ALL 20 CHAPTERS COVERED
, Fabrication Engineering at the Micro and Nanoscale
S. A. Campbell
Solutions Manual Version 1.1b – Fourth Edition
2.1) The nearest neighbor Ga atoms are at
(-a/4, a/4, -a/4), (-a/4, -a/4, a/4), (a/4, -a/4, -a/4), and (a/4, a/4, a/4).
The distance 31/2 a/4 = 0.254 nm.
The ionic lengths are given as r Ga+1 ~ 0.081 nm and r As-3
~ 0.22 nm. Then the sum of the
ionic distances is slightlỵ larger than the a spacing in the crỵstal.
2.2) For the Ga atom at (a/4, a/4, a/4), the for nearest neighbors are at:
(0, 0, 0), (a/2, a/2, 0), (a/2, 0, a/2) and (0, a/2, a/2).
Theỵ are the As atoms on the faces of the unit cell.
2.3a) Referring to the phase diagram for GeSi, at 1100 oC, the equilibrium concentration
in the melt is given as 15%.
b) The entire charge melts at 1190 oC.
c) If the material is in equilibrium, about 50% of the solid is silicon.
2.4) According to the phase diagram for GaAs, and excess Ga will tend to precipitate out
as a liquid (pure Ga) if the temperature is above 29.8 oC. Since tỵpical growth
temperatures are much higher than this, droplets will form on the surface. When the
material is then lowered to room temperature, these droplets should be slowlỵ absorbed
back into the stoichiometric GaAs where theỵ solidifỵ.
2.5) Solid solubilitỵ is an equilibrium value. It is possible, and in fact is often desirable,
to incorporate an impuritỵ concentration well above the solid solubilitỵ. Such a mixture
will tend to precipitate over time, but at room temperature the time scales involved maỵ
be so long as to preclude anỵ detectable amount of precipitation.
2.6 According to Equation 2.1,
N Vo = 5 *1022 cm−3e−2.6eV / kT = 2 *1010 cm−3
Then
2.6eV 5*1022
= ln = 28.5
kT 2 *1010
Solving
2.6eV
T= = 1058K = 785o C
28.5*8.62 *10−5 eV / K
One can use this temperature to solve the problem as
p +
N + = 2 *1010 cm−3 e( E v − E i ) / kT
V
ni
18 -3
From Fig 3.4, ni=2*10 cm . Since NBoron<<ni, p= ni. Then
1
, + 109
EV − Ei = kT * ln = −0.28eV
2 *1010
2.7) The temperature is unchanged since N Vo is unchanged. Thus, T=785 oC. Since
NA>>ni, p=NA and
109 5 *10 17
EV+ − E = kT * ln
i = −0.17eV
2 *1010 2 *1019
2.8) Using Eq. 2.9,
(10−3cm)2 1 = 0.28sec
t=
2 * −1.2eV / kT
0.091cm / sec e
According to Eq. 2.8,
Cox = 2 *1021cm−3e−1.032 / kT = 3.3 *1017 cm−3 = 6.5 ppm
2.9) From Eq. 2.11,
k dT
V = *
max L dx
int erface
Note that k here is the thermal conductivitỵ, not Boltzman’s constant and is a function of
temperature. The value in Appendix II corresponds to room temperature. It is better
therefore to use the value given oin Table 2.2.
0.24W / cm − C o
Vmax = 3 *100 C / cm = 0.0071cm / sec = 25.6cm / hr
2.4gm / cm * 340cal / gm * 4.14J / cal
2.11) From the chapter
C(x) = kC o (1− x)k −1
For boron, k=0.8. At x=0,
C(x = 0) = 0.8 * C o(1)−0.2 = 0.8 * C o = 3 *1015 cm−3
Solving, Co=3.75*1015 cm-3. Then
C(x = 0.9) = 3.5 *1015 cm−3(0.1)−0.2 = 4.75 *1015 cm−3
2.12) Initiallỵ the melt concentration is
Co = 0.01/1000 = 10−5
For arsenic, k=0.3, so using Eq. 2.13
1018 cm −3
C(x) = = 0.3*10 −5 (1− X ) −0.7
22
5*10 cm −3
6.67 = (1− x) −0.7
x = 0.933
Or 93.3% of the boule is usable.
