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Semiconductor Physics and Devices: Basic Principles, 4
th
 edition Chapter 11 
By D. A. Neamen Problem Solutions 
______________________________________________________________________________________ 
 
 






2
1
dTaB xeNQ 
 Then change in the threshold voltage is 
 
 
ox
SD
oc
B
T
C
Q
C
Q
V
max


 
 or 
 
    
ox
dTa
ox
dTa
T
C
xeN
C
xeN
V 
21
 
 which becomes 
 
ox
dTa
T
C
xeN
V 
2
1
 
_______________________________________ 
 
11.23 
 Plot 
_______________________________________ 
 
11.24 
 Plot 
_______________________________________ 
 
11.25 
 
















 1
2
1
j
dTj
ox
dTa
T
r
x
L
r
C
xeN
V 
 
 





























 1
2
1
j
dTj
ox
dT
a
kr
kx
kL
kr
k
C
kx
k
N
e
 
 or 
 TT VkV  
_______________________________________ 
 
11.26 
 
  
8
14
1080
1085.89.3




oxC 
 
710314.4  F/cm 2 
   3758.0
105.1
103
ln0259.0
10
16










fp V 
 
   
  
2/1
1619
14
103106.1
3758.01085.87.114











dTx 
 
51080.1  cm 
 
 
 
 
 
 
 






W
x
C
xeN
V dT
ox
dTa
T

 
 
   
  47
251619
102.210314.4
2
1080.1103106.1












 
 0257.0 TV V 
_______________________________________ 
 
11.27 
 
  
8
14
10120
1085.89.3




oxC 
 
710876.2  F/cm 2 
   3473.0
105.1
10
ln0259.0
10
16









fp V 
 
 
   
  
2/1
1619
14
10106.1
3473.01085.87.114











dTx 
 
5100.3  cm 
 






W
x
C
xeN
V dT
ox
dTa
T

 
 In this case, 1 
 So 
 
    
  W7
251619
10876.2
1030.110106.1
045.0




 
 11.1W m 
_______________________________________ 
 
11.28 
 Plot 
_______________________________________ 
 
11.29 
 






W
x
C
xeN
V dT
ox
dTa
T

 
 Assume that  is a constant, then 
 
 



















kW
kx
k
C
kx
k
N
e
V dT
ox
dT
a
T

 
 or 
 TT VkV  
_______________________________________