Soluções de Problemas de Semicondutores

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Semiconductor Physics and Devices: Basic Principles, 4
th
 edition Chapter 7 
By D. A. Neamen Problem Solutions 
______________________________________________________________________________________ 
 
 1510684.2  dN cm 3 
 1710147.2 aN cm 3 
(b) 
 
 
2/1
12

























dad
aRbis
n
NNN
N
e
VV
x 
 
   








19
14
106.1
10740.01085.87.112
 
 
2/1
1517 10684.210147.2
1
1
80
















 
 
410262.2  cm 
 or 262.2nx m 
 
 
2/1
12

























daa
dRbis
p
NNN
N
e
VV
x 
 
   








19
14
106.1
10740.01085.87.112
 
 
2/1
1517 10684.210147.2
1
80
1
















 
 
61083.2  cm 
 or 0283.0px m 
(c) 
 
W
VV Rbi 
2
max 
 
 
  4100283.0262.2
10740.02


 
 
41038.9  V/cm 
(d) 
  
2/1
2 








daRbi
das
NNVV
NNe
C 
 
   
 






10740.02
1085.87.11106.1 1419
 
 
  
2/1
1517
1517
10684.210147.2
10684.210147.2












 
 
91052.4 C F/cm 2 
_______________________________________ 
 
7.19 
(a)    abiabi NVNV 3 
 
 


















22
ln
3
ln
i
ad
t
i
ad
t
n
NN
V
n
NN
V 
  


























22
lnln3ln
i
ad
t
i
ad
t
n
NN
V
n
NN
V 
 
      3ln0259.03ln  tV 
 02845.0 V 
(b) 
 
2/1
2 








Rbi
as
VV
Ne
C 
 So 
 
 
732.13
33
2/1










a
a
a
a
N
N
NC
NC
 
(c) For a larger doping, the space charge 
width narrows which results in a larger 
capacitance. 
_______________________________________ 
 
7.20 
(a)  
  
  










210
1715
105.1
104104
ln0259.0biV 
 or 
 766.0biV V 
 Now 
 
 
2/1
max
2



















da
da
s
Rbi
NN
NNVVe
 
 or 
     
  








14
19
25
1085.87.11
106.12
103 Rbi VV
 
 
  






1715
1715
104104
104104
 
 or 
  Rbi VV  910 10224.1109 
 so that 
   53.73 Rbi VV V 
 which yields 
 8.72RV V 
(b)  
  
  










210
1716
105.1
104104
ln0259.0biV 
 or 
 826.0biV V 
 We have 
     
  








14
19
25
1085.87.11
106.12
103 Rbi VV
 
 
  






1716
1716
104104
104104
 
 so that 
   008.8 Rbi VV V