Semiconductor Physics and Devices: Basic Principles, 3rd edition {Complete Solution Manual on all Chapters).

Semiconductor Physics and Devices: Basic Principles, 3rd edition Chapter 1
Solutions Manual Problem Solutions


Chapter 1
Problem Solutions FG 4πr IJ 3




1.1
4 atoms per cell, so atom vol. = 4
H3K
(a) fcc: 8 corner atoms × 1/8 = 1 atom Then
6 face atoms × ½ = 3 atoms FG 4πr IJ
3


Total of 4 atoms per unit cell
Ratio =
4
H 3 K × 100% ⇒ Ratio = 74%
(b) bcc: 8 corner atoms × 1/8 = 1 atom
3
16 2 r
1 enclosed atom = 1 atom (c) Body-centered cubic lattice
Total of 2 atoms per unit cell
4
d = 4r = a 3 ⇒ a = r
(c) Diamond: 8 corner atoms × 1/8 = 1 atom 3
6 face atoms × ½ = 3 atoms
F 4 rI 3

4 enclosed atoms = 4 atoms Unit cell vol. = a =
H 3K
3

Total of 8 atoms per unit cell
FG 4πr IJ 3

1.2
(a) 4 Ga atoms per unit cell
2 atoms per cell, so atom vol. = 2
H3K
4 Then
Density =
b −8
g 3
⇒
FG 4πr IJ 3

5.65 x10

Density of Ga = 2.22 x10 cm
−3
Ratio =
H 3 K × 100% ⇒
2
Ratio = 68%
F 4r I
22
3

4 As atoms per unit cell, so that
Density of As = 2.22 x10 cm
22 −3
H 3K
(d) Diamond lattice
(b) 8
8 Ge atoms per unit cell Body diagonal = d = 8r = a 3 ⇒ a = r
8 3
Density =
b −8 3
⇒
g F 8r I 3

5.65 x10 Unit cell vol. = a =
H 3K
3


−3
Density of Ge = 4.44 x10 cm
FG 4πr IJ
22
3




1.3
8 atoms per cell, so atom vol. 8
H3K
(a) Simple cubic lattice; a = 2r Then
Unit cell vol = a = (2 r ) = 8r
3 3 3
FG 4πr IJ 3




1 atom per cell, so atom vol. = (1)
FG 4πr IJ
3
Ratio =
H 3 K × 100% ⇒
8
Ratio = 34%
H3K F 8r I 3



Then H 3K
FG 4πr IJ3




Ratio =
H 3 K × 100% ⇒ Ratio = 52.4%
1.4
From Problem 1.3, percent volume of fcc atoms
3
8r is 74%; Therefore after coffee is ground,
(b) Face-centered cubic lattice Volume = 0.74 cm
3

d
d = 4r = a 2 ⇒ a = =2 2r
2
Unit cell vol = a = 2 2 r
3
c h = 16
3
2r
3




3

,Semiconductor Physics and Devices: Basic Principles, 3rd edition Chapter 1
Solutions Manual Problem Solutions

Then mass density is
1.5 4.85 x10
−23

ρ= ⇒
(a) a = 5.43 A
°
From 1.3d, a =
8
3
r b2.8x10 g−8 3




ρ = 2.21 gm / cm
3

a 3 (5.43) 3 °
so that r = = = 118
. A
8 8
Center of one silicon atom to center of nearest 1.8
(a) a 3 = 2(2.2 ) + 2(1.8) = 8 A
°
°
neighbor = 2r ⇒ 2.36 A
so that
(b) Number density °
8 a = 4.62 A
−3
= ⇒ Density = 5 x10 cm
22


b
5.43 x10
−8 3
g Density of A =
1
⇒ 1.01x10 cm
22 −3


(c) Mass density b4.62 x10 g −8 3




=ρ=
N ( At . Wt .)
=
b5x10 g(28.09) ⇒22
1 −3
Density of B =
b4.62 x10 g ⇒
22
23 −8
1.01x10 cm
NA 6.02 x10
ρ = 2.33 grams / cm
3
(b) Same as (a)
(c) Same material

1.6 1.9
(a) a = 2 rA = 2(1.02) = 2.04 A
°
(a) Surface density
Now 1 1
= 2 = ⇒
2 rA + 2rB = a 3 ⇒ 2rB = 2.04 3 − 2.04
°
a 2 4.62 x10b−8
g 2
2
so that rB = 0.747 A 14
3.31x10 cm
−2


(b) A-type; 1 atom per unit cell
Same for A atoms and B atoms
1 (b) Same as (a)
Density = ⇒
b
2.04 x10
−8 3
g (c) Same material
23 −3
Density(A) = 118
. x10 cm 1.10
B-type: 1 atom per unit cell, so 1
23 −3 (a) Vol density =
. x10 cm
Density(B) = 118 ao
3



