 
p 
 p  p 
  kT ln o 
 kT ln o

n 
n
i

 i 

 p  p 

 kT ln o

 po 
(ii) E F  E Fp
 6  1015  5.05  1014
 0.0259 ln
6  1015

 1.02 1016
 0.0259 ln
6
 1.8  10
 0.58166 eV
 p 
(ii) E Fi  E Fp  kT ln 
 ni 





x (  m)
 2.093 10 3 eV
or
 2.093 meV
_______________________________________
0
1
2
10
20
50
( E Fn  E Fi ) (eV)
-0.581
+0.361
+0.379
+0.420
+0.438
+0.462

lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 6
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Quasi-Fermi level for holes: we have
 p  p 

E Fi  E Fp  kT ln o

 ni 
6.38
 p 
(a) E Fi  E Fp  kT ln 
 ni 
 p

 0.0259 ln

10
 1.5  10 
We have p o  1016 cm 3 and n  p .
We find
x (  m)
p  1011 cm 3 , E Fi  E Fp  0.04914 eV
( E Fi  E Fp ) (eV)
1012
1013
1014
1015
0
+0.58115
50
+0.58140
_______________________________________
6.36
(a) We can write
p 
E Fi  E F  kT ln o 
 ni 
and
 p  p 

E Fi  E Fp  kT ln o

 ni 
so that
E Fi  E Fp  E Fi  E F   E F  E Fp


p 
 p  p 
  kT ln o 
 kT ln o

n 
 i 
 ni 
0.10877
0.16841
0.22805
0.28768
 n  n 

(b) E Fn  E Fi  kT ln o

 ni 
 2  1016  n 

 0.0259  ln
10


 1.5 10
n  1011 cm 3 , E Fn  E Fi  0.365273 eV
1012
1013
1014
0.365274
0.365286
0.365402
0.366536
1015
_______________________________________
or
 p  p 
  0.01kT
E F  E Fp  kT ln o

 po 
Then
p o  p
 exp0.01  1.010
po
or
p
 0.010  low injection, so that
po
p  5  10 cm
(b)
12
E Fn  E Fi
3
 p 
 kT ln 
 ni 
 5  1012
 0.0259 ln
10
 1.5  10




or
E Fn  E Fi  0.1505 eV
_______________________________________
6.37
Plot
_______________________________________
6.39
(a)
R


C n C p N t np  ni2

C n n  n   C p  p  p 
np  n 
2
i
 pO n  n    nO  p  p 
Let n   p   n i . For n  p  0
R
 ni
 ni2

 pO ni   nO n i  pO   nO
(b) We had defined the net generation rate as
g  R  g o  g   R o  R  
where g o  Ro since these are the thermal
equilibrium generation and recombination
rates.
If g   0 , then g  R   R  and
R 
n i
 pO   nO
so that g  R  
ni
 pO   nO
Thus a negative recombination rate implies a
net positive generation rate.
_______________________________________
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 6
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
6.40
We have that
C n C p N t np  n i2
R
C n n  n   C p  p  p 



np  n 
2
i
 pO n  n i    nO  p  n i 
If n  n o  n and p  p o  n , then
n o  n  p o  n   ni2
 pO n o  n  n i    nO  p o  n  n i 
2
n o p o  nn o  p o   n   n i2

 pO n o  n  n i    nO  p o  n  ni 
2
If n  n i , we can neglect n  : also
R
n o p o  n i2
Then
R
nn o  p o 
 pO n o  n i    nO  p o  n i 
(a) For n-type; n o  p O , n o  n i
Then
R
1

 10  7 s 1
n  pO
(b) For intrinsic, n o  p o  n i
Then
2n i
R

n  pO 2n i    nO 2n i 
At x   , p  g  pO so that B  0 ,
Then
x

 Lp 


p  g  pO  A exp
We have
d p 
Dp
 sp 
dx x  0
x 0
We can write
d p 
A
and p   g  pO  A

dx x  0 L p
x 0
Then
Lp
The excess concentration is then

  x 
s

 exp
p  g  pO 1 
 L p 
Dp Lp  s



where


1010 7   10 3 cm
L p  D p pO 
Now
p  10 21 10 7
 


  x 
s

 1 
 exp
3
 L p 


 10 10  s

R
1
1

 7

n  pO   nO 10  5  10  7
R

Solving for A , we find
 sg  pO
A
Dp
s
Lp
or
n

 s g  pO  A

or
 1.67  10  6 s 1

p  1014 1 
(c) For p-type; p o  n o , p o  n i
Then
R
1
1


 2  10  6 s 1
n  nO 5  10  7
_______________________________________
6.41
(a) From Equation (6.56)
p
d 2 p 
Dp
 g
0
2
 pO
dx
Solution is of the form


x
  B exp  x 
p  g  pO  A exp
 Lp 
 Lp 




  x 

 exp
 L p 
10  s


s

(i) For s  0 ,
p  1014 cm 3
(ii) For s  2000 cm/s,

  x 

p  1014 1  0.167 exp
 L p 



(iii) For s   ,

  x 

p  1014 1  exp
 L p 



4
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 6
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(b)
(i) For s  0 ,
p 0   1014 cm 3
(ii) For s  2000 cm/s,
p 0   0.833  1014 cm 3
(iii) For s   ,
p0  0
_______________________________________
6.42
Ln  D n nO 
255 10 7 
 35.4 10 4 cm
(a) At x  0 ,
g  nO  2  10 21 5  10 7  10 15 cm 3
or
n0   g  nO  10 15 cm 3
For x  0
d 2 n  n
d 2 n  n
0



 2 0
Dn
 nO
dx 2
Ln
dx 2
The solution is of the form
x
x

  B exp
n  A exp
L 

L
 n 
 n 
At x  0 ,
n  n0  A  B
At x  W ,
 W 
 W 
  B exp

n  0  A exp

 L 
 Ln 
 n 
Solving these two equations, we find
 n0  exp 2W Ln 
A
1  exp 2W L n 
and
n0
B
1  exp 2W Ln 
Substituting into the general solution, we find
n0
n 
  W 
  W 

  exp
exp
 L 

  Ln 
 n 



   W  x  
  W  x   
 exp 

  exp 
  L n
 Ln
 

which can be written as
W  x 

 Ln 
n0 sinh 
n 
W 
sinh  
 Ln 
where
n0   10 15 cm 3 and Ln  35.4  m
(b) If  nO   , we have
d 2 n 
0
dx 2
so the solution is of the form
n  Cx  D
Applying the boundary conditions, we find
x

n  n01  
 W
_______________________________________
6.43
For  pO   , we have
d 2 p 
0
dx 2
So the solution is of the form
p  Ax  B
At x  W
d p 
 Dp
 s p 
dx x W
x W
or
 D p A  s  AW  B 
which yields
A
B
D p  sW
s
At x  0 , the flux of excess holes is
d p 
1019   D p
 D p A
dx x  0
so that
 1019
A
 1018 cm 4
10
and
1018
10  sW   1018  10  W 
B
s

 s
The solution is now
10 

p  1018 W  x  
s 



lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 6
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(a) For s   ,
p  10
Then
18
20 10
J p  eD p

4
d p 
dx

 x cm
 
3
  1.6  10 19 10   1018

or
J p  1.6 A/cm 2
(b) For s  2 10 3 cm/s,
p  1018 70  10 4  x cm 3
Also
J p  1.6 A/cm 2


_______________________________________
6.44
For W  x  0
d 2 n 
Dn
 G o  0
dx 2
so that
G
d n 
  o x  C1
dx
Dn
and
G
n   o x 2  C1 x  C 2
2 Dn
For 0  x  W ,
d 2 n 
0
dx 2
so that
n  C 3 x  C 4
The boundary conditions are
(1) s  0 at x  W so that
d n 
0
dx x  W
(2) s   at x  W so that
nW   0
(3) n continuous at x  0
d n 
continuous at x  0
(4)
dx
Applying the boundary conditions, we find
G W 2
G W
and C 2  C 4   o
C1  C 3   o
Dn
Dn
Then for W  x  0
G
n  o  x 2  2Wx  2W 2
2 Dn
and for 0  x  W
G W
n  o W  x 
Dn
_______________________________________


6.45
Plot
_______________________________________
6.48
(a) GaAs:
V
2
 10 6 
R 
I 2  10  6
L
and   e  n   p p
R
 A


p  g  p 0  10 21 5  10 8   5  1013 cm 3
For N d  10 16 cm 3 , from Figure 5.3,
 n  7000 cm 2 /V-s,  p  310 cm 2 /V-s



  1.6  10 19 7000  310  5  1013
 0.05848 (  -cm)
Let W  20  m


Then A  Wd  20  10 4 4  10 4
8
 80  10 cm
So R  10 6 

1

2
L
0.05848 80 10 8


2
Which yields L  4.68  10 cm
(b) Silicon:
R  10 6  , p  5  1013 cm 3
For N d  10 16 cm 3 , from Figure 5.3,
 n  1300 cm 2 /V-s,  p  410 cm 2 /V-s



  1.6  10 19 1300  410  5  10 13
 0.01368 (  -cm)
Let W  20  m


Then A  Wd  20  10 4 4  10 4
8
 80  10 cm
So R  10 6 

1

2
L
0.01368 80 10 8

2

Which yields L  1.09  10 cm
_______________________________________
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 7
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Chapter 7
7.1




N N
Vbi  Vt ln a 2 d
 ni
(a)



 2 1015 2  1015 
(i) Vbi  0.0259 ln 

2

 1.5  1010
 0.611 V
 2  1015 2  1016 
(ii) Vbi  0.0259 ln 

2
 1.5  1010

 0.671 V
 2  1015 2  1017 
(iii) Vbi  0.0259 ln 

2
 1.5  1010

 0.731 V
(b)
 2  1017 2  1015 
(i) Vbi  0.0259 ln 

2
 1.5  1010

 0.731 V
 2  1017 2  1016 
(ii) Vbi  0.0259 ln 

2
 1.5  1010

 0.790 V
 2  1017 2  1017 
(iii) Vbi  0.0259 ln 

2
 1.5  1010

 0.850 V
_______________________________________



























7.2
Si: n i  1.5  10 10 cm 3
GaAs: n i  1.8  10 6 cm 3

 and Vt  0.0259 V


(a) N d  1014 cm 3 , N a  10 17 cm 3 '
Then
Si: Vbi  0.635 V
Ge: Vbi  0.253 V
GaAs: Vbi  1.10 V
(c) N d  1017 cm 3 , N a  10 17 cm 3
Si: Vbi  0.814 V
Ge: Vbi  0.432 V
GaAs: Vbi  1.28 V
_______________________________________
7.3
(a) Silicon ( T  300 K)
 Na Nd
Vbi  0.0259 ln 
 1.5  1010


2



3
For N a  N d  10 cm ; Vbi  0.4561 V
14
 1015
;
 0.5754 V
16
 10
;
 0.6946 V
 1017
;
 0.8139 V
(b) GaAs ( T  300 K)
 Na Nd 
Vbi  0.0259 ln 

2
 1.8  10 6 
For N a  N d  10 14 cm 3 ; Vbi  0.9237 V


 1015
;
 1.043 V
16
 10
;
 1.162 V
 1017
;
 1.282 V
(c) Silicon (400 K), kT  0.034533
n i  2.38  10 12 cm 3
For N a  N d  10 14 cm 3 ; Vbi  0.2582 V
Ge: n i  2.4  1013 cm 3
N N
Vbi  Vt ln a 2 d
 ni
(b) N d  5 1016 cm 3 , N a  5 1016 cm 3
Si: Vbi  0.778 V
Ge: Vbi  0.396 V
GaAs: Vbi  1.25 V
;
 1015
 0.4172 V
16
 10
;
 0.5762 V
17
 10
;
 0.7353 V
GaAs(400 K), n i  3.29  10 9 cm 3
For N a  N d  10 14 cm 3 ; Vbi  0.7129 V
;
 1015
 0.8719 V
;
 1016
 1.031 V
;
 1017
 1.190 V
_______________________________________
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 7
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
or
7.4
(a) n-side
N
E F  E Fi  kT ln d
 ni
x p  0.0213  10 4 cm  0.0213  m




We have
 max 




 5  10
 0.0259 ln
10
 1.5  10
15

or
E F  E Fi  0.3294 eV
p-side
N 
E Fi  E F  kT ln a 
 ni 
14
7.5
(a) n-side




N
E F  E Fi  kT ln d
 ni




 2  1016
 0.0259 ln
10
 1.5  10




or
E F  E Fi  0.3653 eV
p-side
N 
E Fi  E F  kT ln a 
 ni 
 

 1017 5  10 15
 0.0259  ln 
2
 1.5  10 10

 2  1016
 0.0259 ln
10
 1.5  10






or
or
Vbi  0.7363 V
(d)
 2  V  N 

1

x n   s bi  a 

 e
 N d  N a  N d 
 211.7  8.85  10 14 0.736 

1.6  10 19

1/ 2

1
 17
 10  5  1015





1/ 2
E Fi  E F  0.3653 eV
(b)
Vbi  0.3653  0.3653
or
Vbi  0.7306 V
(c)
N N 
Vbi  Vt ln a 2 d 
 ni 




or
x n  0.426  10 4 cm  0.426  m
Now
 211.7  8.85  10 14 0.736
xp  
1.6  10 19

Vbi  0.7305 V
(d)


1

 17
 10  5  1015 



 2  1016 2  1016
 0.0259 ln 
2
 1.5  1010
or
 5  1015
 
17
 10
4
15
_______________________________________
E Fi  E F  0.4070 eV
(b)
Vbi  0.3294  0.4070
or
Vbi  0.7364 V
(c)
N N 
Vbi  Vt ln a 2 d 
 ni 

19
 max  3.29  10 4 V/cm
or
 1017
 
15
 5  10
1.6 10 5 10 0.426 10 
11.7 8.85 10 
or
 1017
 0.0259 ln
10
 1.5  10

eN d x n
s
1/ 2
2 V
xn   s bi
 e
 Na

N
 d


1


 N  N 

d 
 a
1/ 2

lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 7
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________


