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Semiconductor Physics and Devices: Basic Principles, 4
th
 edition Chapter 5 
By D. A. Neamen Problem Solutions 
______________________________________________________________________________________ 
 
5.34 
  pLx
ppp e
dx
d
eD
dx
dp
eDJ
/15105

 
 
 
p
Lx
p
L
eeD p/15105


 
(a) (i) 
   
4
1519
1050
10510106.1




pJ 
 6.1 A/cm 2 
(ii) 
   
4
1519
105.22
10548106.1




pJ 
 07.17 A/cm 2 
(b) (i) 
   
4
11519
1050
10510106.1





e
J p 
 589.0 A/cm 2 
(ii) 
   
4
11519
105.22
10548106.1





e
J p 
 28.6 A/cm 2 
_______________________________________ 
 
5.35 
 
dx
dn
eDneJ nnn   
 or 
    










 
 
18
exp10960106.140 1619 x
 
    1619 1025106.1  
 




 









 18
exp
1018
1
4
x
 
 Then 
   




 











 

18
exp22.22
18
exp536.140
xx
 
 We find 
 
 
  




 





 

18
exp536.1
40
18
exp22.22
x
x
 
 or 
   




 

18
exp0.265.14
x
 
_______________________________________ 
 
 
 
 
 
 
5.36 
 (a)  Lx
nnn e
dx
d
eD
dx
dn
eDJ /15102  
 
 
L
eeD Lx
n
/15102 
 
 
   
4
/1519
1015
10227106.1





Lxe
 
 Lxe /76.5  
 (b)  Lx
nTotalp eJJJ /76.510  
  1076.5 /   Lxe A/cm 2 
 (c) We have   opp peJ  
      1619/ 10420106.11076.5 Lxe 
 So  88.1457.8 /   Lxe V/cm 
_______________________________________ 
 
5.37 
(a)  
 
dx
xdn
eDxneJ nn   
 We have 8000n cm 2 /V-s, so that 
    20780000259.0 nD cm 2 /s 
 Then 
      xn128000106.1100 19 
   
 
dx
xdn
207106.1 19 
 which yields 
        
dx
xdn
xn 1714 10312.310536.1100   
 Solution is of the form 
   




 

d
x
BAxn exp 
 so that 
 
 





 

d
x
d
B
dx
xdn
exp 
 Substituting into the differential equation, we 
 have 
   










 
 
d
x
BA exp10536.1100 14 
 
 





 


d
x
B
d
exp
10312.3 17
 
 This equation is valid for all x, so 
  A1410536.1100  
 or 
 
151051.6 A