The document describes 8 problems from an exam on semiconductors. Problem 1 calculates the quantity of metals extracted from various minerals per metric ton. Problem 2 determines film thicknesses and vapor requirements for solid-solid reactions forming InSb, CuSbS2, and CuInSe2. Problem 3 proposes emission lines that could be observed in EDS spectra of the reactants and products. Problem 4 estimates missing values in the reaction Cu + Fe + 2S → CuFeS2 and sketches the expected XRD pattern.

1. Problems 1st Exam
Enrico Castro Grespan
May 23, 2017
Contents
1 Problem 1 2
2 Problem 2 3
3 Problem 3 7
4 Problem 4 8
4.1 a) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
4.2 b) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
5 Problem 5 10
6 Problem 6 10
6.1 a) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
6.2 b) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
6.3 c) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
6.4 d) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
7 Problem 7 12
7.1 a) PbS of crystalline diameter 5 and 8 nm . . . . . . . . . . . . . 12
7.2 b) InSb of crystalline diameter 5 and 8 nm . . . . . . . . . . . . 13
7.3 c) GaSb of crystalline diameter 5 and 8 nm . . . . . . . . . . . . 13
7.4 d) Si of crystalline diameter 2, 3 and 4 nm . . . . . . . . . . . . . 13
8 Problem 8 14
8.1 a) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
8.2 b) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
8.3 c) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 14
9 Problem 9 15
1

2. 1 Problem 1
Many semiconductors occur in nature as minerals galena (PbS), zinc blende
(ZnS), cuprite (Cu20), cassiterite (SnO2), stibnite (Sb2S3) chalcopyrite (CuFeS2),
chalcostibite (CuSbS2). From these metals are extracted by metallurgical pro-
cessing reactions. Consider these examples and determine the quantity of metals
in kg per metric ton of mineral extracted in each case.
1 mol PbS −→ (207.2 + 32.065) g
A ←− 1000 kg
A = 4179.5 mol de PbS
1 mol PbS ←→ 1 mol Pb
4179.4 mol ←→ B
B = 4179.4 mol de Pb
1 mol Pb ←→ 207.3 g
4179.4 mol ←→ C
C = 866 kg de Pb
1 mol S ←→ 32.065 g
4179.4 mol ←→ D
D = 134.01 kg de S
The same method was used for the others minerals and the results are shown
in the next table.
PbS Pb 866 kg S 134 kg
ZnS Zn 670.97 kg S 329.02 kg
Cu2O 2 Cu 888.19 kg O 111.81 kg
SnO2 Sn 787.68 kg 2 O 212.32 kg
Sb2S3 2 Sb 716.87 kg 3 S 283.13 kg
CuFeS2 Cu 346.28 kg Fe 304.32 kg 2 S 349.4 kg
2

3. 2 Problem 2
Semiconducting films may be produced through solid-solid or solid-vapor reac-
tions of the following kind. Determine the thicknesses of films and/or vapor
requirement per cm2
of the films for the following cases:
Sb2S3 + 2In −→ 2 InSb + 3 S ↑
ρ =
m
V
−→ m =
ρ
V
V = 1 cm2
× (thickness in cm)
mSb2S3
= 4.63g cm−3
3 × 10−5
cm3
= 1.389 × 10−4
g
1 mol Sb2S3 −→ 339.7 g
A ←− 1.389 × 10−4
g
A = 4.089 × 10−7
mol
1 mol Sb2S3 −→ 2 mol In
4.089 × 10−7
mol −→ 8.178 × 10−7
mol
1 mol In −→ 114.818 g
8.178 × 10−7
mol −→ 9.389 × 10−5
g
ThicknessIn =
9.389 × 10−5
g
7.29 g cm−3
= 128.8 nm
For InSb
1 mol Sb2S3 −→ 2 mol InSb
4.089 × 10−7
mol −→ 8.178 × 10−7
mol
1 mol InSb −→ 236.58 g
8.178 × 10−7
mol −→ 1.935 × 10−4
g
ThicknessInSb =
1.935 × 10−4
g
5.78 g cm−3
= 334.7 nm
For the second reaction
3

