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Semiconductor Physics and Devices: Basic Principles, 4
th
 edition Chapter 14 
By D. A. Neamen Problem Solutions 
______________________________________________________________________________________ 
 
14.9 
 We have 
 
 
0
2
2

n
p
L
p
n
n
G
dx
nd
D


 
 or 
 
 
n
L
n
pp
D
G
L
n
dx
nd

22
2 
 
 where nnn DL  
 The general solution can be written in the 
 form 
   nL
nn
p G
L
x
B
L
x
Axn  















 sinhcosh 
 For s at 0x means   00 pn . Then 
 nLGA 0 nLGA  
 At Wx  , 
 
 
Wx
po
Wx
p
n ns
dx
nd
D

 

 
 Now 
   








n
nLp
L
W
GWn cosh 
 nL
n
G
L
W
B 







 sinh 
 and 
 
 









 nn
nL
Wx
p
L
W
L
G
dx
nd
sinh

 
 








nn L
W
L
B
cosh 
 so we can write 
 















nn
n
nn
nnL
L
W
L
BD
L
W
L
DG
coshsinh

 
  








n
nLo
L
W
Gs cosh 
 












 nL
n
G
L
W
B sinh 
 Solving for B, we obtain 
 

















































n
o
nn
n
n
no
n
nL
L
W
s
L
W
L
D
L
W
s
L
W
LG
B
sinhcosh
1coshsinh 
 
 
 
 
 
 The solution is then 
   

























nn
nLp
L
x
B
L
x
Gxn sinhcosh1 
 where B was just given. 
_______________________________________ 
 
14.10 
    36
0 1051025   nnn DL  cm 
   7
0 10510  ppp DL  
 310236.2  cm 
 Now 









dp
p
an
n
iS
NL
D
NL
D
enJ 2
 
   21019 105.1106.1   
 
     








 153163 1010236.2
10
10105
25
 
 1010790.1 SJ A/cm 2 
   101079.15  SS AJI 
 
1010950.8  A 
(a) AWeGI LL  
 We find 
  
  
 
6350.0
105.1
1010
ln0259.0
210
1516










biV V 
 
2/1
2















 

da
dabis
NN
NN
e
V
W 
 
   








19
14
106.1
635.01085.87.112
 
 
  
2/1
1516
1516
1010
1010









 
 
 
510508.9 W cm 
 Then 
     52119 10508.95105106.1  LI 
 380.0 A 380 mA 
(b) 








S
L
toc
I
I
VV 1ln 
   







101095.8
380.0
1ln0259.0 
 5145.0ocV V 
(c) 810.0
635.0
5145.0

bi
oc
V
V
 
_______________________________________