Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 1
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Chapter 1
Problem Solutions
Then
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 


Ratio 
100%  52.4%
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
  16 2  r
3
 4 r
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
 4

 r 
Unit cell vol  a  
 3 
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 
Ratio  
 100%  34%
3
 8r 




 3
_______________________________________
1.3
o
(a) a  5.43 A ; From Problem 1.2d,
a
8
r
3
3
3
Ratio 

2 4 r 




3
3
 4 r 3 

2 atoms per cell, so atom vol  2

 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
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 1
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(b) a  21.035  2.07 A
1.4
(a) 4 Ga atoms per unit cell
4
Number density 
3
5.65 10 8

o
(c) A-atoms: # of atoms  8 

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
sin   

  54.74
a
2
3
2
 
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
_______________________________________
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
o
 4.619 A
3
(a) For 1.12(a), A-atoms
1
1
Surface density  2 
2
a
4.619 10 8


 4.687 10 cm
14
2
o
For 1.12(b), B-atoms: a  4.619 A
1
 4.687  10 14 cm 2
a2
For 1.12(a) and (b), Same material
Surface density 
(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  10 14 cm 2
2
a 2
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  10 14 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))
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
_______________________________________
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

2
2
a
4.73 10 8


 4.47 10 cm
14
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
    a 2 2   34 a 2 
2
h2  a 2
2
2


o
6
4.73  5.793 A
So h 
2


 6.32 10 14 cm 2
(iii) (111) plane:
1
3
6
Surface density 
19.3755  10 16
 2.58 10 14 cm 2
(c) fcc
(i) (100) plane:
2
Surface density  2  8.94  10 14 cm 2
a
(ii) (110) plane:
2
Surface density  2
a 2
 6.32 10 14 cm 2
(iii) (111) plane:
1
1
3  3
6
2
Surface density 
19.3755  10 16
 1.03 1015 cm 2
_______________________________________
2
(ii) (110) plane:
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 10 14 cm 2
(b) bcc
(i) (100) plane:
1
Surface density  2  4.47  10 14 cm 2
a
(ii) (110) plane:
2
Surface density  2
a 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
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 
3
(a) #/cm 
3
a3
6.703 10 8

5 1017
100%  10 3 %
22
5 10
2 1015
(b)
100%  4 10 6 %
5 10 22
_______________________________________

1.25
(a) Fraction by weight
2 1016 10.82

 1.542 10 7
22
5 10 28.06
(b) Fraction by weight
1018 30.98

 2.208 10 5
5 10 22 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
h
 8.2099 A
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)
 1.328 10 cm
1
1
4  2
4
2
(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 ao  5.43 A
d
368.4

 67.85
ao
5.43
_______________________________________
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 ao  5.43 A
d
630

 116
a o 5.43
_______________________________________
Then
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
 31
m
9.1110
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 
From Problem 2.3, phase 
2 x

 t
= constant
(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
Then
2 dx
dx
  
    0, 
  p  

 dt
dt
 2 
_______________________________________
or   11.2 A
2.5
(ii) p  2 9.1110 31 12 1.6 10 19
o

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
_______________________________________



 

24


 1.87 10 kg-m/s
6.625 10 34

 3.54 10 10 m
 24
1.8704 10
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
5.915 10  24
o
or   1.12 A


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
 23
2.532 10
o
or   0.262 A
_______________________________________
2.8
E avg 
3
3
kT   0.0259  0.03885 eV
2
2
Now
p avg  2mE avg



 2 9.1110 31 0.03885 1.6 10 19
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
2
1
1
E  m 2  9.1110 31 8.56 10 4
2
2
 3.33 10 21 J
3.334 10 21
or E 
 2.08 10  2 eV
1.6 10 19
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)

or
p avg  1.064  10 25 kg-m/s
p
h






Now
h 6.625 10 34
 
 6.225 10 9 m
p 1.064 10  25




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
p2
h
1  h 
 
 Ee 
E e  e and p e 
e
2m   e 
2m
Then
1  h 
1  10h 
  


 p 2m   e 
2m   p 
which yields
100h
p 
2mc
hc
2
2

Ep  E 

hc
p

hc
2mc 2
 2mc 
100h
100


2
2 9.1110 31 3 10 8
100
15
 1.64  10 J  10.25 keV
_______________________________________


 1.99  10
2
Set E p  Ee and  p  10e
hc

6.625 10 3 10 
34
8
110 10
15
J
Now
E  e V  V 
E 1.99 10 15

e
1.6 10 19
V  1.24 10 4 V  12.4 kV


(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
_______________________________________
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
12 10 10
dE
d  p2 

  p
(ii) E 
 p 
dp
dp  2m 
2p
pp

 p 
2m
m
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
19
 2.095 10 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


 5.06 10 23 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   

m
1500
36
  7 10 m/s
_______________________________________
p 
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
 x, t 
  2  2 1 x, t 

 V x 1 x, t   j 1
2
2m
t
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 1  2 
1
2
2
2m 
x x 
x
x
1 
 2
 V x 1  2   j 1
 2

t
t 

Dividing by 1  2 , we find
  2  1  2 2
1  2 1
2 1 2 






2
2
2m  2 x
1 x
1 2 x x 
 1 2
1 1 
 V x   j 




t

1 t 
 2
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 2
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Since 1 is a solution, then

1  1
1 1


 V x   j 

2
2m 1 x
1 t
Subtracting these last two equations, we have
  2  1  2 2
2 1 2 




2
2m  2 x
1 2 x x 
1 2
 j 

2 t
2.19

2
2

1  2
1 2


 V x   j 

2
2m 2 x
2 t
Subtracting these last two equations, we obtain
 
 2
2

 1  2  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
2
 2
  x 
 dx
P
exp


 a o 
 ao
0 
4

ao
Since 2 is also a solution, we have
2

ao
2
2
    dx  1
Note that

   2a o  
1
  1  1  exp 
P   1exp

 2 
  4a o  
which yields
P  0.393
(b)
 x 
P
 A cos  2 dx  1
2
 x sin x   3
A2  
1
2  1
2
1 / 2

A 2 cos 2 nx dx  1
1 / 2
 x sin 2nx   1 / 2
A2  
1
4n  1 / 2
2
 1  1 
1
A 2       1  A 2  
2
 4  4 
or A  2
_______________________________________
 2
  x 
 dx
exp


 a o 
 a o
4
ao
2

ao a
2
2.18
2
2

ao
1
 3   1 
A      1
 2  2 
1
so A 2 
2
1
or A 
2
_______________________________________
o
0
  2 x  ao 4
2   ao 


 exp

ao  2 
 ao  0
ao
2
 exp a dx
or
2.17
3
  2x 
4

2
  2x 
 exp a dx
o
o
4
  2 x  ao 2
2   ao 


 exp

ao  2 
 a o  ao 4
or

  1 
P   1exp 1  exp 
 2 

which yields
P  0.239
(c)
2
 2
  x 
 dx
P 
exp

 ao
 a o 
0 
ao

ao

  2x 
2
 dx
exp

ao 0
 ao 

  2 x  ao
2   ao 


 exp

ao  2 
 ao  0

  1exp 2  1
which yields
P  0.865
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 2
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
2.20
2.21
P   x  dx

2
a/4
(a) P 
2
2 x 
dx
  cos 
a
2




0
a/4
(a)
 a 
 
   sin   
2 4
 2 
  

 4  
 a  2

 

 a  

 2  a 1a 
   
4 
 a  8
or P  0.409
a/2
(b) P 
2




 2  a sin   
   
 a   8  8  
 

 a 

or P  0.25
a/2
(b) P 

 2x   a / 2
sin 


 2  x
 a 
  
  
 a  2
4   a / 4

a 

2
 x 
  a  cos  a dx
2
a / 2

 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x 




 2  a sin 2   a  sin   
   
 
 a   4  8   8   8  
 
 

 a 
 a 

or P  0.25
a / 2
(c) P 
2
 2x 
  a  sin  a dx
2
a / 2

 4x    a / 2
sin 


 2  x
 a 
  
 2  
 a  2
4
  a / 2

 a  

1 1 
1
 2  0  
4
8 4 

or P  0.0908
(c) P 
2

 4x   a / 2
sin 


 2  x
 a 
  
 2  
 a  2
4
  a/4

 a  

a/4
a / 2
2
  a  sin  a dx
a/4
 x 

 
sin   

 2  a sin   a
 2 
   
 
 a   4  4  8  4  



 

 a 
 a  

 2x 

 4x   a / 4
sin 


2 x
 a 
   
 2  
 a  2
4
  0

 a  

  a  cos  a dx
2
2
0


 2x   a / 4
sin 


2 x
 a 
   
  
 a  2
4   0

a 

2
  a  sin  a dx








sin
2

sin

2

2
a

a
 



   


 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
8
k
8  10
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 2
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
o
or
  78.54 A

 9.1110 27 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
(b) (i)  p  
 10 4 m/s
k  1.5 10 9
or  p  10 6 cm/s

 
2
2

 4.19 10 9 m
9
k
1.5 10

25
 9.11 10 kg-m/s
6.625 10 34

 7.27 10 10 m
9.1110  25
2
k
 8.64  10 9 m 1
7.272  10 10
  8.64 10 9 10 6  8.64 1015 rad/s
_______________________________________

E  2.85 10 4 eV
_______________________________________



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
p 8.537  10  26
2
2
k

 8.097  10 8 m 1

7.76  10 9
  k    8.097 10 8 9.37  10 4 
or   7.586 10 rad/s
_______________________________________
13
2.24
(a)


p  m  9.1110 31 5 10 4
26

or
21

 4.555 10 kg-m/s
h 6.625 10 34
 
 1.454 10 8 m
p 4.555 10  26

2
J




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 1.0698 10
2
En 
2.23
(a) x, t   Ae  j kx t 

n 2 1.054  10 34  2
 2 n 2 2

2
2
2ma
2 9.11 10 31 75  10 10
p  9.1110 27 kg-m/s

 
2.25
En 
(b) E  0.025 1.6 10
 
(b) p  9.1110 31 10 6
or   41.9 A
19
2
 4.32  10 8 m 1
8
1.454  10
 2.16 1013 rad/s
o
(ii)


  k  4.32 10 8 5 10 4 
 
(ii) p  m  9.1110 31 10 4

2
k

E 2  2.67 10 2 eV
E3  6.02 10 2 eV
_______________________________________
2.26
(a) E n 


2
n 2 1.054  10 34  2
 2 n 2 2

2
2ma 2
2 9.11 10 31 10  10 10



J
n 6.018 10 
E 
 n 0.3761 eV
 n 6.018 10
2
20
20
2
2
or
n
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


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


2
n 2 1.054  10 34  2


2 15  10 3 1.2  10  2

15 10 3  n 2 2.538 10 62


2
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

2
2
2ma
2 1.66  10  27 10 14
 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 
E1 
2mE
2
Boundary conditions:
a
a
, x
 x   0 at 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   B2 sin k 2 x
where
2
4 2  2
k2 
 E2 
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   B4 sin k 4 x
where
4
16 2  2
k4 
 E4 
a
2ma 2
_______________________________________
k
 2 n 2 2
(a) E n 
2ma 2
15  10 3 
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
1.054 10  
29.11 10 10 
34 2
 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
2.30
The 3-D time-independent wave equation in
cartesian coordinates for V x, y, z   0 is:
 2 x, y, z   2 x, y, z   2 x, y, z 


x 2
y 2
z 2
2mE
 2   x, y , z   0

Use separation of variables, so let
 x, y, z   X x Y  y Z z 
Substituting into the wave equation, we
obtain
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 2
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
2 X
 2Y
2Z

XZ

XY
x 2
y 2
z 2
2mE
 2 XYZ  0

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 2Z
1  2Y

 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
YZ
or
k x2  k y2  k z2  k 2 
2mE
2
so that
 2 2 2
n x  n 2y  n z2
2ma 2
_______________________________________
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
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
 x, y 
 Ak y sin k x x  cos k y y
y
 2 x, y 
  Ak y2 sin k x x  sin k y y
y
Substituting into the original equation, we
find:
2mE
 k x2  k y2  2  0
(1)

From the boundary conditions,
2
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 2y  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
(b) E 
n 2  n12
2ma 2
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
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
2 1.67  10
 27
 34

2
2
0.5  10 
2 2
 3.939 10 36 J
or
3.939 10 36
 2.46 10 17 eV
1.6 10 19
_______________________________________
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   B2 exp jk 2 x 
where
2m
E  VO 
k2 
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 jk1 x 
 2 x   A2 exp jk 2 x   B2 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 B2  k1 B1
Combining these two equations, we find
 k  k1 
  B2
A2   2
 k 2  k1 
 2k 2 
  B2
B1  
 k 2  k1 
The reflection coefficient is
2
 k  k1 

R
  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
_______________________________________
A2 A2*
2.34
 2 x   A2 exp k 2 x 
P
 x 
 exp 2k 2 x 
A2 A2*
where k 2 

2
2mVo  E 
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 
E
T  16 1   exp 2k 2 a 
 Vo  Vo 
where k 2 


2mVo  E 
2


2 9.1110 31 1.0  0.1 1.6 10 19
1.054 10
34

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 =

10
(a) For a  4 10 m
 0.1  0.1 
9
10
T  16
1 
 exp  2 4.85976 10 4 10
 1.0  1.0 
 0.0295
(b) For a  12 10 10 m
 0.1  0.1 
9
10
T  16
1 
 exp  2 4.85976 10 12 10
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
5

   1.874 10 m/s 1.874 10 7 cm/s

1.2 10
3

 N t 1.6 10

19
2.36
 E 
E 
1 
 exp 2k 2 a 
T  16



 VO  VO 
(a) For m  0.067 mo
2mVO  E 
k2 
2




 31
19 

 20.067  9.1110 0.8  0.2 1.6 10



2
 34


1.054 10




or
k 2  1.027 10 9 m 1
Then
 0.2  0.2 
T  16
1 

 0.8  0.8 
 
T  0.138
(b) For m  1.08m o


1/ 2


or
T  1.27 10 5
_______________________________________
2.37
 E 
E
T  16 1   exp 2k 2 a 
 Vo  Vo 
where k 2 

2mVo  E 
2



2 1.67 10 27 12  110 6  1.6 10 19

34
1.054 10
 7.274 1014 m 1
(a)
1
 1 
T  16 1   exp  2 7.274 1014 10 14
 12  12 
 1.222 exp 14.548
 5.875 10 7
(b)
T  10 5.875 10 7


 

 


 1.222 exp  2 7.274 1014 a


 1.222

2 7.274 1014 a  ln 

6
5
.
875

10


or a  0.842 10 14 m
_______________________________________
2.38

 exp  2 1.027 10 9 15 10 10
or
1/ 2

 
1.874 10 
N t  4.002 10 electrons/cm 3
Density of incident electrons,
4.002 10 8
Ni 
 1.357 1010 cm 3
0.0295
_______________________________________

 exp  2 4.124 10 9 15 10 10
7
8

 31
19 

 21.08 9.1110 0.8  0.2 1.6 10



 34 2


1.054 10


or
k 2  4.124 10 9 m 1
Then
 0.2  0.2 
T  16
1 

 0.8  0.8 

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)
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 2
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
where
2mE
k1 
2
Region II:
 2 x   A2 expk 2 x   B2 exp k 2 x 
where
k2 
2mVO  E 

A1  B1  A2  B2
d  1 d 2


dx
dx
jk1 A1  jk1 B1  k 2 A2  k 2 B2
At x  a :  2   3 
A2 expk 2 a   B2 exp k 2 a 
 A3 exp jk1 a 
d 2 d 3


dx
dx
k 2 A2 expk 2 a   k 2 B2 exp k 2 a 
 jk1 A3 exp jk1 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  k12 expk 2 a   exp k 2 a 
4k 1 k 2


 2 jk1 k 2 expk 2 a   exp k 2 a  
 exp jk1 a 
We then find
A3 A3*
4k1 k 2 2
k  k expk a
2
2
2
1
2
 exp k 2 a 
2
 4k12 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 
A1 A1* 

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 22  k12 exp2k 2 a 
2
4k1 k 2 
Substituting the expressions for k1 and


k 2 , we find
k 12  k 22 
2mV O
2
and
 2mVO  E   2mE 
k12 k 22  
 2 
2

  
2
 2m 
  2  VO  E E 
 

E 
 2m 
E 
  2  VO 1 

 
 VO 
2
Then
2
 2mV O 
A3 A 
 exp2k 2 a 
2
  
A1 A1* 
 2m  2 
E  
E 
16  2  VO 1 

  
 VO  
*
3

A3 A3*
 E 
E 
1 
 exp 2k 2 a 
16



 VO  VO 
Finally,
 E 
A A*
E 
1 
 exp 2k 2 a 
T  3 3*  16



A1 A1
 VO  VO 
_____________________________________
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
2
Region II: V  V1
k1 
 2 2 x 
2mE  V1 
 2 x   0 
x
2
 2 x   A2 exp jk 2 x   B2 exp jk 2 x 
transmitted
reflected
where
2

2mE  V1 
k2 

Region III: V  V2

2 m E  V 2 
x
2
 3 x   A3 exp jk 3 x 
transmitted
where
2
k3 
 3 x   0 
2 m E  V 2 
2
There is no reflected wave in Region III.
The transmission coefficient is defined as:
T
 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
k
4k 1 k 3
4k12
T 3

k1 k1  k 3 2 k1  k 3 2
_______________________________________
2.40
(a) Region I: Since VO  E , we can write
 2 1 x 
 3 A3 A3* k 3 A3 A3*



1 A1 A1* k1 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  B2
 1  2


x
x
k1 A1  k1 B1  k 2 A2  k 2 B2
At x  a :  2   3 
A2 exp jk 2 a   B2 exp jk 2 a 
 A3 exp jk 3 a 
2mVO  E 
 1 x   0
x
2
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 x   A2 sin k 2 x   B2 cosk 2 x 
 3 x   0
where
2
2
 2 3 x 
 2  3


x
x
k 2 A2 exp jk 2 a   k 2 B2 exp jk 2 a 
k1 

2mVO  E 
2
and k 2 

(b) Boundary conditions
At x  0 :  1   2  B1  B2
2mE
2
 1  2

 k 1 B1  k 2 A2
x
x
At x  a :  2   3 
A2 sin k 2 a   B2 cosk 2 a   0
or
B2   A2 tan k 2 a 
(c)
k 
k1 B1  k 2 A2  A2   1  B1
 k2 
and since B1  B2 , then
k 
A2   1  B2
 k2 
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 2
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
From B2   A2 tan k 2 a  , we can write
  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
a o   o 
 o 
P
k 
B2   1  B2 tank 2 a 
 k2 
or
k 
1   1  tan k 2 a 
 k2 
This equation can be written as
 2mE 
V E
1  O
 tan 
 a
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
 100 is independent of  and  , so the wave
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
3
2
 34 2
4 8.85 10
12
2
or
13.58
(eV)
n2
n  1  E1  13.58 eV
En 
equation in spherical coordinates reduces to
1   2   2mo
E  V r   0
 r

r   2
r 2 r 
where
 e2
 2
V r  

4 o r mo a o r
For
 1 
 100 
  
  ao 
Then
1
 100
1  1 

 
r
  a o 
n  2  E 2  3.395 eV
n  3  E 3  1.51 eV
n  4  E 4  0.849 eV
_______________________________________
2.42
We have
r 

exp

 ao 
3/ 2
 1 
r 
  exp

a 
a 
 o
 o 
so
 100
1  1 
r

 
r
  a o 
2
5/2
r 

r 2 exp

 ao 
We then obtain
 1 
 100 
  
  ao 
and
*
P  4 r 2 100 100
1
3/ 2
3
r 

exp

 ao 
  2r 
1  1 

 4 r     exp

  ao 
 ao 
2
or
3/ 2
  2  100   1  1 
r

 
r 
r 
  a o 
5/2

 r   r2 
  r 
    exp

 2r exp



 a 
 ao   ao 
 o 

Substituting into the wave equation, we have
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 2
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
r2

 r  r2
  r 


exp
2r exp



 ao  ao
 a o 
2m 
2 
 2o  E 

mo a o r 
 
 1 
  
  ao 
1
5/ 2
 1   1 
 
 
  
    ao 
3/ 2
r 
0
exp

 ao 
where
 mo e 4
 2
4 o 2 2 2mo ao2
Then the above equation becomes
E  E1 
   r  
1 
r2 
 
2
r

exp



 r 2a
ao 
  a o  
o 

2m    2
 2 
  0
 2o 

  2m o a o m o a o r 
 1 
  
  ao 
3/ 2
 1 
  
  ao 
3/ 2
1

2
or
1
   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.
_______________________________________
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
2
dx
dx
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 
   
 


 2 jk
 
u x 
 E  
exp  j  kx   t 
x
   
 

 
 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
2mV O
2mE

u x   2 u x   0
2


Setting ux   u 2 x  for region II, this
equation becomes
d 2 u 2 x 
du x 
 2 jk 2
2
dx
dx
2mVO 
 2
  k   2 
u 2 x   0
2 

where again
2mE
2  2
Q.E.D.

_______________________________________
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
 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
2
dx
2
  j   k  B exp j   k x
Substituting these equations into the
differential equation, we find
2
   k  A exp j   k x
2

 2 jk

   k  B exp j   k x
 2 jk  j   k A exp j   k x
 j   k B exp j   k x
2


 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.
_______________________________________
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
   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

3.4
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
 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
_______________________________________
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
_______________________________________
sin a
 cos a  cos ka
a
Let ka  y , a  x
Then
sin x
P
 cos x  cos y
x
d
Consider
of this function.
dy
P

2
2m o a
E2 
 2 1.054 10 34 2


2 9.11  10 31 4.2  10 10

2
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

dx
  sin y
dy

dx    1
cos x 
 sin x   sin y
P  2 sin x 

dy   x
x 

For y  ka  n , n  0, 1, 2, ...  sin y  0
So that, in general,
d a  d
dx
0

dy
d ka dk
And
2mE
2
 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 
So
d 1  2mE 
 2m  dE
  2 
 2 
dk 2   
   dk
This implies that
d
dE
n
0
for k 
dk
dk
a
_______________________________________

 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 
1 / 2

2
 a  1.729
2
Then

a  
 2 2
E1 
2m o E 2
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 

 sin x
2 m o E1
 3.4114 10 19 J
From Problem 3.5
 2 a  1.729
3.7

3.8
(a)  1 a  
 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
 6.07 eV
or E 
1.6 10 19
_______________________________________
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 
2 m 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
Eo 
0.859  1.054 10
2 9.11 10 31
10 2

2
E2 
 2.5172 10 19 J
E  E1  E o
 3.4114 10
 2.5172 10
19
 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
2m o E 3
2
 a  2
2  1.054 10
2
E3 




2
E2 
 a  1.729
1.729 2 1.054 10
 31
10 2

2 9.11 10
 1.0198 10
E  E 3  E 2

4.2 10 
18
J
 1.3646 10 18  1.0198 10 18
19
2
 3.4474 10 J
3.4474 10 19
or E 
 2.15 eV
1.6 10 19
_______________________________________
 a  1.515


2 9.11 10 31 4.2  10 10

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
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

1.515 2 1.054 10 34 2
2
 1.3646 10 18 J
At ka   . From Problem 3.5,
 2 a  1.729
2m o E 2

2m o E 3
34 2
2 9.11 10 31 4.2  10 10

2 9.11 10 31 4.2  10 10
2m o E 2

4.2 10 
19
 2 1.054 10 34 2
 3.4114 10 19 J
From Problem 3.6,  2 a  1.515
34 2
2
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
_____________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 3
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
3.11
(a) At ka   ,  1 a  
2 m o E1
E g  1.170 
a  
2
E1 
3.12
For T  100 K,

2
T  300 K, E g  1.125 eV
 3.4114 10 19 J
At ka  0 , By trial and error,
 o a  0.727
2
Eo 
T  400 K, E g  1.097 eV
T  500 K, E g  1.066 eV
T  600 K, E g  1.032 eV
_______________________________________
 a  0.727
0.727 2 1.054 10 34 2

2 9.11 10
 31
 1.8030 10
E  E1  E o
4.2 10 
19
10 2
 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
2

E3 

2 2 1.054 10 34 2


2
 1.3646 10 18 J
At ka   , From Problem 3.6,
 2 a  1.515
2
E2 
 a  1.515
1.515 2 1.054 10
 34
10 2

 7.830 10
E  E 3  E 2
19
 1 d 2E 

m *p   2 

dk 2 

We have that
1
d 2E
d 2E


curve B
curve
A

dk 2
dk 2
so that m *p curve A  m *p curve B 
_______________________________________
34 2
2 9.11 10
1
 1 d 2E 
m   2  2 
  dk 
We have
2
d 2E
curve A  d E2 curve B 
2
dk
dk
so that m * curve A  m * curve B 
_______________________________________
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
3.13
The effective mass is given by
*
J
 3.4114 10 19  1.8030 10 19
2m o E 3
2
T  200 K, E g  1.147 eV

2 9.11 10 31 4.2  10 10
2m o E o
4
636  100
E g  1.164 eV
 2 1.054 10 34 2

4.73 10 100 

4.2 10 
J
 1.3646 10 18  7.830 10 19
 5.816 10 19 J
5.816 10 19
 3.635 eV
or E 
1.6 10 19
_______________________________________
3.15
dE
 0  velocity in -x direction
dk
dE
 0  velocity in +x direction
Points C,D:
dk
Points A,B:
d 2E
0
dk 2
negative effective mass
d 2E
0
Points B,C:
dk 2
positive effective mass
_______________________________________
Points A,D:
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
m 
At k  0.08 10 10 m 1 , E  0.05 eV
10 2
1
 C1  1.25 10 38



1.054 10 34
2
Now m 

2C1
2 1.25 10 38


2
 4.44 10 kg
4.4437 10 31
or m  
 mo
9.1110 31
m   0.488 mo

For B: E  C i k 2
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 20  C1 0.08 1010
Now m  
2



1.054 10 34
2

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 mo
_______________________________________
3.17
For A: E  E  C 2 k 2



  2  1.054 10 34
m 

2C 2
2 6.25 10 39


2

 0.3 1.6 10 19  C 2 0.08 1010
 C 2  7.5 10 38
2

.
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 
2
dk


 8.8873 10 31 kg
 8.8873 10 31
or m  
 mo
9.1110 31
m     0.976 mo

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
For B: E  E  C 2 k 2


 2.705 1014 Hz
c
3 1010
(ii)   
 2.705 1014
 1.109 10 4 cm  1109 nm
_______________________________________
around k  
 0.025 1.6 10 19  C 2 0.08 1010




 C1  1.25 10 37

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 1014
 8.75 10 5 cm  875 nm
E 1.12 1.6 10 19
(b) (i)   
h
6.625 10 34
31



 7.406 10 32 kg
 7.406 10 32
or m  
 mo
9.1110 31
m   0.0813 mo
_______________________________________


 C 0.08 10 
Or E  0.05 1.6 10 19  8 10 21 J
So 8 10 21

  2  1.054 10 34

2C 2
2 7.5 10 38

2
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 3
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Then
E
1
1 d E
 2 2
 12
*
m
 dk k  ko

2
or
2
E1 2
_______________________________________
m* 
3.21


 4 0.082m  1.64m 

(a) mdn
 4 2 / 3 mt  ml
2
1/ 3
2
2/3
1/ 3
o
o

m dn
 0.56mo
(b)
3
2
1
2
1





mcn mt ml 0.082mo 1.64mo

24.39 0.6098

mo
mo

mcn
 0.12mo
_______________________________________
3.22

 m  
 0.45m   0.082m  

 m hh 
(a) m dp
3/ 2 2/3
3/ 2
lh
3/ 2 2/ 3
3/ 2
o
 0.30187  0.02348
2/3
m

dp
o
 mo
 0.473m o
mhh 3 / 2  mlh 3 / 2
mhh 1 / 2  mlh 1 / 2
0.453 / 2  0.0823 / 2  m

0.451 / 2  0.0821 / 2 o

(b) mcp


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 


x 2
y 2
z 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

2 X
 2Y
2Z

XZ

XY
x 2
y 2
z 2
2mE
 2  XYZ  0

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
X x
x 2
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
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 
YZ
2
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 2y  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
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 3
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
1
1 1
1
m
 2m  
 dE  
 dE

2 E
 2E
Substituting these expressions into the density
of states function, we have
 a 3  2mE  1
m
g T E dE  3  2   
 dE
     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
dk 
So
g E  


g E  

1 1 2m n
 
 dE
 2
E

4 2m n
g c E  

4 2m n
gc 
2m n
2a 1
 
 dE
 2
E
Divide by the "volume" a, so
g E  
2m n
1


E

3/ 2
E  Ec

3 / 2 Ec  2 kT
 E  E  dE
c
h3



h3

4 2m n

h

4 2m n
h

Ec
3/ 2
3

Ec  2 kT
2
3/ 2
  E  E c 
3
Ec
3/ 2
3

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 
21 3 / 2
 6.0  10 25 m 3
g c  6.0 1019 cm 3
or
 400 
(ii) At T  400 K, kT  0.0259 

 300 
 0.034533 eV

 0.034533 1.6 10 19

 5.5253 10 21 J
Then
g T E dE 
m 3 J 1
3.26
(a) Silicon, mn  1.08mo
3/ 2
dk 
1.055 1018
E
_______________________________________
4 2m 
g E  
 E
h3
_______________________________________
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


20.067  9.11 10 31
1

1.054  10 34  
E
Then
 

g c  7.953 10 55 2 5.5253 10 21

3/ 2
 9.239 10 25 m 3
or
g c  9.24 1019 cm 3
(b) GaAs, m n  0.067 mo
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

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
 

3/ 2
g   2.3564 10 55 3 4.144 10  21

 9.272 10 23 m 3
 3.266 10 25 m 3
or g c  9.27 1017 cm 3
or g   3.27 1019 cm 3
(ii) At T  400 K, kT  5.5253 10 21 J
(ii)At T  400 K, kT  5.5253 10 21 J
 

g c  1.2288 10 54 2 5.5253 10 21

 

3/ 2
3/ 2
g  2.3564 10 55 3 5.5253 10 21
 1.427 10 24 m 3

3/ 2
 5.029 10 25 m 3
g c  1.43 1018 cm 3
_______________________________________
or g   5.03 1019 cm 3
_______________________________________
3.27
(a) Silicon, m p  0.56m o
3.28
g  E  
g 




4 2m p


3/ 2

4 2m p
h3
4 2m



 E  E  dE
3/ 2

 3/ 2
p
3

E
2
3/ 2

 E   E 
E  3 kT
 3 

2
3/ 2

  3kT 
 3 
6.625 10 
 2.969 10 3kT 

3/ 2
(i)At T  300 K, kT  4.144 10
 

J
g  2.969 10 55 3 4.144 10 21
E  E c  0.3 eV;
 2.614 10 46 m 3 J 1
E  E c  0.4 eV;
 3.018 10 46 m 3 J 1

4 2m p

h3

4 20.48 9.11 10 31
6.625 10 
 2.3564 10 3kT 
3/ 2
E  E

4 20.56  9.11 10 31
6.625 10 

3/ 2
 34 3
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;
_______________________________________
or g   6.34 1019 cm 3
(b) GaAs, m p  0.48m o

 4.454110 55 E  E
 6.337 10 25 m 3

 34 3
55
gc  0
For E  E ;
 
E  Ec
 2.134 10 46 m 3 J 1
3/ 2
or g  4.12 1019 cm 3

6.625 10 
E  E c  0.2 eV;

 4.116 10 25 m 3
g  2.969 10 55 3 5.5253 10 21
3/ 2
 34 3
(b) g 

(ii)At T  400 K, kT  5.5253 10 21 J

E  E c  0.1 eV; g c  1.509 10 46 m 3 J 1

2
3/ 2
 3kT 
3
21
E  Ec
4 21.08 9.11 10 31
For E  E c ;
3/ 2
55
3/ 2
 1.1929 10 56 E  E c
E  3 kT


h3
E
 34 3


E  E
4 20.56  9.11 10 31
g 
4 2m n

h3
h
3/ 2
h3
4 2m p



(a) g c E  
3/ 2
3/ 2
2
3/ 2
 3kT 
3
 
3.29
 
(a)
 
m 
g
(b)

g
m 
3/ 2
gc
m
 1.08 
 n 3/ 2  


g
 0.56 
m
3/ 2
 2.68
p
c

 3/ 2
n
 3/ 2
p
 0.067 


 0.48 
3/ 2
 0.0521
_______________________________________
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!
(b) (i) Wi 

10!12  10! 10!21
 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  
1

1  exp5
f E   6.69 10 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 
E  E c ; f F  exp 
 9.32 10 6

 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
Ec 
; f F  exp 

2
0.0259


 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  
E E
1  exp F

 kT 
(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
_______________________________________
 2.36 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
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 3
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
3.35
  E c  kT  E F  
  E  E F  
f F  exp 
 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  E
 E midgap
Or E F  c
2
_______________________________________
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
E7 
1.054 10  7  
29.11 10 12  10 
34 2
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
_______________________________________


E5 

E13 


2
2
2
2
2
10 2
 31
1.054 10    3  2  3 
29.11 10 12  10 
34 2
 31
2
2
2
2
10 2
 9.194 10 19 J
9.194 10 19
or E13 
 5.746 eV
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 E1  E F  E
being occupied is
1
1
f 1  E1  

 E1  E F 
 E 
1  exp
 1  exp

 kT 
 kT 
The probability of a state at E 2  E F  E
being empty is
1
1  f 2 E 2   1 
 E  EF 
1  exp 2

 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
34 2
 3.76110 19 J
3.76110 19
or E 5 
 2.35 eV
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
3.37
(a) For a 3-D infinite potential well
2mE
 
 n x2  n 2y  n z2  
2
a
th
For 5 electrons, the 5 electron occupies
the quantum state n x  2, n y  2, n z  1 ; so
1.054 10    2  2  1 
29.11 10 12  10 
or
1  f 2 E 2  
1
 E 
1  exp

 kT 
so f 1 E1   1  f 2 E 2 
Q.E.D.
_______________________________________
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

 E  EF 
 E  EF 
exp 1
 1  exp 1

 kT 
 kT 
 0.01
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 ,
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 
kT 
 0.063906  0.0259



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
f E  
 0.1496
 7.15  7.0 
1  exp

 0.08633 
or 14.96%
1
 0.997
(c) f E  
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
  E1  E F  
 exp 

kT
 E  EF 


1  exp 1

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
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 
f E   exp 
  exp 0.0259 
kT




or
f E   8.45 10 13
At E  E 2 ,
  0.4 
 exp

 0.0259 
or 1  f E   1.96 10 7
_______________________________________
3.44

 E  E F 
f E   1  exp

 kT 

so

df E 
 E  E F 
  11  exp

dE
 kT 

  E F  E 2 
1  f E   exp 

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 


or
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 
or
1  f E   1.66 10 19
(b) For E F  E 2  0.4 ,
E1  E F  1.02 eV
At E  E1 ,
  E1  E F 
  1.02 
f E   exp 
  exp 0.0259 
kT




or
f E   7.88 10 18
At E  E 2 ,
1
2
 E  EF 
 1 
   exp

 kT 
 kT 
or
 E  EF 
 1 


 exp
df E   kT 
 kT 

2
dE

 E  E F 
1  exp

 kT 

(a) At T  0 K, For
E  E F  exp    0 
df
0
dE
df
E  E F  exp     
0
dE
df
 
At E  E F 
dE
(b) At T  300 K, kT  0.0259 eV
df
0
For E  E F ,
dE
df
0
For E  E F ,
dE
At E  E F ,
 1 

1
df
 0.0259 

 9.65 (eV) 1
dE
1  12
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
0
For E  E F ,
dE
df
0
For E  E F ,
dE
At E  E F ,
 1 

1
df
 0.04317 

 5.79 (eV) 1
2
dE
1  1
_______________________________________
3.45
(a) At E  Emidgap,
f E  
1
1

 E  EF 
 Eg 
1  exp

 1  exp

kT


 2kT 
Si: E g  1.12 eV,
f E  
or
1
 1.12 
1  exp 

 20.0259 
f E   4.07 10 10
1
 0.66 
1  exp 

 20.0259 
or
f E   2.93 10 6
GaAs: E g  1.42 eV
f E  
or
(a)
or
  E  E F  
f F  exp 

kT


  0.60 
10 8  exp 

 kT 


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  
3.46
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
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 4
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Chapter 4
(b)
4.1
  Eg 

ni2  N c N  exp

 kT 

  2.5 10
  1.12300 
 T 
 2.912 10 
 exp 
0.0259T  
300
ni2  5 1012
2
25
3
38
  Eg 
 T 

 N cO N O 
 exp

 300 
 kT 
3




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
ni (cm 3 )
200
0.01727
7.68 10 4
400
0.03453
2.38 1012
600
0.0518
9.74 1014
T (K)
(b) Germanium
ni (cm 3 )
(c) GaAs
ni (cm 3 )
200
2.16 1010
1.38
400
8.60 10
14
3.28 10
600
3.82 10
16
5.72 1012
 200 
At T  200 K, kT  0.0259 

 300 
 0.017267 eV
 400 
At T  400 K, kT  0.0259 

 300 
 0.034533 eV
7.70 10   3.025 10
n 200  1.40  10 
n i2 400 
2
i
10 2

  Eg 
exp 

 0.034533 


3
  Eg 
 200 

 exp 

 300 
 0.017267 
Eg
 Eg

 8 exp 


 0.017267 0.034533 
3.025  1017  8 exp E g 57.9139  28.9578
 400 


 300 
9
_______________________________________
4.2
Plot
_______________________________________
17
2 2
3


or
4.3
  Eg 

(a) n  N c N  exp

 kT 
2
i
T 
5 10   2.8 10 1.04 10  300

11 2
19
19




 1.12
 exp 

 0.0259T 300 

3
 3.025 1017 
  38.1714
E g 28.9561  ln 

8


or E g  1.318 eV

Now
400 
7.70 10   N N  300

10 2
co
o


  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
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
ni  A
0
.
017267


(b) For T  300 K,
n i B 
  0.20 
4
 exp
  4.43 10
ni  A
 0.0259 
(c) For T  400 K,
ni B 
  0.20 
3
 exp
  3.05 10
ni  A
 0.034533 
_______________________________________
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 
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
1
x1 / 2
kT

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
x
Then g c f F  x exp

 kT 
To find the maximum value:
d g c f F  1 1 / 2
x
 x
exp

dx
2
 kT 
  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
_______________________________________
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
 m *p 
3
E Fi  E midgap  kT ln  * 
 mn 
4


*
n
m n*  0.067 mo
E Fi  Emidgap  0.0382 eV
_______________________________________
4.11
N 
1
kT  ln   
2
 Nc 
19


1
kT  ln  1.04 1019   0.4952kT 
2
 2.8 10 
T (K)
kT (eV)
( E Fi  E midgap)(eV)
200
0.01727
 0.0086
400
0.03453
 0.0171
600
0.0518
 0.0257
_______________________________________
4.12
 m *p 
3
(a) E Fi  E midgap  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
_______________________________________
1
dE
E  EF 

Ec 1  exp
 kT 




E Fi  Emidgap  0.0077 eV
Gallium Arsenide: m *p  0.48m o ,

  E  E F  
 K exp 
dE
kT


Ec
Germanium: m *p  0.37 m o , m n*  0.55mo
E Fi  E midgap 
F

E Fi  Emidgap  0.0128 eV

c
K
Silicon: m  0.56m o , m  1.08mo
*
p
 g E  f E dE
Ec
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  
no  K  kT  exp 

kT


_______________________________________


4.14
Let g c E   C1 E  E c  for E  E c
Then

no 
 g E  f E dE
c
F

E  E c 
Ec
 C1

Ec
 E  EF 
1  exp

 kT 

 C1
dE
  E  E F  
 dE
kT

 E  E  exp 
C
Ec
Let
E  Ec
so that dE  kT  d
kT
We can write
E  E F  E  E c   E c  E F 

Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 4
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Then
The ionization energy is
  E c  E F  
n o  C1 exp 

kT




  E  E c  
 dE
kT

 E  E  exp
c
Ec
or
2
  E c  E F  
n o  C1 exp 

kT


 kT  exp kT d
0
We find that


or
E  0.0053 eV
_______________________________________
4.17


 m *  o 
0.067
  13.6  
13.6
E  
  
m
13.12
 o  s 

 exp  d  exp     1  1
0
0
So
  E c  E F  
2
no  C1 kT  exp 

kT


_______________________________________
4.15
r
m 
We have 1 r  o* 
ao
m 
For germanium, r  16 , m *  0.55mo
Then
 1 
r1  16
a o  290.53
 0.55 
or
o
r1  15.4 A
The ionization energy can be written as
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  Ec  E F 
 1.12  0.2148  0.90518 eV
  E F  E 
(c) 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 
 1.04 1019 

 0.0259 ln 
16 
 2 10

 0.162 eV
(b) Ec  E F  E g  E F  E 
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 mo
*
Then
o
 1 
r1  13.1
0.53  104 A
 0.067 


 2.4110 3 cm 3
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 
 2 1016 

 0.0259 ln 
10 
 1.5 10 
 0.365 eV
_______________________________________
4.19
N 
(a) E c  E F  kT ln  c 
 no 
 2.8 1019 

 0.0259 ln 
5 
 2 10 
 0.8436 eV
E F  E  E g  Ec  E F 
 1.12  0.8436
E F  E  0.2764 eV


  0.27637 
(b) p o  1.04 1019 exp

 0.0259 
 2.414 1014 cm 3
(c) p-type
_______________________________________
4.20
 375 
(a) kT  0.0259
  0.032375 eV
 300 

no  4.7 10
17

 375 


 300 
3/ 2
  0.28 
exp 

 0.032375 
 1.15 1014 cm 3
E F  E  E g  Ec  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  Ec  E F   1.42  0.2154
 1.2046 eV


  1.2046 
p o  7 1018 exp 

 0.0259 
 4.42 10 2 cm 3
_______________________________________
4.21
 375 
(a) kT  0.0259
  0.032375 eV
 300 


 375 
no  2.8 1019 

 300 
3/ 2
  0.28 
exp 

 0.032375 
 6.86 1015 cm 3
E F  E  E g  Ec  E F   1.12  0.28
 0.840 eV


 375 
p o  1.04 1019 

 300 
3/ 2
  0.840 
exp 

 0.032375 
 7.84  10 7 cm 3
N 
(b) E c  E F  kT ln  c 
 no 
 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
Ec  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
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 4
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
4.23
4.25
 E  E Fi 
(a) n o  ni 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  ni exp  Fi

 kT 



 8.80  10 9 cm 3
 E  EF 
p o  ni exp  Fi

 kT 
  0.22 
 1.8 10 6 exp 

 0.0259 


 3.68  10 2 cm 3
_______________________________________
4.24
N 
(a) E F  E  kT ln   
 po 
 1.04 1019 

