Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 1
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
Chapter 1
TYU 1.2
(a) Number of atoms per (100) lattice plane
1
 4  1
4
1
1
Surface Density  2 
2
a
4.65  10 8
Exercise Solutions
Ex 1.1
(a) Number of atoms per unit cell
1
1
 8   6  4
8
2
4
4
(b) Volume Density  3 
a
4.25  10 8



3
 5.21 10 22 cm 3
_______________________________________
Ex 1.2
Intercepts of plane; p=1, q=2, s=2
1 1 1 
Inverse;  , , 
1 2 2 
Multiply by lowest common denominator,
 211 plane
_______________________________________
Ex 1.3
(a) Number of atoms per (100) plane
1
 1 4  2
4
2
2
Surface Density  2 
a
4.25  10 8


2
= 4.62  10 cm
(b) Number of atoms per (110) lattice plane
1
 4  1
4
Surface Density
1
1


a  a 2 4.65 10 8 2 2
14
  

2
 3.27  10 cm
(c) Number of atoms per (111) lattice plane
1 1
 3 
6 2
1
Lattice plane area  bh
2
14
where b  a 2
 

ha 2


2
2
2
1
 
 a 2 
2
 
1/ 2
1/ 2
Test Your Understanding Solutions
1 
3

  2a 2  a 2   a
2 
2

Then lattice plane area
 3
1
  3 a2
 a 2 a


2
2
 2
Surface Density
1
2

 2.67  1014 cm 2
2
3
4.65  10 8
2
_______________________________________
TYU 1.1
TYU 1.3
 1.11 1015 cm 2
(b) Number of atoms per (110) plane
1
1
 2  4  2
4
2
Surface Density
2
2


a  a 2 4.25 10 8
  
 

2
2
2
 7.83  10 cm
_______________________________________
14
1
Number of atoms per unit cell  8   1
8
1
Volume Density  4  10 22  3
a
o
 a  2.92  10 8 cm  2.92 A
o
Radius  r  a 2  1.46 A
_______________________________________


o
(a) For (100) planes, distance  a  4.83 A
(b) For (110) planes, distance
o
a 2 4.83 2


 3.42 A
2
2
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 1
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
TYU 1.4
(a) 8 corner atoms
(b) 6 face-centered atoms
(c) 4 atoms totally enclosed
_______________________________________
TYU 1.5
Number of atoms in the unit cell
1
1
 8  6  4  8
8
2
8
8
Volume Density  3 
a
5.43  10 8


3
 5 10 22 cm 3
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 2
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
Chapter 2
Exercise Solutions
Ex. 2.1
(a) E  h 
hc

(b) E n 
6.625 10 3 10 

34

10
100  10 8
 1.9875  10 17 J
1.9875  10 17
or E 
 124 eV
1.6  10 19
hc 6.625  10 34 3  10 10
(b) E 


4500  10 8
 4.417  10 19 J
4.417  10 19
or E 
 2.76 eV
1.6  10 19
_______________________________________



 2 2 n 2
2ma 2
1.054 10   n
21.67  10 12  10 
34 2
 27
2
2
10 2
 2.28  10 23 n 2 J
2.27967  10 23 n 2
or E n 
1.6  10 19
 1.425  10 4 n 2 eV
Then E1  1.425  10 4 eV
E 2  5.70  10 4 eV
E 3  1.28  10 3 eV
_______________________________________
Ex 2.4
Ex 2.2



 2 9.11 10 31 12  10 3 1.6 10 19

1/ 2
 5.915  10 26 kg-m/s
h 6.625  10 34
 
 1.12  10 8 m
p 5.915  10  26
o
or   112 A
h 6.625  10 34
(c) p  

112  10 10
 5.915  10 26 kg-m/s


26 2
1 p 2 1 5.915  10

2 m 2 2.2  10 31
= 7.952  10 21 J
7.952  10 21
or E 
 4.97  10  2 eV
1.6  10 19
_______________________________________
E
Ex 2.3
1.054 10   n
9.1110 12 10 
34 2
2
 

29.1110 24.555 10 
31
21
1/ 2
1.054 10 34
or
k 2  1.222  10 9 m 1
P  exp 2k 2 d 
o
(a) d  10 A  10  10 10 m
P  exp  2  1.222  10 9 10  10 10
or
P  0.0868  8.68 %
 


o
 2 2 n 2
(a) E n 
2ma 2


2
1
1
m 2  9.11 10 31 10 5
2
2
 4.555  10 21 J
Now
2m
Vo  E  Set Vo  3E
k2 
2
Then
1
k2 
2m2 E 

E
(a) p  2mE
 31
2
2
10 2
 4.179  10 20 n 2 J
4.179  10 20 n 2
or E n 
 0.261n 2 eV
1.6  10 19
Then
E1  0.261 eV, E 2  1.045 eV, E 3  2.351 eV
(b) d  100 A  100 10 10 m
P  exp  2  1.222  10 9 100  10 10
or
P  2.43  10 11  2.43  10 9 %
 


_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 2
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
Ex 2.5

2mVO  E 
(a) k 2 

TYU 2.2
(a) E  0.8 1.6  10 19  1.28  10 19 eV




2 9.11 10 31 1.2  0.12  1.6  10 19
1.054 10 
t 

 34 2
 5.3236  10 m 1
Then
 0.12  0.12 
T  16

1 
1.2 
 1.2 

1.054  10 34


 8.23  10 16 s
E 1.28  10 19
(b) Same as part (a), t  8.23  10 16 s
_______________________________________
2
9
 
 exp  2 5.3236  10
TYU 2.3
(a) k 2 
5 10 
10
9
3
T  7.02  10
 0.12  0.12 
(b) T  16

1 
1.2 
 1.2 
 

2mVO  E 

2









2 9.1110 31 0.8  0.1 1.6  10 19
1.054 10
 34 2
= 4.286  10 m 1
 0.1  0.1 
T  16
1 

 0.8  0.8 
9

 exp  2 5.3236  10 9 25  10 10

T  3.97  10 12
_______________________________________
Ex 2.6
From Example 2.6, we have
13.58
0.0992
En 
eV

2 2
n2
11.7  n
E1  99.2 meV, E 2  24.8 meV,
E 3  11.0 meV
_______________________________________
 
 exp  2 4.2859  10 9 12  10 10
T  5.97  10
(b) k 2 
5


2 9.11 10 31 1.5  0.1 1.6  10 19
1.054 10
 6.061 10 m 1
 0.1  0.1 
T  16
1 

 1.5  1.5 
 34 2
9
 

 exp  2 6.061 10 9 12  10 10

7
T  4.79  10
_______________________________________
Test Your Understanding
TYU 2.1
TYU 2.4
T  5  10 6
 1.054 10 34
(a) p 

x
8  10 10
 1.318  10 25 kg-m/s
 d  p 2 
dE
  p
(b) E 
 p   

dp
 dp  2m 
2p
pp
 p 
2m
m
23
1.2  10
1.318  10 25
E 
9.11 10 31
 1.735  10 18 J or  10.85 eV
_______________________________________




 0.08  0.08 
 16
1 
 exp 2k 2 a 
0.8 
 0.8 
so that exp 2k 2 a   2.88  10 5
2k 2 a  12.571
k2 



2 9.1110 31 0.8  0.08 1.6  10 19
1.054 10 

 34 2
 4.3467  10 9 m 1
Then
12.571
a
 1.446  10 9 m
2 4.3467  10 9


o
or a  14.46 A
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 3
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
Chapter 3
Ex 3.3
Exercise Solutions
4 2m 
h3
3 / 2 2 eV
(a) N 
Ex 3.1
1
m 2
2
E
1
dE  2 m  d  m  d
2




4 2m 
h3
or E 2  7.958  10 19 J.
At ka  2 , we see that  3 a  2 so
2mE 3
2
or
E3
 a  2
or
N  1.28  10 22 cm 3
(b) N 
4 2m 
h3

4.5 10 

Then
E  E 3  E 2

19


4 2m p


3/ 2
E  E  dE
h3

3/ 2
h3
2
3/ 2

   E  E 
 3 

4 20.56  9.11 10 31
E
E  kT

3/ 2
6.625 10 
2
 0.02591.6 10 

 1
3
 34 3

3/ 2

 7.92  10 24 m 3
19
3.929  10
 2.46 eV
1.6  10 19
_______________________________________
3/ 2
19
Or
E 

4 2m p
E  kT
10 2
 1.189  10  7.958  10
 3.929  10 19 J


N  8.29  10 21 cm 3
_______________________________________
J
18

or
2 2 1.054 10 34 2
 1.189  10
1eV
 8.29  10 27 m 3
E
18
2 eV
 1.06286  10
2
 2 3 / 2  13 / 2 1.6  10 19
3
2ma 2

2
  E 3/ 2
3
56
N
2 9.11 10
3/ 2
 1.28  10 28 m 3
Ex 3.4
 31
3/ 2
19
2 2  2


31
 34 3
3/ 2
Ex 3.2
At ka   , we have
sin a
1  8
 cos a
a
From Example 3.2, we have  2 a  5.141 ,
0


6.625 10 
2
 21.6  10 
3

 
 1.76  10 6 m/s
or   1.76  10 4 cm/s
_______________________________________
2 eV
2
  E3/ 2
3
4 2 9.11 10
E 10 12 1.6  10 19

m
9.11 10 31 10 5
 
E  dE
0
3/ 2

So
or

or
N  7.92  10 18 cm 3
_______________________________________
Ex 3.5
gi !
1098!  45
10!