4. Sb2S3(300 nm) + 2Cu + S −→ 2 CuSbS2
For Cu
1 mol Sb2S3 −→ 2 mol Cu
4.089 × 10−7
mol −→ 8.178 × 10−7
mol
1 mol Cu −→ 63.546 g
8.178 × 10−7
mol −→ 5.197 × 10−5
g
ThicknessCu =
5.197 × 10−5
g
8.93 g cm−3
= 58.19 nm
For CuSbS2
1 mol Sb2S3 −→ 2 mol CuSbS2
4.089 × 10−7
mol −→ 8.178 × 10−7
mol
1 mol CuSbS2 −→ 249.45 g
8.178 × 10−7
mol −→ 2.039 × 10−4
g
ThicknessCuSbS2 =
2.039 × 10−4
g
4.99 g cm−3
= 408.8 nm
For S
1 mol Sb2S3 −→ 1 mol S
1 mol S −→ 32.065 g
4.089 × 10−7
mol −→ 113.1 × 10−6
g
S = 13.11 µg cm−2
For the third reaction
Cu(300 nm) + In + 2 Se −→ CuInSe2
ρ =
m
V
−→ m =
ρ
V
V = 1 cm2
× (3 × 10−5
cm)
mCu = 8.93g cm−3
3 × 10−5
cm3
= 2.679 × 10−4
g
4

5. 1 mol Cu −→ 63.546 g
A ←− 2.679 × 10−4
g
A = 4.216 × 10−6
mol
For In
1 mol Cu −→ 1 mol In
1 mol In −→ 114.818 g
4.216 × 10−6
mol −→ 4.840 × 10−4
g
ThicknessIn =
4.840 × 10−4
g
7.29 g cm−3
= 664 nm
For CuInSe2
1 mol Cu −→ 1 mol CuInSe2
1 mol CuInSe2 −→ 336.284 g
4.216 × 10−6
mol −→ 1.418 × 10−3
g
ThicknessCuInSe2
=
1.418 × 10−3
g
5.77 g cm−3
= 2.46 µm
For S
1 mol Cu −→ 1 mol S
1 mol S −→ 32.065 g
4.216 × 10−6
mol −→ 135.2 × 10−6
g
S = 135.2 µg cm−2
For the fourth reaction
Cu(300 nm) + Zn + Sn + 4 S −→ Cu2ZnSnS4
ρ =
m
V
−→ m =
ρ
V
V = 1 cm2
× (3 × 10−5
cm)
mCu = 8.93g cm−3
3 × 10−5
cm3
= 2.679 × 10−4
g
5

6. 1 mol Cu −→ 63.546 g
A ←− 2.679 × 10−4
g
A = 4.216 × 10−6
mol
1 mol Cu −→ 1 mol In
4.216 × 10−6
mol −→ 4.216 × 10−6
mol
For Zn
1 mol Cu −→ 1 mol Zn
1 mol Zn −→ 65.38 g
4.216 × 10−6
mol −→ 2.756 × 10−4
g
ThicknessZn =
2.756 × 10−4
g
7.29 g cm−3
= 378 nm
For Sn
1 mol Cu −→ 1 mol Sn
1 mol Sn −→ 118.71 g
4.216 × 10−6
mol −→ 5.005 × 10−4
g
ThicknessSn =
5.005 × 10−4
g
5.77 g cm−3
= 87 µm
For S
2 mol Cu −→ 4 mol S
1 mol S −→ 64.12 g
4.216 × 10−6
mol −→ 2.70 × 10−4
g
S = 270 µg cm−2
For Cu2ZnSnS4
2 mol Cu −→ 1 mol Cu2ZnSnS4
1 mol Cu2ZnSnS4 −→ 439.442 g
4.216 × 10−6
mol −→ 9.263 × 10−4
g
ThicknessCu2ZnSnS4
=
9.263 × 10−4
g
4.56 g cm−3
= 203 µm
6

7. 3 Problem 3
Propose the energy dispersive emission lines in a graphical plot which may be
observed in the energy Dispersive X-ray spectra (EDS) of the reactant thin films
and in the thin film product in the above reactions
En2 − En1 = 13.6 eV z∗2 1
n2
1
−
1
n2
2
For Kα
n2 = 2 n1 = 1
1
n2
1
−
1
n2
2
=
3
4
z∗
= z − 1
Sb-Kα
En2
− En1
= 13.6 eV (50)2
(3/4) = 25.5 keV
λ = 1240/25500 = 0.049 nm
S-Kα
En2
− En1
= 13.6 eV (15)2
(3/4) = 2.29 keV
λ = 1240/2295 = 0.54 nm
In-Kα
En2 − En1 = 13.6 eV (48)2
(3/4) = 23.5 keV
λ = 1240/23501 = 0.053 nm
Cu-Kα
En2
− En1
= 13.6 eV (28)2
(3/4) = 7.99 keV
λ = 1240/7997 = 0.155 nm
Se-Kα
En2
− En1
= 13.6 eV (33)2
(3/4) = 11.1 keV
λ = 1240/11108 = 0.112 nm
Zn-Kα
En2 − En1 = 13.6 eV (29)2
(3/4) = 8.58 keV
λ = 1240/8578 = 0.144 nm
Sn-Kα
En2
− En1
= 13.6 eV (49)2
(3/4) = 24.5 keV
λ = 1240/24490 = 0.051 nm
7