 211.7  8.85  10 14 0.7305

1.6  10 19

 2  1016
 
16
 2  10
For 300 K;

1

 2  1016  2  1016




1/ 2
x n  0.154  10 4 cm  0.154  m
By symmetry
x p  0.154  10 4 cm  0.154  m
Now
eN d x n
s

7.8
1.6 10 2 10 0.1537 10 
11.7 8.85 10 
19
14
or
_______________________________________

3
xp
Nd

3
Na
xn



which yields N a  7.766  10 15 cm 3
N d  2.33  1016 cm 3

or N a  5.12  1015 cm 3
(c)
 5.12  1015 1.98  1016 
Vbi  0.0259 ln 

2
1.5  1010


 0.695 V
_______________________________________

xn

So N d  3N a

 E  EF 
N a  n i exp Fi

 kT

 0.330 
 1.5  1010 exp

 0.0259 

xp


or N d  1.98  1016 cm 3

0.75 x n  0.25 x p 

 E  E Fi 
(b) N d  n i exp F

 kT

0
 .365 
 1.5  1010 exp

 0.0259 



 Na Nd 
(a) Vbi  0.0259 ln 

2
 1.5  1010 
 3N a2

0.710  0.0259 ln 

2
 1.5  1010 
2
 0.710 
or 3N a2  1.5 1010 exp

 0.0259 
7.6



xn N d  x p N a 
 max  4.75  10 4 V/cm


x n  0.25W  0.25 x n  x p
4
16



or
 max 

 2  1015 4  1016 
Vbi  0.0259 ln 

2


1.8  10 6
 1.157 V
For 400 K;
 2  1015 4  1016 
Vbi  0.034533 ln 

2

 3.28  10 9
 1.023 V
_______________________________________


 2  V
x n   s bi
 e

 Na

N
 d


1


 N  N 
d 
 a
1/ 2

 211.7  8.85  10 14 0.710

1.6  10 19

1
 1 
  
15
3
   4 7.766  10


 

 
1/ 2
6
 x n  9.93  10 cm
7.7
200 K; kT  0.017267 ; n i  1.38 cm 3
n i  1.8  10 6 cm 3
300 K; kT  0.0259 ;
400 K; kT  0.034533 ; n i  3.28  10 9 cm 3
For 200 K;
 2  1015 4 1016 
Vbi  0.017267  ln 

1.382



 1.257 V

or x n  0.0993  m


 211.7  8.85  10 14 0.710
xp  
1.6  10 19

1
 3 
  
1
4
7
.
766
 1015
 


 2.979  10 5 cm
or x p  0.2979  m
 

 

1/ 2
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 7
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________

 max


 211.7  8.85  10 14 0.6350

1.6 10 19

Now
eN d x n

s
1.6 10 2.33 10 0.0993 10 
11.7 8.85 10 
19
 1016
  15
 10
4
16
14
 2  V
x p   s bi
 e
which yields N a  8.127  10 15 cm 3


 213.1 8.85 10 14 1.180
xn  
1.6 10 19

1
 1 
  
15
 3   4 8.127  10

 

 
1/ 2





1/ 2
 

 
eN d x n
s
 max 

1.6 10 10 0.8644 10 
11.7 8.85 10 
19

1/ 2
4
15
14
or
 max  1.34  10 4 V/cm
 3.973  10 5 cm
or x p  0.3973  m
_______________________________________
eN d x n
s
7.10
1.6 10 2.438 10 0.1324 10 

13.18.85 10 
4
16
14
 4.45  10 4 V/cm
_______________________________________

  




1


 N  N 
d 
 a



 2  1017 4  1016 
(a) Vbi  0.0259 ln 

2
 1.5  1010

 0.80813 V
(b) Vbi increases as temperature decreases
At T  300 K, we can write
n i2  1.5 1010
 1016 1015 
(a) Vbi  0.0259 ln 

2
 1.5  1010 
or
Vbi  0.635 V
(b)
 Na

N
 d

1
 16
 10  1015

x p  0.08644  10 4 cm  0.08644  m

1
 3 
  
1
4
8
.
127
 1015
 
 2  V
x n   s bi
 e
1/ 2
(c)

7.9





or
 213.1 8.85  10 14 1.180
xp  
1.6  10 19

19

1

 N  N
d
 a
 1015
  16
 10
5
 max 
 Nd

N
 a
 211.7  8.85  10 14 0.6350

1.6 10 19

N d  2.438  10 16 cm 3
 1.324  10 cm
or x n  0.1324  m
1/ 2
x n  0.8644  10 4 cm  0.8644  m
Now





or
 3.58  10 4 V/cm
(b) From part (a), we can write
2
 1.180 
3N a2  1.8  10 6 exp

 0.0259 


1
 16
 10  1015




2






  1.12 
 K 2.8  1019 1.04  1019 exp

 0.0259 
 K  4.659
At T  287 K, kT  0.024778 eV
1/ 2
 287 
n i2  K 2.8  10 19 1.04  10 19 

 300 
3
  1.12 
 exp

 0.024778 


 4.659  2.5496  10 38 2.3404  10 20
So n  2.780  10
2
i
19

lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 7
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Then


7.12
(b) For N d  10 16 cm 3 ,

 2  1017 4 1016 
Vbi  0.024778 ln 

19
 2.780 10

 0.82494 V
We find
Vbi 287   Vbi 300
 100%
Vbi 300
N
E F  E Fi  kT ln d
 ni
 10 16
 0.0259 ln
10
 1.5  10
0.82494  0.80813
100%  2.08%
0.80813
 2%
_______________________________________

E F  E Fi  0.3473 eV
For N d  10 15 cm 3
 1015
E F  E Fi  0.0259 ln
10
 1.5  10





  1.12 
 K 2.8  10 1.04  10  exp

0.0259
19




 4  10 16 2  10 15
Vbi  0.0259  ln 
2
 1.5  1010
 0.68886 V
For Vbi  0.550 V,  T  300 K
At T  380 K, kT  0.032807 eV
Also

n  4.659  2.8  10
2
i
19





N N
(a) Vbi  Vt ln a 2 d
 ni




  


or

Vbi  0.456 V
(b)
 211.7  8.85  10 14 0.456
xn  
1.6  10 19


1.04 10 
19
7.13
 10 12 10 16 
 0.0259 ln 

2
 1.5  1010 
 380 


 300 

 4  1016 2  1015 
Vbi  0.032807  ln 

24

 4.112  10
V
V
 0.5506
 0.550
_______________________________________

 1012
  16
 10
3
  1.12 
 exp

 0.032807 
 4.112  10 24
Then
E F  E Fi  0.2877 eV
Then
Vbi  0.34732  0.28768
or
Vbi  0.0596 V
_______________________________________
2
19
 K  4.659
At T  300 K,




or




16
15
 T   4  10 2  10 
0.550  0.0259
 ln 

n i2
 300  

Using the procedure from Problem 7.10, we
can write, for T  300 K,
n i2  1.5  10 10




or
7.11
N N
Vbi  Vt ln a 2 d
 ni





1
 12
 10  10 16




1/ 2
or
x n  2.43  10 7 cm
(c)
 211.7  8.85  10 14 0.456
xp  
1.6  10 19



 1016
  12
 10
or
x p  2.43  10 3 cm

1
 12
 10  1016




1/ 2
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 7
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(d)
 max
7.15
eN d x n

s

 max
1.6 10 10 2.43 10 
11.7 8.85 10 
19
7
16
or
 max  3.75  10 2 V/cm
_______________________________________




 11.7  8.85  10 14 0.0259  1.6  10 19

2

1.6  10 19 N d
or







1/ 2
(a) N d  8 1014 cm 3 , L D  0.1447  m
(b) N d  2.2  10 16 cm 3 , L D  0.02760  m
(c) N d  8 1017 cm 3 , L D  0.004577  m
Now
(a) Vbi  0.7427 V
(b) Vbi  0.8286 V
(c) Vbi  0.9216 V
Also
 211.7  8.85  10 14 Vbi 
xn  
1.6  10 19


 8  10 17
 
 Nd


1

 8  1017  N

d





 0.6946 V
 0.7543 V
 0.8139 V
N d  10 14 ;  max  0.443  10 4 V/cm
(ii)
(iii)
(iv)
(b)
(i) For N a
 1015 ;
 1016 ;
 1017 ;
 1.46  10 4 V/cm
 4.60  10 4 V/cm
 11.2  10 4 V/cm
 10 14 , N d  10 14 ; Vbi  0.4561 V
 1015 ;
 1016 ;
 1017 ;
(ii)
(iii)
(iv)
 0.5157 V
 0.5754 V
 0.6350 V
(i) For N a  10 14 ,
N d  1014 ;  max  0.265  10 4 V/cm
 1015 ;
 1016 ;
 1017 ;
(ii)
(iii)
(iv)
1/ 2
Then
(a) x n  1.096  m
(b) x n  0.2178  m
(c) x n  0.02730  m
Now
L
(a) D  0.1320
xn
1/ 2
(i) For N a  10 17 ,
1/ 2





 1015 ;
 1016 ;
 1017 ;
(iii)
(iv)
1/ 2
 1.676  10 5
L D  
Nd


(ii)
7.14
Assume silicon, so




 Na Nd

N N
d
 a
We find
2 1.6  10 19
2e

 3.0904  10  7
s 11.7  8.85  10 14
(a)
(i) For N a  10 17 , N d  10 14 ; Vbi  0.6350 V

14
  kT
L D   2s
 e Nd
 2eVbi

 s
(c)
 0.381 10 4 V/cm
 0.420  10 4 V/cm
 0.443  10 4 V/cm
 max increases as the doping increases,
and the electric field extends further into
the low-doped side of the pn junction.
_______________________________________
7.16

 
 
 5  1016 1015
(a) Vbi  0.0259  ln 
2
 1.5  1010
 0.6767 V

 2  V  V R   N a  N d

(b) W   s bi
 N N
e

a
d

L
(b) D  0.1267
xn
LD
 0.1677
xn
_______________________________________
(c)




1/ 2
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 7
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(i) For V R  0 ,



 5  1016  1015

16
15
 5  10 10

 

 
 4  1016
 
17
 2  10
1/ 2
 
 9.452  10 cm
or W  0.9452  m

 2  V  V R   N a  N d

W   s bi
 N N
e

a
d


 211.7  8.85  10 14 0.6767  5
W 
1.6  10 19


 5  1016  1015  

16
15  
 5  10 10  

1/ 2
 

20.6767 
 1.43  10 4 V/cm
0.9452  10  4
(ii)For V R  5 V,
20.6767  5
 4.15  10 4 V/cm
2.738  10  4
_______________________________________
 max 

 2  1017 4  1016
(a) Vbi  0.0259 ln 
2
 1.5  1010
 0.8081 V
(b)

 2  V  V R   N a

x n   s bi
N
e

 d

2Vbi  V R  20.8081  2.5

W
0.3584  10  4
 1.85  10 5 V/cm








1


 N  N 
d 
 a


 

1/ 2

 
 5.78  10 12 F
or C  5.78 pF
_______________________________________
1/ 2
7.18
N N
(a) Vbi  Vt ln a 2 d
 ni
1/ 2
 0.2987  10 4 cm
or x n  0.2987  m

1

 N  N
d
 a

1/ 2
 1.6  10 19 11.7  8.85  10 14
 2  10  4 
20.8081  2.5



1


 2  1017  4  1016 


 2  V  V R   N d

x p   s bi
N
e

 a


e s N a N d
(d) C  A

 2Vbi  V R N a  N d  
 2  1017 4  1016

17
16
 2  10  4  10
 211.7  8.85  10 14 0.8081  2.5

1.6  10 19

 2  10
 
16
 4  10
1/ 2
(c)  max 
 max 
17

 

 
 0.3584  10 cm
or W  0.3584  m
Also W  x n  x p  0.3584  m
2Vbi  V R 
W
(i)For V R  0 ,

1/ 2
4
 max 






 2  1017  4  1016

17
16
 2  10 4  10
 2.738  10 cm
or W  2.738  m
7.17
1/ 2
 211.7  8.85  10 14 0.8081  2.5

1.6  10 19

4
(c)




1

 2  1017  4  1016

 5.97  10 6 cm
or x p  0.0597  m
5
(ii) For V R  5 V,

 211.7  8.85  10 14 0.8081  2.5

1.6  10 19

 211.7  8.85  10 0.6767 
W 
1.6  10 19

14








 80 N 2 
 Vt ln 2 d 
 ni 
We find
V
80 N d2  n i2 exp bi
 Vt
1/ 2

 1.5  1010





2
 5.762  10 32
 0.740 
exp

 0.0259 
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 7
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
 Vt ln3  0.0259 ln3
 0.02845 V
 N d  2.684  1015 cm 3
N a  2.147  1017 cm 3
(b)
 2  V  V R   N a

x n   s bi
N
e

 d




1


 N  N 
d 
 a
1/ 2

1/ 2
7.20



1


 N  N 
d 
 a
1/ 2
 211.7  8.85  10 14 0.740  10

1.6  10 19

1

 1 
  

17
15 
 80  2.147  10  2.684  10 
3 10 








1/ 2


4 10 4 10 
15
17

4  10  4  1017 
15
9  1010  1.224  10 9 Vbi  V R 
so that
Vbi  V R   73.53 V
which yields
V R  72.8 V



 
 2.147  1017 2.684  1015

17
15
 2.147  10  2.684  10

1/ 2

 
C   4.52 10 9 F/cm 2
_______________________________________
7.19
(a) Vbi 3N a   Vbi N a 
 N 3N  
N N 
 Vt ln  d 2 a   Vt ln  d 2 a 
 ni 
 ni


N N
 Vt ln3  ln  d 2 a

 ni

or
1/ 2
 1.6 10 19 11.7  8.85  10 14

20.740  10


 2 1.6  10 19 Vbi  V R 

14
 11.7  8.85  10

 9.38  10 4 V/cm

 max
5 2


e s N a N d
(d) C   




V
V
N
N
2


R
a
d 
 bi

 2eVbi  V R   N a N d


N N
s

d
 a
or
2Vbi  V R 

W
20.740  10

2.262  0.028310  4


 4  1015 4  1017 
(a) Vbi  0.0259 ln 

10 2

 1.5  10
or
Vbi  0.766 V
Now
1/ 2
 2.83  10 6 cm
or x p  0.0283  m
(c)  max
C 3N a   3N a 

