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Semiconductor Physics and Devices: Basic Principles, 4
th
 edition Chapter 8 
By D. A. Neamen Problem Solutions 
______________________________________________________________________________________ 
 
 Define 
 
kT
eVa
a  and 




 

kT
EE FiFn 
 Then the recombination rate can be written as 
 
  
 iiiiO
iii
neennen
neenen
R
a
a








2
 
 or 
 
 
 

 


eee
en
R
a
a
O
i
2
1
 
 To find the maximum recombination rate, set 
 0
d
dR
 
 
    1
2
1 

 


eee
d
den
a
a
O
i 
 or 
 
 
   2
21
1
0


 


eee
en
a
a
O
i 
    eee a 
 which simplifies to 
 
   
 22
1
0


 






eee
eeeen
a
aa
O
i 
 The denominator is not zero, so we have 
 
  eee a0 
 or 
 aee
 2
 
2
a  
 Then the maximum recombination rate 
 becomes 
 
 
 22max
2
1
aaa
a
eee
en
R
O
i


 


 
 
 
 22
2
1
aa
a
ee
en
O
i


 

 
 or 
 
 
 12
1
2max



a
a
e
en
R
O
i



 
 which can be written as 
 

























1
2
exp2
1exp
max
kT
eV
kT
eV
n
R
a
O
a
i

 
 If  ekTVa  , then we can neglect the (-1) 
 term in the numerator and the (+1) term in the 
 denominator, so we finally have 
 
 
 






kT
eVn
R a
O
i
2
exp
2
max

 
 Q.E.D. 
_______________________________________ 
 
8.35 
 We have 
 
W
gen eGdxJ
0
 
 In this case, 19104 gG cm 3 s 1 and is 
 a constant through the space charge region. 
 Then 
 WgeJ gen
 
 We find 
 









2
ln
i
da
tbi
n
NN
VV 
  
  
  










210
1515
105.1
105105
ln0259.0 
 or 
 659.0biV V 
 Also 
 
 
2/1
2















 

da
daRbis
NN
NN
e
VV
W 
 
   








19
14
106.1
10659.01085.87.112
 
 
  
2/1
1515
1515
105105
105105














 
 or 
 
41035.2 W cm 
 Then 
    41919 1035.2104106.1  genJ 
 or 
 
3105.1 genJ A/cm 2 
_______________________________________