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In this video, I talk about the roadmap to learning semiconductor physics, and what the driving questions we are trying to answer are.
This is part of my series on semiconductor physics (often called Electronics 1 at university). This is based on the book Semiconductor Physics and Devices by Donald Neamen, as well as the EECS 170A/174 courses taught at UC Irvine.
Hope you found this video helpful, please post in the comments below anything I can do to improve future videos, or suggestions you have for future videos.
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Semiconductor Physics and Devices – OptiMa-UFAM
SEMICONDUCTOR PHYSICS & DEVICES: BASIC PRINCIPLES, FOURTH EDITION … Semiconductor physics and devices : basic principles / Donald A. Neamen. — 4th ed.
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Physics of Semiconductor Devices, 4th Edition – Wiley
The Fourth Edition of Physics of Semiconductor Devices remains the standard reference work on the fundamental physics and operational characteristics of all …
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Semiconductor Physics and Devices – Fulvio Frisone
Semiconductor physics and devices : basic principles 1 Donald A. Neamen. -3rd ed. p. cm. Includes bibliographical references and index. ISBN 0-07-232 107- …
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Semiconductor Physics and Devices 4th edition Neaman pdf
Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 1. By D. A. Neamen Problem Solutions. ______. ______. Chapter 1. Problem Solutions.
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Semiconductor Physics Devices Donald Neamen 4th Edition
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주제와 관련된 이미지 semiconductor physics and devices 4th edition
주제와 관련된 더 많은 사진을 참조하십시오 Introduction to Semiconductor Physics and Devices. 댓글에서 더 많은 관련 이미지를 보거나 필요한 경우 더 많은 관련 기사를 볼 수 있습니다.
주제에 대한 기사 평가 semiconductor physics and devices 4th edition
- Author: Jordan Edmunds
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- Date Published: 2018. 3. 25.
- Video Url link: https://www.youtube.com/watch?v=OVnVN0vSXn0
Physics of Semiconductor Devices, 4th Edition
S. M. SZE, P H D, is Honorary Chair Professor, College of Electrical and Computer Engineering, National Chiao Tung University, Taiwan. He has made fundamental and pioneering contributions to semiconductor devices, particularly his co-discovery of the floating-gate memory (FGM) effect that has ushered in the Fourth Industrial Revolution. Dr. Sze has authored, co-authored, and edited more than 400 papers and 16 books. He is a celebrated Member of IEEE, an Academician of Academia Simica, and a member of the US National Academy of Engineering.
YIMING LI, P H D, is Full Professor of Electrical and Computer Engineering at National Chiao Tung University, Taiwan. He has been a Visiting Professor in Stanford University, Grenoble INP, and Tohoku University. He has published more than 300 technical articles in journals, conferences, and book chapters. Dr. Li is an active member of IEEE and has served on technical committees for many international professional conferences including IEDM. He is the recipient of the Pan Wen-Yuan Foundation’s Research Fellowship Award and the Chinese Institute of Electrical Engineering’s Outstanding Young Electrical Engineer Award.
Semiconductor Physics and Devices 4th edition Neaman pdf
Semiconductor Physics and Devices: Basic Principles, 4 edition Chapter 1
By D. A. Neamen Problem Solutions
Chapter 1
Problem Solutions
1.
(a) fcc: 8 corner atoms 8/1 1 atom
6 face atoms 2/1 3 atoms
Total of 4 atoms per unit cell
(b) bcc: 8 corner atoms 8/1 1 atom
1 enclosed atom =1 atom Total of 2 atoms per unit cell
(c) Diamond: 8 corner atoms 8/1 1 atom
6 face atoms 2/1 3 atoms
4 enclosed atoms = 4 atoms Total of 8 atoms per unit cell
1.
