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Semiconductor Physics and Devices: Basic Principles, 4
th
 edition Chapter 6 
By D. A. Neamen Problem Solutions 
______________________________________________________________________________________ 
 
Chapter 6 
 
6.1 
 15105 do Nn cm 3 
 
  4
15
2102
105.4
105
105.1




d
i
o
N
n
p cm 3 
(a) Minority carrier hole lifetime is a 
constant. 
 
7
0 102  ppt  s 
 11
7
4
0
1025.2
102
105.4





p
o
po
p
R

cm 3 s 1 
(b) 
7
144
0 102
10105.4





p
o
po
pp
R


 
 
20105 cm 3 s 1 
_______________________________________ 
 
6.2 
 16102 ao Np cm 3 
 
  4
16
262
1062.1
102
108.1 



o
i
o
p
n
n cm 3 
(a) 21
7
14
0
10
105
105





n
n
R


cm 3 s 1 
(b) 
0n
o
nt
o
pt
o
p
nnp
R

 
 
 
 
 7
4
16
0 105
1062.1
102 




 n
o
o
pt
n
p
 
 
131017.6  s 
_______________________________________ 
 
6.3 
(a) Recombination rates are equal 
 
pO
o
nO
o pn

 
 1610 do Nn cm 3 
 
  4
16
2102
1025.2
10
105.1



o
i
o
n
n
p cm 3 
 Then 
 
6
416
1020
1025.210



nO
 
 which yields 
 61089.8 nO s 
 
 
 
(b) Generation rate = recombination rate 
Then 
 9
6
4
10125.1
1020
1025.2





G cm 3 s 1 
 (c) 
 910125.1 GR cm 3 s 1 
_______________________________________ 
 
6.4 
(a) 
  
10
834
106300
10310625.6







hc
hE 
or 
 191015.3 E J; energy of one photon 
 Now 
 1 W = 1 J/s 
181017.3  photons/s 
 Volume = (1)(0.1) = 0.1 cm 3 
 Then 
 
1.0
1017.3 18
g 
 
191017.3  e-h pairs/cm 3 -s 
 (b) 
   619 10101017.3   gpn 
 or 
 141017.3  pn  cm 3 
_______________________________________ 
 
6.5 
 We have 
 
p
pp
p
gF
t
p



  
 and 
 peDpeJ ppp   
 The hole particle current density is 
 
 
pDp
e
J
F pp
p
p 

  
 Now 
   pDpF ppp    
 We can write 
    ppp 
 and 
 pp 2 
 so 
   pDppF ppp
2  