반도체물성과소자 4판 솔루션 - Semiconductor Physics and Devices Solution PDF

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Semiconductor Physics and Devices Solution...

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Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

Chapter 1 Problem Solutions Then

1.1 (a) fcc: 8 corner atoms 1 / 8  1 atom 6 face atoms 1 / 2  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 1 / 2  3 atoms 4 enclosed atoms = 4 atoms Total of 8 atoms per unit cell _______________________________________

1.2 (a) Simple cubic lattice: a  2r 3 3 Unit cell vol  a 3  2 r   8 r

 4 r 3 1 atom per cell, so atom vol  1   3  Then  4 r 3     3     100%  52.4% Ratio  8r 3 (b) Face-centered cubic lattice d d  4r  a 2  a  2 2r 2



Unit cell vol  a3  2 2  r



3

   

3

 100%  68%

8

Body diagonal  d  8 r  a 3  a 

r

3  8r  Unit cell vol  a     3  

3

3

 4 r3   8 atoms per cell, so atom vol  8   3    Then 3   8 4 r  3   Ratio   100%  34% 3  8r       3 _______________________________________

o

(a) a  5.43 A ; From Problem 1.2d,

a

8

r

3

 16 2  r 3

o a 3  5.43 3   1.176 A 8 8 Center of one silicon atom to center of

Then r 

o

nearest neighbor  2r  2.35 A (b) Number density 8   5 1022 cm 3 3 5.43 10  8 (c) Mass density 22 N At Wt . . 5  10 28.09    NA 6.02 10 23









   2.33 grams/cm 3 _______________________________________

3

3

 4 r 3 2 atoms per cell, so atom vol  2   3 

 4r       3 (d) Diamond lattice

1.3

 4 r 3   4 atoms per cell, so atom vol  4    3  Then  4 r 3   4   3  Ratio  100%  74% 16 2  r 3 (c) Body-centered cubic lattice 4 d 4r  a 3  a  r 3  4  Unit cell vol  a    r   3 

Ratio 

 4 r 3     3 

2

   

Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ o

1.4 (a) 4 Ga atoms per unit cell 4 Number density  5.65 10 8

(b) a  21.035  2.07 A





(c) A-atoms: # of atoms  8 

3

Density 

 Density of Ga atoms  2.22 10 22 cm 3 4 As atoms per unit cell  Density of As atoms  2.22 10 22 cm 3 (b) 8 Ge atoms per unit cell 8 Number density  8 3 5.65 10 



1.5 From Figure 1.15  a 3  (a) d        0.4330a  2   2 

1

2.07 10  8

3

 1.13 10 23 cm 3 1 B-atoms: # of atoms  6   3 2 3 Density  3 2.07 10 8



 Density of Ge atoms  4.44 10 22 cm 3 _______________________________________

1 1 8





 3.38 10 cm 3 _______________________________________ 23

1.9 o

(a) a  2r  4.5 A o

  0.4330 5.65  d  2.447 A

 a (b) d    2   0.7071a  2

# of atoms  8 

1 1 8

Number density  o

  0.70715.65  d  3.995 A _______________________________________

1

4.5 10 

8 3

 1.097 10 22 cm 3 N At.Wt. Mass density    NA

1.0974 10 12.5  22

1.6 a 2  2   2    54.74  sin   3 2  2 a 3 2    109.5 _______________________________________

1.7 o

(a) Simple cubic: a  2r  3.9 A (b) fcc: a  (c) bcc: a 

o

4r 2 4r

 5.515 A

(d) diamond: a 

2 4 r 

o

 9.007 A

3 _______________________________________

1.8 (a) 21.035 2  2 1.035  2rB o

rB  0.4287 A

6.02 1023  0.228 gm/cm3

(b) a 

4r

o

 5.196 A

3

1 # of atoms 8   1  2 8 Number density 

2

5.196 10 

8 3

 1.4257 10 22cm 3 1.4257 10 22 12.5  Mass density    6.02 10 23 3  0.296gm/cm _______________________________________



o

 4.503 A

3





1.10 From Problem 1.2, percent volume of fcc atoms is 74%; Therefore after coffee is ground, Volume = 0.74 cm 3 _______________________________________

Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ o

1.11 o

(b) a  1.8  1.0  2.8 A (c) Na: Density 

1 / 2 

 2.8 10 

8 3

 2.28 10 22cm 3 Cl: Density  2.28 10 22 cm 3 (d) Na: At. Wt. = 22.99 Cl: At. Wt. = 35.45 So, mass per unit cell 1  1  22.99    35.45  2  2   4.85 10 23 6.02 10 23 Then mass density 4.85 10 23   2.21 grams/cm 3 8 3 2.8  10 _______________________________________





1.12 o

(a) a 3  2 2.2   2 1.8   8 A o

Then a  4.62 A Density of A: 1  1.01 10 22 cm 3  8 3 4.62 10 Density of B: 1   1.01 10 22 cm 3 8 3 4.62 10 (b) Same as (a) (c) Same material _______________________________________

 

1.13

 

22.2   21.8 





2

 4.687 10 14 cm 2 o

For 1.12(b), B-atoms: a  4.619 A 1

 4.687 10 14cm 2 a2 For 1.12(a) and (b), Same material

Surface density 

o

For 1.12(b), A-atoms; a  4.619 A Surface density 1  3.315 10 14 cm 2  2 a 2 B-atoms; Surface density 1 14 2  2  3.315 10 cm a 2 For 1.12(a) and (b), Same material _______________________________________

1.14 (a) Vol. Density 

1 a 3o

Surface Density 

1

ao2 2

(b) Same as (a) _______________________________________

1.15 (i) (110) plane (see Figure 1.10(b)) (ii) (111) plane (see Figure 1.10(c))

o

 4.619 A 3 (a) For 1.12(a), A-atoms 1 1 Surface density  2   a 4.619 10 8

a

(b) For 1.12(a), A-atoms; a  4.619 A Surface density 1  3.315 10 14 cm 2  a2 2 B-atoms; Surface density 1   3.315 10 14cm 2 2 a 2

1 1  (iii) (220) plane   , ,    1, 1, 0  2 2  Same as (110) plane and [110] direction 1 1 1 (iv) (321) plane   , ,    2, 3, 6 3 2 1 Intercepts of plane at p  2,q  3,s  6 [321] direction is perpendicular to (321) plane _______________________________________

Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

1.16 (a)



1 1 1  , ,   313  1 3 1 (b)

1 1 1   , ,   121 4 2 4  _______________________________________ 1.17

1 1 1  Intercepts: 2, 4, 3   , ,   2 4 3  (634) plane _______________________________________ 1.18 o

(a) d  a  5.28 A o a 2  3.734 A 2 o a 3 (c) d   3.048 A 3 _______________________________________

(b) d 

1.19

(a) Simple cubic (i) (100) plane: Surface density 

1 1  a2 4.73 108





2

(ii) (110) plane: 1

Surface density  a

2

2

 3.16 10 14cm 2 (iii) (111) plane: 1 Area of plane  bh 2 o

where b  a 2  6.689 A Now a 2  2 2  h  a 2   2   

2



 

3 a 2 4

o 6  4.73  5.793 A So h  2





 6.32 10 14 cm 2 (iii) (111) plane: 1 3 6 Surface density  16 19.3755 10  14  2.58 10 cm 2 (c) fcc (i) (100) plane: 2 Surface density  2  8.94 1014 cm 2 a (ii) (110) plane: 2 Surface density  2 a 2

 6.32 10 14 cm 2 (iii) (111) plane: 1 1 3   3 6 2 Surface density  16 19.3755 10  15  1.03 10 cm 2 _______________________________________

 4.47 10 14 cm 2

 

Area of plane 1  6.68923 108 5.7930410 8 2  19.3755 10 16 cm 2 1 3 6 Surface density  19.3755 10 16  2.58 10 14 cm 2 (b) bcc (i) (100) plane: 1 Surface density  2  4.47 1014 cm 2 a (ii) (110) plane: 2 Surface density  2 a 2

2

1.20 (a) (100) plane: - similar to a fcc: 2 Surface density  2 5.43 10  8





 6.78 10 cm 2 14

(b) (110) plane: Surface density 



4

2 5.43 10 8



2

 9.59 10 14 cm 2

Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (c) (111) plane: Surface density 





