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Semiconductor Physics and Devices: Basic Principles, 4
th
 edition Chapter 9 
By D. A. Neamen Problem Solutions 
______________________________________________________________________________________ 
  dmdmdE nn
** 2
2
1
 
 We can also write 
    FccF EEEEEE  
 nn em   2*
2
1
 
 so that 
 




 









kT
e
h
m
dn nn 
exp2
3
*
 
 

d
kT
mn 2
2*
4
2
exp 







 
 
 We can write 
 
2222
zyx   
 The differential volume element is 
 zyx dddd  24 
 The current is due to all x-directed velocities 
 that are greater than Ox and for all y- and 
 z- directed velocities. Then 
 




 










kT
e
h
m
J nn
ms

exp2
3
*
 
 x
xn
x d
kT
m
Ox











 
 

2
exp
2*
 
 y
yn
d
kT
m












 

2
exp
2*
 
 z
zn d
kT
m












 

2
exp
2*
 
 We can write 
  abiOxn VVem 2*
2
1
 
 Make a change of variables: 
 
 
kT
VV
kT
m abixn 

2
2
2
2*


 
 or 
 
 





 

kT
VVe
m
kT abi
n
x
2
*
2 2
 
 Taking the differential, we find 
  d
m
kT
d
n
xx 








*
2
 
 We may note that when Oxx   , 0 . 
 We may define other change of variables, 
 
 
 
 











2/1
*
2
2*
2
2
n
y
yn
m
kT
kT
m
 
 











2/1
*
2
2*
2
2
n
z
zn
m
kT
kT
m
 
 Substituting the new variables, we have 
 




 




















kT
e
m
kT
h
m
J n
n
n
ms

exp
2
2
2
*
3
*
 
 
     d
kT
VVe abi







 

0
2expexp 
      dd 




 22 expexp 
_______________________________________ 
 
9.20 
 For the Schottky diode, 
    







 
t
a
V
V
exp106101080.0 843 
(a)    
  









84
3
10610
1080.0
ln0259.0SBVa
 
 4845.0 V 
 Then 
   7695.0285.04845.0 pnVa V 
(b)   





 
0259.0
7695.0
exp101080.0 113
pnA 
 
55 1010998.0   pnA cm 2 
_______________________________________ 
 
9.21 
 For the pn junction, 
    16134 104.6108108  sI A 
(a)   












16
6
104.6
10150
ln0259.0aV 
 678.0 V 
(b)   












16
6
104.6
10700
ln0259.0aV 
 718.0 V 
(c)   












16
3
104.6
102.1
ln0259.0aV 
 732.0 V