SOLUTIONS MANUAL
0
,Solutions manual to accompany semiconductordevices physicsandtechnology
3rd edition
TABLE OF CONTENTS
Ch.0 Introduction 0
Ch.1 Energy Bands and Carrier Concentration in Thermal Equilibrium------- 1
Ch.2 Carrier Transport Phenomena 9
Ch.3 p-n Junction 18
Ch.4 Bipolar Transistor and Related Devices---------------------------------------- 35
Ch.5 MOS Capacitor and MOSFET 52
Ch.6 Advanced MOSFET and Related Devices--------------------------------------62
Ch.7 MESFET and Related Devices 68
Ch.8 Microwave Diode, Quantum-Effect and Hot-Electron Devices---------- 76
Ch.9 Light Emitting Diodes and Lasers---------------------------------------------- 81
Ch.10 Photodetectors and Solar Cells 88
Ch.11 Crystal Growth and Epitaxy 96
Ch.12 Film Formation 105
Ch.13 Lithography and Etching 112
Ch.14 Impurity Doping 118
Ch.15 Integrated Devices 126
0
, CHAPTER 1
1. (a) From Fig. 11a, the atom at the center of the cube is surround by four
equidistant nearest neighbors that lie at the corners of a tetrahedron.
Therefore the distance between nearest neighbors in silicon (a = 5.43 Å) is
1/2 [(a/2)2 + ( 2a /2)2]1/2 = 3a /4 = 2.35 Å.
(b) For the (100) plane, there are two atoms (one central atom and 4 corner atoms
each contributing 1/4 of an atom for a total of two atoms as shown in Fig. 4a)
for an area of a2, therefore we have
2/ a2 = 2/ (5.43 × 10-8)2 = 6.78 × 1014 atoms / cm2
Similarly we have for (110) plane (Fig. 4a and Fig. 6)
(2 + 2 ×1/2 + 4 ×1/4) / 2a 2
= 9.6 × 1015 atoms / cm2,
and for (111) plane (Fig. 4a and Fig. 6)
3 2
(3 × 1/2 + 3 × 1/6) / 1/2( )( a ) = = 7.83 × 1014 atoms / cm2.
2a
2 3 2
a
2
2. The heights at X, Y, and Z point are 3 4, 14, and 3 .
4
3. (a) For the simple cubic, a unit cell contains 1/8 of a sphere at each of the eight
corners for a total of one sphere.
∴ Maximum fraction of cell filled
= no. of sphere × volume of each sphere / unit cell volume
= 1 × 4 π (a/2)3 / a3 = 52 %
(b) For a face-centered cubic, a unit cell contains 1/8 of a sphere at each of the
eight corners for a total of one sphere. The fcc also contains half a sphere at
each of the six faces for a total of three spheres. The nearest neighbor
1
, distance is 1/2(a 2 ). Therefore the radius of each sphere is 1/4 (a 2 ).
∴ Maximum fraction of cell filled
= (1 + 3) {4 π [(a/2) / 4 ]} / a3 = 74 %.
(c) For a diamond lattice, a unit cell contains 1/8 of a sphere at each of the eight
corners for a total of one sphere, 1/2 of a sphere at each of the six faces for a
total of three spheres, and 4 spheres inside the cell. The diagonal distance
between (1/2, 0, 0) and (1/4, 1/4, 1/4) shown in Fig. 9a is
1 a 2 a 2 a 2 a
D= = 3
2 2 2 2 4
a
The radius of the sphere is D/2 = 3
8
∴ Maximum fraction of cell filled
3
4 a
= (1 + 3 + 4) 3 / a3 = π = 34 %.
3 8
This is a relatively low percentage compared to other lattice structures.
4. d1 = d2 = d3 = d4 = d
d1 + d 2 + d3 + d 4 = 0
d1 • ( d1 + d 2 + d3 + d 4 ) = d1 • 0 = 0
d1 2 + d1 • d 2 + d 1 • d 3 + d • d = 0
1 4
∴d2+ d2 cos θ 12 + d2cos θ 13 + d2cos θ 14 = d2 +3 d2 cosθ= 0
1
∴ cos θ =
3
1
θ = cos-1 ( ) = 109.470 .
3
5. Taking the reciprocals of these intercepts we get 1/2, 1/3 and 1/4. The smallest
three integers having the same ratio are 6, 4, and 3. The plane is referred to as
(643) plane.
