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10/29/2007
EE 338
Fall 2007
HW#6 Solutions
1
(a)
Using Eq. 4.42 on pg. 151, we can get the electric field for the given nonuniform doping
distribution
0
18
3
00
0
()
1
since the doping distribution is given by
( )
where
10 cm and
0.1
μ
m
If we substitute
( ) into the given equation for electric field, we get
d
x
d
x
x
dd
d
d
x
dN
x
kT
eNx d
x
Nx Ne
N
x
Nx
kT
e
ε
−
−
⎛⎞
=−
⎜⎟
⎝⎠
==
=
0
0
0
0
3
4
1
0.0259V
V
2.6 10
cm
0.1 10 cm
x
x
x
x
x
x
dN
x
N
kT
e
kT
e
N
x
dx
e
N
x
x e
−
−
=
×
×
(b)
We know that diffusion current density is
0
0
0

0
since
holes does not contribute to the diffusion current
diff n
n
d
i
x
x
diff n
n
n
dn
Je
D
N
n
dx
dN
x
N
J
x
eD
eD
e
dx
x
−
=
−
±
(c)
The only other current will rise from the fact that there is a built in electric field due to
the nonuniform doping concentration. The drift current is;
0

0
0
drift n
n
x
x
x
nx d
n
d
n
kT
eN
x
e
N
e
ex
μ
με
−
=
1
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Then the net current density becomes,
00

0
0
0
0
Since we didn't apply any external voltage there shouldn't be any net current density.
0
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This note was uploaded on 02/27/2008 for the course EE 338 taught by Professor Dapkus during the Fall '07 term at USC.
 Fall '07
 Dapkus

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