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Semiconductor Physics and Devices: Basic Principles, 4
th
 edition Chapter 5 
By D. A. Neamen Problem Solutions 
______________________________________________________________________________________ 
 
Chapter 5 
 
5.1 
(a) 
   1519 101300106.1
11


dn Ne
 
  808.4 -cm 
(b) 208.0
8077.4
11


 ( -cm) 1 
_______________________________________ 
 
5.2 
 
ap Ne  
 or 
  380106.1
80.1
19

p
a
e
N


 
 
161096.2  cm 3 
_______________________________________ 
 
5.3 
(a) dn Ne  
   dn N19106.110  
 From Figure 5.3, for 16106dN cm 3 we 
find 1050n cm 2 /V-s which gives 
    1619 1061050106.1   
 08.10 ( -cm) 1 
(b) 
ap Ne

1
 
 
  ap N19106.1
1
20.0

 
 From Figure 5.3, for 1710aN cm 3 we 
find 320p cm 2 /V-s which gives 
 
   
195.0
10320106.1
1
1719




  -cm 
_______________________________________ 
 
5.4 
(a) 
ap Ne

1
 
 
  ap N19106.1
1
35.0

 
 From Figure 5.3, for 16108aN cm 3 we 
find 220p cm 2 /V-s which gives 
 
 
 
 
   1619 108220106.1
1



 
 355.0  -cm 
(b) dn Ne  
   dn N19106.1120  
 From Figure 5.3, for 17102dN cm 3 , 
then 3800n cm 2 /V-s which gives 
    1719 1023800106.1   
 6.121 ( -cm) 1 
_______________________________________ 
 
5.5 
 
 ANe
L
A
L
A
L
R
dn

 
 or 
 RAeN
L
d
n  
 
    1.070102106.1
5.2
1519 


 
 1116 cm 2 /V-s 
_______________________________________ 
 
5.6 
(a) 1610 do Nn cm 3 
and 
 
  4
16
262
1024.3
10
108.1 


o
i
o
n
n
p cm 3 
 (b) 
  onneJ  
 For GaAs doped at 1610dN cm 3 , 
 7500n cm 2 /V-s 
 Then 
     10107500106.1 1619J 
 or 
 120J A/cm 2 
(b) (i) 1610 ao Np cm 3 
 
4
2
1024.3 
o
i
o
p
n
n cm 3 
 (ii) For GaAs doped at 1610aN cm 3 , 
 310p cm 2 /V-s