 0.0259 ln 
15 
 5 10

 0.1979 eV
(b) Ec  E F  E g  E F  E 
 1.12  0.19788  0.92212 eV
  0.92212 
(c) n o  2.8 1019 exp 

 0.0259 


 9.66  10 3 cm 3
(d) Holes
p 
(e) E Fi  E F  kT ln  o 
 ni 
 5 1015 

 0.0259 ln 
10 
 1.5 10 
 0.3294 eV
_______________________________________
3/ 2
 4.3109 1019 cm 3


ni2  4.3109 1019 1.6011019
3
 3.07  10 cm
 E  E Fi 
(b) n o  ni exp  F

 kT 


 400 
N c  2.8 1019 

 300 

 0.22 
 1.8 10 6 exp 

 0.0259 
3/ 2
 1.6011019 cm 3
  0.22 
 1.5 1010 exp 

 0.0259 
6

 400 
N   1.04 1019 

 300 

  1.12 
 exp 

 0.034533 
 5.6702 10 24
 ni  2.3811012 cm 3
N 
(a) E F  E  kT ln   
 po 
 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) no  4.3109 1019 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
(a)


  0.25 
p o  7 1018 exp 

 0.0259 
 4.50 1014 cm 3
E c  E F  1.42  0.25  1.17 eV


  1.17 
n o  4.7 1017 exp 

 0.0259 
 1.13 10 2 cm 3
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 4
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(b) kT  0.034533 eV


 400 
N   7 10 

 300 
18

 8.49  10 9 cm 3
_______________________________________
 1.078 1019 cm 3


 400 
N c  4.7 1017 

 300 
 7.236 10 cm
17
3/ 2
4.28
(a) n o 
3
 1.078 1019 

 0.034533 ln 
14 
 4.50 10 
 0.3482 eV
E c  E F  1.42  0.3482  1.072 eV
  1.07177 
n  7.236 10 exp 
0.034533 
17


4.27

p o  1.04 10
19

E F  E c kT 2

 0 .5
kT
kT
Then F1 / 2  F   1.0
no 

no  7.23 10 cm
(b) kT  0.034533 eV


po 

2

5  10 19 
17
N  F1 / 2  F 
2

1.04 10 F   
19
1/ 2
F
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
 400 


 300 

4.7 10 1.0
So F1 / 2  F   4.26
 400 
N   1.04 1019 

 300 
19
2
4.29
3

19


4
2.8 10 1.0
 3.16 1019 cm 3
2
N c F1 / 2  F 
(b) n o 
  0.870 
n o  2.8 1019 exp 

 0.0259 

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 
F 
 2.40  10 cm
_____________________________________
N c  2.8 10

3
4
(a)
2
For E F  E c  kT 2 ,
N 
E F  E  kT ln   
 po 
o

  0.77175 
no  4.3111019 exp 

 0.034533 
3/ 2
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
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
x 
 E  E c  kT
2
kT
2
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 
no  2.8 1019 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 1016 cm 3
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 
no  4.7 1017 exp 
 3
 0.0259



 4.7 10 exp 3.224
no 
po 
no  1.87 1016 cm 3
Assume E a  E  0.0345 eV
Then
  0.0345 
p o  7 1018 exp 
 3
 0.0259

4.35
(a)


 7 10 exp 4.332
p o  9.20 1016 cm 3
_______________________________________
4.33
Plot
_______________________________________
p o  4 15  1015  3 1015 cm 3
1.5 10   7.5 10 cm
10 2
4
3
12

ni2
1.8 10 6

po
3 1015
10 2
3
3
6 2
4
3
3 1016
(c) no  p o  ni  1.8 10 6 cm 3



3
 375 
(d) ni2  4.7 1017 7.0 1018 

 300 
  1.42300  
 exp 

 0.0259 375 

7.580 10   1.44 10 cm
8 2
2


3


3
 ni  3.853 1010 cm 3
no  N d  1014 cm 3
 ni  7.334 1011 cm 3
3.853 10   1.48 10 cm
p 
10 2
p o  N a  41015 cm 3
7
7.334 10   1.34 10 cm
11 2
8
4 1015

3
 450 
(e) n  2.8 10 1.04 10 

 300 
  1.12300  
 exp 

 0.0259 450  
19
3
4 1015
 450 
(e) ni2  4.7 1017 7.0 1018 

 300 
  1.42300  
 exp 

 0.0259 450  
 375 
(d) ni2  2.8 1019 1.04 1019 

 300 
  1.12300  
 exp 

 0.0259 375 

3
1.8 10   1.08 10 cm
po 
no 
3
3 10
(c) no  p o  ni  1.5 1010 cm 3

2
(b) no  N d  31016 cm 3
16
19
  1.08 10 cm
p o  N a  41015 cm 3
1.5 10   7.5 10 cm
p 
no 
3
 ni  7.580 10 8 cm 3
3 1015
(b) no  N d  31016 cm 3
2
i
13 2
p o  N a  N d  4 1015  1015
no 
or


1.722 10   2.88 10 cm
 31015 cm 3
18
o

1.029 1014
_______________________________________
or
no 
2
 1014 
13 2


 2   1.722 10


 1.029 1014 cm 3
17
4.34
(a)
1014

2
o
3
3
14
10
_______________________________________
4.36
(a) Ge: ni  2.4 1013 cm 3
(i) n o 
 ni  1.722 1013 cm 3

Nd

2
2 1015

2
2
 Nd 

  ni2
 2 
2
 2 1015 
13 2


 2   2.4 10




Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 4
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
or
Then
n o  N d  21015 cm 3

f F E  

13 2
n2
2.4 10
po  i 
no
2 1015
f F E   2.87 10 5
_______________________________________
(ii) p o  N a  N d  1016  7 1015
 31015 cm 3
n2
2.4 1013
no  i 
po
3 1015

4.38
(a) N a  N d  p-type
(b) Silicon:
p o  N a  N d  2.5 1013  11013
or
p o  1.5 1013 cm 3
Then
2
 1.92 1011 cm 3
(b) GaAs: ni  1.8 10 6 cm 3
(i) n o  N d  21015 cm
po 
1.8 10   1.62 10 cm
6 2
3
3
1.8 10   1.08 10 cm
6 2
3
no 
3
3 1015
(c) The result implies that there is only one
minority carrier in a volume of 10 3 cm 3 .
_______________________________________
4.37
(a) For the donor level
nd
1

Nd
 Ed  EF 
1
1  exp

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


2
ni2
1.5 1010

 1.5 10 7 cm 3
po
1.5 1013
Germanium:
2 1015
(ii) p o  N a  N d  31015 cm 3
no 
0.245 

1  exp1 

 0.0259 
or
 2.88 1011 cm 3

1
N  Nd
po  a

2
2
 Na  Nd 

  n i2
2


2
 1.5 1013 
 1.5 1013 
2
 
  2.4 1013
 



2
2




or
p o  3.26 1013 cm 3
Then




2
ni2
2.4 1013

 1.76 1013 cm 3
p o 3.264 1013
Gallium Arsenide:
p o  N a  N d  1.5 1013 cm 3
and
no 


2
ni2
1.8 10 6

 0.216 cm 3
po
1.5 1013
_______________________________________
no 
4.39
(a) N d  N a  n-type
(b) no  N d  N a  2 1015  1.2 1015
 81014 cm 3
po 

ni2
1.5 1010

no
8 1014
  2.8110 cm
2
5
3
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 4
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
p o  N a  N a   N d
(c)
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   5.625 10 cm
n 
10 2
4
3
o
4 1015
_______________________________________
4.44
Plot
_______________________________________
4.45
4.40


no 
2
n2
1.5 1010
no  i 
 1.125 1015 cm 3
po
2 10 5
n o  p o  n-type
_______________________________________
Nd  Na

2
1.1 10 14 
2
 Nd  Na 

  ni2
2


2  10 14  1.2  10 14
2
2
 2  1014  1.2  1014 
  n i2
 

2


4.41



 250 
ni2  1.04 1019 6.0 1018 

 300 
3
1.110  4 10   4 10   n


 0.66
 exp 

 0.0259250 300
ni2
n2
1
 i  n o2  n i2
p o 4n o
4
N a  N d  p-type
Majority carriers are holes
p o  N a  N d  3 1016  1.5 1016

2
 Na 

  ni2
2


N 

 2.752 1012  a 
2 

so ni  5.74 1013 cm 3
 1.5 1016 cm 3
Minority carriers are electrons
Then p o  2.75 1012 cm 3
Na

2
4.9 10 27  1.6 10 27  ni2
4.46
(a)
1
 no  ni
2
no  6.88 1011 cm 3 ,
po 
2
i
po 
 ni  1.376 1012 cm 3
So
13 2
ni2 3.3  10 27

 3  1013 cm 3
n o 1.1 1014
_______________________________________
 1.8936 10 24
no 
13 2
14

2
2
n2
1.5 1010
no  i 
 1.5 10 4 cm 3
16
po
1.5 10
(b) Boron atoms must be added
p o  N a  N a  N d
2
5 1016  N a  3 1016  1.5 1016
N 
  a   1.8936 10 24
 2 


N 
7.5735 10 24  2.752 1012 N a   a 
 2 
So N a  3.5 1016 cm 3
2
N 
  a   1.8936 10 24
 2 
so that N a  2.064 1012 cm 3
_______________________________________
4.42
Plot
_______________________________________
1.5  10   4.5  10 cm
10 2
2
no 
3
3
5  10
_______________________________________
16
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 4
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
4.47
(a)
p o  ni  n-type
(b) p o 
n i2
n2
 no  i
no
po
no 
1.5 10   1.125 10 cm
10 2
16
3
2 10 4
 electrons are majority carriers
p o  210 4 cm 3
 holes are minority carriers
(c) n o  N d  N a
1.125 1016  N d  7 1015
so N d  1.825 1016 cm 3
_______________________________________
N 
(b) E F  E Fi  kT ln  d 
 ni 
 Nd

 0.0259 ln 

10 
1
.
5

10


For 10 14 cm 3 , E F  E Fi  0.2280 eV
10 15 cm 3 , E F  E Fi  0.2877 eV
10 16 cm 3 , E F  E Fi  0.3473 eV
10 17 cm 3 , E F  E Fi  0.4070 eV
_______________________________________
4.50
no  1.05 N d  1.05 1015 cm 3
4.48
p 
E Fi  E F  kT ln  o 
 ni 
For Germanium
T (K)
kT (eV)
ni (cm 3 )
200
0.01727
2.16 1010
400
0.03453
8.60 1014
600
0.0518
3.82 1016
1.05 10  0.5 10 
 0.5 10   n
15 2
Na
N 
  a   ni2
2
 2 
and
N a  1015 cm 3
T (K)
p o (cm 3 )
E Fi  E F  (eV)
200
1.0  10 15
0.1855
400
1.49  10
0.01898
600
3.87 1016
15
15 2
15
0.000674
_______________________________________
4.49
N 
(a) E c  E F  kT ln  c 
 Nd 
 2.8 1019 

 0.0259 ln 

 Nd

14
3
For 10 cm , E c  E F  0.3249 eV



 T 
ni2  2.8 1019 1.04 1019 

 300 
3


 1.12
 exp 

 0.0259T 300 


 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 


10 15 cm 3 , E c  E F  0.2652 eV
 536.5 
N c  2.8 1019 

 300 
10 16 cm 3 , E c  E F  0.2056 eV
 6.696 1019 cm 3
10 17 cm 3 , E c  E F  0.1459 eV
2
i
so ni2  5.25 10 28
Now
2
po 
2
 Nd 

  ni2
 2 
N
(a) n o  d 
2
3/ 2
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 4
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Then, T  200 K, E Fi  E F  0.4212 eV
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.
_______________________________________
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



16
N a  1017 cm 3 , E Fi  E F  0.6408 eV
(b)
N 
 7.0 1018 

E F  E  kT ln     0.0259 ln 

 Na 
 Na

14
3
For N a  10 cm , E F  E  0.2889 eV
3
N a  1015 cm 3 , E F  E  0.2293 eV
16
N a  10 cm 3 , E F  E  0.1697 eV
N a  1017 cm 3 , E F  E  0.1100 eV
_______________________________________
 ni  7.638 10 4 cm 3
At T  400 K,

n  2.8 10
19


 400 
1.04 10 

 300 
19
N 
 Na 
E Fi  E F  kT ln  a   0.0259 ln 

6 
 1.8 10 
 ni 
For N a  1014 cm 3 , E Fi  E F  0.4619 eV
N a  10 cm 3 , E Fi  E F  0.5811 eV
  1.12 
 exp 

 0.017267 
2
i
4.52
(a)
N a  1015 cm 3 , E Fi  E F  0.5215 eV
At T  200 K,
 200 
ni2  2.8 1019 1.04 1019 

 300 
T  400 K, E Fi  E F  0.2465 eV
T  600 K, E Fi  E F  0.0630 eV
_______________________________________
3
4.53
(a) E Fi  E midgap 
  1.12 
 exp 

 0.034533 

 ni  2.38110 cm 3
At T  600 K,
12

n  2.8 10
2
i
19
1.04 10 
19
 600 


 300 
E Fi  Emidgap  0.0447 eV
(b) Impurity atoms to be added so
Emidgap  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

3
 ni  9.740 1014 cm 3
At T  200 K and T  400 K,
p o  N a  31015 cm 3
At T  600 K,

 
 0.4947 
 10 5 exp

 0.0259 
2
 Na 

  ni2
2


3  1015

2
or
2
 3  1015 
14 2


 2   9.740  10


 3.288 1015 cm 3
3
0.0259 ln 10
4
or
  1.12 
 exp 

 0.0518 
N
po  a 
2
 m *p 
3
kT ln  * 
 mn 
4




p o  N a  1.97 1013 cm 3
_______________________________________
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 
  0.1876 eV
 0.0259 ln 
16 
 2 10 
(c) For part (a);
p o  21016 cm 3
or
no 
N d  1.2 1016 cm 3
_______________________________________

ni2
1.5 1010

po
2 1016
For part (b):
n o  21016 cm 3
  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

po 

  0.13346 
N d  4.7 10 exp 

 0.0259 
 2.718 1015 cm 3  N d  1015
17
 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 

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
2
 1.125 10 4 cm 3
4.55
(a) Silicon
N 
(i) E c  E F  kT ln  c 
 Nd 
 2.8 1019 
  0.2188 eV
 0.0259 ln 
15 
 6 10 
(ii) E c  E F  0.2188  0.0259  0.1929 eV


 3.0 1015 cm 3
Add additional acceptor impurities
no  N d  N a
3 1015  7 1015  N a
 N a  41015 cm 3
_______________________________________
4.58
p 
(a) E Fi  E F  kT ln  o 
 ni 
 3 1015 
  0.3161 eV
 0.0259 ln 
10 
 1.5 10 
n 
(b) E F  E Fi  kT ln  o 
 ni 
 3 1016 
  0.3758 eV
 0.0259  ln 
10 
 1.5 10 
(c) E F  E Fi
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

2
5.08  10 15 
4.59
N 
(a) E F  E  kT ln   
 po 
 7.0 1018 
  0.2009 eV
 0.0259 ln 
15 
 3 10 
 7.0 1018 

(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  10 375 300 
 ln 
4  10 15

 0.2526 eV
 450 
(e) E F  E  0.0259

 300 

18

3/ 2



 7.0  10 450 300  
 ln 

1.48  10 7


 1.068 eV
_______________________________________
18
2
 Na 

  ni2
2


5  10 15
2
2
 5  1015 
  n i2
 

2


5.08 10  2.5 10 
 2.5 10   n
15 2
15
15 2
2
i
6.6564 10 30  6.25 10 30  ni2
 ni2  4.064 10 29
  Eg 
ni2  N c N  exp 

 kT 
 350 
kT  0.0259
  0.030217 eV
 300 


2


2
 350 
19
N c  1.2 1019 
  1.633 10 cm 3
 300 
 350 
19
N   1.8 1019 
  2.45 10 cm 3
 300 
Now
4.064 10 29  1.633 1019 2.45 1019



  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
_______________________________________
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 1014 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
no 


2
ni2
1.8 10 6

 5.14 10  4 cm 3
15
po
6.3 10
p 
(d) E Fi  E F  kT ln  o 
 ni 
 6.3 1015 
  0.5692 eV
 0.0259 ln 
6 
 1.8 10 
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 5
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Chapter 5
5.1
1
1

(a)  
e n N d
1.6  10 19 1300  1015
 4.808  -cm

 
1
 0.208 (  -cm) 1
 4.8077
_______________________________________
(b)  
5.2
1

1
1.6  10 220  8  10 16
 0.355  -cm
(b)   e n N d


19





120  1.6 10 19  n N d

From Figure 5.3, for N d  21017 cm 3 ,
then  n  3800 cm 2 /V-s which gives
  1.6 10 19 38002 1017 
  e p N a

1.80
or N a 

e p 1.6 10 19 380
 2.96 1016 cm 3
_______________________________________
5.3
(a)   e n N d
 121.6 (  -cm) 1
_______________________________________
5.5
L
R
A
or  n 


10  1.6 10 19  n N d

From Figure 5.3, for N d  61016 cm 3 we

L
A

L
e n N d A
L
eN d RA
2.5
1.6 10 2 10 700.1
19
15
find  n  1050 cm 2 /V-s which gives
 1116 cm 2 /V-s
_______________________________________
 10.08 (  -cm) 1
1
(b)  
e p N a
5.6
(a) no  N d  1016 cm 3
and
  1.6 10 19 10506 1016 
0.20 
1
1.6 10  N
po 
19
p
a
From Figure 5.3, for N a  1017 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
(a)  
1
e p N a
0.35 

ni2
1.8 10 6

no
1016
  3.24 10 cm
2
4
3
(b)
J  e n n o 
For GaAs doped at N d  1016 cm 3 ,
 n  7500 cm 2 /V-s
Then
J  1.6 10 19 7500 1016 10
or
J  120 A/cm 2


 
(b) (i) p o  N a  1016 cm 3
1
1.6 10  N
no 
19
p
a
From Figure 5.3, for N a  81016 cm 3 we
find  p  220 cm 2 /V-s which gives
ni2
 3.24  10  4 cm 3
po
(ii) For GaAs doped at N a  1016 cm 3 ,
 p  310 cm 2 /V-s
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 5
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
J  e p po 

 1.6 10
19
31010 10
16
or
J  4.96 A/cm 2
_______________________________________
5.7
(a) V  IR  10  0.1R
or
R  100 
(b)
L
L
R
 
A
RA
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 1013 cm 3




(d)   e p po
or
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 1015 cm 3
Note: For the doping concentrations
obtained, the assumed mobility values are
valid.
_______________________________________


5.8
(a) R 
L
A

L
a
For N a  21016 cm 3 , then
 p  400 cm 2 /V-s
R

19
1.6  10
 68.93 
or
0.075
4002 1016 8.5 10 4 
V
2

 0.0290 A
R 68.93
I  29.0 mA
I
V
2

 0.00967 A
R 206.79
or I  9.67 mA
(c) J  ep o d
I
29.0 10 3
 34.12 A/cm 2
4
8.5 10
J
34.12

Then  d 
ep o
1.6  10 19 2  1016
For (a), J 



 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  10 16
4



 3.55 10 cm/s
_______________________________________
3
5.9
(a) For N d  21015 cm 3 , then
 n  8000 cm 2 /V-s
V
5

 200 
I 25  10 3
L
R
e n N d A
R
or L  e n N d RA





 1.6 10 19 8000 2 1015 200 5 10 5
 0.0256 cm
I
(b) J   en o d
A
I
or  d 
Aen o 

e N A
p
(b) R  L  R  68.933  206.79 
25 10 3
1.6 10 19 2 1015
5 10 
5
 1.56 10 cm/s
(c) I  en o d A



6





 1.6 10 19 2 1015 5 10 6 5 10 5
 0.080 A
or I  80 mA
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 5
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(b) N a  N d  1016 cm 3
5.10
(a)  
V 3
  3 V/cm
L 1
 d  n    n 
  n  1250 cm 2 /V-s
 p  410 cm 2 /V-s
d
10 4


3

or
 n  3333 cm 2 /V-s
(c) N a  N d  1018 cm 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
d
10 4
tt 

 8.33 10 11 s
 d 1.2 10 6
Then
10 4
 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
6
7 10
_______________________________________
5.12
1
1

e n no  e p p o e  n   p ni

  n  1350 cm 2 /V-s
 p  480 cm 2 /V-s
1
1350  480 1.5 1010



2
ni2
1.8 10 6

 2.49 10 5 cm 3
po
1.3 1017
(b) Silicon:
1
   e n n o

or
1
1
no 

e n 8 1.6  10 19 1350 
which gives
n o  5.79 1014 cm 3
and




2
ni2
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
 2.28 10  -cm

 p  240 cm 2 /V-s
tt 

10
From Figure 5.3, and using trial and error, we
find
p o  1.3 1017 cm 3 and
Then
1.6 10 
19

no 

1
1.6 10 290  1301.5 10 
5.13
(a) GaAs:
  e p p o  5  1.6  10 19  p p o
 d  9.5 10 6 cm/s
5

 9.92 10 5  -cm
_______________________________________
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,
19

 2.5110  -cm
 d   n   8003


5
(b)

1
1250  410 1.5 1010
1.6 10 
19
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 5
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
5.14
 i  eni  n   p 
(b) R 
Then
10 6  1.6 10 19 1000  600ni
or
ni (300 K)  3.9110 9 cm 3
Now
  Eg 

ni2  N c N  exp

 kT 
or
N N 
E g  kT ln  c 2  
 ni 

(i) Si:

 
R



ni (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
_______________________________________
 i  1.6 10






1.5 10 1350  480


 0.2496 (  -cm) 1
(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 1015 
 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
(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


So   1.6 10 19 1300 1.2 1015
10
or


then  n  1300 cm 2 /V-s



From Figure 5.3, for N d  1.2 1015 cm 3 ,
or
19


 5.15 10 26
5.15
(a) (i) Silicon:  i  eni  n   p

0.25  1.6 10 19  n N d
2


5.16
(a)   e n N d

 1.122
  exp 0.0259
500 300


200 10 4
 1.06 10 6 
2
8
2.23 10 85 10
(iii) GaAs:
200 10 4
R
 9.19 1012 
2.56 10 9 85 10 8
_______________________________________
R
Now
ni2 (500K)  1019
200 10 4
 5.36 10 9 
4.39 10 6 85 10 8
(ii) Ge:
 1019 2 
 0.0259  ln 

2
 3.91 10 9 
which gives
E g  1.122 eV

L
A


 avg 





t
1
1
x
 x dx   o exp
dx
t 0
t 0
 d 

o
t

 d  exp  x 
t
 d 0
 od    t  
exp   1
t   d  
200.3 1  exp  1.5 


1.5 
 0.3 
 3.97 (  -cm) 1
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 5
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
5.18
V
2
 133.3 V/cm
(a)   
L 150  10  4
(b)  x   e n N d x 
T
 avg  e n 



1
x 

2  10 16 1 
dx
T 0
 1.111T 
e n 2 10
T
 x 

x


 21.111T  0

e 2 1016 
T2
 n
T 

T
21.111T 

 e n 2 1016 0.55


16




 1.6 10
19
2
T
V 
L
150 10  4
5
 1.32 10 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
ni  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 1019 11019 exp

 0.0259 
2
1


  Eg 

(a) ni2  N c N  exp

 kT 
7502 10 0.55
(c) I 

5.21
16
 avg  1.32 (  -cm) 1
 avg A
1.327.5 10 4 10 4 
(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 >> ni  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  10
14
N
N 
 d   d   n i2
2
 2 
which becomes
1.05 10  5 10   5 10   n
13 2
14
13 2
2
i
and yields
ni2  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
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 no  e p po 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
which yields
p-type:  p  400 cm 2 /V-s
compensated:  n  1000 cm 2 /V-s
(c) n-type:   e n no

 8.8 (  -cm)
p-type:   e p po

 
p o  n i  n  (Answer to part (b))
 p 


Substituting into the conductivity expression
e n ni2
   min 
1/ 2
ni  n  p
 
 
 e p n i  n  p
 
1/ 2
which simplifies to
 min  2en i  n  p
The intrinsic conductivity is defined as
 i  eni  n   p   eni 
i
n   p
The minimum conductivity can then be
written as
 min 



 1.28 (  -cm)
compensated:   e n no


1
 1.6 10 19 400 2 1016
1/ 2
 

 1.6 10 19 1100  5 1016

1


 1.6 10 19 1000  3 1016

1
 4.8 (  -cm)
J
(d) J     

120
 13.6 V/cm
n-type:  
8 .8
120
 93.75 V/cm
p-type:  
1.28
120
 25 V/cm
compensated:  
4 .8
_______________________________________
5.24
2 i  n  p
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: no  N d  51016 cm 3

n2
1.5 1010
po  i 
no
5 1016
or

2
1

 4.5 10 3 cm 3
p-type: p o  N a  210 cm
16
10 2
4
o
2 10
compensated: n o  N d  N a
16
 5 10  2 10
1.5 10   7.5 10 cm
p 
10 2
3
 T 

 300 
(a) At T  200 K,
3 / 2
 300 

 200 
3/ 2
 n  1300
16
 31016 cm 3
3 1016
3
5.25
16
o
Then
  316 cm 2 /V-s
_______________________________________
3
1.5 10   1.125 10 cm
n 
 0.003167
3
 n  1300
 300 
 1300

 T 
3 / 2
 2388 cm 2 /V-s
(b) At T  400 K,  n  844 cm 2 /V-s
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 5
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
1


1
1

1
2

 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 

5.26
1
1

 0.006
250 500

Then
  167 cm 2 /V-s
_______________________________________
nx1   8.911014 cm 3
_______________________________________
5.27
Plot
_______________________________________
5.32
5.28
Plot
_______________________________________
5.29
 5 1014  n0 
dn

J n  eDn
 eDn 

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 1014 cm 3
_______________________________________




5.30
J n  eD n
dn
n
 eD n
dx
x

16
15
5.31
dn
n
 eD n
dx
x
 1015  nx1  
 2  1.6 10 19 30
4 
 0  20 10 
4 10 3  4.8 10 3  4.8 10 18 nx1 
which yields
nx1   1.67 1014 cm 3
(a) J n  eD n


 eD p 
1016 
x
 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 
4
12 10
 13.3 A/cm 2
(c) For x  12  m,
Jp 0
_______________________________________
5.33
For electrons:
dn
d
J n  eD n
 eD n
10 15 e  x / Ln
dx
dx

 2 10  5 10 
J n  1.6 10 19 27 

0  0.012


J n  5.4 A/cm 2
_______________________________________

2
dp
d  16 
x 
 eD p
10 1   
dx
dx 
 L  
J p  eD p

 
 eD n 1015 e  x / Ln
Ln
At x  0 ,
 1.6 10 19 25 1015
Jn 
 2 A/cm 2
3
2 10
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 1015
 16 A/cm 2
Jp 
4
5 10
J Total  J n x  0  J p x  0

 

 2   16  18 A/cm 2
_______________________________________
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  10 15 e

x / Lp
Lp
1.6 10 105 10 
19
15
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 1015 e 1
(b) (i) J p 
50 10  4
 0.589 A/cm 2
1.6 10 19 48 5 1015 e 1
(ii) J p 
22.5 10  4
 6.28 A/cm 2
_______________________________________

 


 


 

(a) J n  eD n
dn
J n  e n n  eD n
dx



 eD n 2 1015 e  x / L
L
 1.6 10 19 27  2 1015 e  x / L

15 10  4
 5.76e  x / L
(b) J p  J Total  J n  10   5.76e  x / L


 





 5.76e  x / L 10 A/cm 2
(c) We have J p    e p po 
5.76e
x / L




 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

or
  
 1.6 10 19 25 1016
 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1016 exp
 
 18 



dn
d
 eD n
2  10 15 e  x / L
dx
dx
Dn  0.02598000  207 cm 2 /s
Then
100  1.6 10 19 800012nx 
5.35

5.36
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 
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
nx   6.511015  3.255 1015 exp
 cm 3
d


(b)
At x  0 , n0  6.511015  3.255 1015
Or
n0  3.26 1015 cm 3
At x  50  m,


  50 
n50  6.511015  3.255 1015 exp

 21.56 
or
n50  6.19 1015 cm 3
(c)
At x  50  m,
J drf  e n n50




 1.6 10 19 8000 6.19 1015 12
or
J drf x  50  95.08 A/cm 2
Then
J diff x  50  100  95.08
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
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  eD n
dx
  2.5 10 2 
 0.4  2.5 10 2 x 
 exp

 eDn ni 



kT
kT




Assume T  300 K, so kT  0.0259 eV and
ni  1.5 1010 cm 3
Then
 1.6  10 19 25 1.5  1010 2.5  10 2
Jn 
0.0259

 


 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


3
2
(i) At x  0 , J n  2.95 10 A/cm
(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  1016 1  
 L
  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 


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


 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  1019 x
1  10 3 x
 


 


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
J diff  1.243 10 5 exp
 A/cm 2
 L 
(b)
0  J drf  J diff
Now
J drf  e n n

L
(b)     X dx  25.9 L  0 



 x dx






or
0
10 4
 0.0259 10  
3
0
 
dx
1  10 3 x




4
10
 1 
 0.0259 10 3  3  ln 1  10 3 x
0
 10 
 0.0259ln 1  0.1  ln 1
or


   x 
 1.6 10 19 6000 5 1016 exp
 
  L 
10 4
V 
eD n
x
 N do exp

 L 
 L 
 kT 
Dn   n 
  60000.0259
 e 
 X  25.9 V/cm
0
dN d x 
dn
 eD n
dx
dx
We have
0.0259
0.0259


L
10  10  4
or
0.0259
 500 V/cm
L
Which yields L  5.18 10 5 cm
_______________________________________
So  X 
J diff  eD n
 0.0259    1 
   N do e  x / L
N do e  x / L  L 

For N d x   N do e  x / L
5.43
(a) We have
5.40
dN d x 
1
 kT 
(a)  X  




e
N
x
dx

 d
 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 
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
_______________________________________
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) Dn  0.02591150  29.8 cm 2 /s
8
 308.9 cm 2 /V-s
0.0259
35
 1351 cm 2 /V-s
(ii)  p 
0.0259
_______________________________________
5.46
L  10 1 cm, W  10 2 cm, d  10 3 cm


 I X BZ
 1.2 10 3 5 10 2

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
_______________________________________




 250 10 6 5 10 2
5 10 21 1.6 10 19 5 10 5




(c) n 


5
I x Bz
edV H





 0.5 10 3 6.5 10 2
1.6 10 19 5 10 5  0.825 10 3




21
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 
6
19
4
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

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
1.6 10
1.6 10
3
21
n  4.924 10 21 m 3  4.924 1015 cm 3
(d)
IxL
n 
enV xWd
or

0.5 10 10 
6.0110 1510 10 
3
19
or
I x Bz
ned


3
 0.03466 m 2 /V-s
or  n  346.6 cm 2 /V-s
_______________________________________
5.47
(a) V H 

or n  6.011015 cm 3
IX L
(c)  n 
enV X Wd
(b) (i)  p 


 6.0110 m
(ii) Dn  0.02596200   160.6 cm /s
VH 

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
_______________________________________

Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 5
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
5.50
(a) V H  negative  n-type
(b) n 
I x Bz
edV H




 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 1014 cm 3
IxL
(c)  n 
enV xWd





2.5 10 3 0.5 10 2


19
20
8.68 10 2.2 
 1.6 10


1

2
2 
 0.05 10 0.0110 






or
 n  0.8182 m 2 /V-s  8182 cm 2 /V-s
1
(d)    e n n




 1.6  10  19 8182 8.68  10 14
or

  0.88 (  -cm)
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 6
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Chapter 6
6.1
no  N d  51015 cm 3


2
ni2
1.5 1010

 4.5 10 4 cm 3
15
Nd
5 10
(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  1011 cm 3 s 1
2  10  7
p o  p
4.5 10 4  1014

2 10 7
 p0
3
1
6.2
p o  N a  21016 cm 3

n2
1.8 10 6
no  i 
po
2 1016
(a) R  
6.4
(a) E  h 
po

 pt
no
 nt

  1.62 10 cm
2
4
3

34
6300 10 10
n  p  g  3.17 1019 10 10 6 
or
n  p  3.17 1014 cm 3
 


_______________________________________
6.5
 6.17 10 s
_______________________________________
We have
p
p
   F p  g p 
t
p
6.3
(a) Recombination rates are equal
no
p
 o
and
J p  e p p  eD p p
The hole particle current density is
Jp
F p 
  p  D p  p
 e p
Now
  F p   p    p   D p   p
13
 nO
 pO
no  N d  1016 cm 3
po 
Then
1016

ni2
1.5 1010

no
1016
2.25 10
 nO
20 10  6
which yields
 nO  8.89 10 6 s

4
8
(b)
p
2 1016
 pt  o  n0 
 5 10 7
no
1.62 10  4

6.625 10 3 10 
 3.17 10 19 e-h pairs/cm 3 -s
no
 n0


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
n 5 1014

 10 21 cm 3 s 1
 n0 5 10 7
(b) R p 
hc
or
 510 cm s
_______________________________________
20
(b) Generation rate = recombination rate
Then
2.25 10 4
G
 1.125 10 9 cm 3 s 1
6
20 10
(c)
R  G  1.125 10 9 cm 3 s 1
_______________________________________
  2.25 10 cm
2
4
3
We can write
   p     p  p  
and
  p   2 p
so
  F p   p   p  p     D p  2 p
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 6
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Then
p
   p   p  p   
t
 Dp2 p  g p 
p
p
We can then write
D p  2 p   p   p  p   
p
gp 

 p t
_______________________________________
p
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 ,
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) Dn  2 n   n   n  n  
 n 
t
Multiply Equation (1) by  n n and Equation
gR
p
0
For steady-state,
t
Then
0    F p  g p  R p
(2) by  p p , and add the two equations.
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) Dn  2 n   n   n  n  
 gn 
n
n
 n 
t

 n 
t
We find
 n nD p   p pDn  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 
 n 
 g  R  
t
Define
 n nD p   p pD n Dn D p n  p 
D 

nn   p p
Dn n  D p p
and
 
 n  p  p  n
nn   p p
Then we have
 n 
t
Q.E.D.
_______________________________________
D  2 n     n  g  R  
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

From Figure 5.3,  n  1300 cm 2 /V-s
 33.67 cm 2 /s
(c)  nt   n 0  10 7 s
390019001.124 1013  5.124 1013 
39005.124 1013  19001.124 1013 
 nt
p o  N a  71015 cm 3

2
6.11
so  pt  2.18  10 4 s
_______________________________________
6.10
For Ge: ni  2.4 1013 cm 3

4 10
2
 4 10
 
 2
13


2
  2.4 1013





2
ni2
2.4 1013

 1.124 1013 cm 3
no 5.124 1013
(a) We have:
 n  3900 cm 2 /V-s, Dn  101 cm 2 /s
 p  1900 cm /V-s, D p  49.2 cm /s
2
For very, very low injection,
Dn D p n  p 
D 
Dn n  D p p
2
10149.25.124 1013  1.124 1013 

1015.124 1013  49.21.124 1013 
 54.2 cm 2 /s
and
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 no  p  e p  po  p
or
  e n no  e p po  e  n   p p
so
  e  n   p p

2
 5.124 1013 cm 3
po 
  e n n  e p p

2
 Nd 

  ni2
 2 
13
1.124 1013
2 10 6
  nt  9.12 10 6 s
_______________________________________
3.214  10 4 7  1015

 pt
10  7
Nd

2

 nt
 pt
no 
 p0
5.124 1013

 3.2110 4 cm 3
no
p
 o
 nt
nn   p p
 1340 cm 2 /V-s
(b) For holes,  pt   p 0  2  10 6 s
For electrons,
p
n

 kT 
(b) D   Dn  
   n  0.02591300 
 e 
n2
1.5 1010
no  i 
Na
7 1015
 n  p  p  n


In steady-state, p  g  pO
So that
  e  n   p g  pO
_______________________________________

6.12
(a)


p o  N a  1016 cm 3


2
ni2
1.5 1010

 2.25 10 4 cm 3
po
1016
  e n no  n  e p  po  p
no 


 e p po  e  n   p n

Now n  p  g  n 0 1  e
t /  n 0

5 10 1  e
 cm
 4 10 1  e

20
7
14
t /  n 0
 8 10
t /  n 0
3

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
I




 cm
 2  10 1  e
At t  10 s,

p 10 6   2  10 14 1  e 10
p  2 1014 e

 t 10
6
6
3
/ 510
8

 1.4 1016 cm 3
(a) n  p  g  n 0
  n0  2.5 10 7 s

(b) n  p  g  n 0 1  e t /  n 0
For 0  t  10 6 s,
R 
75005 10 
 1.6 10 7500  310

 2  10 1  e
15
For t  10 6 s,

  6.0  0.250e
 (  -cm)

 t 10 6 /  p 0


n
5 10

1  e t /  n 0
 n0 2.5 10 7
14



(c)
t /  p 0
t /  p 0

t /  n 0
 2 10 21 1  e t /  no cm 3 s 1
19


 5 10 1  e
14
14
A
p o  N a  N d  2 1016  6 1015
/  p 0 cm 3
19
t /  p 0
5 1014  2 10 21  n0
  e n no  e n   p p
 6.0  0.250 1  e
0.05
6.15
(b) no  51015 cm 3
  1.6 10
5
or I  2.496  0.218e
mA
_______________________________________
 21014 cm 3
Then for t  10 6 s,
10 10
t /  p 0

t /  p 0
t /  p 0
6
t /  p 0
 2.496 10  2.18 10 4 e
 4  10 21 5  10 8 1  e
14
1.248  0.109e
t /  p 0
3

t /  p 0
14
(ii)     0.690 (  -cm)
_______________________________________
n  p  g  p 0 1  e
t /  p 0
t / 
1
t /  p 0

 4 1014 e p 0 cm 3
  1.6 10 19 1300 8 1015  2 1015
(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
1
(  -cm) 1
_______________________________________



1
(i)   5 1014  5 1014 1  e t /  n 0
4

t   n 0 ln 1.3333  7.19 10 8 s




1
(ii)   5 1014  5 1014 1  e t /  n 0
2
t   n0 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
 p  400 cm 2 /V-s
  e n no  e n   p p



3
(iii)   5 1014  5 1014 1  e t /  n 0
4
t   n0 ln 4  3.47 10 7 s




(iv) 0.95 5 1014  5 1014 1  e t /  n 0

t   n0 ln 20  7.49 10 7 s
_______________________________________
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
no  N d  N a  8 10  2 10
n  5  1014 e  210  / 510 
1
 61015 cm 3
po 

po
(a) Ro 
 p0
  5.4 10 cm
6 2
n
1.8 10

no
6 1015
2
i
4
5.4 10
 4 10 4 
so  p 0  1.35  10 s
 9.16 1012 cm 3
4

 p0
(b) (i) n0  5 10 cm
14

(b) p  g  p 0  2  10
1.35 10 

8
21
_______________________________________
6.17
(a) (i)For 0  t  5 10 7 s
t /  p 0



 5  10 20 5  10 7 1  e

 2.5  10 1  e
14
t /  p 0
t /  p 0
 cm

(b) J n  eD n

 t  510 7 /  pO


cm 3
(ii) p 5 10 7  1.58 1014 cm 3
(b) (i) For 0  t  2 10

6
pt   2.5  10 1  e
14
At t  2 10 6 s,
s
t /  p 0

 cm

6
 2.454 10 cm
pt   2.454 1014 e

7

3
For t  2 10 6 s,


 t  210 6 /  pO

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  n0 e t /  n 0
 
1/ 2

 5 1014 e t /  n 0 cm 3

eD n
2  1014 e  x / Ln
Ln








 1.6 10 19 31.08 2 1014  x / Ln
e
5.575 10 3

6.20
(a) p-type; p pO  10 14 cm 3
and
n pO 
 10 21 5 10 7 e t /  n 0
d n 
d
 eD n
2  10 14 e  x / Ln
dx
dx
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
p  2.5  1014 1  e  210  / 510 
14

(a) nx   px   2 1014 e  x / Ln cm 3
 1.58 1014 cm 3


Ln  Dn n 0  31.08 10 6
 5.575 10 3 cm
p  2.5 1014 1  e 1 / 1 

 31.08 cm 2 /s
3
At t  5 10 s,
pt   1.58 1014 e

6.19
p-type; minority carriers - electrons
 kT 
Dn    n  0.02591200
 e 

7
For t  5 10 7 s

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


n  5 1014  9.16 1012 1  e t /  n 0
 9.16 1012
 4.908 1014 1  e t /  n 0  9.16 1012 cm 3
8

For t  2 10 6 s
3

ni2
1.5 1010

p pO
1014
  2.25 10 cm
2
6
(b) Excess minority carrier concentration
n  n p  n pO
At x  0 , n p  0 so that
n0  0  n pO  2.25 10 6 cm 3
3
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  Dn nO
dx 2
Ln
The general solution is of the form
x
x
  B exp

n  A exp

L 
 Ln 
 n 
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 expsx  ,
 As 2 expsx 
dx
dx 2
Substituting into the differential equation
D p As 2 exp sx    p  o As exp sx 

 
where Ln  Dn n 0  25 10
6

s2 
3
 5 10 cm
d n 
d
J n  eD n
 eD n
5  10 14 e  x / Ln
dx
dx
eD
  n 5  10 14 e  x / Ln
Ln





1.6 10 255 10  e
5 10 
19
14
J n  0.4e  x / Ln A/cm 2
(a) For x  0 ,
n0  5 1014 cm 3
J n 0  0.4 A/cm 2
J p 0  0.4 A/cm 2
nLn   5 10 e
J n Ln   0.4e
1
0
 po
Dp
s
1
0
L2p
The solution for s is
2


 p

1 p
4
s 
o  
 o   2 
 Dp

2 Dp
Lp




Then
(b) For x  Ln  5 10 3 cm,
1
1
 pO
which can be rewritten as
2


  p Lpo 
1   p Lpo


s

 1 
 2D p 
L p  2D p




Define
 p Lpo

2D p
 x / Ln
3
14
0
Dividing by D p , we have
1/ 2

 pO
or
Dp s 2   po s 
nx   5 1014 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  3L n
n3Ln   5 1014 e 3  2.49 1013 cm 3
J n 3Ln   0.4e 3  0.020 A/cm 2
J p 3L n   0.4e 3  0.020 A/cm 2
_______________________________________
s
1 
  1   2 

L p 
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 
s 
  1   2 

L p 
_______________________________________
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
Dn
 no