(a) Simple cubic lattice: a 2 r
Unit cell vol
3 3 3 a 2 r 8 r
1 atom per cell, so atom vol
3
4 1
3 r
Then
Ratio 100 % 52 %4. 8
3
4
3
3
r
r
(b) Face-centered cubic lattice
r
d d r a a 2 2 2
4 2
Unit cell vol
3 3 3 a 2 2 r 16 2 r
4 atoms per cell, so atom vol
3
4 4
3 r
Then
Ratio
100 % 74 % 16 2
3
4 4
3
3
r
r
(c) Body-centered cubic lattice
d r a a r 3
4 43
Unit cell vol
3 3
3
4
a r
2 atoms per cell, so atom vol
3
4 2
3 r
Then
Ratio
100 % 68 %
3
4
3
4 2
3
3
r
r
(d) Diamond lattice
Body diagonal d r a a r 3
8 8 3
Unit cell vol
3 3
3
8
r a
8 atoms per cell, so atom vol
3
4 8
3 r
Then
Ratio
100 % 34 %
3
8
3
4 8
3
3
r
r
1.
(a)
o a .5 43 A ; From Problem 1,
a r 3
8
Then
o
A
a r .1 176 8
.5 43 3
8
3
Center of one silicon atom to center of
nearest neighbor
o 2 r .2 35 A (b) Number density
22 83
5 10 .5 43 10
8
cm 3
(c) Mass density
23
22
.6 02 10
.. 5 10 28. 09
NA
NAtWt
.2 33 grams/cm 3
Semiconductor Physics and Devices: Basic Principles, 4 edition Chapter 1
By D. A. Neamen Problem Solutions
1.
(a) 4 Ga atoms per unit cell
Number density
83 .5 65 10
4
Density of Ga atoms 22 .2 22 10 cm 3
4 As atoms per unit cell
Density of As atoms
22 .2 22 10 cm 3
(b) 8 Ge atoms per unit cell
Number density
83 .5 65 10
8
Density of Ge atoms 22 .4 44 10 cm 3
1.
From Figure 1.
####### (a) a
a d .0 4330 2
3
2
o .0 4330 .5 65 d .2 447 A
(b) a
a d 2 .0 7071 2
o .0 7071 .5 65 d .3 995 A
1.
54. 74 3 2
2
3 2
2 2 2
sin
a
a
109 5.
1.
(a) Simple cubic:
o a 2 r 9 A
(b) fcc:
o A
r a .5 515 2
4
(c) bcc:
o A
r a .4 503 3
4
(d) diamond:
####### o
A
r a .9 007 3
42
1.
####### (a) .12 035 2 .12 035 2 rB
o rB .0 4287 A
(b)
o a .12 035 .2 07 A
(c) A-atoms: # of atoms 1 8
1 8
Density
83 .2 07 10
1
23 .1 13 10 cm 3
B-atoms: # of atoms 3 2
1 6
Density
83 .2 07 10
3
23 .3 38 10 cm 3
1.
(a)
o a 2 r 5 A
8
of atoms 1
1 8
Number density
83 5 10
1
22 .1 097 10 cm 3
Mass density
#######
NA
NAtWt ..
23
22
.6 02 10
.1 0974 10 12 5.
.0 228 gm/cm
3
(b)
o A
r a .5 196 3
4
8
of atoms 1 2
1 8
Number density
83 .5 196 10
2
22 .1 4257 10 cm
3
Mass density
23
22
.6 02 10
.1 4257 10 12 5.
.0 296 gm/cm
3
1. From Problem 1, percent volume of fcc atoms is 74%; Therefore after coffee is ground,
Volume = 0 cm 3
Semiconductor Physics and Devices: Basic Principles, 4 edition Chapter 1
By D. A. Neamen Problem Solutions
1.
(a)
313
1
1 , 3
1 , 1
1
(b)
121
4
1 , 2
1 , 4
1
1.
Intercepts: 2, 4, 3
3
1 , 4
1 , 2
1
(634) plane
1.
(a)
o d a .5 28 A
(b)
o A
a d .3 734 2
2
(c)
o A
a d .3 048 3
3
1.