2

3 2 5.43 10 8



2

 7.83 10 cm 2 _______________________________________ 14

1.21

a

4r 2



4 2.37

o

 6.703 A

2 1 1 8  6  4 8 2 (a) #/cm 3   3 a 6.703 10 8





(a)

3

o a 2 6.703 2   4.74 A 2 2 1 1 (d) # of atoms  3   3   2 6 2 Area of plane: (see Problem 1.19)

(c) d 



3  100%  10  %

 



Volume density 

o 6a  8.2099 A 2

Area 1 1    bh  9.4786 10 8 8.2099 10 8 2 2  15  3.8909 10 cm 2

 

1.26

o

b  a 2  9.4786 A

#/cm 2 

5  1022 2  1015

 

 3.148 10 cm 2



5  1017

1.25 (a) Fraction by weight 2  1016 10.82 7  1.542 10  22 5 10 28.06  (b) Fraction by weight 18 10  30.98 5   2.208 10  5  1022 28.06  _______________________________________







6  100%  4 10  % 5  1022 _______________________________________

(b)

14

h



1.24

 1.328 10 22 cm 3 1 1 4  2  2 4 2 (b) #/cm  a2 2 2  2 6.703 10 8 2



1.23 Density of GaAs atoms 8 4.44 1022 cm 3  3  5.65 10 8 An average of 4 valence electrons per atom, So Density of valence electrons 23  1.77 10 cm 3 _______________________________________

1 d3

 2 10 16cm 3 o

So d  3.684 10  6 cm  d  368.4 A



2

3.8909 10  = 5.14 10 14 cm 2 15

o a 3 6.703 3   3.87 A 3 3 _______________________________________

d

1.22 3 Density of silicon atoms  510 22cm and 4 valence electrons per atom, so 3 Density of valence electrons  210 23 cm _______________________________________

o

We have a o  5.43 A d 368.4   67.85 ao 5.43 _______________________________________ Then

1.27 Volume density 

1 d3

 4 10 15cm 3 o

So d  6.30 10  6 cm  d  630 A o

We have a o  5.43 A 630 d   116 a o 5.43 _______________________________________ Then

Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

Chapter 2 2.6

2.1 Sketch _______________________________________ 2.2 Sketch _______________________________________ 2.3 Sketch _______________________________________

(a) p 

h





6.625 10 34

550 10  27  1.205 10  kg-m/s p 1.2045 10 27    1.32 10 3m/s  m 9.11 10 31 9

or   1.32 10 5 cm/s h 6.625 1034 (b) p    440 10 9  1.506 10  kg-m/s p 1.5057 10 27 3    1.65 10 m/s m 9.11 10 31 or   1.65 10 5 cm/s (c) Yes _______________________________________ 27

2.4 From Problem 2.2, phase 

2 x



t

= constant Then dx 2 dx       0,   p     dt dt   2  2 x From Problem 2.3, phase  t



= constant Then dx 2 dx       0,   p     dt  dt  2  _______________________________________

2.7 (a) (i)







p  2mE  2 9.11 10 31 1.2 1.6 10 19 25

 5.915 10 kg-m/s h 6.625 1034 9    1.12 10 m p 5.915 1025 o

or   11.2 A



 

(ii) p  2 9.11 10 31 12 1.6 10 19

2.5 E  h

hc



 

hc

E Gold: E  4.90 eV   4.90 1.6 10 19 J So, 6.625 10  34 3 10 10   2.54 10 5 cm  4.90 1.6 10 19 or   0.254  m



















Cesium: E  1.90 eV  1.90 1.6  10 19 J So, 6.625 10  34 3 10 10  6.54 10  5 cm  1.90 1.6 10 19 or   0.654  m _______________________________________











 

 1.87 10 6.625 10 34

24

1.8704 10 24



kg-m/s

 3.54 10 10 m

o

or   3.54 A





 5.915 10

 24



(iii) p  2 9.11 10 31 120 1.6 10 19

 