0
dx
 nO
dx 2
or
d 2 n   n
d n  n

o
 2 0
2
Dn
dx
dx
Ln
We have that
Dn  kT 
   so we can define
n  e 
n
o
1
o 

kT e L 
Dn
Then we can write
d 2 n  1 d n  n
 
 2 0
L  dx
dx 2
Ln
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 0

L Ln
which yields
2


L 
1  Ln

  n   1

Ln  2 L 
 2L  


We may note that if  o  0 , then L   
2
1
and  
Ln
(b)
 kT 
Ln  Dn nO where Dn   n  
 e 
so
Dn  12000.0259  31.1 cm /s
and
2
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
o
12
cm
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


or
Dp

g

0
2
2
Dp
dx
dx
For  L  x   L , g   G o = constant. Then
G
d p 
  o x  C1
dx
Dp
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 p 
d p 
 C 3 and
 0 so that
dx
dx 2
p  C 3 x  C 4
For 3L  x   L , g   0 so that
2
d 2 p 
d p 
 C 5 and
dx
 0 so that
dx 2
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)

Go L
3L  x  for 3L  x  L
Dp
_______________________________________
p 
6.27
V
8

 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 

Assume that f x, t   4 Dt 1 / 2 exp

 4Dt 
is the solution to the differential equation
  2 f  f
D 2  
 x  t
To prove: we can write
  x2 
f
1 / 2   2 x 

 4 Dt  
 exp

x
 4 Dt 
 4 Dt 
and
2 f
x
 4 Dt 
2
2
1 / 2
  x2 
  2 x 

 exp


 4 Dt 
 4 Dt 
  x 2 
 2 


 exp

 4 Dt 
 4 Dt 

G L
p  o 3L  x  for L  x  3L
Dp

This value is very close to 0.0259 for
T  300 K.
_______________________________________


D p  10.42 cm /s
2
2
  x2 
f
1 / 2   x   1 
 2  exp

 4 Dt  

 4Dt 
t
 4D  t 


  x2 
  1  3 / 2

exp
 t

 2 
 4Dt 
2 f
Substituting the expressions for
and
x 2
f
into the differential equation, we find
t
0 = 0.
Q.E.D.
(b)
Consider

  x2 
dx
exp

4
Dt



 4 D 
1 / 2

390.6202 9.35 10 6 2
16 32 10 6
Also

Let u  x 2 , then du  2 x  dx or
du
du
dx 

2x 2 u
Let a 
Now
1
4 Dt
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 6
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________


6.31
(a) p-type
  x2 
  x2 
dx  2 exp

exp

 4 Dt dx
 4 Dt 



0



2

 2 u exp au du   u exp au du
1
0

1
0

a
 4Dt
or
Then

p 
E Fi  E F  kT ln  o 
 ni 
 5 1015 

 0.0259 ln 
10 
 1.5 10 
x 
2
4D t
 4Dt exp 4Dt dx  4Dt  1
1

_______________________________________
6.29
Plot
_______________________________________
E Fi  E F  0.3294 eV
(b)
n  p  51014 cm 3
and
no 

 10 cm

2
4
3
 n  n 

E Fn  E Fi  kT ln  o

 ni 
 4.5 10 4  5 1014 

 0.0259 ln 

1.5 1010


n 
(a) E F  E Fi  kT ln  o 
 ni 
15
  4.5 10 cm
Then
6.30
 4 1016 

 0.0259 ln 
10 
 1.5 10 
 0.383225 eV
(b) n  p  g  p 0  2  10 21 5  10 7

ni2
1.5 1010

po
5 1015
or

3
 n  n 

E Fn  E Fi  kT ln  o

 ni 
 4 1016  1015 

 0.0259 ln 
10

 1.5 10

 0.383865 eV
 p  p 

E Fi  E Fp  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
 0.640 meV
or
_______________________________________
E Fn  E Fi  0.2697 eV
and
 p  p 

E Fi  E Fp  kT ln  o

 ni 
 5 1015  5 1014 

 0.0259 ln 

1.5 1010


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 
 ni 
 i 
 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
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 
 2 1014 

 0.0259 ln 
10 
 1.5 10 
 0.2460 eV
_______________________________________
 n 
(a) E Fn  E Fi  kT ln  
 ni 
 E  E Fi 
or n  ni 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 

n n

(a) (i) E Fn  E Fi  kT ln  o

 ni 
 1.02 1016 

 0.0259 ln 
6 
 1.8 10 
 0.58166 eV
 
 p 
(ii) E Fi  E Fp  kT ln  
 ni 
 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
6.33

6.34

 p  p 
p 
  kT ln  o 
 kT ln  o

n 
 ni 
 i 
 p o  p 

 kT ln 

 po 
(ii) E F  E Fp
 6 1015  5.05 1014 

 0.0259 ln 

6 1015


3
 2.093 10 eV
or
 2.093 meV
_______________________________________
 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
no  i 
 3.24 10  4 cm 3
po
1016
We have
 x 
n  1014  
 50 
Then
 3.24 10 4  1014 x 50 
E Fn  E Fi  kT ln 

1.8 10 6


We find
 

x (  m)
0
1
2
10
20
50
( E Fn  E Fi ) (eV)
-0.581
+0.361
+0.379
+0.420
+0.438
+0.462

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)
10 12
0
+0.58115
50
+0.58140
_______________________________________
10 14
10 15
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

10
0.10877
0.16841
0.22805
0.28768
13
 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

 p  p 
p 
  kT ln  o 
 kT ln  o

n 
 ni 
 i 
10 12
0.365274
10
13
0.365286
10
14
0.365402
15
10
0.366536
_______________________________________
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1012 cm 3
(b)
 p 
E Fn  E Fi  kT ln  
 ni 
 5 10 

 0.0259 ln 
10 
 1.5 10 
12
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   ni . For n  p  0
 ni2
 ni

 pO ni   nO ni  pO   nO
(b) We had defined the net generation rate as
g  R  g o  g   R o  R  
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.
_______________________________________
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  ni2
R
C n n  n   C p  p  p 


np  n 
2
i

 pO n  ni    nO  p  ni 
If n  n o  n and p  p o  n , then
no  n p o  n  ni2
 pO no  n  ni    nO  p o  n  ni 
2
no p o  nno  p o   n  ni2

 pO no  n  ni    nO  p o  n  ni 
2
If n  n i , we can neglect n : also
R
no p o  ni2
Then
R
nno  p o 
 pO no  ni    nO  p o  ni 
(a) For n-type; n o  p O , n o  ni
Then
R
1

 10  7 s 1
n  pO
(b) For intrinsic, n o  p o  ni
Then
2ni
R


n  pO 2ni    nO 2ni 
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
 AD p
Lp
R

1
 pO   nO

The excess concentration is then

  x 
s

p  g  pO 1 
 exp
 L p 
Dp Lp  s



where

(c) For p-type; p o  no , p o  ni
Then
R
1
1


 2  10  6 s 1
n  nO 5  10  7
_______________________________________
6.41
(a) From Equation (6.56)
d 2 p 
p
Dp
 g
0
2
 pO
dx
Solution is of the form
x


  B exp  x 
p  g  pO  A exp
 Lp 
 Lp 





L p  D p pO 
Now
p  10 21 10 7
 
1010 7   10 3 cm


  x 
s


 1 

exp
3
 L p 
 10 10  s



1

10 7  5 10 7
R
 1.67  10  6 s 1
n

Solving for A , we find
 sg  pO
A
Dp
s
Lp
or
n

 s g  pO  A

or

p  1014 1 

(i) For s  0 ,
  x 



exp
 L p 
10 4  s


s
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 



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
n 
(ii) For s  2000 cm/s,
p0  0.833 1014 cm 3
(iii) For s   ,
p0  0
_______________________________________
6.42
Ln  Dn nO 
255 10 7 
 35.4 10 4 cm
(a) At x  0 ,
g  nO  2 10 21 5 10 7  1015 cm 3
or
n0  g  nO  1015 cm 3
For x  0
d 2 n  n
d 2 n  n
Dn


0

 2 0
 nO
dx 2
dx 2
Ln
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  x 

 Ln 
W 
sinh  
 Ln 
n0 sinh 


 W 
 W 
  B exp


 L 
L
 n 
 n 
Solving these two equations, we find
 n0 exp 2W Ln 
A
1  exp 2W Ln 
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 
n  0  A exp
   W  x  
  W  x   
 exp 
  exp 

  L n

 Ln
 
which can be written as
where
n0  1015 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
D p  sW 
B
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 

Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 6
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(a) For s   ,
p  1018 20 10 4  x  cm 3
Then
J p  eD p

d p 
dx
 
  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
 Go  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 
(4)
continuous at x  0
dx
Applying the boundary conditions, we find
G W
G W 2
C1  C 3   o
and C 2  C 4   o
Dn
Dn
Then for W  x  0
G
n  o  x 2  2Wx  2W 2
2Dn
and for 0  x  W
G W
n  o W  x 
Dn
_______________________________________


6.45
Plot
_______________________________________
6.48
(a) GaAs:
V
2
R 
 10 6 
I 2  10  6
L
and   e  n   p p
R
 A


p  g  p 0  10 21 5  10 8   5  10 13 cm 3
For N d  1016 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  1016 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 1013
 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
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 7
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Chapter 7
(b) N d  51016 cm 3 , N a  51016 cm 3
7.1
N N 
Vbi  Vt ln  a 2 d 
 ni 
(a)
Si: Vbi  0.778 V
Ge: Vbi  0.396 V



 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 

10 2
 1.5  10






 0.790 V

















7.3
(a) Silicon ( T  300 K)
 Na Nd 
Vbi  0.0259  ln 

10 2
 1.5  10




Si: ni  1.5 1010 cm 3
Ge: ni  2.4 1013 cm 3
GaAs: ni  1.8 10 6 cm 3
N N 
Vbi  Vt ln  a 2 d  and Vt  0.0259 V
 ni 
Then
Si: Vbi  0.635 V
Ge: Vbi  0.253 V
GaAs: Vbi  1.10 V
Si: Vbi  0.814 V
Ge: Vbi  0.432 V
GaAs: Vbi  1.28 V
_______________________________________
7.2
(a) N d  1014 cm 3 , N a  1017 cm 3 '
(c) N d  1017 cm 3 , N a  1017 cm 3

 2 1017 2  1017 

(iii) Vbi  0.0259  ln 
2
 1.5  1010

 0.850 V
_______________________________________

GaAs: Vbi  1.25 V


3
For N a  N d  10 cm ; Vbi  0.4561 V
14
 1015
;
 10
;
16
 0.5754 V
 0.6946 V
 0.8139 V
 10
;
(b) GaAs ( T  300 K)
 Na Nd 
Vbi  0.0259  ln 

2
 1.8  10 6 
17


3
For N a  N d  10 cm ; Vbi  0.9237 V
14
 1.043 V
 10
;
 1.162 V
17
 10
;
 1.282 V
(c) Silicon (400 K), kT  0.034533
 1015
;
16
ni  2.38 1012 cm 3
For N a  N d  1014 cm 3 ; Vbi  0.2582 V
 0.4172 V
 10
 0.5762 V
;
17
 10
 0.7353 V
;
9
GaAs(400 K), ni  3.29 10 cm 3
 1015
;
16
For N a  N d  1014 cm 3 ; Vbi  0.7129 V
 0.8719 V
 10
;
 1.031 V
17
 10
;
 1.190 V
_______________________________________
 1015
16
;
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 7
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
or
7.4
(a) n-side
x p  0.0213  10 4 cm  0.0213  m
N 
E F  E Fi  kT ln  d 
 ni 
 5 1015 

 0.0259 ln 
10 
 1.5 10 
We have
 max 

or
E F  E Fi  0.3294 eV
p-side
N 
E Fi  E F  kT ln  a 
 ni 
 1017 

 0.0259 ln 
10 
 1.5 10 
or
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 
 

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
Vbi  0.7363 V
(d)
1/ 2

 211.7  8.85 10 14 0.736

1.6 10 19

 1017 
1

 17
 

15 
15 
 5  10  10  5  10 
1/ 2
E F  E Fi  0.3653 eV
p-side
N 
E Fi  E F  kT ln  a 
 ni 
 2 1016 

 0.0259 ln 
10 
 1.5 10 
or
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 cm  0.426  m
Now
 211.7  8.85 10 14 0.736
xp  
1.6 10 19

4
Vbi  0.7305 V
(d)

 5  1015 
1

 17
 

17
15 

 10
 10  5  10 

 2 1016 2 1016 
 0.0259  ln 

2
 1.5 1010

or

4
15
_______________________________________
or

19
 max  3.29  10 4 V/cm

 2  V  N 

1

x n   s bi  a 

 N d  N a  N d 
 e
1.6 10 5 10 0.426 10 
11.78.85 10 
or
 10 17 5  10 15 
 0.0259  ln 

2
 1.5  10 10


eN d x n
s
1/ 2
 2  V  N 

1

xn   s bi  a 

 e
 N d  N a  N d 
1/ 2
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

For 300 K;
 2  1016 
1


 

16 
16
16 
 2  10  2  10  2  10 
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

xn  0.25W  0.25 xn  x p
0.75xn  0.25x p 
xn N d  x p N a 
 max  4.75  10 4 V/cm
_______________________________________




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
_______________________________________
xp
xn



3
xp
Nd

3
Na
xn
So N d  3N a
 E  E Fi 
(b) N d  ni 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
10
 1.5 10

or N d  1.98 1016 cm 3
or 3N a2  1.5 1010
7.6





 E  EF 
N a  ni exp Fi

 kT

 0.330 
 1.5 1010 exp

 0.0259 




200 K; kT  0.017267 ; ni  1.38 cm 3
ni  1.8 10 6 cm 3
400 K; kT  0.034533 ; ni  3.28 10 9 cm 3
For 200 K;
 2  1015 4  1016 
Vbi  0.017267  ln 

1.382


 1.257 V


2


3
N d  2.33 1016 cm 3
7.7
300 K; kT  0.0259 ;
710 
 exp 00..0259

which yields N a  7.766 10 cm
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
_______________________________________


15




 2  V  N 

1

x n   s bi  a 

 e
 N d  N a  N d 

1/ 2

 211.7 8.85 10 14 0.710

1.6 10 19

 
1
 1 
  
15  
 3   4 7.766  10  

 x n  9.93 10 6 cm
or x n  0.0993  m

1/ 2


 211.7 8.85 10 14 0.710
xp  
1.6 10 19

 
1
 3 
  
15  
 1   4 7.766  10  

 2.979 10 5 cm
or x p  0.2979  m

1/ 2
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 7
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________

 max 

eN d x n
s
1.6 10 2.33 10 0.0993 10 
11.78.85 10 
19
 1016 
1

  15  16

15 
 10  10  10 
4
16
14

x n  0.8644 10 4 cm  0.8644  m
Now



1
 2  V  N 

x p   s bi  d 

e
N
N

N

d 
 a  a


which yields N a  8.127 1015 cm 3



 213.1 8.85 10 1.180
xn  
1.6 10 19

 
1
 1 
  
15  
 3   4 8.127  10  

 1015 
1

  16  16

15 
 10  10  10 
x p  0.08644  10 4 cm  0.08644  m
(c)


 
1
 3 
  
15  
 1   4 8.127  10  

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 
19
1/ 2
or
 213.1 8.85 10 14 1.180
xp  
1.6 10 19

 max 

1/ 2

 1.324 10 5 cm
or x n  0.1324  m
1/ 2
 211.7 8.85 10 14 0.6350

1.6 10 19

N d  2.438 1016 cm 3
14
1/ 2
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 
4
16
14
 4.45  10 4 V/cm
_______________________________________
7.9

 211.7 8.85 10 14 0.6350

1.6 10 19

Now

ni2  1.5 1010
  


 1016 1015 

(a) Vbi  0.0259  ln 
2
 1.5  1010 
or
Vbi  0.635 V
(b)
 2  V  N 

1

x n   s bi  a 

 e
 N d  N a  N d 



 2 1017 4  1016 

(a) Vbi  0.0259  ln 
2
 1.5  1010

 0.80813 V
(b) V bi increases as temperature decreases
At T  300 K, we can write



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 
ni2  K 2.8 1019 1.04 1019 

 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

Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 7
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Then


7.12
(b) For N d  1016 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 
 1016 

 0.0259 ln 
10 
 1.5 10 
0.82494  0.80813
 100%  2.08%
0.80813
 2%
_______________________________________

or
E F  E Fi  0.3473 eV
For N d  1015 cm 3
 1015 

E F  E Fi  0.0259 ln 
10 
 1.5 10 
7.11
or
N N 
Vbi  Vt ln  a 2 d 
 ni 



16
15
 T   4  10 2  10 
0.550  0.0259 
 ln 

2
ni
 300  

Using the procedure from Problem 7.10, we
can write, for T  300 K,


  1.12 
 K 2.8 10 1.04 10 exp

0.0259
ni2  1.5 1010
2
19
19

 K  4.659
At T  300 K,




n  4.659 2.8 10
19

  


or
Vbi  0.456 V
(b)
 211.7 8.85 10 14 0.456
xn  
1.6 10 19


1.04 10 
 380 


 300 



 4 1016 2 1015 
Vbi  0.032807  ln 

24
 4.112 10

 0.5506 V  0.550 V
_______________________________________

 1012 
1

  16  12

16 
 10  10  10 
3
  1.12 
 exp

 0.032807 
 4.112 10 24
Then
N N 
(a) Vbi  Vt ln  a 2 d 
 ni 


19
7.13
 1012 1016 
 0.0259  ln 

2
 1.5  1010 
 4  1016 2 1015 
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
2
i
E F  E Fi  0.2877 eV
Then
Vbi  0.34732  0.28768
or
Vbi  0.0596 V
_______________________________________
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 
1

  12  12

16 
 10  10  10 
or
x p  2.43  10 3 cm
1/ 2
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 7
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(d)
7.15
1/ 2
eN d x n
 max 
s

1.6 10 10 2.43 10 
11.78.85 10 
19
7
16

14
or
 max  3.75  10 2 V/cm
_______________________________________
7.14
Assume silicon, so
  kT 

L D   2s

 e Nd 

1/ 2



 11.7  8.85 10 14 0.0259  1.6 10 19 


2


1.6 10 19 N d
or

 1.676  10
L D  
Nd

5





 2eVbi  N a N d 


 max  
 s  N a  N d 
We find
2 1.6 10 19
2e

 3.0904 10 7
s 11.7 8.85 10 14
(a)
(i) For N a  1017 , N d  1014 ; Vbi  0.6350 V
(a) N d  81014 cm 3 , L D  0.1447  m


(ii)
 1015 ;
(iii)
 10 ;
(iv)
 1017 ;
16
N d  1014 ;  max  0.443  10 4 V/cm
(ii)
 1015 ;
 1.46 10 4 V/cm
(iii)
 1016 ;
 4.60  10 4 V/cm
 1017 ;
 11.2 10 4 V/cm
(iv)
(b)
(i) For N a  1014 , N d  1014 ; Vbi  0.4561 V
(b) N d  2.2 1016 cm 3 , L D  0.02760  m
(ii)
 1015 ;
(c) N d  81017 cm 3 , L D  0.004577  m
Now
(a) Vbi  0.7427 V
(iii)
 10 ;
(iv)
 1017 ;
(b) Vbi  0.8286 V


 8  1017 
1


 
17



 N d  8 10  N d 
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
L
(b) D  0.1267
xn
LD
 0.1677
xn
_______________________________________
(c)
16
 0.5157 V
 0.5754 V
 0.6350 V
(i) For N a  1014 ,
N d  1014 ;  max  0.265  10 4 V/cm
(c) Vbi  0.9216 V
Also
 211.7  8.85  10 14 Vbi 
xn  
1.6  10 19


 0.6946 V
 0.7543 V
 0.8139 V
(i) For N a  1017 ,
1/ 2
1/ 2

(ii)
 1015 ;
 0.38110 4 V/cm
(iii)
 1016 ;
 0.420 10 4 V/cm
(iv)
 1017 ;
 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.
_______________________________________
(c)
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 

1/ 2
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 7
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(i) For V R  0 ,



 211.7 8.85 10 14 0.6767 
W 
1.6 10 19

 5  10 16  10 15  

16
15  
 5  10 10  

 4 1016 
1



 

17 
17
16 
2

10
2

10

4

10






1/ 2
 

 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  10 16  10 15  

16
15  
 5  10 10  

1/ 2
 


 0.3584 10 cm
or W  0.3584  m
Also W  xn  x p  0.3584  m
20.6767 
 1.43  10 4 V/cm
4
0.9452  10
(ii)For V R  5 V,
20.6767  5
 4.15  10 4 V/cm
2.738  10  4
_______________________________________
 max 
(c)  max 



2Vbi  V R  20.8081  2.5

W
0.3584  10  4
 1.85 10 5 V/cm


e s N a N d
(d) C  A

 2Vbi  V R N a  N d  


 1.6 10 19 11.7 8.85 10 14
 2 10  4 
20.8081  2.5




 2  V  V R   N a 

1



x n   s bi




e

 N d  N a  N d 



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 

 211.7 8.85 10 14 0.8081  2.5

1.6 10 19


 2 10 
1


 

16 
17
16 
 4 10  2 10  4  10 


1/ 2
 2  10 17 4  10 16  

17
16  
 2  10  4  10  
 2 1017 4  1016 

(a) Vbi  0.0259  ln 
2
 1.5  1010

 0.8081 V
(b)
17

1/ 2
4
 max 


 2  10 17  4  10 16  

17
16  
 2  10 4  10  
 2.738 10 cm
or W  2.738  m
2Vbi  V R 
(c)  max 
W
(i)For V R  0 ,

1/ 2
 211.7 8.85 10 14 0.8081  2.5

1.6 10 19

4
7.17
1/ 2
 5.97 10 6 cm
or x p  0.0597  m
 9.452 10 5 cm
or W  0.9452  m
(ii) For V R  5 V,

 211.7 8.85 10 14 0.8081  2.5

1.6 10 19

1/ 2
 0.2987 10 4 cm
or x n  0.2987  m


1
 2  V  V R   N d 

x p   s bi




e

 N a  N a  N d 


1/ 2
 80 N 2 
 Vt ln  2 d 
 ni 
We find
V 
80 N d2  ni2 exp bi 
 Vt 

 1.5 1010
740 
 exp 00..0259

 5.762 10
2

32

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 

1



x n   s bi




e

 N d  N a  N d 

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 
1/ 2
 2.262 10 4 cm
or x n  2.262  m

1/ 2

(c) For a larger doping, the space charge
width narrows which results in a larger
capacitance.
_______________________________________
or


1/ 2
3 10    21.6 10 V  V 
 11.7 8.85  10 
4 10 4 10 

19
bi
R
15
17

4 10  4 1017 
15
or
9 1010  1.224 10 9 Vbi  V R 
so that
Vbi  VR   73.53 V
which yields
V R  72.8 V
 9.38  10 4 V/cm
1/ 2




 1.6 10 19 11.7 8.85 10 14

20.740  10



14
2.262  0.028310  4


 2eVbi  V R   N a N d 


 max  
 N  N 
s

d 
 a

1/ 2
2Vbi  V R 
W
20.740  10 


 4 1015 4  1017 
(a) Vbi  0.0259  ln 

10 2
 1.5 10

5 2


e s N a N d
(d) C   

 2Vbi  V R N a  N d  
 3  1.732
Vbi  0.766 V
Now
 2.83 10 6 cm
or x p  0.0283  m

1/ 2
or
 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
 

1/ 2
C 3 N a   3 N a 


C N a   N a 
7.20


1
 2  V  V R   N d 

x p   s bi




e
N
N

N

d 
 a  a


(c)  max 
 e s N a 
(b) C   

 2Vbi  V R  
 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 3 N  
N N 
 Vt ln  d 2 a   Vt ln  d 2 a 
 n i

 n i 

 N N 
N N 


 Vt ln 3  ln  d 2 a    Vt ln  d 2 a 

 n i  
 n i 




 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  VR   8.008 V




4 10 4 10 
16
17

4 10  4 1017 
16
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 7
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
which yields
V R  7.18 V
(b)


 4  10 4  10
(c) Vbi  0.0259  ln 
10 2
 1.5  10
17

17

2VbiA  V R 
 A 
W  A
W B  VbiA  V R



B  2VbiB  V R  W  A VbiB  V R
W B 


or
 1  5.7543 



 3.13  5.8139 
Vbi  0.886 V
We have
 2 1.6  10 19 Vbi  V R 
2
3  10 5  
14
 11.7  8.85  10




or
 A
 0.316
B 
(c)


4 10 4 10 

17
17

4 10  4 1017 
17
C j  A
so that
Vbi  VR   1.456 V
which yields
V R  0.570 V
_______________________________________
C j B 
1/ 2
or
C j  A
C j B 
1/ 2
1/ 2
 0.319
_______________________________________
or
W  A  VbiA  V R  N a  N dA   N dB 




W B   VbiB  V R  N a  N dB   N dA 
We find
1/ 2
7.22
(a) We have
C j 0 
  


 1018 1015 
VbiA  0.0259  ln 
  0.7543 V
2
 1.5  1010 
  


 1018 1016 
VbiB  0.0259  ln 
  0.8139 V
2
 1.5  1010 
We find
W  A  5.7543  1018  10 15 

 

W B   5.8139  10 18  1016 
 10 16 
  15 
 10 
or


s N a N dB


 2VbiB  V R N a  N dB  
 1015  5.8139  1018  1016 

  16 
 18
15 
 10  5.7543  10  10 
1/ 2
W  A

W B   2  V  V   N  N  1 / 2
s
biB
R
dB
 a


 N N

e

a
dB  


1/ 2
 N  V  V R  N a  N dB 


  dA  biB


 N dB  VbiA  V R  N a  N dA 
7.21
(a)
 2 s VbiA  V R   N a  N dA 



 N N

e

a
dA  





s N a N dA





2
V

V
N

N
R
a
dA 
 biA
W  A
 3.13
W B 
1/ 2
C j 10 
or

 s N a N d



 2Vbi N a  N d  
1/ 2


s N a N d


 2Vbi  V R N a  N d  
C j 0 
 V  VR 

 3.13   bi


C j 10 
 Vbi 
For V R  10 V, we find
3.132 Vbi  Vbi  10
or
Vbi  1.137 V
(b)
x p  0.2W  0.2 x p  xn


1/ 2
1/ 2
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
 0.25 N a2 
1.137  0.0259  ln 

2
 1.8  10 6 
We can then write
 1.137 
1.8 10 6
Na 
exp 

0.25
 20.0259 
which yields
N a  1.23 1016 cm 3
and
N d  3.07 1015 cm 3
_______________________________________

7.23






1.162  V R 2
1.162  0.5
1.162  V R 2
1.50 2 
1.662
which yields V R 2  2.58 V
_______________________________________





e s N a N d
C  AC   A

 2Vbi  V R N a  N d  
15
16
15
16


1/ 2

 2 1015 4  1016 
(b) Vbi  0.0259  ln 

2


1.8  10 6
 1.157 V




e s N a N d
C  AC   A

 2Vbi  V R N a  N d  



1/ 2
 1.6  1019 13.1 8.85  1014
 5  10 4 
21.157  VR 



2 10 4 10  
2 10  4 10 
15
16
15
16
1/ 2
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




1/ 2

 1.6 10 19 11.7 8.85 10 14
 8 10  4 
20.7543  10


1/ 2

0.6889  V R


e s N a N d
(a) C  AC   A

 2Vbi  V R N a  N d  
 2 1015 4  1016 

(a) Vbi  0.0259  ln 
2
 1.5  1010

 0.6889 V


6.2806 10 12
(i) For V R  0 ,
C  7.567 pF
(ii) For V R  5 V,
C  2.633 pF
C
1.50 

C

Vbi  V R 2
C V R1 
So

C V R 2 
Vbi  V R1
7.24

2 10 4 10  

2 10  4 10 

 2  1016 5  1015 
Vbi  0.0259  ln 

2


1.8 10 6
 1.162 V
1
C 
Vbi  V R

 1.6 10 19 11.7 8.85 10 14
 5 10  4 
20.6889  V R 

Then
xp
2 10 5 10  

2 10  5 10 
17
15
17
15
C  4.904 10 12 F
1
1
f 
L
2
C 2 f 
2 LC

1/ 2
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 7
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
L

4.904 10 2 1.25 10 
12
2
6



 1.6 10 19 13.1 8.85 10 14
 10  4 
2Vbi  2

1
4 N 

 3.306 10 3 H  3.306 mH
2
d
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 6 Hz  1.069 MHz
_______________________________________
N a  6.016 1015 cm 3 ,
Vbi  1.10 V
(b) From part (a),
0.6 10 12  2.724 10  20
7.26
 2eVbi  V R N d 
 max  

s


Let Vbi  0.75 V
(a)
Nd
Vbi  2
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 
1/ 2
5 N d 
(b)
(i) For V R  1 V, C  12.14 pF
f 

Nd
Vbi  5
By trial and error,
N d  2.976 1015 cm 3 ,
1/ 2
N a  1.19 1016 cm 3 ,
Vbi  1.135 V
_______________________________________
2.5 10 
 21.6  10 0.75  10 N 

11.7 8.85 10  

5 2
19
7.28
d
14
 N d  1.88 10 cm
16
 
(b) 10
3
5 2


 2 1.6  10 19 0.75  10 N d 


11.7  8.85 10 14




 N d  3.0110 cm
_______________________________________
7.27

x p  0.20W  0.20 xn  x p
0.8x p  0.2xn
xn  4x p


1/ 2

 211.7 8.85 10 14 0.5574

1.6 10 19


 1014 
1

 14
 

15 
15 
 5  10  10  5  10 
 
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 
 2  V  N 

1

x n   s bi  a 

 e
 N d  N a  N d 
 4 N d2

 0.0259  ln 

6 2
 1.8  10 

 

 2  V  N 

1

x p   s bi  d 

 e
 N a  N a  N d 
3
15

 5  1015 1014 

(a) Vbi  0.0259  ln 
2
 1.5  1010

or
Vbi  0.5574 V
(b)



e s N a N d
C  AC   A

 2Vbi  V R N a  N d  
1/ 2
1/ 2
1/ 2
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

7.30
 5  1015 
1

 14
 

14
15 

 10
 10  5  10 
1/ 2

1/ 2


C
(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 
  1.20
(a) Vbi  0.0259  ln 
2
 1.8  10 6 

.20 
8 10 N  1.8 10  exp 01.0259

6 2
16

 N d  5.36 10 cm
19
R
14
which yields
V R  193 V
(b)
xp Nd
N 

 x n  x p  a 
xn
Na
 Nd 
so
 1014 
x n  50 10  4  16 
 10 
4
 0.50 10 cm  0.50  m
(c)
2V
2193.15
 max  R 
W
50.5  10  4
or
 max  7.65  10 4 V/cm

_______________________________________
1/ 2
Vbi  V R
15
4 2

1.287 10 13
d
50 10   211.78.85 10 V
1.6 10 10 



1/ 2
14

 11.7 8.85 10 14 2 1015
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
_______________________________________
or

 1.6 10 19
 10 5 
 2Vbi  V R 
 211.7  8.85  10 14 Vbi  V R 
30  10  4  
1.6  10 19

2 V 
xp   s R 
 eN a 

 e s N d 
(b) C  AC   A  

 2Vbi  V R  
or
x n  2.66 10 4 cm
(c) For x n  30  m, we have

 2 1017 2  1015 
(a) Vbi  0.0259  ln 

2
 1.5 1010

 0.7305 V

3


e s N a N d
(b) C  AC   A

 2Vbi  V R N a  N d  

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  5.36 10 
16
15
 A  7.56 10 5 cm 2
1/ 2




 1.6 10 19
(c) 0.80 10 12  7.56 10 5 
 2Vbi  V R 


13.18.85 10 14 8 1016 5.36 1015 
8 10  5.36 10 
16
1.0582 10 8 
15
1/ 2


2.1585 10 8
Vbi  V R
 Vbi  V R  4.161  1.20  V R
V R  2.96 V
_______________________________________
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
  0 at x   x p  C1  

eN aO
x  xp
s
eN dO x eN dO 
x 

 x n  O 
2 s
s 
2 
_______________________________________
1 
7.34
d 2 x 
 x 
dx 


2
s
dx
dx
For 2  x  1  m,  x   eN d
So
eN d x
d eN d


 C1
dx
s
s
(a)
eN aO x p
s
Then for  x p  x  0

Then for 0  x  x O we have

n-region, 0  x  x O
d 1  x  eN dO


dx
s
2 s
At x  2  m   xO ,   0
So
eN d x O
C1 
s
Then
eN d
x  x O 

s
At x  0 , 0  x  1  , so
0  
or
eN dO x
1 
 C2
2 s
n-region, x O  x  x n
d 2  x  eN dO


dx
s
s
or
2 
eN dO x
 C3
s
We have  2  0 at x  x n
eN x
 C 3   dO n
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 
2 s
s
which gives
eN 
x 
C 2   dO  x n  O 
s 
2 
1.6 10 5 10  110 
11.78.85 10 
19
15
4
14
or
0  7.726 10 4 V/cm
(c) Magnitude of potential difference is
eN d
x  x O dx
  dx 
s




eN d  x 2

 xO  x   C 2

s  2

Let   0 at x   xO , then
0
so that for x O  x  x n , we have
2  
eN d
 1  210  4
s

eN d  x O2
eN d x O2
2



x

C

C

O 
2
2
s  2
2 s

Then we can write
eN d
x  x O 2
 
2 s
At x  1  m
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
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  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
_______________________________________
7.35
(a) V B 
s 
2eN B
2
crit
or



 2
11.7 8.85 10 14 4 10 5
N B  s crit 
2eV B
2 1.6 10 19 40
Then

N B  N a  1.294 10 cm
16
(b) N B 


 2 1.6  10 19 Vbi  V R 
4  10 5  
 11.7  8.85  10 14
or

2


 2  10 2  10  


16
16
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 

2
 1.5  1010

 0.6587 V
Then
19
 2 1.6  10 Vbi  V R 
4  10 5  
 11.7  8.85  10 14

3





 5  10 5  10  


15
11.78.85 10 14 4 10 5 2
21.6 10 19 20
15
1/ 2

15
15 
 5  10  5  10  
 Vbi  V R  207.1
Or N B  N a  2.59 1016 cm 3
_______________________________________
So V R  206 V
_______________________________________
7.36
7.39
For a silicon p  n junction with
Na 



s 
11.7 8.85 10 4 10

2eV B
2 1.6 10 19 80
2
crit
14


5 2
 6.47 10 cm 3
_______________________________________
15
N d  51015 cm 3 and V B  100 V, then,
neglecting V bi we have
 2 s V B 
xn  

 eN d 
7.37
(a) For N d  1016 cm 3 , from Figure 7.15,

 211.7  8.85  10 14 100  


19
5  10 15 
 1.6  10

V B  75 V
(b) For N d  1015 cm 3 ,

1/ 2

1/ 2

or
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
 2eVbi  V R   N a N d 


 max  
 N  N 
s

d 
 a
Set  max   crit and V R  V B


1/ 2

 2 1016 2  1016 
Vbi  0.0259  ln 

10 2
 1.5 10

 0.7305 V


  


 1018 1018 
Vbi  0.0259  ln 
  0.933 V
2
 1.5 1010 
Now
eN d x n
 max 
s
so
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




e

 N d  N a  N d 
Then
 211.7 8.85 10 14
2
6.47 10 6  
1.6 10 19




1/ 2

 10 18 
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


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
At x   x O and x   xO ,   0
So
2
2
ea  x O 
ea  x O 
0

C

C


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   xO , then



0
1/ 2
Assume Vbi  V R

3

eax O3
ea   x O
 x O3   C 2  C 2 

2 s  3
3 s

Then
(a) For x p  75  m
75 10   211.78.85 10 V
1.6 10 10 
 eax O
ea  x 3
  xO2  x  

 3
2 s  3
s

_______________________________________
14
4 2
19
3
 x   
R
15
which yields
V R  4.35 10 3 V
7.44
We have that
(b) For x p  150  m
150 10   211.78.85 10 V
1.6 10 10 
14
4 2
19
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
Impurity gradien
2 1018
a
 10 22 cm 4
4
2 10
From Figure 7.15, V B  15 V
_______________________________________
 ea 2s

C  

12Vbi  V 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
_______________________________________
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 << N d

1/ 2


 3  1017 5  1015 
Then Vbi  0.0259  ln 

2
 1.5  1010

 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


 A  5.3110 5 cm 2
 1.6 10 19
(b) C j  5.309 10 5 
 2Vbi  V R 





 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
C j  1.24 pF
_______________________________________
(ii) For V R  0 ,
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 
V1  V 2 
I f2
 eV 
 kT 

I S exp 2 
 kT 
or
 kT   I f 1 
V1  V 2  
 ln
 e   I f 2 
(a)
I f1
For
 10 , then
I f2
V1 V2  0.0259 ln 10
or
V1 V 2  59.6 mV  60 mV
(b)
I f1
For
 100 , then
I f2
V1 V2  119.3 mV  120 mV
_______________________________________
n po 

  2.8125 10 cm
10 2
ni2
1.5 10

Na
8 1015

n2
1.5 1010
p no  i 
Nd
2 1015
4
  1.125 10 cm
5



  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
_______________________________________

 


  8.110 cm

  3.24  10 cm


ni2
1.8 10 6

Na
4 1016
n2
1.8  10 6
pno  i 
Nd
1016
(a) Va  0.90 V,
2
5
3
2
4
3
 0.90 
p n x n   3.24 10  4 exp

 0.0259 
 4.0 1011 cm 3
3

 

 0.90 
n p  x p  8.110 5 exp

 0.0259 
3
 10.0 1010 cm 3
(b) V a  1.10 V


 1.10 
p n x n   3.24 10  4 exp

 0.0259 
 9.03 1014 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 

 3.95 1012 cm 3
 

 4.69 1013 cm 3
(c) Va  0.55 V
 0.45 
p n x n   1.125 10 5 exp

 0.0259 

 
 0.55 
n p  x p  2.8125 10 4 exp

 0.0259 



 1.88 1014 cm 3
2
V 
p n x n   p no exp a 
 Vt 
V 
n p  x p  n po exp a 
 Vt 
(a) Va  0.45 V,


 0.55 
p n x n   1.125 10 5 exp

 0.0259 
n po 
or

(b) Va  0.55 V,
8.3
V1 V2  0.0259 ln 100
8.2

n p  x p  9.88  10 11 cm 3
(c)
 2.26 1014 cm 3
p n x n   0


np  xp  0
_______________________________________
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 


2
ni2
1.5 1010

Nd
5 1015


10

ni2
1.5 1010

Nd
3 1016

(c) I  I n  I p  1.85  4.52  6.37 mA
_______________________________________
8.6
2



Ln
en
Na

V 
exp a 
 Vt 
Dn
1.6 10 1.8 10 
6 2
19
5  1016
205
5  10 8
 1.10 
 exp

 0.0259 
 1.849 A/cm 2
I n  AJ n  x p   10 3 1.849  A

or I n  1.85 mA

 nO
10 1.6 10 1.5 10 
4
19
10
10 2
16
25
10  6
I S  1.8 10 15 A
(a) For Va  0.5 V,
V 
 exp a 
 no
 Vt 
2
i
Dn
1
Na
or
8.5
eDn n po

or I p  4.52 mA



9.80
10 8
 4.521 A/cm 2
I p  AJ p x n   10 3 4.521 A
I S  Aeni2 
 0.1 7 1015 
 0.0259 ln 
4 
 3.214 10 
 0.6165 V
(ii) p-region - lower doped side
_______________________________________

16
For an n  p silicon diode
2
 7.5 10 3 cm 3
 0.1N a 
(i) V a  Vt ln 

 n po 
(a) J n  x p 
6 2
 1.10 
 exp

 0.0259 
 3.214  10 4 cm 3
p no 
V 
 exp a 
 p0
 Vt 
Dp
19


en i2
Nd
1.6 10 1.8 10 

 4.5 10 4 cm 3
V 
(i) p n x n   p no exp a 
 Vt 
 p x  
or Va  Vt ln  n n 
 p no 
 0.1 5 1015 
 0.0259 ln 

4
 4.5 10 
 0.599 V
(ii) n-region - lower doped side
n2
1.5 1010
(b) n po  i 
Na
7 1015
Lp
V 
exp a 
 Vt 
eD p p no
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
(b) For Va  0.5 V,


   0.5  
I D  1.8 10 15 exp
  1
  0.0259  
or
I D   I S  1.8 10 15 A
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 8
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
8.7
8.9
 1
J s  en i2 
 N a
Dn
1

 no N d


 1.6 10 19 2.4 1013
 1

15
 4  10

Dp 

 p0 

We have
2
48 

2  10  6 
90
1

6
2  10
2  1017
J s  1.568 10 4 A/cm 2
V 
(a) I  AJ s exp a 
 Vt 



 0.25 
 10  4 1.568 10  4 exp

 0.0259 
V  0.0259  ln 1  0.90
 2.44 10 4 A
or I  0.244 mA


(b) I   I s   AJ s   10 4 1.568 10 4

 1.568 10 8 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 
 I 
So Va  Vt ln 

 AJ s 


0.80
 0.0259 ln   4
7 
 10 10 
Va  0.6502 V
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  en i2 
 N a
 V  
I  I S exp   1
  Vt  
or we can write this as
V 
I
 1  exp 
IS
 Vt 
so that
 I

V  Vt ln   1
 IS

In reverse bias, I is negative, so at
I
 0.90 , we have
IS

 0.65 
(iii) I  1.029 10 14 exp

 0.0259 
 8.16 10 4 A
_______________________________________




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
1.20

V 
 0.72 

exp a  exp
 0.0259 
 Vt 
Case 4: I s 
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 n po eD p p no
Jn  J p

Ln
Lp
Dn
 no

Dn
 no


n i2
Na
2
i
or
2
i
1
D p no  N a 


Dn po  N d 
D p no  N a 
1

 
1

Dn po  N d  0.90
Dn 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


Na
 po N d
1
0.90 
8.12
The cross-sectional area is
I 10 10 3
A 
 5 10  4 cm 2
J
20
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  en i2 

 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
Dn 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
N
5 10
5  10  7
d
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 1010

2

1
25
1
10 





7
Nd
5  10
5  10  7 
 14.23N d
We find
N d  7.09 1014 cm 3
and
N a  1.011016 cm 3
_______________________________________
8.13
Plot
_______________________________________
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 
 5 1015 

 0.0259 ln 
10 
 1.5 10 
eD n n po
Jn
Ln

eD n n po eD p p no
Jn  J p

Ln
Lp
Dn
 no

Dn
 no


n i2
Na
D p n i2
ni2


Na
 po N d
1

D p no  N a 


Dn po  N d 
1
We have
Dp  p

1
1
and no 


Dn  n 2.4
 po 0.1
so
Jn

Jn  J p
1
1 1  Na 



2.4 0.1  N d 
1
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
E Fi  E F  0.329 eV
Also on the n-side;
N 
E F  E Fi  kT ln  d 
 ni 
 1017 

