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Semiconductor Physics and Devices: Basic Principles, 4
th
 edition Chapter 12 
By D. A. Neamen Problem Solutions 
______________________________________________________________________________________ 
12.48 
(a) 
 
C
BCB
s
B
pt
N
NNNex
V




2
2
0 
 
  
  14
2419
1085.87.112
1065.0106.1




 
 
  
155
102105102 161516


 
 6.32ptV V 
(b) From Chapter 7, 
 
2/1
max
2


















CB
CB
s
pt
NN
NNeV
 
 
  
  







14
19
1085.87.11
6.32106.12
 
 
  
2/1
1615
1615
102105
102105












 
 
5
max 1001.2  V/cm 
_______________________________________ 
 
12.49 
 
 
C
BCB
s
B
pt
N
NNNex
V




2
2
0 
 
  
  14
2
0
19
1085.87.112
106.1
15




 Bx
 
 
  
15
161516
103
105103105


 
 5
0 10483.1  Bx cm 1483.0 m 
_______________________________________ 
 
12.50 
 We have 
  
 
  









R
F
FCBF
BRC
tCE
II
II
VsatV




1
1
ln 
 We can write 
 
    
    














20.0
99.0
99.01199.0
2.011
0259.0
exp
B
BCE
I
IsatV
 
 or 
 
 
 95.4
01.099.0
8.0
0259.0
exp 















B
BCE
I
IsatV
 
 
 
 
 
 
 
 (a) For   30.0satVCE V, we find 
 5100726.1
0259.0
30.0
exp 





 
  95.4
01.099.0
8.0











B
B
I
I
 
 We find 
 01014.0BI mA 14.10 A 
 (b) For   20.0satVCE V, we find 
 0119.0BI mA 9.11 A 
 (c) For   10.0satVCE V, we find 
 105.0BI mA 105 A 
_______________________________________ 
 
12.51 
 For an npn transistor biased in the active 
 mode, we have 0BCV , so that 
 0exp 







t
BC
V
V
. Now 
  ECBCBE IIIIII  0 
 Then we have 
 

























 CS
t
BE
ESFB I
V
V
II 1exp 
 
























 1exp
t
BE
ESCSR
V
V
II 
 or 
  
















 1exp1
t
BE
ESFB
V
V
II  
   CSR I 1 
_______________________________________ 
 
12.52 
 We can write 
 
















1exp
t
BE
ES
V
V
I 
 E
t
BC
CSR I
V
V
I 
















 1exp 
 Substituting, we find 
 

























 E
t
BC
CSRFC I
V
V
II 1exp 
 
















 1exp
t
BC
CS
V
V
I