(a) Simple cubic
(i) (100) plane:
Surface density
2 82 .4 73 10
1 1
a
14 .4 47 10 cm 2
(ii) (110) plane:
Surface density 2
1 2 a
14 .3 16 10 cm 2
(iii) (111) plane:
Area of plane bh 2
1
where
o b a 2 .6 689 A
Now
2
2 2 2 2 4
3
2
2 2 a
a h a
So
o h .4 73 .5 793 A 2
6
Area of plane
8 8 .6 68923 10 .5 79304 10 2
1
16 19. 3755 10
cm 2
Surface density 16 19. 3755 10
6
1 3
14 .2 58 10 cm 2
(b) bcc (i) (100) plane:
Surface density
14 2 .4 47 10
1 a
cm 2
(ii) (110) plane:
Surface density 2
2 2 a
14 .6 32 10 cm 2
(iii) (111) plane:
Surface density 16 19. 3755 10
6
1 3
14 .2 58 10 cm 2
(c) fcc (i) (100) plane:
Surface density
14 2 .8 94 10
2 a
cm 2
(ii) (110) plane:
Surface density 2
2 2 a
14 .6 32 10 cm 2
(iii) (111) plane:
Surface density 16 19. 3755 10
2
1 3 6
1 3
15 .1 03 10 cm 2
1. (a) (100) plane: – similar to a fcc:
Surface density
82 .5 43 10
2
14 .6 78 10 cm 2
(b) (110) plane:
Surface density
82 .52 43 10
4
14 .9 59 10 cm 2
Semiconductor Physics and Devices: Basic Principles, 4 edition Chapter 1
By D. A. Neamen Problem Solutions
(c) (111) plane:
Surface density
82 .523 43 10
2
14 .7 83 10 cm 2
1.
o
A
r a .6 703 2
.24 37
2
4
(a) #/cm 3
3 83 .6 703 10
2 4
1 6 8
1 8
a
22 .1 328 10 cm 3
(b) #/cm
2
2
2
1 2 4
1 4
2 a
.6 703 10 2
2 82
14 .3 148 10 cm 2
(c)
o
A
a d .4 74 2
.6 703 2
2
2
(d) # of atoms 2 2
1 3 6
1 3
Area of plane: (see Problem 1) o b a 2 .9 4786 A
o A
a h .8 2099 2
6
Area
8 8 .9 4786 10 .8 2099 10 2
1
2
1 bh
15 .3 8909 10 cm 2
#/cm 2 15 .3 8909 10
2
= 14 .5 14 10 cm 2
o
A
a d .3 87 3
.6 703 3
3
3
1.
Density of silicon atoms
22 5 10 cm 3 and
4 valence electrons per atom, so
Density of valence electrons 23 2 10 cm 3
1. Density of GaAs atoms
22 83
.4 44 10 .5 65 10
8
cm 3
An average of 4 valence electrons per atom, So Density of valence electrons 23 .1 77 10 cm 3
1.
(a) 100 % 10 % 5 10
5103 22
17
(b) 100 % 4 10 % 5 10
2106 22
15
1. (a) Fraction by weight
7 22
16 .1 542 10 5 10 28. 06
2 10 10. 82
(b) Fraction by weight
5 22
18 .2 208 10 5 10 28. 06
10 30. 98
1.
Volume density 16 3 2 10
1 d
cm 3
So 6 .3 684 10 d cm
o d 368 4. A
We have
o ao .5 43 A
Then 67. 85 .5 43
368 4. ao
d
1.
Volume density
15 3 4 10
1 d
cm
3
So
6 .6 30 10
d cm
o d 630 A
We have
o ao .5 43 A
Then 116 .5 43
630 ao
d
Semiconductor Physics and Devices: Basic Principles, 4 edition Chapter 2
By D. A. Neamen Problem Solutions
(b)
27 19 .12 67 10 6.12 10
p
23 .2 532 10
kg-m/s
11 23
34 .2 62 10 .2 532 10
.6 62510
m
or
o .0 262 A
2.
.0 0259 .0 03885
2
3
2
3
Eavg kT eV
Now
pavg 2 mEavg
31 19 .92 11 10 .0 03885 6 10
or
25 .1 064 10 pavg kg-m/s
Now
9 25
34 .6 225 10 .1 064 10
.6 62510
p
h m
or o 62. 25 A
2.
p
p p
hc E h
Now
m
p E e e 2
2 and
2
2
1
e
e e
e
h
m
E
h p
Set Ep Ee and p 10 e
Then
2 2 10
2
1
2
1
p e p
h
m
h
m
hc
which yields
mc
h p 2
100
100
2 2 100
2 mc mc h
hc hc E E p
p
100
.92 11 10 3 10
31 82
15 .1 64 10 J 10. 25 keV
2.
