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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

Unit cell vol  a 3  2r   8r 3 3

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



Unit cell vol  a 3  2 2  r



3



3

3

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

3

   

100%  68%

8

Body diagonal  d  8r  a 3  a 

r

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

3

 4 r 3   8 atoms per cell, so atom vol  8   3  Then 3   8 4 r  3  Ratio    100%  34% 3  8r       3 _______________________________________ 1.3 o

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

8

r

3

 16 2  r 3

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

   

Ratio 

2 4 r

3

   

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

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

Then r 

   

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  5.65 10 8



(b) a  21.035  2.07 A o

(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  3 5.65 10 8



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 (a) Simple cubic: a  2r  3.9 A 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 22

6.02 10 23  0.228 gm/cm 3

(b) a 

4r

o

 5.196 A

3 1 # of atoms 8   1  2 8

2

5.196 10 

8 3

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



o

(d) diamond: a 



Number density 

o

1

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

1.6

3

1 1 8

Number density 

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

(c) bcc: a 



23

o

a (b) d    2  0.7071a 2

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  3 2.07  10 8



 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

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 (a) a 3  22.2  21.8  8 A o

o

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

 

1.13 a

 

22.2  21.8





2

 4.687 1014 cm 2 o

For 1.12(b), B-atoms: a  4.619 A 1  4.687  10 14 cm 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 1014 cm 2 2 a 2 B-atoms; Surface density 1   3.315  10 14 cm 2 2 a 2 For 1.12(a) and (b), Same material _______________________________________ 1.14 (a) Vol. Density 

1 a o3 1

Surface Density  a

2 o

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

(b) For 1.12(a), A-atoms; a  4.619 A Surface density 1   3.315 1014 cm 2 2 a 2 B-atoms; Surface density 1   3.315  10 14 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  3.048 A (c) d  3 _______________________________________

(b) d 

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

1 1  2 a 4.73 10 8





2

(ii) (110) plane: 1

Surface density  a

2

2

 3.16 1014 cm 2

(iii) (111) plane:

1 bh 2 o

where b  a 2  6.689 A Now

 

h2  a 2

2

2

 

a 2  3 a 2   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  19.3755  10 16  2.58 10 14 cm 2 (c) fcc (i) (100) plane: 2 Surface density  2  8.94  10 14 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  19.3755  10 16  1.03 1015 cm 2 _______________________________________

 4.47 1014 cm 2

Area of plane 

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  10 14 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 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 a

4r



42.37 

o

 6.703 A

2

2 1 1 8  6 4 8 2  3 (a) #/cm  a3 6.703 10 8



5 1017 100%  10 3 % 22 5 10 2 1015 (b) 100%  4 10 6 % 5 10 22 _______________________________________



1.25 (a) Fraction by weight 2 1016 10.82   1.542 10 7 22 5 10 28.06 (b) Fraction by weight 1018 30.98   2.208 10 5 5 10 22 28.06 _______________________________________



 

 3.148 10 cm 2 14

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 





 



Volume density 

o 6a h  8.2099 A 2

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

 

1.26

o

b  a 2  9.4786 A





(a)

3

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





1.24

3

22

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 10 23 cm 3 _______________________________________

1  2  10 16 cm 3 d3 o

So d  3.684 10 6 cm  d  368.4 A



2 3.8909  10 15 = 5.14 1014 cm 2

#/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 Density of valence electrons  2 10 23 cm 3 _______________________________________

o

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

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 d 630   116 a o 5.43 _______________________________________

Then

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

Chapter 2 2.1

2.6

Sketch _______________________________________

6.625 10 34  550 10 9  1.205 10 27 kg-m/s p 1.2045 10 27    1.32 10 3 m/s  31 m 9.1110 or   1.32 10 5 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 10 3 m/s  31 m 9.1110 or   1.65 10 5 cm/s (c) Yes _______________________________________

2.2 Sketch _______________________________________ 2.3 Sketch _______________________________________ 2.4 From Problem 2.2, phase 

2 x



 t

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

(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, 6.625 10 34 3 1010   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 1010   6.54 10 5 cm 1.90 1.6 10 19 or   0.654  m _______________________________________









24





 1.87 10 kg-m/s 6.625 10 34   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 6.625 10 34   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  23 2.532 10 o

or   0.262 A _______________________________________ 2.8

E avg 

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

Now

p avg  2mE avg





6.625 10 34  85 10 10  7.794 10 26 kg-m/s p 7.794 10 26    8.56 10 4 m/s m 9.1110 31 or   8.56 10 6 cm/s 1 1 E  m 2  9.1110 31 8.56 10 4 2 2 21  3.33 10 J 3.334 10 21 or E   2.08 10  2 eV 1.6 10 19 2 1 (b) E  9.1110 31 8 10 3 2  2.915 10 23 J 2.915 10 23 or E   1.82 10  4 eV 1.6 10 19 p  m  9.1110 31 8 10 3 (a)



 2 9.1110 31 0.03885 1.6 10 19



or p avg  1.064  10 25 kg-m/s

p

h











Now

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





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

(a) E  h 

Now p2 h 1  h   Ee  E e  e and p e  e 2m   e 2m

   

1  h    p 2m   e hc

 1  10h     2m   p   2

2

which yields 100h p  2mc

Ep  E 



hc

p



hc 2mc 2  2mc  100h 100





2

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



 1.99  10

2

Set E p  Ee and  p  10e Then

hc



6.625 10 3 10  34

8

110 10

15

J

Now

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





(b) p  2mE  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 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   x 10  6  1.054 10 28 kg-m/s _______________________________________ 2.13 (a) (i) px  

1.054 10 34  8.783 10  26 kg-m/s 12 10 10 dE d  p2     p (ii) E   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 2.1466 10 24 8.783 10 26 so E  9 10 31 19  2.095 10 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





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







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

2.15 (a) Et   1.054 10 34 t   8.23 10 16 s 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 2m t x and 2 x, t    2  2  2  x, t    V x 2 x, t   j 2 2m t x 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   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   2m x 2   j 1  2  t which can be written as  2 1      2   2 2  2 2 1  2  1 2 2 2m  x x  x x

1   2  V x 1  2   j 1  2  t t   Dividing by 1  2 , we find

 2 2m

 1  2 2 1  2 1 2 1 2        2 2 1 x 1 2 x x   2 x  1 2 1 1   V x   j      t  1 t   2

Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Since 1 is a solution, then

 1  1 1 1    V x   j   2 2m 1 x 1 t Subtracting these last two equations, we have   2  1  2 2 2 1 2      2 2m  2 x 1 2 x x  1 2  j   2 t

2.19 

2

2

 1  2 1 2    V x   j   2 2m 2 x 2 t Subtracting these last two equations, we obtain    2 2   1  2  V x   0 2m 1 2 x x This equation is not necessarily valid, which means that 1 2 is, in general, not a solution to Schrodinger's wave equation. _______________________________________

*

0

Function has been normalized. (a) Now ao

P

2

 2   x   dx exp    a o  a o 

4

 0

ao

Since 2 is also a solution, we have



2 ao

1

2

 x  cos 2  dx  1  2 

P

4 ao

2  ao

1 / 2



A 2 cos 2 nx dx  1

1 / 2

 x sin 2nx   1 / 2 A2   1 4n  1 / 2 2

 1  1  1 A       1  A 2   2  4  4  2

or A  2 _______________________________________



  1   1  1  exp    2   

 2   x   dx exp    a o   a o

2

2.18

  2 x  ao 4    exp    ao  0

2 2



ao

 x sin x   3 A2   1 2  1 2

 3   1  A      1  2  2  1 so A 2  2 1 or A  2 _______________________________________

  ao   2

   2a o P   1exp   4a o which yields P  0.393 (b) ao

A

0

or

2.17 3

  2x   dx a o 

4

 exp

2  ao

2

2

    dx  1

Note that

2 ao

2

  2x   dx a o 

 exp

ao

4

  ao   2

  2 x  ao    exp    a o  ao

2 4

or

   1  P   1exp 1  exp   2   which yields P  0.239 (c) 2

 2   x   dx P  exp   ao  a o  0  ao





2 ao

ao

  2x   dx a o 

 exp 0

  2 x  ao   ao     exp   2   ao  0   1exp 2  1 which yields P  0.865 _______________________________________ 

2 ao

Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 2.20

2.21 P   x  dx



a/4

(a)

 0

2

a/4

(a) P 

2 2 x  dx   cos  a    2 

  2x   a / 4 sin    2 x  a          a  2 4   0  a    a       sin    2 4  2       4    a  2      a   

 2  a 1a      4   a  8 or P  0.409 a/2

(b) P 

2

     2  a sin         a   8  8       a   or P  0.25 a/2

(b) P 

or P  0.25 a / 2

(c) P 

     2   a sin     a  sin           a   4  4   4   4          a   a    or P  1 _______________________________________

 2  2  2x    sin  dx a  a  a / 2



  4x    a / 2 sin     2  x  a      2    a  2 4   a / 2   a   

a / 2

  2x    a / 2 sin     2  x  a      4    a  2    a / 2   a   

 2x   dx  a 

     2  a sin 2   a  sin           a   4  8   8   8         a   a  

1 1  1  2  0   4 8 4   or P  0.0908 2 2  x    cos  dx a  a  a / 2

2

  4x   a / 2 sin     2  x  a      2    a  2 4   a/4   a   

  2x   a / 2 sin     2  x  a         a  2 4   a / 4  a  



2

  a  sin

a/4

a/4

(c) P 

 2x   dx  a 

  4x   a / 4 sin    2 x  a       2    a  2 4   0   a   

 x 

   sin      2  a sin   a  2         a   4  4  8  4          a   a   

2

0

  a  cos  a dx 2

2

  a  sin

        sin 2  sin  2  2 a  a             8    a   4  8   4         a   a    or P  1 _______________________________________ 2.22

or



8 1012  10 4 m/s k 8 10 8  p  10 6 cm/s

(a) (i)  p 





2 2   7.854  10 9 m 8 k 8  10

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

or

  78.54 A



 9.1110 27 kg-m/s 2 1 1 E  m 2  9.1110 31 10 4 2 2  4.555 10 23 J 4.555 10 23 or E   2.85 10  4 eV 1.6 10 19  1.5 1013 (b) (i)  p    10 4 m/s k  1.5 10 9 or  p  10 6 cm/s



 

2 2   4.19 10 9 m 9 k 1.5 10



25

 9.11 10 kg-m/s 6.625 10 34   7.27 10 10 m 9.1110  25 2 k  8.64  10 9 m 1 7.272  10 10   8.64 10 9 10 6  8.64 1015 rad/s _______________________________________



p  9.1110 27 kg-m/s

E  2.85 10 4 eV _______________________________________







For electron traveling in  x direction,

  9.37 10 6 cm/s p  m  9.1110 31  9.37 10 4







 8.537 10 26 kg-m/s h 6.625  10 34    7.76  10 9 m p 8.537  10  26 2

2 k   8.097  10 8 m 1  7.76  10 9

  k    8.097 10 8 9.37  10 4 

or   7.586 10 rad/s _______________________________________ 13

2.24 (a)





p  m  9.1110 31 5 10 4 26





 4.555 10 kg-m/s h 6.625 10 34    1.454 10 8 m p 4.555 10  26



2









2

E n  n 2 1.0698 10 21 J or





n 2 1.0698 10 21 1.6 10 19 or E n  n 2 6.686 10 3 eV Then E1  6.69 10 3 eV



1  m 2 2

1 9.11 10 31  2 2 so   9.37  10 4 m/s  9.37  10 6 cm/s 



n 2 1.054  10 34  2  2 n 2 2  2 2ma 2 9.11 10 31 75  10 10

En 

2.23 (a) x, t   Ae  j kx t 



 

2.25 En 

(b) E  0.025 1.6 10

 

(b) p  9.1110 31 10 6

or   41.9 A

19

2  4.32  10 8 m 1 8 1.454  10

 2.16 1013 rad/s

o

(ii)



   k  4.32 10 8 5 10 4 

 

(ii) p  m  9.1110 31 10 4



2

k



E 2  2.67 10 2 eV E3  6.02 10 2 eV _______________________________________ 2.26 (a) E n 



2





  n 6.018 10    n 0.3761 eV

 n 2 6.018 10 20 J 20

2

En

or



n 2 1.054  10 34  2  2 n 2 2  2ma 2 2 9.11 10 31 10  10 10

2

1.6 10 19

Then E1  0.376 eV E 2  1.504 eV

E 3  3.385 eV

hc E E  3.385  1.504 1.6 10 19

(b)  



 3.0110

19

J





2

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



6.625 10 3 10  34

8

19

3.0110  6.604 10 7 m or   660.4 nm _______________________________________ 2.27





2

n 2 1.054  10 34  2





2 15  10 3 1.2  10  2



15 10 3  n 2 2.538 10 62





2

or n  7.688 10 (b) E n 1  15 mJ (c) No _______________________________________ 29

2.28 For a neutron and n  1 : E1 

 



2

1.054  10 34  2  2 2  2 2ma 2 1.66  10  27 10 14





2

 3.3025 10 13 J

or

3.3025 10 13  2.06 10 6 eV 1.6 10 19 For an electron in the same potential well: E1 

E1 

2mE 2 Boundary conditions: a a , x  x   0 at x  2 2 First mode solution:  1 x   A1 cos k1 x where   2 2 k1   E1  a 2ma 2 Second mode solution:  2 x   B2 sin k 2 x where 2 4 2  2 k2   E2  a 2ma 2 Third mode solution:  3 x   A3 cos k 3 x where 3 9 2  2 k3   E3  a 2ma 2 Fourth mode solution:  4 x   B4 sin k 4 x where 4 16 2  2 k4   E4  a 2ma 2 _______________________________________ k

 2 n 2 2 (a) E n  2ma 2 15  10 3 

so in this region  2 x  2mE  2  x   0 x 2  The solution is of the form  x   A cos kx  B sin kx where

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

 31

2

14 2

 6.0177 10 10 J

or

6.0177 10 10  3.76 10 9 eV 1.6 10 19 _______________________________________ E1 

2.29 Schrodinger's time-independent wave equation  2 x  2m  2 E  V x  x   0 x 2  We know that a a  x   0 for x  and x  2 2 We have a a x V x   0 for 2 2

2.30 The 3-D time-independent wave equation in cartesian coordinates for V x, y, z   0 is:

 2 x, y, z   2 x, y, z   2 x, y, z    x 2 y 2 z 2 2mE  2   x, y , z   0  Use separation of variables, so let  x, y, z   X x Y  y Z z  Substituting into the wave equation, we obtain

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

2 X  2Y 2Z  XZ  XY x 2 y 2 z 2 2mE  2 XYZ  0  2mE Dividing by XYZ and letting k 2  2 , we  find 1  2 X 1  2Y 1  2 Z (1)      k2 0 X x 2 Y y 2 Z z 2 We may set 1 2 X 2 X  2  k x2   k x2 X  0 X x x 2 Solution is of the form X x   A sin k x x   B cosk x x  Boundary conditions: X 0  0  B  0 n  and X x  a   0  k x  x a where n x  1, 2, 3.... Similarly, let 1 2Z 1  2Y   k z2  2  k y2 and Z z 2 Y y Applying the boundary conditions, we find n y , n y  1, 2, 3.... ky  a n k z  z , n z  1, 2, 3... a From Equation (1) above, we have  k x2  k y2  k z2  k 2  0 YZ

or k x2  k y2  k z2  k 2 

2mE 2

so that

 2 2 2 n x  n 2y  n z2 2ma 2 _______________________________________ E  E nx n y nz 

2.31





 2 x, y   2 x, y  2mE (a)   2  x, y   0 x 2 y 2  Solution is of the form:  x, y   A sin k x x  sin k y y We find  x, y   Ak x cos k x x  sin k y y x  2 x, y    Ak x2 sin k x x  sin k y y x 2

 x, y   Ak y sin k x x  cos k y y y

 2 x, y 

  Ak y2 sin k x x  sin k y y

y Substituting into the original equation, we find: 2mE  k x2  k y2  2  0 (1)  From the boundary conditions, 2

o

A sin k x a  0 , where a  40 A n  So k x  x , n x  1, 2, 3, ... a o

Also A sin k y b  0 , where b  20 A So k y 

n y

, n y  1, 2, 3, ... b Substituting into Eq. (1) above 2 2 n 2y  2   2  n x   E nx n y   2m  a 2 b 2  (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. _______________________________________ 2.32 (a) Derivation of energy levels exactly the same as in the text  2 2 2 (b) E  n 2  n12 2ma 2 For n 2  2, n1  1 Then 3 2  2 E  2ma 2





o

(i) For a  4 A E 







2

3 1.054  10 34  2



2 1.67  10  27 4  10 10



2

 6.155 10 22 J 6.155 10 22 or E   3.85 10 3 eV 1.6 10 19

Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (ii) For a  0.5 cm E 





3 1.054  10

2 1.67  10

 27

 34

 2

2

0.5  10 

2 2

 3.939 10 36 J

or

3.939 10 36  2.46 10 17 eV 1.6 10 19 _______________________________________ E 

2.33 (a) For region II, x  0  2 2 x  2m  2 E  VO  2 x   0 x 2  General form of the solution is  2 x   A2 exp jk 2 x   B2 exp jk 2 x  where