  3  1.732
C N a   N a 
(c) For a larger doping, the space charge
width narrows which results in a larger
capacitance.
_______________________________________
1/ 2
 2.262  10 4 cm
or x n  2.262  m
 2  V  V R   N d

x p   s bi
N
e

 a
1/ 2
So
 211.7  8.85  10 14 0.740  10

1.6  10 19

1

 80 
  

17
15 
 1  2.147  10  2.684  10 
 e s N a 
(b) C   

 2Vbi  V R  
 Nd Na 
 
   Vt ln 

2
 ni 
 

 4  1016 4 1017
(b) Vbi  0.0259 ln 
2
 1.5  1010
or
Vbi  0.826 V
We have
 2 1.6  10 19 Vbi  V R 
2
3  10 5  
14
 11.7  8.85  10





so that
Vbi  V R   8.008 V






4 10 4 10 
16
17

4  1016  4  1017 
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 7
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
which yields
V R  7.18 V
(b)


2VbiA  V R 


A
W  A
W B  VbiA  V R



B  2VbiB  V R  W  A VbiB  V R
W B 

 4 1017 4  1017 
(c) Vbi  0.0259 ln 

10 2

 1.5  10
or
Vbi  0.886 V
We have
 2 1.6  10 19 Vbi  V R 
2
3  10 5  
14
 11.7  8.85  10





 1  5.7543 



 3.13  5.8139 

or
 A 
 0.316
 B 
(c)


4 10 4 10 

17
17

4  10  4  1017 
17
C j  A
so that
Vbi  V R   1.456 V
which yields
V R  0.570 V
_______________________________________
C j B 
7.21
(a)
 2 s VbiA  V R   N a  N dA


 N N
e
a
dA


W  A

W B   2  V  V   N  N
s
biB
R
dB
 a

 N N
e
a
dB






1/ 2




1/ 2
or
We find




 1018 1016 
VbiB  0.0259  ln 
  0.8139 V
10 2

 1.5  10
W  A  5.7543  10 18  1015
 

W B   5.8139  10 18  1016




 1016
  15
 10
or
W  A
 3.13
W B 
1/ 2
 VbiB  V R

 V  V
R
 biA
 10 15
  16
 10
 5.8139  10 18  10 16


 5.7543  10 18  10 15


C j  A












1/ 2
1/ 2
 0.319
7.22
(a) We have
C j 0 
C j 10
or

 s N a N d



 2Vbi  N a  N d  
1/ 2


s N a N d







V
V
N
N
2
R
a
d 
 bi
C j 0 
 V  VR
 3.13   bi
C j 10
 Vbi
For V R  10 V, we find
3.132 Vbi
1/ 2
 N a  N dB

 N  N
dA
 a
_______________________________________
  


We find


s N a N dB


 2VbiB  V R N a  N dB  
1/ 2
 1018 1015 
VbiA  0.0259 ln 
  0.7543 V
2
 1.5  1010 
  


1/ 2
 N
  dA
 N dB
C j B 
or
W  A  VbiA  V R   N a  N dA   N dB



W B   VbiB  V R   N a  N dB   N dA



s N a N dA


 2VbiA  V R N a  N dA  
 Vbi  10
or
Vbi  1.137 V
(b)
x p  0.2W  0.2 x p  x n






1/ 2
1/ 2
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 7
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
N
 0.25  d
xn
Na
Now
N N
Vbi  Vt ln a 2 d
 ni
so

16
15




C
2
a


2




 2  1016 5  1015
Vbi  0.0259 ln 
2

1.8  10 6
 1.162 V
1
C 
Vbi  V R











1/ 2





2 10 4 10  
2 10  4 10 
15
16
15
1/ 2
16
6.6457 10 12
1.157  V R
(i) For V R  0 ,
C  6.178 pF
(ii) For V R  5 V,
C  2.678 pF
_______________________________________
7.25


 2 1017 5 1015
Vbi  0.0259 ln 
2
 1.5 1010
 0.7543 V






e s N a N d
(a) C  AC   A

 2Vbi  V R N a  N d  



1/ 2

 1.6  10 19 11.7  8.85  10 14
 8 10  4 
20.7543  10 




e s N a N d
C  AC   A

 2Vbi  V R  N a  N d  



e s N a N d
C  AC   A

 2Vbi  V R  N a  N d  
 1.6  10 19 13.1 8.85  10 14
 5  10  4 
21.157  VR 

C
 2 1015 4  1016
(a) Vbi  0.0259 ln 
2
 1.5  1010
 0.6889 V
16
 2  1015 4  1016
(b) Vbi  0.0259 ln 
2

1.8  10 6
 1.157 V

1.162  V R 2
1.162  0.5
1
.
162
 VR2
1.502 
1.662
which yields V R 2  2.58 V
_______________________________________
1/ 2
0.6889  V R

1.50 

6.2806 10 12
(i) For V R  0 ,
C  7.567 pF
(ii) For V R  5 V,
C  2.633 pF

Vbi  V R 2
C V R1 
So


C V R 2 
Vbi  V R1
7.24

15
We can then write
 1.137 
1.8  10 6
exp 
Na 

0.25
 20.0259  
which yields
N a  1.23  1016 cm 3
and
N d  3.07  10 15 cm 3
_______________________________________


2 10 4 10  

2 10  4 10 
 0.25 N
1.137  0.0259 ln 
6
 1.8  10
7.23

 1.6  10 19 11.7  8.85 10 14
 5  10  4 
20.6889  V R 

Then
xp

1/ 2

2 10 5 10  
2 10  5 10 
17
17
C  4.904  10 12 F
1
1
f 
L
2
C 2 f 
2 LC
15
15

1/ 2
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 7
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
L

4.904 10 2 1.25 10 
2
6



 1.6  10 19 13.1 8.85  10 14
 10  4 
2Vbi  2 

1
12
 3.306  10 3 H  3.306 mH

(b)
(i) For V R  1 V, C  12.14 pF
f 
1

2 3.306  10
3
0.6  10 12  2.724 10  20
12.14 10 
12
1/ 2
1


2 3.306  10 3 6.704  10 12

1/ 2
 1.069 10 Hz  1.069 MHz
_______________________________________
0.6  10 12  2.724  10  20
Vbi  1.135 V
_______________________________________
19
14
 
(b) 10 5
d



7.28
3

 2 1.6  10 19 0.75  10 N d 


11.7  8.85 10 14




2 V
x p   s bi
 e
3
 N d  3.01 10 cm
_______________________________________
15
7.27

x p  0.20W  0.20 x n  x p
0.8x p  0.2x n
xn  4x p
 
 



1


 N  N 

d 
 a
 Nd

N
 a
1/ 2

 10 14
 
15
 5  10
 

 211.7  8.85  10 14 0.5574 

1.6  10 19



1
 14
 10  5  10 15

or
N a x p  N d xn  N d 4x p
x p  5.32  10 6 cm
 N a  4N d
Also
N N 
(a) Vbi  Vt ln  a 2 d 
 n i 
 4 N d2
 0.0259  ln 
6
 1.8  10


 5  10 15 10 14
(a) Vbi  0.0259  ln 
2
 1.5  10 10
or
Vbi  0.5574 V
(b)
2

Nd
Vbi  5
N a  1.19  1016 cm 3 ,
2.5 10 
 21.6  10 0.75  10N

11.7 8.85 10 

 N d  1.88  10 cm
Nd
Vbi  2
By trial and error,
N d  2.976  1015 cm 3 ,
1/ 2
5 2
16
5 N d 
Vbi  1.10 V
(b) From part (a),
7.26
(a)
1/ 2
N a  6.016  1015 cm 3 ,
6
 2eVbi  V R N d 
 max  

s


Let Vbi  0.75 V
2
d
By trial and error,
N d  1.504  1015 cm 3 ,
 7.94 10 Hz  0.794 MHz
(ii) For V R  5 V, C  6.704 pF
5
f 
4 N 


2
2 V
x n   s bi
 e





e s N a N d
C  AC   A






2
V
V
N
N
R
a
d 
 bi
1/ 2
 Na

N
 d


1


 N  N 

d 
 a
1/ 2



1/ 2
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 7
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________


 211.7  8.85  10 14 0.5574 

1.6  10 19

 5  1015
 
14
 10
7.30

1
 14
 10  5  10 15




1/ 2
x n  2.66  10 4 cm

or
50 10 
4 2



V
10 
211.7  8.85  10
1.6 10
 10 14
 16
 10


19
14
R
14




 max 

C

1/ 2
1.287  10 13
Vbi  V R
(i) For V R  1 V, C  9.783  10 14 F
(ii) For V R  3 V, C  6.663  10 14 F
(iii) For V R  5 V, C  5.376  10 14 F
_______________________________________
7.31



 8  10 16 N d
(a) Vbi  0.0259  ln 
2
 1.8  10 6
8 10 N
16
d

 1.8  10 6

2

  1.20

 1.20 
exp

 0.0259 



e s N a N d
(b) C  AC   A






V
V
N
N
2
R
a
d 
 bi

1/ 2

 1.6  10 19
1.10  10 12  A
 21.20  1.0 

13.18.85 10 14 8 1016 5.36 1015 
8 10
16
 5.36  1015
5
 A  7.56  10 cm

1/ 2


2


 1.6  10 19
(c) 0.80  10 12  7.56  10 5 
 2Vbi  V R 





 0.50  10 4 cm  0.50  m
(c)
1/ 2
 N d  5.36  10 15 cm 3
so
x n  50  10



1/ 2
which yields
V R  193 V
(b)
xp Nd
N

 x n  x p  a
xn
Na
 Nd
4


 11.7  8.85  10 14 2  10 15
1/ 2
7.29
An n  p junction with N a  1014 cm 3 ,
(a) A one-sided junction and assume
V R  Vbi . Then



 5  1015 
1

 14
 

14
15 

 10
 10  5  10 
which becomes
9  10 6  1.269  10 7 Vbi  V R 
We find
V R  70.4 V
_______________________________________
2 V 
xp   s R 
 eN a 

 1.6  10 19
 10 5 
 2Vbi  V R 
 211.7  8.85  10 Vbi  V R 
30  10  4  
1.6  10 19

14

 e s N d 
(b) C  AC   A  

 2Vbi  V R  
or
(c) For x n  30  m, we have

 2  10 17 2  1015
(a) Vbi  0.0259  ln 
2
 1.5  1010
 0.7305 V
2V R
2193.15

W
50.5  10  4

13.18.85 10 14 8 1016 5.36 1015 
8 10
16
 5.36  10 15
1.0582  10 8 
or
 max  7.65  10 4 V/cm
_______________________________________

1/ 2


2.1585  10 8
Vbi  V R
 Vbi  V R  4.161  1.20  V R
V R  2.96 V
_______________________________________
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 7
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
7.32
Plot
_______________________________________
7.33
N N
(a) Vbi  Vt ln aO 2 dO
 ni
(c) p-region
eN
d  x 

  aO
dx
s
s
or
eN x
   aO  C1
s
We have




 x 
d 2 x 
dx 


2
s
dx
dx
For 2  x  1  m,  x   eN d
So
eN d x
d eN d


 C1
s
s
dx
(a)
eN aO x p
s
Then for  x p  x  0

eN aO
x  xp
s
eN dO x eN dO 
x 
 x n  O 

2 s
s 
2 
_______________________________________
1 
7.34
  0 at x   x p  C1  

Then for 0  x  x O we have

At x  2  m   x O ,   0
So
eN d x O
C1 
s
Then
eN d
x  x O 

s
At x  0 , 0   x  1  , so
n-region, 0  x  x O
d 1  x  eN dO


dx
s
2 s
0  
or

eN dO x
1 
 C2
2 s
19
15
4
14
0   7.726  10 4 V/cm
(c) Magnitude of potential difference is
eN
  dx  d x  x O dx
s
d 2  x  eN dO


dx
s
s


or
eN dO x
2 
 C3
s
 x2



 2  xO  x   C 2


Let   0 at x   x O , then

We have  2  0 at x  x n
eN dO x n
s
so that for x O  x  x n , we have
2  
1.6 10 5 10  110 
11.7 8.85 10 
or
n-region, x O  x  x n
 C3  
eN d
 1  210  4
s
eN dO
x n  x 
s
We also have  2   1 at x  x O
Then
eN dO x O
eN
 C 2   dO x n  x O 
s
2 s
which gives
eN 
x 
C 2   dO  x n  O 
s 
2 
eN d
s

 x O2
eN d x O2
2
  C2  C2 

x

O

 2
2 s


Then we can write
eN
  d  x  x O 2
2 s
At x  1  m
0
eN d
s
1 
1.6 10 5 10   1  210 
211.7 8.85  10 
19
15
4 2
14
or
1  3.863 V
Potential difference across the intrinsic region
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 7
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
 2  0  d  7.726  10 4 2  10 4 
Then




 2  10 2  10  


 2 1.6  10 19 Vbi  V R 
4  10 5  
 11.7  8.85  10 14
or
 2  15.45 V
By symmetry, the potential difference across
the p-region space-charge region is also
3.863 V. The total reverse-bias voltage is
then
V R  23.863  15.45  23.2 V
_______________________________________
16
 2
(a) V B  s crit
2eN B
or



NB 
s 
11.7  8.85 10 4 10

2eV B
2 1.6  10 19 40 
Then
N B  N a  1.294  10 cm
14

16
Then




3



 2
11.7  8.85 10 14 4 10 5
N a  s crit 
2eV B
2 1.6  10 19 80 




 5  10 5  10  


15
Or N B  N a  2.59  1016 cm 3
_______________________________________
7.36

 2 1.6  10 19 Vbi  V R 
4  10 5  
 11.7  8.85  10 14
5 2
11.7 8.85 10 14 4 10 5 2
21.6  10 19 20 

2
3
 6.47  10 cm
_______________________________________
15

 0.6587 V
2
crit
(b) N B 
1/ 2

16
16 
 2  10  2  10  
 Vbi  V B  51.77 V
So V B  51.04 V
(b)
 5  1015 5  1015 
Vbi  0.0259 ln 