N i ! g i  N i ! 8! 10  8! 8!2 1
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 3
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
Ex 3.6
(a)
  E  E F  
exp 
  0.02
kT


 1 
E  E F  kT ln
  3.9kT
 0.02 
_______________________________________
 7.26  10 6
  E  E F  
f F E   exp 

kT


Test Your Understanding Solutions
  0.30  0.0259 4  
 exp 

0.0259


(b)
Then
  E  E F  
f F E   exp 

kT


  E c  kT 4  E F  
 exp 

kT


  0.30  0.0259  
 exp 

0.0259


TYU 3.1
At ka  2 , we see that  3 a  2 , so
2mE 3
 3.43  10
_______________________________________
or
E3 
Ex 3.7
  E  E F  
f F E   exp 

kT





0
.
30
0
.025 

8  10  6  exp 

kT



  0.325 
5
exp 
  1.25  10
kT


0.325
 ln 1.25 10 5  11.736
kT
0.325
 T 
kT 
 0.02769  0.0259

11.736
 300 


1
 E  EF 
1  exp 

 kT 
 E  EF
  E  E F   
exp 
 1  exp  kT
kT



2ma 2

2 2 1.054 10 34 2

2 9.11 10 31 4.5  10 10

2
2   4 a  3 . Then from
sin  4 a
 cos  4 a
4a
we find, by trial and error,
 4 a  7.870 . Then
1  8
E4 
T  321 K
_______________________________________
1
 E  EF 
1  exp 

 kT 
2 2  2
 1.189  10 18 J
At the other point,  4 a is in the range
so
Ex 3.8
  E  E F  
exp 

kT


 a  2
2
6

7.8702  2
2ma 2
7.8702 1.054 10 34 2


2 9.11 10 31 4.5  10 10

2
 1.8649  10 18 J
Then
E g  E4  E3
 1.8649  10 18  1.189  10 18
 6.762  10 19 J
 0.02
or

   1  0.02

6.762  10 19
 4.23 eV
1.6  10 19
_______________________________________
Eg 
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 3
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
TYU 3.2
From Example 3.2, for ka   ,   1 a  
and E1  2.972  10 19 J.
For 0  a   and ka  0 , we have
sin a
1  8
 cos a
a
By trial and error, a  2.529 rad.
Then
2mE
 a  2.529
2
2.5292  2
E
2ma
2


2 9.11 10
or
m p
mo
 38
 0.2985
_______________________________________
10 2
 1.9258  10 J
Then
E  2.972 10 19  1.9258 10 19
 1.046  10 19 J
or
1.046  10 19
E 
 0.654 eV
1.6  10 19
_______________________________________
TYU3.5
1  f F E   1 
1
 E  EF 
1  exp 

 kT 
  E F  E  
 exp 

kT


  0.35  0.0259 2  
(a) 1  f F E   exp 

0.0259


 8.20  10 7
  0.35  30.0259 2  
(b) 1  f F E   exp 

0.0259


 3.02  10 7
_______________________________________
TYU 3.3
We have E  E c  C1 k 2
E c  0.32  E c 1.6 10 19 



 C1 

10
 10  10 
2
so that C1  5.1876  10 39
We have
m
2
2
m 


2C1
m o 2 m o C1
TYU 3.6
 400 
We find kT  0.0259
  0.034533 eV
 300 
  E c  kT 4  E F  
(a) f F E   exp 

kT


  0.30  0.034533 4  
 exp 

0.034533


1.054 10 
29.11 10 5.1876  10 
34 2
 39
(b)
or

m
 1.175
mo
_______________________________________
TYU 3.4
We have E  E  C 2 k 2
E  0.875  E 1.6 10 19 



 C 2 

10
 12  10 
so that C 2  2.0426  10 38
2
2m o C 2
 31
4.5 10 
 31
mo

34 2
19

m p
1.054 10 

29.11 10 2.0426  10 
2.5292 1.054 10 34 2
 31
We have
2
 1.31 10 4
  0.30  0.034533 
f F E   exp 

0.034533


 6.21 10 5
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 3
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
TYU 3.7
 400 
We find kT  0.0259
  0.034533
 300 
  0.35  0.034533 2  
(a) 1  f F E   exp 

0.034533


 2.41 10 5
  0.35  30.034533 2  
(b) 1  f F E   exp 

0.034533


 8.85  10 6
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 4
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
Chapter 4
Exercise Solutions
Ex 4.1
n i  3.29  10 9 cm 3
For T  250 K,
  E  E F  
f F  exp 

kT


  E c  kT  E F  
 exp 

kT



or
n i  7.13  10 3 cm 3
(b)

n o  3.02  1013 cm 3
_______________________________________
n i 400 3.288  10 9

 4.6110 5
n i 250 7.135  10 3
_______________________________________
Ex 4.4
(a) GaAs
Ex 4.2

(a) N   1.04  10
19

 250 


 300 
E Fi  E midgap 
3/ 2
 7.9115  1018 cm 3
 250 
kT  0.0259 
  0.021583 eV
 300 
 m p
3
kT ln 
 mn
4





3
0.0259 ln 0.48 
4
 0.067 
 38.25 meV

(b) Ge
  E F  E  
p o  N  exp 

kT


 0.27 

 7.9115  1018 exp 

 0.021583 
3
0.0259 ln 0.37 
4
 0.55 
 7.70 meV
_______________________________________
p o  2.919  1013 cm 3
Ex 4.5


2.919  10 13
(b) Ratio 
 4.54  10 3
15
6.43  10
_______________________________________
Ex 4.3
(a) For T  400 K,



 400 
n i2  4.7  10 17 7  10 18 

 300 
3


 1.42
 exp 

 0.0259400 300 
 1.081 10 19
or
3
 5.09  10 7
5
  E c  E F  
n o  N c exp 

kT


0
.
25



 4.7  1017 exp 

 0.0259 




 1.42
 exp 

 0.0259 250 300  
  0.25  0.0259 
 exp 

0.0259


f F  2.36  10

 250 
ni2  4.7  1017 7 1018 

 300 
E Fi  E midgap 
  E F  E  
p o  N  exp 

kT


  0.215 
 1.04  1019 exp 

 0.0259 


 2.58  1015 cm 3
We find E c  E F  1.12  0.215  0.905 eV
  E c  E F  
n o  N c exp 

kT



0
.
905


 2.8  1019 exp 

 0.0259 


 1.87 10 4 cm 3
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 4
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
Ex 4.6
no 
2

pa

po  pa
N c F1 / 2  F 
2
1.5  10 20 

2.8 10 F  
F
So F1 / 2  F   4.748
E  Ec
From Figure 4.10,  F  3.2  F
kT
 E F  E c  3.20.0259  0.08288 eV
_______________________________________

19

 250 


 300 
3/ 2
 2.13  10 cm
19
3
 250 
kT  0.0259
  0.021583 eV
 300 
nd
1

no  nd
Nc
  E c  E d  
1
exp 

2N d
kT



1

 200 
N c  2.8  10 19 

 300 
3/ 2
 
 5.661 1018 cm 3

 
 8.736  10 2
_______________________________________

 1.524  1019 cm 3
 200 
kT  0.0259
  0.017267 eV
 300 
nd
1

no  nd
1.524  1019
  0.045 
1
exp 

16
2 10
 0.017267 

3/ 2
 200 
kT  0.0259
  0.017267 eV
 300 
pa
1

18
po  pa
5.661 10
  0.045 
1
exp 

16
4 10
 0.017267 
no 
 7.50  10 3
(b) T  200 K

 1.75  10 2
(c) Fraction increases as temperature
decreases.
_______________________________________
Ex 4.8
(a) T  250 K


n i 400   2.38  10 12 cm 3
Then
(a) T  250 K
16
N   1.04  10

 200 
N   1.04  1019 

 300 
19
19
16
Ex 4.9
From Example 4.3,
ni 250  7.0  10 7 cm 3
2.13 10  exp  0.045 
1
 0.02158 
210 



