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Semiconductor Physics and Devices: Basic Principles, 4 edition Chapter 1By D. A. Neamen Exercise Solutions ####### Chapter 1Exercise SolutionsEx 1.(a) Number of atoms per unit cell6 421 8 81     (b) Volume Density3 83 25 104 4   a22  21  10 cm 3 Ex 1.Intercepts of plane; p=1, q=2, s=In...

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

Chapter 1 TYU 1.2 (a) Number of atoms per (100) lattice plane 1  4  1 4 1 1 Surface Density  2  2  a 4.65  10 8

Exercise Solutions Ex 1.1 (a) Number of atoms per unit cell 1 1  8  6  4 8 2 4 4 (b) Volume Density  3  a 4.25 10  8







3

 5.21 10 22 cm 3 _______________________________________

Ex 1.2 Intercepts of plane; p=1, q=2, s=2 1 1 1  Inverse;  , ,  1 2 2  Multiply by lowest common denominator,  211 plane _______________________________________ Ex 1.3 (a) Number of atoms per (100) plane 1  1 4  2 4 2 2 Surface Density  2  a 4.25 10  8





2

= 4.62 10 cm (b) Number of atoms per (110) lattice plane 1  4  1 4 Surface Density 1 1    a a 2 4.65 10 8 2 2 14

  



2

 3.27 10 cm (c) Number of atoms per (111) lattice plane 1 1  3  6 2 1 Lattice plane area  bh 2 14

where b  a 2 2  2   1 h  a 2   a 2     2 

 



2

1/ 2

1/ 2

Test Your Understanding Solutions

 1  3  2 a 2  a 2  a 2 2   Then lattice plane area  3 1 3 2  a 2 a   a   2 2  2 Surface Density 1 2   2.67  1014 cm 2 3 8 2 4.65 10 2 _______________________________________

TYU 1.1

TYU 1.3

 1.11 10 cm 2 (b) Number of atoms per (110) plane 1 1  2  4   2 2 4 Surface Density 2 2    a a 2 4.25 10 8 15

  



 

2

2

2

 7.83 10 cm _______________________________________ 14

Number of atoms per unit cell  8  Volume Density  4 10 22 

1 1 8

1 a3 o

8  a  2.92 10  cm  2.92 A o

Radius  r  a 2  1.46 A _______________________________________





o

(a) For (100) planes, distance  a  4.83 A (b) For (110) planes, distance o a 2 4.83 2   3.42 A 2 2 _______________________________________



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

TYU 1.4 (a) 8 corner atoms (b) 6 face-centered atoms (c) 4 atoms totally enclosed _______________________________________ TYU 1.5 Number of atoms in the unit cell 1 1  8   6  4  8 8 2 8 8 Volume Density  3   a 5.43 10 8





3

 5 10 22cm 3 _______________________________________

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

Chapter 2 Exercise Solutions Ex. 2.1 (a)

E  h 

(b) E n 

6.62510 3 10   34

hc

100 10 8  1.9875 10  17J 1.9875 10 17  124 eV or E  1.6 10 19 hc 6.625 10 34 3 1010 (b) E    4500 10 8  4.417 10  19 J







10



2ma

2

1.054 10   n 21.67  10 12 10  34

2

2

2

10 2

27

 2.28 10 n J 2.27967 10 23 n 2 or E n  1.6  10  19  1.425 10  4 n 2eV 23



2

Then E1  1.425  10 4eV 

E 2  5.70 10 4 eV

19

4.417 10  2.76 eV 1.6  1019 _______________________________________ E

or

 2  2 n2

E3  1.28 10  3 eV _______________________________________

Ex 2.4

Ex 2.2







 2 9.11 10 31 12 10 3 1.6 10 19



1/ 2

 5.915 10  26 kg-m/s h 6.625 10  34   1.12  10 8 m p 5.915 10  26





o

or   112 A h 6.625 1034 (c) p    112 10 10  5.915 10  26kg-m/s





 26 2

1 p 1 5.915 10  2 m 2 2.2 10 31 =7.952 10  21J 21 7.952 10  4.97 10 2 eV or E  1.6 10 19 _______________________________________ 2