10 2
 1.5  10


7.35
16
15
1/ 2

15
15 
 5  10  5  10  
 Vbi  V R  207.1
So V R  206 V
_______________________________________
7.39
For a silicon p  n junction with
N d  5 10 15 cm 3 and V B  100 V, then,
neglecting Vbi we have
2 V 
xn   s B 
 eN d 
7.37
(a) For N d  10 16 cm 3 , from Figure 7.15,

1/ 2

 211.7  8.85  10 14 100  


19
5  10 15 
 1.6  10

V B  75 V

1/ 2

(b) For N d  10 15 cm 3 ,
V B  450 V
_______________________________________
x n min   5.09  10 4 cm  5.09  m
_______________________________________
7.38
(a) From Equation (7.36),
7.40
We find
 max
 2eVbi  V R   N a N d


N N
s

d
 a




Set  max   crit and V R  V B


 2  10 16 2  10 16
Vbi  0.0259  ln 
2
 1.5  1010
 0.7305 V


1/ 2


or
  


 10 18 10 18 
Vbi  0.0259 ln 
  0.933 V
10 2
 1.5  10

Now
 max 
eN d x n
s
so
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 7
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
1.6 10 10 x
11.7 8.85 10 
19
10 6 
18
n
14
which yields
x n  6.47  10 6 cm
Now
 2  V  V R   N a 
1


x n   s bi
 N  N  N
e
d
 d  a

Then
 211.7  8.85  10 14
2
6.47  10  6  
1.6  10 19








1/ 2

 1018 
1

 Vbi  V R  18  18

18 
 10  10  10 
which yields
Vbi  V R  6.468 V
or
V R  5.54 V
_______________________________________
7.41
Assume silicon: For an n  p junction
 2  V  V R  
x p   s bi

eN a


Assume Vbi  V R
4 2

150 10 
4 2

At x   x O and x   x O ,   0
So
2
2
ea  x O 
ea  x O 
C
C
0




1
1
s  2 
s  2 
Then
ea

x 2  x O2
2 s
(b)

ea  x 3
2
 x    dx  
  xO  x  C 2
2 s  3

Set   0 at x   x O , then



0
3

eax O3
ea   x O
 x O3   C 2  C 2 

2 s  3
3 s

Then
 x3
 eax O3
2


x
x


O
 3
 3
s


_______________________________________
 x   


1.6 10 10 
211.7  8.85  10 14 V R
19
which yields
V R  4.35  10 3 V
(b) For x p  150  m

1/ 2
(a) For x p  75  m
75 10 
7.43
(a) For the linearly graded junction
 x   eax
Then
d  x  eax


dx
s
s
Now
eax
ea x 2
dx 


 C1
s
s 2
15
ea
2 s
7.44
We have that

V
10 
211.7  8.85  10
1.6 10
19
14
R
15
which yields
V R  1.74  10 4 V
Note: From Figure 7.15, the breakdown
voltage is approximately 300 V. So, in each
case, breakdown is reached first.
_______________________________________
7.42
2  10 18
a
 10 22 cm 4
2  10  4
From Figure 7.15, V B  15 V
_______________________________________
 ea 2s

C  


12
V
V



bi
R 

Then
1/ 3
7.2 10 
 a 1.6  10 11.7 8.85  10 

9 3
19
14
2



120.7  3.5

which yields
a  1.1 10 20 cm 4
_______________________________________
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 7
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
7.45
 e s N a 
(a) C j  AC   A  

 2Vbi  V R  
Let N a  5 1015 cm 3 &lt;&lt; N d

1/ 2


 3  1017 5  10 15 
Then Vbi  0.0259  ln 

10 2
 1.5  10

 0.7648 V
Now
 1.6  10 19
C j  0.45  10 12  A
 20.7648  5






 11.7  8.85  10 14 5  1015

0.45  10 12  A 8.476  10 9
5


 A  5.31 10 cm
 1.6  10 19
(b) C j  5.309  10 5 
 2Vbi  V R 
2





 11.7  8.85  10 14 5  1015
Cj 
1/ 2

1/ 2
1.0805  10 12
Vbi  V R
(i) For V R  2.5 V, C j  0.598 pF
(ii) For V R  0 ,
C j  1.24 pF
_______________________________________
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 8
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Chapter 8
or
8.1
In forward bias
 eV 
I f  I S exp

 kT 
Then
 eV 
I S exp 1 
I f1
 e
 kT 
 exp 

I f2
 eV 2 
 kT
I S exp

 kT 
or
 kT   I f 1 
V1  V 2  
 ln
 e   I f 2 
(a)
I f1
For
 10 , then
I f2

V1  V 2  0.0259 ln100
or
V1  V 2  119.3 mV  120 mV
_______________________________________
n po 
p no



n
1.5  10

Na
8  1015
V
p n x n   p no exp a
 Vt


n p  x p  n po
2
 1.125  10 5 cm 3




V
exp a
 Vt


  0.55 
p n x n   1.125  10 5 exp

 0.0259 
0
  0.55 
n p  x p  2.8125 10 4 exp

 0.0259 
0
_______________________________________

8.3
n po 
 





n i2
1.8  10 6

Na
4 1016
ni2
1.8  10 6

Nd
1016
(a) V a  0.90 V,
p no 

2
 8.1 10 5 cm 3
2
 3.24  10  4 cm 3

 4.0  1011 cm 3
5

 

 0.90 
n p  x p  8.110 5 exp

 0.0259 
 10.0  1010 cm 3
(b) V a  1.10 V






 9.03  1014 cm 3

 3.95  10 12 cm 3
 

 1.10 
p n x n   3.24 10  4 exp

 0.0259 
 0.45 
p n x n   1.125  10 exp

 0.0259 

 
 4.69  1013 cm 3
(c) V a  0.55 V
 2.8125  10 4 cm 3
(a) V a  0.45 V,


 0.55 
n p  x p  2.8125 10 4 exp

 0.0259 
 0.90 
p n x n   3.24 10  4 exp

 0.0259 

n2
1.5  1010
 i 
Nd
2  10 15
 1.88  1014 cm 3


V1  V 2 


V1 V 2  59.6 mV  60 mV
(b)
I f1
 100 , then
For
I f2
10 2

 0.55 
p n x n   1.125  10 5 exp

 0.0259 
or
2
i

(b) V a  0.55 V,
V1 V 2  0.0259  ln 10 
8.2

n p  x p  9.88  1011 cm 3

 0.45 
n p  x p  2.8125  10 4 exp

 0.0259 

 

 1.10 
n p  x p  8.110 5 exp

 0.0259 
 2.26 1014 cm 3
(c) p n x n   0


np  xp  0
_______________________________________
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 8
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
8.4
(a) n po

(b) J p  x n  

10 2
n
1.5  10


Na
5 1016
2
i

 4.5  10 3 cm 3
p no

n2
1.5  1010
 i 
Nd
5 1015

2
 p x  
or V a  Vt ln  n n 
 p no 
(b) n po 

6 2
19
1016
 4.521 A/cm 2







9.80
10 8

or I p  4.52 mA
(c) I  I n  I p  1.85  4.52  6.37 mA
_______________________________________
8.6
For an n  p silicon diode
10 2
 3.214  10 cm

 p0
V
 exp a
 Vt

4
p no 
Dp




I p  AJ p x n   10 3 4.521 A
n
1.5  10

Na
7  1015
I S  Aeni2 
3

10 2
n
1.5  10

Nd
3  1016
2
i
en i2
Nd
V
exp a
 Vt
 1.10 
 exp

 0.0259 
 0.1 5  1015 
 0.0259 ln 

4
 4.5 10

 0.599 V
(ii) n-region - lower doped side
2
i
Lp
1.6 10 1.8 10 

 4.5 10 4 cm 3
V 
(i) p n x n   p no exp a 
 Vt 
eD p p no

 7.5 10 3 cm 3
 0.1N a 
(i) V a  Vt ln 

 n po 
 0.1 7  1015 
 0.0259 ln 
4 
 3.214  10 
 0.6165 V
(ii) p-region - lower doped side
_______________________________________

Dn
1
Na
 nO
10 1.6 10 1.5 10 
4
19
10 2
10 16
25
10  6
or
I S  1.8  10 15 A
(a) For V a  0.5 V,

V
I D  I S exp a
 Vt






 0.5 
 1.8  10 15 exp

 0.0259 
or
I D  4.36  10 7 A
8.5


(a) J n  x p 

eDn n po
Ln
2
i
en
Na
V
exp a
 Vt
V
 exp a
 Vt
Dn
 no
1.6 10 1.8 10 

6 2
19
5  1016
(b) For V a  0.5 V,









205
5  10 8
or
I D   I S  1.8  10 15 A
_______________________________________
 1.10 
 exp

 0.0259 
 1.849 A/cm 2
I n  AJ n  x p  10 3 1.849  A

 

   0.5  
I D  1.8  10 15 exp
  1
  0.0259  

or I n  1.85 mA
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 8
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
8.9
8.7
 1
J s  en i2 
 N a
Dn

 no

1
Nd

 1.6  10 19 2.4  10 13
 1

15
 4  10

Dp 

 p0 

2
90
1

6
2  10
2  10 17
48 

2  10  6 
so that
J s  1.568  10 4 A/cm 2


 I
 1
V  Vt ln

 IS
In reverse bias, I is negative, so at
I
 0.90 , we have
IS




V
(a) I  AJ s exp a
 Vt


 0.25 
 10  4 1.568  10  4 exp

 0.0259 
 2.44  10 4 A
or I  0.244 mA
(b) I   I s   AJ s   10 4 1.568  10 4


V  0.0259 ln1  0.90 

8
 1.568  10 A
_______________________________________
Dn
 no


1
Nd

 1.6 10 19 1.5  1010
 1

17
 5  10

Dp 

 p0 

10 

8  10 8 
J s  5.145  10 11 A/cm 2


I s  AJ s  2  10 4 5.145  10 11
14
 1.029  10
V 
(b) I  I s exp a 
 Vt 


A

 3.6110 7 A

 0.55 
(ii) I  1.029  10 14 exp

 0.0259 
 1.72  10 5 A

8.10




 I s  6.305  10 15 A  6.305  10 12 mA
I s 6.305  10 12

A
2  10  4
 3.153  10 8 mA/cm 2
V 
Case 2: I  I s exp a 
 Vt 
 0.70 
 2  10 12 exp

 0.0259 
or I  1.093 mA
I
2  10 12
Js  s 
A 1 10 3
 2  10 9 mA/cm 2
V 
Case 3: I  AJ s exp a 
 Vt 
Js 

 0.45 
(i) I  1.029  10 14 exp

 0.0259 

V  59.6 mV
_______________________________________
 0.65 
0.50  10 3  I s exp

 0.0259 
2
25
1

7
10
8  10 15
or
V
Case 1: I  I s exp a
 Vt
8.8
 1
(a) J s  eni2 
 N a
We have
 V  
I  I S exp   1
  Vt  
or we can write this as
V 
I
 1  exp 
IS
 Vt 

 0.65 
(iii) I  1.029  10 14 exp

 0.0259 
 8.16  10 4 A
_______________________________________

 I 
So V a  Vt ln 

 AJ s 


0.80
 0.0259 ln   4
7 
 10 10 
V a  0.6502 V



lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 8
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Then



I s  AJ s  10 4 10 7  10 11 mA
I
V
exp a
 Vt
Case 4: I s 





1.20
 0.72 
exp

 0.0259 
I s  1.014  10 12 mA
I s 1.014  10 12

Js
2  10 8
A
 5.07  10 5 cm 2
_______________________________________
8.11
eD n n po
(a)
Jn
Ln

eD
n
eD p p no
Jn  J p
n po

Ln
Lp
Dn
 no

Dn
 no


n i2
Na
or
2
i
2
i
1
D p no  N a

D n po  N d
D p no  N a

D n po  N d





1

 0.90  1

D n po  1

 1

D p no  0.90 
Na

Nd
2510 7  0.1111
105 10 7 

1
25

Na
5  10  7
Dp n
n


 po N d
Na
1
0.90 
8.12
The cross-sectional area is
I 10  10 3
A 
 5  10  4 cm 2
20
J
We have
V 
 0.65 
J  J S exp D   20  J S exp

 0.0259 
 Vt 
which yields
J S  2.522  10 10 A/cm 2
We can write
 1
Dp 
Dn
1


J S  eni2 
 N a  nO N d  pO 
We want
Dn
1

N a  nO
 0.10
Dp
Dn
1
1



N a  nO N d  pO
Na
Nd
 0.07857 or
 12.73
Nd
Na
(b) From part (a),
Na

Nd
D n po  1

 1

D p no  0.20 

2510 7  4
105 10 7 
Na
Nd
 2.828 or
 0.354
Nd
Na
_______________________________________
1
Na
25
1
10


7
Nd
5  10
5  10  7
7.071 10 3
 0.10
Na
3
3
7.071 10 
4.472  10
Nd
which yields
Na
 14.23
Nd
Now
=




J S  2.522  10 10  1.6  10 19 1.5  10 10

2

1
25
1
10 





7
Nd
5  10
5  10  7 
 14.23N d
We find
N d  7.09  10 14 cm 3
and
N a  1.01 10 16 cm 3
_______________________________________
8.13
Plot
_______________________________________
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 8
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
8.14
(a)
8.15
(a) p-side;
N
E Fi  E F  kT ln a
 ni
eD n n po
Jn
Ln

eD n n po eD p p no
Jn  J p

Ln
Lp
Dn
 no

Dn
 no



 5  10 15
 0.0259  ln
10
 1.5  10
n i2
Na
2
i
Dp n
n


Na
 po N d
1
E Fi  E F  0.329 eV
Also on the n-side;
N
E F  E Fi  kT ln d
 ni
E F  E Fi  0.407 eV
(b) We can find
D n  1250 0.0259   32.4 cm 2 /s
D p  320 0.0259   8.29 cm 2 /s
1
1 1  Na


