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ECE 440
Homework VIII
Fall 2008
Due: Monday, October 27, 2008
1.
An abrupt Si pn junction has the following properties at 300 K:
pside
nside
A = 10
3
cm
2
N
a
= 3x10
17
/cm
3
N
d
= 6x10
15
/cm
3
τ
n
=
0.1 μs
τ
p
= 10 μs
μ
p
= 180 cm
2
/Vs
μ
n
= 1160 cm
2
/Vs
μ
n
= 540 cm
2
/Vs
μ
p
= 420 cm
2
/Vs
(a) Draw the band diagram qualitatively under forward and reverse bias showing the quasi
Fermi levels.
(b) Calculate the reverse saturation current due to holes, due to electrons and the total reverse
saturation current.
The equation for reverse saturation current is equation (537b).
(1)
+
=
p
n
n
n
p
p
n
L
D
p
L
D
qA
I
0
For the reverse saturation current due to holes we need the following quantities
(2)
s
cm
u
q
kT
D
p
p
/
87
.
10
420
0259
.
2
=
⋅
=
=
(3)
cm
D
L
p
p
p
0104
.
=
⋅
=
τ
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3
4
2
10
75
.
3

×
=
=
cm
n
n
p
n
i
n
(5)
A
p
L
D
qA
I
n
p
p
sat
p
15
4
3
19
,
10
6.2712
10
75
.
3
0104
.
87
.
10
10
10
6
.
1



×
=
⋅
⋅
⋅
⋅
⋅
=
=
The same for the reverse saturation current due to electrons.
(6)
s
cm
u
q
kT
D
n
n
/
99
.
13
540
0259
.
2
=
⋅
=
=
(7)
cm
D
L
n
n
n
3
10
2
.
1

⋅
=
⋅
=
τ
(8)
3
2
750

=
=
cm
p
n
n
p
i
p
(9)
A
n
L
D
qA
I
p
n
n
sat
n
15
3
3
19
,
10
4
.
1
750
10
2
.
1
99
.
13
10
10
6
.
1




×
=
⋅
⋅
⋅
⋅
⋅
=
=
The total reverse saturation current can be found by evaluating (1) or by summing (5) and
(9) as in (10).
(10)
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