EE211C Introduction to Physical Electronics,
Mid-Term solution
1.
(a) The lattice constant of GaAs is 5.65 . Determine the number of Ga atoms and As atoms per
cm-3. (b) Determine the volume density of germanium atoms in a germanium semiconductor.
The lattice constant of germanium is 5.65 . (10)
Solution) (Error in unit: -0.5 point/ each, calculation mistake: -1point/each)
(a) ∵ 4 Ga atoms per unit cell,
Density of Ga atoms
.
2.22
10
(3 points)
2.22
10
(3 points)
10
(3 points)
∵ 4 As atoms per unit cell,
Density of As atoms
.
(b) ∵ 8 Ge atoms per unit cell, (1point)
Density of Ga atoms
2.
.
4.44
Determine the surface density of atoms for silicon on the (a) (100) plane, (b) (110) plane, and (c)
(111) plane. The lattice constant of silicon is 5.43 . Refer Fig. 1 or Fig. 2. Crystallographic plane
views should be needed for calculating the surface density. (15)
(Fig.1 The zincblende structure)
(Fig.2 The diamond structure)
Page 1 / 6
Solution)
(a) Surface density on (100) plane
4
1
4
1
2
5.43 10
6.78
10
(Calculation: 3 points / Plane view: 2points)
(b) Surface density on (110) plane
1
2
4
√2
4
9.59
1
2
2
4
√2 5.43
10
10
(Calculation: 3 points / Plane view: 2points)
(c) Surface density on (111) plane
3
1
6 3
√3/2
7.83
1
2
2
√3/2 5.43
10
10
(Calculation: 3 points / Plane view: 2points)
3.
The solution to Schrodinger’s wave equation for a particular situation is given by
Ψ x
2/
exp
/
.
Note that the function Ψ x has been normalized between 0 and ∞. Determine the probability of
/4.
finding the particle between the following limit of 0
(10)
Solution)
|ψ
∴P
2
2
/
|
exp
2 /
P=
|ψ
exp
2 /
Correct equation ∶ 5points
/4
0
1 exp
1
2
2
|
|
1
0.393 5points
Page 2 / 6
4.
(a) Calculate the transmission coefficient of an electron with a kinetic energy of 0.1eV impinging
on a potential barrier of height 1.0 eV and a width of 4 . (b) Using the results of (a), determine the
density of electrons per second that impinge the barrier if the tunneling current density is 1.2
mA/cm2. (20)
Solution)
T ≅ 16
1
2
9.11
2
exp
2
10
1
1.054
4.860
4
(a) For a=4
T ≅ 16
0.1
1.0
(b) J
N ev (N
1
10
2
∴
0.1
exp
1.0
10
5points
2 ∙ 4.86
∙ 4
10
10
0.0295 5points
the density of transmitted electrons
2 1.6
9.11
1.6
1.60
,
1
2
E
→
10
0.1
10
10
10
1.2
10
0.1
1.874
10
1.874
10
1.6
10
4.002
10
/
J
1.874
10
10
/
/
3points
3points
⇒
4.002 10
0.0295
1.357
10
4points
Page 3 / 6
5.
Determine the total number (#/cm3) of energy states in silicon between
3kT at T =
and
300 K. (15)
Solution)
∗
Density of states effective mass (Si) of holes,
∗
As g
/
,
the total number of energy states between
4
∗
2
4
2
∗
2
3
∗
4
2
0
3
4 2
0.56 9.11
6.625 10
10
2.60
872.3
10
10
3
Correct equation ∶ 5points
∗
2
3
4
2
2
3
3
∗
2
3
3
3
2.981
10
3
300
4.14
10
When T=300K, kT
1.38
⇒g
10
3
2.981
10
4.13
10
4.13
10
2.981
10
3kT can be calculated as below.
to
2
4
g
g
0.56
J
3
4.14
10
10points
Page 4 / 6
6.
Consider the energy levels shown in Fig. 3. Let T = 300 K. If E1 - EF = 0.30 eV, determine (a) the
probability that an energy state at E = E1 is occupied by an electron and (b) the probability that an
energy state at E = E2 is empty. Here E1 – E2 = 1.42 eV. Refer Fig.3. (15)
(Fig. 3 Energy levels for Problem 6)
Solution)
Using Equation (3.79),
1
1
1
E
1
exp
exp
1
1
exp
1
(a) The probability that an energy state at E
exp
f
E E
kT
≅ exp
E1 E F
kT
≅ exp
exp
E is occupied by an electron
exp
0.3
0.0259
9.323
10 6
Correct equation ∶ 5 points, calculation ∶ 2 points
(b) The probability that an energy state at E
E
1
E
exp
E
E
E
E is empty
E
1.42
1.12
0.0259
0.30
1.658
1.12 eV
10
8points
Correct equation ∶ 6 points, calculation ∶ 2 points
Page 5 / 6
7.
The carrier effective masses in a semiconductor are mn* = 1.21 mo and mp* = 0.70 m0. Determine
the position of the intrinsic Fermi level with respect to the center of the bandgap at T = 300 K. Use
the following given equations. (15)
Solution)
∙
∙
,
For an intrinsic semiconductor,
∙
∙
2∙
2points
2
1
kT ln
2
⟹
∴
3
kT ln
4
2
2
∗
∗
3
4
3
kT ln
4
0.0259
ln
∗
∗
8points
0.70
1.21
10.63 meV 5points
Page 6 / 6