 0.0259 ln 
10 
 1.5 10 
or
E F  E Fi  0.407 eV
(b) We can find
Dn  12500.0259  32.4 cm 2 /s
D p  3200.0259  8.29 cm 2 /s
Now
 1
Dp 
Dn
1

J S  en i2 

 N a  nO N d  pO 

D n no

D p po
Then

2.4
 4.90
0.1

n  p
Jn

Jn  J p
 n  p  4.90


_______________________________________

 1.6 10 19 1.5 1010
1
N 
1  2.04  a 
 Nd 
(b) Using Einstein's relation, we can write
e n ni2

Jn
Ln N a

Jn  J p
e n ni2 e p ni2



Ln N a
Lp Nd

or
 1

15
 5  10

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  en i2 A 



1
Nd
Dp
 po
V 
exp D 
 Vt 
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 8
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________


 1.6 10 19 1.5 1010
 10  1 
2
V 
(e) I p x n   I sp exp a 
 Vt 
4
 10 
17
V 
8.29
exp D 
7
10
 Vt 


 1.3997 10 4 A
I Total  I n  I p
or
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
_______________________________________
 4.198110 5  1.3997 10 4
 1.820 10 4 A
Now
  1 2L p 
1



I p  x n  L p   I p x n  exp


2
Lp





 eD p p no 
Dp n
  eA
(a) I sp  A

 Lp 
 po N d


2
i

 1.6 10
5 10 
4
 8.4896 10 5 A


1.5 1010
10

8 10 8 1.5 1016
2
I sp  1.342  10 14 A
 eD n n po 
Dn n
  eA
(b) I sn  A


 no N a
 Ln 

 1.6 10
5 10 
4
 9.710 10 5 A
_______________________________________

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
Va  0.8Vbi  0.80.746826   0.59746 V
V  n
V 
p n x n   p no exp a  
exp a 
 Vt  N d
 Vt 
2
i

8.17
(a) The excess hole concentration is given by
p n  p n  p no
x
 V  

 p no exp a   1  exp
 Lp 
  Vt  


We find
p no 
 0.0259 
V 
(d) I n  x p  I n x n   I sn exp a 
 Vt 
 0.59746 
 4.025 10 15 exp

 0.0259 
4
3
Then
p n  2.25 10 4 exp


 0.610  
  1
 0.0259  
x


 exp

4
 2.828 10 

 4.198110 5 A
2
 2.828 10 4 cm  2.828  m


  2.25 10 cm
80.0110 6 
L p  D p pO 
1.5 10  exp 0.59746 
1.5 1016

ni2
1.5 1010

Nd
1016
and
10 2
 1.56 1014 cm 3

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
2
i
19

 1 
 1.3997 10  4 exp 
 2 
8.16
19

 0.59746 
 1.342 10 14 exp

 0.0259 
or
p n  3.811014 exp
x

 cm 3
4
 2.828 10 

Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 8
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(b) We have
(b) Problem 8.8
d p n 
J p  eD p
dx


eD p 3.808 1014
2.828 10
4
V 
p n  p no exp a 
 Vt 
 exp
x



4
2
.
828

10


 0.1N d 
p 
or Va  Vt ln  n   Vt ln  2

 p no 
 n i N d 
At x  3  10 4 cm,
1.6 10 19 8 3.808 1014
 3 
J p 3 
exp

4
2.828 10
 2.828 
or
J p 3  0.5966 A/cm 2

 
(c) We have
eDn 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
J no 
 0.1N d2 
 Vt ln 

2
 n i


1.6 10 234.5 10 
19
3
10.72 10  4
 0.610 
 exp

 0.0259 


 0.1 8  1015 2 
 0.0259  ln 

10 2
 1.5  10



 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

  Vt  
 Ln 
The total number of excess electrons is


N p  A n p dx
or
0
J no  0.2615 A/cm
We can also find
J po  1.724 A/cm 2
2
We may note that

x
x
 exp L dx   L exp L   L
n
n
0
n
Then at x  3  m,
Then
or
 V  
N p  ALn n po exp a   1
  Vt  
We find that
Dn  25 cm 2 /s and Ln  50.0  m
Also
J n 3  J no  J po  J p 3
 0.2615  1.724  0.5966
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 i N a 


 0.1N a2 
 Vt ln 

2
 n i

  2.8110 cm
2
4
Then
N p  10 3 50.0  10 4 2.8125  10 4






3

 V  
 exp a   1
  Vt  
or
 0.1 4  10 15 2 
 0.0259  ln 

2
 2.4  10 13

 0.205 V


ni2
1.5 1010

Na
8 1015
n
0
 V  
N p  0.1406exp a   1
  Vt  
Then, we find the total number of excess
electrons in the p-region to be:
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 8
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(a) Va  0.3 V, N p  1.51  10 4
(b) Va  0.4 V, N p  7.17  10 5
(c) Va  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


2
n2
1.5 1010
p no  i 
 2.25 10 4 cm 3
16
Nd
10
Then
 V  
Pn  2.25  10  2 exp a   1
  Vt  

 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 ni2
3
  Eg 
 T 

 C N cO N O 
 exp

 300 
 kT 

So
(a) Va  0.3 V, Pn  2.4110 3
(b) Va  0.4 V, Pn  1.15 10 5
(c) Va  0.5 V, Pn  5.45 10 6
_______________________________________
8.20
  Eg 
V 
 eV 
  exp a 
I  ni2 exp a   exp

V
kT
 kT 
 t 


Then
 eVa  E g 

I  exp

 kT

so
 eV a1  E g1 

exp

kT
I1



I2
 eV a 2  E g 2 

exp

kT


or
 eVa1  eVa 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
or
  Eg 

I S  CT 3 exp

 kT 
(b) Taking the ratio
  Eg 


3 exp
kT2 
I S 2  T2 

  
I S1  T1 
  Eg 

exp

 kT1 

T 
 1
1 

  2   exp  E g 


 T1 
 kT1 kT2 
1
 38.61
For T1  300 K, kT1  0.0259 ,
kT1
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
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 8
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
8.22
Plot
_______________________________________
 n2 
V 
  i  exp a 
 Vt 
 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


 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 

3
3
(a) J p x n  
eD p p no
Wn

Ln
V 
exp a 
 Vt 
V 
D n  n i2 

 exp a 
V 


 no  N a 
 t 

 10 3 1.6  10 19

V 
exp a 
 Vt 

1.5  1010
25

7
5  10
2  1017

2
 0.5516 
 exp

 0.0259 
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 
V 
  i  exp a 
 Vt 
 Na 
1010 7   10 3 cm
or L p  10  m;  Wn  L p
15

  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.
_______________________________________
L p  D p po 
AeDn n po
19


 1.12
 exp 

 0.0259T 300 
8.24
4
In 
 Ae

10 2
19
I p  4.565  10 3 A
10 

10  7 
T 
8.2519 10   2.8 10 1.04 10  300

19
3
 0.5516 
 exp

 0.0259 
or ni2  8.2519 10 27
Now
  Eg 

ni2  N c N  exp

 kT 
27

10 1.6 10 101.5 10 
0.7 10 2 10 

 
25
1

7
5  10
2  1017

AeD p  n i2 
V 

 exp a 
V 


Wn  N d 
 t 
(ii) I p 
1.2 10 6  5 10 4 1.6 10 19 ni2
 1

15
 4  10

 0.1 2  10 15 2 
 0.0259  ln 

2
 1.5  10 10

Va  0.5516 V


V 
(i) p n x n   0.1N d  p no exp a 
 Vt 
 0.1N a2 
or V a  Vt ln 

2
 n i



 0.1 2  10 15 2 
 0.0259  ln 

2
 1.5  10 10

Va  0.5516 V


Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 8
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
AeD p  n i2 
V 

 exp a 
V 
W n  N d 
 t 
(ii) I p 

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

V 
D n  n i2 

 exp a 
V 


 no  N a 
 t 

 10 3 1.6  10 19


1.5  1010
25

7
5  10
2 10 15

2
 0.5516 
 exp

 0.0259 
I n  2.2597 10 4 A
  x n  2W n  
  xn 

  B exp 
  B exp

Lp
L


p


  x 
  2Wn 

 B exp n  1  exp
 L p 
 L p 




We then obtain
 V  
p no exp a   1
  Vt  
B
  x 
  2W n 

exp n  1  exp
 L p 
 L p 




which can be written as
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

  xn 


  B exp  x n 
 A exp
 Lp 
 Lp 




and the boundary condition at x  x n  W n
gives
p n x n  Wn   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


B
 x  Wn 
 V  
p no exp a   1  exp  n

  Vt  
 L p 
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  x 
  x n  W n  x   
n
 exp  n
  exp 

Lp
Lp
 


 
or finally
 x  Wn  x 

sinh  n


L
  Va  
p


p n  p no exp   1 
V


W
  t  
sinh  n 
 Lp 


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 D3  0.5653 V
_______________________________________
Then
W   V  
coth n   exp a   1
 Lp  
Lp

   Vt  
_______________________________________
Jp 
eD p p no
8.26
V 
I D  ni2 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

eV

g
D2 
exp 

kT
2


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
kT
kT1  0.0259 eV, 1  0.0259 V
e
T  310 K ,
kT2
 0.02676 V
kT2  0.02676 eV,
e
T  320 K ,
kT3
 0.02763 V
kT3  0.02763 eV,
e
8.27
(a) We can write
 eV 
I D  C  ni2  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 

 t 
 kT 
  E g  eV 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   C 
Va  E g     ln  1 
 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


1
Nd
Dp 

 p0 

2
25
1

7
10
4  10 16
I s  2.323 10 15 A
Aen i W
(b) I gen 
2 0
We find
10 

10  7 
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 8
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Aeni W
I s  I gen 
 4 1016 4  1016 
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  V  V R   N a  N d 


W   s bi
 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
1.6 10 19

I gen  7.331 10 11 A
1/ 2
16
16 
 4  10  4  10  

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 1010 6.109 10 5
I gen 
7
V 
2 10
 7.33110 11 exp a 
 7.33110 11 A
 2Vt 

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 

Va  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 
3.0545 10 2
so ni 
 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 1014 cm 3
Then
(a)
ni2  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







































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
Va  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
Va  0.6 V, I d  1.73 10 12 A
 1.726 10 12 A

(iii) I D  1.50 10
 22

I rec  6.44 10 9 A, I T  6.44 10 9 A
 0.8 
exp

 0.0259 
Va  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
 8.795 10 A
Aen i W
(b) I gen 
2 0
V a  1.2 V. I d  1.99 10 2 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


I rec  6.90 10 4 A, I T  2.06 10 2 A
_______________________________________

 7 1016 7  1016 
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  10 16  7  10 16  

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
8
 6.049 10
14
A
V 
(ii) I rec  I ro exp a 
 2Vt 




 0.6 

 6 10 14 exp
 20.0259 
 6.436 10 9 A
 0.8 

(iii) I rec  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  ni2
 pO n  n    nO  p  p 
Let  pO   nO   O and n   p   ni
We can write
 E  E Fi 
n  ni exp Fn

kT


and
 E Fi  E Fp 

p  ni exp

kT


We also have
EFn  EFi   EFi  EFp  eVa
so that
E Fi  E Fp  eVa  E Fn  E Fi 
Then
 eV  E Fn  E Fi  
p  ni exp  a

kT






 eV 
  E Fn  E Fi 
 ni exp a   exp 

kT
 kT 


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
ni e ni e a  e   ni2
R
 O ni e  ni  ni e a  e   ni
or
n i e a  1
R
 O 2  e  e a  e 
To find the maximum recombination rate, set
dR
0
d





0




  d 2  e  e  e 
ni e a  1

or

O

 1
a

d
   12  e  e  e 
ni e a  1
O
  2
a


 e  ea  e 
which simplifies to
 n i e a  1
0


O
a
a


 2
8.35
We have
W

J gen  eGdx
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
We find
N N 
Vbi  Vt ln  a 2 d 
 ni 

e
e
a
 



or
Rmax


a
1
O 2e
a 2
 e a 2

ni e


2 e


n i e a  1
a 2




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

 5  1015  5  1015 

 
15
15  
 5  10 5  10 


1/ 2

or
W  2.35 10 4 cm
Then
J gen  1.6  10 19 4  1019 2.35  10 4




or

1
which can be written as
  eV  
ni exp a   1
  kT  
Rmax 
  eV  
2 O exp a   1
  2kT  
O

or
a
2
Then the maximum recombination rate
becomes
ni e a  1
Rmax 
 O 2  ea 2  ea  e a 2

 5  1015 5  1015 
 0.0259  ln 

2
 1.5  1010

The denominator is not zero, so we have
0  e  e a  e 
or
2
ni
 eV 
exp a 
2 O
 2kT 
Q.E.D.
_______________________________________

 e  e  e 
2  e  e  e 

Rmax 
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
J gen  1.5  10 3 A/cm 2
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 8
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
8.36
8.38
 1
J S  en i2 
 N a

Dn
 nO


 1.6  10 19 1.5  10
1
Nd
Dp 

 pO 


 1

16
 3  10
18
10  7
1
10 18
6 

10 7 
10 2

(a) C d 

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
, Cd 
gd 
Vt
2Vt
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
Z 
 d2
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  j7.41
f  10 MHz , Z  2.38  j 7.49
_______________________________________
8.40
Reverse bias


e s N a N d
C j  AC   A  

 2Vbi  V R N a  N d  
(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
_______________________________________


1/ 2

 5  1017 8  1015 
Vbi  0.0259  ln 

2
 1.5  1010

 0.790 V
 1.6 10 19 11.7 8.85 10 14
C j  2 10  4 
2Vbi  V R 










 5  1017 8  1015  

17
15  
 5  10  8  10  

1/ 2
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 8
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Cj 
5.1078  10 12
 pO  20.0259 2.5  10 6 
or
 pO  1.3  10 7 s
F
Vbi  V r
V R (V)
C j (pF)
10
5
3
1
0
0.20
0.40
1.555
2.123
2.624
3.818
5.747
6.650
8.179
At 1 mA,
C d  2.5 10 6 10 3
or
C d  2.5 10 9 F
_______________________________________

Forward bias
For N a  N d  I po  I no
Then
I po p 0 I po 8 10 8
Cd 

 1.544 10 6 I po
2Vt
20.0259

 

I po  Ae
D p  ni2 
V 
  exp a 
 
V 

 p0  N d 
 t 

1.6 10 
 2  10
4


2
V 
 exp a 
 Vt 

I po  1.006 10

V 
exp a  A
 Vt 
+
C j (F)
14
C d (F)
0.20
3.5110 17  6.650 10 12
= C Total (F)
 6.650 10 12
7.92 10 14  8.179 10 12
 8.258 10 12
0.60
1.79 10
10
 ...
 1.79 10 10
_______________________________________
8.41
For a p  n diode, I pO  I nO , then
 1 
 I pO pO
C d  

 2Vt 
Now

 pO
2Vt
Then

 2.5 10 6 F/A
8.42
(a) N a  N d  I po  I no
(i) C d 
or I po 
I po p 0
2Vt
2Vt C d 
 p0


20.0259  10 9
10  7
(ii) I po  Ae
Dp


 p0
V 
ni2
exp a 
Nd
 Vt 


0.518 10 3  5 10  4 1.6 10 19

Va (V)
0.40

 5.18 10 4 A
or I po  0.518 mA
1.5 1010
10

8 10 8
8  1015
19

 10
10 7
1.5 10  exp V 
10 2
a
8 1015
V 
 t 
 0.518 10 3 

Va  0.0259 ln 
14 
 2.25 10 
 0.618 V
V
0.0259
 50 
(iii) rd  t 
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) Va  0.0259 ln 
14 
 2.25 10

 0.5821 V
0.0259
 200 
(iii) rd 
0.1295  10 3
_______________________________________
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
1.6 10 48010 10 
19
16
2
or
R p  26 
n-region:
 L
L
L
Rn  n 

A
 n A e n N d A
so
0.10
Rn 
19
1.6  10 1350  10 15 10  2
or
Rn  46.3 
The total resistance is
R  R p  Rn  26  46.3
or
R  72.3 
(b)
V  IR  0.1  I 72.3
which yields
I  1.38 mA
_______________________________________

8.44
R


 

 n Ln  p L p 
An 

A p 
0.210   0.110 2 
2
2 10
5
2 10
R  150 
We can write
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


 10 2 
V  0.0259 ln  10 
 10 
or V  0.477 V
_______________________________________
8.45
(a) rd 
Vt
V
0.0259
 ID  t 
ID
rd
32
or I D  8.09375 10 4 A
I 
Va  Vt ln  D 
 Is 
 8.09375 10 4 

 0.0259 ln 
12

 5 10

Va  0.4896 V
V
0.0259
 4.3167  10  4 A
(b) I D  t 
rd
60
 4.3167 10 4 

Va  0.0259 ln 
12

 5 10

 0.4733 V
_______________________________________
8.46
5
or

(ii) For I D  10 mA,
(a)
1
V 
1 dI D

 I S   exp a 
rd dVa
V
 t
 Vt 
or
1
10 13
 0.020 

 exp

rd 0.0259
 0.0259 
which yields
rd  1.2 1011 
(b)
1
10 13
  0.02 

 exp

rd 0.0259
 0.0259 
which yields
rd  5.6 1011 
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 8
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
8.47
8.49
I
(a) If R  0.2
IF
Then we have
ts
erf
 pO

IF

IF  IR
1
1

I R 1  0.2
1
IF
C j  4.2 pF at V R  10 V
We have
 nO   pO  10 7 s , I F  2 mA
and
IR 
or
ts
erf
 pO
 pO
(b) If
 0.978 
 I 
 2
t s   pO ln 1  F   10 7 ln 1  
I
 1
R 


ts
 pO
 0.956
IR
 1.0 , then
IF
erf
ts
 pO

V R 10

 1 mA
R 10
So
 0.833
We find
ts
C j  18 pF at V R  0
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
 

 1.1110 7 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
_______________________________________
ts
IF

(a) erf
 p IF  IR
erf 0.3 = erf 0.5477 
 erf 0.55  0.56332
1
Then 0.56332 
I
1 R
IF
IR
1

 1  0.775
I F 0.56332
 t 
exp 2 
  p0 


I 
 1  0.1 R 
 p0
 IF 
 t 
 2 
  p0 


 1 0.10.775
 1.0775
t2
By trial and error,
 0.80
(b) erf
t2

 p0
_______________________________________
8.50


 5  1019 2 
Vbi  0.0259  ln 
  1.136 V
2
 1.5  1010 
We find


 2  V  V a   N a  N d 


W   s bi
 N N 
e

a
d 



1/ 2

 211.7  8.85 10 14 1.136  0.40

1.6 10 19

 5 1019  5 1019 


19 2


5

10



1/ 2

which yields
o
W  6.17 10 7 cm  61.7 A
_______________________________________
8.51
Sketch
_______________________________________
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1017 cm 3
V 
p n x n   0.1N d  9 1014  p no exp a 
 Vt 
p no 

ni2
1.5 1010

Nd
9 1015
  2.5 10 cm
2
4
3
 9 1014 
  0.6295 V
Then Va  0.0259 ln 
4 
 2.5 10 
I
50 10 3
Is 

V 
 0.6295 

exp a  exp
 0.0259 
 Vt 
 1.389 10 12 A
Is 
AeD p p no
Lp
1.389 10

Aeni2
Nd

Dp
 p0
12


A 1.6  10 19 1.5 1010
9 1015


2
1.389 10 12  A 2.828 10 11
2

10
2  10  7
or A  4.9110 cm
_______________________________________
2
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 9
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Chapter 9
9.1
(a) We have
N 
e n  eVt ln  c 
 Nd 
 2.8 1019 
  0.206 eV
 0.0259 ln 
16

 10

(c)
 BO   m    4.28  4.01
or
 BO  0.27 V
and
Vbi   BO   n  0.27  0.206
or
Vbi  0.064 V
Also
2 V 
x d   s bi 
 eN d 
1/ 2


 211.7  8.85  10 14 0.064  


1.6  10 19 10 16



1/ 2
 
or
x d  9.110 6 cm
Then
eN d x d
 max 
s
 Nc 


 Nd 
 2.8 1019 

 0.0259 ln 
15 
 5 10 
=0.2235 V
Vbi  0.65  0.2235  0.4265 V
 n  Vt ln 
 2.8 1019 

(b)  n  0.0259 ln 
16

 10

 0.2056 V
Vbi  0.65  0.2056  0.4444 V
V bi increases,  B 0 remains constant
 2.8 1019 

(c)  n  0.0259 ln 
15

 10

 0.2652 V
Vbi  0.65  0.2652  0.3848 V
V bi 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
9.2
(a) Vbi   B 0   n
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
_______________________________________
 Nc 


 Nd 
 2.8 1019 
  0.2056 V
 0.0259 ln 
16

 10

Vbi  1.09  0.2056  0.8844 V
 n  Vt ln 
 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
5
16
 7.63  10 4 V/cm
14
1/ 2
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 10 16



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  10 15


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  10 15





 7.445 10 cm
or x n  0.7445  m
(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  10 15




1/ 2

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
 4.93  10 4 V/cm
5
15
14
 7.16 10 13 F
or C  0.716 pF
(ii)

C  10 


 
1/ 2


 
1/ 2
 1.6  10 19 11.7  8.85  10 14 10 15 


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)




 
1/ 2


 
1/ 2
 1.6  10 19 11.7  8.85  10 14 10 16 
C  10  4 

20.674  1


9.5


 1.6  10 19 11.7  8.85  10 14 10 15 
C  10  4 

20.615  1


4
 1.309 10 4 cm
or x n  1.309  m
 max 
 2.8 1019 
  0.265 V
15

 10

Vbi  0.88  0.265  0.615 V
 n  0.0259 ln 
1/ 2
5
 max 
 e s N d 
(a) C   

 2Vbi  V R  
We have  B 0  0.88 V
 2.22 10 12 F
or C  2.22 pF
(ii)

C  10 
4
 1.6  10 19 11.7  8.85  10 14 10 16 


20.6745  5


 1.21 10 12 F
or C  1.21 pF
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 9
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
9.7
(a) From the figure, Vbi  0.90 V
(b) We find


19

4
15
14
 9.14  10 4 V/cm
2
 1 
 
3  10 15  0
 C 

 1.034  10 15
V R
2   0.90 
and
2
1.034  1015 
e s N d
We can then write
2
Nd 
19
1.6  10 13.1 8.85  10 14 1.034  1015
or
N d  1.04 1016 cm 3
(c)
N 
 n  Vt ln  c 
 Nd 
 4.7 1017 

 0.0259 ln 
16 
 1.04 10 
or
 n  0.0986 V
(d)
 Bn  Vbi   n  0.90  0.0986
or
 Bn  0.9986 V
_______________________________________

1.6 10 5 10 1.183 10 
11.78.85 10 
 max 
(b)
(i)  

e
4 s



 1.6  10 19 4.66  10 4 

14 
 4 11.7  8.85  10

 0.0239 V


e
16 s 
xm 

1/ 2




1.6  10 19

14
4 
4.66  10 
16 11.7  8.85  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

1/ 2


1.6  10 19
xm  
14
4 
9.14  10 
16 11.7  8.85  10



7
 1.83 10 cm
_______________________________________
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  10 15




e
 x
16 s x
e x  
e2
 ex
16 s x
or
Now
d e x 
 e2
0
 e
dx
16 s x 2
 6.033 10 5 cm
or x n  0.6033  m
 max
  x  
1/ 2

1.6 10 5 10 6.033 10 

11.78.85 10 
Solving for x, we find
 4.66  10 4 V/cm
Substituting this value of x  x m into the
equation for the potential, we find
19
5
15
14


 211.7  8.85  10 0.406  5 
(ii) x n  

1.6  10 19 5  10 15



14

4
 1.183 10 cm
or x n  1.183  m

1/ 2
1/ 2
x  xm 
e
16 s 
e
 
16 s
e
16 s 

e
16 s 
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 10 16




1/ 2
1.6 10 10 3.362 10 
13.18.85 10 
14
e
4 s
1.6 10 
4 13.18.85 10 
0.044 4 13.18.85 10 

19
14
14
2
1.6  10 19
 1.763 10 5 V/cm
Now
1.6 10 10 x
13.18.85 10 
19
  1.763 10 5 
16
n
14
 x n  1.277 10 4 cm
And

x n2  1.277 10 4



 

8.85 10  5.2  4.07   

 10 
25 10 
e
e 
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
5
16
0.0442 

14
 4.64  10 4 V/cm
(b)   0.050.88  0.044 V


1
2 1.6  10 19 13.1 8.85  10 14
 10 13 

e

 e 
 1016  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


or
8

0.83   Bn
 0.038  Bn  0.10  0.2211.13   Bn 
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 eD it  Dit (cm 2 eV 1 )
Then we can write
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 

2 1.6  10 19 11.7  8.85  10 14
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
_______________________________________
 1.296 10 6 A
 V  
(i) I  I sT exp a   1
  Vt  
 10  10 6

V a  0.030217  ln 
 1
6
1.296  10

 0.0654 V
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

 100 10 6

(ii) Va  0.030217  ln 
 1
6
1.296 10

 0.1317 V


10 3

(iii) Va  0.030217  ln 
6 
 1.296 10 
 0.201 V
_______________________________________
9.16
(a)  Bn  0.88 V
(d)
 J 
5

  0.0259 ln 
Va  Vt ln 

8

 1.29 10 
 J sT 
Va  0.512 V
_______________________________________
9.15
(a)  B 0  0.63 V
  0.63 
2
J sT  120300 exp

 0.0259 
  0.88 
2
(b) J sT  1.12300 exp

 0.0259 
 1.768 10 10 A/cm 2
10


(c) Va  0.0259 ln 

10
 1.768 10 
 0.641 V
(d) Va  Vt ln 2  0.0259 ln 2
 0.0180 V
_______________________________________
 2.948 10 4 A/cm 2



I sT  10 4 2.948 10 4  2.948 10 8 A
 I 

(i) Va  Vt ln 

 I sT 
 10 10 6 

 0.0259 ln 
8 
 2.948 10 
 0.151 V
 100 10 6 

(ii) Va  0.0259 ln 
8 
 2.948 10 
 0.211 V


10 3

(iii) Va  0.0259 ln 
8 
 2.948 10 
 0.270 V
9.17
Plot
_______________________________________
9.18
From the figure,  Bn  0.68 V
  
  

J ST  A*T 2 exp Bn   exp

 Vt 
 Vt 
  
  0.68 
2

 120300 exp
  exp

0
.
0259


 Vt 
or
  

J ST  4.277 10 5 exp

 Vt 
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


V
 0.0259 ln 
 0.2056
 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 ST1  4.277 10 5 exp

 0.0259 
dn 
or
For A  10 4 cm 2 , we find


 
 211.7  8.85  10 14 4.4744  
xd  

1.6  10 19 10 16


1/ 2


4 2m n*

3/ 2
3/ 2
E  Ec
h3
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

4 2m n*
  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


Ec
E  Ec
h3
and assuming the Boltzmann approximation
  E  E F 
f F E   exp 

kT


Then
  0.0328 V
Then
I R1  1.52 10 8 A
(b) For V R  4 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


and
m n*
2
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
2kT
 mn 
We can also write
1/ 2
E  E F  E  E c   E c  E F 
 2kT 
m n* z2
  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


 exp
dn  2

h
kT 
m


n 




 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
3
m 
 e n 
 exp
J sm  2


 kT 
 h 
*
n

  m n* x2 
d x
  x exp

 2kT 
 Ox


  m n* y2 
d
 exp
 2kT  y





  m n* z2 
d z
 exp

2
kT




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) Va SB   0.0259 ln   4
8 
 10 6 10 
 0.4845 V
Then
Va  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) Va  0.0259 ln 
16 
 6.4 10 
 0.678 V
 700 10 6 

(b) Va  0.0259 ln 
16 
 6.4 10 
 0.718 V
 1.2 10 3 

(c) Va  0.0259 ln 
16 
 6.4 10 
 0.732 V
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 9
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
For the Schottky junction,
9.23
(a) For I  0.8 mA, we find
I sT  8 10 4 6 10 9  4.8 10 12 A
0.8 10 3
 150 10 6 
J

 1.143 A/cm 2

(a) Va  0.0259 ln 
7 10  4
12 
 4.8 10 
We have
 0.447 V
 J 

Va  Vt ln 
 700 10 6 
J S 

(b) Va  0.0259 ln 

12 
 4.8 10 
For the pn junction diode,
 0.487 V
 1.143 
Va  0.0259  ln 
  0.6907 V
12
 1.2 10 3 
 3 10 

(c) Va  0.0259 ln 
12 
For the Schottky diode,
 4.8 10 
 0.501 V
 1.143 
Va  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  ni2  
 exp

(ii)
 300 
 kT 
3


0.8 10
Then
Va SB   0.0259 ln 
4
9 
3
8

10
6

10
J S 400  400 




 0.490 V
J S 300  300 


0.8 10 3
 Eg
Eg 

Va  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
J S 300 
 t 
Now
V 
 8  10  4 8  10  13 exp a 
I  7 10 4 1.17 10 5 3 10 12
 Vt 
 0.6907 
V 
 exp

 4.8 10 12  6.4 10 16 exp a 
 0.03453 
 Vt 
or
Then
I  120 mA


0.8 10 3
For
the Schottky diode,
Va  0.0259 ln 
12
16 
 4.8 10  6.4 10 
  e BO 
J ST  T 2 exp

Va  0.49032 V
 kT 
Now
 0.49032 
I SB  4.8 10 12 exp

2
0
.
0259
J ST 400  400 




 I SB  0.7998 mA
J ST 300  300 
 0.49032 


  BO

I pn  6.4 10 16 exp

 exp 
 BO 
 0.0259 



0
.
0259
400
300
0
.
0259



























 I pn  0.107  A
_______________________________________
0.82 
 0.82
 1.778 exp 

0
.
0259
0
.
03453 

Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 9
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
or
 5  10 6 120 300 2 
(b)  Bn  0.0259  ln 

J ST 400 
3
0.0259
 4.856  10



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
Rc 10 4

 0.1 
A 10 3
R
10 4
(b) R  c   4  1 
A 10
Rc 10 4
(c) R 

 10 
A 10 5
_______________________________________
(a) R 
9.26
Rc 5 10 5

 5
A
10 5
(i) V  IR  15  5 mV
(ii) V  IR  0.15  0.5 mV
or
 2.8 1019 

0.20  0.0259  ln 

 Nd

which yields
N d  1.24 1016 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 

5
5 10
 50 
10 6
(i) V  IR  150  50 mV
(ii) V  IR  0.150  5 mV
_______________________________________
(b) R 
9.27
 
Vt exp Bn 
 Vt 
Rc 
 2
AT
 R A T 2 
or  Bn  Vt ln  c

 Vt



eN d 
x2 
 xn  x 

s 
2 
At x  x n ,   Vbi , so
  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
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
8 2 
d 2
 x 
d
50  10




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
19
1
.
6

10
N
n
d


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 
N d  3.5 1018 cm 3
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
19
 1.04 10 
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
x P  x 
P 
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
Ec  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 dn x n x 
   dn 
  C3
n 
 2 n
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 

2 n
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 P N aP


Solving for x n , we have
Vbin 


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 dn Vbi
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
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 10
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Chapter 10
10.1
(a) p-type; inversion
(b) p-type; depletion
(c) p-type; accumulation
(d) n-type; inversion
_______________________________________
10.2
N 
(a) (i)  fp  Vt ln  a 
 ni 
 7 1015 

 0.0259 ln 
10 
 1.5 10 
 0.3381 V
 4 s  fp 
x dT  

 eN a 



5
 3.43 10 cm
or x dT  0.343  m
 3 1016 

(ii)  fp  0.03022 ln 
11 
 1.93 10 
 0.3613 V



1/ 2

(a)
 4 s  fn 
 max   eN d x dT  eN d 
Q SD

 eN d 




 350 
(b) kT  0.0259
  0.03022 V
 300 
  Eg 

ni2  N c N  exp

 kT 
 350 
 2.8 1019 1.04 1019 

 300 
so ni  1.93 1011 cm 3
 7 1015 

(i)  fp  0.03022 ln 
11 
 1.93 10 
 0.3173 V
1.25 10 
 1.6 10 N 411.78.85 10 0.30
19
14
d
 N d  7.86 1014 cm 3
2nd approximation:
 7.86 1014 
  0.2814 V
 fn  0.0259 ln 
10 
 1.5 10 
Then
1.25 10 
 1.6 10 N 411.78.85 10 0.2814
8 2
19
14
d
3
  1.12 
 exp

 0.03022 
 3.7110 22
1/ 2
8 2
1/ 2
 1.80 10 cm
or x dT  0.180  m


Then
5

1/ 2
1st approximation: Let  fn  0.30 V
 411.7  8.85  10 14 0.3758 
x dT  

1.6  10 19 3  10 16




10.3
 eN d  4 s  fn

1/ 2
 1.77 10 cm
or x dT  0.177  m
_______________________________________
 3 1016 

(ii)  fp  0.0259 ln 
10 
 1.5 10 
 0.3758 V


1/ 2
5
 3.54 10 5 cm
or x dT  0.354  m



 411.7  8.85  10 14 0.3381 


1.6  10 19 7  10 15




 411.7  8.85  10 14 0.3613 
x dT  

1.6  10 19 3  10 16


1/ 2


 411.7  8.85  10 14 0.3173 
x dT  

1.6  10 19 7  10 15


 N d  8.38 1014 cm 3
 8.38 1014 
  0.2831 V
(b)  fn  0.0259 ln 
10 
 1.5 10 
 s  2 fn  20.2831  0.566 V
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 10
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
10.4
p-type silicon
(a) Aluminum gate

Eg


 ms   m     
  fp 
2e



We have
N 
 fp  Vt ln  a 
 ni 
 6 1015 
  0.334 V
 0.0259 ln 
10 
 1.5 10 
Then
 ms  3.20  3.25  0.56  0.334
or
 ms  0.944 V
(b) n  polysilicon gate
 Eg

 ms  
  fp   0.56  0.334 
 2e

or
 ms  0.894 V
(c) p  polysilicon gate
 Eg

 ms  
  fp   0.56  0.334
 2e

or
 ms  0.226 V
_______________________________________
10.5
 4 1016 
  0.3832 V
 fp  0.0259 ln 
10 
 1.5 10 
Eg


 ms   m     
  fp 
2e


 3.20  3.25  0.56  0.3832
 ms  0.9932 V
_______________________________________
10.6
(a) N d  21017 cm 3
(b) Not possible -  ms is always positive.
(c) N d  21015 cm 3
_______________________________________
10.7
From Problem 10.5,  ms  0.9932 V
Q
V FB   ms  ss
C ox
(a) C ox 

ox 3.9 8.85 10 14

t ox
200 10 8

 1.726 10 7 F/cm 2
5 1010 1.6 10 19
V FB  0.9932 
1.726 10 7
 1.040 V
3.9 8.85 10 14
(b) C ox 
80 10 8
 4.314 10 7 F/cm 2
5 1010 1.6 10 19
V FB  0.9932 
4.314 10 7
 1.012 V
_______________________________________








10.8
(a)  ms  0.42 V
V FB   ms  0.42 V
(b)
C ox 
3.98.85 10 14   1.726 10 7 F/cm 2
200 10 8
Q
4 1010 1.6 10 19
(i) V FB   ss  
C ox
1.726 10 7
 0.0371 V
1011 1.6 10 19
(ii) V FB  
1.726 10  7
 0.0927 V
(c) V FB   ms  0.42 V

 
C ox 



3.98.85 10 14   2.876 10 7 F/cm 2
120 10 8
4 1010 1.6 10 19
(i) V FB  
2.876 10 7
 0.0223 V
1011 1.6 10 19
(ii) V FB  
2.876 10  7
 0.0556 V
_______________________________________

 



Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 10
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
10.9
VTN 
Eg


 ms   m     
  fp 
2e


where
 2 1016 
  0.365 V
 fp  0.0259 ln 
10 
 1.5 10 
Then
 ms  3.20  3.25  0.56  0.365
or
 ms  0.975 V
Now
Q
V FB   ms  ss
C ox
or
Q ss   ms  V FB C ox
We have

3.9 8.85 10 14
C ox  ox 
t ox
450 10 8
or
C ox  7.67 10 8 F/cm 2
So now
Qss   0.975   1 7.67 10 8

(b) p
10.11


 411.7  8.85  10 14 0.3161 
x dT  

1.6  10 19 3  1015




1/ 2

5
 5.223 10 cm
 max   eN d x dT
QSD



 1.6 10 19 3 1015 5.223 10 5
 2 1016 
  0.3653 V
10 
 1.5 10 




 6.958 10 C/cm
3.9 8.85 10 14  2.30110 7 F/cm 2
C ox 
150 10 8
2

 2.507 10 C/cm
3.9 8.85 10 14  2.30110 7 F/cm 2
C ox 
150 10 8
 max   Q ss 
 Q SD
VTP   
   ms  2 fn
C ox



2


1/ 2
 1.6 10 19 2 1016 2.174 10 5
8


 2.507 10 8  1.6 10 19 7 1010 
 

2.30110 7


  ms  20.3161
VTP  0.7898   ms

 2.174 10 cm
 max   eN a x dT
QSD

 3 1015 
  0.3161 V
10 
 1.5 10 
 fn  0.0259 ln 
5

poly gate on p-type:  ms  0.28 V
VTN  0.9843  0.95  0.0343 V
_______________________________________
 fp  0.0259 ln 
8

(c) Al gate on p-type:  ms  0.95 V

 411.7  8.85  10 14 0.3653 
x dT  

1.6  10 19 2  10 16



VTN  0.9843  0.28  1.26 V
10.10


VTN  0.9843  1.12  0.136 V
Q ss
 1.2  10 10 cm 2
e
_______________________________________


(a) n  poly gate on p-type:  ms  1.12 V
or

  ms  2 fp
6.958 10 8  7 1010 1.6 10 19
2.30110 7
  ms  20.3653
 0.9843   ms
 1.92 10 9 C/cm 2

C ox



 max   Q ss
Q SD
(a) n  poly gate on n-type:  ms  0.41 V
VTP  0.7898  0.41  1.20 V
(b) p  poly gate on n-type:  ms  1.0 V
VTP  0.7898  1.0  0.210 V
(c) Al gate on n-type:  ms  0.29 V
VTP  0.7898  0.29  1.08 V
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 10
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Now
10.12
 5 1015 
  0.3294 V
 fp  0.0259 ln 
10 
 1.5 10 
The surface potential is
 s  2 fp  20.3294  0.659 V
We have
Q
V FB   ms  ss  0.90 V
C ox
Now
 max 
Q SD
VT 
  s  V FB
C ox
We obtain
 4 s  fp 
x dT  

 eN a 
 0.3832 V




or


10.13
C ox 

ox 3.9 8.85 10 14

t ox
220 10 8

 1.569 10 7 F/cm 2

Qss  1.6 10
19
4 10 
10
 6.4  10 9 C/cm 2
By trial and error, let N a  41016 cm 3 .