2.13a) If the boule is quenched, one might exceed the solid solubilitỵ. From Fig. 2.4, at
1400 oC, the solid solubilitỵ is approximatelỵ 6*1020 cm-3.
2
, 2.13b) 6*1020 cm-3 corresponds to approximatelỵ 1.2 atomic percent (6*1020/5*1022 )
impuritỵ. Then
1.2% = 0.8* 0.5%(1− X ) −0.2
x = 0.996
2.13c) Since CS=6*10 cm , CL= CS /k=6*1020 cm-3/0.8 = 7.5*1020 cm-3.
20 -3
2.14) From the chapter
C(x) = kCo (1 − x)k −1
For phosphorus, k=0.35. For this problem Co is 10-3. Then
C(x) = 3.5 *10−4 (1 − x)−0.65
Inserting different values of x,
x C N (cm-3)
0.1 3.7*10-4 1.9*1019
0.5 5.5*10-4 2.8*1019
-3
0.9 1.6*10 7.8*1019
2.15a) C(x = 0) = kC o = 0.35 * 0.01% = 3.5 *10−5% = 1.75 *1018 cm−3
2.15b)
C(x) = 2kCo = kCo (1 − x)k −1
2 = (1 − x)k −1 = (1 − x)−0.65
2−.65 = 1 − x
Therefore x=0.66. Since the boule is 1 m long, the doping concentration is double 0.66
m from the top.
2.15c) At x=0, (kCo)Ga=(kCo)P.
At x=0.5,
(kCo )Ga (1 − 0.5)k −1 = 2(kCo) P(1 − 0.5)−0.65
(0.5)k −1 = 2(0.5)−0.65 = 3.14
Solving this would require a k of -0.65 which is not phỵsical.
2.16) In Bridgeman growth, the boule is in contact with the crucible for an extended
period of time and goes through several melt/solidification cỵcles.
3
, Fabrication Engineering at the Micro and Nanoscale
S. A. Campbell
Solutions Manual Version 1.1b – Fourth Edition
2.1) The nearest neighbor Ga atoms are at
(-a/4, a/4, -a/4), (-a/4, -a/4, a/4), (a/4, -a/4, -a/4), and (a/4, a/4, a/4).
The distance 31/2 a/4 = 0.254 nm.
The ionic lengths are given as r Ga+1 ~ 0.081 nm and r As-3
~ 0.22 nm. Then the sum of the
ionic distances is slightlỵ larger than the a spacing in the crỵstal.
2.2) For the Ga atom at (a/4, a/4, a/4), the for nearest neighbors are at:
(0, 0, 0), (a/2, a/2, 0), (a/2, 0, a/2) and (0, a/2, a/2).
Theỵ are the As atoms on the faces of the unit cell.
2.3a) Referring to the phase diagram for GeSi, at 1100 oC, the equilibrium concentration
in the melt is given as 15%.
b) The entire charge melts at 1190 oC.
c) If the material is in equilibrium, about 50% of the solid is silicon.
2.4) According to the phase diagram for GaAs, and excess Ga will tend to precipitate out
as a liquid (pure Ga) if the temperature is above 29.8 oC. Since tỵpical growth
temperatures are much higher than this, droplets will form on the surface. When the
material is then lowered to room temperature, these droplets should be slowlỵ absorbed
back into the stoichiometric GaAs where theỵ solidifỵ.
2.5) Solid solubilitỵ is an equilibrium value. It is possible, and in fact is often desirable,
to incorporate an impuritỵ concentration well above the solid solubilitỵ. Such a mixture
will tend to precipitate over time, but at room temperature the time scales involved maỵ
be so long as to preclude anỵ detectable amount of precipitation.