(a) 10
34
85 10
.6 625 10
h p
26 .7 794 10
kg-m/s
4 31
26 .8 56 10 .9 11 10
.7 794 10
m
p m/s
or 6 .8 56 10 cm/s
2 31 42 .9 11 10 .8 56 10 2
1
2
1 E m
21 .3 33 10 J
or
2 19
21 .2 08 10 6 10
.3 33410
E eV
(b)
31 32 .9 11 10 8 10 2
1
E
23 .2 915 10
J
or 4 19
23 .1 82 10 6 10
.2 91510
E eV
31 3 .9 11 10 8 10 p m
27 .7 288 10 kg-m/s
8 27
35 .9 09 10 .7 288 10
.6 62510
p
h m
or
o 909 A
2.
(a)
10
34 8
1 10
.6 625 10 3 10
hc E h
15 .1 99 10 J
Now
19
15
6 10
.1 99 10
e
E E Ve V
4 V .1 24 10 V 12 4. kV
(b)
31 15 2 .92 11 10 .1 99 10
p mE
23 .6 02 10
kg-m/s Then
11 23
34 .1 10 10 .6 02 10
.6 62510
p
h m
or o .0 11 A
Semiconductor Physics and Devices: Basic Principles, 4 edition Chapter 2
By D. A. Neamen Problem Solutions
2.
6
34
10
.1 054 10
x
p
28 .1 054 10
kg-m/s
2.
(a) (i) xp
26 10
34 .8 783 10 12 10
.1 05410
p kg-m/s
(ii) p m
p
dp
d p dp
dE E
2
2
m
pp p m
p 2
2
Now p 2 mE
31 19 92 10 16 6 10
24 .2 147 10
kg-m/s
so
31
24 26
9 10
.2 1466 10 .8 783 10
E
19 .2 095 10
J
or .1 31 6 10
.2 095 10 19
19
E eV
(b) (i)
26 .8 783 10
p kg-m/s
(ii)
28 19 52 10 16 6 10
p
23 .5 06 10
kg-m/s
28
23 26
5 10
.5 06 10 .8 783 10
E
21 .8 888 10
J
or
2 19
21 .5 55 10 6 10
.8 88810
E eV
2.
32 2
34 .1 054 10 10
.1 05410
x
p
kg-m/s
1500
.1 054 10
32
m
p p m
36 7 10 m/s
2. (a) tE
16 19
34 .8 23 10 6.18 10
.1 05410
t s
(b) 10
34
5 10
.1 054 10
x
p
25 .7 03 10 kg-m/s
2.
####### (a) If 1 , tx and 2 , tx are solutions to
Schrodinger’s wave equation, then
t
tx xV tx j x
tx
m
, ,
,
2
1 2 1
1
2 2
and
t
tx xV tx j x
tx
m
, ,
,
2
2 2 2
2
2 2
Adding the two equations, we obtain
tx tx
m x
, , 2 2 1 2
2 2
xV 1 , tx 2 , tx
tx tx
t
j 1 , 2 ,
which is Schrodinger’s wave equation. So
####### 1 , tx 2 , tx is also a solution.
####### (b) If 1 , tx 2 , tx were a solution to
Schrodinger’s wave equation, then we could write
1 2 1 2
2
2 2
2
xV m x
1 2
t
j
which can be written as
m x x x x
1 2 2
1
2
2 2
2
2
1
2 2 2
t t
xV j 1 2
2 1 2 1
Dividing by 1 2 , we find
m x x x x
1 2
21
2
1
2
1
2
2
2
2
2 1 1 2
2
t t
xV j 1
1
2
2
1 1
Semiconductor Physics and Devices: Basic Principles, 4 edition Chapter 2
By D. A. Neamen Problem Solutions
2.
####### P x dx
2
(a) dx
x
a
a
2
cos
22
4/
4/
40
2 sin
2
2
a
a
a
x
x
a
a
a
a
4
2
sin
2
2 4
4
1
8
2 a a
a
or P .0 409
(b) dx a
x
a
P
a
a
2/
4/
2 cos
2
2/
4 4/
2 sin
2
2
a
a a
a
x
x
a
a
a
a
a
a
4
2
sin
4 8
sin
4
2
4
1
8
1 0 4
1 2
or P .0 0908
(c) dx a
x
a
P
a
a
2
2/
2/
cos
2
2/
2/
4
2 sin
2
2
a
a a
a
x
x
a
a
a
a
a
a
4
sin
4 4
sin
4
2
or P 1
2.