2.4 0.1  N d




Now
 1
J S  en i2 
 N a
or
Jn

Jn  J p
e n N d
L
e n N d  n  e p N a
Lp
We have
 n  e n N d and  p  e p N a
Also
Ln

Lp

D n no

D p po
Then

2. 4
 4.90
0. 1

n  p
Jn

 n  p  4.90
Jn  J p


Dn
 nO


1
Nd
 1.6  10 19 1.5  1010
1
N 
1  2.04  a 
 Nd 
(b) Using Einstein's relation, we can write
e n n i2

Jn
Ln N a

Jn  J p
e n n i2 e p n i2



Ln N a
Lp Nd





or
so
1




 1017
 0.0259  ln
10
 1.5  10




We have
Dp  p

1
1
and no 


 po 0.1
Dn
 n 2.4
Jn

Jn  J p




or
2
i
D p no  N a
1

D n po  N d




 1

15
 5  10

Dp 

 pO 

2
8.29 

10  7 
32.4
1

10  6 10 17
or
J S  4.426  10 11 A/cm 2
Then
I S  AJ S  10 4 4.426  10 11
or
I S  4.426  10 15 A
We find
V 
I  I S exp D 
 Vt 
 0.5 
 4.426  10 15 exp

 0.0259 
or
I  1.07  10 6 A  1.07  A
(c) The hole current is


I p  eni2 A 
_______________________________________



1
Nd
Dp
 po
V
exp D
 Vt




lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 8
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________


 1.6  10 19 1.5  10 10
 10 
2
4
V
8.29
exp D
7
10
 Vt






V 
I p  3.278  10 16 exp D  (A)
 Vt 
Then
I p J p 3.278  10 16


 0.0741
I
J S 4.426  10 15
_______________________________________
I Total


 1.6  10 19 5  10  4
Now
  1 2L p
1


I p  x n  L p   I p x n  exp
 Lp
2





Dp n
  eA


 po N d



 1.6  10 19 5  10  4


1.5  10 10
10

8
8  10
1.5  1016
2

Dn n
  eA



Na
no


1.5  1010
25

2  10  7
5  1016
I sn  4.025  10 15 A



2

 5 1016 1.5  1016 
(c) Vbi  0.0259 ln 

2


1.5  1010
 0.746826 V
V a  0.8Vbi  0.80.746826  0.59746 V
V
p n  x n   p no exp a
 Vt

1.5 10 


 n
 Va 

 
 N exp V 
d

 t 
10 2
1.5  10 16
2
i
 0.59746 
exp

 0.0259 
 1.56  1014 cm 3

(d) I n  x p

V
 I n x n   I sn exp a
 Vt

 4.025  10
15

8.17
(a) The excess hole concentration is given by
p n  p n  p no
 V
 p no exp a
  Vt
We find
p no 
x
 

  1  exp

 Lp 

 




2
n i2
1.5 1010

Nd
1016
and
 2.25  10 4 cm 3
80.0110 6 
L p  D p pO 
 2.828  10 4 cm  2.828  m
Then
p n  2.25 10 4 exp





 0.59746 
exp

 0.0259 
 4.198110 5 A
Then
1
1




I n  x n  L p   I Total  I p  x n  L p 
2
2




 1.820  10 4  8.4896  10 5
 9.710  10 5 A
_______________________________________
2
i


 8.4896  10 5 A
I sp  1.342  10 14 A
 eD n n po
(b) I sn  A
 Ln




 1 
 1.3997 10  4 exp 
 2 
2
i

 1.3997  10 4 A
 In  I p
 4.1981 10 5  1.3997  10 4
 1.820  10 4 A
8.16
(a) I sp

 0.59746 
 1.342  10 14 exp

 0.0259 
or
 eD p p no
 A
 Lp





V
(e) I p x n   I sp exp a
 Vt
1 

17
 10 

 0.610  
  1
 0.0259  
x


 exp

4
 2.828  10 
or
p n  3.81 1014 exp
x

3
 cm
4
 2.828  10 

lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 8
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(b) We have
d p n 
J p  eD p
dx
eD p 3.808  1014
x



exp

4
2.828  10  4
 2.828  10 

 
1.6 10 234.5 10 

19
J no
3
10.72  10  4
 0.610 
 exp

 0.0259 




 0.1N d 
p 
or V a  Vt ln  n   Vt ln  2

p
 n i N d 
 no 
 0.1N d2 
 Vt ln 

2
 n i


(c) We have
eD n n po
V 
J no 
exp a 
Ln
 Vt 
We can determine that
n po  4.5  10 3 cm 3 and Ln  10.72  m
Then
V
p n  p no exp a
 Vt

At x  3  10 4 cm,
1.6  10 19 8 3.808  1014
 3 
J p 3 
exp

2.828  10  4
 2.828 
or
J p 3  0.5966 A/cm 2

(b) Problem 8.8


 0.1 8  1015 2 
 0.0259  ln 

2
 1.5  1010

 0.623 V
_______________________________________


8.19
The excess electron concentration is given by
n p  n p  n po
 V  
x

 n po exp a   1  exp

V
 Ln 
  t  
The total number of excess electrons is


N p  A n p dx
or
0
J no  0.2615 A/cm 2
We can also find
J po  1.724 A/cm 2
We may note that

x
x
dx   L n exp

exp
 L  0  Ln

 Ln 
 n 
0
Then
 V  
N p  AL n n po exp a   1
  Vt  
We find that
D n  25 cm 2 /s and L n  50.0  m
Also

Then at x  3  m,
J n 3  J no  J po  J p 3
 0.2615  1.724  0.5966
or
J n 3  1.39 A/cm 2
_______________________________________
8.18
n po
(a) Problem 8.7
V 
n p  n po exp a 
 Vt 
 np 


  Vt ln  0.1N a 
or V a  Vt ln
2
 n po 
n
N

a 
 i


 0.1N a2 
 Vt ln 

2
 n i


2
 2.81 10 4 cm 3
Then
N p  10 3 50.0  10 4 2.8125  10 4



 V
 exp a
  Vt

 
  1

 
or


 0.1 4  1015 2 
 0.0259  ln 

2
 2.4  1013

 0.205 V


n2
1.5  1010
 i 
Na
8  1015

 V  
N p  0.1406 exp a   1
  Vt  
Then, we find the total number of excess
electrons in the p-region to be:
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 8
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(a) V a  0.3 V, N p  1.51 10 4
(b) V a  0.4 V, N p  7.17  10 5
(c) V a  0.5 V, N p  3.40  10 7
Similarly, the total number of excess holes in
the n-region is found to be
 V  
Pn  AL p p no exp a   1
  Vt  
We find that
D p  10.0 cm 2 /s and L p  10.0  m
Also
p no

n2
1.5  1010
 i 
Nd
10 16

2
 2.25  10 4 cm 3
Then

Pn  2.25  10
2
  Va
exp
  Vt

 
  1

 
So
(a) V a  0.3 V, Pn  2.41 10 3
(b) V a  0.4 V, Pn  1.15  10 5
(c) V a  0.5 V, Pn  5.45  10 6
_______________________________________
8.20
  Eg 
V 
 eV 
  exp a 
I  n i2 exp a   exp

V
kT
 kT 
 t 


Then
 eV a  E g 

I  exp


 kT
so
 eV a1  E g1 

exp

kT
I1



I2
 eVa 2  E g 2 

exp

kT


or
 eV a1  eV a 2  E g1  E g 2 
I1

 exp

I2
kT


We then have
 0.255  0.32  0.525  E g 2 
10  10 3

 exp
6

0.0259
10  10


or
 E g 2  0.59 

10 3  exp

 0.0259 
Then
E g 2  0.59  0.0259  ln 10 3
 
or
E g 2  0.769 eV
_______________________________________
8.21
(a) We have
 1
Dp
Dn
1
I S  Aeni2 

 N a  nO N d  pO
which can be written in the form
I S  C n i2



3
  Eg
 T 
 C N cO N O 
 exp
 300 
 kT




or
  Eg 

I S  CT 3 exp

 kT 
(b) Taking the ratio
  Eg

3 exp
I S 2  T2 
 kT2
   
I S 1  T1 
  Eg
exp
 kT1










 1
1 
  exp  E g 



 kT1 kT2 

1
For T1  300 K, kT1  0.0259 ,
 38.61
kT1
T
  2
 T1
3
For T2  400 K, kT2  0.03453 ,
1
 28.96
kT2
(i) Germanium: E g  0.66 eV
3
I S 2  400 

 exp0.6638.61  28.96
I S1  300 
I
or S 2  1383
I S1
(ii) Silicon: E g  1.12 eV
I S2
I S1
3
 400 
 exp1.12 38.61  28.96 
 
 300 
I S2
 1.17  10 5
I S1
_______________________________________
or
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 8
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
8.22
Plot
_______________________________________
 n2
  i
 Nd
8.23
First case:
If
V 
 exp a 
Is
 Vt 
Va
0.50
or Vt 

 0.05049 V
If
ln 2  10 4
ln
Is

 0.1N d2 
or V a  Vt ln 

2
 n i



or n i2  8.2519  10 27
Now
  Eg
n i2  N c N  exp
 kT



3
L p  D p po 
1010
7
  10
or L p  10  m;  W n  L p
(a) J p  x n  
eD p p no
Wn
V
exp a
 Vt




3
10 2
4
15

V
exp a
 Vt





V
 exp a
V

 t





AeDn n po
Ln
D n  n i2

 no  N a

 10 3 1.6  10 19


1.5  1010
25

7
5  10
2  10 17

2
 0.5516 
 exp

 0.0259 

  1.12 300  
 T 
2.8337  10 11  
 exp 

300


 0.0259T  
By trial and error,
T  502 K
The reverse-bias current is limiting factor.
_______________________________________
3




19
19


 1.12
 exp 




T
0
.
0259
300


8.24

V
 exp a
V

 t

3
In 






I p  4.565  10 3 A



 Ae
19

 0.5516 
 exp

 0.0259 
 
10
10  7




10 1.6 10 101.5 10 
0.7 10 2 10 

T 
8.2519 10   2.8 10 1.04 10  300

27
AeD p  n i2

Wn  N d
(ii) I p 
25
1

7
5  10
2  1017

V
 exp a
V

 t



1.2  10 6  5  10 4 1.6  10 19 n i2
 1

15
 4  10




 0.1 2 1015 2 
 0.0259 ln 

2
 1.5  1010

V a  0.5516 V

 T 
Now 0.05049  0.0259 

 300 
 T  584.8 K
Second case:
 1
Dp
Dn
1
I s  Aeni2 

 N a  no N d  po
V
(i) p n  x n   0.1N d  p no exp a
 Vt
I n  2.26  10 6 A
I  In  I p
 2.26  10 6  4.565  10 3
 4.567  10 3 A
or I  4.567 mA
V
(b) (i) n p  x p  0.1N a  n po exp a
 Vt


 n2
  i
 Na
cm
 0.1N a2 
or V a  Vt ln 

2
 n i



V
 exp a
V

 t


 0.1 2  1015 2 
 0.0259  ln 

2
 1.5  1010

V a  0.5516 V










lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 8
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
AeD p  n i2

Wn  N d
(ii) I p 


V
 exp a
V

 t

Then, from the first boundary condition, we
obtain
 V  
p no exp a   1
  Vt  




10 1.6 10 101.5 10 
0.7 10 2 10 
3
10 2
19
4
17
 0.5516 
exp

 0.0259 
I p  4.565  10 5 A
I n  Ae

 10
3
D n  ni2

 no  N a
1.6 10 
19

V
 exp a
V

 t






1.5  1010
25

7
5  10
2  10 15

2
I n  2.2597  10 4 A
I  In  I p
 2.2597  10 4  4.565  10 5
 2.716  10 4 A
or I  0.2716 mA
_______________________________________
8.25
(a) We can write for the n-region
d 2 p n  p n
 2 0
dx 2
Lp
The general solution is of the form


x
  B exp  x 
p n  A exp
 Lp 
 Lp 




The boundary condition at x  x n gives
 Va  
  1

 Vt  
p n x n   p no exp





We then obtain
 0.5516 
 exp

 0.0259 

  xn
  x n  2W n  
  B exp 
  B exp
Lp


 Lp
  2W n 
  x 

 B exp n  1  exp
 L p 
 L p 






  xn 
  B exp  x n 
 A exp
 Lp 
 Lp 




and the boundary condition at x  x n  Wn
gives
p n x n  W n   0


 x  Wn 
  B exp   x n  W n  
 A exp n


 Lp 
Lp




From this equation, we have
  2x n  W n  
A   B exp 

Lp


 V  
p no exp a   1
  Vt  
B
  2W n
  x 
exp n  1  exp
 Lp
 L p 



which can be written as




 x  Wn 
 V  
p no exp a   1  exp  n

  Vt  
 L p 
B
W 
  Wn 

exp n   exp
 Lp 
 Lp 




We can also find
  x n  W n  
 V  
 p no exp a   1  exp 

Lp
  Vt  


A
W 
  Wn 

exp n   exp
 Lp 
 Lp 




The solution can now be written as
 V  
p no exp a   1
  Vt  
p n 
W 
2 sinh  n 
 Lp 


  x  W n  x 
   x n  W n  x   
 exp  n
  exp 

Lp
Lp
 


 
or finally
 x  Wn  x 

sinh  n


Lp
  Va  


p n  p no exp   1 
W 
  Vt  
sinh  n 
 Lp 


lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 8
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(b)
d p n 
J p  eD p
dx
x  xn
 V  
 eD p p no exp a   1
  Vt  
=
W 
sinh  n 
 Lp 


 1 


 cosh  x n  W n  x 

 Lp 

 x  xn
Lp




For T  310 K ,
1.12  0.60 1.12  V D 2

0.0259
0.02676
which yields
V D 2  0.5827 V
For T  320 K ,
1.12  0.60 1.12  V D 3

0.0259
0.02763
which yields
V D 3  0.5653 V
_______________________________________
Then
W   V  
coth n   exp a   1
 Lp  
Lp
   Vt  