8. Figure 1:
4 Problem 4
Consider the reaction for the formation of chalcopyrite:
Cu (300nm) + Fe (...nm) + 2 S (...µg/s) −→ CuFeS2 (...nm)
4.1 a)
Estimate the values of the missing thicknesses. CuFeS2 has tetragonal structure
with a = 5.29 ˚A and c = 10 ˚A and 4 formula units per cell.
ρ =
m
V
−→ m =
ρ
V
V = 1 cm2
× (3 × 10−5
cm)
mCu = 8.93g cm−3
3 × 10−5
cm3
= 2.679 × 10−4
g
1 mol Cu −→ 63.546 g
A ←− 2.679 × 10−4
g
A = 4.216 × 10−6
mol
1 mol Cu −→ 1 mol Fe
4.216 × 10−6
mol −→ 4.216 × 10−6
mol
For Fe
1 mol Cu −→ 1 mol Fe
8

9. 1 mol Fe −→ 55.845 g
4.216 × 10−6
mol −→ 2.354 × 10−4
g
ThicknessF e =
2.354 × 10−4
g
7.874 g cm−3
= 299 nm
For CuFeS2
1 mol Cu −→ 1 mol CuFeS2
1 mol CuFeS2 −→ 183.511 g
4.216 × 10−6
mol −→ 7.737 × 10−4
g
ρCuF eS2 =
183.511 ∗ 1.66 × 10−24
g
(5.29 × 10−8 cm)2(10−8 cm)
= 10.88g cm−2
ThicknessCuF eS2
=
7.737 × 10−4
g
10.88 g cm−3
= 711 nm
For S
1 mol Cu −→ 2 mol S
1 mol S −→ 32.065 g
8.432 × 10−6
mol −→ 2.703 × 10−4
g
S = 270.4 µg cm−2
4.2 b)
Propose the XRD pattern in the 2θ interval 20o
60o
for the reacting thin film
stack and for the product for Cu-Kα radiation (1.5406 ˙A).
2θ = 2 Arcsin
λ
√
h2 + l2 + k2
2 a0
9

10. Fe Cu FeCuS2
2 θ BCC FCC Tetragonal
a0 = 2.87˚A a0 = 3.61˚A a0 = 5.39˚A
(110) (111) (112)
1 44.61 43.38 40.98
(200) (200) (200)
2 64.95 50.52 33.21
(211) (220) (220)
3 74.25 47.68
(220) (311) (312)
4 64.65
(310) (400) (222)
5 59.35
5 Problem 5
In an experiment in Photovoltaic Lab II, Pb-S-Se thin films are produced
by chemical bath deposition from solution mixtures containing lead nitrate,
thiourea and selenosulfate. Three possibilities exist in the composition of the
resulting thin film:
a) Mixture of PbS and PbSe
b) Solid solution of the type PbS0.5Se0.5 or variation of such composition
c) A mixture of PbS, PbSe and PbS0.5Se0.5 Propose the XRD pattern for
the thin film produced in the three cases for Cu-K radiation (1.5406 ) in the 2θ
interval 20o
60o
. PbSe (a = 6.117˚A)
For c) the lattice constant is 6.0266 ˚A and all three are rock salt so the
possibles planes are (111), (200), (220), (311) and (222). Using the relation
2θ = 2 Arcsin
λ(
√
h2 + l2 + k2)2
2 a0
PbS PbSe PS0.5Se0.5
2 θ Rock salt Rock salt Rock salt
a0 = 5.9362˚A a0 = 6.117˚A a0 = 6.0266˚A
(111) 25.98 25.19 25.57
(200) 30.08 29.17 29.61
(220) 43.05 41.72 42.37
(311) 50.97 49.36 50.15
(222) 53.41 51.71 52.55
6 Problem 6
A crystalline semiconductor solid solution is formed of GaAsxSb1−x with its
band gap equal to that of Si (1.12 eV). Estimate the following:
10