7.912 10  exp  0.045 
 0.021583 
410 



Ex 4.7
(a) T  250 K
N c  2.8  10
1
 3.91 10 2
(b) T  200 K
19
1/ 2
1
18
Nd  Na
 N  Na
  d
2
2

2

  n i2

7  1015  3  1015
2
 7  10 15  3  10 15
 
2

n o  4  1015 cm 3

2

  7  10 7




2

2
n i2
7  10 7

 1.225 cm 3
no
4  10 15
(b) T  400 K
7  1015  3  1015
no 
2
po 
 7  10 15  3  10 15
 
2

2

  2.38  1012




2
n o  4  10 15 cm 3

 250 


 300 
3/ 2
 7.912  10 cm
 250 
kT  0.0259
  0.021583 eV
 300 
18
3
2.38 10 
12 2
 1.416  10 9 cm 3
4  10 15
_______________________________________
po 
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 4
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
Ex 4.10
For T  250 K,
po 



 250 
n i2  1.04  10 19 6  10 18 

 300 
3
 4  1016  8  1015
 
2



 0.66
 exp 

 0.0259 250 300  
p o  3.2  10 16 cm 3
 1.894  10 24
or n i  1.376  10 12 cm 3
For T  350 K,

no 


 350 
n i2  1.04  10 19 6  10 18 

 300 
3

 2  10 14
 
 2
2



2
 9.47  10 cm
3

2
n o  3.059  1014 cm 3
1.80 10 

14 2
3
 1.059  10 cm
3.059  1014
_______________________________________
po
14
Ex 4.11
(a) p o 
Na  Nd
2
 N  Nd
  a
2

2

  n i2

2




Ex 4.13
9


 2.8  1019 

 0.0259  ln
15 
 3  10 
E c  E F  0.2368 eV
_______________________________________
2

  1.80  10 14


2
n i2
1.5  1010
 7.03  10 3 cm 3

po
3.2 1016
N
E c  E F  kT ln c
 no

  1.376  10 12


n2
1.376  10 12
po  i 
no
2  1014
(b) T  350 K
2  10 14
no 
2

n o  3 10 15 cm 3
 3.236  10
or n i  1.80  10 14 cm 3
(a) T  250 K
2  10 14
no 
2
2

Ex 4.12
n o  N d  N a  8  10 15  5  10 15
28
n o  2 10 14 cm 3

2

  1.5  1010


(b) p o  n o  n i  1.5  10 10 cm 3
_______________________________________


 0.66
 exp 

 0.0259  350 300 
 2  10 14
 
 2
4  1016  8  1015
2
N 
E c  E F  kT ln c 
 no 
We have E c  E F  E c  E d   E d  E F 
 0.05  3kT  0.05  30.0259
 0.1277 eV
N 
So 0.1277  0.0259  ln c 
 no 
Nc
Or
 exp4.9305  138.45
no
Then n o 
Nc
2.8  1019

138.45
138.45
n o  2.02  1017 cm 3
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 4
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
Test Your Understanding Solutions
n i 450 1.7224  1013

 2.26  10 8
n i 200 7.6334  10 4
_______________________________________
TYU 4.1
(c)
  E c  E F  
n o  N c exp 

kT


  0.22 
 2.8  1019 exp

 0.0259 
TYU 4.4

 5.73  10 cm
Now
E F  E  1.12  0.22  0.90 eV

3


 1.42
 exp 

 0.0259450 300 
n i 450 3.853  1010

 2.81 1010
1.369
n i 200
_______________________________________
 6.53  1013 cm 3
(c)
Now
E c  E F  1.42  0.30  1.12 eV
  1.12 
n o  4.7  1017 exp

 0.0259 
TYU 4.5
 0.0779 cm 3
_______________________________________


 200 
(a) n i2  2.8  1019 1.04  1019 

 300 
3
 4.639  10 20
 5.827  10 9
 n i  7.63  10 4 cm 3




 0.66
 exp 

 0.0259200 300 
3


 1.12
 exp 

 0.0259200 300 
 450 
(b) n i2  2.8  1019 1.04  1019 

 300 

 200 
(a) n i2  1.04  1019 6.0  1018 

 300 
TYU 4.3
 n i  2.15  10 10 cm 3



 450 
(b) n i2  1.04  10 19 6.0  1018 

 300 
3


 0.66
 exp 




0
.
0259
450
300


3


 1.12
 exp 




 0.0259 450 300 
 2.967  10 26
 n i  1.72  10 13 cm 3

 1.485  10 21
 n i  3.85  10 10 cm 3
  0.30 
p o  7.0  1018 exp

 0.0259 


 450 
(b) n i2  4.7  1017 7.0  1018 

 300 
TYU 4.2

3
 1.874
 n i  1.37 cm 3
 8.43  10 3 cm 3
_______________________________________




 1.42
 exp 

 0.0259200 300 
  E F  E  
p o  N  exp 

kT



0
.
90


 1.04  1019 exp

 0.0259 


 200 
(a) n i2  4.7  1017 7.0  1018 

 300 
3
15
 8.820  10 30
 n i  2.97  10 15 cm 3
n i 450 2.9699  1015

 1.38  10 5
n i 200 2.1539  1010
_______________________________________
(c)
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 4
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
TYU 4.6
 m p
3
 kT ln 
 mn
4

E Fi  E midgap
(b) p o 




3
0.0259 200  ln 0.56 
4
 300   1.08 
 8.505 meV
3
 400   0.56 
(b) E Fi  E midgap  0.0259
 ln

4
 300   1.08 
 17.01 meV
_______________________________________
po 

TYU 4.9
pa

po  pa
31
12




1
2


1

4
12
TYU 4.8
2

N c F1 / 2  F 
For E c  E F ,  F  0 and F1 / 2 0   0.65
no 
2

2.8 10 0.65
19
n o  2.05  10 19 cm 3
  0.045 
exp

 0.0259 
3/ 2
For T  100 K, N c  5.389  1018 cm 3
2
 1.0184  10 21 J
or E  6.37 meV
Also
r1 116

 133.3
0.12
ao
_______________________________________
(a) n o 

 T 
N c  2.8  1019 

 300 
 4 16.08.85 10 
 34 2
 
TYU 4.10
We have
2 
 1
 3


 1.64 0.082 
 0.12 9.11 10 31 1.6  10 19
1.04  10
4 1017
pa
 0.179
po  pa
_______________________________________
19 4
2
 1
m ce
2
 3

mo
 m1 m 2
 0.120
2 1.054  10
1
19
or
1.6 10 
 4 13.18.85 10 
 0.067  9.11 10

1
N
   E a  E  
1   exp 

kT
4N a


1
 8.4823  10 22 J
or E  5.30 meV
Also
r1
 m  113.1
r  o  
 195.5
0.067
ao
m 
(b) Ge:
Conductivity effective mass
E
1.04 10 0.65
19


 m 
m o e 4
 

m
 o
E
2
2 2 4 
2 1.054  10 34
2
p o  7.63  1018 cm 3
_______________________________________
TYU 4.7
(a) GaAs:


For E  E F ,  F  0 and F1 / 2 0   0.65
(a) E Fi  E midgap 

N  F1 / 2  F 
2
T  200 K, N c  1.524  10 19 cm 3
T  300 K, N c  2.8  10 19 cm 3
T  400 K, N c  4.311 10 19 cm 3
 T 
kT  0.0259

 300 
For T  100 K, kT  0.008633 eV
T  200 K, kT  0.01727 eV
T  300 K, kT  0.0259 eV
T  400 K, kT  0.03453 eV
Fraction of ionized impurity atoms
nd
 1
no  nd
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 4
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
(a) For T  100 K,
nd
1

no  nd
N
  E c  E d  
1  c exp 

2N d
kT


1

5.389  1018
  0.045 
1
exp

2 1015
 0.008633 
 0.0638
Now, percentage ionized atoms
 1  0.0638  100  93.62 %
(b) For T  200 K,
nd
1

no  nd
Nc
  E c  E d  
1
exp 

2N d
kT


 
1

1.524  1019
  0.045 
exp

15
2 10
 0.01727 
 0.001774
Now, percentage ionized atoms
 1  0.001774 100  99.82 %
(c) For T  300 K,
nd
1

no  nd
Nc
  E c  E d  
1
exp 

2N d
kT


1
 
1
2.8  1019
  0.045 
1
exp

15
2 10
 0.0259 
 0.000406
Now, percentage ionized atoms
 1  0.000406 100  99.96 %
(d) For T  400 K,
nd
1

no  nd
N
  E c  E d  
1  c exp 

2N d
kT



 
1

4.311 10
  0.045 
exp

2 1015
 0.03453 
 0.000171
Now, percentage ionized atoms
 1  0.000171  100  99.98 %
_______________________________________
1
19
 
