EMSE 201 — Introduction to Materials Science & Engineering
3 December 2008
Name:
SOLUTION
Department of Materials Science and Engineering
1 of 5
Case Western Reserve University
Test #3 — 75 minutes; 100 points; 6 questions; 5 pages; 10% of course grade
Partial credit
will be given for correct set-ups and reasoning. Give
units
on numerical answers where
appropriate. Please write all answers on these pages; use the backs if needed.
Constants
:
Wiedemann-Franz constant,
£
=
π
2
k
B
2
/(3q
e
2
)
=
2.445
×
10
-8
W
Ω
K
-2
Boltzmann’s constant,
k
B
= 1.381
×
10
-23
J K
-1
= 8.620
×
10
-5
eV K
-1
gas constant,
R
: 8.314 J mol
-1
K
-1
Charge on an electron,
q
e
= –1.602
×
10
-19
C
Faraday’s constant,
F
= 96,500 C mol
–1
1) State whether each of the following elements will act as a
donor
or
acceptor
when added to the
indicated semiconducting material. Briefly
justify
your answer. Assume that the added elements act as
substitutional impurites.
a)
(2 points)
B added to Si
Boron is in column IIIA, whereas Si is in column IVA. With one fewer valence electron
to contribute to the covalent bonding of the Si crystal, an empty state (hole) will
result in the valence band (
1 pt
), making boron an
acceptor
dopant (
1 pt
) in silicon.
b)
(4 points)
Zn added to GaAs
Zinc will most likely substitute for Ga, being just 1 lower in atomic number than Ga
(
1.5 pt
). Zinc sits in column IIB, whereas Ga is in column IIIA. With one fewer valence
electron than Ga to contribute to the covalent bonding of the GaAs crystal, an empty
state (hole) will result in the band structure (
1.5 pt
), making Zn an
dopant
(
1 pt
) in GaAs.
c)
(4 points)
In added to CdTe
Indium will most likely substitute for Cd, being just 1 higher in atomic number than Cd
and being more cationic/metallic in character than anionic (
1.5 pt
). Indium sits in
column IIIA, whereas Cd is in column IIB. With one more valence electron than Cd to
contribute to the covalent bonding of the CdTe crystal, an extra electron will result in
the conduction band (
1.5 pt
), making In a
donor
dopant (
1 pt
) in CdTe.
2)
(8 points)
The room-temperature electrical conductivity of a silicon specimen is 5.93
×
10
–3
(
Ω
–1
m
–1
).
The hole concentration is known to be 7.0
×
10
17
m
–3
. Given that the electron and hole mobilities are 0.14
and 0.050 m
2
V
–1
s
–1
respectively,
compute the electron concentration.
State whether the specimen is
intrinsic, n-type, or p-type. Justify
your answer.
To solve for
n
, use Equation 12.13, which, after rearrangement, gives:
n
=
σ
−
pq
e
μ
h
q
e
e
(
3 pts
)
= 1.4
×
10
16
m
–3
(
2 pt
answer
+ 1 pt
units)
This material is
p-type
(
1 pt
) since
p
(7.0
×
10
17
m
–3
) is greater than
n
(
1 pt
).