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Semiconductor Physics and Devices: Basic Principles, 4
th
 edition Chapter 12 
By D. A. Neamen Problem Solutions 
______________________________________________________________________________________ 
 
 
 
510175.1
300
400

BO
BO
n
n
 
 or 
     35 105.410175.1400 BOn 
 81029.5  cm 3 
 Then 
 
   
   8.15739.0tanh108.15
1029.525106.1
4
819




sOJ 
 or 
 510865.2 sOJ A/cm 2 
 Finally 
 
  



 







3004000259.02
55.0
exp
10865.2
105
1
1
5
8
 
 or 
 9999994.0 
_______________________________________ 
 
12.31 
 Plot 
_______________________________________ 
 
12.32 
 Plot 
_______________________________________ 
 
12.33 
 Plot 
_______________________________________ 
 
12.34 
 Plot 
_______________________________________ 
 
12.35 
(a)  
 
C
ACE
oACE
o
C
I
VV
rVV
r
I


1
 
 (i) 67.101
2.1
1202


or k 
 (ii) 00984.0
67.101
11

o
o
r
g (k ) 1 
   161084.9
  
(iii) 22.1
667.101
1204


CI mA 
 
 
 
 
 
 
(b) 
 (i) 648
25.0
1602





C
ACE
o
I
VV
r k 
 (ii) 00154.0
648
11

o
o
r
g (k ) 1 
   161054.1
  
 (iii) 253.0
648
1604


CI mA 
_______________________________________ 
 
12.36 
 
180
25






o
EC
C
C
EC
o
r
V
I
I
V
r 
 01667.0 CI mA 67.16 A 
_______________________________________ 
 
12.37 
 
 
 
2/1
12


















CBB
CCBbis
dB
NNN
N
e
VV
x 
 
   









19
14
106.1
1085.87.112 CBbi VV
 
 
 
2/1
161516
15
102102
1
102
102














 
     2/111108832.5 CBbi VV   
 Now 
 









2
ln
i
CB
tbi
n
NN
VV 
  
  
  










210
1615
105.1
102102
ln0259.0 
 6709.0 V 
 (i) For 4CBV V, 1658.0dBx m 
 (ii) For 8CBV V, 2259.0dBx m 
 (iii) For 12CBV V, 2730.0dBx m 
 Neglecting the B-E space charge width, 
 (i) For 4CBV V, 
 6842.01658.085.0 Bx m 
 (ii) For 8CBV V, 
 6241.02259.085.0 Bx m 
 (iii) For 12CBV V, 
 5770.02730.085.0 Bx m