_______________________________________
Jp 
eD p p no
8.26
V 
I D  n i2 exp D 
 Vt 
For the temperature range 300  T  320 K,
neglect the change in N c and N  .
Then
  Eg 
 eV 
  exp D 
I D  exp

 kT 
 kT 


  E g  eV D 
 exp 

kT


Taking the ratio of currents, but maintaining
I D a constant, we have


  E g  eV D1 
exp 

kT1


1
  E g  eV D 2 
exp 

kT2


We then have
E g  eV D1 E g  eV D 2

kT1
kT2
We have
T  300 K , V D1  0.60 V and

kT1  0.0259 eV,

kT1
 0.0259 V
e
T  310 K ,
kT2  0.02676 eV,
kT2
 0.02676 V
e
T  320 K ,
kT3  0.02763 eV,
kT3
 0.02763 V
e
8.27
(a) We can write
 eV 
I D  C  n i2  exp a 
 kT 
where C is a constant, independent of
temperature.
As a first approximation, neglect the
variation of N c and N  with temperature
over the range of interest. We can then write
  Eg 
V 
  exp a 
I D  C1  exp
V 

kT
 t 


E
eV



g
a 
 C1  exp 

kT


where C1 is another constant, independent of
temperature. We find
E g  eV a
C 
 ln 1 
kT
 ID 


or
 kT   C1 

Va  E g  
  ln
 e   ID 
_______________________________________
8.28
 1
(a) I s  Aeni2 
 N a


Dn
 n0

 10 4 1.6  10 19 1.5  1010
 1

16
 4  10
Aeni W
2 0
We find

1
Nd
Dp 

 p0 

2
25
1

10  7 4  10 16
I s  2.323  10 15 A
(b) I gen 

10 

10  7 
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 8
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Aeni W
I s  I gen 
16
16
 4  10 4  10 
2 0
Vbi  0.0259  ln 

4
10 2
10 1.6  10 19 4.734  1014 6.109  10 5
 1.5  10


2 10  7
 0.7665 V
and
Then
1/ 2
I s  I gen  2.314  10 6 A
 2 s Vbi  V R   N a  N d 


W 
 N N 
or I s  I gen  2.314  A
e

a
d 

(b) From Problem 8.28
 211.7  8.85  10 14 0.7665  5

I s  2.323  10 15 A
19

1
.
6
10

I gen  7.331 10 11 A
1/ 2
 4  10 16  4  10 16  

V 
V 
16
16  
So I  I s exp a   I gen exp a 
 4  10 4  10  
 Vt 
 2Vt 
W  6.109  10 5 cm
V 
Then
2.323  10 15 exp a 
 Vt 
10 4 1.6  10 19 1.5  10 10 6.109  10 5
I gen 
7
V 
2 10
 7.331 10 11 exp a 
11
 2Vt 
 7.331 10 A

11
I gen 7.331 10
V 
(c)

 3.16  10 4
exp a 
15
11
Is
2.323  10
 Vt  7.331 10

_______________________________________
 V  2.323  10 15
exp a 
 2Vt 
8.29
(a) Set I S  I gen ,
V 
exp a   3.1558  10 4
 1
 2Vt 
D p  Aeni W
Dn
1


Aeni2 
V a  2Vt ln 3.1558  10 4
2 0
 N a  n 0 N d  p 0 
 0.5366 V
 1
25
1
10 
_______________________________________
ni 


16
10  7 4  10 16 10  7 
 4  10
8.30
W
6.109  10 5


 kT 
Dn  
   n  0.0259 5500 
2 0
2 10  7
 e 
2
3.0545  10
so n i 
 142.5 cm 2 /s
3.9528  10 13  2.50  10 13
D p  0.0259 220   5.70 cm 2 /s
 4.734  10 14 cm 3
Then
(a)
n i2  2.2407  10 29
 1
Dp 
Dn
1
3

(i) I s  Aeni2 

19
19  T 
 2.8  10 1.04  10 

 N a  n 0 N d  p 0 
 300 
2
3
 2  10 4 1.6  10 19 1.8  10 6
  1.12 300  
 T 
10
7.6947  10  
 exp 

 1
142.5
1
5.70 
 300 
 0.0259 T  



16
8
16
By trial and error,
2  10
7  10
2  10 8 
 7  10
T  567 K
I s  1.50  10 22 A
We have







































lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 8
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
V
(ii) I D  I s exp a
 Vt





8.31
Using results from Problem 8.30, we find
V a  0.4 V, I d  7.64  10 16 A,

 0.6 
 1.50  10  22 exp

 0.0259 
I rec  1.35  10 10 A, I T  1.35  10 10 A
V a  0.6 V, I d  1.73  10 12 A
 1.726  10 12 A

I rec  6.44  10 9 A, I T  6.44  10 9 A

 0.8 
(iii) I D  1.50  10  22 exp

 0.0259 
V a  0.8 V, I d  3.90  10 9 A
I rec  3.06  10 7 A, I T  3.10  10 7 A
 3.896  10 9 A


V a  1.0 V, I d  8.80  10 6 A
 1.0 
(iv) I D  1.50  10  22 exp

 0.0259 
I rec  1.45  10 5 A, I T  2.33  10 5 A
6
 8.795  10 A
Aeni W

2 0
(b) I gen

V a  1.2 V. I d  1.99  10 2 A

I rec  6.90  10 4 A, I T  2.06  10 2 A
_______________________________________

 7  10 16 7  10 16 
Vbi  0.0259  ln 

2


1.8  10 6
 1.263 V
 213.1 8.85  10 14 1.263  3
W 
1.6  10 19




8.32
Plot
_______________________________________

 7  1016  7  1016

16
16
 7  10 7  10


 

 
8.33
Plot
_______________________________________
1/ 2

8.34
We have that
5
 4.201 10 cm
(i)Then
I gen 
2 10 1.6 10 1.8 10 4.20110 
22  10 
4
19
 6.049  10
14
A
V
(ii) I rec  I ro exp a
 2Vt
(iii) I rec
6
8








 0.6 

 6  10 14 exp
 20.0259  
 6.436  10 9 A
 0.8 

 6  10 14 exp
 20.0259  
 3.058  10 7 A
 1.0


(iv) I rec  6  10 14 exp
 20.0259  
 1.453  10 5 A
_______________________________________


5
R
np  n i2
 pO n  n    nO  p  p 
Let  pO   nO   O and n   p   n i
We can write
 E  E Fi 

n  ni exp Fn
kT


and
 E Fi  E Fp 

p  n i exp

kT


We also have
E Fn  E Fi   E Fi  E Fp  eVa


so that
E Fi  E Fp  eVa  E Fn  E Fi 


Then
 eV  E Fn  E Fi  
p  n i exp  a

kT


  E Fn  E Fi  
 eV 
 ni exp a   exp 

kT
kT




lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 8
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Define
eV
 E  E Fi 
 a  a and    Fn

kT
kT


Then the recombination rate can be written as
n i e  n i e  a  e   n i2
R
 O n i e  ni  ni e a  e   n i
or
ni e a  1
R
 O 2  e  e a  e 
To find the maximum recombination rate, set
dR
0
d






or
0





  d 2  e
d
ni e a  1
O

   12  e
n i e a  1
O

 e  a  e 

 e a  e 

1

 e  e
2  e  e
O
a


e
a


e
e
a
 

or
R max 

0
In this case, G  g   4 1019 cm 3 s 1 and is
a constant through the space charge region.
Then
J gen  eg W




 O 2  e a 2  e a

n i e a  1
2






or

 2

ni e a  1




 5  1015 5  1015
 0.0259  ln 
2
 1.5  1010
Vbi  0.659 V
Also
 2  V  V R   N a  N d

W   s bi
 N N
e
a
d









1/ 2
 211.7  8.85  10 14 0.659  10 

1.6  10 19

a


e

N N
Vbi  Vt ln a 2 d
 ni
2
Then the maximum recombination rate
becomes
n i e a  1
R max 
 O 2  e a 2  e a  e  a 2

W
J gen  eGdx
We find
e
The denominator is not zero, so we have
0  e   e  a  e 
or
2
8.35
We have
2

ni
 eV 
exp a 
2 O
 2kT 
Q.E.D.
_______________________________________

 e  e a  e 
which simplifies to
 ni ea  1
0

R max 
 5  10 15  5  10 15
 
15
15
 5  10 5  10







1/ 2
or
W  2.35  10 4 cm
Then
J gen  1.6  10 19 4  10 19 2.35  10 4




or

2 O  a 2  1
which can be written as
  eV  
n i exp a   1
  kT  
R max 
  eV  
2 O exp a   1
  2kT  
J gen  1.5  10 3 A/cm 2
_______________________________________
If V a  kT e  , then we can neglect the (-1)
term in the numerator and the (+1) term in the
denominator, so we finally have
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 8
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
8.36
8.38
 1
J S  eni2 
 N a

Dn
 nO


 1.6  10 19 1.5  10
1
Nd
Dp 

 pO 


 1

16
 3  10
10 2

1
10 18
(a) C d 

18
10  7
6 

10  7 
J S  1.638  10 11 A/cm 2
Now
V 
J D  J S exp D 
 Vt 
We want
J  0  JG  J D
or
V 
0  25  10 3  1.638  10 11 exp D 
 Vt 
which can be written as
V 
25  10 3
exp D  
 1.526  10 9
11
V
1
.
638
10

 t 
We find
V D  Vt ln 1.526  10 9
or
V D  0.548 V
_______________________________________

8.37
(a) rd 
Cd 
Vt
0.0259

 21.6 
I DQ 1.2  10 3
I DQ 0
2Vt

1.2 10 0.5 10 
3
6
20.0259 
8
 1.16  10 F
or C d  11.6 nF
0.0259
 216 
0.12  10 3
0.12  10 3 0.5  10 6
Cd 
20.0259 




10
 5.79  10 C
(b) For I D  0.12 mA

Q  C d  V  1.158  10 9 50  10 3
11
 5.79  10 C
_______________________________________
8.39
For a p  n diode
I DQ
I DQ pO
gd 
, Cd 
2Vt
Vt
Now
10 3
gd 
 3.86  10  2 S
0.0259
and
10 3 10 7
Cd 
 1.93  10 9 F
20.0259 
We have
g  jC d
1
1
 d2
Z 
Y g d  jC d
g d   2 C d2



where   2 f
We obtain
f  10 kHz , Z  25.9  j 0.0814
f  100 kHz , Z  25.9  j 0.814
f  1 MHz , Z  23.6  j 7.41
f  10 MHz , Z  2.38  j 7.49
_______________________________________
8.40
Reverse bias
1/ 2
(b) rd 


Q  C d  V  1.158  10 8 50  10 3
or

Q
, For I D  1.2 mA
V

 1.16  10 9 F
or C d  1.16 nF
_______________________________________


e s N a N d
C j  AC   A  

 2Vbi  V R N a  N d  
 5  10 17 8  10 15 
Vbi  0.0259  ln 

2
 1.5  10 10

 0.790 V
 1.6  10 19 11.7  8.85  10 14
C j  2  10  4 
2Vbi  V R 












 
 5  10 17 8  10 15

17
15
 5  10  8  10

 

1/ 2
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 8
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Cj 
5.1078  10 12
Vbi  V r
V R (V)
 pO  20.02592.5  10 6 
F
or
 pO  1.3 10 7 s
C j (pF)
10
5
3
1
0
0.20
0.40
At 1 mA,
C d  2.5  10 6 10 3
or
C d  2.5  10 9 F
_______________________________________

1.555
2.123
2.624
3.818
5.747
6.650
8.179
Then
I po p 0 I po 8  10 8

 1.544  10  6 I po
Cd 
2Vt
20.0259 
I po
 
D p  n i2
 Ae

 p 0  N d

 2  10
4


V
  exp a
V

 t

1.6 10 






V
 exp a
 Vt

I po  1.006  10
V a (V)
0.20
0.40
0.60
C d (F)
14

V
exp a
 Vt
+
2




C j (F)
= C Total (F)
8.41
For a p  n diode, I pO  I nO , then
 pO
2Vt
Then


 2.5  10  6 F/A
2Vt
2Vt C d 
 p0


20.0259  10 9
10  7

 5.18  10 4 A
 0.518 mA
(ii) I po  Ae
Dp

 p0
V 
n i2
exp a 
Nd
 Vt 


0.518  10 3  5  10  4 1.6  10 19

 1.79  10 10
_______________________________________

 I pO pO


or I po 
I po p 0

 A


3.51 10 17  6.650  10 12
 6.650  10 12
14
7.92  10  8.179  10 12
 8.258  10 12
1.79  10 10  ...
 1
C d  
 2Vt
Now
(i) C d 
or I po
1.5  10 10
10

8  10 8
8  10 15
19

8.42
(a) N a  N d  I po  I no
Forward bias
For N a  N d  I po  I no


1.5 10 
10 2
8  10
15

10
10  7
V
exp a
 Vt
 0.518  10 3 

V a  0.0259  ln
14 
 2.25  10 
 0.618 V
V
0.0259
(iii) rd  t 
 50 
I D 0.518  10 3
(b)
2V C  20.0259  0.25  10 9
(i) I po  t d 
 p0
10  7






 1.295  10 4 A
or I po  0.1295 mA
 0.1295  10 3 

(ii) V a  0.0259  ln
14 
 2.25  10

 0.5821 V
0.0259
(iii) rd 
 200 
0.1295  10 3
_______________________________________
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 8
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(b) Set R  0
(i) For I D  1 mA,
8.43
(a) p-region:
pL
L
L
Rp 


A
 p A e p N a A

 10 3
V  0.0259  ln 10
 10
or V  0.417 V

so
Rp 
0.2
480 1016 10  2
1.6 10 
19
 

(ii) For I D  10 mA,
or
 10 2 
V  0.0259  ln 10 
 10 
or V  0.477 V
_______________________________________
R p  26 
n-region:
Rn 
n L
A