TYU 4.11
p o  N a  N d  2  1016  5  10 15
 1.5  1016 cm 3
Then
no 


ni2
1.8  10 6

po
1.5  1016
2
 2.16  10 4 cm 3
_______________________________________
TYU 4.12
(b) n 
Nd
N
  d
2
 2
2

  ni2

Then
5 10 
14 2
1.1 1015  5  1014 
 n i2
which yields
n i2  1.1 10 29
Now
  Eg
ni2  N c N  exp
 kT







 T 
1.1 10 29  2.8  1019 1.04  1019 

 300 
3


 1.12
 exp 

 0.0259 T 300  
By trial and error,
T  552 K
_______________________________________
TYU 4.13
At T  550 K,



 550 
ni2  2.8  1019 1.04  1019 

 300 
3


 1.12
 exp 

 0.0259550 300 
 1.0236  10 29
or n i  3.20  10 14 cm 3
N
N
n o  d   d
2
 2
Set n o  1.05 N d
2

  ni2

Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 4
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
Then
2
1.05  0.5N d    N d   ni2
 2 
0.22913N d  ni
2
3.20  1014
 1.40  1015 cm 3
0.22913
_______________________________________
or N d 
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 5
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
Chapter 5
Exercise Solutions
Ex 5.1
Ex 5.5
J drf  e p p o   e p N a 


75  1.6  10 19 480N a 120
which yields N a  8.14  1015 cm 3
_______________________________________
Ex 5.2 Using Figure 5.2;
(a) T  25 C,
(i) N a  1016 cm 3 ,   p  410 cm 2 /V-s
(ii) N a  1018 cm 3 ,   p  130 cm 2 /V-s
(i) T  0 C,   p  550 cm 2 /V-s
(ii) T  100 C,   p  300 cm 2 /V-s
_______________________________________
Ex 5.3
(a) For N I  N a  N d  2.8  10 17  8  10 16
 3.6  1017 cm 3 ,
  p  200 cm 2 /V-s
(b)   e p N a  N d 



 1.6  10 19 200  2  10 17
  6.4 (  -cm)
1

1
1
 0.156  -cm
 6.4
_______________________________________
(c)  


 
 

3
(b) N a  10 cm ,
14
dp
dx
d
x L
 eD p
1016 e p
dx
 1  x Lp
e
 eD p 10 16 
 Lp 


16
 eD p 10
x L

e p
Lp
J p  eD p
1.6 10 810  e
19
16
x Lp
4
2  10
x

J p  64 exp
 Lp 


(a) For x  0 ,
J p  64 A/cm 2
(b) For x  2 10 4 cm,
  2  10 4
J p  64 exp
4
 2  10

  23.54 A/cm 2


(c) For x  10 3 cm,
  10 3 
  0.431 A/cm 2
J p  64 exp
4 

2
10


_______________________________________

Ex 5.4
5
V

 2500 
I 2  10 3
RA 2500  10 6
(b)  

 2.083  -cm
L
1.2  10 3
1
1
(c)   
 0.480 (  -cm) 1
 2.083
 e p N a
(a) R 


0.48
 3.00  1018
1.6  10 19
Using Figure 5.3 and trial and error,
N a  7.3  10 15 cm 3
Then  p N a 
(d)  p  410 cm 2 /V-s
_______________________________________
Ex 5.6
dN d  x 
10 16  x L

e
dx
L
So
  10 16   x L

e


 kT   L 
 x  

16  x L
 e  10 e
 
 kT  1  0.0259

  
2
 e  L  2  10
or  x  1.295 V/cm
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 5
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
(b)   e n N d  N a 
Ex 5.7
 1.6  10 19 1000  3  10 16
Dn
215
n 

or
 kT  0.0259


  4.8 (  -cm) 1
 e 
We find
 n  8301 cm 2 /V-s
1
1
 0.208 (  -cm)
 
_______________________________________
 4.8
_______________________________________
Ex 5.8
From Equation (5.59),
TYU 5.3
 p epV xWd 
1
Ix 
  e n N d 
L

320 1.6 10 19 1016 10 10 2 8 10 4
So

1
0.2
1.6  10 19  n N d 
 10
0.1
I x  2.048  10 4 A




 

  


or I x  0.2048 mA
From Equation (5.53),
I B
2.048  10 4 5  10 2
VH  x z 
epd
1.6  10 19 10 22 8  10  6







4
 8  10 V
or V H  0.80 mV
_______________________________________


 n N d  6.25  1019
Using Figure 5.3 and trial and error,
N d  6  10 16 cm 3 and
 n  1050 cm 2 /V-s
_______________________________________
TYU 5.4
J diff  eDn
 1015
dn
 eDn   4
dx
 10
x


 exp
L 

 n 

We have D n  25 cm 2 /s
Ln  10 4 cm  1  m
Test Your Understanding Solutions
Then
TYU 5.1
n o  N d  N a  1015  1014  9  1014 cm
Also


2
n2
1.5  1010
po  i 
 2.5  10 5 cm 3
14
no
9  10
Now
J drf  e  n n o   p p o   e n n o 



 1.6  10
19
13509 10 35
(a) x  0 , J diff  40 A/cm 2
(b) x  1  m, J diff  14.7 A/cm 2
(c) x   , J diff  0
_______________________________________
14
or
TYU 5.5
J diff  eD p
J drf  6.80 A/cm 2
_______________________________________
dp
dx
So

 0 0p.010
20   1.6  10 19 10
TYU 5.2
(a) For N I  7  1016 cm 3 ,
 n  1000 cm /V-s;  p  350 cm /V-s
2
x
2
J diff  40 exp
 A/cm
 1 
2
 p  1.25  1017  4  1017  p
And
p x  0.01  2.75  10 17 cm 3
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 6
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
Chapter 6
Exercise Solutions
Ex 6.4
Ex 6.1
R n 
n 1015 e t / 10

 no
10  6
R n 
For t  0 ,
t  1  s,
6
cm 3 s 1
15
10
 10 21 cm 3 s 1
10  6
10 15 e 1 / 1
 3.68  10 20 cm 3 s 1
10  6
t  4  s,
R n 
1015 e 4 / 1
 1.83  1019 cm 3 s 1
10  6
t  10  s,
R n 
1015 e 10 / 1
 4.54  10 16 cm 3 s 1
10  6
_______________________________________
R n 
Ex 6.2
(a) 10 14 e t / 50  1014 e 1
 t  50 ns
(b) 1014 e t / 50  1013
 1014 
t  50  10 9 ln 13   1.15  10  7 s
 10 
or t  115 ns
_______________________________________


Ex 6.3

(a) pt   g  po 1  e

 t /  po




 5  10 1  e

 t /  po
(i) p 0   5  10 1  e
14

(ii) p 10

7
0
 t /  po


  5 10 1  e 
(iii) p 5  10
7

1 / 1
 3.16  1014 cm 3
 5  10 14 1  e 5 / 1


 4.966  1014 cm 3
(iv) p    5  10 14 1  e 

1/ 2
 3.536  10 3 cm
or L  35.36  m
(b) n  1015 e  x / Ln ( x  0 )
or n  1015 e  x / Ln ( x  0 )
(i) n  1015 e 0  1015 cm 3
(ii) n  1015 e 30 / 35.36  4.28 1014 cm 3
(iii) n  1015 e 50 / 35.36  2.43  1014 cm 3
(iv) n  1015 e 85 / 35.36  9.04 1013 cm 3
(v) n  1015 e 120 / 35.36  3.36  1013 cm 3
_______________________________________
Ex 6.5


  x   p 0t 2 


exp
1/ 2
4D p t
4D p t


(a)  p  0 t  400100  10 7  4  10 3 cm
px, t  

e
t /  p 0



or  40  m
(i) x  20  m.