(a) dx a
x
a
P
a
2 sin
22
4/
4/
2 4
4 sin
2
2
a
a
a
x
x
a
a
a
a
8
sin
8
2
or P .0 25
(b) dx a
x
a
P
a
a
2 sin
22
2/
4/
2/
4/
2 4
4 sin
2
2
a
a a
a
x
x
a
a
a
a
a
a
8
sin
8 8
sin 2
4
2
or P .0 25
(c) dx a
x
a
P
a
a
2 sin
22
2/
2/
2/
2/
2 4
4 sin
2
2
a
a a
a
x
x
a
a
a
a
a
a
8
sin 2
8 4
sin 2
4
2
or P 1
2.
(a) (i) 4 8
12 10 8 10
8 10
k
p
m/s
or 6 p 10 cm/s
9 8 .7 854 10 8 10
22
k
m
Semiconductor Physics and Devices: Basic Principles, 4 edition Chapter 2
By D. A. Neamen Problem Solutions
or
o 78. 54 A
(ii)
31 4 .9 11 10 10
p m 27 .9 11 10 kg-m/s
2 31 42 .9 11 10 10 2
1
2
1 E m
23 .4 555 10 J
or 4 19
23 .2 85 10 6 10
.4 55510
E eV
(b) (i) 4 9
13 10 5 10
5 10
k
p
m/s
or
6 p 10 cm/s
9 9 .4 19 10 5 10
22
k
m
or
o 41 9. A
(ii)
27 .9 11 10
p kg-m/s 4 .2 85 10
E eV
2.
(a)
j kx t tx Ae ,
(b)
19 2 2
1 E .0 025 6 10 m
31 2 .9 11 10 2
1
so
4 .9 37 10 m/s 6 .9 37 10 cm/s
For electron traveling in x direction, 6 .9 37 10 cm/s
31 4 .9 11 10 .9 37 10 p m
26 .8 537 10 kg-m/s
9 26
34 .7 76 10 .8 537 10
.6 62510
p
h m
8 9 .8 097 10 .7 76 10
2 2
k m 1
8 4 k .8 097 10 .9 37 10
or 13 .7 586 10 rad/s
2.
(a)
31 4 .9 11 10 5 10
p m 26 .4 555 10
kg-m/s
8 26
34 .1 454 10 .4 555 10
.6 62510
p
h m
8 8 .4 32 10 .1 454 10
2 2
k m 1
8 4 k .4 32 10 5 10 13 .2 16 10 rad/s
(b)
31 6 .9 11 10 10
p 25 .9 11 10 kg-m/s
10 25
34 .7 27 10 .9 11 10
.6 62510
m
9 10 .8 64 10 .7 272 10
2
k m 1
9 6 15 .8 64 10 10 .8 64 10 rad/s
2.
31 102
2 342 2
2
222
.92 11 10 75 10
.1 054 10
2
n
ma
n En
2 21 .1 0698 10 En n J
or
19
2 21
6 10
.1 0698 10
n En
or
2 3 .6 686 10
En n eV
Then 3 1 .6 6910
E eV 2 2 .2 6710
E eV 2 3 .6 0210
E eV
2.
(a)
31 102
2 342 2
2
222
.92 11 10 10 10
.1 054 10
2
n
ma
n En
2 20 .6 018 10 n J
or
.0 3761
6 10
.6 018102 19
2 20 n
n En
eV
Then E 1 .0 376 eV
E 2 .1 504 eV
E 3 .3 385 eV
(b) E
hc
19 .3 385 .1 504 6 10
E 19 .3 01 10
J
Semiconductor Physics and Devices: Basic Principles, 4 edition Chapter 2
By D. A. Neamen Problem Solutions
2
2
2
2
2
2
z
Z XY y
Y XZ x
X YZ
2 2 XYZ
mE
Dividing by XYZ and letting 2
2 2
mE k , we
find
(1) 0
1112 2
2
2
2
2
2
k z
Z
y Z
Y
x Y
X
X
We may set
12 2
2 2 2
2
k X x
X k x
X
X
x x
Solution is of the form
####### xX A sin xxk B cos xxk
####### Boundary conditions: X 0 0 B 0
####### and
a
n xX a k
x x
0
where nx …,2,
Similarly, let
2 2
2 1 ky y
Y
Y
and
2 2
2 1 kz z
Z
Z
Applying the boundary conditions, we find
a
n k
y y
, ny …,2,
a
n k
z z
, nz 3,2, …
From Equation (1) above, we have
2 2 2 2 kx ky kz k
or
2
2 2 2 2 2
mE kx ky kz k
so that
2 2 2 2
22
2
nnn nx ny nz ma
E E zyx
2.
