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AN INTRODUCTION TO SEMICONDUCTOR DEVICES SOLUTION
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'AN INTRODUCTION TO SEMICONDUCTOR DEVICES' neamen 著
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An Introduction to Semiconductor Devices Chapter 1
Solutions Manual Problem Solutions
______________________________________________________________________________________
2
Chapter 1
Problem Solutions
1.1
(a) fcc: 8 corner atoms × 1/8 = 1 atom
6 face atoms
× ½ = 3 atoms
Total of 4 atoms per unit cell
(b) bcc: 8 corner atoms × 1/8 = 1 atom
1 enclosed atom = 1 atom
Total of 2 atoms per unit cell
(c) Diamond: 8 corner atoms
×
1/8 = 1 atom
6 face atoms × ½ = 3 atoms
4 enclosed atoms = 4 atoms
Total of 8 atoms per unit cell
_______________________________________
1.2
4 Ga atoms per unit cell
Density
=⇒
−
4
565 10
8
3
. x
b
g
Density of Ga
=
−
2.22 10
22 3
xcm
4 As atoms per unit cell, so that
Density of As
=
−
2.22 10
22 3
xcm
_______________________________________
1.3
8 Ge atoms per unit cell
Density
=⇒
−
8
565 10
8
3
. x
b
g
Density of Ge =
4.44 10
22 3
xcm
−
_______________________________________
1.4
(a) Simple cubic lattice; a
r
= 2
Unit cell vol
== =
()
arr
3
3
3
28
1 atom per cell, so atom vol.
=
()
F
H
G
I
K
J
1
4
3
3
π
r
Then
Ratio
r
r
=×⇒
F
H
G
I
K
J
4
3
8
100%
3
3
π
Ratio
=
52.4%
(b)
Face-centered cubic lattice
dra== ⇒42 a
d
r==
2
22
Unit cell vol
== =ar r
3
3
3
22 162
ch
4 atoms per cell, so atom vol.
=
F
H
G
I
K
J
4
4
3
3
π
r
Then
Ratio
r
r
=×⇒
F
H
G
I
K
J
4
4
3
16 2
100%
3
3
π
Ratio = 74%
(c)
Body-centered cubic lattice
dra a r== ⇒=43
4
3
Unit cell vol.
==
F
H
I
K
ar
3
3
4
3
2 atoms per cell, so atom vol.
=
F
H
G
I
K
J
2
4
3
3
π
r
Then
Ratio
r
r
=×⇒
F
H
G
I
K
J
F
H
I
K
2
4
3
4
3
100%
3
3
π
Ratio = 68%
(d)
Diamond lattice
Body diagonal
== = ⇒=dra a r83
8
3
Unit cell vol.
==
F
H
I
K
a
r
3
3
8
3
8 atoms per cell, so atom vol.
8
4
3
3
π
r
F
H
G
I
K
J
Then
Ratio
r
r
Ratio=×⇒=
F
H
G
I
K
J
F
H
I
K
8
4
3
8
3
100% 34%
3
3
π
_______________________________________
An Introduction to Semiconductor Devices Chapter 1
Solutions Manual Problem Solutions
______________________________________________________________________________________
3
1.5
We have 83ra
o
=
so that
rA==
°
565 3
8
1223
.
.
Then
dr A==
°
22.45
_______________________________________
1.6
From Problem 1.4, percent volume of fcc atoms
is 74%. Therefore after coffee is ground,
Volume =
074
3
. cm
_______________________________________
1.7
(a) aA=
°
543. From 1.3d, ar=
8
3
so that
r
a
A== =
()
°
3
8
543 3
8
118
.
.
Center of one silicon atom to center of nearest
neighbor
=⇒2r 2.36 A
°
(b)
Number density
=⇒
−
8
543 10
8
3
. x
bg
Density =
−
510
22 3
xcm
(c)
Mass density
== = ⇒
()
()
ρ
NAtWt
N
x
x
A
..
.
.
510 2809
602 10
22
23
bg
ρ
= 2.33
3
grams cm/
_______________________________________
1.8
(a) ar A
A
== =
()
°
221022.04.
Now
22 322.0432.04rra r
AB B
+= ⇒= −
so that
rA
B
=
°
0747.