E

Ex 2.3

 

1 1 2 9.11 10 31 105 m  2 2  4.555 10 21J Now E

(a) p  2mE

k2 

2m Vo  E  2

2

Set Vo  3E

Then k2  

1 2 m2 E  

2 9.11 10 24.555 10   31

1.054 10

 21 1 / 2

 34

or k2  1.222 10 9 m 1

P  exp 2k 2 d o

 (a) d  10 A  10 10 10 m P  exp   2 1.222 109 10 10 10 or P  0.0868  8.68%

 





o

 2  2n 2 (a) E n  2ma 2

1.054 10   n 2  9.11 10 12 10  34 2



31

2

2

10 2

 4.179 10  20 n 2 J  20

or E n 

4.179 10 n 2  0.261n eV 19 1.6  10  2

Then E1  0.261eV, E 2  1.045 eV, E 3  2.351 eV

 (b) d  100 A  100 10 10 m P  exp  2 1.222 109 100 10 or P  2.43 10 11  2.43 10 9 %

 



10



_______________________________________

Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 2 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________ TYU 2.2 (a) E  0.8 1.6 10 19 1.28 10 19 eV

Ex 2.5



2mV O  E (a) k2  2





  1.054 10   8.23 10 16 s 19 E 1.28 10 (b) Same as part (a), t  8.23 10  16 s _______________________________________ 34



 

2 9.11 10 31  1.2  0.12 1.6 10  19





1.054 10

 t



34 2

 5.3236 10 m 1 Then  0.12  0.12  T  16  1   1.2   1.2   9

TYU 2.3 (a) k2 

 



 exp 2 5.3236 109 5 10 10



3

T  7.02  10  0.12  0.12  (b) T  16 1   1.2   1.2 

 



2m V O  E  2





 







 



2 9.11 10 31 0.8  0.1 1.6 10 19

1.054 10

 34 2

= 4.286 10 m 1  0.1  0.1  T  16  1    0.8  0.8  9



 exp 2 5.3236 109 25 10 10



 12

T  3.97 10 _______________________________________ Ex 2.6

From Example 2.6, we have 13.58 0.0992 En   eV 2 2 n2 11.7  n E 1  99.2 meV, E 2   24.8meV, E 3  11.0 meV _______________________________________

 

 exp 2 4.2859 109 12 10 10 5

T  5.97 10

(b) k 2 





2 9.11 10 31 1.5  0.11.6 10 19

1.054 10

 6.061 10 m 1  0.1  0.1  1   T  16   1.5  1.5 

 34 2

9

 



 exp 2 6.061 109 12 10 10



7

T  4.79 10 _______________________________________

Test Your Understanding TYU 2.1 34

 1.054 10  8  1010 x  1.318 10 25 kg-m/s  d  p 2  dE   p (b) E    p     dp  dp  2m  p p 2p   p  2m m 23 25 1.2  10  1.318 10  E   31 9.11 10  1.735 10 18 J or  10.85 eV _______________________________________

(a)  p 







TYU 2.4 T  5  10  6

 0.08  0.08   16 1   exp  2 k 2 a 0.8   0.8  so that exp  2 k2 a   2.88 10 5 2k 2a  12.571 k2 







2 9.11 10 31  0.8  0.08 1.6 10  19

1.054 10 



34 2

 4.3467 10 m 1 Then 12.571 a  1.446 10 9 m 2 4.3467 109 9





o

or a  14.46 A _______________________________________

Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 3 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________

Chapter 3 Ex 3.3

Exercise Solutions

4 2m 3 / 2

(a) N 

Ex 3.1 1 E  m 2 2

1 dE  2 m   d  m   d 2 or  

 





