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Semiconductor Physics and Devices: Basic Principles, 4
th
 edition Chapter 12 
By D. A. Neamen Problem Solutions 
______________________________________________________________________________________ 
 


























E
Et
BE
E
EOE
pE
L
xV
V
L
peD
J
tanh
1
1exp 
 Now 
 
  2
18
2102
1025.2
10
105.1



E
i
EO
N
n
p cm 3 
 and 
   8108  EOEE DL  
 410828.2  cm 
 Then 
 
   
4
219
10828.2
1025.28106.1




pEJ 
 



















828.2
8.0
tanh
1
1
0259.0
60.0
exp 
 or 
 04251.0pEJ A/cm 2 
 We can find 
 
B
BOB
nC
L
neD
J  
 
















































B
B
B
B
t
BE
L
x
L
x
V
V
tanh
1
sinh
1exp
 
 
   
4
319
1066.8
105.415106.1




 
 








































66.8
7.0
tanh
1
66.8
7.0
sinh
1
0259.0
60.0
exp
 
 or 
 773.1nCJ A/cm 2 
 The recombination current density is 
 








t
BE
roR
V
V
JJ
2
exp 
  
 




 
0259.02
60.0
exp103 8 
 or 
 310218.3 RJ A/cm 2 
 
 
 
 
 (b) Using the calculated current densities, we 
 find 
 
04251.0779.1
779.1




pEnE
nE
JJ
J
 
 or 
 9767.0 
 We also find 
 
779.1
773.1

nE
nC
T
J
J
 
 or 
 9966.0T 
 Also 
 
pERnE
pEnE
JJJ
JJ


 
 
04251.0003218.0779.1
04251.0779.1


 
 or 
 9982.0 
 Then 
    9982.09966.09767.0  T 
 or 
 9716.0 
 Now 
 
9716.01
9716.0
1 





 
 or 
 2.34 
_______________________________________ 
 
12.24 
 (a) We have 
 
E
B
B
E
E
B
E
B
B
E
E
B x
x
D
D
N
N
x
x
D
D
N
N


 1
1
1
 
 or 
 
E
B
N
N
K  1 
 (i) Now 
 
 
 
K
N
N
K
N
N
A
B
E
BO
E
BO



1
2
1


 
 















 K
N
N
K
N
N
E
BO
E
BO 1
2
1 
 K
N
N
K
N
N
E
BO
e
BO 
2
1 
 or finally