L
nA





L
e n N d A
so
8.45
Rn 
0.10
1350 1015 10 2
1.6 10 
19
 

(a) rd 
or
or I D  8.09375  10 4 A
R n  46.3 
The total resistance is
R  R p  R n  26  46.3
I 
V a  Vt ln D 
 Is 
 8.09375  10 4
 0.0259  ln
12
 5  10
V a  0.4896 V
or
R  72.3 
(b)
V  IR  0.1  I 72.3
which yields
I  1.38 mA
_______________________________________
8.44
R

 n Ln 
An 
0.210

2
2  10 5
 p L p 
A p 
  0.110 
2
 1
1 dI D

 I S 
rd dV a
 Vt
V

 exp a
V

 t





or
I 
V  I D R  Vt ln D 
 IS 
(a) (i) For I D  1 mA,
 10 3
V  10 3 150   0.0259  ln 10
 10
or V  0.567 V
(ii) For I D  10 mA,




 10 2
V  10  2 150   0.0259  ln 10
 10
or V  1.98 V






Vt 0.0259

 4.3167  10  4 A
rd
60
 4.3167  10 4 

V a  0.0259  ln
12


 5  10
 0.4733 V
_______________________________________
(a)
R  150 
We can write

(b) I D 




8.46
2  10 5
or

Vt
V
0.0259
 ID  t 
ID
rd
32
1
10 13
 0.020 

 exp

rd 0.0259
 0.0259 
which yields
rd  1.2  10 11 
(b)
1
10 13
  0.02 

 exp

rd 0.0259
 0.0259 
which yields
rd  5.6  10 11 
_______________________________________
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 8
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
8.47
8.49
C j  18 pF at V R  0
I
(a) If R  0.2
IF
Then we have
ts
erf
 pO
C j  4.2 pF at V R  10 V
IF


IF  IR
1
1

I R 1  0.2
1
IF
We have
 nO   pO  10 7 s , I F  2 mA
and
IR 
or
ts
erf
 pO
So
 0.833
 I 
 2
t s   pO ln1  F   10  7 ln1  
I
 1
R 


We find
ts
 pO
(b) If
 0.978 
ts
 pO
 0.956
IR
 1.0 , then
IF
erf
ts
 pO

V R 10

 1 mA
R 10
1
 0.50
11

or
t s  1.1 10 7 s
Also
18  4.2
C avg 
 11.1 pF
2
The time constant is
 S  RC avg  10 4 11.110 12
 

7
 1.11 10 s
Now, the turn-off time is
t off  t s   S  1.1  1.11  10 7
which yields
ts
 0.228
 pO
_______________________________________
8.48
or
t off  2.21 10 7 s
_______________________________________
(a) erf
ts
p
IF

IF  IR
8.50
erf 0.3 = erf 0.5477 
 erf 0.55  0.56332
Then 0.56332 
1
I
1 R
IF
(b) erf
t2
 p0

 t
 2
  p0

By trial and error,
 p0


1/ 2





I 
 1  0.1 R 
 IF 




 1 0.10.775
 1.0775
t2

 2  V  V a   N a  N d 


W   s bi
 N N 
e

a
d 


 211.7  8.85  10 14 1.136  0.40

1.6  10 19

IR
1

 1  0.775
I F 0.56332
 t
exp 2
  p0


 5  1019 2 
Vbi  0.0259 ln 
  1.136 V
2
 1.5  1010 
We find
 0.80

 5  1019  5  10 19

2

5  10 19







1/ 2
which yields
o
W  6.17  10 7 cm  61.7 A
_______________________________________
8.51
Sketch
_______________________________________
_______________________________________
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 8
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
8.53
From Figure 7.15, N d  9 1015 cm 3
Let N a  5 10 17 cm 3
V
p n x n   0.1N d  9  1014  p no exp a
 Vt
p no 

n i2
1.5  1010

Nd
9  1015





2
 2.5  10 4 cm 3
 9  1014
Then V a  0.0259 ln
4
 2.5  10
I
50  10 3
Is 

 0.6295 
V 

exp a  exp
 0.0259 
 Vt 

  0.6295 V


 1.389  10 12 A
Is 
AeD p p no
Lp
1.389  10 12

Aeni2
Nd



Dp
 p0
A 1.6  10 19 1.5  10 10
9  10 15


2
1.389  10 12  A 2.828  10 11
2

10
2  10  7
or A  4.91 10 cm
_______________________________________
2
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 9
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Chapter 9
9.1
9.2
(a) Vbi   B 0   n
(a) We have
N
e n  eVt ln c
 Nd




N 
 n  Vt ln c 
N
 2.8  10 19
 0.0259  ln
16
 10

  0.206 eV


(c)
 BO   m    4.28  4.01
or
and
Vbi   BO   n  0.27  0.206
or
Vbi  0.064 V
Also
1/ 2


 211.7  8.85  10 14 0.064  


1.6  10 19 1016



1/ 2
 
or
x d  9.1 10 6 cm
Then
eN d x d
 max 
s

 2.8  1019 

 0.0259 ln
15 
 5  10 
=0.2235 V
Vbi  0.65  0.2235  0.4265 V
 2.8  1019 

(c)  n  0.0259  ln
15


 10
 0.2652 V
Vbi  0.65  0.2652  0.3848 V
Vbi decreases,  B 0 remains constant
_______________________________________
9.3
(a)  B 0   m    5.1  4.01  1.09 V
(b) Vbi   B 0   n
1.6 10 10 9.1`10 

11.7 8.85 10 
19
d
 2.8  1019 

(b)  n  0.0259  ln
16

 10

 0.2056 V
Vbi  0.65  0.2056  0.4444 V
Vbi increases,  B 0 remains constant
 BO  0.27 V
2 V 
x d   s bi 
 eN d 

6
16
14
or
 max  1.41 10 4 V/cm
(d)
Using the figure,  Bn  0.55 V
So
Vbi   Bn   n  0.55  0.206
or
Vbi  0.344 V
We then find
x n  2.11 10 5 cm
and
 max  3.26  10 4 V/cm
_______________________________________
N 
 n  Vt ln c 
 Nd 
 2.8  10 19 
  0.2056 V
 0.0259  ln
16

 10

Vbi  1.09  0.2056  0.8844 V
 2  V  V R  
(c) x n   s bi

eN d


1/ 2



 
 211.7  8.85  10 14 0.8844  1 
(i) x n  

1.6  10 19 10 16


 4.939 10 5 cm
or x n  0.4939  m
 max 

eN d x n
s
1.6 10 10 4.939 10 
11.7 8.85 10 
19
 7.63  10 4 V/cm
5
16
14
1/ 2
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 9
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________



 
 211.7  8.85  10 14 0.8844  5 
(ii) x n  

1.6  10 19 1016



5

1/ 2

 1.292 10 cm
or x n  1.292  m
1.6 10 10 8.727 10 
11.7 8.85 10 
19

4
 8.727  10 cm
or x n  0.8728  m
 max 

 213.1 8.85  10 14 0.7623  5 
(ii) x n  

1.6 10 19 5 1015


1/ 2
5
16
 max 
14
1.6 10 5 10 1.292 10 
13.18.85 10 
19
4
15
14
 1.35 10 5 V/cm
_______________________________________
 8.92 10 4 V/cm
_______________________________________
9.4
(a)  B 0   m    5.1  4.07  1.03 V
9.6
1/ 2
 4.7  1017 
  0.1177 V
(b)  n  0.0259 ln
15 
 5 10 
(c) Vbi  1.03  0.1177  0.9123 V
(d)


 213.1 8.85 10 14 0.9123  1 
(i) x n  

1.6  10 19 5  1015





(i)
1.6 10 5 10 7.445 10 
13.18.85 10 
19
5
15
14
 5.14  10 V/cm
4


 213.1 8.85  10 14 0.9123  5 
(ii) x n  

1.6 10 19 5 1015



 2.8  1019 
  0.265 V
15


 10
Vbi  0.88  0.265  0.615 V
 n  0.0259 ln
1/ 2
 7.445  10 5 cm
or x n  0.7445  m
 max 
 e s N d 
(a) C   

 2Vbi  V R  
We have  B 0  0.88 V

1/ 2


C  10
 max 

C  10
1.6 10 5 10 1.309 10 
13.18.85 10 
19
4
15
14
 9.03  10 4 V/cm
_______________________________________
 213.1 8.85  10 14 0.7623  1 
(i) x n  

1.6  10 19 5  10 15




 max 
1/ 2

 7.147  10 5 cm
or x n  0.7147  m
1.6 10 5 10 7.147 10 
13.18.85 10 
19
5
15
14

 
1/ 2



 
1/ 2
 1.6  10 19 11.7  8.85  10 14 1015 


20.615  5


 3.84  10 13 F
or C  0.384 pF
 2.8 1019 
  0.206 V
(b)  n  0.0259 ln
16

 10

Vbi  0.88  0.206  0.674 V
(i)

(b)  n  0.1177 V
(c) Vbi  0.88  0.1177  0.7623 V
(d)

4




 
1/ 2


 
1/ 2
 1.6  10 19 11.7  8.85  10 14 1016 
C  10  4 

20.674  1


9.5


 1.6  10 19 11.7  8.85 10 14 1015 


20.615  1


 7.16 10 13 F
or C  0.716 pF
(ii)
4
 1.309  10 cm
or x n  1.309  m
4
 2.22  10 12 F
or C  2.22 pF
(ii)

 1.6  10 19 11.7  8.85  10 14 1016 
C  10  4 

20.6745  5



 1.21 10 12 F
or C  1.21 pF
_______________________________________
 4.93 10 4 V/cm
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 9
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
9.7
 max 
(a) From the figure, Vbi  0.90 V
(b) We find




 4.7  10
 0.0259 ln
16
 1.04  10
or
14
e
4 s
(i)  




 1.6  10 19 4.66  10 4 

14 
 4 11.7  8.85  10

V
 0.0239

xm 

1/ 2
e
16 s 





1.6  10 19

14
4 


16
11
.
7
8
.
85
10
4
.
66
10






1/ 2

or
x m  2.57  10 7 cm




 1.6  10 19 9.14  10 4 
(ii)   
14 

 4 11.7  8.85  10
 0.0335 V
 n  0.0986 V

1/ 2


1.6 10 19
xm  
4 
14
9.14 10 
16 11.7  8.85  10
(d)

 Bn  Vbi   n  0.90  0.0986
or
4
15
(b)




17
19
 9.14 10 4 V/cm
2
 1 
 
3  10 15  0
 C 
 1.034  1015

2   0.90 
V R
and
2
1.034  1015 
e s N d
We can then write
2
Nd 
1.6  10 19 13.1 8.85  10 14 1.034  1015
or
N d  1.04  10 16 cm 3
(c)
N 
 n  Vt ln c 
 Nd 
1.6 10 5 10 1.183 10 
11.7 8.85 10 


7
 1.83 10 cm
_______________________________________
 Bn  0.9986 V
_______________________________________
9.9
9.8
We have
From Figure 9.5,  BO  0.63 V
 2.8  10 
  0.224 V
(a)  n  0.0259 ln
15 
 5  10 
Vbi   B 0   n  0.63  0.224  0.406 V
19


 211.7  8.85 10 14 0.406  1 
(i) x n  

1.6  10 19 5  1015




1/ 2

 6.033  10 5 cm
or x n  0.6033  m
 max 
1.6 10 5 10 6.033 10 
11.7 8.85 10 
19
5
15
14
 4.66 10 4 V/cm


 211.7  8.85 10 0.406  5 
(ii) x n  

1.6 10 19 5  1015



1/ 2
 1.183 10 4 cm
or x n  1.183  m
14


1/ 2
  x  
e
 x
16 s x
e x  
e2
 ex
16 s x
or
Now
d e  x 
 e2
 e
0
dx
16 s x 2
Solving for x, we find
e
x  xm 
16 s 
Substituting this value of x  x m into the
equation for the potential, we find
e
e

 
16

s 
e
16 s
16 s 
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 9
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
which yields
(b) We have
E g  e O  e Bn
e
 
1
4 s

2e s N d  Bn   n 
eDit
_______________________________________


9.10
From Figure 9.5,  BO  0.88 V
 4.7  1017 
  0.0997 V
(a)  n  0.0259 ln
16

 10

Vbi   B 0   n  0.88  0.0997  0.780 V


 213.1 8.85  10 14 0.780 
xn  

1.6  10 19 1016




1/ 2
1.6 10 10 3.362 10 
13.18.85 10 
1.6 10 
4 13.18.85  10 
0.044 4 13.18.85 10 

19
0.0442 
14
14
2
1.6  10
 1.763  10 V/cm
19
5
1.6 10 10 x
13.18.85 10 
19
16
n
14
 x n  1.277  10 4 cm
And

x n2  1.277 10 4

14
1/ 2

2


1.6 10 10 
213.1 8.85 10 14 0.780  V R 
19
16
 V R  10.5 V
_______________________________________
9.11
Plot
_______________________________________
9.12
(a)  BO   m    5.2  4.07
or
 BO  1.13 V
Bn
13

or
8

0.83   Bn
 0.038  Bn  0.10  0.2211.13   Bn 
e
4 s
  1.763  10 5 




 

8.85 10  5.2  4.07   

 10 
25  10 
e
e 
5
16
14
Now
21.6 10 13.18.85 10 
19
 1013
e
 e
14
 4.64  10 4 V/cm
(b)   0.050.88  0.044 V

1
 10 16  Bn  0.10 
 3.362  10 5 cm
or x n  0.3362  m
 max 
i
 m     Bn 
eDit 
which becomes
e1.43  0.60   Bn 
 
19

We find
 Bn  0.858 V
(c)
If  m  4.5 V, then
 BO   m    4.5  4.07
or
 BO  0.43 V
From part (b), we have
0.83   Bn
 0.038  Bn  0.10  0.2214.5  4.07   Bn 
We then find
 Bn  0.733 V
With interface states, the barrier height is less
sensitive to the metal work function.
_______________________________________
9.13
We have that
E g  e O  e Bn


1
eDit

2e s N d  Bn   n 

i
 m     Bn 
eDit 
Let eDit  Dit (cm 2 eV 1 )
Then we can write
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 9
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
e1.12  0.230  0.60 
 350 
(b) kT  0.0259 
  0.030217 eV
1
300 