2m E  VO  k2  2 Term with B 2 represents incident wave and term with A2 represents reflected wave. Region I, x  0  2 1 x  2mE  2  1 x   0 x 2  General form of the solution is  1 x   A1 exp jk1 x   B1 exp jk1 x  where 2mE 2 Term involving B1 represents the transmitted wave and the term involving A1 represents reflected wave: but if a particle is transmitted into region I, it will not be reflected so that A1  0 . Then  1 x   B1 exp jk1 x   2 x   A2 exp jk 2 x   B2 exp jk 2 x  (b) Boundary conditions: (1)  1 x  0   2 x  0

Combining these two equations, we find  k  k1    B2 A2   2  k 2  k1 

 2k 2    B2 B1    k 2  k1  The reflection coefficient is

2.34

 2 x   A2 exp k 2 x  P

 x 

2



 exp 2k 2 x 

A2 A2*

2mVo  E 

where k 2 

2







2 9.1110 31 3.5  2.8 1.6 10 19

1.054 10 k 2  4.286 10 9 m 1



34

o

(a) For x  5 A  5 10 10 m P  exp 2k 2 x 

 

k1 

 1  2  (2) x x 0 x x 0 Applying the boundary conditions to the solutions, we find B1  A2  B 2 k 2 A2  k 2 B2  k1 B1

2

 k  k1   R   2 * B 2 B 2  k 2  k 1  The transmission coefficient is 4k 1 k 2 T  1 R  T  k1  k 2 2 _______________________________________ A2 A2*



 exp  2 4.2859 10 9 5 10 10  0.0138



o

(b) For x  15 A  15 10 10 m

 



P  exp  2 4.2859 10 9 15 10 10  2.6110



6

o

(c) For x  40 A  40 10 10 m

 



P  exp  2 4.2859 10 9 40 10 10



15

 1.29  10 _______________________________________

2.35

E T  16  Vo where k 2 





 E 1   V o 

  exp 2k 2 a   

2mVo  E  2





2 9.1110 31 1.0  0.1 1.6 10 19 1.054 10

34



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

k 2  4.860  10 9 m 1

k2 =



10

(a) For a  4 10 m  0.1  0.1  9 10 T  16 1   exp  2 4.85976 10 4 10  1.0  1.0   0.0295 (b) For a  12 10 10 m  0.1  0.1  9 10 T  16 1   exp  2 4.85976 10 12 10 1 . 0 1 . 0   

 





 





 1.24 10 5 (c) J  N t e , where N t is the density of transmitted electrons. E  0.1 eV  1.6 10 20 J 1 1  m 2  9.1110 31  2 2 2 5     1.874 10 m/s 1.874 10 7 cm/s



1.2 10

3



 N t 1.6 10



19

2.36

 E  E 1  T  16  V V O  O  (a) For m  0.067 mo

  exp 2k 2 a   

2mVO  E 

k2 

2







 31 19   20.067  9.1110 0.8  0.2 1.6 10  2  1.054 10 34 







    

or

k 2  1.027 10 9 m 1 Then  0.2  0.2  T  16 1    0.8  0.8 

 

T  0.138 (b) For m  1.08m o





1/ 2

 





or T  1.27 10 5 _______________________________________

2.37

E T  16  Vo where k 2 



 E 1   V o 

  exp 2k 2 a   

2mVo  E  2







2 1.67 10 27 12  110 6  1.6 10 19



34

1.054 10  7.274 1014 m 1 (a) 1  1  T  16 1   exp  2 7.274 1014 10 14  12  12   1.222 exp 14.548  5.875 10 7 (b) T  10 5.875 10 7





 



 





 1.222 exp  2 7.274 1014 a





 1.222  2 7.274 1014 a  ln   6 5 . 875  10   or a  0.842 10 14 m _______________________________________ 2.38



 exp  2 1.027 10 9 15 10 10 or

1/ 2



 

1.874 10 

N t  4.002 10 electrons/cm 3 Density of incident electrons, 4.002 10 8 Ni   1.357 1010 cm 3 0.0295 _______________________________________



 exp  2 4.124 10 9 15 10 10

7

8



 31 19   21.08 9.1110 0.8  0.2 1.6 10  2  1.054 10 34  or k 2  4.124 10 9 m 1 Then  0.2  0.2  T  16 1    0.8  0.8 



Region I x  0 , V  0 ;

Region II 0  x  a  , V  VO

Region III x  a  , V  0 (a) Region I:  1 x   A1 exp jk1 x   B1 exp jk1 x  (incident) (reflected)

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

2mE k1  2 Region II:  2 x   A2 expk 2 x   B2 exp k 2 x  where k2 

2mVO  E  

A1  B1  A2  B2 d  1 d 2   dx dx jk1 A1  jk1 B1  k 2 A2  k 2 B2

At x  a :  2   3 

A2 expk 2 a   B2 exp k 2 a 

 A3 exp jk1 a 

d 2 d 3   dx dx k 2 A2 expk 2 a   k 2 B2 exp k 2 a 

 jk1 A3 exp jk1 a  The transmission coefficient is defined as A A* T  3 3* A1 A1 so from the boundary conditions, we want to solve for A3 in terms of A1 . Solving

for A1 in terms of A3 , we find  jA3 k 22  k12 expk 2 a   exp k 2 a  4k 1 k 2





 2 jk1 k 2 expk 2 a   exp k 2 a  

 exp jk1 a 

We then find

A3 A3*

4k1 k 2 2

k

2 2



 k12 expk 2 a   exp k 2 a 

2

 4k12 k 22 expk 2 a   exp k 2 a 

2

We have k2 

2

Region III:  3 x   A3 exp jk1 x   B3 exp jk1 x  (b) In Region III, the B3 term represents a reflected wave. However, once a particle is transmitted into Region III, there will not be a reflected wave so that B3  0 . (c) Boundary conditions: At x  0 :  1   2 

A1 

A1 A1* 



2mVO  E 

2 If we assume that VO  E , then k 2 a will be large so that expk 2 a   exp k 2 a  We can then write A3 A3* 2 A1 A1*  k 2  k 2 expk 2 a  4k1 k 2 2 2 1





 4k12 k 22 expk 2 a 

2



which becomes A3 A3* A1 A1*  k 22  k12 exp2k 2 a  2 4k1 k 2  Substituting the expressions for k1 and





k 2 , we find k 12  k 22 

2mV O 2

and

 2mVO  E   2mE  k12 k 22    2  2     2

 2m    2  VO  E E     E  2m    2  VO 1  V    O  2

 E   

Then 2

 2mV O  A3 A   exp2k 2 a  2    A1 A1*   2m  2  E   E  16  2  VO 1       VO   * 3



A3 A3*  E 16  VO

 E 1   V O 

  exp 2k 2 a   

Finally,  E  A A* E  1   exp 2k 2 a  T  3 3*  16    A1 A1  VO  VO  _____________________________________

Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 2.39 Region I: V  0  2 1 x  2mE  2  1 x   0  x 2   1 x   A1 exp jk1 x   B1 exp jk1 x  incident reflected where

2mE 2 Region II: V  V1 k1 

 2 2 x 

2mE  V1 

 2 x   0  x 2  2 x   A2 exp jk 2 x   B2 exp jk 2 x  transmitted reflected where 2



2mE  V1 

k2 

 Region III: V  V2 

2 m E  V 2 

x 2  3 x   A3 exp jk 3 x  transmitted where 2

k3 

 3 x   0 

2 m E  V 2 

2 There is no reflected wave in Region III. The transmission coefficient is defined as: T

 k 3 A3 exp jk 3 a 

But k 2 a  2n 

exp jk 2 a   exp jk 2 a   1

Then, eliminating B1 , A2 , B 2 from the boundary condition equations, we find k 4k 1 k 3 4k12 T 3  k1 k1  k 3 2 k1  k 3 2 _______________________________________ 2.40 (a) Region I: Since VO  E , we can write

 2 1 x 

 3 A3 A3* k 3 A3 A3*    1 A1 A1* k1 A1 A1*

From the boundary conditions, solve for A3 in terms of A1 . The boundary conditions are: At x  0 :  1   2  A1  B1  A2  B2  1  2   x x k1 A1  k1 B1  k 2 A2  k 2 B2

At x  a :  2   3  A2 exp jk 2 a   B2 exp jk 2 a 

 A3 exp jk 3 a 

2mVO  E 

 1 x   0 x 2 Region II: V  0 , so  2 2 x  2mE  2  2 x   0 x 2  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   B1 expk1 x   2 x   A2 sin k 2 x   B2 cosk 2 x   3 x   0 where 2

2

 2 3 x 

 2  3   x x k 2 A2 exp jk 2 a   k 2 B2 exp jk 2 a 

k1 



2mVO  E  2

and k 2 

 (b) Boundary conditions At x  0 :  1   2  B1  B2

2mE 2

 1  2   k 1 B1  k 2 A2 x x At x  a :  2   3 

A2 sin k 2 a   B2 cosk 2 a   0

or

B2   A2 tan k 2 a 

(c)

k  k1 B1  k 2 A2  A2   1  B1  k2  and since B1  B2 , then k  A2   1  B2  k2 

Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 2 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ From B2   A2 tan k 2 a  , we can write

  2r    r 2 exp  a o   ao  To find the maximum probability dP r  0 dr   2r  4   2  2   r exp   3    a  a a o   o   o  P

k  B2   1  B2 tank 2 a   k2  or

k  1   1  tan k 2 a   k2  This equation can be written as  2mE  V E 1  O  tan   a 2 E    or  2mE  E   tan   a 2 VO  E    This last equation is valid only for specific values of the total energy E . The energy levels are quantized. _______________________________________

4

3

 

  2r    2r exp   a o 

which gives r 0  1  r  ao ao or r  a o is the radius that gives the greatest probability. _______________________________________ 2.43

 100 is independent of  and  , so the wave

2.41

En   

 mo e 4

(J)

4 o 2 2 2 n 2  mo e 3

4 o 2 2 2 n 2



(eV)

   21.054 10  n

 9.11 10 31 1.6  10 19

4 8.85 10

12

2

3

 34 2

2

equation in spherical coordinates reduces to 1   2   2mo E  V r   0  r  r   2 r 2 r  where  e2  2 V r    4 o r mo a o r For

or 13.58 (eV) n2 n  1  E1  13.58 eV En 

 100

 100 1  r 

n  3  E 3  1.51 eV

2.42 We have

 100

 1      ao 1

   

3/ 2

r   exp   ao 

and * P  4 r 2 100 100

1  1  4 r      ao 2

or

3

   2r   exp    a    o 

   

3/ 2

r   exp   ao 

Then

n  2  E 2  3.395 eV n  4  E 4  0.849 eV _______________________________________

 1      ao 1

 1    ao

   

3/ 2

 1  r    exp  a  a   o  o 

so  100 1  1 r   r   a o We then obtain 2

  2  100   1 r  r  r  

   

5/2

r   r 2 exp   ao 

 1    ao

   

5/2

  r   r2    r      exp   2r exp     a   ao   ao   o   Substituting into the wave equation, we have

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

r2

  r  r2   r    exp 2r exp     ao  ao  a o  2m  2   2o  E   mo a o r   

 1     ao

1

   

5/ 2

 1   1          ao

   

3/ 2

r  0 exp   ao 

where

 mo e 4

 2 4 o 2 2 2 2mo ao2 Then the above equation becomes E  E1 

   r   1  r2    2 r  exp     r 2a ao    a o   o   2m    2  2    0  2o     2m o a o m o a o r 

 1     ao

   

3/ 2

 1     ao

   

3/ 2

1



or

1

   r   exp    a o 

  2 1   1 2    2   2  0  a o r a o  a o a o r  which gives 0 = 0 and shows that  100 is indeed a solution to the wave equation. _______________________________________ 2.44 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, 4th edition Chapter 3 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

Chapter 3 3.1 If a o 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 a o were to decrease, the bandgap energy would increase and the material would begin to behave more like an insulator. _______________________________________ 3.2 Schrodinger's wave equation is:   2  2   x, t   V  x     x, t  2m x 2

 x, t  t Assume the solution is of the form:    E    x, t   u x  exp  j  kx    t         j

Region I: V x   0 . Substituting the assumed solution into the wave equation, we obtain:   2    E    jkux  exp  j  kx   t  2m x        

  u x   E    exp  j  kx   t   x       

    jE   E    j    u x  exp  j  kx   t           which becomes   2   E   2   jk  u x  exp  j  kx   t  2m         2 jk



  u x   E   exp  j  kx   t  x      

   2 u x   E    exp  j  kx   t   2 x       

   E     Eux  exp  j  kx   t        This equation may be written as u x   2 u x  2mE  k 2 u x   2 jk   2 u x   0 x x 2 

Setting u x   u1 x  for region I, the equation becomes: d 2 u1  x  du x   2 jk 1  k 2   2 u1 x   0 2 dx dx where 2mE 2  2 Q.E.D.  In Region II, V x   VO . Assume the same form of the solution:    E    x, t   u x  exp  j  kx    t        Substituting into Schrodinger's wave equation, we find:   2   E   2   jk  u x  exp  j  kx   t  2m       



 2 jk





  u x   E   exp  j  kx   t  x      

   2 u x   E    exp  j  kx   t   2 x           E    VO u x  exp  j  kx   t           E    Eux  exp  j  kx   t       

This equation can be written as: u x   2 u x   k 2 u x   2 jk  x x 2 2mV O 2mE  u x   2 u x   0 2   Setting ux   u 2 x  for region II, this equation becomes d 2 u 2 x  du x   2 jk 2 2 dx dx 2mVO   2   k   2  u 2 x   0 2   where again 2mE 2  2 Q.E.D.  _______________________________________

Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 3 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 3.3 We have d 2 u1  x 

du1 x   k 2   2 u1  x   0 dx dx Assume the solution is of the form: u1 x   A exp j   k x  B exp j   k x The first derivative is du1 x   j   k A exp j   k x  dx  j   k B exp j   k x and the second derivative becomes d 2 u1  x  2   j   k  A exp j   k x 2 dx 2   j   k  B exp j   k x Substituting these equations into the differential equation, we find 2    k  A exp j   k x 2



 2 jk



   k  B exp j   k x  2 jk  j   k A exp j   k x  j   k B exp j   k x 2





 k 2   2 A exp j   k x  B exp j   k x  0 Combining terms, we obtain   2  2k  k 2  2k   k   k 2   2  A exp j   k x















   2  2k  k 2  2k   k   k 2   2  B exp j   k x  0 We find that Q.E.D. 00 For the differential equation in u 2 x  and the proposed solution, the procedure is exactly the same as above. _______________________________________

We have the solutions u1 x   A exp j   k x

 B exp j   k x for 0  x  a and u 2 x   C exp j   k x

 D exp j   k x

for b  x  0 . The first boundary condition is u1 0  u 2 0

   k D  0 The third boundary condition is u1 a   u 2  b  which yields A exp j   k a  B exp j   k a  C exp j   k  b 

 D exp j   k  b  and can be written as A exp j   k a  B exp j   k a  C exp j   k b

 D exp j   k b  0 The fourth boundary condition is du1 du  2 dx x  a dx x  b which yields j   k A exp j   k a   j   k B exp j   k a



3.4

which yields A B C  D  0 The second boundary condition is du1 du  2 dx x  0 dx x  0 which yields   k A    k B    k C

 j   k C exp j   k  b 

 j   k D exp j   k  b  and can be written as   k A exp j  k a    k B exp j   k a

   k C exp j   k b

   k D exp j   k b  0 _______________________________________

3.5

(b) (i) First point: a   Second point: By trial and error, a  1.729 (ii) First point: a  2 Second point: By trial and error, a  2.617 _______________________________________

Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 3 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 3.6

(b) (i) First point: a   Second point: By trial and error, a  1.515 (ii) First point: a  2 Second point: By trial and error, a  2.375 _______________________________________ 3.7 sin a  cos a  cos ka a Let ka  y , a  x Then sin x P  cos x  cos y x d Consider of this function. dy P







d 1 P   x  sin x  cos x   sin y dy We find  dx dx  2 1 P  1x  sin x   x  cos x   dy dy   dx  sin x   sin y dy Then  dx    1 cos x   sin x   sin y P  2 sin x   dy   x x  

For y  ka  n , n  0, 1, 2, ...  sin y  0 So that, in general, d a  d dx 0  dy d ka dk And



2mE 2

3.8 (a)  1 a  

2 m o E1 2

 2 2

E1 

2m o a 2

1 / 2

d 1  2mE   2m  dE   2   2  dk 2       dk This implies that d dE n 0 for k  dk dk a _______________________________________



 2 1.054 10 34 2





2 9.11  10 31 4.2  10 10



2

 3.4114 10 19 J From Problem 3.5  2 a  1.729

2m o E 2

 a  1.729

2 E2 

1.729 2 1.054 10 34 2





2 9.11 10 31 4.2  10 10



2

 1.0198 10 18 J E  E 2  E1  1.0198 10 18  3.4114 10 19  6.7868 10 19 J 6.7868 10 19 or E   4.24 eV 1.6 10 19 (b)  3 a  2

2m o E 3 2 E3 



 a  2

2 2 1.054 10 34 2



2 9.11 10 31 4.2  10 10



2

 1.3646 10 18 J From Problem 3.5,  4 a  2.617

2m o E 4 2 E4 

So

a  

 a  2.617

2.617 2 1.054 10 34 2





2 9.11 10 31 4.2  10 10



2

 2.3364 10 18 J E  E 4  E 3  2.3364 10 18  1.3646 10 18

 9.718 10 19 J 9.718 10 19  6.07 eV or E  1.6 10 19 _______________________________________

Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 3 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 3.9 (a) At ka   ,  1 a  

2 m o E1 2

 E1 

3.10 (a)  1 a  

2 m o E1

a  

2

 2 1.054 10 34 2





2 9.11 10 31 4.2  10 10

E1 



2

 3.4114 10 19 J At ka  0 , By trial and error,  o a  0.859

2

10 2

 31

E2 

 2.5172 10 19 J E  E1  E o  3.4114 10

19

 2.5172 10

19

 8.942 10 20 J 8.942 10 20 or E   0.559 eV 1.6 10 19 (b) At ka  2 ,  3 a  2

2m o E 3 2 E3 

 a  2

1.054 10  29.11 10 4.2  10  2 

34 2

2

 1.3646 10 18 J At ka   . From Problem 3.5,  2 a  1.729

2 E2 



2 9.11 10

 31

 1.0198 10 E  E 3  E 2

 

34 2

10 2

J

 1.3646 10 18  1.0198 10 18 19

2

 a  1.515

1.515 2 1.054 10 34 2





2 9.11 10 31 4.2  10 10



2

 7.830 10 19 J E  E 2  E1  7.830 10 19  3.4114 10 19

 4.4186 10 19 J 4.4186 10 19 or E   2.76 eV 1.6 10 19 (b)  3 a  2

2m o E 3 2 E3 



 a  2

2 2 1.054 10 34 2



2 9.11 10 31 4.2  10 10

2m o E 4

4.2 10

18





2

 1.3646 10 18 J From Problem 3.6,  4 a  2.375

 a  1.729

1.729 2 1.054 10



10 2

 31

2m o E 2



2 9.11 10 31 4.2  10 10

2m o E 2

1.054 10  29.11 10 4.2  10  0.859 

Eo 

 2 1.054 10 34 2

 3.4114 10 19 J From Problem 3.6,  2 a  1.515

34 2

2

a  

 3.4474 10 J 3.4474 10 19 or E   2.15 eV 1.6 10 19 _______________________________________

2 E4 

 a  2.375

2.375 2 1.054 10 34 2





2 9.11 10 31 4.2  10 10



2

 1.9242 10 18 J E  E 4  E 3  1.9242 10 18  1.3646 10 18

 5.597 10 19 J 5.597 10 19 or E   3.50 eV 1.6 10 19 _____________________________________

Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 3 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 3.11 (a) At ka   ,  1 a  

2 m o E1

E g  1.170 

a  

2 E1 

3.12 For T  100 K,



2

Eo 

T  400 K, E g  1.097 eV T  500 K, E g  1.066 eV T  600 K, E g  1.032 eV _______________________________________

 a  0.727

0.727 2 1.054 10 34 2



2 9.11 10

 31

 1.8030 10 E  E1  E o

4.2 10 

19

10 2

 1.6084 10 19 J 1.6084 10 19 or E   1.005 eV 1.6 10 19 (b) At ka  2 ,  3 a  2 2

 E3 



2 2 1.054 10 34 2





2

 1.3646 10 18 J At ka   , From Problem 3.6,  2 a  1.515

2 E2 

 a  1.515

1.515 2 1.054 10



2 9.11 10

 7.830 10 E  E 3  E 2

 34

19

1

 1 d 2E  m   2  2    dk  We have 2 d 2E curve A  d E2 curve B  2 dk dk so that m * curve A  m * curve B  _______________________________________

3.14 The effective mass for a hole is given by

 a  2

2 9.11 10 31 4.2  10 10

2m o E 2

3.13 The effective mass is given by *

J

 3.4114 10 19  1.8030 10 19

2m o E 3



T  300 K, E g  1.125 eV

 3.4114 10 19 J At ka  0 , By trial and error,  o a  0.727

2

2

T  200 K, E g  1.147 eV



2 9.11 10 31 4.2  10 10

2m o E o

4

636  100 E g  1.164 eV

 2 1.054 10 34 2



4.73 10 100

 1 d 2E m *p   2   dk 2  We have that

   

1

d 2E d 2E   curve B curve A  dk 2 dk 2 so that m *p curve A  m *p curve B  _______________________________________



34 2

4.2 10 

10 2

J

 1.3646 10 18  7.830 10 19

 5.816 10 19 J 5.816 10 19  3.635 eV or E  1.6 10 19 _______________________________________

3.15 dE  0  velocity in -x direction dk dE  0  velocity in +x direction Points C,D: dk

Points A,B:

d 2E 0 dk 2 negative effective mass d 2E 0 Points B,C: dk 2 positive effective mass _______________________________________ Points A,D:

Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 3 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 3.16 For A: E  C i k 2

m 

At k  0.08 10 10 m 1 , E  0.05 eV Or E  0.05 1.6 10 19  8 10 21 J So 8 10 21

10 2

1



 

1.054 10 34 2 Now m   2C1 2 1.25 10 38 



2

 4.44 10 kg 4.4437 10 31 or m    mo 9.1110 31 m   0.488 mo



For B: E  C i k 2 At k  0.08 10 10 m 1 , E  0.5 eV Or E  0.5 1.6 10 19  8 10 20 J





So 8 10 20  C1 0.08 1010

Now m  

2



 

1.054 10 34 2  2C1 2 1.25 10 37



2

32

 4.44 10 kg 4.4437 10 32 or m    mo 9.1110 31 m   0.0488 mo _______________________________________

3.17 For A: E  E  C 2 k 2







  2  1.054 10 34 m   2C 2 2 6.25 10 39





2



 0.3 1.6 10 19  C 2 0.08 1010  C 2  7.5 10 38

2



. 2 _______________________________________ E  E O  E1 cos k  k O  Then dE   E1    sin  k  k O  dk   E1 sin  k  k O  and d 2E  E1 2 cos k  k O  2 dk

 

 8.8873 10 31 kg  8.8873 10 31 or m    mo 9.1110 31 m     0.976 mo



3.19 (c) Curve A: Effective mass is a constant Curve B: Effective mass is positive around k  0 , and is negative

3.20

 C 2  6.25 10 39

For B: E  E  C 2 k 2





 2.705 1014 Hz c 3 1010 (ii)     2.705 1014  1.109 10 4 cm  1109 nm _______________________________________

around k  

 0.025 1.6 10 19  C 2 0.08 1010









 C1  1.25 10 37



2

3.18 (a) (i) E  h E 1.42 1.6 10 19 or    h 6.625 10 34  3.429 1014 Hz hc c 3 1010 (ii)     E  3.429 1014  8.75 10 5 cm  875 nm E 1.12 1.6 10 19 (b) (i)    h 6.625 10 34

31



 

 7.406 10 32 kg  7.406 10 32 or m    mo 9.1110 31 m   0.0813 mo _______________________________________

   C 0.08 10 

 C1  1.25 10 38



  2  1.054 10 34  2C 2 2 7.5 10 38



2

Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 3 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Then 2

1 1 d E  2 2 * m  dk



E1 2

k  ko

or

2 E1 2 _______________________________________ m* 

3.21

  0.082m  1.64m 

 (a) mdn  4 2 / 3 mt  ml 2

 42 / 3

1/ 3 2

1/ 3

o

o

 m dn  0.56mo

(b)

3 2 1 2 1      mcn mt ml 0.082mo 1.64mo 

24.39 0.6098  mo mo

 mcn  0.12mo _______________________________________

3.22

  m    0.45m   0.082m  

  m hh  (a) m dp

3/ 2 2/3

3/ 2

lh

3/ 2 2/ 3

3/ 2

o

 0.30187  0.02348

2/3

m

 dp

o

 mo

 0.473m o

mhh 3 / 2  mlh 3 / 2 mhh 1 / 2  mlh 1 / 2 0.453 / 2  0.0823 / 2  m  0.451 / 2  0.0821 / 2 o

 (b) mcp 

 m cp  0.34m o

_______________________________________ 3.23 For the 3-dimensional infinite potential well, V x   0 when 0  x  a , 0  y  a , and 0  z  a . In this region, the wave equation is:  2 x, y, z   2 x, y, z   2 x, y, z    x 2 y 2 z 2 2mE   x, y , z   0 2 Use separation of variables technique, so let  x, y, z   X x Y  y Z z  Substituting into the wave equation, we have 

2 X  2Y 2Z  XZ  XY x 2 y 2 z 2 2mE  2  XYZ  0  Dividing by XYZ , we obtain 1  2 X 1  2 Y 1  2 Z 2mE       2 0 X x 2 Y y 2 Z z 2  Let 1 2 X 2 X  2  k x2   k x2 X  0 X x x 2 The solution is of the form: X x   A sin k x x  B cos k x x Since  x, y, z   0 at x  0 , then X 0   0 so that B  0 . Also,  x, y, z   0 at x  a , so that X a   0 . Then k x a  n x  where n x  1, 2, 3, ... Similarly, we have 1 2Z 1  2Y   k z2  2  k y2 and Z z 2 Y y From the boundary conditions, we find k y a  n y  and k z a  n z  YZ

2

where n y  1, 2, 3, ... and n z  1, 2, 3, ... From the wave equation, we can write 2mE  k x2  k y2  k z2  2  0  The energy can be written as

2 2   n x  n 2y  n z2   2m a _______________________________________ E  E nx n y nz 





2

3.24 The total number of quantum states in the 3-dimensional potential well is given (in k-space) by  k 2 dk 3 g T k dk  a 3



where 2mE 2 We can then write k2 

2mE  Taking the differential, we obtain k

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

1 1 1 1 m  2m    dE    dE  2 E  2E Substituting these expressions into the density of states function, we have  a 3  2mE  1 m g T E dE  3  2     dE      2E Noting that h  2 this density of states function can be simplified and written as 4 a 3 2m3 / 2  E  dE g T E dE  h3 Dividing by a 3 will yield the density of states so that dk 

So g E  





g E  

 1 1 2m n    dE  2 E



4 2m n

g c E  



4 2m n

gc 

2m n 2a 1    dE  2 E Divide by the "volume" a, so g E  

2m n 1   E



3/ 2

E  Ec



3 / 2 Ec  2 kT



h3

  

h3



4 2m n



h

h





3/ 2

2 3/ 2   E  E c  3



 3/ 2 n 3

4 2m

E  E c  dE

Ec

3



Ec  2 kT Ec

2 3/ 2   2kT  3

4 21.08 9.11 10 31

6.625 10   7.953 10 2kT 



3/ 2

2 3/ 2   2kT  3

 34 3

3/ 2

55

(i) At T  300 K, kT  0.0259 eV



 0.0259 1.6 10 19

 4.144 10



Then g c  7.953 10

55

21



J

24.144 10 

21 3 / 2

 6.0  10 25 m 3

g c  6.0 1019 cm 3

or

 400  (ii) At T  400 K, kT  0.0259    300   0.034533 eV



 0.034533 1.6 10 19



 5.5253 10 21 J

Then g T E dE 

m 3 J 1

3.26 (a) Silicon, mn  1.08mo

3/ 2

dk 

1.055 1018

E _______________________________________

4 2m  g E    E h3 _______________________________________

3.25 For a one-dimensional infinite potential well, 2m n E n 2  2   k2 2 a2 Distance between quantum states      k n 1  k n  n  1   n    a   a a Now 2  dk g T k dk      a Now 1 k   2m n E 



20.067  9.11 10 31 1  1.054  10 34   E

Then

 



g c  7.953 10 55 2 5.5253 10 21



3/ 2

 9.239 10 25 m 3

or

g c  9.24 1019 cm 3

(b) GaAs, m n  0.067 mo gc 





4 20.067  9.11 10 31

6.625 10   1.2288 10 2kT 

 34 3

54

3/ 2



3/ 2

2 3/ 2   2kT  3



Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 3 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (i) At T  300 K, kT  4.144 10 21 J

 



g c  1.2288 10 54 2 4.144 10 21



(i)At T  300 K, kT  4.144 10 21 J

 



3/ 2

g   2.3564 10 55 3 4.144 10  21

 9.272 10 23 m 3

 3.266 10 25 m 3

or g c  9.27 1017 cm 3

or g   3.27 1019 cm 3

(ii) At T  400 K, kT  5.5253 10 21 J

 



g c  1.2288 10 54 2 5.5253 10 21





3/ 2

(ii)At T  400 K, kT  5.5253 10 21 J

 



3/ 2

g  2.3564 10 55 3 5.5253 10 21

 1.427 10 24 m 3



3/ 2

 5.029 10 25 m 3

g c  1.43 1018 cm 3 _______________________________________

or g   5.03 1019 cm 3 _______________________________________

3.27 (a) Silicon, m p  0.56m o

3.28

g  E   g 

 





4 2m p





3/ 2

h3



4 2m p h3

4 2m h



3/ 2

E  E

h3

4 2m p









E  E  dE

3/ 2



 3/ 2 p

3



2 3/ 2   E   E   3 



2 3/ 2    3kT   3 

4 20.56  9.11 10 31

6.625 10   2.969 10 3kT 



3/ 2

 34 3

E

(i)At T  300 K, kT  4.144 10

 



g  2.969 10 55 3 4.144 10 21

E  E c  0.3 eV;

 2.614 10 46 m 3 J 1

E  E c  0.4 eV;

 3.018 10 46 m 3 J 1

(b) g 

(b) GaAs, m p  0.48m o



4 20.48 9.11 10 31



h3

6.625 10   2.3564 10 3kT 



3/ 2

E  E



4 20.56  9.11 10 31

6.625 10 



3/ 2

 34 3

E  E

g  0



3/ 2

E  E  0.2 eV;

 7.968 10 45 m 3 J 1

E  E  0.3 eV;

 9.758 10 45 m 3 J 1

 1.127 10 46 m 3 J 1 E  E  0.4 eV; _______________________________________



 34 3

55



4 2m p

E  E  0.1 eV; g   5.634 10 45 m 3 J 1

or g   6.34 1019 cm 3



gc  0

For E  E ;

 

E  Ec

 34 3

 4.454110 55 E  E

 6.337 10 25 m 3

g 

6.625 10 

3/ 2

or g  4.12 1019 cm 3

g  2.969 10 55 3 5.5253 10 21

3/ 2

 2.134 10 46 m 3 J 1



(ii)At T  400 K, kT  5.5253 10 21 J





E  E c  0.2 eV;



 4.116 10 25 m 3





E  E c  0.1 eV; g c  1.509 10 46 m 3 J 1



J

E  Ec

4 21.08 9.11 10 31

For E  E c ;

E  3 kT

2 3/ 2  3kT  3 21

3/ 2

 1.1929 10 56 E  E c

3/ 2

55



h3

E

E  3 kT





4 2m n

(a) g c E  

3/ 2

2 3/ 2  3kT  3  

3.29

  m  m

(a)

gc m  n g m p

(b)

gc g

3/ 2

   

3/ 2 3/ 2

 3/ 2 n  3/ 2 p

 1.08     0.56 

3/ 2

 0.067     0.48 

 2.68 3/ 2

 0.0521

_______________________________________

Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 3 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 3.30 Plot _______________________________________ 3.31 (a) Wi 

(b)

gi! 10!  N i ! g i  N i ! 7!10  7 !

