ECE 440
Homework VIII Solutions
Spring 2006
Due: Friday, March 03, 2006
1.
An abrupt Si pn junction is formed by alloying a uniformly doped ntype silicon bar of
which N
d
= 4x10
16
/cm
3
. During the alloying process, a uniform counter doping of acceptors
of N
a
= 1.5x10
17
/cm
3
is introduced in the region for x<0. For x>0, the doping remains to be
N
d
= 4x10
16
/cm
3
. So, x<0 is the pside and x>0 is the nside.
(a) Calculate the Fermi level positions at 300 K in the p and n regions.
Treat each side separately (as if they were unconnected) to calculate the Fermi level
positions far from the junction.
pside
:
eV
n
N
N
kT
n
p
kT
E
E
i
d
a
i
f
i
409
.
0
10
5
.
1
10
4
10
5
.
1
ln
0259
.
0
ln
ln
10
16
17
0
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
×
×
−
×
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
−
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
=
−
nside
:
eV
n
N
kT
n
n
kT
E
E
i
d
i
i
f
383
.
0
10
5
.
1
10
4
ln
0259
.
0
ln
ln
10
16
0
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
×
×
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
=
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛
=
−
(b) Draw an equilibrium band diagram for the junction and determine the contact potential
V
o
from the diagram.
Looking at the band diagram we can calculate qV
0
geometrically. The contact potential is
the difference between the Fermi level positions on the n and p sides of the junction.
eV
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 Spring '11
 Leburton
 Electrostatics, Electric charge, Ion, Pn junction, nd

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