34

kg-m/s

6.625 10 10  1.12 10  m 5.915 10  24 o

or   1.12 A





Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (b)



p  2 1.67 10

27

2.10

1.21.6 10  19

E avg 

3  3 kT    0.0259  0.03885 eV 2  2

Now



6.625 10  34

or

  8.56 10 6 cm/s

o

2.8

h

85 10 10 26  7.794 10  kg-m/s p 7.794 10 26     8.56 10 4 m/s 31 m 9.11 10

 2.532 10  23kg-m/s 6.625 10 34  11  2.62 10 m  2.532 10 23 or   0.262 A _______________________________________

p

(a)







p avg  2mE avg



 2 9.11 10

  0.038851.6 10 

31

19

or p avg  1.064 10  25 kg-m/s





Now



1 1 m 2  9.11 10 31 8.56 10 4 2 2 21  3.33 10  J 21 3.334 10 or E   2.08 10 2 eV 19 1.6  10 1 2 (b) E  9.11 10 31 8  103 2 23  2.915 10  J 23 2.915 10  1.82 10 4 eV or E  1.6  10 19 p  m  9.11 10 31 8 103

E

34 h 6.625 10 9  6.225 10  m   25 p 1.064 10



2







 27

 7.288 10 kg-m/s h 6.625 10 35  9.09 10 8 m   p 7.288 10 27

or o

  62.25 A

o

_______________________________________

or   909 A _______________________________________

2.9

E p  h p 

2.11

hc

p

Now 2 h 1  h  pe  Ee    and pe  e 2m   e  2m Set E p  E e and  p  10 e

Then 2

2

1  h  1  10h        p 2 m  e  2m   p  which yields 100h p  2 mc hc hc 2mc2   2mc  Ep  E   p 100h 100 



 31



2 9.11 10 3 10 100

8



2

 1.64 10  15J  10.25 keV _______________________________________





 6.625 10  3 10  34

1 10

8

10

 1.99 10  J 15

2

Ee 

hc

hc

(a) E  h  

Now

E 1.99 10 15  19 e 1.6  10 4 V  1.24 10 V  12.4 kV E  e V  V 





(b) p  2 mE  2 9.11 10 31 1.99 10 15



23

 6.02 10 kg-m/s Then h 6.625 10 34    1.10 10 11 m 23 p 6.02 10 or o

  0.11 A _______________________________________

Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

2.12  1.054 10 34 p   x 10 6

 1.054 10  kg-m/s _______________________________________ 28

2.13 (a) (i) p x   p 

1.054 10 12 1010

34

 8.783 10  26 kg-m/s

 p2     2 m   p   p p 2p   p  m 2m

dE d  p  (ii) E  dp dp

Now p  2mE



 

 2 9 10 31 16 1.6 10 19



24

 2.147 10 kg-m/s 2.1466 1024 8.783 1026 so E  9  10 31  2.095 10  19 J 2.095 1019 or E   1.31 eV 1.6  10 19  (b) (i) p  8.783 10 26 kg-m/s







 

(ii) p  2 5 10 28 16 1.6 10 19





 23

 5.06 10 kg-m/s 5.06 10 23 8.783 10 26 E  5  10 28  8.888 10  21 J 8.888 10 21  5.55 10  2 eV or E  1.6 10 19 _______________________________________







2.14 34  1.054 10   1.054 10 32 kg-m/s 2 x 10 p 1.054 10 32 p  m      m 1500

p 

   7 10  m/s _______________________________________ 36

2.15 (a) E t   1.054 10 34   8.23 10 16 s t  0.8 1.6  10 19





 1.054 10 34  x 1.5 10 10  7.03 10  25 kg-m/s _______________________________________

(b) p 

2.16 (a) If 1 x, t  and 2 x, t  are solutions to Schrodinger's wave equation, then   x , t    2 2 1 x, t  V x 1 x , t   j  1  2 t 2m x and 2 x , t    2 2 2 x, t   V x  2 x , t   j   t 2m x 2 Adding the two equations, we obtain  2 2 1 x , t   2 x , t   2m x 2  V x  1 x, t  2 x, t     x, t   2  x, t  t 1 which is Schrodinger's wave equation. ...