10987!  1098  120  7!3! 321 121110! 12! (i) Wi   10!12  10! 10!21  66

12111098! 12! (ii) Wi   8!12  8! 8!4321  495 _______________________________________ 3.32 f E  

1  E  EF  1  exp   kT 

(a) E  E F  kT , f E   f E   0.269

1  1  exp1

(b) E  E F  5kT , f E  

f E   6.69 10

1  1  exp5

3

(c) E  E F  10kT , f E  

1  1  exp10 

f E   4.54 10 5 _______________________________________

3.34

  E  E F   f F  exp   kT     0.30  E  E c ; f F  exp   9.32 10 6   0.0259 

(a)

Ec 

kT   0.30  0.0259 2  ; f F  exp   2 0.0259  

 5.66 10 6   0.30  0.0259 E c  kT ; f F  exp   0.0259    3.43 10 6 3kT   0.30  30.0259 2 Ec  ; f F  exp   2 0.0259    2.08 10 6   0.30  20.0259   E c  2kT ; f F  exp   0.0259    1.26 10 6 1 (b) 1  f F  1   E  EF  1  exp    kT    E F  E   exp   kT  

  0.25  5 E  E ; 1  f F  exp    6.43 10 0 . 0259   kT   0.25  0.0259 2 E  ; 1  f F  exp   2 0.0259    3.90 10 5   0.25  0.0259 E  kT ; 1  f F  exp   0.0259  

3.33 1 1  f E   1   E  EF 1  exp  kT or 1 1  f E   E E 1  exp F   kT 

 2.36 10 5   

(a) E F  E  kT , 1  f E   0.269

(b) E F  E  5kT , 1  f E   6.69 10 3

(c) E F  E  10kT , 1  f E   4.54 10 5 _______________________________________

E 

3kT ; 2

  0.25  30.0259 2  1  f F  exp   0.0259    1.43 10 5

E  2kT ;

  0.25  20.0259 1  f F  exp   0.0259  

 8.70 10 6 _______________________________________

Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 3 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 3.35

  E c  kT  E F     E  E F   f F  exp   exp    kT kT     and   E F  E  1  f F  exp   kT     E F  E  kT    exp   kT     E c  kT  E F  So exp   kT  

  E F  E  kT   exp   kT   Then E c  kT  E F  E F  E  kT E  E  E midgap Or E F  c 2 _______________________________________ 3.36

 2 n 2 2 2ma 2 For n  6 , Filled state En 

E6 

1.054 10  6   29.11 10 12  10  34 2

2

2

10 2

 31

 1.5044 10 18 J 1.5044 10 18 or E 6   9.40 eV 1.6 10 19 For n  7 , Empty state

E7 

1.054 10  7   29.11 10 12  10  34 2

2

2

10 2

 31

 2.048 10 18 J 2.048 10 18 or E 7   12.8 eV 1.6 10 19 Therefore 9.40  E F  12.8 eV _______________________________________





E5 



E13 





2

2

2

2

2

10 2

 31

1.054 10    3  2  3  29.11 10 12  10  34 2

 31

2

2

2

2

10 2

 9.194 10 19 J 9.194 10 19 or E13   5.746 eV 1.6 10 19 The 14th electron would occupy the quantum state n x  2, n y  3, n z  3 . This state is at the same energy, so E F  5.746 eV _______________________________________

3.38 The probability of a state at being occupied is 1 f 1  E1    E1  E F 1  exp  kT

E1  E F  E

1    E   1  exp   kT   The probability of a state at E 2  E F  E being empty is 1 1  f 2 E 2   1   E  EF  1  exp 2   kT 

  E  exp  1  kT   1    E    E  1  exp  1  exp   kT   kT 

2

2 2   n x  n 2y  n z2   2m a

34 2

 3.76110 19 J 3.76110 19 or E 5   2.35 eV 1.6 10 19 For the next quantum state, which is empty, the quantum state is n x  1, n y  2, n z  2 . This quantum state is at the same energy, so E F  2.35 eV (b) For 13 electrons, the 13th electron occupies the quantum state n x  3, n y  2, n z  3 ; so

3.37 (a) For a 3-D infinite potential well

2mE    n x2  n 2y  n z2   2 a th For 5 electrons, the 5 electron occupies the quantum state n x  2, n y  2, n z  1 ; so

1.054 10    2  2  1  29.11 10 12  10 

or

1  f 2 E 2  

1  E  1  exp   kT  so f 1 E1   1  f 2 E 2  Q.E.D. _______________________________________

Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 3 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 3.39 (a) At energy E1 , we want 1 1   E  EF   E  EF exp 1  1  exp 1  kT   kT 1  E  EF  1  exp 1   kT  This expression can be written as  E  EF  1  exp 1   kT   1  0.01  E1  E F  exp   kT  or  E  EF  1  0.01 exp 1   kT  Then E1  E F  kT ln 100 or E1  E F  4.6kT (b) At E  E F  4.6kT ,

  

 0.01

f  E1  

1 1   E  E F  1  exp4.6  1  exp 1   kT  which yields f E1   0.00990  0.01 _______________________________________

3.40 (a)

  E  E F     5.80  5.50 f F  exp    exp   kT 0.0259      9.32 10

6

 700  (b) kT  0.0259   0.060433 eV  300    0.30  3 f F  exp    6.98 10 0 . 060433     E F  E  (c) 1  f F  exp   kT     0.25  0.02  exp    kT 

1   0.25  or exp    50   kT  0.02 0.25  ln 50  kT or 0.25  T  kT   0.063906  0.0259    ln 50  300  which yields T  740 K _______________________________________ 3.41 (a)

f E  

1  0.00304  7.15  7.0  1  exp   0.0259  or 0.304% (b) At T  1000 K, kT  0.08633 eV Then 1 f E    0.1496  7.15  7.0  1  exp   0.08633  or 14.96% 1  0.997 (c) f E   6  .85  7.0  1  exp   0.0259  or 99.7% (d) 1 At E  E F , f E   for all temperatures 2 _______________________________________ 3.42 (a) For E  E1

f E  

1  E  EF 1  exp 1  kT

  E1  E F    exp   kT     

Then

  0.30  6 f E1   exp   9.32 10  0.0259  For E  E 2 , E F  E 2  1.12  0.30  0.82 eV Then 1 1  f E   1    0.82  1  exp   0.0259  or

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

  E F  E 2  1  f E   exp   kT  

   0.82  1  f E   1  1  exp   0.0259     0.82  14  exp   1.78 10  0.0259  (b) For E F  E 2  0.4 eV, E1  E F  0.72 eV At E  E1 ,

  E1  E F    0.72  f E   exp    exp 0.0259  kT    

or

f E   8.45 10 13 At E  E 2 ,

  0.4   exp   0.0259 

or 1  f E   1.96 10 7 _______________________________________ 3.44   E  EF f E   1  exp  kT 

 df E   E  E F    11  exp  dE  kT    E  EF   1     exp   kT   kT  2

  0.4   exp   0.0259  or

1  f E   1.96 10 7 _______________________________________ 3.43 (a) At E  E1

  E1  E F    0.30  f E   exp   exp   kT  0.0259   

or

f E   9.32 10 6 At E  E 2 , E F  E 2  1.42  0.3  1.12 eV So   E F  E 2  1  f E   exp   kT  

  1.12   exp   0.0259  or

1  f E   1.66 10 19 (b) For E F  E 2  0.4 , E1  E F  1.02 eV

  E1  E F    1.02  f E   exp    exp 0.0259  kT    

or

f E   7.88 10 18 At E  E 2 ,

1

so

  E F  E 2  1  f E   exp   kT  

At E  E1 ,

  

or

 E  EF   1     exp df E   kT   kT   2 dE   E  E F  1  exp   kT   (a) At T  0 K, For E  E F  exp    0 

df 0 dE df E  E F  exp      0 dE df   At E  E F  dE (b) At T  300 K, kT  0.0259 eV df 0 For E  E F , dE df 0 For E  E F , dE At E  E F ,

 1   1 df  0.0259    9.65 (eV) 1 dE 1  12

Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 3 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (c) At T  500 K, kT  0.04317 eV df 0 For E  E F , dE df 0 For E  E F , dE At E  E F ,

 1   1 df  0.04317    5.79 (eV) 1 2 dE 1  1 _______________________________________ 3.45 (a) At E  Emidgap,

f E  

1 1   E  EF   Eg  1  exp   1  exp  kT    2kT 

Si: E g  1.12 eV,

f E  

or

1  1.12  1  exp    20.0259 

f E   4.07 10 10

1  0.66  1  exp    20.0259 

or

f E   2.93 10 6 GaAs: E g  1.42 eV f E  

or

(a)

or

  E  E F   f F  exp   kT     0.60  10 8  exp    kT 





0.60  ln 10  8 kT 0.60 kT   0.032572 eV ln 10 8

 

 T  0.032572  0.0259   300  so T  377 K   0.60  (b) 10  6  exp    kT 





0.60  ln 10  6 kT 0.60 kT   0.043429 ln 10 6

 

 T  0.043429  0.0259   300  or T  503 K _______________________________________ 3.47 (a) At T  200 K,

Ge: E g  0.66 eV

f E  

3.46

1  1.42  1  exp    20.0259 

f E   1.24 10 12 (b) Using the results of Problem 3.38, the answers to part (b) are exactly the same as those given in part (a). _______________________________________

 200  kT  0.0259   0.017267 eV  300  1 f F  0.05   E  EF  1  exp   kT  1  E  EF  exp  1  19   kT 0 . 05   E  E F  kT ln 19  0.017267  ln 19  0.05084 eV By symmetry, for f F  0.95 , E  E F  0.05084 eV Then E  20.05084  0.1017 eV (b) T  400 K, kT  0.034533 eV For f F  0.05 , from part (a), E  E F  kT ln 19  0.034533 ln 19  0.10168 eV Then E  20.10168  0.2034 eV _______________________________________

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

Chapter 4 (b)

4.1

  Eg ni2  N c N  exp  kT

   

   2.5 10   1.12300   T   2.912 10   exp  0.0259T   300

ni2  5 1012

2

25

3

  Eg  T   N cO N O   exp  300   kT 3

38

   

    By trial and error, T  417.5 K _______________________________________

where N cO and N O are the values at 300 K.

4.4

T (K)

kT (eV)

(a) Silicon ni (cm 3 )

200

0.01727

7.68 10 4

400

0.03453

2.38 1012

600

0.0518

9.74 1014

T (K)

(b) Germanium ni (cm 3 )

200

2.16 10

400 600

 200  At T  200 K, kT  0.0259    300   0.017267 eV  400  At T  400 K, kT  0.0259    300   0.034533 eV

10

7.70 10  200  1.40 10 

n i2 400 

(c) GaAs ni (cm 3 )

n

2 i

8.60 10

3.28 10

3.82 10

16

5.72 1012

2 2

 400     300 

9

_______________________________________ 4.2 Plot _______________________________________

 3.025  10 17

  Eg  exp    0.034533    3   Eg   200    exp    300   0.017267  Eg  Eg   8 exp     0.017267 0.034533 

1.38

14

10 2



3





3.025  1017  8 exp E g 57.9139  28.9578

or

4.3

  Eg (a) n  N c N  exp  kT 2 i

   

T  5 10   2.8 10 1.04 10  300  11 2

19

19





   1.12  exp    0.0259T 300 



38



Now

7.70 10 

10 2

 400   N co N o    300 

3

  1.318   exp   0.034533 

3

 T  2.5 10  2.912 10    300    1.12300   exp    0.0259T   By trial and error, T  367.5 K 23

3

 3.025 1017    38.1714 E g 28.9561  ln   8   or E g  1.318 eV



5.929 10 21  N co N o 2.370 2.658 10 17 6



so N co N o  9.4110 cm _______________________________________ 37

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

4.5

  1.10  exp  n i B    0.20   kT    exp  n i  A   0.90   kT  exp   kT  For T  200 K, kT  0.017267 eV For T  300 K, kT  0.0259 eV For T  400 K, kT  0.034533 eV (a) For T  200 K, n i B    0.20  6  exp   9.325 10 ni  A 0 . 017267   (b) For T  300 K, n i B    0.20  4  exp   4.43 10 ni  A  0.0259  (c) For T  400 K, ni B    0.20  3  exp   3.05 10 ni  A  0.034533  _______________________________________ 4.6

  E  E F  (a) g c f F  E  E c exp   kT     E  E c   E  E c exp   kT  

  E F  E    exp   kT  

Let E  E  x

x Then g  1  f F   x exp   kT  To find the maximum value d g  1  f F  d    x     0  x exp dx dx   kT  Same as part (a). Maximum occurs at kT x 2 or kT E  E  2 _______________________________________ 4.7

n  E1   nE 2 

  E c  E F    exp   kT  



E1  E c  4kT and E 2  E c 

Then nE1   nE 2 

kT 2

  E1  E 2   exp   kT kT   2

4kT

  1   2 2 exp   4    2 2 exp 3.5 2   

1 1/ 2 x  x exp 0 kT  kT 

which yields 1 x1 / 2 kT  x 1/ 2 kT 2 2x The maximum value occurs at kT E  Ec  2

   E1  E c   E1  E c exp   kT     E 2  E c   E 2  E c exp   kT  

where

Let E  E c  x

x Then g c f F  x exp   kT  To find the maximum value: d g c f F  1 1 / 2 x  x exp  dx 2  kT 

  E F  E   g  1  f F   E  E exp   kT     E  E   E  E exp   kT  

or

n  E1   0.0854 n E 2  _______________________________________

4.8 Plot _______________________________________

Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 4.9 Plot _______________________________________

4.13 Let g c E   K  constant Then 

no 

4.10 E Fi  E midgap

 m *p 3  kT ln  *  mn 4 

   

* n

E Fi  Emidgap  0.0128 eV

Gallium Arsenide: m *p  0.48m o ,

m n*  0.067 mo E Fi  Emidgap  0.0382 eV _______________________________________ 4.11



T (K)

   

19   1 kT  ln  1.04 1019   0.4952kT  2  2.8 10 

kT (eV)

( E Fi  E midgap)(eV)  0.0086

200

0.01727

400

0.03453

 0.0171

600

0.0518

 0.0257

_______________________________________

1 E  EF  Ec 1  exp  kT 

 m *p 3  kT ln  *  mn 4 

   

3  0.70   0.0259 ln   4  1.21   10.63 meV 3  0.75  (b) E Fi  E midgap  0.0259 ln   4  0.080   43.47 meV _______________________________________

  

dE



Let E  Ec so that dE  kT  d kT We can write E  E F   E c  E F   E  E c  so that   E c  E F     E  E F   exp    exp     exp  kT kT     The integral can then be written as



  E c  E F   n o  K  kT  exp   exp  d kT  0 which becomes   E c  E F   no  K  kT  exp   kT   _______________________________________ 



4.14 Let g c E   C1 E  E c  for E  E c Then 

4.12 (a) E Fi  E midgap





E Fi  Emidgap  0.0077 eV

N 1 kT  ln   2  Nc

E  f F E dE

  E  E F    K exp  dE kT   Ec

Germanium: m *p  0.37 m o , m n*  0.55mo

E Fi  E midgap 

c



K

Silicon: m  0.56m o , m  1.08mo * p

g

Ec

no 

g

c

E  f F E dE

Ec

E  E c 



 C1



Ec

 E  EF 1  exp  kT 

 C1

 E  E

Ec

C

  

dE

 E  E F    dE kT  

 exp 

Let E  Ec so that dE  kT  d kT We can write E  E F  E  E c   E c  E F 



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

The ionization energy is

  E c  E F   n o  C1 exp   kT   



c

Ec

or

  E  E c    dE kT 

 E  E  exp

  E c  E F   n o  C1 exp   kT  

 kT  exp kT d 0

We find that 



 o     s

2

 0.067  13.6   13.6  13.12 

or

E  0.0053 eV _______________________________________ 4.17





 m* E    mo



 exp  d  exp     1  1 0

0

So

  E c  E F   2 no  C1 kT  exp   kT   _______________________________________ 4.15

r m  We have 1 r  o*  ao m  For germanium, r  16 , m *  0.55mo Then  1  r1  16 a o  290.53  0.55  or o

r1  15.4 A The ionization energy can be written as 2

 m *  o    13.6  eV E      m o  s  0.55 13.6  E  0.029 eV  162 _______________________________________

