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Semiconductor Physics and Devices: Basic Principles, 4
th
 edition Chapter 9 
By D. A. Neamen Problem Solutions 
______________________________________________________________________________________ 
 
 
  
  8
14
19
1050
1085.87.11
106.1
35.0 





 n
d x
N
 
 
 






2
1050
28
 
 or 
  814 10251073.735.0   nd xN 
 We have 
 
  
 
2/1
19
14
106.1
1085.87.112













d
bi
n
N
V
x 
 and 
 nbiV  70.0 
 By trial and error, we find 
 18105.3 dN cm 3 
_______________________________________ 
 
9.30 
 (b) 








a
tpBO
N
N
V  ln 
   










16
19
105
1004.1
ln0259.0 
 or 
 138.0BO V 
_______________________________________ 
 
9.31 
 Sketches 
_______________________________________ 
 
9.32 
 Sketches 
_______________________________________ 
 
9.33 
 Electron affinity rule 
  pnc eE   
 For GaAs, 07.4 and for AlAs, 5.3 . 
 If we assume a linear extrapolation between 
 GaAs and AlAs, then for 
 Al 3.0 Ga 7.0 As 90.3 
 Then 
 17.090.307.4  cE eV 
_______________________________________ 
 
 
 
 
 
 
9.34 
 Consider an n-P heterojunction in thermal 
 equilibrium. Poisson's equation is 
 
 
dx
dx
dx
d 




2
2
 
 In the n-region, 
 
 
n
dn
n
n eNx
dx
d




 
 
 For uniform doping, we have 
 1C
xeN
n
dn
n 

 
 The boundary condition is 
 0 n at nxx  , so we obatin 
 
n
ndn xeN
C

1 
 Then 
  n
n
dn
n xx
eN


 
 In the P-region, 
 
P
aPp eN
dx
d



 
 which gives 
 2C
xeN
P
aP
P 

 
 We have the boundary condition that 
 0 P at Pxx  , so that 
 
P
PaP xeN
C

2 
 Then 
  xx
eN
P
P
aP
P 

 
 Assuming zero surface charge density at 
 0x , the electric flux density D is 
 continuous, so    00 PPnn  , which 
 yields 
 PaPndn xNxN  
 We can determine the electric potential as 
   dxx nn  
 3
2
2
C
xxeNxeN
n
ndn
n
dn 