11. 6.1 a)
The value of x and the value of lattice constant by considering proportionality
EgSi = x EgGaAs + (1 − x) EgGaSb
x =
EgSi − EgGaSb
EgGaAs − EgGaSb
=
1.12 − 0.72
1.42 − 0.72
= 0.57
a = x aGaAs + (1 − x) aGaSb = 0.57(5.6533) + (1 − 0.57)(6.0959) = 5.8434 ˚A
6.2 b)
Mass density of the semiconductor formed using the atomic mass and the lattice
constant calculated, calculate the mass of GaAs0.57Sb0.43. GaSb=191.483 g/mol
GaAs=144.6446 g/mol
GaAs0.53Sb0.43 = x mGaAs+(1−x) mGaSb = 0.57(144.6446)+(1−0.57)(191.483) = 164.7851 g mol−1
ρ =
4 atoms(164.7851 g/mol)
6.023 × 1023 atoms/mol(5.8434 × 10−8 cm)3
= 5.4849g cm−3
6.3 c)
EDX spectrum
Ga-Kα, z∗
= 30
En2
− En1
= 13.6 eV (30)2
(3/4) = 9.18 keV
λ = 1240/9180 = 0.135 nm
As-Kα, z∗
= 32
En2 − En1 = 13.6 eV (32)2
(3/4) = 10.44 keV
λ = 1240/10445 = 0.119 nm
Sb-Kα, z∗
= 30
En2
− En1
= 13.6 eV (50)2
(3/4) = 25.5 keV
λ = 1240/25500 = 0.49 nm
11

12. 6.4 d)
XRD pattern for the thin film for Cu-Kα radiation (1.5406 ); 2 θ interval 20o
60o
.
Both GaAs and GaSb have Zincblende structure, so GaAs0.57Sb0.43 also has a
zincblende structure. The Peaks of difraction are located in planes (111), (220),
(311) and a weak peak in plane (200). Using the lattice constant calculated is
possible calculate the angle 2 θ as it is shown.
2 θ GaAs0.53Sb0.47
a0 = 5.8434˚A
(111) 26.39
(220) 43.77
(311) 51.84
(200) 30.56
7 Problem 7
Consider Bohr radius of excitons and examine whether in the following cases the
Bohr radius exceeds the crystal diameter. If so, strong quantum confinement
of exciton is expected and hence the optical band gap exceeds that of the bulk
material. Thus, estimate the optical band gap of crystalline semiconductors of
the following specifications:
rB = 0.0529177 nm



r
z∗
m
m∗
e



Eg(eV )nm = EgA +
h2
2(a nm)2(m∗)me q
7.1 a) PbS of crystalline diameter 5 and 8 nm
me = 0.25 and mh = 0.25
1
m∗
e
=
1
0.25
+
1
0.25
= 8 m∗
e = 0.125 me
rB = 0.0529177 nm (17/0.125) = 7.19 nm
Eg5 nm = 0.41 +
h2
2(5 nm)2(0.125)me q
= 0.892 eV
Eg8 nm = 0.41 +
h2
2(8 nm)2(0.125)me q
= 0.598 eV
12

13. 7.2 b) InSb of crystalline diameter 5 and 8 nm
me = 0.0145 and mh = 0.4
1
m∗
e
=
1
0.0145
+
1
0.4
= 71.43 m∗
e = 0.014 me
rB = 0.0529177 nm (17.7/0.014) = 66.9 nm
Eg5 nm = 0.17 +
h2
2(5 nm)2(0.014)me q
= 4.47 eV
Eg8 nm = 0.17 +
h2
2(8 nm)2(0.014)me q
= 1.85 eV
7.3 c) GaSb of crystalline diameter 5 and 8 nm
me = 0.042 and mh = 0.4
1
m∗
e
=
1
0.042
+
1
0.4
= 26.31 m∗
e = 0.038 me
rB = 0.0529177 nm (15.7/0.038) = 21.86 nm
Eg5 nm = 0.72 +
h2
2(5 nm)2(0.038)me q
= 2.30 eV
Eg8 nm = 0.72 +
h2
2(8 nm)2(0.038)me q
= 1.34 eV
7.4 d) Si of crystalline diameter 2, 3 and 4 nm
me = 0.19 and mh = 0.49
1
m∗
e
=
1
0.19
+
1
0.49
= 7.30 m∗
e = 0.137 me
rB = 0.0529177 nm (11.9/0.137) = 4.60 nm
Eg2 nm = 1.12 +
h2
2(2 nm)2(0.137)me q
= 3.87 eV
Eg3 nm = 1.12 +
h2
2(3 nm)2(0.137)me q
= 2.34 eV
Eg4 nm = 1.12 +
h2
2(4 nm)2(0.137)me q
= 1.81 eV
13

