EE 331
Devices and Circuits I
Autumn 2007
Problem Set #2 Solution
2.18
Since Ge is from column IV, acceptors come from column III and donors come from column
V.
(a) Acceptors: B, Al, Ga, In, Tl
(b) Donors: N, P, As, Sb, Bi
2.23
N
A
>
N
D
: N
A
−
N
D
=
10
15
−
10
14
=
9
x
10
14
/cm
3
If we assume
N
A
−
N
D
>>
2
n
i
=
10
14
/
cm
3
:
p
=
N
A
−
N
D
=
9
x
10
14
/cm
3

n
=
n
i
2
p
=
2510
26
9
x
10
14
=
2
.
78
x
10
12
/cm
3
If
we use Eq. 2.12 :
p
=
9
x
10
14
±
9
x
10
14
(
)
2
+
4 5
x
10
13
(
)
2
2
=
9
.
03
x
10
14
and
n
=
2
.
77
x
10
12
/cm
3
.
The answers are essentially the same.
2.29
(a) Arsenic is a donor, and boron is an acceptor.
N
D
= 2 x 10
18
/cm
3
, and N
A
= 8 x 10
18
/cm
3
.
Since N
A
> N
D
, the material is ptype.
3
3
18
6
20
2
3
18
i
3
18
3
10
i
7
16
10
6
10
and
10
6
So
2n
>>
/
10
6
and
/
10
n
re,
temperatu
room
At
(b)
/cm
.
/cm
x
/cm
p
n
n
/cm
x
p
cm
x
N
N
cm
i
D
A
=
=
=
=
=
−
=
2.33
Indium is from column 3 and is an acceptor.
N
A
= 7 x 10
19
/cm
3
.
Assume N
D
= 0, since it is not
specified.
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