4 2 9.11 10

N  1.28 10 22 cm 3 (b) N 

4 2m 3 / 2 2 3 / 2  E h3 3





 1.189 10 Then E  E3  E 2



4 2m p



2

 4.5 10 



10 2



4 2m p h

J 

3/ 2

E  E  dE



3/ 2

3

E   2 3/ 2   E   E   E kT  3 



4 20.56 9.11 10  31



3/ 2

 6.625 10    2    1 0.02591.6 10  3  34 3

 19

 3.929 10  J 19



Or 





h3

E  kT

 18

3.929 1019

 2.46eV

1.6  10 19 _______________________________________



3/ 2

N  8.29 10 21 cm 3 _______________________________________

 1.189 10 18  7.958 10 19

E 



or

2 ma2

2 9.11 10



 8.29 10 27 m 3

E

31

1eV

 1.06286 10 2 3/ 2 3/ 2 2 1 1.6 10 19  3

N

2 2 1.054 10 34 

2eV

56

2 2 2  

m 3

or

Ex 3.4

or

E3

28

 1.28 10

2mE 3  a  2 2

3/ 2

 19

 1.76  10  6 m/s

or E 2  7.958 10 19J. At ka  2 , we see that  3 a  2 so

0

3/ 2

34 3

  

Ex 3.2 At ka   , we have sin a 1  8  cos  a a From Example 3.2, we have  2 a  5.141,

2 eV

  6.625 10  2  21.6  10  3 31

19

or    1.76  10 4 cm/s _______________________________________

E  dE

0

4 2m 3 / 2 2 3 / 2  E 3 h3

10  1.6  10  E  31 5 m 9.1110 10 12



h3



So

2 eV

3/ 2



 7.92 10 24 m 3 or N  7.92 10 18 cm  3 _______________________________________

Ex 3.5

g i! 10 9 8!   45 10!   Ni ! gi  N i ! 8!10  8!  8! 21 _______________________________________

Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 3 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________

Ex 3.6

   E  EF  (a) f F E   exp  kT     E c  kT 4  E F   exp   kT      0.30  0.0259 4  exp   0.0259    7.26 10  6    E  EF  (b) f F E   exp  kT  

   0.30  0.0259   exp   0.0259    3.43 10  6 _______________________________________

Then

  E  E F   exp   0.02 kT  

 1    3.9kT E E F  kT ln  0.02  _______________________________________

Test Your Understanding Solutions TYU 3.1 At ka  2 , we see that  3 a  2 , so 2 mE3



2

or

E3 

Ex 3.7

  E  E F  f F E   exp   kT  

  0.30  0.025  8 10 6  exp   kT     0.325 5 exp    1.25 10  kT 





0.325  ln 1.25 105  11.736 kT 0.325  T  kT   0.02769  0.0259   11.736  300  so T  321 K _______________________________________ Ex 3.8    E  EF  exp  kT  

1 E EF  1  exp   kT 

 0.02 1  E  EF  1  exp    kT    E  E F   E  E F  exp  1  exp   1  0.02 kT    kT 

 a  2

2 2  2 2 ma 2

 2 2 1.054 10 34 

2







2 9.11 10 31 4.5  1010



2

 1.189 10  18 J At the other point,  4 a is in the range 2   4 a  3 . Then from sin  4 a 1  8  cos  4 a 4 a we find, by trial and error,  4a  7.870. Then E4 

7.8702 2 2ma 2

 7.8702 1.054 1034 

2







2 9.11 10 31 4.5  1010



2

 1.8649 10  18J Then Eg  E 4  E 3  1.8649 10 18 1.189 10 18  6.762 10  19 J or 6.762 10  4.23eV 19 1.6 10  _______________________________________ 19

Eg 

Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 3 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________

TYU 3.2 From Example 3.2, for ka   ,   1 a   and E1  2.972  10 19 J. For 0  a   and ka  0 , we have sin  a 1  8  cos a a By trial and error,  a  2.529 rad. Then 2mE  a  2.529 2 E

2.5292  2 2ma 2

or



2 9.11 10

4.5 10 

10 2

 1.9258 10 19J Then E  2.972 10 19 1.9258 10 19  1.046 10 19 J or 19 1.046 10 E   0.654eV 19 1.6  10 _______________________________________

