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Semiconductor Physics and Devices: Basic Principles, 4
th
 edition Chapter 5 
By D. A. Neamen Problem Solutions 
______________________________________________________________________________________ 
 
5.10 
(a) 3
1
3

L
V
V/cm 
  nd 
3
104


 d
n

 
 or 
 3333n cm 2 /V-s 
 (b) 
   3800 nd  
 or 
 3104.2 d cm/s 
_______________________________________ 
 
5.11 
(a) Silicon: For 1 kV/cm, 
 6102.1 d cm/s 
 Then 
 11
6
4
1033.8
102.1
10 




d
t
d
t

s 
 For GaAs: 6105.7 d cm/s 
 Then 
 11
6
4
1033.1
105.7
10 




d
t
d
t

s 
(b) Silicon: For 50 kV/cm, 
 6105.9 d cm/s 
 Then 
 11
6
4
1005.1
105.9
10 



tt s 
 For GaAs: 6107d cm/s 
 Then 
 11
6
4
1043.1
107
10 



tt s 
_______________________________________ 
 
5.12 
 
  ipnopon nepene 





11
 
(a) 1410 da NN cm 3 
 1350 n cm 2 /V-s 
 480p cm 2 /V-s 
 
   1019 105.14801350106.1
1



 
  51028.2 -cm 
 
 
 
(b) 1610 da NN cm 3 
 1250 n cm 2 /V-s 
 410p cm 2 /V-s 
 
   1019 105.14101250106.1
1



 
  51051.2 -cm 
(c) 1810 da NN cm 3 
 290 n cm 2 /V-s 
 130p cm 2 /V-s 
 
   1019 105.1130290106.1
1



 
  51092.9 -cm 
_______________________________________ 
 
5.13 
(a) GaAs: 
   opop ppe  19106.15  
 From Figure 5.3, and using trial and error, we 
 find 
 17103.1 op cm 3 and 
 240p cm 2 /V-s 
 Then 
 
  5
17
262
1049.2
103.1
108.1 



o
i
o
p
n
n cm 3 
(b) Silicon: 
 onne

 
1
 
 or 
 
   1350106.18
11
19

n
o
e
n

 
 which gives 
 141079.5 on cm 3 
 and 
 
  5
14
2102
1089.3
1079.5
105.1




o
i
o
n
n
p cm 3 
 Note: For the doping concentrations obtained 
 in part (b), the assumed mobility values are 
 valid. 
_______________________________________