(b)
A-type; 1 atom per unit cell
Density
=⇒
−
1
2.04 10
8
3
x
bg
Density(A) =
118 10
23 3
. xcm
−
B-type: 1 atom per unit cell, so
Density (b) =
118 10
23 3
. xcm
−
_______________________________________
1.9
Simple cubic; ar A
o
==
°
24.2
bcc:
43
42.1
3
4.85ra a A
oo
=⇒= =
()
°
fcc:
42
42.1
2
594ra a A
oo
=⇒= =
()
°
.
diamond:
83
82.1
3
ra a
oo
=⇒=
()
or
aA
o
=
°
9.70
_______________________________________
1.10
(b)
a
=
+
⇒18 10.. aA=
°
2.8
(c)
Na: Density
=
−
12
2.8 10
8
3
x
b
g
=
−
2.28 10
22 3
xcm
Cl: Density (same as Na)
=
−
2.28 10
22 3
xcm
(d)
Na: At.Wt. = 22.99
Cl: At. Wt. = 35.45
So, mass per unit cell
=
+
=
()()
−
1
2
22.99
1
2
3545
602 10
4.85 10
23
23
.
. x
x
Then mass density is
ρ
=⇒
−
−
4.85 10
2.8 10
23
8
3
x
x
bg
ρ
= 2.21
3
gm cm/
_______________________________________
1.11
(a)
aA322.2 218 8=+=
() ()
°
.
so that
aA=
°
4.62
Density of A
=⇒
−
1
4.62 10
8
3
x
b
g
101 10
22 3
. xcm
−
An Introduction to Semiconductor Devices Chapter 1
Solutions Manual Problem Solutions
______________________________________________________________________________________
4
Density of B
=⇒
−
1
4.62 10
8
x
b
g
101 10
22 3
. xcm
−
(b)
Same as (a)
(c)
Same material
_______________________________________
1.12
(a) Surface density
== ⇒
−
1
2
1
4.62 10 2
2
8
2
a
x
bg
331 10
14 2
. xcm
−
Same for A atoms and B atoms
(b)
Same as (a)
(c)
Same material
_______________________________________
1.13
(a) Vol density
=
1
3
a
o
Surface density
=
1
2
2
a
o
(b)
Same as (a)
_______________________________________
1.14
(100) plane: Density =
No atoms
a
o
.
2
==
−
−
2
543 10
678 10
8
2
14 2
.
.
x
xcm
bg
(110) plane: Density =
No atoms
a
o
.
2
2
==
−
−
4
543 10 2
9.59 10
8
2
14 2
. x
xcm
b
g
(111) plane: Density =
No atoms
a
o
.
2
32
ch
=
⋅+ ⋅
=
−
−
16 3 12 3
543 10 3 2
7.83 10
8
2
14 2
a
f
a
f
b
g
ch
. x
xcm
(a)
(110) = highest density
(b)
(100) = lowest density
_______________________________________
1.15
(a) Planes intercept at:
(i)
pq s
=
=
∞
=
∞1, ,
(ii)
pqs
=
=
=
∞
31,,
(iii)
pq s
=
=
∞
=
32,,
(b)
Vector components:
(i)
pqs
=
=
=
110,,
(ii)
pqs
=
=
=
311,,
(iii)
pq s
=
=
=
123,,
_______________________________________
1.16
(a)
1
1
1
3
1
1
,,
F
H
I
K
⇒
()313
(b)
1
4
1
2
1
4
121,,
F
H
I
K
()
⇒
_______________________________________
1.17
(a) da A
o
==
°
525.
(b)
d
a
A
o
==
°
2
2
371.
(c)
d
a
A
o
==
°
3
3
303.
_______________________________________
1.18
(a) da A
o
==
°
520.
(b)
d
a
A
o
== =
()
°
2
2
520 2
2
368
.
.
(c)
d
a
A
o
== =
()
°
3
3
520 3
3
300
.
.
_______________________________________
1.19
a
r
A
o
== =
(
)
°
4
3
42.25
3
5196.
(a)
Density =
22
5196 10
143 10
3
8
3
22 3
a
x
xcm
o
==
−
−
.
.
bg
An Introduction to Semiconductor Devices Chapter 1
Solutions Manual Problem Solutions
______________________________________________________________________________________
5
(b)
d
a
A
o
== =
°
2
2
5196 2
2
367
.
.
(d)
Surface density
==
−
2
2
2
5196 10 2
2
8
2
a
x
o
.
bg
or
=
−
524 10
14 2
. xcm
_______________________________________
1.20
Density of silicon atoms =
−
510
22 3
xcm and 4
valence electrons per atom, so
Density of valence electrons =
210
23 3
xcm
−
_______________________________________
1.21
Density of GaAs atoms
==
−
−
8
565 10
4.44 10
8
3
22 3
atoms
x
xcm
.
b
g
An average of 4 valence electrons per atom,
Density of valence electrons =
177 10
23 3
. xcm
−
_______________________________________
1.22
Silver: Group 1 = 1 valence electron per atom
At.Wt. = 107.88, Mass density = 1050
3
./gm cm
Mass per unit cell,
==
−
107.88
602 10
179 10
23
22
.