(a)
, 0
, , 2 2 2
2
2
2
yx
mE
y
yx
x
yx
Solution is of the form:
, yx A sin xxk sin yyk
We find
#######
Ak xk yk x
yx x cos x sin y
,
Ak xk yk x
yx x sin x sin y
, 2 2
2
#######
Ak xk yk y
yx y sin x cos y
,
Ak xk yk y
yx y sin x sin y
, 2 2
2
Substituting into the original equation, we find:
(1) 0
2 2
2 2
mE kx ky
From the boundary conditions,
A sin xak 0 , where
o a 40 A
So a
n k
x x
, nx …,3,2,
Also A sin ybk 0 , where
o b 20 A
So b
n k
y y
, ny …,3,2,
Substituting into Eq. (1) above
2
22
2
2 22
2 b
n
a
n
m
E x y nn yx
(b)Energy is quantized – similar to 1-D result. There can be more than one quantum state per given energy – different than 1-D result.
2. (a) Derivation of energy levels exactly the same as in the text
(b)
2 1
2 2 2
22
2
n n ma
E
For n 2 ,2 n 1 1
Then
2
22
2
3
ma
E
(i) For
o a 4 A
27 102
342 2
.12 67 10 4 10
.13 054 10
E
22 .6 155 10
J
or
3 19
22 .3 85 10 6 10
.6 15510
E eV
Semiconductor Physics and Devices: Basic Principles, 4 edition Chapter 2
By D. A. Neamen Problem Solutions
(ii) For a 5 cm
27 22
34 2 2
.12 67 10 5 10
.13 054 10
E
36 .3 939 10
J or
17 19
36 .2 46 10 6 10
.3 93910
E eV
2.
(a) For region II, x 0
0
2 2 2 2
2
2
E V x
m
x
x O
General form of the solution is
####### 2 x A 2 exp jk 2 x B 2 exp jk 2 x
where
E VO
m k 22
2
Term with B 2 represents incident wave and
term with A 2 represents reflected wave.
Region I, x 0
0
2 2 2 1
1
2
x
mE
x
x
General form of the solution is
####### 1 x A 1 exp jk 1 x B 1 exp jk 1 x
where
12
2
mE k
Term involving B 1 represents the
transmitted wave and the term involving A 1
represents reflected wave: but if a particle is transmitted into region I, it will not be reflected so that A 1 0.
Then
####### 1 x B 1 exp jk 1 x
####### 2 x A 2 exp jk 2 x B 2 exp jk 2 x
(b) Boundary conditions:
####### (1) 1 x 0 2 x 0
(2) 0
2
1
x x x x
Applying the boundary conditions to the
solutions, we find
B 1 A 2 B 2
Ak 22 Bk 22 Bk 11
Combining these two equations, we find
2 2 1
2 1 2 B k k
k k A
2 2 1
2 1
2 B k k
k B
The reflection coefficient is 2
2 1
2 1 * 22
22
k k
k k
BB
AA R
The transmission coefficient is
2 1 2
421 1 k k
kk T R T
2.
