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Semiconductor Physics and Devices: Basic Principles, 4
th
 edition Chapter 8 
By D. A. Neamen Problem Solutions 
______________________________________________________________________________________ 
 
      





 
17
421019
10
1
10105.1106.1 
 










t
D
V
V
exp
10
29.8
7
 
 or 
 







 
t
D
p
V
V
I exp10278.3 16 (A) 
 Then 
 0741.0
10426.4
10278.3
15
16






S
pp
J
J
I
I
 
_______________________________________ 
 
8.16 
(a) 
d
i
po
p
p
nop
sp
N
nD
eA
L
peD
AI
2











 
    
16
210
8
419
105.1
105.1
108
10
105106.1






 
 
1410342.1 spI A 
(b) 
a
i
no
n
n
pon
sn
N
nD
eA
L
neD
AI
2











 
    
16
210
7
419
105
105.1
102
25
105106.1







 1510025.4 snI A 
(c)  
  
  










210
1616
105.1
105.1105
ln0259.0biV 
 746826.0 V 
      59746.0746826.08.08.0  bia VV V 
   
















t
a
d
i
t
a
nonn
V
V
N
n
V
V
pxp expexp
2
 
 
 









0259.0
59746.0
exp
105.1
105.1
16
210
 
 
141056.1  cm 3 
(d)     








t
a
snnnpn
V
V
IxIxI exp 
   





 
0259.0
59746.0
exp10025.4 15 
 
5101981.4  A 
 
 
 
 
 
 
(e)   








t
a
spnp
V
V
IxI exp 
   





 
0259.0
59746.0
exp10342.1 14 
 4103997.1  A 
 
pnTotal III  
 45 103997.1101981.4   
 410820.1  A 
 Now 
  
 







 







p
p
nppnp
L
L
xILxI
21
exp
2
1
 
   




 
 
2
1
exp103997.1 4 
 
5104896.8  A 
 Then 
 











 pnpTotalpnn LxIILxI
2
1
2
1
 
 
54 104896.810820.1   
 
510710.9  A 
_______________________________________ 
 
8.17 
 (a) The excess hole concentration is given by 
 nonn ppp  
 







 


















pt
a
no
L
x
V
V
p exp1exp 
 We find 
 
  4
16
2102
1025.2
10
105.1



d
i
no
N
n
p cm 3 
 and 
   61001.08  pOpp DL  
 
410828.2  cm 828.2 m 
 Then 
   











 1
0259.0
610.0
exp1025.2 4
np 
 








410828.2
exp
x
 
 or 
   








4
14
10828.2
exp1081.3
x
pn cm 3