6. (a) The lattice constant for GaAs is 5.65 Å, and the atomic weights of Ga and As
are 69.72 and 74.92 g/mole, respectively. There are four gallium atoms and
2
0
,Solutions manual to accompany semiconductordevices physicsandtechnology
3rd edition
TABLE OF CONTENTS
Ch.0 Introduction 0
Ch.1 Energy Bands and Carrier Concentration in Thermal Equilibrium------- 1
Ch.2 Carrier Transport Phenomena 9
Ch.3 p-n Junction 18
Ch.4 Bipolar Transistor and Related Devices---------------------------------------- 35
Ch.5 MOS Capacitor and MOSFET 52
Ch.6 Advanced MOSFET and Related Devices--------------------------------------62
Ch.7 MESFET and Related Devices 68
Ch.8 Microwave Diode, Quantum-Effect and Hot-Electron Devices---------- 76
Ch.9 Light Emitting Diodes and Lasers---------------------------------------------- 81
Ch.10 Photodetectors and Solar Cells 88
Ch.11 Crystal Growth and Epitaxy 96
Ch.12 Film Formation 105
Ch.13 Lithography and Etching 112
Ch.14 Impurity Doping 118
Ch.15 Integrated Devices 126
0
, CHAPTER 1
1. (a) From Fig. 11a, the atom at the center of the cube is surround by four
equidistant nearest neighbors that lie at the corners of a tetrahedron.
Therefore the distance between nearest neighbors in silicon (a = 5.43 Å) is
1/2 [(a/2)2 + ( 2a /2)2]1/2 = 3a /4 = 2.35 Å.
(b) For the (100) plane, there are two atoms (one central atom and 4 corner atoms
each contributing 1/4 of an atom for a total of two atoms as shown in Fig. 4a)
for an area of a2, therefore we have
2/ a2 = 2/ (5.43 × 10-8)2 = 6.78 × 1014 atoms / cm2
Similarly we have for (110) plane (Fig. 4a and Fig. 6)
(2 + 2 ×1/2 + 4 ×1/4) / 2a 2
= 9.6 × 1015 atoms / cm2,
and for (111) plane (Fig. 4a and Fig. 6)
3 2
(3 × 1/2 + 3 × 1/6) / 1/2( )( a ) = = 7.83 × 1014 atoms / cm2.
2a
2 3 2
a
2
2. The heights at X, Y, and Z point are 3 4, 14, and 3 .
4
3. (a) For the simple cubic, a unit cell contains 1/8 of a sphere at each of the eight
corners for a total of one sphere.
∴ Maximum fraction of cell filled
= no. of sphere × volume of each sphere / unit cell volume
= 1 × 4 π (a/2)3 / a3 = 52 %
(b) For a face-centered cubic, a unit cell contains 1/8 of a sphere at each of the
eight corners for a total of one sphere. The fcc also contains half a sphere at
each of the six faces for a total of three spheres. The nearest neighbor
1
, distance is 1/2(a 2 ). Therefore the radius of each sphere is 1/4 (a 2 ).
∴ Maximum fraction of cell filled
= (1 + 3) {4 π [(a/2) / 4 ]} / a3 = 74 %.
(c) For a diamond lattice, a unit cell contains 1/8 of a sphere at each of the eight
corners for a total of one sphere, 1/2 of a sphere at each of the six faces for a
total of three spheres, and 4 spheres inside the cell. The diagonal distance
between (1/2, 0, 0) and (1/4, 1/4, 1/4) shown in Fig. 9a is
1 a 2 a 2 a 2 a
D= = 3
2 2 2 2 4
a
The radius of the sphere is D/2 = 3
8
∴ Maximum fraction of cell filled
3
4 a
= (1 + 3 + 4) 3 / a3 = π = 34 %.
3 8
This is a relatively low percentage compared to other lattice structures.
4. d1 = d2 = d3 = d4 = d
d1 + d 2 + d3 + d 4 = 0
d1 • ( d1 + d 2 + d3 + d 4 ) = d1 • 0 = 0
d1 2 + d1 • d 2 + d 1 • d 3 + d • d = 0
1 4
∴d2+ d2 cos θ 12 + d2cos θ 13 + d2cos θ 14 = d2 +3 d2 cosθ= 0
1
∴ cos θ =
3
1
θ = cos-1 ( ) = 109.470 .
3
5. Taking the reciprocals of these intercepts we get 1/2, 1/3 and 1/4. The smallest
three integers having the same ratio are 6, 4, and 3. The plane is referred to as
(643) plane.
6. (a) The lattice constant for GaAs is 5.65 Å, and the atomic weights of Ga and As
are 69.72 and 74.92 g/mole, respectively. There are four gallium atoms and
2
