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Semiconductor Physics and Devices: Basic Principles, 4
th
 edition Chapter 6 
By D. A. Neamen Problem Solutions 
______________________________________________________________________________________ 
 
6.23 
 Plot 
_______________________________________ 
 
6.24 
(a) From Equation (6.55) 
 
   
0
2
2

nO
onn
n
dx
nd
dx
nd
D




 
 or 
 
   
0
22
2

n
o
n
n
L
n
dx
nd
Ddx
nd 
 
 We have that 
 






e
kTD
n
n

 so we can define 
 
  LekTD
o
o
n
n




1
 
 Then we can write 
 
   
0
1
22
2



nL
n
dx
nd
Ldx
nd 
 
 The solution is of the form 
    xnn   exp0 where 0 
 Then 
 
 
 n
dx
nd


 and 
 
 n
dx
nd

 2
2
2
 
 Substituting into the differential equation, we 
 find 
      0
1
2
2 


nL
n
n
L
n

 
 or 
 0
1
2
2 


nLL

 
 which yields 
 

















 1
22
1
2
L
L
L
L
L
nn
n
 
 We may note that if 0 o , then L 
 and 
nL
1
 
 (b) 
 nOnn DL  where 






e
kT
D nn  
 so 
    1.310259.01200 nD cm 2 /s 
 and 
    47 104.391051.31  nL cm 
 
 
 or 
 4.39nL m 
 For 12o V/cm, then 
 
  4106.21
12
0259.0 


o
ekT
L cm 
 and 
 21075.5  cm 1 
(c) Force on the electrons due to the electric 
field is in the negative x-direction. Therefore, 
the effective diffusion of the electrons is 
reduced and the concentration drops off faster 
with the applied electric field. 
_______________________________________ 
 
6.25 
 p-type so the minority carriers are electrons 
 and 
    
 
t
nn
gnnD
nO
nn






2
 
 Uniform illumination means that 
     02  nn  . For nO , we are 
 left with 
 
 
g
dt
nd


 which gives 1Ctgn  
 For 0t , 00 1  Cn 
 Then 
 tGn o
 for Tt 0 
 For Tt  , 0g so that 
 
0
dt
nd 
 
 And 
 TGn o
 (no recombination) 
_______________________________________ 
 
6.26 
 n-type, so minority carriers are holes and 
    
 
t
pp
gppD
pO
pp






2 
 We have pO , 0 , and 
 
 
0


t
p
(steady-state). Then we have 
 
 
0
2
2
 g
dx
pd
D p

 or 
 
pD
g
dx
pd 

2
2 
 
 For LxL  , oGg  = constant. Then 
 
 
1Cx
D
G
dx
pd
p
o 