 1.008 10 C/cm 2
 ms  0.94 V
Then
 max   Q ss
Q SD
VTN 
  ms  2 fp
C ox
1.008 10 7  6.4 10 9
1.569 10 7
 0.94  20.3832
Then VTN  0.428 V  0.45 V
_______________________________________


We also find

3.9 8.85 10 14
C ox  ox 
t ox
400 10 8
or
C ox  8.629 10 8 F/cm 2
Then
3.304 10 8
VT 
 0.659  0.90
8.629 10 8
or
VT  0.142 V
_______________________________________


7
 max   3.304  10 8 C/cm 2
Q SD

1/ 2
 1.6 10 19 4 1016 1.575 10 5



 1.575 10 cm
 max 
QSD
or


5
1/ 2
x dT  0.413 10 4 cm
Then
 max   1.6  10 19 5  1015 0.413  10 4
Q SD

 411.7  8.85  10 14 0.3832  
x dT  

1.6  10 19 4  1016


1/ 2
 411.7  8.85  10 14 0.3294  


1.6  10 19 5  10 15



 4 1016 

10 
 1.5 10 
 fp  0.0259 ln 
10.14
C ox 

ox 3.9 8.85 10 14

t ox
180 10 8

 1.9175 10 7 F/cm 3


Qss  1.6 10 19 4 1010
9
 6.4  10 C/cm

2
By trial and error, let N d  51016 cm 3
Now
 5 1016 

 fn  0.0259 ln 
10 
 1.5 10 
 0.3890 V


 411.7  8.85  10 14 0.3890  
x dT  

1.6  10 19 5  1016




1/ 2

5
 1.419 10 cm

QSD max 



 1.6 10 19 5 1016 1.419 10 5
7
 1.135 10 C/cm
 ms  1.10 V
Then
3

Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 10
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
VTP  

 Q  max   Q  
SD
Now
  ms  2 fn
ss
C ox
V FB   ms 
1.135 10  6.4 10 
7
9
1.9175 10 7
 1.10  20.3890
Then VTP  0.303 V, which is within the
specified value.
_______________________________________
10.15
We have C ox  1.569 10 7 F/cm 2
 1.03 







 9.456 10 C/cm
VTP  


 
 
 1.6 10 19 1015 8.630 10 5
8
 1.38110 C/cm
Now
VTN 
 max 
Q SD
C ox
1/ 2

2
 V FB  2 fp
1.38110 8
 1.08  20.2877 
1.9175 10 7
or VTN  0.433 V
_______________________________________
10.17
(a) We have n-type material under the gate, so
 Q  max   Q  
ss
C ox
 4 s  fn 
x dT  t C  

 eN d 
where
 1015 
  0.288 V
 fn  0.0259 ln 
10 
 1.5 10 
Then
1/ 2
  ms  2 fn
 9.456 10 9  6.4 10 9 

 

1.569 10 7


 0.33  20.2697 
 0.970 V
Then VTP  0.970 V  0.975 V which
meets the specification.
_______________________________________
10.16
(a)  ms  1.03 V
C ox 

2
 ms  0.33 V
SD

 8.630 10 5 cm
 max 
QSD
1/ 2
 1.6 10 19 5 1014 1.182 10 4
Then
1.9175 10

 1.182 10 cm
 max 
QSD
9
10
7
 411.7  8.85  10 14 0.2877  
x dT  

1.6  10 19 10 15


4

19
 1015 

(b)  fp  0.0259 ln 
10 
 1.5 10 
 0.2877 V
By trial and error, let N d  51014 cm 3
Now
 5 1014 

 fn  0.0259 ln 
10 
 1.5 10 
 0.2697 V

1.6 10 6 10 
V FB  1.08 V
Qss  6.4 10 9 C/cm 2
 411.7  8.85  10 14 0.2697  
x dT  

1.6  10 19 5  10 14


Q ss
C ox
3.98.85 10 14 
180 10 8
 1.9175 10 7 F/cm 2


 411.7  8.85  10 14 0.288 
x dT  

1.6  10 19 10 15



1/ 2
 
or
x dT  t C  0.863 10 4 cm  0.863  m
(b)
t 
 max   Qss  ox    ms  2 fn
VT   QSD
 ox 
For an n  polysilicon gate,
Eg

 ms  
  fn   0.56  0.288
 2e



or
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 10
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
 ms  0.272 V
Now
 max   1.6  10 19 1015 0.863  10 4
Q SD

or
 

We have
Qss  1.6 10 19 1010  1.6 10 9 C/cm 2
We now find
 1.38 10 8  1.6 10 9
VT 
500 10 8
14
3.9 8.85 10
 0.272  20.288
or
VT  1.07 V
_______________________________________
 





10.18
Eg


(b)  ms   m     
  fp 
2e


where
 m     0.20 V
and
 1016 
  0.3473 V
 fp  0.0259 ln 
10 
 1.5 10 
Then
 ms  0.20  0.56  0.3473
or
 ms  1.107 V
(c) For Q ss  0


 411.7  8.85  10 0.3473 
x dT  

1.6  10 19 10 16



or
 max   4.797  10 8 C/cm 2
Q SD
Then
 1.107  20.3473
VT  0.00455 V  0 V
_______________________________________
10.19
Plot
_______________________________________
10.20
Plot
_______________________________________
10.21
Plot
_______________________________________

ox 3.9 8.85 10 14

t ox
120 10 8

 2.876 10 7 F/cm 2
ox
 
C FB
  V 
t ox   ox  t s
 s  eN a

3.98.85 10 14 
14
 3.9  0.0259 11.7 8.85  10 
120  10 8 


 11.7 
1/ 2
x dT  0.30 10 4 cm  0.30  m
Now
 max   1.6  10 19 1016 0.30  10 4
Q SD

14
C ox 
 

8
10.23
(a) For f  1 Hz (low freq),
or

8
10.22
Plot
_______________________________________
t 
 max   ox    ms  2 fp
VTN  QSD
 ox 
We find
14
4.797 10 300 10 
3.98.85 10 
or
 max   1.38  10 8 C/cm 2
Q SD

VT 

1.6 10 10 
19
16
  1.346 10 7 F/cm 2
C FB
ox
 
C min
 
t ox   ox   x dT
 s 
Now
 1016 
  0.3473 V
 fp  0.0259 ln 
10 
 1.5 10 


 411.7  8.85  10 14 0.3473 
x dT  

1.6  10 19 10 16



5
 3.00 10 cm
 
1/ 2
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 10
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Then
 
C min
3.98.85 10
14


 3. 9 
5
120  10 8  
 3.00  10
11
.
7




 
C min
f  1 MHz (high freq),
C ox  2.876 10 7 F/cm 2 (unchanged)
  1.346 10 F/cm
C FB
2
(c) V FB   ms  1.10 V
C ox
 

2
 max 
Q SD
C ox
 V FB  2 fn
Now
 max   eN d x dT
QSD



 1.6 10 19 5 1014 1.182 10 4
9
 9.456 10 C/cm

2
Then
9.456 10 9
 0.95  20.2697 
2.876 10  7
VTP  0.378 V
_______________________________________
VTP  
f  1 Hz (low freq),

ox 3.9 8.85 10 14

t ox
120 10 8

3.98.85 10 14 
14
 3.9  0.0259 11.7 8.85  10 
120  10 8 


 11.7 
f  1 MHz (high freq),
VTP  
 2.876 10 7 F/cm 2
ox
 
C FB
  V 
t ox   ox  t s
 s  eN a

(b)
(c) V FB   ms  0.95 V
 4.80 10 C/cm
4.80 10 8
VTN 
 1.10  20.3473
2.876 10 7
VTN  0.2385 V
_______________________________________
C ox 
C  (inv)  C ox  2.876 10 7 F/cm 2
  8.504 10 9 F/cm 2
C  (inv)  C min
 1.6 10 19 1016 3.00 10 5
10.24
(a)

  8.504 10 9 F/cm 2 (unchanged)
C min
 V FB  2 fp
8

 3.9 
4
120  10 8  
 1.182  10
 11.7 
  4.726 10 8 F/cm 2 (unchanged)
C FB
Now
 max   eN a x dT
QSD

3.98.85 10 14 
C ox  2.876 10 7 F/cm 2 (unchanged)
  3.083 10 8 F/cm 2
C  (inv)  C min
 max 
Q SD

 8.504 10 9 F/cm 2
(unchanged)
  3.083 10 8 F/cm 2 (unchanged)
C min
VTN 

1/ 2
 1.182 10 cm
Then
C  (inv)  C ox  2.876 10 7 F/cm 2
7

4
 3.083 10 8 F/cm 2
(b)

 411.7  8.85  10 14 0.2697  
x dT  

1.6  10 19 5  10 14


1.6 10 5 10 
19
14
  4.726 10 8 F/cm 2
C FB
ox
 
C min
 
t ox   ox   x dT
 s 
Now
 5 1014 
  0.2697 V
 fn  0.0259 ln 
10 
 1.5 10 
10.25
The amount of fixed oxide charge at x is
 x x C/cm 2
By lever action, the effect of this oxide charge
on the flatband voltage is
1  x 
   x x
V FB  
C ox  t ox 
If we add the effect at each point, we must
integrate so that
x x 
1
dx
C ox 0 t ox
t ox
V FB  

_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 10
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
10.26
(a) We have  x  
1
QSS
1

z
F
IJF I
G
z
HK
b gHK
t ox
t ox
Cox tox  t
t ox
QSS
t ox
t
SS
ox
ox
ox
SS
ox
19
z
t ox
8
10
14
gb8  10 g
10
8
2
1
Cox 0 t ox
 x  O
xI
F
G
Ht J
Kdx
ox
1

Cox

af z
t ox
O
t ox
2
x dx
2
0
which becomes
ox
FB
1
19
200  10
O
dx
SS
ox
SS
VFB  
FQ I t  at  t f   Q
C H
t K
C
or
Ft I
V   Q  G J
H K
b
1.6  10 g
b8  10 gb200  10 g

3.9b
8.85  10 g
1

O
or  O  1.28  10
Now
x x 
dx
Cox 0 t ox
1
ox
2
t
Then
VFB  
b
t   Q   
2 1.6  10
VFB  
1

O

2
x
3
t ox
 O t ox
2

 I af
3
3
F
t
G
J
Ht K
Then
b
1.28  10 g
b200  10 g
V 
33.9b
8.85  10 g
ox
0
ox
ox
ox
2
8
2
14
FB
or
VFB  0.0742 V
or V FB  0.0494 V
_______________________________________
(b)
We have
10.27
Sketch
_______________________________________
 x  
QSS
b1.6  10 gb8  10 g
19

10
200  10
t ox
8
 6.4  10   O
3
Now
x x 
 ox
VFB  
dx   O
xdx
Cox 0 t ox
Cox t ox 0
or
1
z
z
t ox
10.28
Sketch
_______________________________________
t
10.29
(b)
N N 
V FB  Vbi  Vt ln  a 2 d 
 ni 
 O t ox
2
VFB  

b
gb200  10 g
23.9 b
8.85  10 g
 6.4  10
3
8
14
or
VFB  0.0371 V
(c)
Fx I
 x    GJ
Ht K
O
ox
We find
  


 1016 1016 
 0.0259  ln 

2
 1.5  1010 
2 ox
2
or
V FB  0.695 V
(c) Apply VG  3 V, Vox  3 V
For VG  3 V,

d

dx
s
n-side:   eN d
eN
eN x
d
  d     d  C1
dx
s
s
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 10
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
have
eN d x n
  0 at x   x n , then C1  
2Vn  Vox  Vbi
s
which
can be written as
so
eN d x n2 eN d x n t ox
eN

 Vbi
   d x  x n  for  x n  x  0
s
s
s
or
In the oxide,   0 , so
V 
d
x n2  x n t ox  bi s  0
 0    constant. From the
eN d
dx
boundary conditions, in the oxide
eN x
 d n
s
In the p-region,
eN a
eN a x

d



 C2
dx
s
s
s


eN
t  x   x


Solving for x n , we obtain
Vbi  VG , so
  0 at x  t ox  x p , then
t
x n  x p   ox 
2
a
ox
p
eN a x p
s
So that N a x p  N d x n

eN d x n
s
Since N a  N d , then x n  x p
The potential is

   dx
For zero bias, we can write
Vn  Vox  V p  Vbi
xn  x p  
2
across
the n-region, the oxide, and the p-region,
respectively. For the oxide:
eN d x n t ox
V ox    t ox 
s
For the n-region:

eN d  x 2

Vn x  
 x n  x   C 

s  2

Arbitrarily, set V n  0 at x   x n , then
eN d x n2
C 
so that
2 s
eN d
 x  x n 2
2 s
eN d x n2
At x  0 , V n 
which is the voltage
2 s
drop across the n-region. Because of
symmetry, Vn  V p . Then for zero bias, we
500  10 8
2


 500 10 8 
11.7 8.85 10 14 3.695
 
 

2
1.6 10 19 1016


which yields
x n  x p  4.646  10 5 cm
2

 
Now
V ox 
where Vn , Vox , V p are the voltage drops
V n x  
 t ox  s Vbi  VG 
  
eN d
 2 
We find
s
At x  t ox ,   
2
 t ox  s Vbi
  
eN d
 2 
If we apply a voltage V G , then replace V bi by
t
x n   ox 
2

eN d x n t ox
s
1.6 10 10 4.646 10 500 10 
11.78.85 10 
19
5
16
8
14
or
Vox  0.359 V
We also find
eN d x n2
Vn  V p 
2 s
1.6 10 10 4.646 10 

211.7 8.85 10 
19
5 2
16
14
or
Vn  V p  1.67 V
_______________________________________
10.30
(a) n-type
(b) We have
200 10 12
C ox 
 110 7 F/cm 2
3
2 10
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 10
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Also
C ox 

ox

3.9 8.85 10 14
 t ox  ox 
t ox
C ox
110 7


Qn  C ox  VGS  V DS 

 max   Q ss
 Q SD



  ms  2 fp  
C
ox

 

or
o
6
t ox  3.45 10 cm  34.5 nm  345 A
(c)
Q
V FB   ms  ss
C ox
or
Q
 0.80  0.50  ss 7
10
which yields
Qss  3 10 8 C/cm 2  1.875 1011 cm 2
(d)
ox
 
C FB
    kT  s 

t ox   ox  


 s   e  eN d 
 
Using the definition of threshold voltage VT ,
we have
Qn  C ox VGS  V DS   VT 
At saturation
V DS  V DS sat   VGS  VT
which then makes Q n equal to zero at the
drain terminal.
_______________________________________
10.33
0.025911.7 8.85 10 14  
1.6 10 2 10 
19

16
which yields
  7.82 10 8 F/cm 2
C FB
or
C FB  156 pF
_______________________________________
10.31
(a) Point 1: Inversion
2: Threshold
3: Depletion
4: Flat-band
5: Accumulation
_______________________________________
10.32
We have
Qn  Cox VGS  Vx    ms  2 fp



 max 
 Q ss  Q SD
Now let V x  V DS , so
 max   Q ss
 Q SD



  ms  2 fp  
C

 
ox

 max  is a
For a p-type substrate, Q SD
negative value, so we can write

10.34



k p W
2

2V SG  VT V SD  V SD
2 L
 0.10 
2

15 20.8  0.40.25  0.25
 2 
(a) I D 

Qn  C ox VGS  V DS 



 
 3.9 8.85 10 14  3.45 10 6
 3.9 


 11.7 
k n W
2

2VGS  VT V DS  V DS
2 L
 0.18 
2

8 20.8  0.40.2  0.2
 2 
 0.0864 mA
k W
2
(b) I D  n  VGS  VT 
2 L
 0.18 
2

80.8  0.4
2


 0.1152 mA
(c) Same as (b), I D  0.1152 mA
k W
2
(d) I D  n  VGS  VT 
2 L
 0.18 
2

81.2  0.4
 2 
 0.4608 mA
_______________________________________
(a) I D 


I D  0.103 mA
k p W
2
 V SG  VT 
(b) I D 
2 L
 0.10 
2

150.8  0.4
 2 
 0.12 mA


Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 10
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
k p W
2
 V SG  VT 
2 L
 0.10 
2

151.2  0.4
 2 
 0.48 mA
(d) Same as (c), I D  0.48 mA
_______________________________________
(c) I D 
10.37
C ox 
3.98.85 10 14   3.138 10 7 F/cm 2
110 10 8
 C W 425 3.138  10 7 20
K n  n ox 
2L
21.2


 1.11110 3 A/V 2 =1.111 mA/V 2
(a) VGS  0 , I D  0
VGS  0.6 V, V DS sat   0.15 V,
10.35
I D sat   1.1110.6  0.452
 0.025 mA
VGS  1.2 V, V DS sat   0.75 V,
k W
2
(a) I D  n  VGS  VT 
2 L
 0.6  W 
2
1.0  
 1.4  0.8
 2  L 
W
 9.26
L
 0.6 
2
(b) I D  
9.261.85  0.8
2


 3.06 mA
k W
2
2VGS  VT V DS  V DS
(c) I D  n 
2 L
 0.6 
2

9.26 21.2  0.80.15  0.15
 2 
 0.271 mA
_______________________________________
I D sat   1.1111.2  0.45
 0.625 mA
VGS  1.8 V, V DS sat   1.35 V,





I D sat   1.1111.8  0.45
 2.025 mA
VGS  2.4 V, V DS sat   1.95 V,
2
I D sat   1.1112.4  0.45
 4.225 mA
(c) I D  0 for VGS  0.45 V
VGS  0.6 V,
2

2

2

2
I D  1.111 20.6  0.450.1  0.1
 0.0222 mA
VGS  1.2 V,
10.36
(a) Assume biased in saturation region
k p W
2
ID 
 V SG  VT 
2 L
 0.12 
2
0.10  
200  VT 
 2 
 VT  0.289 V
I D  1.111 21.2  0.450.1  0.1
 0.156 mA
VGS  1.8 V,
Note: V SD  1.0 V  V SG  VT  0  0.289 V
So the transistor is biased in the saturation
region.
 0.12 
2
(b) I D  
200.4  0.289
 2 
 0.570 mA
 0.12 
(c) I D  
2020.6  0.2890.15
 2 
 0.15
2
2

or
I D  0.293 mA
_______________________________________
I D  1.111 21.8  0.450.1  0.1
 0.289 mA
VGS  2.4 V,





I D  1.111 22.4  0.450.1  0.1
 0.422 mA
_______________________________________
10.38
C ox 
2

ox 3.9 8.85 10 14

t ox
110 10 8

 3.138 10 7 F/cm 2
 p C ox W
Kp 
2L
210 3.138 10 7 35

21.2


 9.6110 4 A/V 2 =0.961 mA/V 2
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 10
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(a) V SG  0 , I D  0
10.40
Sketch
_______________________________________
V SG  0.6 V, V SD sat   0.25 V
I D sat   0.9610.6  0.35
 0.060 mA
V SG  1.2 V, V SD sat   0.85 V
2
I D sat   0.9611.2  0.35
 0.694 mA
V SG  1.8 V, V SD sat   1.45 V
10.41
Sketch
_______________________________________
2
10.42
We have
V DS sat   VGS  VT  V DS  VT
so that
V DS  V DS sat   VT
I D sat   0.9611.8  0.35
 2.02 mA
V SG  2.4 V, V SD sat   2.05 V
2
I D sat   0.9612.4  0.35
 4.04 mA
(c) I D  0 for V SG  0.35 V
V SG  0.6 V
2

I D  0.961 20.6  0.350.1  0.1
 0.0384 mA
V SG  1.2 V

I D  0.961 21.2  0.350.1  0.1
 0.154 mA
V SG  1.8 V

2

2
I D  0.961 21.8  0.350.1  0.1
 0.269 mA
V SG  2.4 V

2
2



I D  0.961 22.4  0.350.1  0.1
 0.384 mA
_______________________________________
10.39
(a) From Problem 10.37, K n  1.111 mA/V 2
For VGS  0.8 V, I D  0
VGS  0 , V DS sat   0.8 V
I D sat   1.1110  0.8
 0.711 mA
VGS  0.8 V, V DS sat   1.6 V
2
I D sat   1.1110.8  0.8
 2.84 mA
VGS  1.6 V, V DS sat   2.4 V
2
I D sat   1.1111.6  0.8
 6.40 mA
_______________________________________
2
Since V DS  V DS sat  , the transistor is always
biased in the saturation region. Then
2
I D  K n VGS  VT 
where, from Problem 10.37,
K n  1.111 mA/V 2 and VT  0.45 V
Then
V DS  VGS
I D (mA)
0
0
1
0.336
2
2.67
3
7.22
4
14.0
5
23.0
_______________________________________
10.43
From Problem 10.38, K p  0.961 mA/V 2

2
I D  K p 2V SG  VT V SD   V SD

I D
 2K p VSG  VT 
VSD VSD 0
For V SG  0.35 V, g d  0
For V SG  0.35 V,
g d  20.961V SG  0.35
For V SG  2.4 V,
g d  20.9612.4  0.35
 3.94 mA/V
_______________________________________
gd 
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 10
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
10.44
(a) g m 

I D
V GS
 

2
K n 2VGS  VT V DS   V DS
VGS

 K n 2V DS 
1.25  K n 20.05
 K n  12.5 mA/V 2
(b) I D  12.520.8  0.30.05 0.05
 0.594 mA
2

(c) I D  12.50.8  0.3
 3.125 mA
_______________________________________
2
where
C ox 


I D sat   18  A
(d)
V DS  V DS sat 

I D sat   0.033 ,
then
W n C ox
 3  0.2
2L
W n C ox
 0.139 10 3
2L
or


1
10 n 8.12 10 8  0.139 10 3
2
which yields
 n  342 cm 2 /V-s
_______________________________________
10.46
(a)
V DS sat   VGS  VT
or
4  VGS  0.8  VGS  4.8 V


C ox  8.12 10 8 F/cm 2
We are given W L  10 . From the graph, for
VGS  3 V, we have
or
2
or


2

or
I D  42.5  A
_______________________________________
10.47
(a) C ox 
3.98.85 10 14 
180 10 8
 1.9175 10 7 F/cm 2
or
0.033 

I D sat   1.25 10 5 2  0.8
 1.25 10 5 23  0.81  1
W n C ox
 VGS  VT 
2L
ox 3.9 8.85 10 14

t ox
425 10 8
so V DS  V DS sat 
2
I D  K n 2VGS  VT V DS  V DS
10.45
We find that VT  0.2 V
Now
I D sat  
(b)
2
2
sat 
I D sat   K n VGS  VT   K nV DS
so
2
2 10 4  K n 4
which yields
K n  12.5  A/V 2
(c)
V DS sat   VGS  VT  2  0.8  1.2 V

(i) k n   n C ox  450 1.9175 10 7
5
 8.629 10 A/V
or k n  86.29  A/V 2

2
 k   W 
2
(ii) I D sat    n  VGS  VT 
 2  L 
 0.08629  W 
2
0.8  
 2  0.4
2

 L 
W
 7.24
L
(b) (i) k p   p C ox  210  1.9175  10 7


5
 4.027 10 A/V
or k p  40.27  A/V 2
2
 k p  W 
2
(ii) I D sat     VSG  VT 
2
L
  
 0.04027  W 
2
0.8  
 2  0.4
2
L

 

W
 15.5
L

Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 10
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
_______________________________________
 1.135 10 7 C/cm 2
10.48
From Problem 10.37, K n  1.111 mA/V 2
(a) g mL 
 

2
K n 2VGS  VT V DS   V DS
VGS



2
K n VGS  VT 
VGS
(b) g ms 

 2 K n VGS  VT   21.1111.5  0.45
 3.452 10 4 A/V 2
or K n  0.3452 mA/V 2
For I D  0.5  0.3452VGS  0.7713
 

2
K p 2V SG  VT V SD   V SD
V SG

(b) g ms 


2
K p V SG  VT 
V SG
(c) (i) For V SB  0 , VT  VTO  0.7713 V
(ii) V SB  1 V,

VT  0.5594 20.389  1

 20.389

 0.2525 V
VT  0.7713  0.2525  1.024 V
 2K p VSG  VT   20.9611.5  0.35
or g ms  2.21 mA/V
_______________________________________
(iii) V SB  2 V,

VT  0.5594 20.389  2
 20.389
10.50
C ox
Now C ox 
3.98.85 10 14 
(iv) V SB  4 V,


 20.389


2 1.6 10 19 11.7 8.85 10 14 5 1016
2.30110 7
  0.5594 V


 411.7  8.85  10 14 0.3890  
(i) x dT  

1.6  10 19 5  10 16



 1.419 10 5 cm
 max 
QSD



10.51

1016 
  0.3473 V
10 
 1.5 10 
 fp  0.0259 ln 
 2  V  2 
 0.12 20.3473  2.5
1/ 2
VT  

fp
SB
fp
 20.3473

 1.6 10 19 5 1016 1.419 10 5


 0.7294 V
VT  0.7713  0.7294  1.501 V
_______________________________________
1/ 2
 5 1016 
  0.3890 V
(b)  fp  0.0259 ln 
10 
 1.5 10 


VT  0.5594 20.389  4
150 10 8
 2.30110 7 F/cm 2
Then

 0.4390 V
VT  0.7713  0.4390  1.210 V
2e s N a
(a)  

2
 VGS  1.975 V
 K p 2VSD   0.96120.1
or g mL  0.192 mA/V

For I D  0 , VGS  VTO  0.7713 V
10.49
From Problem 10.38, K p  0.961 mA/V 2
(a) g mL 
 V FB  2 fp
1.135 10 7
 0.5  20.3890
2.30110 7
 0.7713 V
 C W
K n  n ox
2L

450  2.301  10 7 8

21.2 

so g ms  2.33 mA/V
_______________________________________
C ox

 K n 2V DS   1.11120.1
so g mL  0.222 mA/V
 max 
Q SD
VTO 
or
VT  0.114 V

Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 10
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Now VT  VTO  VT
0.5  VTO  0.114
 VTO  0.386 V
_______________________________________
10.52
(a) C ox 
3.98.85 10 14 
C ox  8.63 10 8 F/cm 2
We find
Qss  1.6 10 19 5 1010  8 10 9 C/cm 2
Then
 max   Q ss
Q SD
VT 
  ms  2 fp
C ox



2e s N d
C ox

or
VT  0.357 V


2 1.6 10 19 11.7 8.85 10 14 5 1015

1.726 10 7
  0.2358 V 1 / 2
(b) For NMOS, apply V SB and VT shifts in a
positive direction, so for VT  0 , we want
VT  0.357 V.
So
 5 1015 
  0.3294 V
(b)  fn  0.0259 ln 
10 
 1.5 10 
 2  V  2 
 0.22  0.2358 20.3294  V
VT  
fn
BS
BS


1015 
  0.288 V
10 
 1.5 10 
 fp  0.0259 ln 

1/ 2
or
or
 max   1.38  10 8 C/cm 2
Q SD
Also
C ox 
or

ox 3.9 8.85 10 14

t ox
400 10 8

 
8



0.357  0.211 0.576  VSB  0.576
which yields
VSB  5.43 V
_______________________________________
W n C ox
VGS  VT 
L
W n ox
VGS  VT 

Lt ox
gm 
or
 
fp

8.63 10

 


SB
 20.288  VSB  20.288
10.55
(a)
1/ 2
x dT  0.863 10 4 cm
Now
 max   1.6  10 19 1015 0.863  10 4
Q SD
fp
10.54
Plot
_______________________________________
also


 2  V  2 
2 1.6 10 19 11.7 8.85 10 14 1015
 0.357 
10.53
(a) n  poly-to-p-type   ms  1.0 V

C ox
or
 V BS  2.39 V
_______________________________________
 411.7  8.85  10 14 0.288 


1.6  10 19 10 15


2e s N a
VT 
fn
 20.3294
 4 s  fp 
x dT  

 eN a 

 1.38 10 8  8 10 9 
  1.0  20.288
 

8.63 10 8


200 10 8
 1.726 10 7 F/cm 2




104003.98.85 10 14  5  0.65
475 10 8
or
g m  1.26 mS
Now
gm
g
1
g m 
 m  0.8 
1  g m rs
gm
1  g m rs
which yields
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 10
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
rs 
1  1
1  1


 1 
 1


g m  0.8  1.26  0.8 
Then
C M  1.035 10 14


 

 1  0.8974 10 3 10 10 3
or
rs  0.198 k 

or
(b) For VGS  3 V, g m  0.683 mS
Then
0.683
g m 
 0.602 mS
1  0.6830.198
or
g m
0.602

 0.88
g m 0.683
which is a 12% reduction.
_______________________________________
C M  1.032 10 13 F
Now
C gsT  C ox L  0.75  10 4 W 


3.98.85 10
14


8
500 10
 2 10 4  0.75 10 4 20 10 4



or
C gsT  3.797  10 14 F
We now find
10.56
(a) The ideal cutoff frequency for no overlap
capacitance is,
gm
 V  V 
fT 
 n GS 2 T
2 C gs
2 L

400 4  0.75

2 2  10
or



3.98.85 10

14


500 10
 0.75 10 4 20 10 4



C gdT  1.035  10 14 F
W n C ox
VGS  VT 
L
20 10 4 4003.9 8.85 10 14

2 10  4 500 10 8
 4  0.75
gm 







20 10 3.98.85 10 4 10 
4
14
500 10
6
8
or
g m  0.5522 10 3 S
We have
CM  C gdT 1  g m RL 

 1.035 10 14
 


 1  0.5522 10 3 10 10 3
or
g m  0.8974 10 3 S

We find
g m  WC ox ds
or



8


0.8974 10 3
2 3.797 10 14  1.032 10 13

We find
C gdT  C ox 0.75  10 4 20  10 4
Also

10.57
(a) For the ideal case

4 10 6
f T  ds 
2 L 2 2 10  4
or
f T  3.18 GHz
(b) With overlap capacitance (using the
values from Problem 10.56),
gm
fT 
2 C gdT  C M
where
CM  C gdT 1  g m RL 

2 C gsT  C M
f T  1.01 GHz
_______________________________________



gm
or
4 2
f T  5.17 GHz
(b) Now
gm
fT 
2 C gsT  C M
fT 
or
C M  6.750 10 14 F

Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 10
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Then
fT 
0.5522 10 3
2 3.797 10 14  6.75 10 14


or
f T  0.833 GHz
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Chapter 11
11.1
(a)
11.3
 V

I D  10 15 exp GS 
 2.1Vt 
For VGS  0.5 V,
 2 1016 
  0.3653 V
10 
 1.5 10 
 fp  0.0259 ln 
We find that


0.5
I D  10 exp 

 2.10.0259 
I D  9.83 10 12 A
For VGS  0.7 V,
15
I D  3.88 10 10 A
For VGS  0.9 V,
 


 2.544 10 cm/V 1 / 2
L 
   V    V sat 
2 s
eN a
fp
DS

L  2.544 10 5
DS


 0.3653  2  0.3653  0.6

L  1.413 10 5 cm  0.1413  m
(b) V DS sat   VGS  VT  1.0  0.4  0.6 V

L  2.544 10 5


 0.3653  4  0.3653  0.6

L  2.816 10 cm  0.2816  m
5
(c) V DS sat   VGS  VT  2.0  0.4  1.6 V

L  2.544 10 5


 0.3653  2  0.3653  1.6

L  3.46110 6 cm  0.0346  m
(d) V DS sat   VGS  VT  2.0  0.4  1.6 V

L  2.544 10 5
I 
VGS2  VGS1  nVt ln  D 2 
 I D1 
(a) VGS 2  VGS 1  0.0259  ln 10
 0.0596 V
(b) VGS 2  VGS 1  1.50.0259  ln 10
 0.0895 V
(c) VGS 2  VGS 1  2.10.0259  ln 10
 0.125 V
_______________________________________
fp
(a) V DS sat   VGS  VT  1.0  0.4  0.6 V
11.2
V 
exp GS2 
 V  VGS1  
I D2
 nVt 

 exp  GS2

I D1
nVt
V 


exp GS1 
nV
 t 


5
8
I D  1.54 10 A
Then the total current is:
I T  I D 10 6
For VGS  0.5 V, I T  9.83  A
For VGS  0.7 V, I T  0.388 mA
For VGS  0.9 V, I T  15.4 mA
(b)
Power: P  I T  V DD
Then
For VGS  0.5 V, P  49.2  W
For VGS  0.7 V, P  1.94 mW
For VGS  0.9 V, P  77 mW
_______________________________________

211.7  8.85  10 14
1.6  10 19 2  1016
2 s

eN a


 0.3653  4  0.3653  1.6 
L  1.749 10 5 cm  0.1749  m
_______________________________________
11.4
 2 1016 
  0.3653 V
10 
 1.5 10 
 fp  0.0259 ln 
We find that
2 s

eN a

211.7  8.85  10 14
1.6  10 19 2  1016




 2.544 10 5 cm/V 1 / 2
V DS sat   VGS  VT  2.0  0.4  1.6 V
L 
2 s
eN a
   V    V sat 
fp
DS
fp
DS
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________

(a) L  2.544 10 5


 0.3653  3  0.3653  1.6

5
L  1.10 10 cm  0.110  m



L
0.2326
 0.10 
L
L
 L  2.326  m
_______________________________________
Now


3.9 8.85 10
C ox  ox 
t ox
120 10 8
14




1/ 2

 1.6 10 19 4 1016 1.575 10 5
7
 1.008 10 C/cm

2
So
1.008 10 7
 0.5223  20.3832
2.876 10 7
 0.595 V
V DS sat   VGS  VT  1.25  0.595  0.655 V
VT 
L  7.35 10 cm  0.0735  m


(ii) L  1.799 10 5

 0.3832  0.655  2  0.3832  0.655

L  1.303 10 cm  0.1303  m
5



L
0.2205
 0.12 
L
L
L  1.84  m
_______________________________________
2.876 10 7


6
(b)
 411.7  8.85  10 0.3832  
x dT  

1.6  10 19 4  10 16





L  2.205 10 cm  0.2205  m
19
 1.575 10 5 cm
 max 
QSD
  fp  V DS sat 
5
10

DS
 0.3832  0.655  1  0.3832  0.655

4 10 1.6 10 


5
DS
 0.3832  0.655  4  0.3832  0.655
V FB  0.5223 V
Now
 max 
Q SD
VT 
 V FB  2 fp
C ox
We find
 4 1016 
  0.3832 V
 fp  0.0259 ln 
10 
 1.5 10 

fp
(iii) L  1.799 10 5
 2.876 10 7 F/cm 2
Q
V FB   ms  ss
C ox
14
   V sat   V
(i) L  1.799 10
L  2.326 10 5 cm  0.2326  m
 0.5 
2 s
eN a
(a) L 
 0.3653  5  0.3653  1.6
11.5

 1.799 10 cm/V 1 / 2
L
0.110
 0.10 
L
L
 L  1.10  m




5
Now
(b) L  2.544 10 5

211.7  8.85  10 14
1.6  10 19 4  1016
2 s

eN a
11.6
 3 1016 
  0.3758 V
10 
 1.5 10 
 fp  0.0259 ln 

211.7  8.85  10 14
1.6  10 19 3  10 16
2 s

eN a




5
 2.077 10 cm/V 1 / 2
(a) Ideal,
k W
2
I D  n  VGS  VT 
2 L
 0.05  15 
2


1.0  0.4
2
0
.
80



 0.16875 mA
(i) V DS sat   1.0  0.4  0.6 V

L  2.077 10 5


 0.3758  2  0.3758  0.6
 1.150 10 cm  0.115  m
5
 L 
I D  
 ID
 L  L 

Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
0.80



0.16875
 0.80  0.115 
 0.19708 mA

(ii) L  2.077 10 5

11.7
(a)

 0.3758  4  0.3758  0.6

 2.293 10 cm  0.2293  m
5
(ii) I D  I D 1  V DS 
 L 
I D  
 ID
 L  L 
 75.93751  0.021.5
 78.22  A
0.80



0.16875
0
.
80

0
.
2293


 0.23655 mA
(iii) ro 
1
 0.23655  0.19708  10 3 


42


 ro  5.07 10 4   50.7 k 
(c) V DS sat   VGS  VT  2.0  0.4  1.6 V

(i) L  2.077 10 5


 0.3758  2  0.3758  1.6

6
 2.819 10 cm  0.02819  m
 L 
I D  
 ID
 L  L 
0.80



0.16875
0
.
80

0
.
02819


 0.17491 mA
(ii) L  2.077 10 5



 0.3758  4  0.3758  0.6

1
1
(b)
 0.075 
2
(i) I D  
101.25  0.35
2


 0.30375 mA
(ii) I D  0.303751  0.021.5
 0.3129 mA
1
(iii) ro 
 165 k 
0.020.30375
_______________________________________
11.8
Plot
_______________________________________
11.9
(a) Assume V DS sat   1 V. Then

 sat 
 1.425 10 5 cm  0.1425  m
We find
 L 
I D  
 ID
 L  L 
L (  m)
1
0.5
0.25
0.13
1
 0.20532  0.17491 10 3 


42


V DS sat 
L
3
0.80



0.16875
 0.80  0.1425 
 0.20532 mA
 I D 

ro  

 V DS 
1
 I D 0.02 75.94 
 0.658 M   658 k 
(b)
 I D 

ro  

 V DS 
k n W
2
 VGS  VT 
2 L
 0.075 
2

100.8  0.35
2


 0.07594 mA  75.94  A
(i) I D 
1
ro  6.577 10 4   65.77 k 
_______________________________________
 sat (V/cm)
3.33 10 3
110 4
210 4
410 4
7.69 10 4
(b)
Assume  n  500 cm 2 /V-s, we have
   n  sat
Then
For L  3  m,   1.67 10 6 cm/s
For L  1  m,   510 6 cm/s
For L  0.5  m,   10 7 cm/s
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
11.10
k n   n C ox 
(a)
 n ox
IDL
1
L


ro L  L 2 V DS
t ox
4253.98.85 10 14 


8
110 10
 1.334 10 4 A/V 2  0.1334 mA/V 2
 k   W 
2
I D   n  VGS  VT 
2
L
  
fp

We find
2 s

eN a
11.11
(a)



DS
 0.3758  0.35
 1.660 10 cm  0.166  m
2
(mA)
1 / 3
Where  O  1000 cm 2 /V-s and

5
I D sat   0.173VGS  1
and
  eff 

Let  eff   O 

 C 


 0.3758  2.35
or
(b)
 3 1016 
  0.3758 V
10 
 1.5 10 
V DS  V DS sat   V DS  VGS  VT  V DS
 0.8  0.45  2  2.35 V
Now
2.077 10 5
L

 6.290 10 6 cm/V
V DS 2 0.3758  2.35


I D sat   0.173 VGS  1 (mA) 1 / 2
 fp  0.0259 ln 
L  2.077  10 5

W n C ox
VGS  VT 2
2L
 10 
2
  500 6.9 10 8 VGS  1
2

 2.077 10 5 cm/V 1 / 2


I D sat  

211.7  8.85  10 14
1.6  10 19 3  10 16


 ro  5.865 10 4   58.65 k 
_______________________________________
2 s 1
1 / 2
  fp  V DS
eN a 2
L

V DS

 1.705 10 5
   V    V sat 
DS
6
4 2

Now
fp
4
 ro  1.04 10 5   104 k 
(b)
0.1362 10 3 0.8 10 4
1

 6.290 10 6
ro
0.8  0.16610 4 2
L 
2 

 I D L 1L  L   

 V DS 
IDL
L


2
L  L V DS
L 
3
 9.615 10 6
 0.1334  20 
2


0.8  0.45
 2  1.2 
 0.1362 mA
I D

1
  L 


 ID 

ro V DS V DS  L  L 


L  L1
 IDL
V DS
2 s
eN a
0.1362 10 1.2 10   6.290 10 
1.2  0.16610 

 C  2.5 10 4 V/cm
V
Let  eff  GS
t ox
We find


3.9 8.85 10 14
C ox  ox  t ox  ox 
t ox
C ox
6.9 10 8
or

o
t ox  500 A
Then

Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
VGS
 eff
 eff
1
2
3
-410 5
-397
6  10 5
347
8 10
4
5
5
10  10 5
I D sat 
0
0.370
0.692
315
0.989
292
1.27

11.12
Plot
_______________________________________
(a) C ox 
3.98.85 10 14 


2


2

(ii) I D  0.410 221.0  1.0
 1.23 mA

(iii) I D  0.410 221.25  1.25
 1.409 mA

(iv) I D  0.410 222  2
 1.64 mA
(b) I D  WC ox VGS  VT  ds
2
2


10
ds
7
ds

 0.5 
6
(i) For V DS  0.5 V,  ds  
 4 10
1
.
25



 1.6 10 6 cm/s



11.15
(a) Non-saturation region
1
W 
2
I D   n C ox   2VGS  VT V DS  V DS
2
L
We have

C
C ox  ox  ox
t ox
k
and
W  kW , L  kL
also
VGS  kVGS , V DS  kVDS
So
1  C  kW 
I D   n  ox 

2  k  kL 

 

 3.452 10  A
 3.452 10  mA


I D  3.452 10 7 4 10 6
 1.38 mA
(iv) For V DS  2 V,  ds  410 6 cm/s
I D  1.38 mA
(c) For part (a), V DS sat   2 V
For part (b), V DS sat   1.25 V
_______________________________________
 2kVGS  VT kVDS  kVDS 
 10 3 1.726 10 7 2 ds
I D  3.452 10 7 1.6 10 6
 0.552 mA


 4.10 10 4 A/V 2  0.410 mA/V 2
For VGS  VT  2 V, V DS sat   2 V
(i) I D  0.410 220.5  0.5
 0.7175 mA
 3.2 10 6 cm/s
11.14
Plot
_______________________________________
200 10 8
 1.726 10 7 F/cm 2
 C W
K n  n ox
2L
475 1.726 10 7 10

21.0 



I D  3.452 10 7 3.2 10 6
 1.10 mA
(iii) For V DS  1.25 V,  ds  410 6 cm/s
(c) The slope of the variable mobility curve is
not constant, but is continually decreasing.
_______________________________________
11.13

 1.0 
6
(ii) For V DS  1.0 V,  ds  
 4 10
 1.25 


2

Then
I D   kI D
In the saturation region,
1  C  kW 
2
I D   n  ox 
kVGS  VT 
2  k  kL 
Then
I D   kI D
(b)
P  I DV DD  kI D kVDD   k 2 P
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
11.16
I D sat   WC ox VGS  VT  sat
C 
 kW  ox kVGS  VT  sat
 k 
or
I D sat    kI D sat 
_______________________________________
11.17
(a)
k n W
2
  VGS  VT 
2 L
 0.15  6 
2


3  0.45
 2  1.2 
2.438 mA
(ii) Scaled device:
V D  VGS  k 3  0.653  1.95 V
(i) I D max  
 0.15   0.15 
k n  

  0.2308 mA/V 2
 k   0.65 
L  k 1.2  0.651.2  0.78  m
W  k 6  0.656  3.90  m
Then
 0.2308  3.9 
2
I D max   

1.95  0.45
 2  0.78 
 1.298 mA
(b) (i) Pmax   I D max V D  2.4383
 7.314 mW
(ii) Pmax   1.2981.95
 2.531 mW
_______________________________________
11.18
C ox 

ox 3.9 8.85 10 14

t ox
120 10 8

 5 1016 
  0.3890 V
10 
 1.5 10 
 411.7  8.85  10 14 0.3890  
x dT  

1.6  10 19 5  10 16



 1.419 10 5 cm

11.19
C ox 

1/ 2



3.98.85 10 14 
80 10 8
 4.314 10 7 F/cm 2
 2 1016 
  0.3653 V
 fp  0.0259 ln 
10 
 1.5 10 


 411.7  8.85  10 14 0.3653 
x dT  

1.6  10 19 2  10 16




1/ 2

5
 2.174 10 cm
1.6 10 19 2 1016 2.174 10 5
VT  
4.314 10  7
 0.30 

 1  20.2174   1



0.30

 0.70 
VT  0.0391 V




VT  VTO  VT
0.35  VTO  0.0391
 VTO  0.389 V
_______________________________________
C ox 
 fp  0.0259 ln 


11.20
 2.876 10 7 F/cm 2


eN a x dT  r j 
2x

1  dT  1

C ox  L 
rj

 
19
16
1.6 10
5 10 1.419 10 5

2.876 10 7
 0.25 

 1  20.1419   1


0.25
 0.80 

VT  0.0569 V
_______________________________________
VT  
3.98.85 10 14 
200 10 8
 1.726 10 7 F/cm 2
 3 1016 
  0.3758 V
 fp  0.0259 ln 
10 
 1.5 10 


 411.7  8.85  10 14 0.3758 
x dT  

1.6  10 19 3  10 16



 1.80 10 5 cm


1/ 2
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
VT  0.15

1.6 10 3 10 1.80 10 
19
16
5
1.726 10 7
 0.30 

 1  20.18  1


0.30
 L 

 0.30 
0.15  0.5006
0.4832
 L 
 L  0.484  m
_______________________________________
11.21
We have
L   L  a  b 
and from the geometry
2
2
2
 r j  x dS 
(1) a  r j   x dT
and
2
2
2
 r j  x dD 
(2) b  r j   x dT
From (1)
a  r j 2  r j  x dS 2  x dT2
so that
a
r  x   x
2
j
dS
2
dT
 rj
which can be written as
2
2


 x dS   x dT 

 
  1
a  r j  1 


r j   r j 



or
2
2


2 x dS  x dS   x dT 

a  rj  1


 1
 rj   rj 
rj

 



Define
x2  x2
 2  dS 2 dT
rj
We can then write


2x
a  r j  1  dS   2  1
rj


Similarly from (2), we will have


2x
b  r j  1  dD   2  1
rj


where
x2  x2
 2  dD 2 dT
rj
The average bulk charge in the trapezoid (per
unit area) is
 L  L 
QB  L  eN a x dT 

