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Semiconductor Physics and Devices: Basic Principles, 4
th
 edition Chapter 12 
By D. A. Neamen Problem Solutions 
______________________________________________________________________________________ 
 Now 
 








t
BE
B
BB
C
V
V
x
neD
J exp0 
 where 
 
 
16
2102
0
102
105.1



B
i
B
N
n
n 
 
410125.1  cm 3 
 so 
 
   








0259.0
650.0
exp
10125.125106.1 419
B
C
x
J 
 
Bx
3105686.3 
 A/cm 2 
 (i)For 4CBV V, 16.52CJ A/cm 2 
 (ii)For 8CBV V, 18.57CJ A/cm 2 
 (iii)For 12CBV V, 85.61CJ A/cm 2 
(b) 
ACE
C
CE
C
VV
J
V
J




 
 
AV



650.04
16.52
412
16.5285.61
 
 4.38 AV V 
_______________________________________ 
 
12.38 
 We find 
 
  3
16
2102
105.7
103
105.1




B
i
BO
N
n
n cm 3 
 and 
   








t
BE
BOB
V
V
nn exp0 
   






0259.0
7.0
exp105.7 3 
 or 
   151010.40 Bn cm 3 
 We have 
 
 
B
BBB
B
x
neD
dx
dn
eDJ
0
 
 
   
Bx
1519 1010.420106.1 


 
 or 
 
Bx
J
210312.1 
 A/cm 2 
 Neglecting the space charge width at the B-E 
 junction, we have 
 
 
 
pBOB xxx  
 Now 
  
  
  










210
1516
105.1
105103
ln0259.0biV 
 or 
 705.0biV V 
 Also 
 
 
2/1
12

























CBB
CCBbis
p
NNN
N
e
VV
x 
 
   









19
14
106.1
1085.87.112 CBbi VV
 
 
2/1
161516
15
103105
1
103
105




















 
 or 
     2/11110163.6 CBbip VVx  
 
 For 5CBV V, 1875.0px m 
 For 10CBV V, 2569.0px m 
 (a) For 0.1BOx m 
 For 5CBV V, 
 8125.01875.00.1 Bx m 
 Then 
 5.161
108125.0
10312.1
4
2






J A/cm 2 
 For 10CBV V, 
 7431.02569.00.1 Bx m 
 and 
 6.176
107431.0
10312.1
4
2






J A/cm 2 
 We can write 
  ACE
CE
VV
V
J
J 


 
 where 
 
510
5.1616.176








CBCE V
J
V
J
 
 02.3 A/cm 2 /V 
 Then 
   AV 7.502.35.161 
 which yields 
 8.47AV V