19
14

2 1.6  10 11.7  8.85  10
Dit
  0.63 
2
I sT  10  4 120350 exp

1/ 2
16
 0.030217 
 5  10 0.60  0.164 






8.85 10  4.75  4.01  0.60
D  20  10 
14
8
it
We then find
Dit  4.97  1011 cm 2 eV 1
_______________________________________
9.14
 2.8  1019 
  0.224 V
(a)  n  0.0259  ln
15 
 5  10 
(b) Vbi   Bn   n  0.89  0.224  0.666 V
  e Bn 
(c) J sT  A T 2 exp

 kT 
  0.89 
2
 120 300  exp

 0.0259 
J sT  1.29 10 8 A/cm 2
 J 
5

  0.0259 ln
V a  Vt ln

8

 1.29 10 
 J sT 
V a  0.512 V
_______________________________________
9.15
(a)  B 0  0.63 V
  0.63 
2
J sT  120300 exp

 0.0259 
 2.948  10 4 A/cm 2
 10 4 2.948  10 4  2.948  10 8 A


 I
(i) V a  Vt ln
 I sT

 1.296  10 6 A
 V  
(i) I  I sT exp a   1
  Vt  

 10  10 6
 1
V a  0.030217  ln 
6

1.296  10
 0.0654 V
 100  10 6

 1
(ii) V a  0.030217  ln 
6
1
.
296
10



 0.1317 V


10 3

(iii) V a  0.030217  ln
6 
1
.
296
10



 0.201 V
_______________________________________
9.16
(a)  Bn  0.88 V
(d)
I sT








 10  10 6 

 0.0259  ln
8 
 2.948 10 
 0.151 V
 100  10 6 

(ii) V a  0.0259 ln
8 
 2.948  10 
 0.211 V


10 3

(iii) V a  0.0259 ln
8 
 2.948  10 
 0.270 V
  0.88 
2
(b) J sT  1.12300  exp

 0.0259 
 1.768  10 10 A/cm 2
10


(c) V a  0.0259 ln

10
1
.
768

10


 0.641 V
(d) V a  Vt ln2  0.0259 ln2
 0.0180 V
_______________________________________
9.17
Plot
_______________________________________
9.18
From the figure,  Bn  0.68 V
 
J ST  A *T 2 exp Bn
 Vt
  


  exp
V 

 t 

  
  0.68 
2

 120 300 exp
  exp

 0.0259 
 Vt 
or
  

J ST  4.277  10 5 exp

 Vt 
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 9
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
We have
Also
e
 
1.6  10 19 1016 0.761 10 4
4 s
 max 
11.7  8.85 10 14
Now
or
N 
 n  Vt ln c 
 max  1.176  10 5 V/cm
 Nd 
and
 2.8  1019 
1/ 2
  0.2056 V
 0.0259  ln
 1.6  10 19 1.176  10 5 
16




10




14

 4 11.7  8.85  10
and
or
Vbi   Bn   n  0.68  0.2056  0.4744 V
  0.03803 V
(a) We find for V R  2 V,
Then
1/ 2
 2 s Vbi  V R  
 0.03803 
xd  
J ST 2  4.277  10 5 exp


eN d
 0.0259 


1/ 2
or
 211.7  8.85  10 14 2.4744  


J ST 2  1.86  10 4 A/cm 2
1.6  10 19 10 16


Finally,
or
I R 2  1.86  10 8 A
x d  0.566  10 4 cm  0.566  m
_______________________________________
Then
eN d x d
9.19
 max 
s
We have that

 








1.6 10 10 0.566 10 
11.7 8.85 10 
19
4
16
14


 1.6  10 19 8.745  10 4 
  

14

 4 11.7  8.85  10


1/ 2

 0.0328 
J ST 1  4.277  10 5 exp

 0.0259 

 211.7  8.85  10 4.4744 
xd  

1.6  10 19 1016




4 2m n*

3/ 2

4 2m n*

3/ 2
E  Ec
h3
  E  E F  
 exp 
 dE
kT


If the energy above E c is kinetic energy, then
J ST 1  1.52  10 4 A/cm 2
14

Ec
dn 
or
For A  10 4 cm 2 , we find
I R1  1.52  10 8 A
(b) For V R  4 V, then

E  Ec
h3
and assuming the Boltzmann approximation
  E  E F  
f F E   exp 

kT


Then
  0.0328 V
Then


J s m   x dn
g c E  
or


The incremental electron concentration is
dn  g c E  f F E dE
where
 max  8.745  10 4 V/cm



 
or
Now


 
1/ 2
1 * 2
m n  E  E c
2
We can then write
E  Ec  
or
x d  0.761 10 4 cm  0.761  m
and
m n*
2
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 9
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
1/ 2
1
m n* y2
 2kT 
dE  m n*  2d  m n*d
2







y
2
 m*   
2kT
 n 
We can also write
1/ 2
*
2
E  E F   E  E c   E c  E F 
 2kT 
m n z
  2   z   *   
1
2kT
 mn 
 m n* 2  e n
2
Substituting the new variables, we have
so that
3
2
3
 m *   2kT 
  e n 
 m n* 
 e n 
J s m  2 n    *   exp




dn  2
exp

h   mn 
kT 




 kT 
 h 

  eVbi  V a  
2
  m n* 2 
 exp 
   exp   d
  4  2 d
 exp
kT
 0


 2kT 


We can write
 exp   2 d  exp   2 d
 2   x2   y2   z2


_______________________________________
The differential volume element is


  


 
4  2 d  d x d y d z
The current is due to all x-directed velocities
that are greater than  Ox and for all y- and
z- directed velocities. Then
J

s m
 m*
 2 n
 h
3

 e n
 exp

 kT





  m n* x2
  x exp
 2kT
 Ox

d x




  m n* y2
 exp
 2kT




d
 y


  m n* z2
 exp
 2kT



d z


We can write
1 * 2
m n Ox  eVbi  V a 
2
Make a change of variables:
m n* x2
2Vbi  V a 
2 
2kT
kT
or
eV  V a  
2kT 
 x2  *  2  bi

kT
mn 

Taking the differential, we find
 2kT 
 x d x   * d
 mn 
We may note that when  x   Ox ,   0 .
We may define other change of variables,
9.20
For the Schottky diode,
V
0.80  10 3  10  4 6 10 8 exp a
 Vt







 0.80  10 3 
(a) V a SB   0.0259 ln   4
8 
 10 6  10 
 0.4845 V
Then
V a  pn   0.4845  0.285  0.7695 V





 0.7695 
(b) 0.80  10 3  A pn 10 11 exp

 0.0259 
 A pn  0.998 10 5  10 5 cm 2
_______________________________________
9.21
For the pn junction,
I s  8  10  4 8  10  13  6.4  10  16 A



 150  10 6
(a) V a  0.0259 ln
16
 6.4 10
 0.678 V
 700 10 6
(b) V a  0.0259 ln
16
 6.4 10
 0.718 V
 1.2  10 3
(c) V a  0.0259 ln
16
 6.4 10
 0.732 V












lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 9
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
For the Schottky junction,
9.23
4
9
12
(a) For I  0.8 mA, we find
I sT  8  10 6  10  4.8  10 A
0.8  10 3
 150  10 6 

 1.143 A/cm 2
J

(a) V a  0.0259 ln
7  10  4
12 
 4.8 10 
We have
 0.447 V
 J 

V a  Vt ln
 700 10 6 


(b) V a  0.0259 ln
 JS 
12 
 4.8 10 
For the pn junction diode,
 0.487 V
 1.143 
V a  0.0259 ln
  0.6907 V
12
 1.2  10 3 
 3  10 

(c) V a  0.0259 ln
12 
 4.8 10 
For the Schottky diode,
 0.501 V
 1.143 
V a  0.0259 ln
  0.4447 V
8
_______________________________________
 4  10 
(b) For the pn junction diode,
9.22
3
  Eg 
 T 
(a) (i) I  0.80 mA in each diode

J S  n i2  
 exp

(ii)
 300 
 kT 
3


0.8  10
Then
V a SB   0.0259  ln 
4
9 
3
J S 400   400 
 8  10 6  10 


 0.490 V
J S 300   300 


0.8  10 3
 Eg
Eg 

V a  pn   0.0259 ln 
4
13 
 exp 



 8  10 8  10
 0.0259 400 300  0.0259 
 0.721 V
1.12 
 1.12
(b) Same voltage across each diode
 2.37 exp 

0
.
0259
0
.
03453 

I  0.8  10 3  I SB  I pn
or
 Va 
4
9
J S 400 
 8  10 6  10 exp 
 1.17  10 5
V
t
 
J S 300 
Now
V 
 8  10  4 8  10  13 exp a 
I  7  10 4 1.17  10 5 3  10 12
 Vt 















V
 4.8  10 12  6.4  10 16 exp a
 Vt







Then


0.8  10 3
V a  0.0259 ln 
12
16 
 4.8  10  6.4  10 
V a  0.49032 V




 0.49032 
I SB  4.8  10 12 exp

 0.0259 
 I SB  0.7998 mA
 0.49032 
I pn  6.4  10 16 exp

 0.0259 
 I pn  0.107  A
_______________________________________



 0.6907 
 exp

 0.03453 
or
I  120 mA
For the Schottky diode,
  e BO
J ST  T 2 exp
 kT
Now
J ST 400   400 


J ST 300   300 



2


  BO

 exp 
 BO 
 0.0259400 300  0.0259 
0.82 
 0.82
 1.778 exp 

0
.
0259
0
.
03453 

lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 9
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
or
 5  10 6 120 300 2 
(b)
0
.
0259
ln




J ST 400 


Bn
0.0259
 4.856  10 3


J ST 300 
 0.198 V
Then
_______________________________________
I  7  10 4 4.856  10 3 4  10 8
9.28
 0.4447 
 exp

(b) We need  n   m    4.2  4.0  0.20 V
 0.03453 
And
or
N 
I  53.3 mA
 n  Vt ln c 
_______________________________________
 Nd 





9.24
Plot
_______________________________________
9.25
R c 10 4

 0.1 
A 10 3
R
10 4
(b) R  c   4  1 
A 10
R
10 4
(c) R  c  5  10 
A 10
_______________________________________
(a) R 
9.26
R c 5  10 5

 5
A
10 5
(i) V  IR  15  5 mV
(ii) V  IR  0.15  0.5 mV
5
5  10
 50 
10  6
(i) V  IR  150   50 mV
(ii) V  IR  0.150   5 mV
_______________________________________
9.27
 
Vt exp Bn 
 Vt 
Rc 
 2
AT
 R A T 2 
or  Bn  Vt ln  c

 Vt


or
 2.8  1019 

0.20  0.0259  ln

 Nd

which yields
N d  1.24  10 16 cm 3
(c)
Barrier height = 0.20 V
_______________________________________
9.29
We have that
eN d

x n  x 
s
Then
eN 
x2 
   dx  d  x n  x    C 2
s 
2 
Let   0 at x  0  C 2  0 , so

(a) R 
(b) R 


x2 

 xn  x 

2 

At x  x n ,   Vbi , so

eN d
s
  Vbi 
or
xn 

 5 10 5 1203002 
(a)  Bn  0.0259  ln 

0.0259


 0.258 V
eN d x n2

s
2
2 s Vbi
eN d
Also
Vbi   BO   n
where
N 
 n  Vt ln c 
 Nd 
Now for

0.70
  BO 
 0.35 V
2
2
we have
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 9
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
9.34
1.6  10 19 N d
0.35 
x n 50  10 8
Consider an n-P heterojunction in thermal
14
11.7  8.85 10
equilibrium. Poisson's equation is
2
 x 
d 2
d
50  10 8 




2

dx
2
dx

In the n-region,
or
d n  x  eN dn


0.35  7.73  10 14 N d x n  25  10 8
dx
n
n
We have
For uniform doping, we have
1/ 2
 211.7  8.85  10 14 Vbi 
eN dn x
xn  

n 
 C1
1.6  10 19 N d


n
and
The boundary condition is
Vbi  0.70   n
 n  0 at x   x n , so we obatin
By trial and error, we find
eN dn x n
C1 
18
3
N d  3.5 10 cm
n
_______________________________________
Then
eN dn
x  x n 
n 
9.30
n
N 
In the P-region,
(b)  BO   p  Vt ln  
 Na 
d p
eN
  aP
 1.04 1019 
dx
P

 0.0259 ln
16

which gives

 5  10
eN x
or
 P   aP  C 2
 BO  0.138 V
P
_______________________________________
We have the boundary condition that
 P  0 at x  x P , so that
9.31
eN aP x P
Sketches
C2 
P
_______________________________________
Then
9.32
eN aP
P 
x P  x 
Sketches
P
_______________________________________
Assuming zero surface charge density at
x  0 , the electric flux density D is
9.33
continuous,
so n  n 0 P  P 0 , which
Electron affinity rule
yields
E c  e  n   p
N dn x n  N aP x P
For GaAs,   4.07 and for AlAs,   3.5 .
We can determine the electric potential as
If we assume a linear extrapolation between




 











GaAs and AlAs, then for
Al 0.3 Ga 0.7 As    3.90
Then
E c  4.07  3.90  0.17 eV
 n x     n dx

 eN x 2 eN x x 
   dn  dn n   C 3
n 
 2 n
_______________________________________
lOMoARcPSD|13114859
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 9
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Now
Vbin   n 0    n  x n 

eN x 2 eN dn x n2 
 C 3  C 3  dn n 

n 
2 n

or
eN dn x n2
2 n
Similarly on the P-side, we find
eN aP x P2
VbiP 
2 P
We have that
eN dn x n2 eN aP x P2
Vbi  Vbin  VbiP 

2 n
2 P
We can write
N 
x P  x n  dn 
 N aP 
Substituting and collecting terms, we find
2
 e  N N  e n N dn
 2
Vbi   P dn aP
  xn


2
N
n P
aP


Vbin 
Solving for x n , we have


2 n P N aP Vbi
xn  

 eN dn P N aP  n N dn  
Similarly on the P-side, we have
1/ 2
1/ 2


2 n P N dnVbi
xP  

 eN aP P N aP  n N dn  
The total space charge width is then
W  xn  x P
Substituting and collecting terms, we obtain
 2 n P Vbi N aP  N dn  
W 

 eN dn N aP P N aP  n N dn  
_______________________________________
1/ 2