 2 
or
 L  L 
QB  eN a x dT 

 2L 
We can write
L  L 1 L 1 1
L  a  b
 
 
2L
2 2L 2 2L
which is
L  L
ab
 1
2L
2L

 max  in the
Now, Q B replaces QSD
threshold equation. Then
 max 
Q B
Q SD
VT 

C ox
C ox

eN a x dT  a  b  eN a x dT
1

C ox 
2L 
C ox
or
eN a x dT a  b 

C ox
2L
Then substituting, we obtain

rj 
eN x
2 x dS

VT   a dT 
  2  1
 1 
C ox
2 L 
rj


VT  

 
2x
  1  dD   2  1 
rj

 
Note that if x dS  x dD  x dT , then     0
and the expression for VT reduces to that
given in the text.
_______________________________________
11.22
We have L   0 , so Equation (11.27)
becomes
L  L
L
1


2L
2L 2


2x
 rj 

 1   1  dT  1 
L 
rj
 




or
 1
rj 
2x
 1  dT  1 
L 
rj
 2

Then Equation (11.28) is
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
1
Q B  eN a x dT  
2
Then change in the threshold voltage is
 max 
Q B
Q SD
VT 

C oc
C ox
or
1 2eN a x dT  eN a x dT 
VT 

C ox
C ox
which becomes
1 eN x
VT    a dT
2 C ox
_______________________________________
VT 
1.6 10 3 10 1.80 10   2 
 

4.314 10 2.2 10 
19
11.25
VT  

eN a x dT 
2 x dT
rj 

 1 
  1
C ox  L 
rj
 
 

16
5 2
7
4
VT  0.0257 V
_______________________________________
11.27
C ox 
3.98.85 10 14 
120 10 8
 2.876 10 7 F/cm 2
 1016 
  0.3473 V
 fp  0.0259 ln 
10 
 1.5 10 
11.23
Plot
_______________________________________
11.24
Plot
_______________________________________
eN a x dT   x dT 


C ox  W 


 411.7  8.85  10 14 0.3473 
x dT  

1.6  10 19 10 16



1/ 2
 
5
 3.0 10 cm
eN a x dT   x dT 
VT 


C ox  W 
In this case,   1
So
1.6 10 10 1.0310 
0.045 
2.876 10 W 
19
5 2
N 
e a kxdT  
 
2kxdT
 krj 
 k 

 1 
 1

kr j
 C ox 
 
 kL 


k


W  1.11  m
_______________________________________
VT  kVT
_______________________________________
11.28
Plot
_______________________________________
11.26
11.29
or
C ox
3.98.85 10 14 

80 10 8
 4.314 10 7 F/cm 2
 3 1016 
  0.3758 V
 fp  0.0259 ln 
10 
 1.5 10 


 411.7  8.85  10 14 0.3758 
x dT  

1.6  10 19 3  10 16



5
 1.80 10 cm

16
7
eN a x dT   x dT 


C ox  W 
Assume that  is a constant, then
VT 
N 
e a kxdT 
  kxdT 
 k 
VT 


 C ox 
 kW 


 k 
1/ 2

or
VT  kVT
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Now
11.30
(a)
ID
V
(i)  ox  G
t ox
6  10 
6
10
10
VG
10
200  10 8
 VG  12 V
10
8
7
6
5
10 4
12
4V
(ii) VG 
3
(b)
VG
(i) 6  10 6 
80  10 8
 VG  4.8 V
10
3
11.31
(a) VG  83  24   ox t ox  6 10 6 t ox 


t ox  4 10 6 cm
o
or t ox  40 nm  400 A


(b) VG  123  36   ox t ox  6 10 6 t ox 
6
t ox  6 10 cm
o
or t ox  60 nm  600 A
_______________________________________
11.32
Snapback breakdown means M  1 , where
 IO 
  0.18 log 10 

9 
 3 10 
and
1
M 
m
V 
1   CE 
 V BO 
Let V BO  15 V and m  3 . Now when

3
0.0941
14.5
0.274
13.5
0.454
12.3
0.634
10.7
0.814
8.6
0.994
2.7
11.33
One Debye length is
 kT e  
LD   s

 eN a 
4.8
 1.6 V
3
_______________________________________
V 
1   CE 
 15 
we can write this as
VCE
_______________________________________
(ii) VG 
M  1 




 11.7  8.85  10 14 0.0259  


1.6  10 19 10 16


1/ 2
 
or
LD  4.09 10 6 cm
Six Debye lengths is then
6L D  0.246 10 4 cm  0.246  m
From Example 11.5, we have
x dO  0.336  m, which is the zero-biased
source-substrate junction width.
At near punch-through, we will have
x dO  6 L D  x d  L
where x d is the reverse-biased drainsubstrate junction width. Now
0.336  0.246  x d  1.2
or
x d  0.618  m
Then, at near punch-through we have
 2  V  V DS  
x d   s bi

eN a


1/ 2
or
Vbi  V DS 
3
V 
1   CE     VCE  15  3 1  
 15 
1/ 2

x d2 eN a
2 s
0.618 10  1.6 10 10 
211.78.85 10 
which yields
Vbi  V DS  2.95 V
4 2
19
14
16
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
From Example 11.5, we have Vbi  0.874 V,
so
V DS  2.08 V
which is the near punch-through voltage. The
ideal punch-through voltage was
V DS  4.9 V
_______________________________________
11.35
With a source-to-substrate voltage of 2 volts,
 2  V  V SB  
x dO   s bi

eN a



1/ 2

 211.7  8.85  10 14 0.902  2  


1.6  10 19 3  10 16




1/ 2

or
11.34
 


 10 3  10 
Vbi  0.0259  ln 
  0.902 V
2
 1.5  1010

The zero-biased source-substrate junction
width is given by
19
2 V 
x dO   s bi 
 eN a 
16










1/ 2

11.36
or
C ox 
6
LD  2.36 10 cm
so that
6L D  0.142 10 4 cm  0.142  m
Now
x dO  6 L D  x d  L

1/ 2

 211.7  8.85  10 14 0.902  5 


1.6  10 19 3  10 16




3.98.85 10 14   2.876 10 7 F/cm 2
120 10 8
eD I
VT 
C ox
Implant acceptor ions for a positive threshold
voltage shift.
VT C ox  0.80 2.876 10 7
DI 

e
1.6 10 19
12
2
 1.438 10 cm
_______________________________________

We have for V DS  5 V,
 2  V  V DS  
x d   s bi

eN a



x d  0.584 10 4 cm  0.584  m
Then
L  x dO  6 L D  x d
 0.354  0.142  0.584
or
L  1.08  m
_______________________________________
1/ 2
 11.7  8.85  10 14 0.0259  


1.6  10 19 3  10 16



1/ 2
or
x dO  0.197 10 4 cm  0.197  m
The Debye length is
 kT e  
LD   s

 eN a 

1/ 2
 211.7  8.85  10 14 0.902  5  2  


1.6  10 19 3  10 16


1/ 2
or

 2  V  V DS  V SB  
x d   s bi

eN a


1/ 2
 211.7  8.85  10 14 0.902  


1.6  10 19 3  10 16



x dO  0.354 10 4 cm  0.354  m
We have 6 L D  0.142  m from the previous
problem.
Now
1/ 2

or
x d  0.505 10 4 cm  0.505  m
Then
L  0.197  0.142  0.505
or
L  0.844  m
_______________________________________
11.37
C ox 

3.98.85 10 14 
180 10 8
 1.9175 10 7 F/cm 2
eD I
VT 
C ox
Implant donor ions for a negative threshold
voltage shift.
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
DI 
VT C ox
e

0.601.9175 10 7 
1.6  10
V FB   ms 
19
2
 7.19 10 cm
_______________________________________
11
 1.08 
3.98.85 10 14 


19
10

2.30110


 3.797 10 5 cm
 max 
QSD




 3.645 10 C/cm
 max 
Q SD
VTO 
 V FB  2 fp
C ox
2
3.645 10
 1.115  20.3341
2.30110 7
VTO  0.2884 V
(b) For a positive threshold voltage shift, add
acceptor ions.
VT  VT  VTO  0.50   0.2884
 0.788 V
Then
VT C ox 0.788 2.30110 7
DI 

e
1.6 10 19
12
2
 1.13 10 cm
_______________________________________

C ox 
3.98.85 10 14 
180 10 8
 1.9175 10 7 F/cm 2





 6.958 10 C/cm
 max 
Q SD
VTO  
 V FB  2 fn
C ox
2
6.958 10 8
 0.9966  20.3653
1.9175 10 7
VTO  0.0969 V
(b) For a negative threshold voltage shift,
add donor ions.
VT  VT  VTO  0.40   0.0969
 0.3031 V
VT C ox
Then D I 
e
0.3031 1.9175 10 7

1.6 10 19
 3.63 1011 cm 2
_______________________________________

8
11.39
(a)  ms  1.08 V
1/ 2

1/ 2
 1.6 10 19 6 1015 3.797 10 5


8
 6 1015 
  0.3341 V
10 
 1.5 10 
8

 1.6 10 19 2 1016 2.1744 10 5
7
 411.7  8.85  10 14 0.3341 
x dT  

1.6  10 19 6  10 15




 2.1744 10 cm
 max 
QSD
 fp  0.0259 ln 

7
5
5 10 1.6 10 
 1.115 V
19
11
 411.7  8.85  10 14 0.3653 
x dT  

1.6  10 19 2  10 16


150 10 8
 2.30110 7 F/cm 2
Q
V FB   ms  ss
C ox
 1.08 
10 1.6 10 
1.9175 10
 0.9966 V
 2 1016 
  0.3653 V
 fn  0.0259 ln 
10 
 1.5 10 
11.38
(a)  ms  1.08 V
C ox 
Q ss
C ox

11.40
 4 1015 
  0.3236 V
(a)  fp  0.0259 ln 
10 
 1.5 10 
3.9 8.85 10 14
C ox 
80 10 8
 4.314 10 7 F/cm 2




 411.7  8.85  10 14 0.3236  
x dT  

1.6  10 19 4  10 15




 4.576 10 5 cm
 max 
QSD



8
2
1/ 2

 1.6 10 19 4 1015 4.576 10 5
 2.929 10 C/cm

Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
VTO 
 max 
Q SD
C ox
Then
 V FB  2 fp
2.929 10 8
 1.25  20.3236
4.314 10 7
 0.5349 V
(b) For a positive threshold voltage shift, add
acceptor ions.
VT  VT  VTO  0.40   0.5349
 0.9349 V
VT C ox
Then D I 
e
0.9349 4.314 10 7

1.6 10 19
 2.52 1012 cm 2
(c) Add acceptor ions.
VT  VT  VTO  0.40   0.5349
 0.1349 V
0.1349 4.314 10 7
Then D I 
1.6 10 19
 3.64 1011 cm 2
_______________________________________





11.41
The total space charge width is greater than
xi , so from Chapter 10
VT 
2e s N a
 2  V  2 
fp
C ox
SB
fp
Now

1014 
  0.228 V
10 
 1.5 10 
 fp  0.0259 ln 
and
C ox
3.98.85 10 14 

21.6 10 11.78.85 10 10 
19
14
14 1 / 2
6.90 10 8

 20.228  VSB  20.228
or

11.42
(a)

1017 
  0.407 V
10 
 1.5 10 
 fn  0.0259 ln 
and
VT  0.0834 0.456  VSB  0.456




 411.7  8.85  10 14 0.407  
x dT  

1.6  10 19 10 17



1/ 2
 
5
 1.026 10 cm
n poly on n-type  ms  0.32 V
We have
 max   1.6  10 19 1017 1.026  10 5
Q SD





 1.64 10 7 C/cm 2
Now
VTP   1.64 10 7  1.6 10 19 5 1010



80 10
8
3.98.85 10 14 


 0.32  20.407 
or
VTP  1.53 V (Enhancement PMOS)
(b) For VT  0 , shift threshold voltage in
positive direction, so implant acceptor ions.
VT C ox
eD I
VT 
 DI 
C ox
e
so
1.533.9 8.85 10 14
DI 
80 10 8 1.6 10 19
or
DI  4.13 1012 cm 2
_______________________________________

500 10 8
 6.90 10 8 F/cm 2
Then
VT

V SB (V)
VT (V)
1
0.0443
3
0.0987
5
0.1385
_______________________________________




Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
11.43
The areal density of generated holes is
 8 1012 10 5 750 10 8  6 1012 cm 2
The equivalent surface charge trapped is
 0.10 6 1012  6 1011 cm 2
Then
Q
VT   ss
C ox

 




6 10 1.6 10 750 10 
3.98.85 10 
19
11
8
14
or
VT  2.09 V
_______________________________________
11.44
The areal density of generated holes is
61012 cm 2 . Now

3.9 8.85 10 14
C ox  ox 
t ox
750 10 8


 4.6 10 8 F/cm 2
Then

 

Qss
6 1012 x  1.6 10 19

C ox
4.6 10 8
where x is the fraction of holes that may be
trapped. For VT  0.50 V we find
x  0.024  x  2.4%
_______________________________________
VT  
11.45
We have the areal density of generated holes
as
 g  t ox 
where g is the generation rate and  is the
radiation dose. The equivalent charge trapped
is
 xg t ox
where x is the fraction of generated holes
trapped.
Then
 exg 
Q
exg t ox
t 2
VT   ss  
 
ox t ox   ox  ox
C ox
or
2
VT   t ox 
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Chapter 12
Then
12.1
Sketch
_______________________________________
12.2
Sketch
_______________________________________

5 10 3 

13 
 1.738 10 
 BE  0.0259 ln 
  BE  0.6237 V
_______________________________________
12.5
12.3
(a)  
eD n ABE n B 0
(a) I S 
xB

1.6 10 185 10 4 10 
19
5
3
0.80 10  4
 7.2 10 15 A
 
(b) I C  I S exp BE 
 Vt 
 0.58 
(i) I C  7.2 10 15 exp

 0.0259 


0.9850

 65.7
1   1  0.9850
(b)
(i) For I C  38.27  A,
IB 
IC
IB 
0.571
 0.008695 mA
65.67
 8.695  A
38.27
 0.5828  A
 65.67
I
38.27
IE  C 
 38.85  A
 0.9850
(ii) For I C  0.571 mA,

 3.827 10 5 A  38.27  A


 0.65 
(ii) I C  7.2 10 15 exp

 0.0259 
0.571
 0.5797 mA
0.9850
(iii) For I C  8.519 mA,
IE 
 5.710 10 4 A  0.571 mA
 0.72 
(iii) I C  7.2 10 15 exp

 0.0259 


12.4
iC 
 
eDn ABE
 n B 0  exp BE 
xB
 Vt 
2 10 3 
1.6 10 22A
BE
0.80 10  4

 0.60 
 2 10 4 exp

 0.0259 
 ABE  1.975 10 4 cm 2
(b) 5 10 3 
1.6 10 221.975 10 
19
4
0.80 10
4
 

 2 10 exp BE 
 0.0259 
 

5 10 3  1.738 10 13 exp BE 
 0.0259 


4


38.27
 0.2310  A
165.7
38.27
IE 
 38.50  A
0.9940
(ii) For I C  0.571 mA,
IB 
19

8.519
 0.1297 mA
65.67
8.519
IE 
 8.649 mA
0.9850
0.9940
 165.7
(c)  
1  0.9940
(i) For I C  38.27  A,
IB 
 8.519 10 3 A  8.519 mA
_______________________________________
(a)

IB 
0.571
 0.003446 mA
165.7
 3.446  A
IE 
0.571
 0.5744 mA
0.9940
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(iii) For I C  8.519 mA,
(b) For VCE  3 V, I C  0
IB 
8.519
 0.05141 mA
165.7
 51.41  A
8.519
IE 
 8.570 mA
0.9940
_______________________________________
(i) VCE  VCC  I C RC
0.2  3  I C 10  , I C  0.28 mA
(ii) For VCB  0  VCE  V BE  0.65 V
0.65  3  I C 10  , I C  0.235 mA
_______________________________________
12.9
12.6
I
0.625
 148.8
(a)   C 
I B 0.0042
(a)
148.8


 0.9933
1   149.8
I
0.625
IE  C 
 0.6292 mA
 0.9933
I
1.254
 0.9851
(b)   C 
I E 1.273

0.9851

 66.0
1   1  0.9851
I
1.254
IB  C 
 0.0190 mA

66
 19.0  A
iC   i B  100 0.05
iC  5 mA
We have
 CE  VCC  iC R  10  51
or
 CE  5 V
_______________________________________
12.8
(a) For VCE  3 V, I C  0
(i) VCE  VCC  I C RC
0.2  3  I C 25 , I C  0.112 mA
(ii) For VCB  0  VCE  V BE  0.65 V
0.65  3  I C 25 , I C  0.094 mA

 1.125 10 4 cm 3

ni2
1.5 1010

NC
1015
pC 0 

2

2
 2.25  10 5 cm 3
V 
(b) n B 0  n B 0 exp BE 
 Vt 


 0.640 
 1.125 10 4 exp

 0.0259 
 6.064 1014 cm 3
V 
p E 0  p E 0 exp BE 
 Vt 
I E  1   I B  1510.065  9.815  A
_______________________________________
or
2
ni2
1.5 1010

NB
2 1016
n B0 
150
 0.99338
(c)  
151
I C   I B  1500.065  9.75  A
12.7
(c) For i B  0.05 mA,

 2.8125 10 2 cm 3



ni2
1.5 1010

NE
8 1017
pE0 


 0.640 
 2.8125 10 2 exp

 0.0259 
 1.516 1013 cm 3
_______________________________________
12.10
(a) n E 0 

ni2
1.5 1010

NE
5 1017

 4.5 10 2 cm 3
p B0 

ni2
1.5 1010

NB
1016
2

 2.25  10 4 cm 3
nC 0 

ni2
1.5 1010

NC
1015
 2.25  10 5 cm 3
2

2
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
At x  0 ,
V 
(b) p B 0  p B 0 exp EB 
 n BO
d n B 
  V BE  
 Vt 

 1
exp
dx x  0
 x B    Vt  
 0.615 
4

L B sinh 
 2.25 10 exp

 LB 
 0.0259 
 4.62 1014 cm 3
 x  
 cosh B   1
V 
 L B  
n E 0  n E 0 exp EB 
At x  x B ,
 Vt 




 0.615 
 4.5 10 2 exp

 0.0259 
 9.24 1012 cm 3
_______________________________________
12.11

n2
1.5 1010
(a) n B 0  i 
NB
2 1016

V 
 nB 0 exp BE 
 Vt 
 2 1015 

Then V BE  0.0259 ln 
4 
 1.125 10 
 0.6709 V


2
 2.8125 10 2 cm 3
V 
p E 0  p E 0 exp BE  +
 Vt 


 0.6709 
 2.8125 10 exp

 0.0259 
2
 5.0 1013 cm 3
_______________________________________
12.12
We have
d n B 

dx
 x 
 cosh B 
 L B 
Taking the ratio
d n B 
dx x  x B
Now
n B 0  0.1N B   2 1015 cm 3
ni2
1.5 1010

NE
8 1017

  V BE  
exp
  1
 xB  
  Vt  



L B sinh 
 LB 
 n BO
2
 1.125 10 4 cm 3
(b) p E 0 
d n B 

dx x  x B
  V BE  
  1
exp
 x B    Vt  

sinh 
 LB 
n BO
 1 
x x 1
 x 
 cosh B
 

 
cosh
 LB 
 LB  LB
 L B 
d n B 
dx x  0
  V BE  
x 
  1  cosh B 
exp
L 

  Vt  
 B

  V BE  
x 
  1  cosh B   1
exp
L 

 B
  Vt  
1
x 
cosh B 
 LB 
x
(a) For B  0.1  Ratio  0.9950
LB

(b) For
xB
 1.0  Ratio  0.648
LB
xB
 10  Ratio  9.08 10 5
LB
_______________________________________
(c) For
12.13
In the base of the transistor, we have
d 2 n B x  n B x 
DB

0
 BO
dx 2
or
d 2 n B x  n B x 

0
dx 2
L2B
where LB  DB BO
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
The general solution to the differential
The general solution is of the form
equation is of the form
 x 
x

p B x   A exp   B exp
 x 
x

 LB 
 LB 
n B x   A exp   B exp
 LB 
 LB 
From the boundary conditions, we can write
From the boundary conditions, we have
p B 0  A  B  p B 0  p BO
n B 0  A  B  n B 0  n BO
 V  
 p BO exp EB   1
  V BE  
  1
 n BO exp
  Vt  

V
  t  
Also
Also
x 
x 
p B x B   A exp B   B exp B 
 xB 
  xB 
 LB 
 LB 

n B x B   A exp   B exp
 LB 
 LB 
  p BO
 n BO
From the first boundary condition equation,
From the first boundary condition, we can
we find
write
 V  
A  p BO exp EB   1  B
  V BE  
  1  B
A  n BO exp
  Vt  

  Vt  
Substituting into the second boundary
Substituting into the second boundary
equation, we obtain
condition, we find
 V  
x 
p BO exp EB   1  exp B   p BO
  xB 
  x B 
  exp

 LB 
B exp
  Vt  
B

L
  L B 

 B 
x 
2 sinh  B 
  V BE  
 xB 
 LB 
  1  exp
  n BO
 n BO exp


and then we obtain
  Vt  
 LB 
 V  
Solving for B, we find
x 
 p BO exp EB   1  exp B   p BO
  V BE  
x 
  Vt  
 LB 
  1  exp B   n BO
n BO exp
A
L 

V
 B
x 
  t  
B
2 sinh  B 
x 
 LB 
2 sinh  B 
Substituting the expressions for A and B into
 LB 
the general solution and collecting terms, we
We then find
obtain
  V BE  
  xB 
  1  exp
 n BO exp
  V EB  
p BO
 L   n BO

  1
p B x  
 B 
exp
  Vt  
A
 x B    Vt  

sinh 
x 
2 sinh  B 
L B 

 LB 
x x
 x 
_______________________________________
  sinh 

 sinh  B
 LB 
 L B 
12.14
_______________________________________
In the base of the pnp transistor, we have
d 2 p B x  p B x 
12.15
DB

0
For the idealized straight line approximation,
 BO
dx 2
the total minority carrier concentration is
or
given by
d 2 p B x  p B x 


0
  V   x  x 
dx
L2B

nB x   nBO exp BE    B

  Vt   xB 
where LB  DB BO
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
The excess carrier concentration is
and
n B x   n B x   n BO
x 
sinh  B   1.1752
so for the idealized case, we can write
 LB 
  V BE   x B  x  
Then
  
  1
n BO x   n BO exp
  x
V
x 
x 
  t   B  
n BO  B   n B  B 
 2 
 2 
1
At x  x B , we have
x 
2
n BO  B 
 2 
 1   V  
x 
n BO  B   n BO  exp BE   1
  V BR 
 2 
 2   Vt  
 0.50  0.4434   1.0  0.8868
exp

  Vt 
For the actual case, we have

  V BE  
V 
1
n BO
x 
exp BE   1
  1
n B  B  
exp

2
 x    Vt  
 Vt 
 2 
sinh  B  
V 
 LB 
Again assume that exp BE   1 . Then the
 Vt 
 x 
 x 
 sinh  B   sinh  B 
ratio becomes
 2LB 
 2 L B 
0.0566

 0.1132  11.32%
xB
0.50
 0.10 , we have
(a) For
LB
_______________________________________
 xB 
  0.0500208
sinh 
12.16
 2LB 
(a) p B x B   5 10 3   p B 0
and
 p B 0  510 3 cm 3
 xB 
  0.100167
sinh 
n2
 LB 
p B0  i
NB
Then
 xB 
x 
  n B  B 
 2 
 2 
n BO 
 xB 

 2 
n BO 
  V BE 
0.50  0.49937   1.0  0.99875
exp

  Vt 

V 
1
exp BE   1
2
 Vt 
V 
If we assume that exp BE   1 , then we
 Vt 
find that the ratio is
0.00063

 0.00126  0.126%
0.50
x
(b) For B  1.0 , we have
LB
 x 
sinh  B   0.5211
 2LB 
 NB 

ni2
1.5 1010

p B0
5 10 3

2
 4.5  10 16 cm 3
V 
p B 0  p B 0 exp EB 
 Vt 
 p 0 
 V EB  Vt ln  B 
 p B0 
 1015 

 0.0259 ln 
3 
 5 10 
 0.6740 V
(b) Using the linear approximation,
V 
d p B x  eD B p B 0
J  eD B

exp EB 
dx
xB
 Vt 
 J
Since x B  L B , J
x0
Then
x  xB
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
12.17
1.6  1019 10 5  103
 0.674 
J 
exp

(a) For an npn transistor biased in saturation,
4
0.8  10
 0.0259 
the excess minority carrier electron
 20.0 A/cm 2
concentration in the base is found from
(c) Using Equation (12.15a),
d 2 n B x  n B x 
DB

0
 p B0
d p B 0 
 BO
dx 2

dx
x 
or
L B  sinh  B 
L
d 2 n B x  n B x 
 B

0
dx 2
L2B

 xB  x 
 x 
  V EB  





 exp
  1 cosh L   cosh L 
where LB  DB BO

B


 B 
  Vt  

The
general solution is of the form
Now
 x 
x
eD B p B 0
d p B x 

n B x   A exp   B exp
J  eD B

LB 
L B 
dx
 xB 



L B sinh 
If x B  L B , then also x  L B , so that
 LB 


x 
x 
  V  
 x x
 x 
  B1 

n B x   A1 
  cosh

 exp EB   1 cosh B
 LB 
 LB 
  Vt  
 LB 
 L B 
 x 
For x  0 , sinh 1  1.1752 , cosh1  1.5431

  A  B    A  B 
 LB 
cosh0  1.0
which can be written as
Then
19
3
 x 
1.6 10 10 5 10
n B x   C  D 
J

x 0
10 10  4 1.1752
 LB 
The boundary conditions are
  0.6740  

 exp
  1 1.5431  1.0 
 V  
  0.0259  

n B 0  C  n BO exp BE   1
  Vt  
J
 2.1042 A/cm 2
and
x 0
For x  x B ,
 V  
x 
n B x B   C  D B   n BO exp BC   1
19
3
1.6 10 10 5 10
 LB 
  Vt  
J

4
x  xB
10 10 1.1752
The coefficient D can be written as
  V  
  0.6740  

L 
 exp
  1 1.0  1.5431
D   B n BO exp BC   1
  0.0259  

  Vt  
 xB 
2
J
 1.3636 A/cm
  V   
x  xB
 exp BE   1 
(d) For part (b),
  Vt   
The excess electron concentration is then
J
given by
x  xB
 1 .0
  V   
x 
n B x   n BO exp BE   1  1  
J
x 0
  Vt    x B 
For part (c),
  V    x 

 exp BC   1  
J
Vt    x B 



x  xB
1.3636

 0.648
2.1042
J

 



 


 




x 0
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(b) The electron diffusion current density is
V 
4.86111011  5.4662 1011  exp BC 
d n B x 
J n  eD B
 Vt 
dx
V BC  0.0259 ln 6.0511010
  V BE     1 
 0.6430 V
  1  

 eD B n BO exp


  Vt    x B 
(b) VCE sat   V BE  V BC  0.70  0.6430
 0.057 V
  V    1 

 exp BC   1  
(c) We have
  Vt    x B 
Qn
 V 
n x  V 
 B 0 B exp BE   exp BC 
or
e
2   Vt 
 Vt 
 V BC 
eD B n BO   V BE 
3

4
  exp

Jn  
exp
4.5  10 0.7  10
 V 
xB
  Vt 

 t 
2
(c) The total excess charge in the base region
  0.70 
 0.643 
is
 exp
  exp

xB
0
.
0259

 0.0259 
 
QnB  e n B x dx
 0.1575 5.466 1011  6.052 1010




  V   
x2 

 en BO exp BE   1   x 
2 x B 
  Vt   
xB
  V    x 2 

 exp BC   1  

 0
  Vt    2 x B 

which yields
 en BO x B   V BE  
  1
exp

2
  Vt  
  V   
 exp BC   1 
  Vt   
_______________________________________
12.18
(a) Using the linear approximation, we can
write
 V 
eD B n B 0   V BE 
  exp BC 
Jn 
exp
 V 
x B   Vt 
 t 

n2
1.5 1010
n B0  i 
NB
5 1016
 4.5 10 cm
3
Then
125 


0
Q nB 


2
Qn
 9.56  10 10 cm 2
e
(d) In the collector,
  V     x  

p C x   p C 0 exp BC   exp

  Vt    LC 
Now
Qp 
 p C x dx 
e
0

  V 
   x   

 p C 0 exp BC   LC exp

  Vt 
  LC  0
V 
 p C 0 LC exp BC 
 Vt 
We find
pC 0 

ni2
1.5 1010

NC
1015

2
 2.25  10 5 cm 3
Then
Qp



 0.643 
 2.25 10 5 35 10  4 exp

e
 0.0259 
3
1.6 10 254.5 10 
19

3
0.7 10  4
  0.70 
 V BC 

 exp
  exp

 Vt 
  0.0259 
 4.77 1013 cm 2
_______________________________________
12.19
(b)

n2
1.5 1010
n BO  i 
NB
1017
and
  2.25 10 cm
2
3
3
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
2
12.20
ni2
1.5 1010
4
3
cm
pCO 


3
.
21

10
Low-injection limit is reached when
NC
7 1015
p C 0  0.10N C  0.10 5 1014
At x  x B ,
or
 V BC 
p C 0  5 1013 cm 3

n B x B   n BO exp

We have
 Vt 






 0.565 
 2.25 10 3 exp

 0.0259 
or n B x B   6.7 1012 cm 3
At x   0 ,
V 
p C 0  p CO exp BC 
 Vt 


 0.565 
 3.2110 4 exp

 0.0259 
or p C 0  9.56 1013 cm 3
(c) From the B-C space charge region,
 1017 7  1015 
Vbi1  0.0259  ln 

2
 1.5  1010

 0.745 V
Then
 211.7 8.85 10 14 0.745  0.565
x p1  
1.6 10 19

 





 7  10 15 
1


 

17
15
17 

 10
 7  10  10 
1/ 2
6
or x p1  1.23  10 cm
From the B-E space charge region
 1019 1017 
Vbi 2  0.0259  ln 

2
 1.5 1010 
 0.933 V
Then
 211.7 8.85 10 14 0.933  2
x p2  
1.6 10 19

  




 1019 
1

  17  19

17 
 10  10  10 
1/ 2
or
x p 2  1.94  10 5 cm
Now
x B  x BO  x p1  x p 2
 1.20  0.0123  0.194
or
x B  0.994  m
_______________________________________


2
n2
1.5 1010
p CO  i 
 4.5 10 5 cm 3
14
NC
5 10
Also
V 
p C 0  p CO exp CB 
 Vt 
or
 p 0 
VCB  Vt ln  C 
 p CO 
 5 1013 

 0.0259  ln 
5 
 4.5 10 
or
VCB  0.48 V
_______________________________________
12.21
(a)
(i)  
1
(ii)  T 
(iii)  
1

I pE
I nE
1
 0.99305
0.0035
1
0.50
I nC 0.495

 0.990
I nE
0.50
I nE  I pE
I nE  I R  I pE
0.50  0.0035
 0.990167
0.50  0.005  0.0035
(iv)    T   0.993050.9900.990167 
 0.97345

0.97345

 36.7
(v)  
1   1  0.97345

120

(b) For   120   
1   121
  0.991736
Then    T    0.997238

 T  0.997238 
I nC
I
 nC
I nE 0.50
 I nC  0.4986 mA
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
1
1
1
  0.997238 

(b)  
I pE
I pE
N D x
1 B  E  B
1
1
N E DB x E
I nE
0.50
 I pE  0.00138 mA  1.38  A

I nE  I pE
I nE  I R  I pE
0.997238 
0.50  0.00138
0.50  I R  0.00138
 I R  0.00139 mA  1.39  A
_______________________________________
12.22
(a) Using Equation (12.37)
eD B p B 0 ABE
I nC 
LB
   V EB  

  1
 exp


   Vt  

1



 xB  
 sinh  x B 


tanh
L 
 L 

 B
 B 

Now

n2
1.5 1010
p B0  i 
NB
1016
12.23
(a) We have

   V BE  

  1
 exp



eD B n BO    Vt  
1
J nE 



LB 
 xB 
 xB  




tanh
sinh


 LB 
 LB  

2
105 10 7 
 2.236 10 3 cm
We find
 0.70 10 4 
x 

sinh  B   sinh 
3 
 LB 
 2.236 10 
 0.03131
 0.70 10 4 
 xB 

  tanh 
tanh 
 2.236 10 3 
 LB 


 0.03130
Then
1.6 10 19 10 2.25 10 4 5 10 4
I nC 
2.236 10 3
   0.550  

  1
 exp

1 
   0.0259  



0.03131
0.03130 






 

I nC  4.29 10 4 A  0.429 mA
16
(c) I C  I E  0.95442125  119.3  A
_______________________________________
 2.25  10 4 cm 3
L B  D B B 0 
1
 0.95969
 10
 15  0.7 


1 
 

17 
 5  10  10  0.5 
1
1
T 

x 
 0.70  10  4 

cosh B  cosh
3 
 LB 
 2.236  10 
 0.99951
   T   0.959690.999510.995
 0.95442

0.95442


 20.94
1   1  0.95442
Then
I C  I B  20.940.80  16.75  A

We find that
n BO 

ni2
1.5 1010

NB
5 1016
and
L B  D B BO 
  4.5 10 cm
2
3
3
155 10 8 
 8.660 10 4 cm
Then
J nE 

1.6 10 154.5 10 
19
3
4
8.66 10


 0.60 

 exp

1


 0.0259 



0
.
70
0
.
70



 tanh 
 sinh 


 8.66 
 8.66  
or
J nE  1.779 A/cm 2
We also have
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(b) Using the calculated current densities, we
eD E p EO   V BE  
1
  1 
J pE 
find
exp

LE
  Vt   tanh  x E 
J nE
1.779


L 
 E
J nE  J pE 1.779  0.04251
Now
or

ni2
1.5 1010

NE
1018
p EO 
and
L E  D E EO 
  2.25 10 cm
2
3
810 8 
 2.828 10 4 cm
Then
J pE 
1.6 10 82.25 10 
19
2
2.828 10  4
  0.60  
 exp
  1 
  0.0259  
1
 0.8 
tanh 

 2.828 
or
J pE  0.04251 A/cm 2
We can find
eD B n BO
J nC 
LB
   V BE  

  1
 exp


   Vt  

1



 xB  
 sinh  x B 


tanh 
L 


 B
 LB  


1.6 10 154.5 10 
19
  0.9767
2
We also find
J
1.773
 T  nC 
J nE 1.779
or
 T  0.9966
Also
J nE  J pE

J nE  J R  J pE

1.779  0.04251
1.779  0.003218  0.04251
or
  0.9982
Then
   T   0.9767 0.99660.9982
or
  0.9716
Now

0.9716


1   1  0.9716
or
  34.2
_______________________________________
3
8.66 10
4
   0.60  

  1
 exp

1
   0.0259  




 0.7  
 sinh  0.7 
tanh 


 8.66 
 8.66  

or
J nC  1.773 A/cm
The recombination current density is
V 
J R  J ro exp BE 
 2Vt 
 0.60 
 3 10 8 exp 

 20.0259
or
J R  3.218 10 3 A/cm 2
12.24
(a) We have

N D x
1
 1 B  E  B
N B DE x B
N E DB x E
1


N E DB x E
or
  1 K 
2


NB
NE
(i) Now
2 N BO
K
NE
 B 

N
  A
1  BO  K
NE
1
 2 N BO
 N

 1 
 K 1  BO  K 
NE
NE



2 N BO
N BO
 1
K 
K
Ne
NE
or finally
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(c) Neglect any change in space charge width.
N
D x
 B 
 1  BO  E  B
1
  A
N E DB x E

  V BE 
J
(ii)

1  rO  exp

We have
J sO
 2Vt 
N B DE x B
xB
  1


 1 K 
 V 
J
K
N E DB x E
xE
 1  rO  exp BE   1 
J sO
J sO
 2Vt 
Then
(i)
x
1  K   BO
K
1
2x E
 C 


J sOB
 B 
K 
K 
1 

x BO

 1 
  A



1 K 
K
  A
J sOB  J sOA 

1

xE
J sOA

x BO 
x BO 
K
K
1  K  

 1  K  
 1

2 x E 
x E 

J sOB J sOA
x BO
x BO
Now
 1 K 
 K 
2xE
xE
n2
J sO  n BO  i
x
NB
 1  K   BO
2xE
so
2 N BO K N BO K
N K
or
 B 
 1

 1  BO
N
D x
 C 
  A
C
C
C
 1  BO  E  BO
  A
N E DB 2x E
Then finally
(b) (i) We find
  V BE 

J rO exp

 T B 
2
V



B
t


1

1

 T  A
  A
 eD B n BO 


(ii)
x
B


2
 1  x

BO 2   
1  
(ii) We find

 

  V BE 
 T C   2  L B  

J rO exp

2
2Vt 
 T  A
 C 
 1 x  

 1
1   BO 
  A
 eD B n BO 
 2  L B  




 xB 
2
 1 x  
BO
(d) Device C has the largest  . The emitter
1 

 8  L B  
injection efficiency, base transport, and



recombination
factors all increase.
 1  x 2 
_______________________________________
BO
1 

 2  L B  


12.25
 1  x  2  1  x  2 
 1   BO  1   BO  
 8  L B   2  L B  



2
1 x 
1 x 
 1   BO    BO 
8  LB 
2  LB 
or finally
 T C 
3 x 
 1   BO 
 T  A
8  LB 
(a) We have

2
1
1

N B DE x B
N
1


1 K  B
N E DB x E
NE
or
NB
NE
Then
  1 K 
2
(i)
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(c) Neglect any change in space charge width.
NB
1 K 
1
2 N EO
 B 


  V BE 
J
N
  A

1  rO exp
1 K  B

J sO
N EO
 2Vt 
1
K

N B 
N 

 1
1  K  B 
 1  K 



K
J
sO
2 N EO 
N EO 
1

J sO
NB
N
 1 K 
K B
(i)
2 N EO
N EO
K
1
NB
J sOB 
 B 
K 
K 
 1 K 
1 


 1 
2 N EO
 J

K
  A
J
sOB 
sOA 

1
or
J sOA
N B DE x B
 B 
 1


K
K
 1

  A
2 N EO D B x E
J sOB J sOA
(ii) Now
Now
x
1

 1 K  B
1
J sO 
x
xE
1 K  B
N E xE
xE
so
Then
 B 
 1  K 2 N EO   K N EO 
xB
1 K 
  A
x EO 2
 C 
or

xB
  A
 B 
1 K 
 1  K   N EO
x EO
  A

Recombination factor decreases
2 x 
x 
 1  K   B 1  K   B 
(ii) We have
x EO 
x EO 

 C 
x 
x
x
 1  K  EO   K x EO 
 1  2K   B  K   B
  A
 2 
x EO
x EO
or
x
 C 
1
 1 K  B
 1  K   x EO
x EO
  A
2
or finally
Recombination factor increases
N D
x
 C 
_______________________________________
 1 B  E  B
  A
N E D B x EO
12.26
(b) We have
(b)
2
1  xB 
2

 T  1  
n2
1.5 1010
2  L B 
n BO  i 
 2.25 10 3 cm 3
NB
1017
(i)
Then
 T B 
1
V 
 T  A
n B 0  n BO exp BC 
 Vt 
(ii)
 T C 
 0.6 
 2.25 10 3 exp
1

 T  A
 0.0259 




 2.59 1013 cm 3
Now
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(b) For D E  D B , L E  L B , x E  x B , we
eD B n B 0 
J nC 
have
xB
1
1
19
13
1.6 10 20 2.59 10






1

p
n
1

N
EO
BO
B NE 
10  4

or
and  
2
1 
J nC  0.828 A/cm

 

Assuming a long collector
V 
eD p
J pC  C nO exp BC 
LC
 Vt 
where
p cO 

ni2
1.5 1010

NC
1016
and
LC  DC  CO 
  2.25 10 cm
2
4
3
J pC 
152 10 7 
1.6 10 152.25 10 
19
4
1.732 10 3
 0.6 
 exp

 0.0259 
or
J pC  0.359 A/cm 2
The collector current is
I C  J nC  J pC  A



 0.828  0.359 10 3


0.01
0.10
1.0
10.0
0.990
0.909
0.50
0.0909
99
9.99
1.0
0.10
(c) For x B L B  0.10 , the value of  is
unreasonably large, which means that the
base transport factor is not the limiting factor.
For x B L B  1.0 , the value of  is very
small, which means that the base transport
factor will probably be the limiting factor.
 1.732 10 3 cm
Then
NB NE

If N B N E  0.01 , the emitter injection
efficiency is probably not the limiting factor.
If, however, N B N E  0.01 , then the current
gain is small and the emitter injection
efficiency is probably the limiting factor.
_______________________________________
12.28
We have
or
I C  1.19 mA
The emitter current is
I E  J nC  A  0.828 10 3
or
I E  0.828 mA
_______________________________________


12.27
(a)
0.01
0.10
1.0
10.0
Now
n BO 
eD B n BO
L B tanh x B L B 

ni2
1.5 1010

NB
1017
and
L B  D B BO 
  2.25 10 cm
2
3
2510 7 
 15.8 10 4 cm
T
1
T 
and  
coshx B L B 
1T
x B LB
J sO 
T
0.99995
0.995
0.648
0.0000908

19,999
199
1.84
0
Then
J sO 
1.6 10 252.25 10 
15.8 10  tanh0.7 15.8
19
4
or
J sO  1.287 10 10 A/cm 2
Now
3
3
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
1
1

0.99656 
16
  V BE 
 2  10  8  0.80 
J

 
1  rO exp
1  



J sO
 2Vt 
 N E  23  0.35 
1
 N E  4.611018 cm 3

  V BE 
_______________________________________
2 10 9
1
exp 



2
0
.
0259
1.287 10 10


12.30
or
(a) We have J rO  5 10 8 A/cm 2
(a)
We find
1

2
n2
1.5 1010
  V BE 
n BO  i 
 4.5 10 3 cm 3
1  15.54  exp

16
N
0
.
0518
5

10


B
and
and
(b)
L B  D B BO  25 10 7


 15.8 10 4 cm
1 
Then
Now
eD B n BO
V BE


J sO 
L B tanh x B L B 
0.20
0.7535
3.06
1.6 10 19 25 4.5 10 3
0.40
0.99316
145

0.60
0.999855
6,902
15.8 10  4 tanh x B L B 
or
(c) If V BE  0.4 V, the recombination factor is
1.139 10 11
J sO 
likely the limiting factor in the current gain.
tanh x B L B 
_______________________________________
We have
1
12.29

  V BE 

150
J



 0.993377
1  rO exp

1   151
J sO
 2Vt 
   T 
For T  300 K and V  0.55 V.





