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Semiconductor Physics and Devices: Basic Principles, 4
th
 edition Chapter 12 
By D. A. Neamen Problem Solutions 
______________________________________________________________________________________ 
 







 


t
BE
sO
rO
V
V
J
J
2
exp1
1
 
 
 



 






0259.02
exp
10287.1
102
1
1
10
9
BEV
 
 or 
 (a) 
 
  




 


0518.0
exp54.151
1
BEV
 
 and 
(b) 
 





1
 
 Now 
BEV   
0.20 
0.40 
0.60 
0.7535 
0.99316 
0.999855 
3.06 
145 
6,902 
 
(c) If 4.0BEV V, the recombination factor is 
likely the limiting factor in the current gain. 
_______________________________________ 
 
12.29 
 993377.0
151
150
1





 
  T 
   9975.0993377.0 T 
 995867.0 T 
 Let 80.0Bx m 
   7
0 10223  BBB DL  
 
310145.2  cm 
 Then 
 






















3
4
10145.2
1080.0
cosh
1
cosh
1
B
B
T
L
x
 
 99930.0 
 Now 
 
  
99656.0
9975.099930.0
993377.0




T
 
 
E
B
B
E
E
B
x
x
D
D
N
N


1
1
 
 
 



















 


35.0
80.0
23
8102
1
1
99656.0
16
EN
 
 181061.4  EN cm 3 
_______________________________________ 
 
12.30 
 (a) We have 8105 rOJ A/cm 2 
 We find 
 
  3
16
2102
105.4
105
105.1




B
i
BO
N
n
n cm 3 
 and 
   71025  BOBB DL  
 
4108.15  cm 
 Then 
 
 BBB
BOB
sO
LxL
neD
J
tanh
 
 
   
   BB Lxtanh108.15
105.425106.1
4
319




 
 or 
 
 BB
sO
Lx
J
tanh
10139.1 11
 
 We have 
 







 


t
BE
sO
rO
V
V
J
J
2
exp1
1
 
 For 300T K and 55.0BEV V. 
 
 995.0 
 





 





















0518.0
55.0
exptanh
10139.1
105
1
1
11
8
B
B
L
x
 
 which yields 
 0468.0
B
B
L
x
 
 or 
    739.08.150468.0 Bx m 
 (b) For 400T K and 8105 rOJ A/cm 2 , 
 
 
 
  





 





 







0259.0
exp
3004000259.0
exp
300
400
300
400
3
g
g
BO
BO
E
E
n
n
 
 For 12.1gE eV,