4.16

r1 m  r  o*  ao m  For gallium arsenide, r  13.1 , We have

m  0.067 mo *

Then o  1  r1  13.1 0.53  104 A  0.067 

N (a) E c  E F  kT ln  c  no

     2.8 1019    0.0259 ln  15   7 10   0.2148 eV (b) E F  E  E g  Ec  E F   1.12  0.2148  0.90518 eV   E F  E  (c) p o  N  exp   kT  





  0.90518   1.04 1019 exp    0.0259   6.90  10 3 cm 3 (d) Holes n  (e) E F  E Fi  kT ln  o   ni   7 1015    0.0259 ln  10   1.5 10   0.338 eV _______________________________________

4.18

N (a) E F  E  kT ln    po

     1.04 1019    0.0259 ln  16   2 10   0.162 eV (b) Ec  E F  E g  E F  E   1.12  0.162  0.958 eV   0.958  (c) n o  2.8 1019 exp   0.0259 





 2.4110 3 cm 3

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

p  (d) E Fi  E F  kT ln  o   ni   2 1016    0.0259 ln  10   1.5 10   0.365 eV _______________________________________ 4.19

N (a) E c  E F  kT ln  c  no

     2.8 1019    0.0259 ln  5   2 10   0.8436 eV E F  E  E g  Ec  E F   1.12  0.8436 E F  E  0.2764 eV





  0.27637  (b) p o  1.04 1019 exp   0.0259   2.414 1014 cm 3 (c) p-type _______________________________________

4.20

 375  (a) kT  0.0259   0.032375 eV  300 



no  4.7 10

17



 375     300 

3/ 2

  0.28  exp    0.032375 

 1.15 1014 cm 3 E F  E  E g  Ec  E F   1.42  0.28

 1.14 eV





 375  p o  7 1018    300 

3/ 2

  1.14  exp    0.032375 

 4.99  10 3 cm 3

 4.7 1017   (b) E c  E F  0.0259 ln  14   1.15 10   0.2154 eV E F  E  E g  Ec  E F   1.42  0.2154  1.2046 eV



p o  7 1018



  1.2046  exp    0.0259 

 4.42 10 2 cm 3 _______________________________________

4.21

 375  (a) kT  0.0259   0.032375 eV  300 





 375  no  2.8 1019    300 

3/ 2

  0.28  exp    0.032375 

 6.86 1015 cm 3 E F  E  E g  Ec  E F   1.12  0.28

 0.840 eV





 375  p o  1.04 1019    300 

3/ 2

  0.840  exp    0.032375 

 7.84  10 7 cm 3

N (b) E c  E F  kT ln  c  no

     2.8 1019    0.0259 ln  15   6.862 10   0.2153 eV E F  E  1.12  0.2153  0.9047 eV





  0.904668  p o  1.04 1019 exp    0.0259   7.04  10 3 cm 3 _______________________________________

4.22 (a) p-type Eg

1.12  0.28 eV 4 4   E F  E  p o  N  exp   kT  

(b) E F  E 







  0.28   1.04 1019 exp    0.0259   2.10 1014 cm 3 Ec  E F  E g  E F  E 

 1.12  0.28  0.84 eV   E c  E F   n o  N c exp   kT  





  0.84   2.8 1019 exp    0.0259   2.30  10 5 cm 3 _______________________________________

Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 4.23

4.25

 E  E Fi  (a) n o  ni exp  F   kT   0.22   1.5 1010 exp    0.0259 



 400  kT  0.0259   0.034533 eV  300 





 7.33 1013 cm 3  E  EF  p o  ni exp  Fi   kT    0.22   1.5 1010 exp    0.0259 







 8.80  10 9 cm 3  E  EF  p o  ni exp  Fi   kT    0.22   1.8 10 6 exp    0.0259 





 3.68  10 2 cm 3 _______________________________________

4.24

N (a) E F  E  kT ln    po

     1.04 1019    0.0259 ln  15   5 10   0.1979 eV (b) Ec  E F  E g  E F  E   1.12  0.19788  0.92212 eV   0.92212  (c) n o  2.8 1019 exp    0.0259 





 9.66  10 3 cm 3 (d) Holes p (e) E Fi  E F  kT ln  o  ni

     5 1015    0.0259 ln  10   1.5 10   0.3294 eV _______________________________________

3/ 2

 4.3109 1019 cm 3





ni2  4.3109 1019 1.6011019

 3.07  10 cm  E  E Fi  (b) n o  ni exp  F   kT 





 400  N c  2.8 1019    300 

3

 0.22   1.8 10 6 exp    0.0259 

3/ 2

 1.6011019 cm 3



6



 400  N   1.04 1019    300 



  1.12   exp    0.034533   5.6702 10 24

 ni  2.3811012 cm 3

N  (a) E F  E  kT ln     po   1.6011019    0.034533 ln  15   5 10   0.2787 eV (b) E c  E F  1.12  0.27873  0.84127 eV





  0.84127  (c) no  4.3109 1019 exp    0.034533   1.134 10 9 cm 3 (d) Holes p  (e) E Fi  E F  kT ln  o   ni 

 5 1015    0.034533 ln  12   2.38110   0.2642 eV _______________________________________ 4.26 (a)





  0.25  p o  7 1018 exp    0.0259   4.50 1014 cm 3 E c  E F  1.42  0.25  1.17 eV





  1.17  n o  4.7 1017 exp    0.0259   1.13 10 2 cm 3

Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (b) kT  0.034533 eV





 400  N   7 10    300  18



 8.49  10 9 cm 3 _______________________________________

 1.078 1019 cm 3





 400  N c  4.7 1017    300   7.236 10 cm 17

3/ 2

4.28 (a) n o 

3

N E F  E  kT ln    po

no 







4.27



p o  1.04 10







po 



no  7.23 10 cm (b) kT  0.034533 eV



N   1.04 1019

 400     300 



 400     300 

4.7 10 1.0 17

2



N  F1 / 2  F  2



1.04 10 F    19

1/ 2

F

So F1 / 2  F   4.26 E  E F kT E  E F  3.00.0259  0.0777 eV _______________________________________

We find  F  3.0 

3/ 2

 1.6011019 cm 3 19



5  10 19 

3



2

4.29

  0.870  n o  2.8 1019 exp    0.0259  4

19

 5.30 1017 cm 3 _______________________________________

  0.25  exp    0.0259 

 6.68 1014 cm 3 E c  E F  1.12  0.25  0.870 eV



2.8 10 1.0

 3.16 1019 cm 3 2 N c F1 / 2  F  (b) n o 

 2.40  10 cm _____________________________________



2

3

19

N c F1 / 2  F 

E F  E c kT 2   0 .5 kT kT Then F1 / 2  F   1.0

17

N c  2.8 10



F 

  1.07177   7.236 10 exp  0.034533  4

(a)

2

For E F  E c  kT 2 ,

   

 1.078 1019    0.034533 ln  14   4.50 10   0.3482 eV E c  E F  1.42  0.3482  1.072 eV no



  0.77175  no  4.3111019 exp    0.034533 

3/ 2

3/ 2

4.30

E F  E c 4kT  4 kT kT Then F1 / 2  F   6.0

(a)  F 

 4.3111019 cm 3 N  E F  E  kT ln     po 

 1.60110  0.034533 ln  14  6.68 10  0.3482 eV E c  E F  1.12  0.3482  0.7718 eV

19

   

no 

2



2





N c F1 / 2  F 

2.8 10 6.0 19

 1.90 10 20 cm 3

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

4.7 10 6.0

2

p E  

17



 3.18 1018 cm 3 _______________________________________

4.31 For the electron concentration nE   g c E  f F E  The Boltzmann approximation applies, so n E  

or

nE  



4 2m h

E  Ec

  E  E F    exp   kT  



4 2m n* h



* 3/ 2 n 3



3/ 2

3

 kT

  E c  E F   exp   kT  

E  Ec   E  E c   exp   kT kT  

Define E  Ec x kT Then

nE   nx   K x exp x 

To find maximum nE   nx  , set

dnx  1  0  K  x 1 / 2 exp x  dx 2  x 1 / 2  1 exp x  

or

1  0  Kx 1 / 2 exp x   x 2  which yields 1 E  Ec 1 x   E  E c  kT 2 kT 2 For the hole concentration pE   g  E 1  f F E  Using the Boltzmann approximation

pE  

or



4 2m *p h3



3/ 2

E  E   E F  E    exp   kT  



4 2m *p h



3

 kT

3/ 2

   E F  E   exp   kT  

E  E   E  E  exp   kT kT  

Define x 

E  E kT

Then

px   K  x  exp x 

To find maximum value of pE   px  , set dp x   0 Using the results from above, dx  we find the maximum at 1 E  E  kT 2 _______________________________________ 4.32 (a) Silicon: We have   E c  E F   n o  N c exp   kT   We can write E c  E F  E c  E d   E d  E F  For E c  E d  0.045 eV and E d  E F  3kT eV we can write   0.045  no  2.8 1019 exp   3  0.0259 

   2.8 10 exp 4.737 19

or

n o  2.45 1017 cm 3 We also have   E F  E  p o  N  exp   kT   Again, we can write E F  E  E F  E a   E a  E  For E F  E a  3kT and E a  E  0.045 eV Then 0.045   p o  1.04 1019 exp  3  0.0259  

   1.04 10 exp 4.737  19

or

p o  9.12 1016 cm 3

Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (b) GaAs: assume E c  E d  0.0058 eV Then   0.0058  no  4.7 1017 exp   3  0.0259 

   4.7 10 exp 3.224 17

4.35 (a)

   7 10 exp 4.332

p o  9.20 1016 cm 3 _______________________________________ 4.33 Plot _______________________________________

p o  4 15  1015  3 1015 cm 3

1.5 10 

10 2

 7.5 10 4 cm 3 3 1015 (b) no  N d  31016 cm 3 po

 7.5 10 cm 3







3

 ni  7.334 1011 cm 3

7.334 10  4 1015





 1.34 10 8 cm 3

19

19



3

2

 1.08 10 3 cm 3

(b) no  N d  31016 cm 3

1.8 10 

6 2

 1.08 10  4 cm 3 3 1016 (c) no  p o  ni  1.8 10 6 cm 3 po 







3

 375  (d) ni2  4.7 1017 7.0 1018    300    1.42300    exp    0.0259 375   ni  7.580 10 8 cm 3 p o  N a  41015 cm 3

7.580 10 

8 2

4 1015

 1.44 10 2 cm 3





3

 450  (e) ni2  4.7 1017 7.0 1018    300    1.42300    exp    0.0259 450    ni  3.853 1010 cm 3 no  N d  1014 cm 3

po

 450  (e) n  2.8 10 1.04 10    300    1.12300    exp    0.0259 450   2 i



3.853 10  

10 2

p o  N a  41015 cm 3

no 



ni2 1.8 10 6  po 3 1015



 375  (d) ni2  2.8 1019 1.04 1019    300    1.12300    exp    0.0259 375 

11 2

13 2

no 

3

3 10 (c) no  p o  ni  1.5 1010 cm 3 16

2

p o  N a  N d  4 1015  1015

no 

or

10 2



1.722 10 

 31015 cm 3

18

1.5 10  



 2.88 1012 cm 3 1.029 1014 _______________________________________

no  1.87 1016 cm 3 Assume E a  E  0.0345 eV Then   0.0345  p o  7 1018 exp   3  0.0259 

no 

 1014  1014   1.722 1013    2 2  

 1.029 1014 cm 3

po 

or

4.34 (a)

2

no 

14

 1.48 10 7 cm 3

10 _______________________________________ 4.36 (a) Ge: ni  2.4 1013 cm 3 (i) n o 

 ni  1.722 1013 cm 3 

Nd N   d 2  2

2

   ni2 

 2 1015 2 1015   2  2

2

   2.4 1013  





2

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

Then

n o  N d  21015 cm 3



f F E  



13 2

n2 2.4 10 po  i  no 2 1015

(ii) p o  N a  N d  1016  7 1015  31015 cm 3

n2 2.4 1013 no  i  po 3 1015



2

 1.92 1011 cm 3

(b) GaAs: ni  1.8 10 6 cm 3 (i) n o  N d  21015 cm

1.8 10 

6 2

 1.62 10 3 cm 3 2 1015 (ii) p o  N a  N d  31015 cm 3

po 

1.8 10 

6 2

 1.08 10 3 cm 3 3 1015 (c) The result implies that there is only one minority carrier in a volume of 10 3 cm 3 . _______________________________________ no 

4.37 (a) For the donor level nd 1  Nd  Ed  EF 1 1  exp 2  kT



0.245   1  exp1    0.0259 

or

 2.88 1011 cm 3



1

  

1 1  0.20  1  exp  2  0.0259 

or nd  8.85  10  4 Nd (b) We have 1 f F E    E  EF  1  exp   kT  Now E  E F  E  E c   E c  E F  or E  E F  kT  0.245

f F E   2.87 10 5 _______________________________________ 4.38 (a) N a  N d  p-type (b) Silicon: p o  N a  N d  2.5 1013  11013 or p o  1.5 1013 cm 3 Then





2

ni2 1.5 1010   1.5 10 7 cm 3 po 1.5 1013 Germanium: no 

N  Nd  N  Nd po  a   a 2 2 

 1.5 1013   1.5 1013       2 2    or p o  3.26 1013 cm 3 Then



2

2

   n i2 

   2.4 1013  





2



2

ni2 2.4 1013   1.76 1013 cm 3 p o 3.264 1013 Gallium Arsenide: p o  N a  N d  1.5 1013 cm 3 and no 





2

ni2 1.8 10 6   0.216 cm 3 po 1.5 1013 _______________________________________ no 

4.39 (a) N d  N a  n-type (b) no  N d  N a  2 1015  1.2 1015  81014 cm 3

po 



ni2 1.5 1010  no 8 1014



2

 2.8110 5 cm 3

Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ p o  N a  N a   N d

(c)

4 10

15

 N a  1.2 10  2 10 15

4.43 Plot _______________________________________

15

 N a  4.8 1015 cm 3

1.5 10  

10 2

3

 5.625 10 cm 4 1015 _______________________________________ no

4

4.44 Plot _______________________________________ 4.45

4.40





no 

2

n2 1.5 1010 no  i   1.125 1015 cm 3 po 2 10 5 n o  p o  n-type _______________________________________

Nd  Na  N  Na   d 2 2 

1.1 10 14 

4.41







 250  ni2  1.04 1019 6.0 1018    300 

3

1.110

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

ni2 n2 1  i  n o2  n i2 p o 4n o 4

2

 ni2

14

 4 1013

  4 10 

13 2

so ni  5.74 1013 cm 3

N a  N d  p-type Majority carriers are holes p o  N a  N d  3 1016  1.5 1016  1.5 1016 cm 3 Minority carriers are electrons

Then p o  2.75 1012 cm 3



2

Na N    a   ni2 2  2 

N    2.752 1012  a  2  

   n i2  

4.9 10 27  1.6 10 27  ni2

4.46 (a)

1  no  ni 2 no  6.88 1011 cm 3 ,

po 

2

 2  1014  1.2  1014   2 

po 

 ni  1.376 1012 cm 3

So

2  10 14  1.2  10 14 2

ni2 3.3  10 27   3  1013 cm 3 n o 1.1 1014 _______________________________________

 1.8936 10 24

no 

2

   ni2 



2

n2 1.5 1010 no  i   1.5 10 4 cm 3 16 po 1.5 10 (b) Boron atoms must be added p o  N a  N a  N d

2

5 1016  N a  3 1016  1.5 1016

2

N   a  2

   1.8936 10 24 





N  7.5735 10 24  2.752 1012 N a   a   2 

So N a  3.5 1016 cm 3

2

N    a   1.8936 10 24  2  so that N a  2.064 1012 cm 3 _______________________________________ 4.42 Plot _______________________________________

1.5  10 

10 2

2

no 

 4.5  10 3 cm 3

5  10 _______________________________________ 16

Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 4.47 (a)