   2  10 3
0.36788
exp

6
3.545  10 3
 4  10
 38.18
(ii) x  40  m,
p 



2
0.36788
exp0
3.545  10 3
 103.8
(iii) x  60  m,

 5 1014 cm 3
(b) p max   5  1014  0.01N d
Yes, low-injection condition is met.
_______________________________________




  2  10 3
0.36788
exp

6
3.545 10 3
 4  10
p  38.18
(b) x  40  m
p 
 0
14

p 
 5  10 21 10 7 1  e
14
 
(a) Ln  D n no  25 5 10 7
(i) t  5  10 8 s,
  2  10 3
0.60653
exp 
3
6
2.50663  10
 2  10
 32.75
p 
2
2






Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 6
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
(ii) t  10 7 s,
0.36788
p 
exp0
3.545  10 3
 103.8
(iii) t  2  10 7 s,
   4 10 3 2 
0.1353
p 
exp


6
5.013 10 3

 8  10
 3.65
_______________________________________

Figure 5.3,  n  7500 cm 2 /V-s.
  e n N d  1.6  10 19 75005  1015 
1

 1.93  10
13

(b) For N a  2  10 cm
3
in silicon, from
  e p N a  1.6  10 19 4002  1016 
1
 1.28 (  -cm)
 11.7  8.85  10 14
Then  d  

1.28




10 2
n
1.5  10

 3.214  10 4 cm 3
po
7 1015
(a) In thermal equilibrium,
p 
E Fi  E F  kT ln o 
 ni 
 7  1015
 0.0259 ln
10
 1.5  10
 0.33808 eV

n

 n  p o  p   n i2
 po n o  n  n i    no  p o  p  n i 
o
  

 5  10 10  10   5  10 10 
 1015  1014 1014  1.5 1010
7
15
2
7
14
14
R  1.83  10 20 cm 3 s 1
_______________________________________
Ex 6.7
p o  N a  N d  10 16  3  1015
no 
R
or
 8.09 10 13 s
or  d  0.809 ps
_______________________________________
2
i
Ex 6.8
n-type; n o  10 15 cm 3 , p o  2.25  10 5 cm 3
We have
Figure 5.3,  p  400 cm 2 /V-s.
 7  1015 cm 3
 3.214  10 4  4  1014 

 0.0259 ln

1.5 1010


 0.26395 eV
_______________________________________
 no   po  5  10 7 s
s
16




n  p  1014 cm 3 ,
 d  0.193 ps
or
 7  1015  4 1014
 0.0259  ln
1.5  1010

 0.33952 eV
 n  n 

E Fn  E Fi  kT ln o

 ni 

Ex 6.6
(a) For N d  5 1015 cm 3 in GaAs, from
 6 (  -cm)
 13.1 8.85  10 14
Then  d  

6
(b) Quasi-Fermi levels,
 p  p 

E Fi  E Fp  kT ln o

 ni 




Ex 6.9
  p0s 
0
(a) For  p 0 s  0  p s  p B 
  p0 


From Equation (6.109),
 x / Lp
x / Lp
px   g  p 0  Ae
 Be
As x  , p  g  p 0  1014 cm 3
 A0
As x  0, p  0  B   g  p 0

Then px   g  p 0 1  e
(b) px  0  0
(c) R  
x / Lp

p0 p0

 R  
 p0s
0
Note: p0  0 is a result of R    .
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 6
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
Ex 6.10
TYU 6.4
n-type; Minority carriers = holes
dp
d px 
J diff  eD p
 eD p
dx
dx
19
15
1.6  10 10 10
  10 
exp


4
 (31.6  10 )
 31.6 
Then hole diffusion current density
J diff  0.369 A/cm 2
 x / Lp


se

(a) p x   g  p 0 1 

D p L p  s 

(i) For s   ,


px   g  p 0 1  e

x / Lp


(ii) For s  0 ,
px   g  p 0
(b)
(i) For s   , p0  0
(ii) For s  0 , p0  g  p 0 ,
We have
J diff (electrons)   J diff (holes)
 p x   constant
_______________________________________
Test Your Understanding Solutions
TYU 6.1
(a) p-type; Minority carriers = electrons
 t 

(b) nt   n0 exp

  no 
Then
 t 
3
nt   1015 exp
 cm
6
 5  10 
_______________________________________



 t
 10 20 5  10  6 1  exp
6
 5  10


 t
or nt   5  1014 1  exp
6
 5  10







TYU 6.3
x

 Lp 


nx   px   n0  exp
1010 6   31.6 10 4 cm
Then
x

3
 cm
4
 31.6  10 
_______________________________________

nx   px   1015 exp
_______________________________________
TYU 6.5
p 

exp  t  po
4 D t 

1/ 2
p
(a)
(b)
(c)
Now
(c) As t   , n   5 1014 cm 3
_______________________________________
L p  D p po 
Then electron diffusion current density
J diff  0.369 A/cm 2
(d)
TYU 6.2
(a) p-type; Minority carriers = electrons

  t 

(b) nt   g  no 1  exp


  no 

  
exp 1 5
 p  73.0
4 1010 
6
1/ 2
exp 5 5
4 105 10 
6
1/ 2
 p  14.7
exp 15 5
4 1015 10 
 p  1.15
4 1025 10 
 p  0.120
6
1/ 2
exp 25 5
6
1/ 2
x   p  o t  386 10  t
Then
(a) x  38.6  m
(b) x  193  m
(c) x  579  m
(d) x  965  m
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 6
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
(c) (i) x   p  o t
TYU 6.6
Using the results from TYU 6.5, we find
  x   pot 2 
exp  t  p



p 
exp
1/ 2
4D p t
4 D p t


(a) (i) x   p  o t






 1.093  10 2  386 10  10 6
 7.07  10 3

exp 1 5




1/ 2







 3.21 10 3  386 10  10 6
 7.07  10 3
   7.07  10 3
p  73.0 exp 
6
 410 10
or
p  20.9
(b) (i) x   p  o t




2






 1.22  10 2  386 10  5  10 6
3
 7.1 10
   7.1 10 3 2 
p  14.7 exp 

6
 410 5  10

or
p  11.4










 5.08  10 2  386 10  15  10 6

3


   7.1 10 3 2 

6
 410 15  10 
p  1.15 exp 
or








p  1.06
_______________________________________
3


p  1.06
(ii) x   p  o t
 7.1 10

 2.64  10 2  38610 5  10 6
 7.1 10
  7.1 10 3 2
p  14.7 exp 
6
 410 5  10
or
p  11.4
(ii) x   p  o t




3
  7.1 10 3 2
p  1.15 exp 
6
 410 15  10
or
  7.07  10
 exp 
6
 410 10
4 10 10  6
  7.07  10 3 2 
 73.0 exp 

6
 410 10

or
p  20.9
(ii) x   p  o t
p 
 7.1 10

3 2

 6.50  10 2  386 10  15  10 6



Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 7
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
Chapter 7
Exercise Solutions

Ex 7.1
N N
(a) Vbi  Vt ln a 2 d
 ni





 

 5  1015 1017 
(i) Vbi  0.0259 ln 

2

 1.5  1010
 0.736 V
 2  1016 2  1015 
(ii) Vbi  0.0259 ln 

2
 1.5  1010

 0.671 V
(b)
 5  1015 1017 
(i) Vbi  0.0259  ln 

2
 1.8  10 6

 1.20 V
 2 1016 2 1015 
(ii) Vbi  0.0259 ln 

2


1.8 10 6
 1.14 V
_______________________________________



















 5  10 16 5  10 15
 0.0259  ln 
2
 1.5  10 10
 0.7184 V




 4.52  10 5 cm
Now
 max 

1

 5  1015  5  10 16


eN d x n
s
1.6 10 5 10 4.1110 
11.7 8.85 10 
19
6
16
14
 3.18  10 4 V/cm
_______________________________________
Ex 7.3
N N
(a) Vbi  Vt ln a 2 d
 ni






 5  1015
 
16
 5  10


 x n  4.11 10 6 cm
Now
W  x n  x p  4.11 10 6  4.11 10 5








1


 N  N 
d 
 a
1/ 2
 211.7  8.85  10 14 0.7184  4

1.6  10 19

1/ 2
 5  10 15
 
16
 5  10
1/ 2
 x p  4.11 10 5 cm
 2  V  V R   N a

x n   s bi
N
e

 d
 2  V  N 

1

x n   s bi  a 

 e
 N d  N a  N d 
 211.7  8.85  10 14 0.7184 

1.6  10 19



1


 5  1015  5  10 16 


 5 1015 5  1016
 0.0259 ln 
2
 1.5 1010
 0.718 V
Then


 5  10 16
 
15
 5  10
 

Ex 7.2
N N
Vbi  Vt ln a 2 d
 ni


 211.7  8.85  10 14 0.7184
xp  
1.6  10 19




1/ 2

1

 5  1015  5  1016




1/ 2
 1.054  10 5 cm
or x n  0.1054  m
 2  V  V R   N d

x p   s bi
N
e

 a



1


 N  N 
d 
 a
1/ 2

 211.7  8.85  10 14 0.7184  4

1.6  10 19

 5  10 16
 
15
 5  10

1

 5  10 15  5  10 16

 1.054  10 4 cm



1/ 2
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 7
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
or x p  1.054  m
 2  V  V R   N a  N d

