Semiconductor physics and devices 4th sol PDF

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Semiconductor Physics and Devices: Basic Principles, 4 edition Chapter 1By D. A. Neamen Problem Solutions Chapter 1Problem Solutions1.(a) fcc: 8 corner atoms 8/1  1 atom6 face atoms  2/1  3 atomsTotal of 4 atoms per unit cell(b) bcc: 8 corner atoms 8/1  1 atom1 enclosed atom =1 atom Total of 2...

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

Chapter 1 Problem Solutions 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 ce ll _______________________________________ 1.2 (a) Simple cubic lattice: a  2r Unit cell vol  a 3  2r 3  8r 3

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



Unit cell vol  a 3  2 2  r



3

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

Ratio 

 4r     3 (d) Diamond lattice

3

100%  68%

8

Body diagonal  d  8 r  a 3  a   8r  3  Unit cell vol  a     3

r

3

3

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

o

(a) a  5.43 A ; From Problem 1.2d, a

8

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  10 22 cm 3  3 5.43 10 8 (c) Mass density N At .Wt . 5  10 22 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 

 4 r 3     3 

2 

1.3

 16 2  r 3

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

Then

   

Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 1 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 1.4 (a) 4 Ga atoms per unit cell 4 Number density  8 5.65  10



o

(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   0.4330 a (a) d     2  2 



 3.38 10 cm 3 _______________________________________

# of atoms  8 

o

a 2  2   2 sin      54.74  a 2 3 2   3 2    109.5 _______________________________________ 1.7 o

(a) Simple cubic: a  2r  3.9 A

3

o

 5.515 A  4.503 A 2 4r 

o

 9.007 A

3 _______________________________________

1.8 (a)

21.035 2  21.035  2rB o

rB  0.4287 A

4.510 

8 3





1. 0974 10 12.5  23 6.02 10  0.228 gm/cm3 

(b) a 

4r 3

22

o

 5.196 A

1 # of atoms 8   1  2 8

Number density 

2

5.196 10 

8 3

 1.4257 1022 cm 3 22 1. 4257 10 12.5  Mass density    23 6.02 10 3  0.296 gm/cm _______________________________________



o

(d) diamond: a 

1

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

1.6

(c) bcc: a 

1 1 8

Number density 

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

2



23

o

 a (b) d    2  0.7071a  2

4r

8 3

(a) a  2r  4.5 A

 0.4330 5.65   d  2.447 A

4r

2.07 10 

1.9

o

(b) fcc: a 

1

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



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

1 1 8



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 ______________________________________________________________________________________ 1.11

o

o

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

1 / 2 

2.810 

8 3

 2.28 10 22 cm 3 3 22 Cl: Density  2.28 10 cm (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 23 4.85 10  2.21 grams/cm 3  3 2.8 10 8 _______________________________________





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  3 4.62 10 8 (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 1014 cm 2 2 a For 1.12(a) and (b), Same material

Surface density 

o

For 1.12(b), A-atoms; a  4.619 A Surface density 1  2  3.315 10 14 cm 2 a 2 B-atoms; Surface density 1  2  3.315 1014 cm 2 a 2 For 1.12(a) and (b), Same material _______________________________________ 1.14 (a) Vol. Density 

1 ao3

Surface Density 

1 2 o

a 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  2  3.315 10 14 cm 2 a 2 B-atoms; Surface density 1   3.315 1014 cm 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  2 a 4.73 10 8





2

 4.47 10 14 cm 2 1 2

a 2  3.16 1014 cm 2 (iii) (111) plane: 1 Area of plane  bh 2 o

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

 

2

2

  3 a 2  4 

o 6 So h  4.73   5.793 A 2

 





 6.32 1014 cm 2 (iii) (111) plane: 1 3 6 Surface density  19.3755  10 16  2.58 1014 cm 2 (c) fcc (i) (100) plane: 2 14 Surface density  2  8.94 10 cm 2 a (ii) (110) plane: 2 Surface density  2 a 2  6.32 1014 cm 2 (iii) (111) plane: 1 1 3   3 6 2 Surface density  19.3755  10 16  1.03 10 15 cm 2 _______________________________________

(ii) (110) plane: Surface density 

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 1014 cm 2 (b) bcc (i) (100) plane: 1 Surface density  2  4.47 1014 cm 2 a (ii) (110) plane: 2 Surface density  a2 2

2

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





 6.78 10 cm 2 14

(b) (110) plane: Surface density 



4

2 5.43  10  8



2

 9.59 1014 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 1014 cm 2 _______________________________________

1.21

o 4r 4 2.37    6.703 A 2 2 1 1 8   6 4 8 2 (a) #/cm 3  3 a 6.703 10  8

a



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



8 2



5  10 17 100%  10 3 % 5  10 22 2 10 15 100 %  4 10 6 % (b) 5  10 22 _______________________________________

3

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

 

2

 3.148 10 cm o a 2 6.703 2   4.74 A (c) d  2 2 1 1 (d) # of atoms  3  3  2 6 2 Area of plane: (see Problem 1.19)



o



#/cm 2 



14



1  2 10 16 cm 3 3 d o

6 So d  3.684 10  cm  d  368.4 A



2 3.8909 10 15

= 5.14 10

 

