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Semiconductor Physics and Devices: Basic Principles, 4
th
 edition Chapter 9 
By D. A. Neamen Problem Solutions 
______________________________________________________________________________________ 
 
9.7 
 (a) From the figure, 90.0biV V 
 (b) We find 
 
 
15
15
2
10034.1
90.02
0103
1













RV
C
 
 and 
 
ds Ne

2
10034.1 15
 
 We can then write 
    151419 10034.11085.81.13106.1
2


dN 
 or 
 161004.1 dN cm 3 
 (c) 
 








d
c
tn
N
N
V ln 
   










16
17
1004.1
107.4
ln0259.0 
 or 
 0986.0n V 
 (d) 
 0986.090.0  nbiBn V  
 or 
 9986.0Bn V 
_______________________________________ 
 
9.8 
 From Figure 9.5, 63.0BO V 
(a)   224.0
105
108.2
ln0259.0
15
19










n V 
 406.0224.063.00  nBbiV  V 
 (i)
   
  
2/1
1519
14
105106.1
1406.01085.87.112











nx 
 
510033.6  cm 
 or 6033.0nx m 
 
   
  14
51519
max
1085.87.11
10033.6105106.1




 
 
41066.4  V/cm 
 (ii)
   
  
2/1
1519
14
105106.1
5406.01085.87.112











nx 
 
410183.1  cm 
 or 183.1nx m 
 
 
   
  14
41519
max
1085.87.11
10183.1105106.1




 
 41014.9  V/cm 
(b) 
 (i) 
s
e





4
 
 
  
  
2/1
14
419
1085.87.114
1066.4106.1












 
 0239.0 V 
 


s
m
e
x
16
 
 
 
   
2/1
414
19
1066.41085.87.1116
106.1












 
 
or 
 mx
71057.2  cm 
 (ii) 
  
  
2/1
14
419
1085.87.114
1014.9106.1












 
 0335.0 V 
 
   
2/1
414
19
1014.91085.87.1116
106.1












mx 
 
71083.1  cm 
_______________________________________ 
 
9.9 
 We have 
   x
x
e
x
s






16
 
 or 
   ex
x
e
xe
s





16
2
 
 Now 
 
  
e
x
e
dx
xed
s




2
2
16
0


 
 Solving for x, we find 
 


s
m
e
xx
16
 
 Substituting this value of mxx  into the 
 equation for the potential, we find 
 





s
s
s
e
e
e




16
16
16