W   s bi
 N N
e

a
d








Now 7.2  10 4
1/ 2


 

 





1

 5  10 15  5  10 16







1

 5  10 15  5  10 16

 1.432  10 4 cm
or x p  1.432  m




 

 



 5  1015 3  10 16
 0.0259 ln 
2
 1.8  10 6
 1.173 V


 2eVbi  V R   N a N d


N N
s

d
 a






 

 


1/ 2

15
15

16




5 10 2 10  

5 10  2 10 
15
15
1/ 2
16
C   8.48  10 9 F/cm 2
 1.6  10 19 13.1 8.85  10 14
(c) C   
21.162  8

16
1/ 2
16
C   6.36  10 9 F/cm 2
_______________________________________
Ex 7.6 For a one-sided junction
1/ 2
 e s N a 
C  

 2Vbi  V R  
C  A  C   10 5 C 

 10
5

12




 1.6  10 19 11.7  8.85  10 14 N a 


23  0.765


0.105 10   10  2.20 10 N
12 2
1/ 2

5 10 2 10  

5 10  2 10 
0.105  10














 1.6  10 19 13.1 8.85  10 14

21.162  4 

Ex 7.4
 max



e s N a N d
(b) C   

 2Vbi  V R N a  N d  
1/ 2
 1.576  10 4 cm
or W  1.576  m
_______________________________________


16
 2  1016 5  10 15 
 0.0259 ln 

2
1.8  10 6


 1.16 V
1/ 2
 211.7  8.85  10 14 0.7184  8
W 
1.6  10 19

 5  1015  5  10 16

15
16
 5  10 5  10
N N
(a) Vbi  Vt ln a 2 d
 ni
1/ 2
 211.7  8.85  10 14 0.7184  8
xp  
1.6  10 19

N N
Vbi  Vt ln a 2 d
 ni
15
Vbi  V R  1.173  V R  4.382
Then V R  3.21 V
_______________________________________

 1.432  10 5 cm
or x n  0.1432  m
 5  10 16
 
15
 5  10

Ex 7.5

 5  10
 
16
 5  10

5.184  10 9  1.1829  10 9 Vbi  V R 
1/ 2
 211.7  8.85  10 14 0.7184  8
xn  
1.6  10  10 19

15

 2 1.6  10 19 Vbi  V R 

 13.1 8.85  10 14

4
 1.159  10 cm
or W  1.159  m
(b) Vbi  0.718 V
2
 5  10 3  10

15
16
 5  10  3  10
 211.7  8.85  10 14 0.7184  4

1.6  10 19

 5  1015  5 1016

15
16
 5 10 5 10

5 2
So N a  5.01 1015 cm 3
32
a
1/ 2
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 7
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
N N
We have Vbi  Vt ln a 2 d
 ni
V
n2
Then N d  i exp bi
Na
 Vt

1.5 10 
10 2
5.01 1015
 3.240  10 5 cm
or W  0.3240  m





 0.765 
exp

 0.0259 
N d  3.02  10 17 cm 3
_______________________________________

 2  V   N
x n   s bi  a
e

 Nd


1


 N  N 
d 
 a







1

 4  1015  3  1016





1/ 2





1

 4  1015  3  1016

 3  1016
 
15
 4  10
1/ 2

1


 2  1017  1016 


 4.469  10 5 cm
or x p  0.4469  m

1/ 2

 211.7  8.85  10 14 0.6994
W 
1.6  10 19



1


 N  N 
d 
 a
1/ 2
 4  1015  3  1016  

15
16  
 4  10 3  10  


 211.7  8.85  10 0.7722

1.6  10 19

14
 1016
 
17
 2  10

 211.7  8.85  10 14 0.6994
xp  
1.6  10 19

 3.085  10 5 cm
or x n  0.3085  m


 5.96  10 6 cm
or x n  0.0596  m
1/ 2

 2  V   N
x p   s bi  d
e

 Na
4
16
14
 4  1015
 
16
 3  10
 211.7  8.85  10 14 0.7722

1.6  10 19

 2  1017
 
16
 10
19
 4.77  10 4 V/cm
 4  1015 3  1016
(b) Vbi  0.0259  ln 
2
 1.5  1010
 0.699 V
 211.7  8.85  10 14 0.6994
xn  
1.6  10 19

 
 
 2  1017 1016
(a) Vbi  0.0259 ln 
2
 1.5  1010
 0.772 V

1.6 10 10 0.3085 10 
11.7 8.85 10 

Test Your Understanding Solutions
TYU 7.1
eN d x n
s
 max 




 5.064  10 cm
or W  0.5064  m



1/ 2
 1.54  10 6 cm
or x p  0.0154  m






1.6 10 3 10 5.96 10 
11.7 8.85 10 
19
6
16
14
 2.76  10 4 V/cm
_______________________________________


 5  1016 5  1015
Vbi  0.0259  ln 
2
 1.8  10 6
 1.186 V
 211.7  8.85  10 14 0.7722

1.6  10 19

 2  10  10

17
16
 2  10 10

 max 
TYU 7.2
1/ 2
17

5

1

 2  1017  1016

 2  V   N  N d
W   s bi  a
e

 Na Nd

1/ 2
16
 

 
 
1/ 2

 2  V   N
x n   s bi  a
e

 Nd





1


 N  N 
d 
 a
1/ 2
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 7
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________



 5  1015
 
16
 5  10




1

 5  1015  5  1016

1/ 2
 5.590  10 6 cm
or x n  0.05590  m






1/ 2




1/ 2

1

 5  1016  5  1015








1

 5  1015  5  1016


 

 
1/ 2

 211.7  8.85  10 14 0.718  12 
xn  
1.6  10 19


 5  1016
 
15
 5  10

1

 5  1015  5  1016

or x n  1.73  10 4 cm

eN d x n
s
1.6 10 5 10 5.59 10 
13.18.85 10 
 5  1015
 
16
 5  10
14

1

 5  1015  5  1016

or x p  1.73  10 5 cm
TYU 7.3
Also
W  x n  x p  1.90  10 4 cm



Now



1


 N  N 
d 
 a



1/ 2
20.718  12
1.90  10  4
 1.34  10 5 V/cm
_______________________________________
 max 
1/ 2
1/ 2

 3.86  10 4 V/cm
_______________________________________
 5  1016 5  1015
(a) Vbi  0.0259 ln 
2
 1.5  1010
 0.718 V
Now for V R  8 V,



 211.7  8.85  10 14 0.718  12
xp  
1.6 10 19

6
16
 2  V  V R   N a

x n   s bi
N
e

 d
or
1/ 2
2Vbi  V R  20.718  8

W
1.575  10  4
 1.11 10 5 V/cm
(b) For V R  12 V
 6.149  10 5 cm
or W  0.6149  m





Now
W  x n  x p  W  1.58  10 4 cm
1/ 2
 5  1015  5  1016

15
16
 5  10 5  10

1/ 2
 max 

19




 211.7  8.85  10 14 0.718  8
xp  
1.6  10 19

Also
 213.1 8.85  10 14 1.186
W 
1.6  10 19




 x p  1.432  10 5 cm

1

 5  1015  5  1016

 2  V   N  N d
W   s bi  a
e

 Na Nd
 max 
 5  1016
 
15
 5  10
 5  1015
 
16
 5  10
 5.590  10 5 cm
or x p  0.5590  m
 max 

 x n  1.432  10 4 cm
 2  V   N 
1
x p   s bi  d 
e
N
N


 a  a N d
 213.1 8.85  10 14 1.186 
xp  
1.6  10 19

 5  1016
 
15
 5  10

 211.7  8.85  10 14 0.718  8
xn  
1.6  10 19

 213.1 8.85  10 14 1.186
xn  
1.6  10 19

Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 7
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
TYU 7.4


 3  1016 8  1015
Vbi  0.0259 ln 
2
 1.5 1010
 0.717 V




1/ 2
 e s
 N a N d 


C  A
 2Vbi  V R   N a  N d 
 1.6  10 19 11.7  8.85  10 14
 5  10 5 
2Vbi  V R 







 
 3 1016 8  1015

16
15
 3  10  8  10

1/ 2

 
or
 5.232 10 16 
C  5  10 5 

 Vbi  V R 
(a) For V R  2 V,



C  5  10
5
1/ 2
 5.232  10 16 


 0.717  2 

1/ 2
 6.94 10 13 F  0.694 pF
(b) For V R  5 V,
 5.232  10 16 
C  5 10 5 

 0.717  5 


1/ 2
 4.78 10 13 F  0.478 pF
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 8
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
Chapter 8
Exercise Solutions
Ex 8.1



