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Semiconductor Physics and Devices: Basic Principles, 4
th
 edition Chapter 12 
By D. A. Neamen Problem Solutions 
______________________________________________________________________________________ 
 
50.0
1
1
1
1
997238.0
pE
nE
pE I
I
I



 
 00138.0 pEI mA 38.1 A 
 
pERnE
pEnE
III
II


 
 
00138.050.0
00138.050.0
997238.0



RI
 
 00139.0 RI mA 39.1 A 
_______________________________________ 
 
12.22 
(a) Using Equation (12.37) 
B
BEBB
nC
L
ApeD
I 0 
 
















































B
B
B
B
t
EB
L
x
L
x
V
V
tanh
1
sinh
1exp
 
 Now 
 
 
16
2102
0
10
105.1 

B
i
B
N
n
p 
 
41025.2  cm 3 
   7
0 10510  BBB DL  
 
310236.2  cm 
 We find 
 



















3
4
10236.2
1070.0
sinhsinh
B
B
L
x
 
 03131.0 
 



















3
4
10236.2
1070.0
tanhtanh
B
B
L
x
 
 03130.0 
 Then 
 
    
3
4419
10236.2
1051025.210106.1




nCI 
 




























03130.0
1
03131.0
1
0259.0
550.0
exp
 
 
41029.4 nCI A 429.0 mA 
 
 
 
(b) 
E
B
B
E
E
B
x
x
D
D
N
N


1
1
 
 95969.0
5.0
7.0
10
15
105
10
1
1
17
16























 
 






















3
4
10236.2
1070.0
cosh
1
cosh
1
B
B
T
L
x
 
 99951.0 
    995.099951.095969.0  T 
 95442.0 
 94.20
95442.01
95442.0
1







 
 Then 
     75.1680.094.20  BC II A 
(c)     3.11912595442.0  EC II A 
_______________________________________ 
 
12.23 
 (a) We have 
 
















































B
B
B
B
t
BE
B
BOB
nE
L
x
L
x
V
V
L
neD
J
sinh
1
tanh
1exp
 
 We find that 
 
  3
16
2102
105.4
105
105.1




B
i
BO
N
n
n cm 3 
 and 
   810515  BOBB DL  
 
410660.8  cm 
 Then 
 
   
4
319
1066.8
105.415106.1




nEJ 
 


































66.8
70.0
sinh
1
66.8
70.0
tanh
0259.0
60.0
exp
 
 or 
 779.1nEJ A/cm 2 
 We also have