  0.995
  T  0.995867
L B  D B B 0 
232 10 7 
 2.145 10 3 cm
Then
1
1

x 
 0.80  10  4 

cosh B  cosh
3 
 LB 
 2.145  10 
 0.99930
T 
Now



0.993377

 0.99656
 T  0.99930 0.9975
1
N B DE x B
1


N E DB x E

BE
0.993377   T 0.9975
Let x B  0.80  m
 


1
 5  10 8 
x 
 0.55 
  tanh  B   exp
1  

11 
L 
0
1
.
139

10
 .0518 
 B


which yields
xB
 0.0468
LB
or
x B  0.046815.8  0.739  m

(b) For T  400 K and J rO  5 10 8 A/cm 2 ,
 Eg


exp 

n BO 400  400 
 0.0259400 300 

 
n BO 300  300 
  Eg 
exp 

 0.0259 
For E g  1.12 eV,
3
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(b)
n BO 400 
 1.175  10 5
V  V A 2  160
n BO 300 

 648 k 
(i) ro  CE
IC
0.25
or


n BO 400  1.175 10 5 4.5 10 3
 5.29  10 cm
8
Then
J sO 

3
1.6 10 255.29 10 
15.8 10  tanh0.739 15.8
19
8
4
or
J sO  2.865 10 5 A/cm 2
Finally
1

8


5 10
 0.55
1
 exp 

5
2.865 10
 20.0259400 300 
or
  0.9999994
_______________________________________
12.31
Plot
_______________________________________
12.32
Plot
_______________________________________
12.33
Plot
_______________________________________
12.34
Plot
_______________________________________
12.35
(a) I C 
V  V 
1
VCE  V A   ro  CE A
ro
IC
2  120
 101.67 k 
1 .2
1
1

 0.00984 (k  ) 1
(ii) g o 
ro 101.67
(i) ro 
 9.84 10 6 
1
4  120
 1.22 mA
(iii) I C 
101.667
(ii) g o 
1
1

 0.00154 (k  ) 1
ro 648
 1.54 10 6 
1
4  160
 0.253 mA
648
_______________________________________
(iii) I C 
12.36
ro 
V EC
V EC 5  2
 I C 

I C
ro
180
I C  0.01667 mA  16.67  A
_______________________________________
12.37
 2  V  VCB   N C
 
1
x dB   s bi



e

 N B N B  N C   

1/ 2

 211.7  8.85  10 14 Vbi  VCB 


1.6  10 19
 2  1015
 
1


16
15
16  
2  10  2  10  
 2  10

 5.8832 10 V  V 
1/ 2

1/ 2
11
bi
CB
Now
N N 
Vbi  Vt ln  B 2 C 
 ni 



 2 1015 2 1016 
 0.0259  ln 

2
 1.5  1010

 0.6709 V
(i) For VCB  4 V, x dB  0.1658  m


(ii) For VCB  8 V, x dB  0.2259  m
(iii) For VCB  12 V, x dB  0.2730  m
Neglecting the B-E space charge width,
(i) For VCB  4 V,
x B  0.85  0.1658  0.6842  m
(ii) For VCB  8 V,
x B  0.85  0.2259  0.6241  m
(iii) For VCB  12 V,
x B  0.85  0.2730  0.5770  m
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Now
x B  x BO  x p
 V BE 
eD B n B 0
Now

JC 
exp

xB
V
 3  1016 5  10 15 
 t 
Vbi  0.0259  ln 

2
where
 1.5  1010

2
ni2
1.5 1010
or
n B0 

NB
2 1016
Vbi  0.705 V
4
3
Also
 1.125 10 cm
1/ 2
so
 2 s Vbi  VCB   N C 

1
19
4




xp  
1.6 10 25 1.125 10
 0.650 
 N  N  N 
e
JC 
exp


C 
 B  B

xB
 0.0259 
14

211.7  8.85  10 Vbi  VCB 
3.5686 10 3

A/cm 2

1.6  10 19

xB





 
(ii)For VCB  8 V, J C  57.18 A/cm 2
(iii)For VCB  12 V, J C  61.85 A/cm 2
or
  7.5 10 cm
2
3
3
and
V 
n B 0  n BO exp BE 
 Vt 

 0.7 
 7.5 10 3 exp

 0.0259 
or
n B 0  4.10 1015 cm 3
We have
dn B eD B n B 0 
J  eD B

dx
xB

1.6 10 204.10 10 
19
15
xB
or
1.312 10 2
A/cm 2
xB
Neglecting the space charge width at the B-E
junction, we have
J

1/ 2
For VCB  10 V, x p  0.2569  m
 V A  38.4 V
_______________________________________


For VCB  5 V, x p  0.1875  m
61.85  52.16
52.16

12  4
4  0.650  V A


x p  6.163 10 11 Vbi  VCB 
J C
JC

(b)
VCE VCE  V A
ni2
1.5 1010

NB
3 1016

 5  1015 
1


 

16 
15
16 
 3  10  5  10  3  10 
(i)For VCB  4 V, J C  52.16 A/cm 2
n BO 




12.38
We find

(a) For x BO  1.0  m
For VCB  5 V,
x B  1.0  0.1875  0.8125  m
Then
1.312 10 2
J
 161.5 A/cm 2
0.8125 10  4
For VCB  10 V,
x B  1.0  0.2569  0.7431  m
and
1.312 10 2
J
 176.6 A/cm 2
4
0.743110
We can write
J
VCE  V A 
J
VCE
where
J
J
176.6  161.5


VCE
VCB
10  5
 3.02 A/cm 2 /V
Then
161.5  3.025.7  V A 
which yields
V A  47.8 V
1/ 2
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
12.39
(b) For x BO  0.80  m
(a)
For V  5 V,
CB
x B  0.80  0.1875  0.6125  m
Then
1.312 10 2
J
 214.2 A/cm 2
0.6125 10  4
For VCB  10 V,
x B  0.80  0.2569  0.5431  m
and
1.312 10 2
J
 241.6 A/cm 2
0.543110  4
Now
J
J
241.6  214.2


VCE
VCB
10  5
 5.48 A/cm 2 /V
We can write
J
VCE  V A 
J
VCE
or
214.2  5.485.7  V A 
which yields
V A  33.4 V
(c) For x BO  0.60  m
For VCB  5 V,
x B  0.60  0.1875  0.4125  m
Then
1.312 10 2
J
 318.1 A/cm 2
0.4125 10  4
For VCB  10 V,
x B  0.60  0.2569  0.3431  m
and
1.312 10 2
J
 382.4 A/cm 2
4
0.343110
Now
J
J
382.4  318.1


VCE VCB
10  5
 12.86 A/cm 2 /V
We can write
J
VCE  V A 
J
VCE
or
318.1  12.865.7  V A 
which yields
V A  19.0 V
_______________________________________
 2  V  V BC   N C
 
1
x dB   s bi



e

 N B N B  N C   

1/ 2

 211.7  8.85  10 Vbi  V BC 


1.6  10 19
14
1015
 
1
  16  15
16  
10  10  
10



1/ 2


 1.1766 10 10 Vbi  VBC 
Now
N N 
Vbi  Vt ln  B 2 C 
 ni 
1/ 2
  


 1015 1016 
 0.0259  ln 

10 2
 1.5  10

 0.6350 V
For V BC  1 V, x dB  0.1387  m
For V BC  5 V, x dB  0.2575  m
Then x dB  0.2575  0.1387  0.1188  m
V 
eD B p B 0 ABE
exp EB 
xB
 Vt 
(b) I C 
We find

n2
1.5 1010
p B0  i 
NB
1016

2
 2.25  10 4 cm 3
Then
IC 
1.6 10 102.25 10 10 
19
4
4
xB
 0.625 
 exp

 0.0259 

1.0874 10 7
A
xB
For V BC  1 V, I C 
1.0874 10 7
0.70  0.138710 4
 1.937 10 3 A  1.937 mA
1.0874 10 7
For V BC  5 V, I C 
0.70  0.257510 4
 2.456 10 3 A  2.456 mA
Then
I C  2.456  1.937  0.519 mA
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
12.41
I C
IC

(c)
(a) We have
V BC V EC  V A
1

0.519 10 3
1.937 10 3
p
D
L
tanh x B L B 

1  EO E B 
5 1
1  0.625  V A
n BO D B L E tanh x E L E 
V A  13.3 V
For x B  x E , L B  L E , D B  D E , we obtain
V EC  V A 1.625  13.3
1

(d) ro 

IC
1.937  10 3
 E g 
N

1  B exp
 7.705 10 3   7.705 k 

NE
 kT 
_______________________________________
For N E  1019 cm 3 , we have E g  80 meV
12.40
Then
Let x B  x E , L B  L E , D B  D E
1
0.996 
Then the emitter injection efficiency is
NB
 0.080 
1  19
exp

1
1
0.0259 
10



p
n2 N
which yields
1  EO 1  iE  B
n BO
N E niB2
N  1.83 1015 cm 3
B
where niB2  ni2 .
For no bandgap narrowing, niE2  ni2 .
With bandgap narrowing,
 E g 
2

niE
 ni2 exp

 kT 
Then
1

 E g 
N

1  B exp

NE
 kT 
(a) No bandgap narrowing, so E g  0
   T    0.9952 . We find
NE


10 17

0.5
0.495
0.980
18
0.909
0.8999
8.99
10 19
0.990
0.980
49.3
10 20
0.9990
0.989
90.2
10
(b) Taking into account bandgap narrowing,
we find
NE
E g (meV)

10 17
0


0.5
0.495
0.98
3.63
18
25
0.792
0.784
10 19
80
0.820
0.812
4.32
20
230
0.122
0.121
0.14
10
10
_______________________________________
(b) Neglecting bandgap narrowing, we would
have
1
1

 0.996 
NB
NB
1
1  19
NE
10
which yields
N B  4.02 1016 cm 3
_______________________________________
12.42
(a)
 Length   S 2
Area
 x B L 
S 2

e p N B x B L
(i) R 

1
1.6 10 2502 10 
19

16
5  10 4
0.65  10 25  10 
4
4
R  3.846 10 3   3.846 k 


(ii) V  I B 2R  5 10 6 3.846 10 3
 0.01923 V

Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Now
V x  S 2 
exp  BE

S 2
1
n B x  S 2
 0.0259 
R


(iii)
e p N B x B L
n B x  0
V x  0 
exp  BE

1
 0.0259 
545.8 
19

1
.
6

10
250  2  1016
 0.60 
exp 

S 2
 0.0259 


0.65  10  4 25  10  4
 0.60  V 
exp 

 S  1.42 10 4 cm  1.42  m
 0.0259 
_______________________________________
  V 
 exp 

 0.0259 
12.44
(a)
  0.01923 
 exp 

  ax 
 0.0259 

N B  N B 0 exp
Then
 xB 
n B x  S 2
where
 0.476
n B x  0
 N 0 
a  ln  B
0
(b)
 N B x B  
1
and is a constant. In thermal equilibrium
(i) R 
19
1.6  10 250  2  1016
dN B
J p  e p N B   eD p
0
1.5 10 4
dx

so that
0.65 10  4 25 10  4
3
D p 1 dN B  kT  1 dN B
 1.154 10   1.154 k 







N B dx
e  N B dx
6
3

p
(ii) V  I B 2R  5 10 1.154 10
which becomes
 0.005769 V
 a
  ax 
 kT  1
  V 
n B x  S 2
  exp

   
 N B 0 
(iii)
 exp 

e
N
x
x


B
B
B




n B x  0
 Vt 
 kT    a  1
  0.005769 
 
    
 NB
 exp 

 e   xB  N B
 0.0259 
or
Then
 a  kT 
n B x  S 2 

  
 0.80

n B x  0
 x B  e 
_______________________________________
which is a constant.














12.43
  V 
n B x  S 2
 0.90  exp 

n B x  0
 Vt 
Then


(b) The electric field is in the negative
x-direction which will aid the flow of
minority carrier electrons across the base.
(c)
 1 
 1 
V  Vt ln 
  0.0259  ln 

 0.90 
 0.90 
V  0.002729 V  I B 2R


 5 10 6 R
R  545.8 

dn
dx
Assuming no recombination in the base, J n
will be a constant across the base. Then
J
dn   n 
dn   
  n  n 
 
 n 

dx  Dn 
eDn dx
 Vt 
J n  e n n  eD n
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
12.46
 kT 
where Vt  

We want BVCEO  60 V
 e 
Then
The homogeneous solution to the differential
BVCBO
BVCBO
equation is found from
BVCEO 
 60 
3
n 
dn H
50
 An H  0
dx
which yields
BVCBO  221 V

where A 
For this breakdown voltage, we need
Vt
N C  1.5 1015 cm 3
The solution is of the form
n H  n H 0 exp Ax
The depletion width into the collector at this
voltage is
The particular solution is found from
xC  x n
nP  A  B
Jn
eD n
The particular solution is then
 Jn 




J
B  eD n  J nVt
nP  

 n
A
eD n  e n 

 
V 
 t
The total solution is then
Jn
n
 n H 0  exp  Ax
e n 
and
V 
V 
ni2
n0  n pO exp BE  
exp BE 


V
N
0
B
 t 
 Vt 
Then
V 
ni2
J
n H 0 
exp BE   n
N B 0
 Vt  e n 
_______________________________________
where B 
12.45
(a) For N C  21015 cm 3 ,
BVBC 0  180 V

0.9930

 141.86
1   1  0.9930
BVBC 0
180
BVEC 0 

 34.5 V
3
n

141.86
 2  V  V BC   N B 

1



  s bi




e

 N C  N B  N C 
We find
 1.5  1015 1016 
Vbi  0.0259  ln 
  0.646 V
2
 1.5  1010

and
V BC  BVCEO  60 V
so that
 211.7  8.85 10 14 0.646  60
xC  
1.6 10 19

1/ 2

(c) For N B  510 cm 3 ,
BVEB  19 V
_______________________________________



 1016 
1

 16
 

15 
15 
 1.5  10  10  1.5  10 
1/ 2
or
x C  6.75 10 4 cm  6.75  m
_______________________________________
12.47
(a) For N C  81015 cm 3 ,
BVCB 0  64 V
(b) V pt 
(b)  
16
 


ex
N N  N B 
 B C
2 s
NC
2
B0
1.6 10 0.50 10 
211.78.85 10 
5 10 8 10  5 10 

4 2
19
14
16
15
16
8 1015
V pt  70.0 V
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
12.48
(a) For VCE sat   0.30 V, we find
ex B2 0 N B N C  N B 
 0.30 
5

(a) V pt 
exp 
  1.0726 10
2 s
NC
 0.0259 
2
1.6 10 19 0.65 10 4
 0.8  I B 

4.95
 
14
211.7 8.85 10
 0.99 I B  0.01 
We find
2 1016 5 1015  2 1016

I B  0.01014 mA  10.14  A
5 15








(b) For VCE sat   0.20 V, we find
V pt  32.6 V
(b) From Chapter 7,
I B  0.0119 mA  11.9  A
 2eV pt  N B N C 


 max  
 s  N B  N C 


 2 1.6 10 19 32.6

14
 11.7  8.85 10


(c) For VCE sat   0.10 V, we find
1/ 2
I B  0.105 mA  105  A
_______________________________________



 5  1015 2  1016  


 5 1015  2  1016  
1/ 2
 max  2.01 10 5 V/cm
_______________________________________
12.49
V pt 
ex B2 0 N B N C  N B 

2 s
NC
1.6 10 x 
211.7 8.85 10 
5 10 3 10  5 10 

19
15 
2
B0
14
16
15
16
3 1015
 x B 0  1.483 10 5 cm  0.1483  m
_______________________________________
12.50
We have
 I 1   R   I B  F 
VCE sat   Vt  ln  C


 F I B  I C 1   F   R 
We can write
11  0.2  I B  0.99 
V sat  
exp  CE




 0.0259  0.99I B  11  0.99  0.20 
or
V sat   0.8  I B 
4.95
exp  CE
  
 0.0259   0.99 I B  0.01 
12.51
For an npn transistor biased in the active
mode, we have V BC  0 , so that
V 
exp BC   0 . Now
 Vt 
I E  I B  I C  0  I B  I C  I E 
Then we have


 V  


I B   F I ES exp BE   1  I CS 
V


  t  



  V   
   R I CS  I ES exp BE   1 

  Vt   
or
 V  
I B  1   F I ES exp BE   1
  Vt  
 1   R I CS
_______________________________________
12.52
We can write
 V  
I ES exp BE   1
  Vt  
 V  
  R I CS exp BC   1  I E
  Vt  
Substituting, we find


 V  


I C   F  R I CS exp BC   1  I E 
V




  t  


 V  
 I CS exp BC   1
  Vt  
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
From the definition of currents, we have
For VCB  V BC  0.5 V
I E   I C for the case of I B  0 . Then
I C  2.5277 10 7  2.4214 10 5
 V  
 2.396 10 5 A  24  A
I C   F  R I CS exp BC   1   F I C
  Vt  
For VCB  V BC  0.25 V
 V  
I C  2.5277 10 7  1.556110 9
 I CS exp BC   1
  Vt  
 2.5110 7 A  0.251  A
When a C-E voltage is applied, then the B-C
For VCB  V BC  0 V
 V BC 
I C  2.5277 10 7 A  0.2528  A
0.
becomes reverse biased, so exp

V
 t 
(c) For V BE  0.6 V,
Then
I C  5.7063 10 4
I C   F  R I CS   F I C  I CS
  V
 
Finally, we find
 10 13 exp BC   1
I 1   F  R 
  0.0259  
I C  I CEO  CS
1 F
For VCB  V BC  0.5 V
_______________________________________
I  5.7063 10 4  2.4214 10 5


C
12.53
 V  
(a) I C   F I ES exp BE   1
  Vt  
 V  
 I CS exp BC   1
  Vt  
For V BE  0.2 V,


  0.20  
I C  0.992 5 10 14 exp
  1
  0.0259  
  V
 
 10 13 exp BC   1
  0.0259  


 1.1197 10 10
  V
 
 10 13 exp BC   1
  0.0259  
For VCB  V BC  0.5 V


I C  1.1197 10 10  2.4214 10 5
 2.42110 5 A  24.21  A
For VCB  V BC  0.25 V
I C  1.1197 10 10  1.556110 9
 1.44 10 9 A
For VCB  V BC  0 V
I C  2.5277 10 7

 10
I C  5.7063 10 4 A  0.5706 mA
_______________________________________
12.54
 I 1   R   I B  F 
VCE sat   Vt ln  C


 F I B  1   F I C  R 

51  0.15  I B
 0.975 
 0.0259 ln 


 0.975I B  1  0.9755  0.150 


4.25  I B
6.5
 0.0259 ln 
 0.975I B  0.125

I B  0.15 A, VCE sat   0.187 V
I B  0.25 A, VCE sat   0.143 V
I B  0.50 A, VCE sat   0.115 V
I B  1.0 A, VCE sat   0.0956 V
_______________________________________
12.55
(a) (i) re 
Vt
0.0259

 0.1036 k 
IE
0.25
 e  reC je  103.6 0.35  10 12 
I C  1.1197 10 10 A
(b) For V BE  0.4 V,
13
 5.464 10 4 A  0.5464 mA
For VCB  V BC  0.25 V
  V BC  
  1
exp
  0.0259  

 3.626 10 11 s  36.26 ps
(ii)  b 

x B2
0.65 10 4

2 Dn
225

2
 8.45 10 11 s  84.5 ps
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
12.57
x
2.2 10 4
(iii)  d  dc 
We have
7
s
10
 ec   e   b   d   c
 2.2 10 11 s  22 ps
We are given
(iv)  c  rc C   C s
 b  100 ps and  e  25 ps
12
We find
 180.020  0.02010
13
x
1.2 10 4
 7.2 10 s  0.72 ps
d  d 
 1.2 10 11 s
s
10 7
(b)  ec   e   b   d   c
or
 36.26  84.5  22  0.72  143.48 ps
 d  12 ps
1
1

(c) f T 
12
Also
2  ec 2 143.48  10
 c  rc C c  10 0.110 12  10 12 s
 1.109 10 9 Hz  1.109 GHz
or
f T 1.109  10 9
 c  1 ps

(d) f  

125
Then
 8.87 10 6 Hz  8.87 MHz
 ec  25  100  12  1  138 ps
_______________________________________
We obtain
1
1
fT 

12.56
2


2

138
 10 12
2
ec
x B2
0.5 10 4
b 

 6.25 10 11 s
 1.15 10 9 Hz
2DB
220
or
We have
f T  1.15 GHz
 b  0.2 ec
_______________________________________
so that
 ec  3.125 10 10 s
Then
1
1
fT 

2  ec 2 3.125  10 10
or
f T  5.09 10 8 Hz  509 MHz
_______________________________________












Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 13
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Chapter 13
13.1
Sketch
_______________________________________
(c) VDS sat   V pO  Vbi  VGS 
(i) V DS sat   3.312  1.328  0
 1.984 V
(ii) V DS sat   3.312  1.328   1.0
 0.984 V
_______________________________________
13.2
Sketch
_______________________________________
13.3
(a) V pO 
(i) V pO 
ea 2 N d
2 s
13.4
1.6 10 0.40 10  3 10 
213.18.85 10 
4 2
19
16
14
 3.312 V


(i) V pO 






1/ 2

 213.1 8.85  10 14 1.328  0.5   0.5 


1.6  10 19 3  10 16







h2  4.57 10 5 cm  0.457  m
h2  a  a  h2  0




1/ 2


1/ 2

h2  2.42 10 cm  0.242  m
a  h2  0.40  0.242  0.158  m
(ii) h 2
5
1/ 2


 213.1 8.85  10 14 1.328  2.5   0.5 


1.6  10 19 3  10 16




 211.7  8.85  10 14 0.860  0   0.5 


1.6  10 19 3  10 16



 211.7  8.85  10 14 0.860  0.5   0.5 


1.6  10 19 3  10 16


h2  3.35 10 5 cm  0.335  m
a  h2  0.40  0.335  0.065  m
(iii) h 2


 2  V  V DS  VGS  
(b) h2   s bi

eN d


(i) h 2


h2  2.97 10 5 cm  0.297  m
a  h2  0.40  0.297  0.103  m
(ii) h 2

16
14
 2.849 V
1/ 2
 213.1 8.85  10 14 1.328  0   0.5 


1.6  10 19 3  10 16


4 2
19
 3  1016 2  1018 
(ii) Vbi  0.0259  ln 

2
 1.5  1010

 0.860 V
V p  Vbi  V pO  0.860  3.709
 1.984 V
 2  V  V DS  VGS  
(b) h2   s bi

eN d


(i) h 2
1.6 10 0.40 10  3 10 
211.78.85 10 
 3.709 V
 3  1016 2  1018 
(ii) Vbi  0.0259  ln 

2


1.8  10 6
 1.328 V
V p  Vbi  V pO  1.328  3.312

ea 2 N d
2 s
(a) V pO 
1/ 2


1/ 2

h2  2.83 10 5 cm  0.283  m
a  h2  0.40  0.283  0.117  m
(iii) h 2


 211.7  8.85  10 14 0.860  2.5   0.5 


1.6  10 19 3  10 16



5


h2  4.08 10 cm  0.408  m
h2  a  a  h2  0
1/ 2
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 13
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(c) VDS sat   V pO  Vbi  VGS 
or
(i) V DS sat   3.705  0.860  0
 2.845 V
(ii) V DS sat   3.705  0.860   1.0
 1.845 V
_______________________________________
(a) V pO 
Na 
ea N a
 Na 
2 s

19
2 s V pO
4 2
 

14
GS
19
9
GS
 VGS  0.8178 V
 2  V  V SD  VGS  
(d) h2   s bi

eN a


4 2
14
0.50 10 
 213.18.85  10 1.28  0  V  

1.6 10 9.433 10  

2.5 10  1.5363 10 1.28  V 
SD
19
14
15
9
15
4 2
GS
19
1/ 2
0.65 10 
 211.7 8.85  10 0.8095  V  

1.6 10 8.425 10  

0.65 10   1.536 10 0.8095  V 
1/ 2
4 2
9
15
9
 1.47 V
(c) a  h2  0.15  0.65  h2
h2  0.50  m
 2  V  V SD  VGS  
h2   s bi

eN a


1/ 2
0.50 10 
 211.7 8.85  10 0.8095  0  V  


1.6 10 8.425 10 


2.5 10  1.536 10 0.8095  V 
 9.433 1015 cm 3
 9.433  1015 1018 

(b) Vbi  0.0259  ln 
2


1.8  10 6
 1.280 V
V p  V pO  Vbi  2.75 1.280


 2  V  V SD  VGS  
h2   s bi

eN a


14
4 2

 

 h2  0.50  m
ea 2
2.75
0.65 10 
213.1 8.85 10
1.6 10

 8.425  1015 1018 
(b) Vbi  0.0259  ln 

2


1.5  10 10
 0.8095 V
V p  V pO  Vbi  2.75  0.8095
 1.9405 V
(c) a  h2  0.15  0.65  h2
13.5
2
N a  8.425 1015 cm 3
9
SD
 V SD  1.94 V
_______________________________________
GS
 VGS  0.347 V
 2  V  V SD  VGS  
(d) h2   s bi

eN a


0.65 10 
13.7
1/ 2
N N 
(a) Vbi  Vt ln  a 2 d 
 ni 
4 2


 213.1 8.85  10 1.28  V SD  


1.6  10 19 9.433  10 15



14


0.65 10   1.5363 10 1.28  V 
4 2
9
SD
 V SD  1.47 V
_______________________________________
13.6
(a) N a 

2 s V pO
ea 2
211.7 8.85 10 14 2.75


1.6 10 0.65 10 
19
4 2



 2  1016 3  1018 
 0.0259  ln 

2
 1.5  1010

 0.860 V
V p  V pO  Vbi


3.0  V pO  0.860  V pO  3.86 V
 2 s V pO 
Now a  

 eN a 

1/ 2

 211.7  8.85  10 14 3.86  


19
2  10 16
 1.6  10


5


 5.0  10 cm  0.50  m
1/ 2
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 13
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(b) V pO  3.86 V
13.10
(c) VSD sat   V pO  Vbi  VGS 
(i) VSD sat   3.86  0.86  3.0 V
(ii) V SD sat   3.86  0.86  1.5  1.5 V
_______________________________________

 

 5  1015 1018 
(a) Vbi  0.0259  ln 

2
 1.8  10 6

 1.264 V
VSD sat   V pO  Vbi  VGS 

3.5  V pO  1.264  1.0  V pO  5.764 V
13.8
 2 s V pO 
a

 eN a 
N N 
(a) Vbi  Vt ln  a 2 d 
 ni 




3.0  V pO 1.328  V pO  4.328 V
 2 s V pO 
a

 eN a 
1/ 2




(b)
(i) V pO  5.764 V
(ii) V p  V pO  Vbi  5.764 1.264
1/ 2

13.11
(a)
I P1 
(c) VSD sat   V pO  Vbi  VGS 

(i) VSD sat   4.328  1.328  0  3.0 V
(ii) V SD sat   4.328  1.328  1.5  1.5 V
_______________________________________
13.9
(a) VDS sat   V pO  Vbi  VGS 
Now
 4 1016 4  1018 
Vbi  0.0259  ln 

2
 1.5  1010

 0.886 V
We find
5  V pO  0.886  V pO  5.886 V
 2 s V pO 
a

 eN a 




 211.7  8.85  10 5.886  


1.6  10 19 4  10 16




1/ 2
14


 1.293 10 cm  1.293  m
(b) V pO  4.328 V


1/ 2
4
 5.60 10 5 cm  0.560  m


 4.5 V
_______________________________________
 213.1 8.85  10 14 4.328 


1.6  10 19 2  10 16




 213.1 8.85  10 14 5.764  


1.6  10 19 5  10 15



 2  1016 3  1018 
 0.0259  ln 

2
 1.8  10 6

 1.328 V
V p  V pO  Vbi
1/ 2
1/ 2

 n eN d 2 Wa 3
6 s L
10001.6 10 19 1016 2
611.78.85 10 14 

400 10 0.5 10 
20 10  4
or
I P1  1.03 mA
(b)
ea 2 N d
V PO 
2 s



or
V PO  1.93 V
Also
  


 1019 1016 
Vbi  0.0259  ln 
  0.874 V
10 2
 1.5 10

Now
V DS sat   V PO  Vbi  VGS 
(b) (i) V pO  5.886 V
 1.93  0.874  VGS
(ii) V p  Vbi  V pO  0.886  5.886  5.0 V
_______________________________________
 

 1.6  10 19 0.5  10 4 2 1016 


211.7  8.85  10 14


5
 4.36 10 cm  0.436  m
4 3
4
or
V DS sat   1.056  VGS
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 13
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
We have
V P  Vbi  V PO  0.874  1.93  1.056 V
Then
(i) For VGS  0 , V DS sat   1.06 V
(ii) For VGS 
1
V P  0.264 V,
4
V DS sat   0.792 V
(iii) For VGS 
1
V P  0.528 V,
2
V DS sat   0.528 V
(iv) For V P 
3
V P  0.792 V,
4
V DS sat   0.264 V
(c)

 V  VGS 

I D1 sat   I P1 1  3 bi


 V PO 
 2 Vbi  VGS 

 1 
 3
V PO 



 0.874  VGS 
 1.031  3

1.93



 2 0.874  VGS 

 1 
 3

1.93


(i) For VGS  0 , I D1 sat   0.258 mA
(ii) For VGS  0.264 V,
I D1 sat   0.141 mA
Vbi  VGS  / V PO
VGS
g d (mS)
0
0.453
0.523
-0.264
0.590
0.371
-0.528
0.726
0.237
-0.792
0.863
0.114
-1.056
1.0
0
_______________________________________
13.13
n-channel JFET - GaAs
(a)
e N Wa
GO1  n d
L
1.6 10 19 8000  2 1016

10 10  4
 30 10 4 0.35 10 4
or
GO1  2.69 10 3 S
(b)
V DS sat   V PO  Vbi  VGS 
We have
ea 2 N d
V PO 
2 s


13.12
  V  V 1 / 2 
GS
 
g d  GO1 1   bi
  V PO  



3I P1 3 1.03 10 3

V PO
1.93


1.6 10 0.35 10  2 10 
213.18.85 10 
4 2
19
16
14
V PO  1.69 V
We find



 5  1018 2  1016 
Vbi  0.0259  ln 

2


1.8  10 6


or
I D1 sat   0.0148 mA
_______________________________________
GO1 

or
I D1 sat   0.0608 mA
where


(iii) For VGS  0.528 V,
(iv) For VGS  0.792 V,


or
GO1  1.60 10 3 S  1.60 mS
Then
Vbi  1.34 V
Then
V P  Vbi  V PO  1.34  1.69  0.35 V
We then obtain
V DS sat   1.69  1.34  VGS   0.35  VGS
For VGS  0 , V DS sat   0.35 V
For VGS 
1
V P  0.175 V,
2
V DS sat   0.175 V
(c)

 V  VGS 

I D1 sat   I P1 1  3 bi

 V PO 

 2 Vbi  VGS 

 1 
 3
V PO 


Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 13
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
where
 n eN d 2 Wa 3
I P1 

6 s L
80001.6 10 19 2 1016 2
613.18.85 10 14 
30 10 0.35 10 

4
4 3
10 10  4
or
I P1  1.515 mA
Then

 1.34  VGS 
I D1 sat   1.5151  3

 1.69 

 2 1.34  VGS 
 (mA)
 1 
 3

1
.
69


For VGS  0 , I D1 sat   0.0506 mA
and
For VGS  0.175 V,
I D1 sat   0.0124 mA
_______________________________________
13.14
g mS 
V  VGS 
3I P1 
1  bi
V PO 
V PO 
We have
I P1  1.03 mA, V PO  1.93 V, Vbi  0.874 V
The maximum transconductance occurs when
VGS  0 . Then
g mS max  
31.03 
0.874 
1

1.93 
1.93 
or
g mS max   0.524 mS
For W  400  m, we have
g mS max  
0.524
400  10  4
or
g mS max   13.1 mS/cm = 1.31 mS/mm
_______________________________________
13.15
The maximum transconductance occurs for
VGS  0 , so we have
(a) g mS max  
Vbi 
3I P1 
1

V PO 
V PO 
which can be written as

Vbi 
g mS max   G O1 1 

V PO 

We found
GO1  2.69 mS, Vbi  1.34 V, V PO  1.69 V
Then

1.34 
g mS max   2.69 1 

1.69 

or
g mS max   0.295 mS
This is for a channel length of L  10  m.
(b) If the channel length is reduced to
L  2  m, then
 10 
g mS max   0.2947    1.47 mS
2
_______________________________________
13.16
n-channel MESFET - GaAs
(a)
ea 2 N d
V PO 
2 s

1.6 10 0.5 10  1.5 10 
213.18.85 10 
19
4 2
16
14
or
V PO  2.59 V
Now
Vbi   Bn   n
where
N 
 4.7 1017 

 n  Vt ln  c   0.0259 ln 
16 
 Nd 
 1.5 10 
or
 n  0.0892 V
so that
Vbi  0.90  0.0892  0.811 V
Then
VT  Vbi  V PO  0.811  2.59
or
VT  1.78 V
(b )If VT  0 for an n-channel device, the
device is a depletion mode MESFET.
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 13
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(b) VT  Vbi  V pO
We find
N 
 4.7 1017 

 n  Vt ln  c   0.0259 ln 
16 
 Nd 
 2 10 
 0.0818 V
Vbi   Bn   n  0.87  0.0818  0.788 V
Then
VT  0.788  1.5  0.712 V
13.17
n-channel MESFET - GaAs
(a) We want VT  0.10 V
Then
VT  Vbi  V PO   Bn   n  V PO
so
 N  ea 2 N d
VT  0.10  0.89  Vt ln  c  
 N d  2 s
which can be written as
17


0.0259 ln  4.7 10 
 Nd

 2  V  V DS  VGS  
(c) h2   s bi

eN d



1/ 2



0.788  0  0.4
1.6 10 0.35 10  N  0.89  0.10

213.18.85 10 
 213.1 8.85 10 14 Vbi  V DS  VGS  


1.6  10 19 2 1016


or
(i) h2  7.246 10 10
4 2
19
d
14



1/ 2



 Nd

 8.453 10 17 N d  0.79
By trial and error,
N d  8.11015 cm 3
(b) At T  400 K
(ii)
N c 400  400 

  1.54
N c 300  300 
Then
N c 400  4.7 1017 1.54
h2  7.246 10 10 0.788  4.0  0.4
0.0259 ln  4.7 10
17


3/ 2

 400 
Vt  0.0259
  0.03453
 300 
Then
 7.24 1017 

VT  0.89  0.03453 ln 
15 
 8.110 
 8.453 10 17 8.11015
which becomes
VT  0.050 V
_______________________________________



13.18
1/ 2

 213.1 8.85  10 14 1.5 


19
16
 1.6  10 2  10






h 2  7.246  10  10 0.788  1.0  0.4
1/ 2
 3.17110 5 cm  0.3171 m
a  h2  0.330  0.3171  0.0129 m
(iii)


1/ 2
h2  a  a  h2  0
_______________________________________
Also

a  h2  0.330  0.1677  0.1623  m
 5.64 10 5 cm  0.564  m
 7.24 10 17 cm 3
 2 s V pO 
(a) a  

 eN d 
 1.677 10 5 cm  0.1677  m


1/ 2
1/ 2

 3.30 10 5 cm  0.330  m
13.19
(a) V pO 
ea 2 N d
2 s
1.6 10 0.50 10  5 10 

213.18.85 10 
19
4 2
14
 0.8626 V
We find
 4.7 1017 

15 
 5 10 
 n  0.0259 ln 
 0.1177 V
Vbi   Bn   n  0.87  0.1177
 0.7523 V
VT  Vbi  V pO  0.7523  0.8626
 0.1103 V
15
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 13
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
 4.7 1017 
  0.0713 V
(b)  n  0.0259 ln 
16 
 3 10 
Vbi   Bn   n  0.87  0.0713
 0.7987 V
VT  Vbi  V pO
and
Vbi   Bn   n  0.82  0.206  0.614 V
With V DS  0 and VGS  0.35 V, we find
a  h  0.075 10 4
 2  V  VGS  
 a   s bi

eN d


or V pO  Vbi  VT  0.7987   0.1103
 0.909 V
so that
a  0.075 10 4
Then
 2 s V pO 
a

 eN d 

1/ 2



1/ 2

 2.095 10 5 cm  0.2095  m
_______________________________________
13.20
VT  Vbi  V PO   Bn   n  V PO
a  0.26 10 4 cm  0.26  m
Now
ea 2 N d
VT  Vbi  V PO  0.614 
2 s
or
4 2
We obtain
VT  0.092 V
(b)
V DS sat   V PO  Vbi  VGS 
 Vbi  VT   Vbi  VGS 
or
V DS sat   VGS  VT  0.35  0.092
which yields
V DS sat   0.258 V
_______________________________________
4 2
19
16
14
T
1.6 10 0.25 10  N
213.18.85 10 
d
14

or
19
0.5  0.85   n  V PO
Now
 4.7 1017 

 n  0.0259 ln 

 Nd

and
ea 2 N d
V PO 
2 s
or

1/ 2
1.6 10 0.26 10  10 
V  0.614 
211.78.85 10 
We want VT  0.5 V, so


 
 211.7  8.85  10 14 0.614  0.35 


1.6  10 19 10 16


 213.1 8.85  10 14 0.909  


1.6  10 19 3  10 16



1/ 2

13.22
V PO  4.3110 17 N d
Then
 4.7 1017 

0.5  0.85  0.0259 ln 

 Nd

17
 4.3110 N d
By trial and error
N d  5.45 1015 cm 3
_______________________________________


13.21
n-channel MESFET - silicon
(a) For a gold contact,  Bn  0.82 V.
We find
 2.8 1019 
  0.206 V
 n  0.0259 ln 
16

 10

 4.7 1017 

(a)  n  0.0259  ln 
16 
 2 10 
 0.0818 V
(i) Vbi   Bn   n  0.90  0.0818
 0.818 V
ea 2 N d
(ii) V pO 
2 s

1.6 10 0.65 10  2 10 
213.18.85 10 
19
4 2
14
 5.83 V
(iii) VT  Vbi  V pO  0.818  5.83
 5.012 V
16
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 13
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(b) VDS sat   V pO  Vbi  VGS 
(i) V DS sat   5.83  0.818   1.0
 4.01 V
(ii) V DS sat   5.83  0.818   2.0
 3.01 V
(iii) V DS sat   5.83  0.818   3.0
 2.01 V
_______________________________________
13.23
(a) k n 
 n s W
2aL
650013.18.85 10 14 12 10 4 



2 0.25 10  4 1.5 10  4
3

 1.206 10 A/V  1.206 mA/V 2
2
(b) I D1 sat   k n VGS  VT 
2
2

I D

2

k n VGS  VT 
VGS VGS

 2k n VGS  VT 
1.25  2k n 0.45  0.15
 k n  2.083 mA/V
kn 
2.083 10 3 
V pO 

ea 2 N d
2 s
1.6 10 0.50 10  310 

211.78.85 10 
4 2
19
16
14


1/ 2
 211.7  8.85  10 14 10  4.953 


1.6  10 19 3  10 16




1/ 2

5
L  4.666 10 cm
Now
1 2L  0.90
L
 1
L
L
L
4.666  10 5
L

20.10 
20.10 
(b) VDS sat   V pO  Vbi  VGS 
650013.18.85 10 14 W

2 0.25 10
4
1.5 10 
4
3
 W  2.073 10 cm  20.73  m
(b) I D1 sat   k n VGS  VT 

L  2.333 10 4 cm  2.333  m
2
 n s W
2aL
 

 1018 3  1016 
Vbi  0.0259  ln 

2
 1.5  1010

 0.8424 V
 2  V  V DS sat  
L   s DS

eN d


(ii) V DS sat   0.45  0.15  0.30 V
_______________________________________
(a) g ms 
13.27
 5.795  0.8424  4.953 V
2
(i) V DS sat   0.25  0.15  0.10 V
13.24
13.26
Plot
_______________________________________
 5.795 V
(a) VDS sat   V pO  Vbi  VGS 
(i) I D1 sat   1.2060.25  0.15
 0.01206 mA  12.06  A
(ii) I D1 sat   1.2060.45  0.15
 0.1085 mA
(c) V DS sat   VGS  VT
13.25
Plot
_______________________________________
2
(i) I D1 sat   2.0830.25  0.15
 0.02083 mA  20.83  A
2
(ii) I D1 sat   2.0830.45  0.15
 0.1875 mA
_______________________________________
2
 5.795  0.8424  3
 1.953 V


 211.7  8.85  10 14 10  1.953 
L  

1.6  10 19 3  10 16




1/ 2

5
 5.892 10 cm
Then
L
L
5.892  10 5

20.10 
20.10 
 2.946 10 4 cm  2.946  m
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 13
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
13.28
We have that


L

I D 1  I D1 
 L  1 2L 
Assuming that we are in the saturation region,
then I D 1  I D 1 sat  and I D1  I D1 sat  .
We can write
1
I D 1 sat   I D1 sat  
1 L
1 
2 L
If L  L , then
 1 L 
I D 1 sat   I D1 sat 1  

 2 L 
We have that
 2  V  V DS sat  
L   s DS

eN d


1/ 2
 2  V  V sat  

  s DS 1  DS
V DS 
 eN d 
which can be written as

 2  V  V DS  VGS  
h2  h sat   s bi

eN d


and
 5  1018 4  1016 
Vbi  0.0259  ln 

2
 1.5  1010

or
Vbi  0.8915 V
1/ 2







 211.7  8.85  10 14 0.8915  2  
h sat  

1.6  10 19 4  10 16



1/ 2
1  2 s  V DS sat  
1 


2 L  eN d V DS 
V DS 
 
For VGS  0 , we obtain
1/ 2
 2 s  V DS sat  
1 

L  V DS 

V DS 
 eN d V DS 
If we write
I D 1 sat   I D1 sat 1  V DS 
then by comparing equations, we have

13.29
(a) Saturation occurs when   110 4 V/cm.
As a first approximation, let
V
  DS
L
Then
V DS    L  10 4 2 10 4  2 V
(b) We have that
So that  is nearly a constant.
_______________________________________

or
hsat  0.306 10 4 cm  0.306  m
(c) We then find
I D1 sat   eN d  sat a  hsat W


 
 0.50  0.30610 30 10 
 1.6 10 19 4 1016 10 7
4
1/ 2
The parameter  is not independent of V DS .
Define
V DS
x
V DS sat 
and consider the function
1 1
f  1  
x x
which is directly proportional to  . Then
x
f x 
1.5
0.222
1.75
0.245
2.0
0.250
2.25
0.247
2.50
0.240
2.75
0.231
3.0
0.222

1/ 2
or
4
I D1 sat   3.72 mA
(d) For VGS  0 , we have

 V  2 Vbi 

I D1 sat   I P1 1  3 bi 1 

 V PO  3 V PO 
Now
 eN d 2 Wa 3
I P1  n
6 s L

10001.6 10 19 4 1016 2
611.7 8.85 10 14 

or
I P1  12.36 mA
30 10 0.5 10 
2 10 
4 3
4
4
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 13
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Also
V PO 