10 2
n po 
n i2
1.5  1010

Na
5  1016
p no 
n
1.5  10

Nd
2  1016

2
i
n po 
V
exp a
 Vt

n p  x p  n po

2
 4.05  10 4 cm 3
 4.5  10 3 cm 3

p no 
 1.125  10 4 cm 3

 3.57  1014 cm 3
V 
p n x n   p no exp a 
 Vt 
2

2
 
  1
 
1.6  10 19 210  4.05  10 4

4.583  10 3
  1.05  
 exp
  1
  0.0259  





J n  x p  1.20 A/cm 2

 0.650 
 1.125  10 exp

 0.0259 
eD p p no   eVa  
 1
exp
L p   kT  
1.6  10 19 8 1.62  10 4

6.325  10  4
  1.05  
 exp
  1
  0.0259  
J p x n  
 8.92  10 14 cm 3
We have that
n p  x p  N a and p n x n   N d






n i2
1.8 10 6

Nd
2  1016

 1.62 10 4 cm 3
eDn n po   eV a
Jn  xp 
exp
Ln   kT




 0.650 
 4.5  10 3 exp

 0.0259 
4

ni2
1.8  10 6

Na
8 1015


so low injection applies.
_______________________________________
 

J p x n   0.1325 A/cm 2
Ex 8.2
 1
J s  en i2 
 N a
Dn
 no


1
Nd

 1.6 10 19 1.8  10 6
 1

15
 8  10

Dp 

 po 

18

2
210
1

7
10
2  10 16
8
5  10 8



J s  3.30  10 A/cm
_______________________________________
2
Ex 8.4
In the n-region, for N d  2  1016 cm 3 ,
 n  6000 cm 2 /V-s
or
21010
1.3325
J

e n N d
1.6  10 19 6000 2  1016
 0.0694 V/cm
In the p-region, for N a  8 1015 cm 3 ,
n 
Ln  D n no
7
  4.583 10
3
cm
L p  D p po


J  e n N d  n
Ex 8.3
We find

The total current density is:
J T  J n  x p  J p x n 
 1.20  0.1325
J T  1.33 A/cm 2
_______________________________________




 p  320 cm 2 /V-cm
85 10 8   6.325 10 4 cm
J  e p N a  p
or
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 8
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
p 
J
e p N a

1.3325
1.6 10 3208 10
19
J gen 
15
 3.25 V/cm
_______________________________________


 1.6  10
 1

15
 2  10
19
1
Nd
1.8 10 
or J s  1.677  10
A/cm

9.8
5 10 8

 2  V  V R   N a  N d

W   s bi
 N N
e

a
d







or W  2.141 10 4 cm

10 2
19
2  10
15
25
5  10  7
or
I Sn  1.273  10 13 A
en i2
Nd
I Sp  A 
Dp
 po
 1.6 10 1.5 10 
10 2
19
8  10 16



Then
I S  I Sn  I Sp  1.318  10 13 A
V
(a) I D  I S exp a
 Vt







 0.550 
 1.318  10 13 exp

 0.0259 
 2.2  10 4 A
1/ 2
 2 1015  8 1016  

15
16  
 2  10 8 10  

 1.6 10 1.5 10 

 10 3 



 0.610 
(b) I D  1.318  10 13 exp

 0.0259 
 2.23  10 3 A
 213.1 8.85  10 1.174  5

1.6  10 19

14
 no
I Sp  4.5  10 15 A
 2  1015 8  1016 
(b) Vbi  0.0259 ln 

6 2
 1.8  10

 1.174 V

Dn
or
2

eni2
Na
I Sn  A 

Dp 

 p0 

207
1

5  10 8 8  1016

 10 3 
6 2
17
8
Ex 8.7
Ex 8.6
 n0
4
6
6.166  10 10
 3.68  10 7
Js
1.677  10 17
_______________________________________
and V a1  1.050 V.
Then
1.42  V a 2 1.42  1.050

310
300
which yields
V a 2  1.0377 V
so V  1.0377  1.050  0.0123 V
or V  12.3 mV per 10 C increase in
temperature.
_______________________________________

19
J gen
(c)
Let T2  310 K, T1  300 K, E g  1.42 eV,
Dn
1.6 10 1.8 10 2.14110 
25  10 
J gen  6.166  10 10 A/cm 2
Ex 8.5
From Example 8.5, we have
E g  eVa 2 E g  eVa1

kT2
kT1
 1
(a) J s  eni2 
 N a
eni W
2 0
Now
1/ 2
(a) rd 
Vt
0.0259

 118 
I D 2.2  10  4
(b) rd 
0.0259
 11.6 
2.23  10 3
10
10  7
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 8
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
We find
TYU 8.2
I pO  I Sp
V
exp a
 Vt
V 

 , I nO  I Sn exp a 
V 

 t 

(a) I n  A 
Then
(a) I pO  7.511 10 6 A;

 10
eni2
Na
3
Dn
 no
5  1016

5
(b) I pO  7.617  10 A;
(b) I p  A 
We find

 I pO pO  I nO nO




7

 
 2.155  10 5  10 
1
7.617  10 5 10  7
20.0259 
7
or C d  2.09  10 8 F  20.9 nF
_______________________________________
Test Your Understanding Solutions
n po 


 6.48  10 5 cm 3


 6.48  10  4 cm 3
n i2
1.8  10 6

Na
5  1016
n i2
1.8  10 6

Nd
5  1015
For V a max  ,
p no 
2




10 2
11016
10
 0.625 
 exp

10  7
 0.0259 
or I p  1.09  10 3 A  1.09 mA
(c) I Total  I n  I p
 1.538  10 4  1.087 10 3
 1.24  10 3 A
or I Total  1.24 mA
_______________________________________
TYU 8.3
From TYU 8.2, I n  0.154 mA
Now
eD p p no
V 
 exp a 
I p  A
Wn
 Vt 
We find
p no 
Then


n i2
1.5 1010

Nd
1016

2
 2.25  10 4 cm 3
 1.6 10 102.25 10 
I p  10 3 

V
 exp a
 Vt
19

2
V 
p n x n   p no exp a 
 Vt 
so that
 0.1N d 
 p x  
V a  Vt ln n n   Vt ln 

 p no 
 p no 
 0.1 5  10 15 
 0.0259  ln 
4 
 6.48  10 
or V a max   1.067 V
_______________________________________

 po
 1.6 10 1.5 10 


 
 2.125  10 5  10 
3
Dp
 10 3 
or C d  2.07  10 9 F  2.07 nF
TYU 8.1
en i2
Nd
1
7.511 10  6 10  7
20.0259 
4
(b) C d 
25
 0.625 
 exp

7
5  10
 0.0259 
or I n  1.54  10 4 A  0.154 mA
I nO  2.155  10 3 A
(a) C d 
10 2
19
I nO  2.125  10 A
So




 1.6 10 1.5 10 
4
 1
C d  
 2Vt
V
 exp a
 Vt
19
2  10
4
4
 0.625 
 exp

 0.0259 
or I p  5.44 mA
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 8
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
TYU 8.4
V
(a) J  J s exp a
 Vt
 1
J s  en i2 
 N a
I Sp  A 




Dn
1

Nd
 no


 1.6  10 19 1.5  1010
 1

15
 2  10

2
10 

10  7 
 1.045 
(b) I D  5.338  10  21 exp

 0.0259 



 2  10 15  8  10 16

15
16
 2  10 8  10

 

 
1/ 2

5

J rec

 1.78  10 3 A
Now
(a) rd 
Vt
0.0259

 264 
I D 9.828  10 5
(b) rd 
0.0259
 14.6 
1.779  10 3
We have
V 
V 
I pO  I Sp exp a  , I nO  I Sn exp a 
 Vt 
 Vt 
We find
(a) I pO  1.181 10 6 A; I nO  9.71 10 5 A
(b) I pO  2.137  10 5 A;




1.5  10 10 4.865  10 5
2 10  7
1.6 10 
19


I nO  1.757  10 3 A

 0.35 
 exp 

 20.0259 
 5.020  10 4 A/cm 2
4
J rec 5.020  10

 2.35
J
2.137  10  4
_______________________________________
(c)
Now
 1
C d  
 2Vt
So
(a) C d 

 I pO pO  I nO nO





 
 9.71 10 5  10 
1
1.181 10  6 10  7
20.0259 
5
7
 9.40  10 10 F  0.940 nF
1
(b) C d 
2.137  10 5 10  7
20.0259


