Semiconductor Physics and Devices: Basic Principles (3rd Edition) by Donald A. Neamen – Solution Manual | Complete Worked Solutions for Semiconductor Device Analysis

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


Chapter 1
Problem Solutions F 4 r I 3




1.1
4 atoms per cell, so atom vol. = 4G
H 3 JK
(a) fcc: 8 corner atoms  1/8 = 1 atom Then
6 face atoms  ½ = 3 atoms F4r IJ
4G
3


Total of 4 atoms per unit cell
Ratio =
H3 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
6 face atoms  ½ = 3 atoms F I
4
3 3




H 3 rK F 4r I
4 enclosed atoms = 4 atoms
Unit cell vol. = a =
3
Total of 8 atoms per unit cell 3




1.2 2 atoms per cell, so atom vol. = 2G
H J
3 K
F 4r I
(a) 4 Ga atoms per unit cell
4 Then 3

Density = 
b5.65x10
−
8


Density of Ga = 2.22 x10 cm
g3


−3
2G
H 3 JK  100%  Ratio = 68%
F4r I
22
Ratio =
3

4 As atoms per unit cell, so that
−3
Density of As = 2.22 x10 cm
22

(d) Diamond lattice
(b) 8
Body diagonal = d = 8r = a3 3  a =
F I
8 Ge atoms per unit cell r
Density = 8 8r 3

b5.65x10 g −8 3

Unit cell vol. = a =
H 3 K F 4r I
3
−3
Density of Ge = 4.44 x10 cm
22 3




1.3
8 atoms per cell, so atom vol. 8G
H 3 JK
a = (2ra) ==2r8r
8GF JI
(a) Unit
Simple
cell cubic
vol =lattice;
3 3 3 Then 4r
3



F 4r I
Ratio = H
3K
3

() Ratio = 34%
1G
H 3 JK
 100% 
1 atom per cell, so atom vol. =
Then
F 8r I 3




FG 4r IJ
3




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 =
3
c2 2 rh = 16 2 r 3
3




3

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

Then mass density is
−23
1.5 4.85x10
= 
(a)
a = 5.43 A
 From 1.3d, a =
8
3
r b
2.8x10
−8
g3




 = 2.21 gm / cm
3

a 3 (5.43) 3 
so that r = = = 1.18 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 
a = 4.62 A
b g
8 −3
= 5.43x10−8  Density = 5x10 cm
3 22

1 22 −3
Density of A =  1.01x10 cm
(c) Mass density
b g(28.09)  b4.62 x10 g −8 3



=  = N ( At.Wt.) =
22
5x10 −3
Density of B = 
22
1 1.01x10 cm
NA 6.02 x10
23
b4.62 x10 g −8



 = 2.33 grams / cm (b) Same as (a)
3

(c) Same material

1.6 1.9
(a) a = 2rA = 2(1.02) = 2.04 A

(a) Surface density
Now 1 1
= 2 = 
2rA + 2rB = a 3  2rB = 2.04 3 − 2.04 a 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
Density = (b) Same as (a)
b g

−8 3 (c) Same material
2.04 x10
23 −3
Density(A) = 1.18x10 cm 1.10
B-type: 1 atom per unit cell, so 1
23 −3
(a) Vol density =
Density(B) = 1.18x10 cm 3
ao
1
1.7 Surface density = 2
(b) ao 2
 (b) Same as (a)
a = 1.8 + 1.0  a = 2.8 A
(c) 1.11
12 −3 Sketch
Na: Density = = 2.28x10 cm
22



1.12

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


4

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

b4.50x10
2 atomsg
= 
−8 14 −2
1.13 9.88x10 cm

(a) Distance between nearest (100) planes is: 2

d = a = 5.63 A
(b) Distance between nearest (110) planes is: (ii) (110) plane, surface density,
1 2 atoms −2
=  6.99 x10 cm
14
a 5.63
d= a 2= =
2 2 2
F (111) 1I
or (iii) plane, surface density,

d = 3.98 A 1


=H 6 2K =
(c) Distance between nearest (111) planes is: 3  + 3 4
1 a 5.63 3
d= a 3= = 2
a
3 3 3 2
15 −2
or or 1.14 x10 cm

d = 3.25 A
1.15
1.14 (a)
(a) (100) plane of silicon – similar to a fcc,
 2 atoms
Simple cubic: a = 4.50 A
b g
surface density = 
−8 2
(i) (100) plane, surface density, 5.43x10
1 atom −2 −2


b g
=  4.94 x10 cm
14 14
6.78x10 cm
−8 2
4.50x10 (b)
(ii) (110) plane, surface density, (110) plane, surface density,
−2 −2
 3.49 x10 cm =  9.59 x10 cm
14 14
= 1 atom 4 atoms

b4.50x10 g
2
−8 2



(iii) 1(111) plane, surface1density, (c)
3
F I6Katoms
H 2 1
(111) plane, surface density,
4 atoms 7.83x10 cm
14 −2

= 
= 1
c h
a 2 (x)
=
1
a 2 
a 3
=
3a
2
3 b5.43x10 g −8 2



2 2 2
1 1.16
−2
=  2.85x10 cm
14

d = 4r = a 2
then
(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 4.94x10 cm
4 atoms
Volume Density = 6.364 x10−8
b g
(ii) (110) plane, surface density, 3

2 atoms −2


b g
=  6.99 x10 cm
14

−8 2 22 −3
2 4.50x10 1.55x10 cm
(iii) (111) plane, surface density, (b)
14 −2 Distance between (110) planes,
Same as (a),(iii), surface density 2.85x10 cm
1
(c) = a 2 = a = 6.364 
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
(c)
1.20
b5x10 g(30.98) 16




b5x10 g(28.06) 
Surface density
(a) Fraction by weight  22
2 atoms 2
= =
2 −6
2a 1.10x10
(b) Fraction by weight
b g
or
10 (10.82)
14 −2 18
3.49 x10 cm

b
5x10
16
g(30.98) + 5x1022 b g(28.06) 
1.17 −6

−3
7.71x10
Density of silicon atoms = 5x10 cm and 4
22


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
−6 
Density of GaAs atoms d = 7.94 x10 cm = 794 A
8 atoms −3
= 4.44 x10 cm
b g
= We have a O = 5.43 A

22
3
−8
5.65x10 So
An average of 4 valence electrons per atom, d 794 d
23 −3 =  = 146
Density of valence electrons 1.77 x10 cm aO 5.43 aO

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




6