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ECE331 Homework #6 Solution
1.
(a)
Since n0 << ni, so this is the minority carrier concentration. This is a ptype
semiconductor.
(b)
The minority carrier concentration n
0
= 4.5×10
4
cm
3
.
The majority carrier concentration p
0
,
3
15
4
2
10
10
0
.
5
10
5
.
4
)
10
5
.
1
(
0
2
0
−
×
=
×
×
=
=
cm
n
ni
p
(c)
We have both donors and acceptors in the material. The donor concentration is
given. Based on the space charge neutrality, the acceptor concentration,
Na = p0 + Nd – n0 = 1.0×10
16
cm
3
.
(d)
The carrier mobilities for electrons and holes are 1330 cm
2
/Vs and 495 cm
2
/Vs,
respectively (Anderson Text p. 116). So the conductivity is
cm
S
Vs
cm
cm
C
qp
qp
qn
p
p
n
/
396
.
0
495
10
5
10
6
.
1
2
3
15
19
=
×
×
×
×
×
=
≈
+
=
−
−
μ
σ
2.
(a)
The intrinsic carrier concentration at 300 K is
3
9
10
473
.
8
)
2
/
exp(
−
×
=
−
=
cm
kT
E
N
N
n
g
v
c
i
The current density at 300K,
2
14
19
n
1
/
6
.
1
100
1000
10
10
6
.
1
E
)
(
cm
A
E
qN
qn
E
qp
qn
E
J
n
d
p
n
=
×
×
×
×
=
≈
=
+
=
=
−
(b)
Since J
2
= 1.05J
1
, if we assume the mobility is independent of temperature, then
n
02
= 1.05n
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 Spring '09
 RAJAN
 carrier concentration, minority carrier concentration

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