####### 2 x A 2 exp 2 xk
xk
AA
x P 2 * 22
2
exp 2
where
#######
22
2
Vm E k o
34
31 19
.1 054 10
.92 11 10 5 6.18 10
9 k 2 .4 286 10 m 1
(a) For 10 5 5 10
o x A m
####### P exp 2 2 xk
9 10 exp .42 2859 10 5 10
.0 0138
(b) For 10 15 15 10
o x A m
9 10 exp .42 2859 10 15 10 P
6 .2 61 10
(c) For 10 40 40 10
o x A m
9 10 exp .42 2859 10 40 10 P
15 .1 29 10
2.
ak
V
E
V
E T o o
161 exp 22
where
#######
22
2
Vm E k
o
34
31 19
.1 054 10
.92 11 10 0 6.11 10
Semiconductor Physics and Devices: Basic Principles, 4 edition Chapter 2
By D. A. Neamen Problem Solutions
where
12
2
mE k
Region II:
####### 2 x A 2 exp 2 xk B 2 exp 2 xk
where
22
2
Vm E k
O
Region III:
####### 3 x A 3 exp jk 1 x B 3 exp jk 1 x
(b)
In Region III, the B 3 term represents a
reflected wave. However, once a particle is transmitted into Region III, there will
not be a reflected wave so that B 3 0.
(c) Boundary conditions: At x 0 : 1 2
A 1 B 1 A 2 B 2
dx
d
dx
d 1 2
jkA 11 jkB 11 Ak 22 Bk 22
At x a : 2 3
####### A 2 exp 2 ak B 2 exp 2 ak
####### A 3 exp jk 1 a
dx
d
dx
d 2 3
####### Ak 22 exp 2 ak Bk 22 exp 2 ak
####### jkA 31 exp jk 1 a
The transmission coefficient is defined as
11
33
AA
AA T
so from the boundary conditions, we want to solve for A 3 in terms of A 1. Solving
for A 1 in terms of A 3 , we find
k k ak ak
kk
jA A 2 2 2 1
2 2 21
3 1 exp exp 4
2 jkk 21 exp 2 ak exp 2 ak
####### exp jk 1 a
We then find
k k ak
kk
AA AA 2
2 1
2 2 2 21
33 11 exp 4
2 exp 2 ak
2 2 2
2 2
2 41 kk exp ak exp ak
We have
#######
22
2
Vm E k O
If we assume that VO E , then 2 ak will
be large so that
####### exp 2 ak exp 2 ak
We can then write
2 2
2 1
2 2 2 21
33 11 exp 4
k k ak kk
AA AA
2 2
2 2
2 41 kk exp ak
which becomes
k k ak
kk
AA AA 2
2 1
2 2 2 21
33 11 exp 2 4
Substituting the expressions for k 1 and
k 2 , we find
2
2 2
2 1
2
mVO k k
and
2 2
2 2
2 1
22
Vm E mE kk O
V E E
m O
2
2
2
E
V
E V
m
O
O
1
2
2
2
Then
E V
E V
m
ak
mV AA
AA
O
O
O
1
2 16
exp 2
2
2
2
2
2
2
33
11
ak
V
E
V
E
AA
O O
2
33
16 1 exp 2
Finally,
ak
V
E
V
E
AA
AA T O O
2 11
33 16 1 exp 2
Semiconductor Physics and Devices: Basic Principles, 4 edition Chapter 2
By D. A. Neamen Problem Solutions
2.
Region I: V 0
0
2 2 2 1
1
2 x
mE
x
x
####### 1 x A 1 exp jk 1 x B 1 exp jk 1 x
incident reflected
where
12
2
mE k
Region II: V V 1
0
2 2 2
1 2
2
2 x
mE V
x
x
####### 2 x A 2 exp jk 2 x B 2 exp jk 2 x
transmitted reflected
where
#######
2
1 2
2
mE V k
Region III: V V 2
#######
#######
0
2 2 3
2 2
3
2 x
mE V
x
x
####### 3 x A 3 exp jk 3 x
transmitted
where
#######
2
2 3
2
mE V k
There is no reflected wave in Region III.