 
5 10 
 1.757  10 3
TYU 8.5
I Sn  A 
2
i
en
Na
 no
6 2
19
2  10
or I Sn  5.274  10 21 A
7
 1.70  10 8 F  17.0 nF
_______________________________________
Dn
1.6 10 1.8 10 
 10 
3
9.8
10  7
 9.83  10 5 A

 4.865  10 cm
Then
V
en W
J rec  i exp a
2 o
 2Vt
8  1016
 0.970 
(a) I D  5.338  10  21 exp

 0.0259 
 2.137  10 4 A/cm 2
 2  10 15 8  1016 
(b) Vbi  0.0259 ln 

2
 1.5  10 10

 0.7068 V
We find
 211.7  8.85  10 14 0.7068  0.35
W 
1.6  10 19


6 2
So I S  I Sn  I Sp  5.338  10 21 A


19
or I Sp  6.415  10 23 A
 0.35 
Then J  2.891 10 10 exp

 0.0259 

 po
 1.6 10 1.8 10 

J s  2.891 10 10 A/cm 2

Dp
 10 3 
Dp 

 po 

25
1

7
10
8  1016

eni2
Nd
15
207
5  10  7
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 8
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
TYU 8.6
From Figure 5.3, for N d  8  1016 cm 3 ,
(b) erf
  n  900 cm 2 /V-s
t2
 pO

In the n-region,
  e n N d

 1.6  10
19
9008 10 
16
 11.52 (  -cm)
Then
Rn 
l
A

1
0.01  0.868 
11.5210 3 
In the p-region,
  e p N a



 1.6  10 19 480  2  10 15

1
 0.1536 (  -cm)
Then
0.01  65.1 
l

Rp 
A 0.1536 10 3
The total resistance is
R  R n  R p  66 


_______________________________________
TYU 8.7
(a) erf
ts
 pO

IF
IF  IR
Now
IR 
VR 2
  0.5 mA
RR 4
erf
 pO
So
ts

1.75
 0.778
1.75  0.5
From Appendix G,
ts
 pO
 0.864
So that
2
t s  0.864  10 7  0.746  10 7 s


 t 2  pO

I
 1  0.1 R
 IF



 0.5 
 1  0.1
  1.0286
 1.75 
For N a  2  1015 cm 3 ,
  p  480 cm 2 /V-s

exp  t 2  pO
By trial and error
t2
 1.25
 pO


 t 2  1.25 10 7  1.25  10 7 s
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 9
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
Chapter 9
Exercise Solutions
Ex 9.3
Ex 9.1
 2  V  V R  
x n   s bi

eN d


 B 0   m    4.55  4.07  0.48 V
N 
 4.7  10 17 
  0.0259 ln


 5  10 15 



 0.1177 V
Vbi   B 0   n  0.48  0.1177  0.3623 V
 n  Vt ln c
 Nd
2 V 
x n   s bi 
 eN d 


 213.1 8.85  10 0.3623 


1.6  10 19 5  1015






 max 



19
5
16
14

2
13.1 8.85 10 14 2.335 1012

1/ 2
1.6 10 10 8.309 10 
11.7 8.85 10 
 1.284 10 5 V/cm
2
1

2
e s
 1 
 
 C 
V R
3

x n  8.309  10 5 cm
2

e
4 s
 1.6  10 19 6.42  10 4 

14 
 4 11.7  8.85  10

  0.0281 V
(b) V R  5 V,
 1 
 
8.5  1012
 C 

 2.335  10 12
V R
3  0.64

or N d  4.62  10 cm
_______________________________________
18
5
16
14
Then  
Ex 9.2
From Figure 9.3,
Vbi  0.64 V
1.6 10 
19
 6.42  10 4 V/cm
5
 2.24  10 4 V/cm
_______________________________________
19
1.6 10 10 4.155 10 
11.7 8.85 10 

14

eN d x n
s
 max 
1.6 10 5 10 3.24 10 

13.18.85 10 
Then N d 
1/ 2
 x n  4.155  10 5 cm
1/ 2

15
1/ 2
 
V  V 
 1.294  10 9 bi
R
(a) V R  1 V, Vbi  0.334 V
 3.24  10 5 cm
eN d x n
 max 
s
19

 211.7  8.85  10 14 Vbi  V R  




1.6  10 19 1016
1/ 2
14


1/ 2


1/ 2
 1.6  10 19 1.284  10 5 
  

14

 4 11.7  8.85  10
  0.0397 V
_______________________________________


Ex 9.4
A 
4 em n k 2
h3
Assume m n  m o , then
A 




4 1.6  10 19 9.1110 31 1.38  10 23
6.625 10

2
 34 3
 1.20  10 6 A/K 2 -m 2
 A   120 A/K 2 -cm 2
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 9
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
p
N
eVbi  E  kT ln Po  n
p
N
P
 no
Ex 9.5
V
I  AJ s exp a
 Vt




 I 

so that V a  Vt ln

 AJ s 
For the pn junction:


10  10  6

V a  0.0259  ln  4
11 
 10 3.66  10

 0.5628 V
For the Schottky junction:


10 10  6

V a  0.0259 ln  4
5 
 10 5.98  10 
 0.1922 V
_______________________________________








N 
(b)  n  Vt ln c 
 Nd 
 2.8  10 19 
  0.237 V
 0.0259  ln
15 
 3  10 
Vbi   Bo   n  0.49  0.237  0.253 V
 2  V  V R  
(c) x n   s bi

eN d


 I S  4.66  10 A
_______________________________________
1/ 2


1/ 2

6
 1.287  10 cm
o
or x n  128.7 A
_______________________________________
Ex 9.8
From Example 9.8, E  0.70 eV.
We find
Now


1.6 10 3 10 1.505 10 
11.7 8.85 10 
19
4
15
14
or  max  6.98  10 4 V/cm
 213.1 8.85 10 14 0.80 


19
7  1018
 1.6  10

p no 

1/ 2
or x n  1.505  10 cm
Then
eN d x n
 max 
s



 211.7  8.85 10 14 0.253  5 


1.6 10 19 3 1015


Ex 9.7
We have


1/ 2
4
12
2 V 
x n   s bi 
 eN d 

TYU 9.1
(a)  Bo  4.5  4.01  0.49 V



Test Your Understanding Solutions
Then
  0.3 
I S  5  10  7 exp

 0.0259 

 1015 6  1018

 0.70  0.0259  ln 
11
18 


5
.
76
10
7
10


or Vbi  0.889 V
_______________________________________
Ex 9.6
 V  0.3 

I ST exp a

  0.3 
I
 Vt 

 1  ST exp
IS
Vt 
 Va 

I S exp 
 Vt 
 




n i2
2.4  1013

Nd
1015

2
 5.76  10 11 cm 3
 e s N d 
(d) C   

 2Vbi  V R  


1/ 2



 1.6 10 19 11.7  8.85 10 14 3  1015 


20.253  5


2
9
or C   6.88  10 F/cm
_______________________________________
TYU 9.2
(a)  Bo  5.12  4.07  1.05 V
 4.7  1017 
  0.131 V
(b)  n  0.0259 ln
15 
 3  10 
Vbi  1.05  0.131  0.919 V
1/ 2
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 9
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________


 213.1 8.85  10 14 0.919  5 
(c) x n  

1.6  10 19 3  1015




1/ 2

4
or x n  1.69  10 cm
Now
1.6  10 19 3  1015 1.69  10 4
 max 
13.1 8.85 10 14


or  max  7  10 V/cm




4



 1.6  10 19 13.1 8.85  10 14
(d) C   
20.919  5


 3 1015


1/ 2
or C   6.86  10 9 F/cm 2
_______________________________________
TYU 9.3
 
e
4 s




 1.6  10 19 6.98 10 4 

14 
 4 11.7  8.85  10

or   0.0293 V
xm 

1/ 2
e
16 s 


1.6  10 19

4 
14
6.98  10 
16 11.7  8.85  10


1/ 2

o
or x m  2.10  10 7 cm  21.0 A
_______________________________________
TYU 9.4
V 
I  I S exp a 
 Vt 
Then
 I 
V a  Vt ln 
 IS 
(a) For the pn junction diode
 100 10 6
V a  0.0259 ln
14
 10
For the Schottky diode
 100 10 6
V a  0.0259 ln
9
 10

  0.596 V



  0.298 V


(b) For the pn junction diode
 10 3 
V a  0.0259 ln 14   0.656 V
 10 
For the Schottky diode
 10 3 
V a  0.0259 ln 9   0.358 V
 10 
_______________________________________