 Ec  0.32  E c 1.6 10 19      C1   10  10 10

2

so that C1  5.1876 10 39 We have 2 2 m   m  2 C1 m o 2 mo C 1

1.054 10  2 9.11 10 5.1876 10  34 2

31

39

or m  1.175 mo _______________________________________

TYU 3.4 We have E  E  C 2 k 2

E   0.875  E  1.6 10 19      C 2   10  12 10 

so that C 2  2.0426 10 38

 2 mo C2

1.054 10  2 9.11 10 2.0426 10 

m p mo

 31

 38

 0.2985

TYU3.5 1  E  EF  1 exp   kT    E F  E   exp  kT      0.35   0.0259 2 (a) 1 f F  E  exp   0.0259   1  f F E   1 

 8.20 10  7    0.35  3 0.0259 2  (b) 1 f F  E  exp   0.0259    3.02 10 7 _______________________________________

TYU 3.3 We have E  E c  C1 k 2



mo

2



_______________________________________

 2.5292 1.054 10 34   31

m p

34 2



2



We have

2

TYU 3.6 400  We find kT  0.0259   0.034533eV 300    E c  kT 4  E F  (a) f F E   exp   kT     0.30  0.034533 4   exp  0.034533    1.31 10 4    0.30 0.034533 (b) f F E   exp   0.034533    6.21 10 5 _______________________________________

Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 3 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________

TYU 3.7  400  We find kT  0.0259    0.034533  300    0.35  0.034533 2   (a) 1 f F  E  exp  0.034533    2.41 10 5    0.35 3 0.034533 2  (b) 1 f F  E  exp  0.034533    8.85 10  _______________________________________ 6

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

Chapter 4 Exercise Solutions Ex 4.1

ni  3.29 10 9 cm 3 For T  250 K,

  E  E F  f F  exp   kT     E c kT  E F   exp   kT     0.25  0.0259    exp   0.0259   f F  2.36 10





 5.09 10 7

5

or ni  7.13 10 3 cm 3

(b)



n o  3.02 10 13 cm 3 _______________________________________

n i 400 3.288 109  4.61 10 5  n i 250 7.135 103 _______________________________________ Ex 4.4 (a) GaAs

Ex 4.2





 250  (a) N   1.04  1019    300 

 mp  3 kT ln    m  4  n  3  0.48    0.0259 ln  4  0.067    38.25meV

E Fi  E midgap 

3/ 2

 7.9115 10 18 cm 3  250  kT   0.0259   0.021583eV  300 

(b) Ge

  E F  E    p o  N exp  kT    0.27    7.9115 10 18 exp    0.021583

3  0.0259 ln 0.37  4  0.55    7.70meV _______________________________________

p o  2.919 10 13cm 3

Ex 4.5





2.919 1013 3 (b) Ratio   4.54  10  6.43 10 15 _______________________________________

Ex 4.3 (a) For T  400 K,







3

 400  n i2  4.7 1017 7  1018    300 

  1.42  exp    0.0259 400 300   1.081 10 19 or



  1.42  exp    0.0259250 300  

  E c  E F   n o  N c exp  kT    0 . 25    4.7 10 17 exp    0.0259 



3

 250  ni2  4.7 1017 7  1018    300 

E Fi  E midgap 

  E F  E  p o  N  exp  kT     0.215  1.04 10 19 exp   0.0259 





 2.58 10 15 cm 3 We find E c  E F  1.12  0.215  0.905eV    E c  E F  n o  N c exp  kT     0.905   2.8 10 19 exp   0.0259 





 1.87 10 4 cm 3 _______________________________________

Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 4 By D. A. Neamen Exercise Solutions ______________________________________________________________________________________ Ex 4.6 no 

2



1.5 10

20

pa  po  p a

N c F1 / 2  F 



2



2.8 10 F   19

1/ 2

F

So F1 / 2  F   4.748 From Figure 4.10,  F  3.2 

E F  Ec

kT  E F  E c  3.2 0.0259  0.08288 eV _______________________________________

 250 ...