N (b) E F  E Fi  kT ln  d  ni

p o  ni  n-type

(b) p o 

n i2 n2  no  i no po

1.5 10 

10 2

 1.125 1016 cm 3 2 10 4  electrons are majority carriers

no 

p o  210 4 cm 3  holes are minority carriers (c) n o  N d  N a

1.125 1016  N d  7 1015 so N d  1.825 1016 cm 3 _______________________________________

     Nd   0.0259 ln   10  1 . 5  10   For 10 14 cm 3 , E F  E Fi  0.2280 eV 10 15 cm 3 , E F  E Fi  0.2877 eV 10 16 cm 3 , E F  E Fi  0.3473 eV 10 17 cm 3 , E F  E Fi  0.4070 eV _______________________________________

4.50

p  E Fi  E F  kT ln  o   ni  For Germanium T (K) kT (eV)

ni (cm 3 )

200

0.01727

2.16 1010

400

0.03453

8.60 1014

600

0.0518

3.82 1016

1.05 10

15

 0.5 1015





2

 0.5 1015

Na N    a   ni2 2  2 

and

N a  1015 cm 3

E Fi  E F  (eV)

T (K)

p o (cm 3 )

200

1.0  10 15

0.1855

400

1.49  10

0.01898

600

3.87 1016

0.000674

_______________________________________ 4.49

N (a) E c  E F  kT ln  c  Nd

     2.8 1019    0.0259 ln    Nd  14 3 For 10 cm , E c  E F  0.3249 eV 10 15 cm 3 , E c  E F  0.2652 eV 10 16 cm 3 , E c  E F  0.2056 eV 10 17 cm 3 , E c  E F  0.1459 eV



2

 ni2

so ni2  5.25 10 28 Now







 T  ni2  2.8 1019 1.04 1019    300 

3

   1.12  exp    0.0259T 300 

2

15

   ni2 

no  1.05 N d  1.05 1015 cm 3

4.48

po 

2

N N (a) n o  d   d 2  2





 T  5.25 10 28  2.912 10 38    300 

3

  12972.973   exp   T   By trial and error, T  536.5 K (b) At T  300 K, N  E c  E F  kT ln  c   no   2.8 1019   E c  E F  0.0259 ln  15   10   0.2652 eV At T  536.5 K,  536.5  kT  0.0259   0.046318 eV  300 





 536.5  N c  2.8 1019    300   6.696 1019 cm 3

3/ 2

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

N E c  E F  kT ln  c  no

Then, T  200 K, E Fi  E F  0.4212 eV T  400 K, E Fi  E F  0.2465 eV

   

 6.696 1019   E c  E F  0.046318 ln  15   1.05 10   0.5124 eV then E c  E F   0.2472 eV (c) Closer to the intrinsic energy level. _______________________________________ 4.51

p  E Fi  E F  kT ln  o   ni  At T  200 K, kT  0.017267 eV T  400 K, kT  0.034533 eV T  600 K, kT  0.0518 eV







16

N a  1017 cm 3 , E Fi  E F  0.6408 eV (b)

N   7.0 1018   E F  E  kT ln     0.0259 ln    Na   Na  14 3 For N a  10 cm , E F  E  0.2889 eV

3

N a  1015 cm 3 , E F  E  0.2293 eV 16

N a  10 cm 3 , E F  E  0.1697 eV

N a  1017 cm 3 , E F  E  0.1100 eV _______________________________________

 ni  7.638 10 4 cm 3 At T  400 K,



n  2.8 10

19





 400  1.04 10    300  19

N   Na  E Fi  E F  kT ln  a   0.0259 ln   6   1.8 10   ni  For N a  1014 cm 3 , E Fi  E F  0.4619 eV N a  10 cm 3 , E Fi  E F  0.5811 eV

  1.12   exp    0.017267 

2 i

4.52 (a)

N a  1015 cm 3 , E Fi  E F  0.5215 eV

At T  200 K,

 200  ni2  2.8 1019 1.04 1019    300 

T  600 K, E Fi  E F  0.0630 eV _______________________________________

3

4.53 (a) E Fi  E midgap 

  1.12   exp    0.034533 



 ni  2.38110 cm 3 At T  600 K, 12



n  2.8 10 2 i

19

1.04 10  19

 600     300 

 

 0.4947   10 5 exp   0.0259 

2

N N  p o  a   a   ni2 2  2 

 3.288 1015 cm 3

3 0.0259 ln 10 4

E Fi  Emidgap  0.0447 eV (b) Impurity atoms to be added so Emidgap  E F  0.45 eV (i) p-type, so add acceptor atoms (ii) E Fi  E F  0.0447  0.45  0.4947 eV Then  E  EF  p o  ni exp Fi   kT 

3

 ni  9.740 1014 cm 3 At T  200 K and T  400 K, p o  N a  31015 cm 3 At T  600 K,

 3  1015 3  1015   2  2

   

or

  1.12   exp    0.0518 



 m *p 3 kT ln  *  mn 4 

or 2

   9.740  1014  





2

p o  N a  1.97 1013 cm 3 _______________________________________

Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 4.54

  E c  E F   n o  N d  N a  N c exp   kT  

so





  0.215  N d  5 1015  2.8 1019 exp   0.0259   5 1015  6.95 1015

N  (b) E F  E Fi  kT ln  c   Nd   2.8 1019    0.1876 eV  0.0259 ln  16   2 10  (c) For part (a); p o  21016 cm 3

or

no 

N d  1.2 1016 cm 3 _______________________________________



ni2 1.5 1010  po 2 1016

For part (b): n o  21016 cm 3

  E c  E F   N d  N c exp   kT     0.1929   2.8 1019 exp    0.0259 



 N d  1.0311016 cm 3 Additional donor atoms (b) GaAs  4.7 1017   (i) E c  E F  0.0259  ln  15   10   0.15936 eV (ii) E c  E F  0.15936  0.0259  0.13346 eV



po 



  0.13346  N d  4.7 10 exp    0.0259   2.718 1015 cm 3  N d  1015 17

 N d  1.718 1015 cm 3 Additional donor atoms _______________________________________ 4.56

N  (a) E Fi  E F  kT ln     Na   1.04  1019    0.1620 eV  0.0259 ln  16   2  10 



ni2 1.5 1010  no 2 1016



2

 1.125 10 4 cm 3 _______________________________________

4.57

 E  E Fi  n o  ni exp  F   kT   0.55   1.8 10 6 exp    0.0259 





N d  1.6311016 cm 3  N d  61015

2

 1.125 10 4 cm 3

4.55 (a) Silicon

N  (i) E c  E F  kT ln  c   Nd   2.8 1019    0.2188 eV  0.0259 ln  15   6 10  (ii) E c  E F  0.2188  0.0259  0.1929 eV





 3.0 1015 cm 3 Add additional acceptor impurities no  N d  N a

3 1015  7 1015  N a  N a  41015 cm 3 _______________________________________ 4.58

p  (a) E Fi  E F  kT ln  o   ni   3 1015    0.3161 eV  0.0259 ln  10   1.5 10  n  (b) E F  E Fi  kT ln  o   ni   3 1016    0.3758 eV  0.0259  ln  10   1.5 10  (c) E F  E Fi

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

p  (d) E Fi  E F  kT ln  o   ni 

4.60 n-type

15   375   4 10   0.0259  ln  11  300    7.334 10   0.2786 eV n  (e) E F  E Fi  kT ln  o   ni  14  450   1.029 10   0.0259  ln  13  300   1.722 10   0.06945 eV _______________________________________

n  E F  E Fi  kT ln  o   ni   1.125  10 16    0.3504 eV  0.0259  ln    10  1.5  10  ______________________________________ 4.61 2

po 

Na N    a   ni2 2  2 

5.08  10 15 

4.59

N  (a) E F  E  kT ln     po   7.0 1018    0.2009 eV  0.0259 ln  15   3 10   7.0 1018   (b) E F  E  0.0259  ln  4   1.08 10   1.360 eV  7.0 1018   (c) E F  E  0.0259  ln  6   1.8 10   0.7508 eV  375  (d) E F  E  0.0259   300 





 7.0  10 375 300   ln  4  10 15   0.2526 eV  450  (e) E F  E  0.0259   300 



18



3/ 2

  

 7.0  10 450 300    ln   1.48  10 7    1.068 eV _______________________________________ 18

5.08 10

5  10 15 2

 5  1015    2 15

2

   n i2  

  2.5 10 

 2.5 1015

2

15 2

 ni2

6.6564 10 30  6.25 10 30  ni2

 ni2  4.064 10 29

  Eg  ni2  N c N  exp    kT   350  kT  0.0259   0.030217 eV  300 





2





2

 350  19 N c  1.2 1019    1.633 10 cm 3  300   350  19 N   1.8 1019    2.45 10 cm 3  300  Now 4.064 10 29  1.633 1019 2.45 1019







  Eg   exp    0.030217 

3/ 2

So







 1.633 1019 2.45 1019  E g  0.030217  ln   4.064 10 29    E g  0.6257 eV _______________________________________

Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 4 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 4.62 (a) Replace Ga atoms  Silicon acts as a donor N d  0.05 7 1015  3.5 1014 cm 3 Replace As atoms  Silicon acts as an acceptor N a  0.95 7 1015  6.65 1015 cm 3









(b) N a  N d  p-type (c) p o  N a  N d  6.65 1015  3.5 1014  6.3 1015 cm 3

no 





2

ni2 1.8 10 6   5.14 10  4 cm 3 15 po 6.3 10

p  (d) E Fi  E F  kT ln  o   ni   6.3 1015    0.5692 eV  0.0259 ln  6   1.8 10  _______________________________________

Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 5 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

Chapter 5 5.1 1 1  (a)   e n N d 1.6  10 19 1300  1015  4.808  -cm



 

1  0.208 (  -cm) 1  4.8077 _______________________________________ (b)  

5.2

1



1 1.6  10 220  8  10 16  0.355  -cm (b)   e n N d





19











120  1.6 10 19  n N d



From Figure 5.3, for N d  21017 cm 3 , then  n  3800 cm 2 /V-s which gives

  1.6 10 19 38002 1017 

  e p N a  1.80 or N a   e p 1.6 10 19 380

 2.96 1016 cm 3 _______________________________________

5.3 (a)   e n N d

 121.6 (  -cm) 1 _______________________________________ 5.5

L

R

A

or  n 





10  1.6 10 19  n N d



From Figure 5.3, for N d  61016 cm 3 we



L

A



L

e n N d A

L

eN d RA 2.5

1.6 10 2 10 700.1 19

15

find  n  1050 cm 2 /V-s which gives

 1116 cm 2 /V-s _______________________________________

 10.08 (  -cm) 1 1 (b)   e p N a

5.6 (a) no  N d  1016 cm 3 and

  1.6 10 19 10506 1016 

0.20 

1

1.6 10  19

p

po 

Na

From Figure 5.3, for N a  1017 cm 3 we find  p  320 cm 2 /V-s which gives



1

1.6 10 32010  19

17

 0.195  -cm

_______________________________________ 5.4 (a)  

1 e p N a

0.35 



ni2 1.8 10 6  no 1016



2

 3.24 10  4 cm 3

(b) J  e n n o  For GaAs doped at N d  1016 cm 3 ,

 n  7500 cm 2 /V-s Then J  1.6 10 19 7500 1016 10 or J  120 A/cm 2





 

(b) (i) p o  N a  1016 cm 3

1

1.6 10  19

p

Na

From Figure 5.3, for N a  81016 cm 3 we find  p  220 cm 2 /V-s which gives

no 

ni2  3.24  10  4 cm 3 po

(ii) For GaAs doped at N a  1016 cm 3 ,

 p  310 cm 2 /V-s

Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 5 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

J  e p po 



 1.6 10

19

31010 10 16

or

J  4.96 A/cm 2 _______________________________________ 5.7 (a) V  IR  10  0.1R or R  100  (b) L L R   A RA or 10 3   0.01 (  -cm) 1 100 10 3 (c)   e n N d or 0.01  1.6 10 19 1350N d Then N d  4.63 1013 cm 3









(d)   e p po or 0.01  1.6 10 19 480 p o Then p o  1.30 1014 cm 3  N a  N d So N a  1.30 1014  1015  1.13 1015 cm 3 Note: For the doping concentrations obtained, the assumed mobility values are valid. _______________________________________





5.8 (a) R 

L

A



e

L p Na A



For N a  21016 cm 3 , then

 p  400 cm 2 /V-s R



19

1.6  10  68.93 

V 2   0.0290 A R 68.93 I  29.0 mA

I

or

0.075 4002 1016 8.5 10 4 

(b) R  L  R  68.933  206.79  V 2   0.00967 A R 206.79 or I  9.67 mA (c) J  ep o d I

29.0 10 3  34.12 A/cm 2 4 8.5 10 J 34.12  Then  d  ep o 1.6  10 19 2  1016 For (a), J 







 1.066 10 cm/s 9.67 10 3 For (b), J   11.38 A/cm 2 8.5 10  4 11.38 d  1.6  10 19 2  10 16 4







 3.55 10 cm/s _______________________________________ 3

5.9 (a) For N d  21015 cm 3 , then

 n  8000 cm 2 /V-s V 5   200  I 25  10 3 L R e n N d A R

or L  e n N d RA











 1.6 10 19 8000 2 1015 200 5 10 5  0.0256 cm I (b) J   en o d A I or  d  Aen o 



25 10 3 1.6 10 19 2 1015

5 10  5

 1.56 10 cm/s (c) I  en o d A







6











 1.6 10 19 2 1015 5 10 6 5 10 5  0.080 A or I  80 mA _______________________________________

Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 5 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (b) N a  N d  1016 cm 3

5.10 (a)  

V 3   3 V/cm L 1

 d  n    n 

  n  1250 cm 2 /V-s

 p  410 cm 2 /V-s

d

10 4   3



or

 n  3333 cm 2 /V-s

(c) N a  N d  1018 cm 3   n  290 cm 2 /V-s

or

 p  130 cm 2 /V-s

 d  2.4 10 3 cm/s _______________________________________ 5.11 (a) Silicon: For   1 kV/cm,  d  1.2 10 6 cm/s Then d 10 4 tt    8.33 10 11 s  d 1.2 10 6

Then

10 4  1.05 10 11 s 9.5 10 6 For GaAs:  d  7 10 6 cm/s Then 10 4 tt   1.43 10 11 s 6 7 10 _______________________________________ 5.12

1 1  e n no  e p p o e  n   p ni



  n  1350 cm 2 /V-s

 p  480 cm 2 /V-s 1 1350  480 1.5 1010







2

ni2 1.8 10 6   2.49 10 5 cm 3 po 1.3 1017 (b) Silicon: 1    e n n o  or 1 1 no   e n 8 1.6  10 19 1350  which gives n o  5.79 1014 cm 3 and









2

ni2 1.5 1010   3.89 10 5 cm 3 no 5.79 1014 Note: For the doping concentrations obtained in part (b), the assumed mobility values are valid. _______________________________________ po 

(a) N a  N d  1014 cm 3

 2.28 10  -cm



 p  240 cm 2 /V-s

tt 



10

From Figure 5.3, and using trial and error, we find p o  1.3 1017 cm 3 and

Then

1.6 10 

19



no 



1

1.6 10 290  1301.5 10 

5.13 (a) GaAs:   e p p o  5  1.6  10 19  p p o

 d  9.5 10 6 cm/s

5



 9.92 10 5  -cm _______________________________________

For GaAs:  d  7.5 10 6 cm/s Then d 10 4 tt    1.33 10 11 s  d 7.5 10 6 (b) Silicon: For   50 kV/cm,

19



 2.5110  -cm

 d   n   8003





5

(b)



1 1250  410 1.5 1010

1.6 10  19

Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 5 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 5.14

 i  eni  n   p 

(b) R 

Then 10 6  1.6 10 19 1000  600ni or ni (300 K)  3.9110 9 cm 3 Now   Eg   ni2  N c N  exp   kT  or N N  E g  kT ln  c 2    ni 





 

 1019 2  0.0259  ln   3.91 10 9 which gives E g  1.122 eV





2

  

 

2

ni (500 K)  2.27 1013 cm 3 Then  i  1.6 10 19 2.27 1013 1000  600 which gives  i (500 K)  5.8110 3 (  -cm) 1 _______________________________________

5.15 (a) (i) Silicon:  i  eni  n   p

 i  1.6 10

















1.5 10 1350  480

then  n  1300 cm 2 /V-s







So   1.6 10 19 1300 1.2 1015



 0.2496 (  -cm) 1 (b) Using Figure 5.2, (i) For T  250 K ( 23 C),   n  1800 cm 2 /V-s