The transmission coefficient is defined as:
11
33
1
3 * 11
33
1
3
AA
AA
k
k
AA
AA T
From the boundary conditions, solve for A 3
in terms of A 1. The boundary conditions are:
At x 0 : 1 2
A 1 B 1 A 2 B 2
x x
1 2
Ak 11 Bk 11 Ak 22 Bk 22
At x a : 2 3
####### A 2 exp jk 2 a B 2 exp jk 2 a
####### A 3 exp jk 3 a
x x
2 3
####### Ak 22 exp jk 2 a Bk 22 exp jk 2 a
####### Ak 33 exp jk 3 a
But 2 ak 2 n
####### exp jk 2 a exp jk 2 a 1
Then, eliminating B 1 , A 2 , B 2 from the
boundary condition equations, we find
#######
2 1 3
31 2 1 3
2 1
1
3 4 4
k k
kk
k k
k
k
k T
2. (a) Region I: Since VO E , we can write
0
2 2 2 1
1
2
x
Vm E
x
x O
Region II: V 0 , so
0
2 2 2 2
2
2
x
mE
x
x
Region III: V 3 0
The general solutions can be written, keeping in mind that 1 must remain
finite for x 0 , as
####### 1 x B 1 exp 1 xk
####### 2 x A 2 sin 2 xk B 2 cos 2 xk
####### 3 x 0
where
#######
12
2
Vm E k O and 22
2
mE k
(b) Boundary conditions
At x 0 : 1 2 B 1 B 2
1122
1 2 Bk Ak x x
At x a : 2 3
####### A 2 sin 2 ak B 2 cos 2 ak 0
or
####### B 2 A 2 tan 2 ak
(c)
1 2
1 11 22 2 B k
k Bk Ak A
and since B 1 B 2 , then
2 2
1 2 B k
k A
Semiconductor Physics and Devices: Basic Principles, 4 edition Chapter 2
By D. A. Neamen Problem Solutions
o o o ao
r
a
r
a
r r r a
2 exp exp
1 1
2/5 2
2
m ra
E
m
oo
o
2
2
2
exp 0
1 1
2/
o ao
r
a
where
2
2
22
4
1 4 o 22 oo
o
ma
em E E
Then the above equation becomes
o o o ao
r r a ar
r
a
2
2
2/
2
1 exp
1 1
0 2
2 2 2 2
ma m ra
m
oo oo
o
or
o ao
r
a
exp
1 1
2/
2 1 1 2 2 2
ora ao ao ora
which gives 0 = 0 and shows that 100 is
indeed a solution to the wave equation.
2.
All elements are from the Group I column of
the periodic table. All have one valence
electron in the outer shell.
Semiconductor Physics and Devices: Basic Principles, 4 edition Chapter 3
By D. A. Neamen Problem Solutions
Chapter 3
3.
If ao were to increase, the bandgap energy
would decrease and the material would begin
to behave less like a semiconductor and more
like a metal. If ao were to decrease, the
bandgap energy would increase and the
material would begin to behave more like an
insulator.
3.
Schrodinger’s wave equation is:
xV tx
x
tx
m
,
,
2 2
2 2
#######
t
tx j
,
Assume the solution is of the form:
#######
t
E tx xu jkx
, exp
####### Region I: xV 0. Substituting the
assumed solution into the wave equation, we
obtain:
#######
t
E jkux jkx m x
exp 2
2
#######
t
E jkx x
xu
exp
#######
t
E xu jkx
jE j
exp
which becomes
#######
t
E jk xu jkx m
exp 2
2
2
#######
t
E jkx x
xu jk
2 exp
#######
t
E jkx x
xu
exp 2
2
#######
t
E Eux j kx
exp
This equation may be written as
0
2 2 2 2
2 2
xu
mE
x
xu
x
xu xuk jk
####### Setting xu u 1 x for region I, the equation
becomes:
2 1 0
1 2 2 2
1
2 k u x dx
du x jk dx
ud x
where
2
2 2
mE Q.E.
####### In Region II, xV VO. Assume the same
form of the solution:
#######
t
E tx xu jkx
, exp
Substituting into Schrodinger’s wave equation, we find:
#######
t
E jk xu jkx m
exp 2
2
2
#######
t
E jkx x
xu jk
2 exp
#######
t
E j kx x
xu
exp 2
2
#######
t
E VO xu jkx
exp
#######
t
E Eux jkx
exp
This equation can be written as:
2
2 2 2 x
xu
x
xu xuk jk
####### 0
2 2 2 2 xu
mE xu
mVO
####### Setting xu u 2 x for region II, this
equation becomes
dx
du x jk dx
ud x 2 2
2
2 2
0
2 2 2
2 2
u x
mV k
O
where again
2
2 2
mE Q.E.
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