1
1.7 Surface density = 2
ao 2
(b)
° (b) Same as (a)
. + 1.0 ⇒ a = 2.8 A
a = 18
(c) 1.11
12 −3 Sketch
Na: Density = = 2.28 x10 cm
22


b2.8x10 g −8 3

1.12
Cl: Density (same as Na) = 2.28 x10 cm
22 −3
(a)
F 1 , 1 , 1I ⇒ (313)
(d)
Na: At.Wt. = 22.99 H 1 3 1K
Cl: At. Wt. = 35.45 (b)
So, mass per unit cell
F 1 , 1 , 1 I ⇒ (121)
1 1
(22.99) + (35.45) H 4 2 4K
= 2 2 −23
23
= 4.85 x10
6.02 x10


4

,Semiconductor Physics and Devices: Basic Principles, 3rd edition Chapter 1
Solutions Manual Problem Solutions

1.13 2 atoms −2
= ⇒ 9.88 x10 cm
14
(a) Distance between nearest (100) planes is:
d = a = 5.63 A
° b4.50x10 g −8 2



(ii) (110) plane, surface density,
(b)Distance between nearest (110) planes is:
2 atoms −2
1 a 5.63 = ⇒ 6.99 x10 cm
14

d= a 2=
2 2
=
2 b
2 4.50 x10
−8 2
g
or (iii) (111) plane, surface density,
d = 3.98 A
°
F1 1 I
(c) Distance between nearest (111) planes is: =
H
3⋅ + 3⋅
6 2
=
K4
1
d= a 3=
a
=
5.63 3 2
a
3 4.50 x10
−8 2
b g
3 3 3 2
or or . x10 cm
114
15 −2

°
d = 3.25 A
1.15
1.14 (a)
(a) (100) plane of silicon – similar to a fcc,
Simple cubic: a = 4.50 A
° 2 atoms
surface density = ⇒
(i) (100) plane, surface density,
1 atom
5.43 x10 b
−8 2
g
−2 14 −2
= ⇒ 4.94 x10 cm
14
6.78 x10 cm
b
4.50 x10
−8 2
g (b)
(ii) (110) plane, surface density, (110) plane, surface density,
1 atom −2 4 atoms
⇒ 3.49 x10 cm −2
14
= ⇒ 9.59 x10 cm
14
=
b
2 4.50 x10
−8 2
g b
2 5.43 x10
−8 2
g
(iii) (111) plane, surface density, (c)
1 FI 1 (111) plane, surface density,
3
6 HK
atoms
2 1 =
4 atoms
⇒ 7.83 x10 cm
14 −2
=
1
c h
a 2 ( x)
=
1
⋅a 2 ⋅
a 3
=
3a
2
b
3 5.43 x10
−8 2
g
2 2 2
1 1.16
−2
= ⇒ 2.85 x10 cm
14


b
3 4.50 x10
−8
g 2

then
d = 4r = a 2

(b) 4r 4(2.25) °
Body-centered cubic a= = = 6.364 A
(i) (100) plane, surface density, 2 2
14 −2 (a)
Same as (a),(i); surface density 4.94 x10 cm
4 atoms
(ii) (110) plane, surface density, Volume Density = ⇒

=
2 atoms
⇒ 6.99 x10 cm
14 −2
6.364 x10
−8
b g 3




b
2 4.50 x10
−8 2
g . x10 cm
155
22 −3



(iii) (111) plane, surface density, (b)
14 −2 Distance between (110) planes,
Same as (a),(iii), surface density 2.85 x10 cm
1 a 6.364
(c) = a 2 = = ⇒
Face centered cubic 2 2 2
(i) (100) plane, surface density or




5

, Semiconductor Physics and Devices: Basic Principles, 3rd edition Chapter 1
Solutions Manual Problem Solutions

°
4.50 A
1.20
(c)
Surface density b5x10 g(30.98) ⇒ 16




=
2 atoms
=
2
(a) Fraction by weight ≈
b5x10 g(28.06) 22



2a
2
b
2 6.364 x10
−8
g 2
. x10
110
−6



or (b) Fraction by weight
3.49 x10 cm
14 −2
b10 g(10.82) 18

≈
b5x10 g(30.98) + b5x10 g(28.06) ⇒
16 22


1.17 −6
−3
7.71x10
Density of silicon atoms = 5 x10 cm
22
and 4
valence electrons per atom, so 1.21
23 −3
Density of valence electrons 2 x10 cm 1 −3
Volume density = = 2 x10 cm
15
3
d
1.18 So
Density of GaAs atoms −6
d = 7.94 x10 cm = 794 A
°

8 atoms −3
= = 4.44 x10 cm
22 °
We have a O = 5.43 A
b
5.65 x10
−8 3
g So
An average of 4 valence electrons per atom, d 794 d
23 −3 = ⇒ = 146
Density of valence electrons 1.77 x10 cm a O 5.43 aO

1.19
16
2 x10
(a) Percentage = 22
x100% ⇒
5 x10
−5
4 x10 %
15
1x10
(b) Percentage = 22
x100% ⇒
5 x10
−6
2 x10 %




6