13.31
(a)
ea 2 N d
2 s
   n   80005 10 3   4 10 7 cm/s
1.6 10 0.5 10  4 10 
211.78.85 10 
4 2
19
16
Then
14
td 
or
V PO  7.726 V
Then
or
I D1 sat   9.05 mA
_______________________________________
13.30
(a) If L  1  m, then saturation will occur
when
V DS    L  10 4 110 4  1 V
We find
 

 2  V  V DS  VGS  
h2  h sat   s bi

eN d


We have Vbi  0.8915 V and for VGS  0 , we
obtain
(b) Assume    sat  10 7 cm/s
Then
L
2 10 4
td 

 2 10 11 s
 sat
10 7
or
t d  20 ps
_______________________________________
13.32
(a)
   n   100010 4   10 7 cm/s
Then
td 
1/ 2


 211.7  8.85  10 0.8915  1 
h sat  

1.6  10 19 4  10 16


14


4
hsat  0.247 10 cm  0.247  m
Then
I D1 sat   eN d  sat a  hsat W

 
 0.50  0.247 10 30 10 
 1.6 10 19 4 1016 10 7
4
4
I D1 sat   4.86 mA
If velocity saturation did not occur, then from
the previous problem, we would have
2
I D1 sat   9.05   18.1mA
1
(b) If velocity saturation occurs, then the
relation I D1 sat   1 L  does not apply.
_______________________________________
L


2 10 4
 2 10 11 s
7
10
or
t d  20 ps
(b) For    sat  10 7 cm/s
td 
1/ 2
or
or
2 10 4
 5 10 12 s
4 10 7
t d  5 ps
0.8915 

7.726 
 2
 1 
 3




or

 0.8915 
I D1 sat   12.361  3

 7.726 


L
L
 sat

2 10 4
 2 10 11 s
10 7
or
t d  20 ps
_______________________________________
13.33
The reverse-bias current is dominated by the
generation current. We have
V P  Vbi  V PO
We find
 5  1018 3  1016 
Vbi  0.0259  ln 

2
 1.5  1010

or
Vbi  0.884 V

Also V PO 

ea 2 N d
2 s







 1.6  10 19 0.3  10 4 2 3  10 16 


211.7  8.85  10 14




Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 13
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
or
13.34
(a) The ideal transconductance for VGS  0 is
V PO  2.086 V
Then
V P  0.884  2.086  1.20 V
Let VGS  1.20 V
Now
 2  V  V DS  VGS  
x n   s bi

eN d



 211.7 8.85 10 14

1.6 10 19



or

1/ 2


3 10 
1/ 2


16


x n  4.314 10 10 2.084  VDS 
(a) For V DS  0 , x n  0.30  m
(b) For V DS  1 V, x n  0.365  m
(c) For V DS  5 V, x n  0.553  m
1/ 2

The depletion region volume at the drain is
L
Vol  a  W   x n 2a W 
2
 2.4 10 4 
 30 10  4
 0.3 10  4 

2


4
 x n  0.6 10 30 10 4



or





Vol  10.8 10 12  x n 18 10 8
(a) For V DS  0 , Vol  1.62 10

11
cm 3
(b) For V DS  1 V, Vol  1.737 10 11 cm 3
(c) For V DS  5 V, Vol  2.075 10 11 cm 3
The generation current at the drain is
 n 
I DG  e i  Vol
 2 O 
 1.5 1010 
 1.6 10 19 
Vol
8 
 2 5 10 
or
I DG  2.4 10 2 Vol
(a) For V DS  0 , I DG  0.39 pA








0.884  V DS   1.20 


Vbi 
g mS  GO1 1 

V PO 

where
e N Wa
GO1  n d
L
1.6 10 19 4500 7 1016

1.5 10  4
 5 10 4 0.3 10 4
or
GO1  5.04 mS
We find
ea 2 N d
V PO 
2 s


(b) For V DS  1 V, I DG  0.42 pA
(c) For V DS  5 V, I DG  0.50 pA
_______________________________________

1.6 10 0.310  7 10 
213.18.85 10 
19
4 2
16
14
or
V PO  4.347 V
We have
 4.7 1017 
  0.049 V
16 
 7 10 
 n  0.0259 ln 
so that
Vbi   Bn   n  0.89  0.049  0.841 V
Then

0.841 
g mS  5.04 1 

4.347 

or
g mS  2.82 mS
(b) With a source resistance
gm
g
1
g m 
 m 
1  g m rs
g m 1  g m rs
For
g m
1
 0.80 
gm
1  2.823rs
we obtain
rs  88.6 
(c)
L
L L

rs 

A A e n n A

Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 13
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
so



L  88.56 1.6 10 19 4500 7 1016



 0.3 10 4 5 10 4

or
13.36
(a) For constant mobility
e N a 2
fT  n d 2
2 s L
L  0.67 10 4 cm  0.67  m
_______________________________________
13.35
Considering the capacitance charging time,
we have
gm
fT 
2 C G
where
 WL
CG  s
a

13.18.85 10 14 5 10 4 1.5 10 4 
0.3 10


13.37
CG  2.9 10 F
We must use g m , so we obtain

2.82 10 0.80  1.238 10 Hz
2 2.9 10 
3
11


e n N d a 2
2 s L2
1.6 10 10002 10 0.40 10 
2 11.78.85 10 L
19
4 2
16
14
2
786.975
L2
786.975
fT 
2
3 10  4
fT 
15
 1.629 10 11 s
Taking into account the channel transit time
and the capacitance charging time, we find
1
1
fT 

2  2 1.629  10 11
or
f T  9.77 10 9 Hz  9.77 GHz
_______________________________________

f T  1.33 10 Hz  13.3 GHz
_______________________________________
fT 

4 2
14
10
4
We can also write
1
1
fT 
C 
2  C
2 f T
so
1
C 
 1.285  10 12 s
2 1.238  1011
The channel transit time is
1.5 10 4
tt 
 1.5 10 11 s
7
10
The total time constant is
  1.5 10 11  1.285 10 12
4 2
16
(b) For saturation velocity model

10 7
f T  sat 
2 L 2 1.2 10  4
15

19
f T  4.12 1011 Hz  412 GHz
or
fT 
1.6 10 75004 10 0.30 10 
2 13.18.85  10 1.2  10 
(a)


 8.74  10 9 Hz  8.74 GHz
786.975
fT 
2
1.5 10  4
(b)


 3.50 10 Hz  35.0 GHz
_______________________________________
10
13.38
fT 
e p N a a 2
2 s L2
or
 e p N a a 2 
L

 2 s f T 
1/ 2






 1.6 10 19 420 2 1016 0.40 10  4 2 


2 11.7  8.85 10 14 f T


L
18.18
fT

1/ 2
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 13
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(a) L 
 g mS 

  5.02 S/cm  502 mS/mm
 W 
(b) At Vg  0 , we obtain
18.18
5 10
9
 2.57 10 4 cm  2.57  m
(b) L 
N
I D sat 

 V  V 
d  d  off O s
W

18.18
12  10
9
 1.66 10 4 cm  1.66  m
_______________________________________
13.39
(a)

I D sat 
 5.37 A/cm  537 mA/mm
W
_______________________________________
13.41
1.6 10 3 10 350 10 
212.28.85 10 
19
8 2
18
14
or
N
V g  Voff
ed  d 
For Vg  0 , we have


12.28.85 10 14  2.07
1.6 10 19 350  8010 8 
or
n S  3.25 1012 cm 2
_______________________________________
13.40
(a) We have
I D sat  
N W
V  V  V 
d  d  g off O s


We find
 g mS 
  I D sat   N  s

 


 W  V g  W  d  d 

0.30  0.85  0.22  V P 2
V P 2  0.93 V
We have
eN d d d2
VP2 
2 N
or
2 N V P 2
d d2 
eN d
(b)
nS 
E c
 VP2
e
We want Voff  0.3 V, so
V off   B 
or
V P 2  2.72 V
Then
Voff  0.89  0.24  2.72
or
Voff  2.07 V
or
12.28.85  10  12 
2.07  12  10 7 
8
350  8010 
or
E c
V off   B 
 VP2
e
where
eN d d d2
VP2 
2 N
nS 


12.28.85 10 14 2 10 7 
350  8010 8 


212.2 8.85 10 14 0.93
1.6 10 19 2 1018
We then obtain




o
d d  2.5110 6 cm  251 A
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 14
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Chapter 14
14.1
(b)  
1.24
 max 
m
Eg
1.24
 1.11  m
1.12
1.24
 1.88  m
(b) Ge:  max 
0.66
1.24
 0.873  m
(c) GaAs:  max 
1.42
1.24
 0.919  m
(d) InP:  max 
1.35
_______________________________________
(a) Si:  max 
14.2
(a) For   480 nm,
1.24 1.24
E

 2.58 eV

0.480
For   725 nm,
1.24
E
 1.71 eV
0.725
(b) For E  0.87 eV,
1.24 1.24


 1.43  m
E
0.87
For E  1.32 eV,
1.24

 0.939  m
1.32
For E  1.90 eV,
1.24

 0.653  m
1.90
_______________________________________
14.3
(i) From Figure 14.4,   910 3 cm 1
I d 
 exp  d 
(ii) 
I 0

(i) From Figure 14.4,   2.6 10 4 cm 1
I d 
 exp  d 
(ii) 
I 0



 exp  2.6 10 4 0.80 10 4
 0.125
Fraction absorbed  1  0.125  0.875
_______________________________________
14.4
g 
 I x 
h
For h  1.3 eV,  
1.24
 0.95  m
1 .3
For silicon:   310 2 cm 1
Then for I x   10 2 W/cm 2 , we obtain
g 
3 10 10 
1.6 10 1.3
2
2
19
or
g   1.44 1019 cm 3 s 1
The excess concentration is
n  g   1.44 1019 10 6
or
n  1.44 1013 cm 3
_______________________________________


14.5
(a) p  g  p 0  g  

p
 p0
5 1015
 2.5 10 22 cm 3 s 1
2 10 7
For h  1.65 eV,
1.24

 0.752  m
1.65
g 
1.24
 0.752  m
(a)  
1.65

4

 exp  9 10 1.2 10
 0.340
Fraction absorbed  1  0.34  0.66
3
1.24
 0.653  m
1.90
From Figure 14.4,   910 3 cm 1
g h 
I 0 


2.5 10 1.6 10 1.65
22
9 10 3
 0.733 W/cm 2
19
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 14
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(b)
I  d 
 0.1  exp  d 
I 0

where L p  D p p
 
0.1  exp  9 10 3 d 
d
1
 1 
ln 

3
9 10
 0.1 
 2.56 10 4 cm  2.56  m
_______________________________________
14.6
1.24 1.24


 0.886  m
E
1.40
From Figure 14.4,   4.5 10 2 cm 1
I d 
 0.1  exp  d 
(a) 
I 0
1
 1 
 1 
d  ln 
ln 


2
  0.1  4.5 10  0.1 
1
 5.12 10 3 cm  51.2  m
(b) d 
1
 1 
ln 

2
4.5 10
 0.3 
 2.68 10 3 cm  26.8  m
_______________________________________
14.7
GaAs:
For x  1  m  10 4 cm, we have 50%
absorbed or 50% transmitted, then
I x 
 0.50  exp  x 
IO
We can write
1  1   1 
     ln 
    4   ln 2
 x   0.5   10 
or
  0.69 10 4 cm 1
This value corresponds to
  0.75  m , E  1.65 eV
_______________________________________
14.8
The ambipolar transport equation for minority
carrier holes in steady state is
d 2 p n 
p
Dp
 GL  n  0
2
p
dx
or
d 2 p n  p n
G
 2  L
2
Dp
dx
Lp
The photon flux in the semiconductor is
x    O exp  x 
and the generation rate is
G L   x     O exp  x 
so the differential equation becomes
d 2 p n  p n
 O
 2 
exp  x 
Dp
dx 2
Lp
The general solution is of the form
x


  B exp  x 
p n x   A exp
 Lp 
 Lp 




  O p
 2 2
 exp  x 
 L p 1
As x   , p n  0 so that B  0 . Then
  x    O p

 exp  x 
 L p   2 L2  1
p


At x  0 , we have
d p n 
Dp
 sp n
dx x 0
x 0
so we can write
  O p
p n
 A 2 2
x 0
 L p 1
p n x   A exp
and
2
d p n 
A   O p

 2 2
dx x  0
L p  L p 1
Then we have
AD p  2  O p D p
s  O  p


 sA  2 2
2 2
Lp
 L p 1
 L p 1
Solving for A, we find
  O p  s   D p 
A 2 2


 L p  1  s  D p L p 
The solution can now be written as
  O p
p n x   2 2
 L p 1


 s  Dp

x
  exp  x 

 exp
 Lp 
 s  D p L p



_______________________________________


Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 14
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
14.9
We have
d 2 n p
n p
Dn
 GL 
0
2
n
dx
or
d 2 n p n p
G
 2  L
Dn
dx 2
Ln
 
 
where Ln  Dn n
The general solution can be written in the
form
 x 
 x 
n p x   A cosh   B sinh    G L n
 Ln 
 Ln 
For s   at x  0 means n p 0  0 . Then
0  A  G L n  A  G L n
At x  W ,
d n p
 Dn
 s o n p
dx x W
x W
Now
W 
n p W   G L n cosh 
 Ln 
 
W 
 B sinh    G L n
 Ln 
and
d n p
 
dx

x W
W 
G L n
sinh  
Ln
 Ln 

W 
B
cosh 
Ln
 Ln 
so we can write
 W  BDn
W 
G L n Dn
sinh   
cosh 
Ln
L
L
n
 n
 Ln 
W 
 s o  G L  n cosh 
 Ln 

W 
  G L n 
 B sinh 

 Ln 

Solving for B, we obtain


W 
 W   
  1 
G L  L n sinh    s o n cosh


 Ln 
 L n   

B
W 
W 
Dn
cosh   s o sinh  
Ln
L
 n
 Ln 
The solution is then

 x 
 x 
n p x   G L n 1  cosh   B sinh  
 L n 
 Ln 

where B was just given.
_______________________________________
14.10
L n  D n n 0 
2510 6   5 10 3 cm
105 10 7 
L p  D p p 0 
 2.236 10 3 cm
 Dn
Dp 


Now J S  en i2 
 Ln N a L p N d 




 1.6 10 19 1.5 1010

2


25
10


3
16
3
15 
2.236 10 10 
 5 10 10
10
J S  1.790 10 A/cm 2

  

 
I S  AJ S  5 1.79 10 10

 8.950 10 10 A
(a) I L  eG L AW
We find
 1016 1015 
Vbi  0.0259  ln 
  0.6350 V
2
 1.5 1010 
  


 2  V  N  N d 

W   s bi  a

 e
 N a N d 

1/ 2

 211.7 8.85 10 14 0.635

1.6 10 19

 10 16  1015  
  16
15  
 10 10  
1/ 2
  
W  9.508 10 5 cm
Then
I L  1.6 10 19 5 10 21 5 9.508 10 5
 0.380 A  380 mA


 

 I 
(b) Voc  Vt ln 1  L 
 IS 
0.380 

 0.0259 ln 1 

10
 8.95 10 
Voc  0.5145 V
Voc 0.5145

 0.810
(c)
Vbi
0.635
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 14
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
14.11
From Problem 14.10, I S  8.95 10 10 A
 I 
(a) Voc  Vt ln 1  L 
 IS 

120 10 3 

 0.0259 ln 1 
10 
 8.95 10 
 0.4847 V
 V  
(b) I  I L  I S exp   1
  Vt  
100 10 3  120 10 3
 V  
 8.95  10 10 exp   1
  Vt  
 V  0.4383 V
 V 
V 
I
(c) 1  m  exp m   1  L
IS
 Vt 
 Vt 
120 10 3
 1
8.95 10 10
 1.34110 8
By trial and error, V m  0.412 V
Now
 V  
I m  I L  I S exp m   1
  Vt  
 120  10 3


  0.412  
 8.95 10 10 exp
  1
  0.0259  
 I m  112.75 10 3 A  112.75 mA
Pm  I mVm  112.750.412
 46.5 mW
V
0.412
(d) V m  I m R L  R L  m 
I m 0.11275
R L  3.65 
_______________________________________
14.12
From Problem 14.10, I S  8.95 10 10 A
(a)
 I 
(i) Voc  Vt ln 1  L 
 IS 

10 10 3 

 0.0259 ln 1 
10 
 8.95 10 
 0.420 V
 V 
V 
I
(ii) 1  m  exp m   1  L
IS
 Vt 
 Vt 
10 10 3
 1
8.95 10 10
 1.117 10 7
By trial and error, V m  0.351 V
Now
 V  
I m  I L  I S exp m   1
  Vt  
 10 10 3


  0.351  
 8.95 10 10 exp
  1
  0.0259  
I m  9.3110 3 A  9.31 mA
Then
Pm  I mVm  9.310.351  3.27 mW
(b)

100 10 3 

(i) Voc  0.0259 ln 1 
10 
 8.95 10 
 0.480 V
 V 
V 
I
(ii) 1  m  exp m   1  L
IS
 Vt 
 Vt 
100 10 3
 1
8.95 10 10
 1.117 10 8
By trial and error, Vm  0.407 V
Now
 V  
I m  I L  I S exp m   1
  Vt  
 100  10 3


  0.407  
 8.95 10 10 exp
  1
  0.0259  
I m  9.40 10 2 A  94.0 mA
Then
Pm  I mVm  94.00.407   38.3 mW
P
38.3
 11.7
(c) m 2 
Pm1 3.27
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 14
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
 V  
I  50  10 3  4.579  10 12 exp   1
  Vt  
14.13

 J 
VOC  Vt ln 1  L 
 JS 
 30 10 3 

 0.0259 ln 1 

JS


where
 1 D
Dp 
1
n

J S  en i2 

Nd  p 
 N a  n

which becomes


J S  1.6 10 19 1.8 10 6
 1

 N a
We see that when I  0 , V  VOC  0.599 V.
We find
2

7
8 
5  10 
or
 6.708 10 4

J S  5.184 10 7 
 1.183 10 15 
N
a


Then
J S (A/cm 2 )
N a (cm 3 )
VOC (V)


10 15
3.477 10 17
0.891
10 16
3.478 10 18
0.950
17
19
10
3.484 10
14.14
(a)
I L  J L  A  25 10 3 2  50 10 3 A
We have
 1 D
Dp 
1
n

J S  en i2 

Nd  p 
 N a  n

or




J S  1.6 10 19 1.5 1010
(b) The voltage at the maximum power point
is found from
 Vm 
 Vm 
I
  1 L
1 
  exp

IS
 Vt 
 Vt 
50  10 3
4.58 10 12
 1.092 1010
 1
1.01
10 18
3.539 10 20
1.07
_______________________________________

2
 1
18
1

 19
16
6
3

10
5

10
10

which becomes
J S  2.289 10 12 A/cm 2
or
I S  4.579 10 12 A
We have
 V  
I  I L  I S exp   1
  Vt  
or

6
7 
5  10 
I (mA)
50
50
50
50
49.98
49.84
48.89
42.36
33.46
14.19
V (V)
0
0.1
0.2
0.3
0.4
0.45
0.50
0.55
0.57
0.59

225
1
 19
8
5  10
10

By trial and error,
V m  0.520 V
At this point, we find
I m  47.6 mA
so the maximum power is
Pm  I mVm  47.60.520
or
Pm  24.8 mW
(c) We have
V V
0.520
V  IR  R   m 
I
I m 47.6  10 3
or
R  10.9 
_______________________________________
14.15
 180 10 3 

(a) Voc  0.0259  ln 1 
2 10 9 

 0.474 V
 V 
V 
I
(b) 1  m  exp m   1  L
IS
 Vt 
 Vt 
180 10 3
 9  10 7
 1
9
2 10
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 14
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
By trial and error, V m  0.402 V
V 
I m  I L  I S exp m 
 Vt 


1
 1.69 10 A  169 mA
Pm  I mVm  1690.402  67.9 mW
V m 0.402

 2.379 
I m 0.169
(d) R L  1.52.379  3.568 
Now
V 
V
I
 I L  I S exp 
RL
 Vt 
V
 V 
 180 10 3  2 10 9 exp

3.568
 0.0259 
By trial and error, V  0.444 V
V
0.444

 0.1244 A
Then I 
R L 3.568


P  IV  124.40.444  55.2 mW
_______________________________________
14.16
 100 10 3 

(a) Voc  0.0259  ln 1 
10 10 

 0.5367 V
 V 
V 
I
(b) 1  m  exp m   1  L
IS
 Vt 
 Vt 
100 10 3
 1
10 10
9
 10
By trial and error, V m  0.461 V
Then
 0.461 
I m  100 10 3  10 10 exp

 0.0259 


 9.463 10 2 A  94.63 mA
Pm  I mVm  94.630.461  43.62 mW
10
 21.7  n  22 cells
0.461
(d) Now V  220.461  10.14 V
P  IV
5.2  I 10.14  I  0.5128 A
(c) n 
Then n  
V
10.14

 17.86 
I 0.5678
_______________________________________
So R L 
 0.402 
 180 10 3  2 10 9 exp

 0.0259 
(c) R L 
(e) Then I  60.09463  0.5678 A
0.5128
 5.42  n   6
0.09463
14.17
Let x  0 correspond to the edge of the space
charge region in the p-type material. Then in
the p-region
d 2 n p
n p
Dn
 GL 
0
2
n
dx
or
d 2 n p n p
G
 2  L
2
Dn
dx
Ln
 
 
where
G L   x     O exp  x 
Then we have
d 2 n p n p
 O
 2 
exp  x 
2
Dn
dx
Ln
 
The general solution is of the form
x
x
  B exp

n p x   A exp

L 
 Ln 
 n 
 
 2 O2 n exp  x 
 Ln  1
As x   , n p  0 so that B  0 . Then
  x    O n
 2 2
 exp  x 

 Ln   Ln  1
 
We also have n p 0  0  A  2 O2 n ,
 Ln  1
which yields
 
A  2 O2 n
 Ln  1
We then obtain

     x 
  exp  x 
n p x   2 O2 n exp
 L n  1   L n 

n p x   A exp
where  O is the incident flux at x  0 .
_______________________________________
14.18
For 90% absorption, we have
x 
 exp  x   0.10
O
Then
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 14
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
exp   x  
14.20
n-type, so holes are the minority carrier
(a)
p  G L p  10 21 10 8
1
 10
0 .1

or
1
x     ln 10
 

or
p  n  1013 cm 3
For h  1.7 eV ,   10 4 cm 1
Then
 1 
x   4   ln 10  2.3 10  4 cm
 10 
or
x  2.3  m
(b)
  ep  n   p
14.19
(a) no  N d  51015 cm 3
I  e n no A



 1.6 10 19 1200  5 1015



14
 2.56 10 2 (  -cm) 1
(d) I L   A


3


 2.56 10  2 5 10  4 

4
 120 10 
3
 3.2 10 A  3.2 mA
IL
(e)  ph 
eG L AL
3.2 10 3

1.6 10 19 10 21 5 10  4 120 10  4
ph  3.33
_______________________________________
 

10 8000  250
13


p


  1.32 10 2 (  -cm) 1
(c)
 AV
I L  J L  A   A 
L
2
1.32 10 10 4 5

100 10  4
or
I L  0.66 mA
(d)
IL
 ph 
eG L AL

  
(c)   ep   
 1.6 10 10 1200  400
19
19
or
3


 5 10  4 

4
 120 10 
I  0.12 A  120 mA
(b) p  GL p 0  10 21 10 7  1014 cm 3
n

 1.6 10
and for h  2.0 eV,   10 5 cm 1 .
Then
 1 
x   5   ln 10  0.23 10  4 cm
 10 
or
x  0.23  m
_______________________________________





0.66 10 3
1.6 10 19 10 21 10  4 100 10  4

 


or
ph  4.125
_______________________________________
14.21
x    O exp  x 
The electron-hole generation rate is
g    x     O exp  x 
and the excess carrier concentration is
p   p x
Now
  ep  n   p
and
J L   
The photocurrent is now found from

IL 

W
xO
0
0
    dA   dy     dx

xO
 
 We  n   p  p  dx
0
Then
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 14
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________


xO

0



 1.6 10
 We  n   p   O p
which becomes
I L  We n   p O p 1  exp  xO 
Now
I L  50 10 4 1.6 10 19 1200  45050




 10 2 10 1  exp 5 10 10 
7
16
or
I L  0.131  A
_______________________________________
14.22
 

 10 2  10
Vbi  0.0259  ln 
2
 1.5  1010
 0.6530 V
16
15

 10 3 cm


1/ 2
 211.7  8.85 10 14 0.653  5

1.6 10 19


 1016  2  1015  
  16
15  
 10 2  10  

1/ 2

Then
W  2.095 10 4 cm
(a) I L1  eWG L A

 
 1.6 10 19 2.095 10 4 10 21 10 3
 3.352 10 A  33.52  A
(b) In n-region,
p  G L p 0  10 21 10 7  1014 cm 3
5



In p-region,
n  G L n0  10 21 5 10 7
 
Dp
or
d 2 p n 
p
 GL  n  0
2

p
dx
d 2 p n  p n
G
 2  L
2
Dp
dx 
Lp
which yields
G L L2p
p np 
 G L p
Dp

 
 2.095  35.36  10.010 4
The general solution is found to be
  x 


  B exp  x  
p nh x   A exp
 Lp 
 Lp 




The particular solution is found from
p np
G
 2  L
Dp
Lp
1010 7 
 2  V  V R   N a  N d 


W   s bi
 N N 
e

a
d 

3
positive in the negative x direction. The
homogenerous solution is found from
d 2 p nh  p nh
 2 0
dx  2
Lp
 
 3.536 10 3 cm
L p  D p p 0 
10 10 
where L p  D p p and where x  is
255 10 7 
L n  D n n 0 

21
14.23
In the n-region under steady state and for
  0 , we have
4
4
19
I L  7.593 10 4 A  0.7593 mA
_______________________________________
 1
 xO
  exp  x 
 
 0


(c) I L  eGL A W  Ln  L p
I L  We  n   p   O p exp  x dx

 51014 cm 3

The total solution is the sum of the
homogeneous and particular solutions, so we
have
  x 


  B exp x    G L p
p n x   A exp
 Lp 
 Lp 




One boundary condition is that p n remains
finite as x    which means that B  0 .
Then at x   0 , p n 0  0  p n 0  p nO , so
that p n 0   p nO .
We find that
A   p nO  GL p
The solution is then written as


  x 

 Lp 


p n x   G L p  G L p  p nO  exp
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 14
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
The diffusion current density is found as
d p n x 
J p  eD p
dx
x  0
But
d p n 
d p n 

dx
dx 
since x and x  are in opposite directions.
So
d p n 
J p  eD p
dx  x0

 eD p  GL p  p nO
14.25
(a) G L 0 
I 0
10 3 0.080

h
1.6 10 19 1.5
 3.33 10 20 cm 3 s 1
Then
G L x   G L 0 exp x 
   

 3.33 10 20 exp  10 3 x 
(b) J L  e o 1  exp W 
1  exp W 

1.6 10 19 3.333 10 20

10 3


eG L 0

 1 


 exp  x  

 Lp 
 L p  x 0






 
 1  exp 10 100 10 
4
3
Finally
J p  eG L L p 
eD p p nO
J L  5.33 10 2 A/cm 2  53.3 mA/cm 2
_______________________________________
Lp
_______________________________________
14.26
(a) J L  eWG L
14.24
(a) J L  e o 1  exp W 

5 10 
 1  exp 10 2 10 
Diode A: J L  1.6 10
19
17
4
4
J L  6.92 10 2 A/cm 2

5 10 
 1  exp 10 10 10 
Diode B: J L  1.6 10
19
17
4
4
J L  8.0 10 2 A/cm 2



 1  exp 10 80 10 
Diode C: J L  1.6 10 19 5 1017
4
4
J L  8.0 10 2 A/cm 2
(b) J L  e o 1  exp W 



 1  exp 5 10 2 10 
Diode A: J L  1.6 10 19 5 1017
2
4
J L  7.613 10 3 A/cm 2



 1  exp 5 10 10 10 
Diode B: J L  1.6 10 19 5 1017
2
4
J L  3.148 10 2 A/cm 2



 1  exp 5 10 80 10 
Diode C: J L  1.6 10 19 5 1017
2
4
J L  7.853 10 2 A/cm 2
_______________________________________


 
 1.6 10 19 20 10 4 10 21
 0.32 A/cm
(b) J L  e o 1  exp W 
2
1  exp W 

1.6 10 19 10 21

10 3

eG L 0

 
 
 1  exp 10 20 10 
3
4
J L  0.138 A/cm 2
_______________________________________
14.27
The minimum  occurs when   1  m
which gives   10 2 cm 1 . We want
x 
 exp  x   0.10
O
which can be written as
1
exp  x  
 10
0.10
Then
1
1
x  ln 10   2 ln 10   2.30  10  2 cm

10
or
x  230  m
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 14
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
14.28
For Al x Ga1 x As system, a direct bandgap for
0  x  0.45 , we have
E g  1.424  1.247 x
At x  0.45 , E g  1.985 eV, so for the direct
bandgap
1.424  E g  1.985 eV
which yields
0.625    0.871  m
_______________________________________
14.29
(a) From Figure 14.24, E g  1.64 eV

1.24 1.24

 0.756  m
Eg
1.64
(b) From Figure 14.24, E g  1.78 eV
1.24 1.24

 0.697  m
Eg
1.78
_______________________________________

14.33
We can write the external quantum efficiency
as
 ext  T1T2
where T1  1  R1 and where R1 is the
reflection coefficient (Fresnel loss), and the
factor T 2 is the fraction of photons that do not
experience total internal reflection. We have
 n  n1 

R1   2
 n 2  n1 
so that
2
 n  n1 

T1  1  R1  1   2
 n 2  n1 
which reduces to
4n1 n 2
T1 
n1  n 2 2
Now consider the solid angle from the source
point. The surface area described by the solid
angle is  p 2 . The factor T1 is given by
T1 
14.30
Eg 
1.24
1.24

 1.85 eV

0.670
From Figure 14.23, x  0.35
_______________________________________
14.31
1.24
 1.85 eV

0.670
From Figure 14.24, x  0.38
_______________________________________
Eg 
1.24

14.32
(a) For GaAs, n 2  3.66 and for air, n1  1.0 .
The critical angle is
n 
 1 
 C  sin 1  1   sin 1 
  15.86
 3.66 
 n2 
The fraction of photons that will not
experience total internal reflection is
2 C 215.86 

 8.81%
360
360
(b)Fresnel loss:
2
 n  n1 
 3.66  1 
  
R   2
  0.3258
n

n
 3.66  1 
1 
 2
The fraction of photons emitted is then
0.08811  0.3258  0.0594  5.94%
_______________________________________
2
2
 p2
4 R 2
From the geometry, we have
  p 2
 
sin  C  
 p  2R sin  C 
R
 2 
 2 
Then the area is
 
A   p 2  4 R 2  sin 2  C 
 2 
Now
 p2
 
T1 
 sin 2  C 
4 R 2
 2 
From a trig identity, we have
  1
sin 2  C   1  cos C 
 2  2
Then
1
T1  1  cos C 
2
The external quantum efficiency is now
4n1 n 2
1
 ext  T1T2 
 1  cos C 
2
n1  n2  2
or
 ext 
2n1 n 2
n1  n 2 2
1  cos C 
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 14
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
14.34
For an optical cavity, we have

N   L
2
If  changes slightly, then N changes slightly
also. We can write
N 1 1 N 1  1 2

2
2
Rearranging terms, we find
N 1 1 N 1  1 2 N 1 1 N 1  2  2




0
2
2
2
2
2
If we define   1   2 , then we have
N1

  2
2
2
We can approximate  2   , then
N 1
2L
 L  N1 
2

Then
1 2L


 
2 
2
which yields
 
2
2L
_______________________________________
14.35
For GaAs:
h  1.42 eV   
Then
 
2
2L

1.24
 0.873  m
1.42
0.873 10   5.08 10 cm
275 10 
4 2
7
4
or
  5.08 10 3  m
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 15
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Chapter 15
15.1 See diagrams in Figure 8.29
_______________________________________
15.2
V
0.60  0.15

 25 
I
2  2010 3
_______________________________________
R
15.6
s
10 7
 5 10 9 Hz
4
2 L 2 10 10
 5 GHz
_______________________________________
f 

15.7
(a) n po 
15.3
fr 
1
2 R min C j
R min
1
Rp
1

2 10  2  10 9


10
1
1
 2.39  10 Hz  23.9 MHz
_______________________________________
7
15.4
(a) no L  1012 cm 2
1012
 10 3 cm  10  m
15
10
L
10 3
(ii)  

 6.667 10 11 s
 d 1.5 10 7
1
1
(iii) f  
 6.667  10 11
 1.5 1010 Hz  15 GHz
(b)
1012
(i) L  16  10  4 cm  1  m
10
L
10 4
(ii)  

 6.667 10 12 s
 d 1.5 10 7
1
1
(iii) f  
 6.667  10 12
 1.5 1011 Hz  150 GHz
_______________________________________
(i) L 
15.5
(a)  
V
9

 6  10 3 V/cm
L 15  10  4
(b)  d  1.5 10 7 cm/s
d
1.5 10 7
 11010 Hz
L 15 10  4
 10 GHz
_______________________________________
(c)
f 




ni2
1.5 1010

NB
8 1015

2
 2.8125 10 4 cm 3
V 
(i) n p 0  n po exp BE 
 Vt 
 n p 0  

 V BE  Vt ln 
 n po 




1014

 0.0259  ln 
4 
 2.8125 10 
 0.5696 V
(ii) Neglecting any recombination in the base
eD B n po A
V 
IC 
exp BE 
xB
 Vt 
1.6 10 19 20 2.8125 10 4 0.4

2 10  4
 0.5696 
 exp

 0.0259 
I C  0.640 A

 

(b) n p 0   0.1N B  8  1014 cm 3
 8 1014 

(i) V BE  0.0259  ln 
4 
 2.8125 10 
 0.6234 V
1.6 10 19 20 2.8125 10 4 0.4
(ii) I C 
2 10  4
 0.6234 
 exp

 0.0259 
I C  5.12 A
_______________________________________

 

Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 15
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
15.8
(a) From Figure 7.15, BVBC  450 V
(b) V pt 
15.10
(a) BVCEO 
N N  N B 
ex
 B C
2 s
NC
2
B
1.6 10 2 10 

211.7 8.85 10 
8 10 6 10  8 10 

4 2
19
14
15
14
15
6 1014
V pt  354.4 V
(c) From Figure 7.15, BVBE  65 V
_______________________________________
15.9
From the junction breakdown curve, for
BVCBO  1000 V, we need the collector doping
concentration to be N C  21014 cm 3 .
Depletion width into the base (neglect V bi ).
 2  V  N 

1

x p   s BC  C 

e
 N B  N B  N C 



1/ 2

 211.7 8.85 10 14 1000

1.6 10 19


 2  1014 
1


 

15 
15
14 
 5  10  5  10  2  10 
1/ 2
or
x p  3.16  10 4 cm  3.16  m
(Minimum base width)
Depletion width into the collector
 2  V  N 

1

x n   s BC  B 

e
 N C  N B  N C 



1/ 2

 211.7 8.85 10 14 1000

1.6 10 19


 5  1015 
1


 

14 
15
14 
 2  10  5  10  2  10 
BVCBO
n

300
(i) BVCEO 
3
(ii) BVCEO 
3
10
300
 139 V
 81.4 V
50
(b)
125
(i) BVCEO 
3
(ii) BVCEO 
3
 58.0 V
10
125
 33.9 V
50
_______________________________________
15.11
(a) We have
 eff   A  B   A   B
so
180  25 B  25   B
or
155  26 B
which yields
 B  5.96
(b) We have
 B i EA  iCB
or
 1  A 
  iCA  iCB
 B 
 A 
so
5.96 1  25   iCA  20
 25 
which yields
i CA  3.23 A
_______________________________________
15.12
1/ 2
or
x n  78.9 10 4 cm  78.9  m
(Minimum collector width)
_______________________________________

1

(b) PT    I C , max  VCEQ
2


1

30    I C , max 60
2


 I C ,max  1.0 A
RL 
VCEQ
I CQ

60
 120 
0.5
VCE,max  120 V
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 15
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________

1

(c) PT    I C , max  VCEQ
2


2
30   VCEQ
2
 VCEQ  30 V
RL 
VCEQ
I CQ

30
 30 
1
VCE,max  2VCEQ  230  60 V
(d) Same as part (b)
_______________________________________
15.13
V 
(a) PT  VCEQI CQ   CC   I CQ
 2 
10  6I CQ  I CQ  1.667 A
RL 
VCEQ
I CQ

6
 3.60 
1.667
(b) I C ,max  2I CQ  21.667  3.333 A
_______________________________________
15.14
If VCC  25 V, then
V
25
I C max   CC 
 0.25 A  I C , rated
R L 100
The power
P  I C VCE  I C VCC  I C R L 
Now, to find the maximum power point
dP
 0  VCC  2 I C R L  25  I C 2100 
dI C
which yields
I C  0.125 A
So
Pmax   0.12525  0.125100
or
Pmax   1.56 W  PT
So maximum VCC is VCC  25 V
_______________________________________
15.15
V DS
ID
Power dissipated in the transistor
V2
P  I DV DS  DS
Ron
Now R on 
We have
200  V DS
ID 
100
so we can write
V2
 200  V DS 
P  
  V DS  DS
Ron
 100 
For T  25 C, Ron  2  .
Then
V2
 200  V DS 

  V DS  DS
2
 100 
which yields
V DS  3.92 V
The power is
 200  3.92 
P
3.92  7.69 W
 100

We then have
V DS (V)
R on (  )
T (C)
P (W)
25
2.0
3.92
7.69
50
2.33
4.56
8.91
75
2.67
5.19
10.1
100
3.0
5.83
11.3
_______________________________________
15.16
(a) We have, for three devices in parallel,
V V V
 
 5  V 1.51  5
1 .8 2 2 .2
or
V  3.311 V
V
Then, I  , so that
R
I 1  1.839 A
I 2  1.656 A
I 3  1.505 A
Now, P  IV , so
P1  6.09 W
P2  5.48 W
P3  4.98 W
(b) Now
1
1 
 1
V


  5  V  3.882 V
 1.8 3.6 2.2 
Then
I 1  2.157 A, P1  8.37 W
I 2  1.078 A, P2  4.19 W
I 3  1.765 A, P3  6.85 W
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 15
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
15.17
(a) Let the n-drift region doping
concentration be N d  1014 cm 3 .
15.18
(b) In the saturation region,
2
2
I D  K n VGS  VT   0.20VGS  2
V DS  V DD  I D R L  60  I D 10
  


 1014 1015 
Vbi  0.0259  ln 

2
 1.5  1010 
 0.516 V
For the base region,
For VGS  4 V, I D  0.8 A, V DS  52 V
P  I DV DS  0.852  41.6 W
For VGS  6 V, I D  3.2 A, V DS  28 V
 2  V  V R   N d 

1



x p   s bi




e

 N a  N a  N d 

For VGS  8 V, transistor biased in the
nonsaturation region.
60  V DS
2
 0.20  28  2 V DS  V DS
10

 211.7 8.85 10 14 0.516  200

1.6 10 19


 1014 
1

  15  14

15 
 10  10  10 


1/ 2
For VGS  6 V, P  PT so transistor may
be damaged.
_____________________________________

 211.7  8.85 10 14 0.516  200
xn  
1.6 10 19

15.19
1/ 2
x n  4.86 10 3 cm  48.6  m
= drift region width
(b) Assume N d  1014 cm 3
Vbi  0.516 V
 2  V  V R   N d 

1



x p   s bi




e

 N a  N a  N d 

1/ 2

 1014 
1


  15  14

15 
10
10

10






V DD
60

 20 
I D, max
3
 V   I D, max 

(b) P   DD   

 2   2 
V
V
R L  DD  10  I D, max  DD
I D, max
10
Then
1/ 2
x p  3.08  10 4 cm  3.08  m
= channel length
 211.7 8.85 10 14 0.516  80
xn  
1.6 10 19


 1015 
1


  14  14

15 
10
10

10






1
 1

(a) P    V DD   I D , max 
2
2



I
60


 
45    D   I D, max  3 A
 2  2 
RL 
 211.7 8.85 10 14 0.516  80

1.6 10 19


2
We obtain 2.0V DS
 25V DS  60  0
 P  3.245.676   18.39 W
= channel length
 1015 
1


  14  14

15 
10
10

10





 V DS  3.24 V, I D  5.676 A
x p  4.86  10 4 cm  4.86  m

P  3.228  89.6 W
1/ 2
1/ 2
x n  3.08 10 3 cm  30.8  m
= drift region width
_______________________________________
V  V 
P   DD    DD 
 2   20 
Or
V2
45  DD
40
 V DD  42.4 V
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 15
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
15.20
We have  1   2  1 . Now
1 
1
1  1
and  2 
2
1  2
so

2
1   2  1 
1
1  1 1   2
which can be written as
 1   2    2 1   1 
1 1
1   1 1   2 
or
1  1 1   2   1 1   2    2 1  1 
Expanding, we find
1  1   2  1  2
 1  1  2   2  1  2
which yields
1  2  1
_______________________________________
15.21
The reverse-biased p-well to substrate junction
corresponds to the J 2 junction in an SCR. The
photocurrent generated in this junction will be
similar to the avalanche generated current in an
SCR, which can trigger the device.
_______________________________________
15.22
Case 1: Terminal 1(+), terminal 2(-), and I G
negative: this triggering was discussed in the
text.
Case 2: Terminal 1(+), terminal 2(-), and I G
positive: the gate current enters the P2 region
directly so that J3 becomes forward biased.
Electrons are injected from N2 and diffuse into
N1, lowering the potential of N1. The junction
J2 becomes more forward biased, and the
increased current triggers the SCR so that
P2N1P1N4 turns on.
Case 3: Terminal 1(-), terminal 2(+), and I G
positive: the gate current enters the P2 region
directly so that the J3 junction becomes more
forward biased. More electrons are injected
from N2 into N1 so that J1 also becomes more
forward biased. The increased current triggers
the P1N1P2N2 device into its conducting state.
Case 4: Terminal 1(-), terminal 2(+), and I G
negative: in this case, the J4 junction becomes
forward biased. Electrons are injected from N3
and diffuse into N1. The potential of N1 is
lowered which increases the forward biased
potential of J1. This increased current then
triggers the P1N1P2N2 device into its
conducting state.
_______________________________________