Soluções de Problemas em Física de Semicondutores

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Semiconductor Physics and Devices: Basic Principles, 4
th
 edition Chapter 13 
By D. A. Neamen Problem Solutions 
______________________________________________________________________________________ 
 
 or 
 086.2POV V 
 Then 
 20.1086.2884.0 PV V 
 Let 20.1GSV V 
 Now 
 
 
2/1
2





 

d
GSDSbis
n
eN
VVV
x 
 
  
 







19
14
106.1
1085.87.112
 
 
  
 
2/1
16103
20.1884.0





 DSV
 
 or 
     2/110 084.210314.4 DSn Vx   
 (a) For 0DSV , 30.0nx m 
 (b) For 1DSV V, 365.0nx m 
 (c) For 5DSV V, 553.0nx m 
 
 The depletion region volume at the drain is 
        WaxW
L
aVol n 2
2






 
    4
4
4 1030
2
104.2
103.0 

 






 
 
    44 1030106.0   nx 
 or 
  812 1018108.10   nxVol 
 (a) For 0DSV , 
111062.1 Vol cm 3 
 (b) For 1DSV V, 
1110737.1 Vol cm 3 
 (c) For 5DSV V, 
1110075.2 Vol cm 3 
 
 The generation current at the drain is 
 Vol
n
eI
O
i
DG 








2
 
  
 
Vol










8
10
19
1052
105.1
106.1 
 or 
   VolI DG  2104.2 
 (a) For 0DSV , 39.0DGI pA 
 (b) For 1DSV V, 42.0DGI pA 
 (c) For 5DSV V, 50.0DGI pA 
_______________________________________ 
 
 
13.34 
 (a) The ideal transconductance for 0GSV is 
 









PO
bi
OmS
V
V
Gg 11 
 where 
 
L
WaNe
G dn
O

1 
 
   
4
1619
105.1
1074500106.1




 
   44 103.0105   
 or 
 04.51 OG mS 
 We find 
 
s
d
PO
Nea
V


2
2
 
 
    
  14
162419
1085.81.132
107103.0106.1




 
 or 
 347.4POV V 
 We have 
   049.0
107
107.4
ln0259.0
16
17










n V 
 so that 
 841.0049.089.0  nBnbiV  V 
 Then 
  









347.4
841.0
104.5mSg 
 or 
 82.2mSg mS 
 (b) With a source resistance 
 
smm
m
sm
m
m
rgg
g
rg
g
g






1
1
1
 
 For 
 
  sm
m
rg
g
823.21
1
80.0



 
 we obtain 
  6.88sr 
 (c) 
 
A
L
A
L
rs



 Ane
L
n