  1.6 10 19 18001.2 1015 

 0.346 (  -cm) 1 (ii) For T  400 K ( 127 C),   n  670 cm 2 /V-s

  1.6 10 19 6701.2 1015 

 0.129 (  -cm) 1 _______________________________________ 5.17 t

 i  4.39 10 6 (  -cm) 1 (ii) Ge:  i  1.6 10 19 2.4 1013 3900  1900 or  i  2.23 10 2 (  -cm) 1 (iii) GaAs:  i  1.6 10 19 1.8 10 6 8500  400 or  i  2.56 10 9 (  -cm) 1





10

or





200 10 4  1.06 10 6  2 8 2.23 10 85 10 (iii) GaAs: 200 10 4 R  9.19 1012  2.56 10 9 85 10 8 _______________________________________ R







(ii) Ge:

From Figure 5.3, for N d  1.2 1015 cm 3 ,

or

19

200 10 4  5.36 10 9  4.39 10 6 85 10 8



   1.122 exp    0.0259500 300 



R

0.25  1.6 10 19  n N d

 5.15 10 26



(i) Si:

5.16 (a)   e n N d

Now

ni2 (500K)  1019

L A





 avg 







 

t

1 1 x  x dx   o exp dx t 0 t 0  d 



o t



 d  exp  x 

t

 d 0

 od    t   exp   1 t   d  

200.3 1  exp  1.5    1.5   0.3 

 3.97 (  -cm) 1 _______________________________________

Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 5 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 5.18 V 2  133.3 V/cm (a)    L 150  10  4 (b)  x   e n N d x 

 avg  e n 

1 T



T

x    2 10 1  1.111T dx 16

0

e n 2 10 T

 x 

 x    21.111T  0  e 2 1016  T2  n T   T 21.111T    e n 2 1016 0.55 



16









 1.6 10

19

2

T

V  L 150 10  4 5  1.32 10 A or I  13.2  A (d) Top surface;   1.6 10 19 750 2 1016















5.19 Plot _______________________________________ 5.20 (a)   10 V/cm so  d   n   135010  1.35 10 4 cm/s or  d  1.35 10 2 m/s Then 1 T  m n* d2 2 2 1  1.08 9.1110 31 1.35 10 2 2 or T  8.97 10 27 J  5.60 10 8 eV





 7.18 1019

ni  8.47 10 9 cm 3

 0.24 (  -cm) 1 J    0.24133.3  32 A/cm 2 _______________________________________





   

or

 2.4 (  -cm) J    2.4133.3  320 A/cm 2 Bottom surface:   1.6 10 19 750 2 1015





  1.10   2 1019 11019 exp   0.0259 

2

1





  Eg (a) ni2  N c N  exp  kT

7502 10 0.55

(c) I 



5.21

16

 avg  1.32 (  -cm) 1  avg A 1.327.5 10 4 10 4 

(b)   1 kV/cm  d  13501000  1.35 10 6 cm/s or  d  1.35 10 4 m/s Then 2 1 T  1.08 9.1110 31 1.35 10 4 2 or T  8.97 10 23 J  5.60 10 4 eV _______________________________________



For N d  1014 cm 3 >> ni  n o  1014 cm 3 Then J    e n n o 





 

 1.6 10 19 1000  1014 100

or

J  1.60 A/cm 2 (b) A 5% increase is due to a 5% increase in electron concentration, so 2

n o  1.05  10

14

N N   d   d   n i2 2  2 

which becomes

1.05 10

14

 5 1013

  5 10  2

13 2

 ni2

and yields ni2  5.25 10 26 3   Eg   T    2 1019 11019   exp   300   kT 







or

   1.10  T  2.625 10 12    exp    300   0.0259T 300  By trial and error, we find T  456 K _______________________________________ 3

Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 5 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (b) From Figure 5.3, n-type:  n  1100 cm 2 /V-s

5.22

n2 (a)   e n no  e p po and n o  i po Then e n 2   n i  e p p o po To find the minimum conductivity, set  1e n ni2 d 0  e p dp o p o2 which yields

p-type:  p  400 cm 2 /V-s compensated:  n  1000 cm 2 /V-s (c) n-type:   e n no



 8.8 (  -cm) p-type:   e p po



  p o  n i  n  (Answer to part (b))  p    Substituting into the conductivity expression e n ni2    min  1/ 2 ni  n  p

 

 

 e p n i  n  p

  1/ 2

which simplifies to

 min  2en i  n  p The intrinsic conductivity is defined as

 i  eni  n   p   eni 

i

n   p

The minimum conductivity can then be written as

 min 







 1.28 (  -cm) compensated:   e n no





1

 1.6 10 19 400 2 1016

1/ 2

 



 1.6 10 19 1100  5 1016



1





 1.6 10 19 1000  3 1016



1

 4.8 (  -cm) J (d) J       120  13.6 V/cm n-type:   8 .8 120  93.75 V/cm p-type:   1.28 120  25 V/cm compensated:   4 .8 _______________________________________ 5.24

2 i  n  p

1

n   p



_______________________________________



1

1



1

2



1

3

1 1 1   2000 1500 500  0.00050  0.000667  0.0020



5.23 (a) n-type: no  N d  51016 cm 3



n2 1.5 1010 po  i  no 5 1016

or



2

1



 4.5 10 3 cm 3

p-type: p o  N a  210 cm 16

1.5 10  

10 2

no

3

 5 10  2 10 16

5.25 16

 31016 cm 3

1.5 10  

10 2

po

3 1016

Then   316 cm 2 /V-s _______________________________________

 1.125 10 4 cm 3

2 10 compensated: n o  N d  N a 16

 0.003167

 7.5 10 3 cm 3

 T    300  (a) At T  200 K,

 n  1300

 300    200 

 n  1300

3 / 2

 300   1300   T 

3 / 2

3/ 2

 2388 cm 2 /V-s

(b) At T  400 K,  n  844 cm 2 /V-s _______________________________________

Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 5 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

1





1

1



1

2



 1015  nx1   (b)  2  1.6 10 19 230 4   0  20 10  4  103  3.68  102  3.68  1017 nx1 



5.26 1 1   0.006 250 500



Then   167 cm 2 /V-s _______________________________________

nx1   8.911014 cm 3 _______________________________________

5.27 Plot _______________________________________

5.32

5.28 Plot _______________________________________ 5.29

 5 1014  n0  dn  J n  eDn  eDn   dx  0.01  0   5 1014  n0   0.19  1.6 10 19 25  0.010   Then 0.190.010  5 1014  n0 1.6  10 19 25 which yields n0  0.25 1014 cm 3 _______________________________________









5.30 J n  eD n

dn n  eD n dx x



16

15

5.31 dn n  eD n dx x  1015  nx1    2  1.6 10 19 30 4   0  20 10  4 10 3  4.8 10 3  4.8 10 18 nx1  which yields nx1   1.67 1014 cm 3

(a) J n  eD n





 eD p 

1016  x  21   L  L

(a) For x  0 ,  1.6 10 19 10 1016 2 Jp  12 10  4  26.7 A/cm 2 (b) For x  6  m,



  



  

6   1.6 10 19 10 1016 21    12  Jp  4 12 10  13.3 A/cm 2 (c) For x  12  m,

Jp 0 _______________________________________ 5.33 For electrons: dn d J n  eD n  eD n 10 15 e  x / Ln dx dx



 2 10  5 10  J n  1.6 10 19 27   0  0.012   J n  5.4 A/cm 2 _______________________________________



2 dp d  16  x   eD p 10 1    dx dx   L  

J p  eD p



 

 eD n 1015 e  x / Ln Ln At x  0 ,  1.6 10 19 25 1015 Jn   2 A/cm 2 3 2 10 For holes: dp d  x / Lp J p  eD p  eD p 5  10 15 e dx dx 



  









 eD p 5  10 15 e



 x / Lp

Lp

For x  0 ,  1.6 10 19 10 5 1015  16 A/cm 2 Jp  4 5 10 J Total  J n x  0  J p x  0



 



 2   16  18 A/cm 2 _______________________________________

Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 5 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 5.34



dp d x / Lp J p  eD p  eD p 5  10 15 e dx dx



(a) (i) J p 





eD p 5  10 15 e



x / Lp

Lp

1.6 10 105 10  19

15

50 10  4  1.6 A/cm 2 1.6 10 19 48 5 1015 (ii) J p  22.5 10  4  17.07 A/cm 2 1.6 10 19 10 5 1015 e 1 (b) (i) J p  50 10  4  0.589 A/cm 2 1.6 10 19 48 5 1015 e 1 (ii) J p  22.5 10  4  6.28 A/cm 2 _______________________________________



 





 





 



(a) J n  eD n

dn J n  e n n  eD n dx







 eD n 2 1015 e  x / L L  1.6 10 19 27  2 1015 e  x / L  15 10  4  5.76e  x / L (b) J p  J Total  J n  10   5.76e  x / L 



 











 5.76e  x / L 10 A/cm 2 (c) We have J p    e p po  5.76e

x / L









 10  1.6 10

19



42010  16

So   8.57e  x / L 14.88 V/cm _______________________________________ 5.37 (a) J  e n nx   eD n

dnx  dx

We have  n  8000 cm 2 /V-s, so that



or

  

 1.6 10 19 25 1016

 1  4  18 10

 x  exp    18 

Then

   x  x  40  1.536exp   22.22 exp  18   18    We find 22.22 exp  x   40  18   1.536  exp  x   18  or x   14.5  26.0 exp   18  _______________________________________







 1.6  10 19 207 



   x   40  1.6 10 19 9601016 exp    18  





dn d  eD n 2  10 15 e  x / L dx dx

Dn  0.02598000  207 cm 2 /s Then 100  1.6 10 19 800012nx 

5.35



5.36

dnx  dx

which yields







100  1.536  10 14 nx   3.312  10 17

 dndxx 

Solution is of the form x nx   A  B exp   d  so that dnx   B x  exp  dx d  d  Substituting into the differential equation, we have    x  100  1.536 10 14  A  B exp   d  







3.312 10  B exp  x  17

d This equation is valid for all x, so 100  1.536 10 14 A or A  6.511015





 d 

Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 5 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Also

x 1.536 10 14 B exp   d  

3.312 10  B exp  x   0 17

d  d  which yields d  2.156 10 3 cm At x  0 , e n n0  50 so that 50  1.6 10 19 800012 A  B  which yields B  3.255 1015 Then x nx   6.511015  3.255 1015 exp  cm 3 d   (b) At x  0 , n0  6.511015  3.255 1015 Or n0  3.26 1015 cm 3 At x  50  m,





  50  n50  6.511015  3.255 1015 exp   21.56  or n50  6.19 1015 cm 3 (c) At x  50  m,

J drf  e n n50









 1.6 10 19 8000 6.19 1015 12 or

J drf x  50  95.08 A/cm 2 Then J diff x  50  100  95.08 or J diff x  50  4.92 A/cm 2 _______________________________________ 5.38

 E  E Fi  n  ni exp F   kT  (a) E F  E Fi  ax  b , b  0.4





0.15  a 10 3  0.4 which yields a  2.5 10 2

Then E F  E Fi  0.4  2.5 10 2 x so  0.4  2.5 10 2 x   n  ni exp  kT   dn (b) J n  eD n dx

  2.5 10 2   0.4  2.5 10 2 x   exp   eDn ni     kT kT     Assume T  300 K, so kT  0.0259 eV and ni  1.5 1010 cm 3 Then  1.6  10 19 25 1.5  1010 2.5  10 2 Jn  0.0259



 





 0.4  2.5 10 2 x    exp  0.0259   or

 0.4  2.5 10 2 x   J n  5.79 10  4 exp  0.0259   3 2 (i) At x  0 , J n  2.95 10 A/cm (ii) At x  5  m, J n  23.7 A/cm 2 _______________________________________ 5.39 (a) J n  e n n  eD n





dn dx

 

 x  80  1.6 10 19 1000  1016 1    L

  1016    1.6 10 19 25.9   L  where L  10 10 4  10 3 cm We find  x   80  1.6  1.6 3   41.44  10  or x  80  1.6  1  41.44 L  Solving for the electric field, we find 24.1  V/cm x    1 L 





Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 5 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (b) For J n  20 A/cm 2

5.42

x  20  1.6  1  41.44 L  Then 13.3  V/cm  x 1    L _______________________________________







or





 25.9 10 10 4  0.0259 V or   25.9 mV _______________________________________ 5.41 From Example 5.6 0.0259 1019  0.0259 10 3 x  1016  1019 x 1  10 3 x

  



  



D n  155.4 cm 2 /s Then  1.6 10 19 155.4 5 1016 x J diff  exp  0.110  4  L  or x J diff  1.243 10 5 exp  A/cm 2  L  (b) 0  J drf  J diff Now J drf  e n n



L

(b)     X dx  25.9 L  0 







 x dx













or

0

10 4

dx  0.0259 10   1  10 x  3

3

0

 



 1   0.0259 10 3  3  ln 1  10 3 x  10   0.0259ln 1  0.1  ln 1 or





   x   1.6 10 19 6000 5 1016 exp     L 

10 4

V 

eD n x  N do exp   L   L 

 kT  Dn   n    60000.0259  e 

 X  25.9 V/cm

0

dN d x  dn  eD n dx dx

We have

0.0259 0.0259   L 10  10  4

or

0.0259  500 V/cm L Which yields L  5.18 10 5 cm _______________________________________ So  X 

J diff  eD n

 0.0259    1     N do e  x / L N do e  x / L  L 



For N d x   N do e  x / L

5.43 (a) We have

5.40

dN d x  1  kT  (a)  X       e N x dx   d  0.0259  d   N do e  x / L N do e  x / L dx

dN d x  1  kT   x      dx  e  N d x 



104 0

V  2.73 mV _______________________________________

   x  J drf  48exp     L  We have J drf   J diff so   48exp  x   1.243 10 5 exp  x   L    L  which yields   2.59 10 3 V/cm _______________________________________

Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 5 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 5.44 Plot _______________________________________

5.48 (a) V H  0  n-type (b) n 

5.45 (a) (i) Dn  0.02591150  29.8 cm 2 /s

8  308.9 cm 2 /V-s 0.0259 35  1351 cm 2 /V-s (ii)  p  0.0259 _______________________________________

5.46 L  10 1 cm, W  10 2 cm, d  10 3 cm





 I X BZ  1.2 10 3 5 10 2  ned 2 10 22 1.6 10 19 10 5









 1.875 10 3 V or V H  1.875 mV (b) V  1.875 10 3 H  H   0.1875 V/cm W 10  2 _______________________________________









 250 10 6 5 10 2 5 10 21 1.6 10 19 5 10 5









(c) n 





5

I x Bz edV H











 0.5 10 3 6.5 10 2 1.6 10 19 5 10 5  0.825 10 3









21

0.5 10 0.5 10  4.924 10 1.255 10 5 10  3

1.6 10

19

21

2

4

5

or

 n  0.1015 m 2 /V-s  1015 cm 2 /V-s _______________________________________

250 10 10  5 10 0.12 10 5 10  6

19

4

5.49 (a) V H   H W   16.5 10 3 5 10 2 or V H  0.825 mV (b) V H  negative  n-type



V H  0.3125 mV (b) V  0.3125 10 3 H  H  W 2 10  2 or  H  1.56 10 2 V/cm (c) IxL n  enV xWd

1.6 10

1.6 10

3

21

n  4.924 10 21 m 3  4.924 1015 cm 3 (d) IxL n  enV xWd

or



0.5 10 10  6.0110 1510 10  3

19

or

I x Bz ned





3

 0.03466 m 2 /V-s or  n  346.6 cm 2 /V-s _______________________________________

5.47 (a) V H 



or n  6.011015 cm 3 IX L (c)  n  enV X Wd

(b) (i)  p 





 6.0110 m

(ii) Dn  0.02596200   160.6 cm /s

VH 



21

2

(a)



 I X BZ  0.50 10 3 0.10  edV H 1.6 10 19 10 5  5.2 10 3

3

4

5

or

 n  0.3125 m 2 /V-s  3125 cm 2 /V-s _______________________________________



Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 5 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 5.50 (a) V H  negative  n-type (b) n 

