Soluções de Problemas de Semicondutores

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Semiconductor Physics and Devices: Basic Principles, 4
th
 edition Chapter 4 
By D. A. Neamen Problem Solutions 
______________________________________________________________________________________ 
 
or 
 15102 do Nn cm 3 
 
 
15
2132
102
104.2



o
i
o
n
n
p 
 111088.2  cm 3 
 (ii) 1516 10710  dao NNp 
 15103 cm 3 
 
 
15
2132
103
104.2



o
i
o
p
n
n 
 111092.1  cm 3 
(b) GaAs: 6108.1 in cm 3 
 (i) 15102 do Nn cm 
  3
15
26
1062.1
102
108.1 


op cm 3 
 (ii) 15103 dao NNp cm 3 
 
  3
15
26
1008.1
103
108.1 


on cm 3 
(c) The result implies that there is only one 
minority carrier in a volume of 
310 cm 3 . 
_______________________________________ 
 
4.37 
 (a) For the donor level 
 





 


kT
EEN
n
Fdd
d
exp
2
1
1
1
 
 








0259.0
20.0
exp
2
1
1
1
 
 or 
 
41085.8 
d
d
N
n
 
 (b) We have 
  





 


kT
EE
Ef
F
F
exp1
1
 
 Now 
    FccF EEEEEE  
 or 
 245.0 kTEE F 
 
 
 
 
 
 Then 
  








0259.0
245.0
1exp1
1
Ef F 
 or 
   51087.2 Ef F 
_______________________________________ 
 
4.38 
(a)  da NN p-type 
(b) Silicon: 
 1313 101105.2  dao NNp 
 or 
 13105.1 op cm 3 
 Then 
 
  7
13
2102
105.1
105.1
105.1




o
i
o
p
n
n cm 3 
 Germanium: 
 2
2
22
i
dada
o n
NNNN
p 




 


 
  213
2
1313
104.2
2
105.1
2
105.1







 







 
 
 or 
 131026.3 op cm 3 
 Then 
 
  13
13
2132
1076.1
10264.3
104.2




o
i
o
p
n
n cm 3 
 Gallium Arsenide: 
 13105.1  dao NNp cm 3 
 and 
 
 
216.0
105.1
108.1
13
262




o
i
o
p
n
n cm 3 
_______________________________________ 
 
4.39 
(a)  ad NN n-type 
(b) 1515 102.1102  ado NNn 
 
14108 cm 3 
 
  5
14
2102
1081.2
108
105.1




o
i
o
n
n
p cm 3