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Semiconductor Physics and Devices: Basic Principles, 4
th
 edition Chapter 8 
By D. A. Neamen Problem Solutions 
______________________________________________________________________________________ 
 
8.7 
 
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   21319 104.2106.1   
 
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 617615 102
48
102
1
102
90
104
1
 
 410568.1 sJ A/cm 2 
(a) 
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
t
a
s
V
V
AJI exp 
    
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



 
0259.0
25.0
exp10568.110 44 
 
41044.2  A 
 or 244.0I mA 
(b)   44 10568.110   ss AJII 
 
810568.1  A 
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8.8 
(a) 
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   21019 105.1106.1   
 
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 815717 108
10
108
1
10
25
105
1
 
 1110145.5 sJ A/cm 2 
   114 10145.5102   ss AJI 
 
1410029.1  A 
(b) 
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
t
a
s
V
V
II exp 
 (i)   
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


 
0259.0
45.0
exp10029.1 14I 
 
71061.3  A 
 (ii)   





 
0259.0
55.0
exp10029.1 14I 
 
51072.1  A 
 (iii)   
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


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 
0259.0
65.0
exp10029.1 14I 
 
41016.8  A 
_______________________________________ 
 
 
 
8.9 
 We have 
 
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 1exp
t
S
V
V
II 
 or we can write this as 
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
tS V
V
I
I
exp1 
 so that 
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


 1ln
S
t
I
I
VV 
 In reverse bias, I is negative, so at 
 90.0
SI
I
, we have 
    90.01ln0259.0 V 
 or 
 6.59V mV 
_______________________________________ 
 
8.10 
 Case 1: 
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
t
a
s
V
V
II exp 
 





 
0259.0
65.0
exp1050.0 3
sI 
 1510305.6  sI A
1210305.6  mA 
 
4
12
102
10305.6





A
I
J s
s 
 
810153.3  mA/cm 2 
 Case 2: 

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

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

t
a
s
V
V
II exp 
   
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



 
0259.0
70.0
exp102 12 
 or 093.1I mA 
 
3
12
101
102





A
I
J s
s 
 
9102  mA/cm 2 
 Case 3: 
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
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


t
a
s
V
V
AJI exp 
 So 
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
s
ta
AJ
I
VV ln 
  
  
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


 74 1010
80.0
ln0259.0 
 6502.0aV V