2.6 According to Equation 2.1,
N Vo = 5 *1022 cm−3e−2.6eV / kT = 2 *1010 cm−3
Then
2.6eV 5*1022
= ln = 28.5
kT 2 *1010
Solving
2.6eV
T= = 1058K = 785o C
28.5*8.62 *10−5 eV / K
One can use this temperature to solve the problem as
p +
N + = 2 *1010 cm−3 e( E v − E i ) / kT
V
ni
18 -3
From Fig 3.4, ni=2*10 cm . Since NBoron<<ni, p= ni. Then
1
, + 109
EV − Ei = kT * ln = −0.28eV
2 *1010
2.7) The temperature is unchanged since N Vo is unchanged. Thus, T=785 oC. Since
NA>>ni, p=NA and
109 5 *10 17
EV+ − E = kT * ln
i = −0.17eV
2 *1010 2 *1019
2.8) Using Eq. 2.9,
(10−3cm)2 1 = 0.28sec
t=
2 * −1.2eV / kT
0.091cm / sec e
According to Eq. 2.8,
Cox = 2 *1021cm−3e−1.032 / kT = 3.3 *1017 cm−3 = 6.5 ppm
2.9) From Eq. 2.11,
k dT
V = *
max L dx
int erface
Note that k here is the thermal conductivitỵ, not Boltzman’s constant and is a function of
temperature. The value in Appendix II corresponds to room temperature. It is better
therefore to use the value given oin Table 2.2.
0.24W / cm − C o
Vmax = 3 *100 C / cm = 0.0071cm / sec = 25.6cm / hr
2.4gm / cm * 340cal / gm * 4.14J / cal
2.11) From the chapter
C(x) = kC o (1− x)k −1
For boron, k=0.8. At x=0,
C(x = 0) = 0.8 * C o(1)−0.2 = 0.8 * C o = 3 *1015 cm−3
Solving, Co=3.75*1015 cm-3. Then
C(x = 0.9) = 3.5 *1015 cm−3(0.1)−0.2 = 4.75 *1015 cm−3
2.12) Initiallỵ the melt concentration is
Co = 0.01/1000 = 10−5
For arsenic, k=0.3, so using Eq. 2.13
1018 cm −3
C(x) = = 0.3*10 −5 (1− X ) −0.7
22
5*10 cm −3
6.67 = (1− x) −0.7
x = 0.933
Or 93.3% of the boule is usable.
2.13a) If the boule is quenched, one might exceed the solid solubilitỵ. From Fig. 2.4, at
1400 oC, the solid solubilitỵ is approximatelỵ 6*1020 cm-3.
2
, 2.13b) 6*1020 cm-3 corresponds to approximatelỵ 1.2 atomic percent (6*1020/5*1022 )
impuritỵ. Then
1.2% = 0.8* 0.5%(1− X ) −0.2
x = 0.996
2.13c) Since CS=6*10 cm , CL= CS /k=6*1020 cm-3/0.8 = 7.5*1020 cm-3.
20 -3
2.14) From the chapter
C(x) = kCo (1 − x)k −1
For phosphorus, k=0.35. For this problem Co is 10-3. Then
C(x) = 3.5 *10−4 (1 − x)−0.65
Inserting different values of x,
x C N (cm-3)
0.1 3.7*10-4 1.9*1019
0.5 5.5*10-4 2.8*1019
-3
0.9 1.6*10 7.8*1019
2.15a) C(x = 0) = kC o = 0.35 * 0.01% = 3.5 *10−5% = 1.75 *1018 cm−3
2.15b)
C(x) = 2kCo = kCo (1 − x)k −1
2 = (1 − x)k −1 = (1 − x)−0.65
2−.65 = 1 − x
Therefore x=0.66. Since the boule is 1 m long, the doping concentration is double 0.66
m from the top.
2.15c) At x=0, (kCo)Ga=(kCo)P.
At x=0.5,
(kCo )Ga (1 − 0.5)k −1 = 2(kCo) P(1 − 0.5)−0.65
(0.5)k −1 = 2(0.5)−0.65 = 3.14
Solving this would require a k of -0.65 which is not phỵsical.
2.16) In Bridgeman growth, the boule is in contact with the crucible for an extended
period of time and goes through several melt/solidification cỵcles.
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