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March 11, 2008
ECE 315
Prelim # 1 Solution
Spring 2008
1.
Consider a silicon p
+
n junction sample where the acceptor density far exceeds that of
the donor density.
N
a
= 10
19
cm

3
N
d
= 10
16
cm

3
a)(8 pts)
Using the concept of overall charge neutrality, show that we can ignore the de
pletion region thickness on the p side of the diode.
In the depletion region the charge on each side is equal and opposite, then:
qN
d
x
n
=
qN
a
x
p
→
x
p
=
N
d
N
a
x
n
=
x
n
1000
≈
0
.
What can be said!
b)(7 pts)
Compute the diode’s built in potential (I call this
φ
B
).
The built in potential (back of exam) is given by:
φ
B
=
kT
q
ln
N
a
N
d
n
2
i
= 25
.
86(10

3
) ln
10
19
10
16
[1
.
08(10
10
)]
2
= 0
.
8892
volts .
c)(7 pts)
At what applied voltage will the depletion region be 1
μ
m thick. Neglecting the
depletion on the pside, then from back of exam:
x
d
≈
x
n
=
s
2
±
s
(
φ
B

V
A
)
qN
d
N
a
N
d
+
N
a
≈
s
2
±
s
(
φ
B

V
a
)
qN
d
→
V
a
=
φ
B

qN
d
x
2
d
2
±
.
V
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