Volume density 

o 6a  8.2099 A 2

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

 

1.26

b  a 2  9.4786 A h



(a)

3

2

14



1.24

2

6.703 10 

1.23 Density of GaAs atoms 8  4.44 10 22 cm 3  8 3 5.65  10 An average of 4 valence electrons per atom, So Density of valence electrons  1.77 1023 cm 3 _______________________________________

cm 2

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

d

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

o

We have ao  5.43 A 368.4 d   67.85 5.43 ao _______________________________________

Then

1.27 Volume density 

1  4 10 15 cm 3 d3 o

So d  6.30 10  6 cm  d  630 A o

We have ao  5.43 A 630 d  116 ao 5.43 _______________________________________

Then

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

Chapter 2 2.1 Sketch _______________________________________ 2.2 Sketch _______________________________________ 2.3 Sketch _______________________________________ 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  _______________________________________

2.6

6.625  1034  550 10 9  1.205 10  27 kg-m/s p 1.2045 1027  1.32  103 m/s   m 9.11 10  31 5 or   1.32 10 cm/s h 6.625 10 34 (b) p    440  10 9  1.506 10  27 kg-m/s p 1.5057  10 27  1.65 103 m/s   m 9.11 10  31 5 or   1.65 10 cm/s (c) Yes _______________________________________ (a) p 

h



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  1.12 109 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 hc

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



















Cesium: E  1.90 eV  1.90 1.6 10 19 J So, 6.625 10 34 3  1010  6.54  10  5 cm  1.90 1.6 10 19 or   0.654 m _______________________________________













24

 1.87 10 kg-m/s 34 6.625 10   3.54 10 10 m   24 1.8704 10 o

or   3.54 A







(iii) p  2 9.1110 31 120 1.6 10 19 24

 5.915 10 kg-m/s 34 6.625 10  1.12  10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

 2.532 10  23 kg-m/s 6.625 10  34  2.62  10 11 m  2.532 10  23 o

or   0.262 A _______________________________________

or

3 3 kT    0. 0259  0.03885 eV 2 2

or

Now

pavg  2mEavg

(b)







 2 9.1110 31 0.03885  1.6 10 19



or

or

pavg  1.064  10 25 kg-m/s

h





2.8

E avg 

6.625 10 34 10  85 10 26  7.794 10 kg-m/s p 7.794 10 26  8.56 104 m/s   m 9.11 10 31   8.56 106 cm/s 1 1 E  m  2  9.11 10 31 8.56 10 4 2 2  3.33  10 21 J 3.334  1021 E  2.08  10 2 eV 19 1.6 10  1 2 E  9.11 10 31 8 10 3 2  2.915 10 23 J 2.915  1023 E  1.82  10 4 eV 19 1.6 10  31 8  10 3 p  m   9.1110

(a) p 







Now

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





2







 7.288 10 27 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

(a) E  h  

Now pe2 1  h h  Ee   and pe  2m e 2m   e Set E p  Ee and  p  10 e Then Ee 

1  h    p 2m   e hc

2

 1  10h     2m   p  

   

which yields 100 h p  2 mc

Ep E 



hc

p



2 hc 2 mc  2 mc  100h 100





2

31 8 2 9.11 10  3 10 100  1.64 10 15 J  10.25 keV _______________________________________







 1.99 10

2

2

hc

6.625 10 3 10   34

8

1 10 10

15

J

Now

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





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



 23

 6.02  10 kg-m/s Then  h 6.625 10 34  1.10 1011 m   p 6.02  10 23 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   6 x 10  28 1.054 10 kg-m/s _______________________________________ 2.13 (a) (i) px  

1.054 10  34  8.783 10  26 kg-m/s 10 12 10  2 dE d p  (ii) E     p  p  dp dp  2m  2p pp   p  2m m p 

Now p  2mE



 



 2 9 10 31 16  1.6 10 19 24

 2.147 10 kg-m/s 24 26 2.1466 10  8.783 10  so E   31 9 10 19  2.095 10 J 19 2.095 10  or E   1.31 eV 19 1.6  10   (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 23 26 5.06 10 8.783  10 E  28 5  10  21  8.888  10 J 8.888 10 21 or E   5.55  10 2 eV 19 1.6  10  _______________________________________







2.14

 1.054  10 34   1.054 10 32 kg-m/s x 10  2 p 1.054  10 32  p  m    m 1500   7  10 36 m/s _______________________________________ p 

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  2m t x 2 and 2  2 x , t    2  2  x, t  V x 2  x, t  j  2 t 2m x Adding the two equations, we obtain  2  2 1 x ,t   2 x ,t   2 m x 2  V x 1 x , t   2 x , t   1 x, t  2  x, t  t which is Schrodinger's wave equation. So 1 x, t   2 x, t  is also a solution.  j

(b) If 1 x , t  2  x, t were a solution to Schrodinger's wave equation, then we could write 2   2 1   2  V  x1   2   2 m x 2   j 1  2  t which can be written as 2   2   2 1   2   2 2 1     1  2 2 2 x x  2m  x x 1   2  2  V x 1   2   j 1 t  t  Dividing by 1  2 , we find

 2 2m

2  1  2 2 1  1 2 ...