.
x
x
Now
ρ
=⇒=
−−
F
H
G
I
K
J
179 10 179 10
1050
22
3
22
13
..
.
/
x
a
a
x
o
o
or
aA
o
=
°
2.57
Then, density of valence electrons
=
−
1
2.57 10
8
3
x
bg
or
Density =
589 10
22 3
. xcm
−
_______________________________________
1.23
We find cos
θ
2
4
34
1
3
F
H
I
K
==
a
a
o
o
ch
so that
θ
2
54.74=°
or
θ
=
°109.5
_______________________________________
1.24
(a) Ratio
x
x
x=⇒
−
410
510
810
16
22
5
%
(b)
Ratio
x
x
x=⇒
−
210
510
410
15
22
6
%
_______________________________________
1.25
(a) Fraction by weight
≈⇒
()
()
510 3098
510 2806
16
22
x
x
b
g
b
g
.
.
110 10
4
.%
x
−
(b)
Fraction by weight
≈
+
⇒
()
() ()
10 10 82
5 10 30 98 5 10 28 06
18
16 22
b
g
b
g
b
g
.
..xx
7.71 10
4
x
−
%
_______________________________________
1.26
Volume density ==
−
1
210
3
15 3
d
xcm
So
dxcmA==
−°
7.94 10 794
6
We have
aA
O
=
°
543.
So
d
a
d
a
oo
=⇒=
794
543
146
.
_______________________________________
An Introduction to Semiconductor Devices Chapter 2
Solutions Manual Problem Solutions
______________________________________________________________________________________
8
Chapter 2
Problem Solutions
2.1
Eh
hc hc
E
==⇒=
ν
λ
λ
Gold:
EeV xJ==
()
−
4 90 4 90 1 6 10
19
...
bg
So
λ
=⇒
−
−
()
6 625 10 3 10
4.90 16 10
34 10
19
.
.
xx
x
b
g
b
g
b
g
2.54 10
5
xcm
−
or
λ
μ
= 0 254. m
Cesium:
EeV xJ==
()
−
190 190 16 10
19
...
bg
So
λ
=⇒
−
−
()
−
6 625 10 3 10
190 16 10
654 10
34 10
19
5
.
..
.
xx
x
xcm
bgbg
b
g
or
λ
μ
= 0 654. m
_______________________________________
2.2
(a) Electron: (i) K.E.
== =
−
TeV x J11610
19
.
pmT x x==
−−
2 2 9.11 10 16 10
31 19
bgbg
.
or
pxkgms=−
−
54 10
25
./
λ
== ⇒
−
−
h
p
x
x
6625 10
54 10
34
25
.
.
or
λ
=
°
12.3 A
(ii) K.E.
== =
−
TeVxJ100 16 10
17
.
pmT=⇒2
pxkgms=−
−
54 10
24
./
λ
=⇒
h
p
λ
=
°
123. A
(b)
Proton: K.E. == =
−
TeV x J11610
19
.
pmT x x==
−−
22167101610
27 19
..
bgbg
or
pxkgms=−
−
2.31 10
23
/
λ
== ⇒
−
−
h
p
x
x
6625 10
2.31 10
34
23
.
or
λ
=
°
0 287. A
_______________________________________
2.3
EkT
avg
== ⇒
()
3
2
3
2
00259.
or
EeV
avg
= 0 01727.
Now
pmE
avg avg
= 2
=
−−
()
2 9.11 10 0 01727 16 10
31 19
xx
bgbg
..
or
pxkgms
avg
=−
−
7.1 10
26
/
Now
λ
== ⇒
−
−
h
p
x
x
6625 10
7.1 10
34
26
.
or
λ
=
°
933. A
_______________________________________
2.4
(a) Emv x x==
−
1
2
1
2
9.11 10 2 10
2314
2
bgbg
or
ExJ=⇒
−
1822 10
22
. ExeV=
−
114 10
3
.
Also
pmv x x== ⇒
−
9.11 10 2 10
31 4
b
g
b
g
pxkgms=−
−
1822 10
26
./
Now
λ
== ⇒
−
−
h
p
x
x
6625 10
1822 10
34
26
.
.
λ
=
°
364 A
(b)
p
hx
x
== ⇒
−
−
λ
6625 10
125 10
34
10
.
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