ECE 440
HW3 Solutions
Summer 2009
Tue, Jun 30, 2009
1
1.
Calculate both the electron and hole concentrations for a germanium sample doped
with 2x10
16
arsenic atoms/cm
3
(a) at room temperature (b) at 450 K.
Doping germanium with arsenic will make the sample ntype. Thus, N
D
= 2x10
16
cm
3
. To
find the concentrations, one should solve equations (327) and (324) in S&B
simultaneously.
(1)
2
i
o
o
n
p
n
=
(2)
D
o
A
o
glyph817
p
glyph817
n
+
=
+
(a) At room temperature
Combining (1) and (2) along with n
i
from Figure (317) is enough to find the electron
concentration
(3)
3
16
16
2
13
10
2
10
2
)
10
5
.
2
(
−
=
+
=
cm
x
x
n
x
n
o
o
The hole concentration follows from equation (1):
(4)
3
10
2
13
10
125
.
3
)
10
5
.
2
(
−
=
=
cm
x
n
x
p
o
o
(b) At 450 K
Combining (1) and (2) along with n
i
from Figure (317) is enough to find the electron
concentration
(4)
3
16
16
2
16
10
41
.
2
10
1
)
10
1
(
−
=
+
=
cm
x
x
n
x
n
o
o
The hole concentration follows from equation (1):
(5)
3
15
2
13
10
14
.
4
)
10
5
.
2
(
−
=
=
cm
x
n
x
p
o
o
2.
A germanium sample is doped with 1x10
14
/cm
3
acceptors and 4x10
13
/cm
3
donors. (a)
What are the electron and hole concentrations respectively at room temperature under
equilibrium conditions? (b) What are the electron and hole concentrations respectively
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Summer '09
 Lie
 carrier concentration, hole concentrations, S&B

Click to edit the document details