14. 8 Problem 8
Intrinsic temperature (Ti) of an extrinsic semiconductor is the temperature at
which the intrinsic carrier concentration reaches the value of electron- or hole-
concentration estimated at room temperature for an extrinsic semiconductor
prepared by doping with donor/acceptor atoms. See p. 20 Book of S. M. Sze.
Evaluate the value of Ti for the following cases:
ni(T) =
1
2π2
√
π
2
2 m kB T
2
3/2
(me mh)3/4
e(Eg q)/kBT
8.1 a)
p-Si of electrical resistivity 10 Ωcm → 0.1 Ωm
pp =
σp
q µp
=
1
0.10(1.602 × 10−19)(0.045)
= 1.39 × 1021
m−3
ni(T) =
1
2π2
√
π
2
2 m kB T
2
3/2
(0.19 × 0.49)3/4
e1.12/kBT
= pp
using Mathematica for the calculation Ti = 645.5 K
8.2 b)
n-GaAs of electrical resistivity 1 Ωcm → 0.01 Ωm
nn =
σp
q µn
=
1
0.01(1.602 × 10−19)(0.85)
= 7.34 × 1020
m−3
ni(T) =
1
2π2
√
π
2
2 m kB T
2
3/2
(0.067 × 0.04)3/4
e1.42/kBT
= nn
using Mathematica for the calculation Ti = 991.3 K
8.3 c)
p-Ge of electrical resistivity 1 Ωcm → 0.01 Ωm
pp =
σp
q µp
=
1
0.01(1.602 × 10−19)(0.19)
= 3.28 × 1021
m−3
ni(T) =
1
2π2
√
π
2
2 m kB T
2
3/2
(0.082 × 0.28)3/4
e0.66/kBT
= pp
using Mathematica for the calculation Ti = 546.2 K
14

15. 9 Problem 9
Estimate and draw to the same scale the energy level diagrams of the above
three cases of semiconductors. Also draw the energy level diagram for p-Si/n-
GaAs and p-Ge/n-GaAs heterojunctions of materials of the above specifications,
clearly indicating the built-in voltage Vbi and the depletion layer width of the
junctions
p-Si of electrical resistivity 10 Ωcm → 0.1 Ωm; pp = 1.387 × 1021
m−3
EF p =
Eg
2
−
kBT
q
ln
pp
ni
=
1.12 eV
2
−0.0259 eV ln
1.39 × 1021
m−3
1.45 × 1016 m−3
= 0.26 eV
n-GaAs of electrical resistivity 1 Ωcm→ 0.01 Ωm; nn = 7.344 × 1020
m−3
EF n =
Eg
2
+
kBT
q
ln
nn
ni
=
1.42 eV
2
+0.0259 eV ln
7.344 × 1020
m−3
1.79 × 1012 m−3
= 1.22 eV
p-Si/n-GaAs
Vbi = EF n − EF p = 1.22 eV − 0.26 eV = 0.96 eV
W =
2 0 r (Vbi − 2kBT/q)
pp q
=
2(8.854 × 10−12
)(11.9) (0.96 − 2 × 0.0259) eV
1.387 × 1021m−3 (1.602 × 10−19)
= 928 nm
p-Ge of electrical resistivity 1 Ωcm→ 0.01 Ωm; pp = 3.28 × 1021
m−3
EF p =
Eg
2
−
kBT
q
ln
pp
ni
=
0.66 eV
2
−0.0259 eV ln
3.28 × 1021
m−3
2.4 × 1019 m−3
= 0.20 eV
p-Ge/n-GaAs
Vbi = EF n − EF p = 1.22 eV − 0.20 eV = 1.02 eV
W =
2 0 r (Vbi − 2kBT/q)
pp q
=
2(8.854 × 10−12
)(16) (1.02 − 2 × 0.0259) eV
3.28 × 1021m−3 (1.602 × 10−19)
= 722 nm
15