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Semiconductor Physics and Devices: Basic Principles, 4
th
 edition Chapter 12 
By D. A. Neamen Problem Solutions 
______________________________________________________________________________________ 
 The general solution to the differential 
 equation is of the form 
   






 









BB
B
L
x
B
L
x
Axn expexp 
 From the boundary conditions, we have 
     BOBB nnBAn  00 
 
















 1exp
t
BE
BO
V
V
n 
 Also 
   






 









B
B
B
B
BB
L
x
B
L
x
Axn expexp 
 BOn 
 From the first boundary condition, we can 
 write 
 B
V
V
nA
t
BE
BO 
















 1exp 
 Substituting into the second boundary 
 condition, we find 
 















 








B
B
B
B
L
x
L
x
B expexp 
 BO
B
B
t
BE
BO n
L
x
V
V
n 
























 exp1exp 
 Solving for B, we find 
 


































B
B
BO
B
B
t
BE
BO
L
x
n
L
x
V
V
n
B
sinh2
exp1exp
 
 We then find 
 















 



















B
B
BO
B
B
t
BE
BO
L
x
n
L
x
V
V
n
A
sinh2
exp1exp
 
_______________________________________ 
 
12.14 
 In the base of the pnp transistor, we have 
 
    
0
2
2

BO
BB
B
xp
dx
xpd
D


 
 or 
 
    
0
2
2

B
BB
L
xp
dx
xpd 
 
 where BOBB DL  
 
 
 The general solution is of the form 
   






 









BB
B
L
x
B
L
x
Axp expexp 
 From the boundary conditions, we can write 
     BOBB ppBAp  00 
 
















 1exp
t
EB
BO
V
V
p 
 Also 
   






 









B
B
B
B
BB
L
x
B
L
x
Axp expexp 
 BOp 
 From the first boundary condition equation, 
 we find 
 B
V
V
pA
t
EB
BO 
















 1exp 
 Substituting into the second boundary 
 equation, we obtain 
 


































B
B
BO
B
B
t
EB
BO
L
x
p
L
x
V
V
p
B
sinh2
exp1exp
 
 and then we obtain 
 















 



















B
B
BO
B
B
t
EB
BO
L
x
p
L
x
V
V
p
A
sinh2
exp1exp
 
 Substituting the expressions for A and B into 
 the general solution and collecting terms, we 
 obtain 
  




























 1exp
sinh
t
EB
B
B
BO
B
V
V
L
x
p
xp 
 



















 

BB
B
L
x
L
xx
sinhsinh 
_______________________________________ 
 
12.15 
 For the idealized straight line approximation, 
 the total minority carrier concentration is 
 given by 
   






 


















B
B
t
BE
BOB
x
xx
V
V
nxn exp