I x Bz edV H









 2.5 10 3 2.5 10 2 1.6 10 19 0.0110  2  4.5 10 3









or n  8.68 10 20 m 3  8.68 1014 cm 3 IxL (c)  n  enV xWd







  2.5 10 3 0.5 10 2   19 20 8.68 10 2.2   1.6 10   1  2 2   0.05 10 0.0110 













or

 n  0.8182 m 2 /V-s  8182 cm 2 /V-s 1 (d)    e n n 







 1.6  10  19 8182 8.68  10 14 or



  0.88 (  -cm) _______________________________________

Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

Chapter 6 6.1

no  N d  51015 cm 3





2

ni2 1.5 1010   4.5 10 4 cm 3 15 Nd 5 10 (a) Minority carrier hole lifetime is a constant.  pt   p 0  2  10 7 s po 

R po 

po

 p0

(b) R po 



4.5  10 4  2.25  1011 cm 3 s 1 2  10  7

p o  p

4.5 10 4  1014  2 10 7

 p0

3

1

6.2

p o  N a  21016 cm 3



n2 1.8 10 6 no  i  po 2 1016 (a) R  

(a) E  h 

po



 pt

no

 nt



2

 1.62 10  4 cm 3



6.625 10 3 10  34

6300 10 10

(b)

n  p  g  3.17 1019 10 10 6 

or

  



_______________________________________ 6.5

 6.17 10 s _______________________________________

We have p p    F p  g p  t p

6.3 (a) Recombination rates are equal no p  o

and J p  e p p  eD p p The hole particle current density is Jp F p    p  D p  p  e p Now   F p   p    p   D p   p

13

 nO

 pO

no  N d  1016 cm 3

po  Then 1016



ni2 1.5 1010  no 1016

2.25 10  nO 20 10  6 which yields  nO  8.89 10 6 s 

4



8

n  p  3.17 1014 cm 3

p 2 1016  o  n 0   5 10 7 no 1.62 10  4







 3.17 10 19 e-h pairs/cm 3 -s

no

 n0

hc

E  3.15 10 19 J; energy of one photon Now 1 W = 1 J/s  3.17 1018 photons/s Volume = (1)(0.1) = 0.1 cm 3 Then 3.17 1018 g 0.1

n 5 1014   10 21 cm 3 s 1  n0 5 10 7

(b) R p 

 pt



6.4

or

 510 cm s _______________________________________ 20

(b) Generation rate = recombination rate Then 2.25 10 4 G  1.125 10 9 cm 3 s 1 6 20 10 (c) R  G  1.125 10 9 cm 3 s 1 _______________________________________

2

 2.25 10 4 cm 3

We can write    p     p  p   and   p   2 p so   F p   p   p  p     D p  2 p

Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Then p    p   p  p    t

 Dp2 p  g p 

p

p

We can then write D p  2 p   p   p  p   

p gp    p t _______________________________________ p

By charge neutrality, n  p  n  n  p  and  n   p    2 n    2 p  and t t Also p n  R gn  g p  g ,

p

n

Then we have (1) D p  2 n    p   n   p   gR

6.6 From Equation (6.18), p p    F p  g p  t p

and (2) Dn  2 n   n   n  n  

 n  t Multiply Equation (1) by  n n and Equation gR

p 0 For steady-state, t Then 0    F p  g p  R p

For a one-dimensional case, dF p  g p  R p  10 20  2 1019 dx or dF p  81019 cm 3 s 1 dx _______________________________________ 6.7 From Equation (6.18), dF p 0  0  2 1019 dx or dF p  21019 cm 3 s 1 dx _______________________________________ 6.8 We have the continuity equations (1) D p  2 p    p   p   p  

gp 

p

p



p  t

and (2) Dn  2 n   n   n  n    gn 

n

n

 n  t



 n  t

(2) by  p p , and add the two equations. We find  n nD p   p pDn  2 n 

  n  p  p  n  n





  n n   p p g  R



 nn   p p





 tn 

Divide by  n n   p p , then   n nD p   p pD n  2   n   nn   p p      n  p  p  n      n    n n   p p   n   g  R   t Define  n nD p   p pD n Dn D p n  p  D   nn   p p Dn n  D p p

and

 

 n  p  p  n nn   p p

Then we have

 n  t Q.E.D. _______________________________________

D  2 n     n  g  R  

Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 6.9

 

p-type material; minority carriers are electrons (a)     n



From Figure 5.3,  n  1300 cm 2 /V-s

 33.67 cm 2 /s   n 0  10 7 s

390019001.124 1013  5.124 1013  39005.124 1013  19001.124 1013 

 nt

p o  N a  71015 cm 3



n2 1.5 1010 no  i  Na 7 1015

2

  nt  9.12 10 6 s _______________________________________

 pt

6.11

so  pt  2.18  10 4 s _______________________________________ 6.10 For Ge: ni  2.4 1013 cm 3



Nd N   d 2  2

4 10 2

13

13



   2.4 1013  





2



2

ni2 2.4 1013   1.124 1013 cm 3 no 5.124 1013 (a) We have:  n  3900 cm 2 /V-s, Dn  101 cm 2 /s

 p  1900 cm /V-s, D p  49.2 cm /s 2

For very, very low injection, Dn D p n  p  D  Dn n  D p p

2

10149.25.124 1013  1.124 1013   1015.124 1013  49.21.124 1013   54.2 cm 2 /s

and

With excess carriers n  n o  n and p  p o  p For an n-type semiconductor, we can write n  p  p Then   e n no  p  e p  po  p or   e n no  e p po  e  n   p p so   e  n   p p



2

 5.124 1013 cm 3

po 

  e n n  e p p



2

   ni2 

 4 10    2

1.124 1013 2 10 6



 nt

3.214  10 4 7  1015   pt 10  7

no 

 p0

5.124 1013



 3.2110 4 cm 3 no p  o

 nt

nn   p p

 1340 cm 2 /V-s (b) For holes,  pt   p 0  2  10 6 s For electrons, p n 

 kT  (b) D   Dn      n  0.02591300   e  (c)  nt

 n  p  p  n





In steady-state, p  g  pO So that   e  n   p g  pO _______________________________________



6.12 (a)





p o  N a  1016 cm 3





2

ni2 1.5 1010   2.25 10 4 cm 3 po 1016   e n no  n  e p  po  p no 





 e p po  e  n   p n



Now n  p  g  n 0 1  e



 8 10



5 10 1  e 1  e  cm

20

 4 1014

t /  n 0

7

t /  n 0

t /  n 0

3



Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

 38010   1.6 10 900  380    4 10 1  e  (  -cm)   0.608  0.08191  e

Then

  1.6 10

19

16

where p  g  p 0 e



t /  n 0

t /  n 0

1



    1.6 10 1300  400  4  10 e 19

  1.248  0.109e I



    cm  2  10 1  e

At t  10 s,



p 10 6   2  10 14 1  e 10

p  2 1014 e



 t 10

6

6

3

/ 510

8



 1.4 1016 cm 3 (a) n  p  g  n 0

  n0  2.5 10 7 s



(b) n  p  g  n 0 1  e t /  n 0

For 0  t  10 6 s,

R 

75005 10   1.6 10 7500  310   2  10 1  e 15

For t  10 6 s,

   6.0  0.250e

 (  -cm)



 t 10 6 /  p 0





n 5 10  1  e t /  n 0  n0 2.5 10 7 14







(c)

t /  p 0

t /  p 0



t /  n 0

 2 10 21 1  e t /  no cm 3 s 1

19





 5 10 1  e 14

14

A

p o  N a  N d  2 1016  6 1015

/  p 0 cm 3

19

t /  p 0

5 1014  2 10 21  n0

  e n no  e n   p p

 6.0  0.250 1  e

0.05

6.15

(b) no  51015 cm 3

  1.6 10

5

or I  2.496  0.218e mA _______________________________________

 21014 cm 3

Then for t  10 6 s,

10 10

t /  p 0



t /  p 0

t /  p 0

6

t /  p 0

 2.496 10  2.18 10 4 e

 4  10 21 5  10 8 1  e 14

1.248  0.109e

t /  p 0

3



t /  p 0

14

(ii)     0.690 (  -cm) _______________________________________

n  p  g  p 0 1  e

t /  p 0

t / 

1

t /  p 0



 4 1014 e p 0 cm 3   1.6 10 19 1300 8 1015  2 1015

(b) (i)  0  0.608 (  -cm) 1

6.13 (a) For 0  t  10 6 s,



 8  10 20 5  10 7 e

19

14

t /  p 0

1

(  -cm) 1 _______________________________________







1 (i)   5 1014  5 1014 1  e t /  n 0 4



t   n 0 ln 1.3333  7.19 10 8 s









1 (ii)   5 1014  5 1014 1  e t /  n 0 2 t   n0 ln 2  1.73 10 7 s

6.14 L V ; R A R A I  V L For N I  N d  N a  8 1015  2 1015

I

 1016 cm 3

Then,  n  1300 cm 2 /V-s

 p  400 cm 2 /V-s   e n no  e n   p p







3 (iii)   5 1014  5 1014 1  e t /  n 0 4 t   n0 ln 4  3.47 10 7 s









(iv) 0.95 5 1014  5 1014 1  e t /  n 0



t   n0 ln 20  7.49 10 7 s _______________________________________

Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 6.16 At t  2 10 6 s, 15 15 6 7 no  N d  N a  8 10  2 10 n  5  1014 e  210  / 510  1

 61015 cm 3

po 



po

(a) Ro 

 p0

 9.16 1012 cm 3



6 2

n 1.8 10  no 6 1015 2 i

5.4 10

 4 10 4 

so  p 0  1.35  10 s



 p0

 4.908 1014 1  e t /  n 0 (b) (i) n0  5 10 cm 14



(b) p  g  p 0  2  10

1.35 10 



8

21

_______________________________________ 6.17 (a) (i)For 0  t  5 10 7 s t /  p 0







 5  10 20 5  10 7 1  e



 2.5  10 1  e 14

t /  p 0

t /  p 0

 cm



(b) J n  eD n



 t  510 7 /  pO





cm 3

(ii) p 5 10 7  1.58 1014 cm 3 (b) (i) For 0  t  2 10



6

pt   2.5  10 1  e 14

At t  2 10 6 s,

s t /  p 0

 cm





 2.454 10 cm

pt   2.454 1014 e





3

For t  2 10 6 s,





 t  210 6 /  pO



cm 3

(ii) p 2 10 6  2.454 1014 cm 3 _______________________________________ 6.18 (a) For 0  t  2 10 6 s nt   g  n0 e t /  n 0

 

1/ 2



 5 1014 e t /  n 0 cm 3



eD n 2  1014 e  x / Ln Ln











 



 1.6 10 19 31.08 2 1014  x / Ln e 5.575 10 3



6.20 (a) p-type; p pO  10 14 cm 3 and





2

ni2 1.5 1010   2.25 10 6 cm 3 p pO 1014 (b) Excess minority carrier concentration n  n p  n pO n pO 

 10 21 5 10 7 e t /  n 0

d n  d  eD n 2  10 14 e  x / Ln dx dx

J n  0.1784e  x / Ln A/cm 2 Holes diffuse at same rate as minority carrier electrons, so J p  0.1784e  x / Ln A/cm 2 _______________________________________

3

6 7 p  2.5  1014 1  e  210  / 510 

14



(a) nx   px   2 1014 e  x / Ln cm 3

 1.58 1014 cm 3





Ln  Dn n 0  31.08 10 6  5.575 10 3 cm

p  2.5 1014 1  e 1 / 1 





 31.08 cm 2 /s

3

At t  5 10 s,

pt   1.58 1014 e

3

6.19 p-type; minority carriers - electrons  kT  Dn    n  0.02591200  e 



7

For t  5 10 7 s

 9.16 1012  9.16 1012 cm 3

(iii) n   51014 cm 3 _______________________________________

 2.7  10 cm (c)    p 0  1.35  10 8 s

pt   g  p 0 1  e





(ii) n 2 10 6  9.16 1012 cm 3

3

13





n  5 1014  9.16 1012 1  e t /  n 0

4

8



For t  2 10 6 s

 5.4 10  4 cm 3

At x  0 , n p  0 so that

n0  0  n pO  2.25 10 6 cm 3

Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (c) For the one-dimensional case, d 2 n  n Dn  0  nO dx 2 or d 2 n  n  2  0 where L2n  Dn nO dx 2 Ln The general solution is of the form x x   B exp  n  A exp  L   Ln   n  For x   , n remains finite, so B  0 . Then the solution is x  n  n pO exp   Ln  _______________________________________ 6.21

6.22 n-type, so we have d 2 p  d p  p Dp   po  0 dx  pO dx Assume the solution is of the form p  A expsx  Then d p  d 2 p   As expsx  ,  As 2 expsx  dx dx 2 Substituting into the differential equation D p As 2 exp sx    p  o As exp sx 



 

where Ln  Dn n 0  25 10

6



s2 

3

 5 10 cm d n  d J n  eD n  eD n 5  10 14 e  x / Ln dx dx eD   n 5  10 14 e  x / Ln Ln











1.6 10 255 10  e 5 10  19

14

 x / Ln

3

J n  0.4e  x / Ln A/cm 2 (a) For x  0 , n0  5 1014 cm 3



J p 0  0.4 A/cm 2

nLn   5 10 e J n Ln   0.4e

1

0

 po Dp

s

1 0 L2p

 p Lpo 2D p

Then

(b) For x  Ln  5 10 3 cm, 1

1

 pO

The solution for s is 2    p  1 p 4 s  o    o   2   Dp  2 Dp Lp     which can be rewritten as 2     p Lpo  1   p Lpo   s   1   2D p  L p  2D p     Define

J n 0  0.4 A/cm 2

14

0

Dividing by D p , we have

1/ 2



 pO

or

Dp s 2   po s 

nx   5 1014 e  x / Ln cm 3

A expsx 

 1.84 10 cm 14

3

 0.147 A/cm 2

J p L n   0.4e 1  0.147 A/cm 2

(c) For x  15 10 3 cm  3L n

n3Ln   5 1014 e 3  2.49 1013 cm 3 J n 3Ln   0.4e 3  0.020 A/cm 2

J p 3L n   0.4e 3  0.020 A/cm 2

_______________________________________

s

1 Lp

  1   2   

In order that p  0 as x   , use the minus sign for x  0 and the plus sign for x  0 . Then the solution is p  A exps  x  for x  0 p  A exps  x  for x  0 where 1  s    1   2   L p  _______________________________________

Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 6.23 Plot _______________________________________ 6.24 (a) From Equation (6.55) d 2 n d n  n Dn  no  0 dx  nO dx 2 or d 2 n   n d n  n  o  2 0 2 Dn dx dx Ln We have that Dn  kT     so we can define n  e 

n

o 1 o   kT e L  Dn Then we can write d 2 n  1 d n  n    2 0 L  dx dx 2 Ln

The solution is of the form n  n0 exp x  where   0 Then d n  d 2 n    n  and   2 n  dx dx 2 Substituting into the differential equation, we find 1 n  2 n    n  2  0 L Ln or



1    2 0  L Ln which yields 2   L  1  Ln    n   1  Ln  2 L   2L     We may note that if  o  0 , then L    2

1 and   Ln (b)

 kT  Ln  Dn nO where Dn   n    e  so

Dn  12000.0259  31.1 cm /s and 2

Ln 

31.15 10 7   39.4 10 4 cm

or

Ln  39.4  m

For  o  12 V/cm, then L 

kT e  0.0259  21.6 10  4 o

12

cm

and

  5.75 10 2 cm 1 (c) Force on the electrons due to the electric field is in the negative x-direction. Therefore, the effective diffusion of the electrons is reduced and the concentration drops off faster with the applied electric field. _______________________________________ 6.25 p-type so the minority carriers are electrons and n n  D n  2 n    n   n   g     nO t Uniform illumination means that n    2 n   0 . For  nO   , we are left with d n   g  which gives n  g t  C1 dt For t  0 , n  0  C1  0 Then n  G o t for 0  t  T For t  T , g   0 so that

d n  0 dt

And n  G o T (no recombination) _______________________________________ 6.26 n-type, so minority carriers are holes and p p  D p  2 p    p   p   g     pO t We have  pO   ,   0 , and

 p   0 (steady-state). Then we have t d 2 p  d 2 p  g   or Dp  g  0 2 2 Dp dx dx

For  L  x   L , g   G o = constant. Then G d p    o x  C1 dx Dp

Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ and

We find D p 10.42   0.02668 V  p 390.6

G p   o x 2  C1 x  C 2 2D p

For L  x  3L , g   0 so we have

d p 

d p   C 3 and  0 so that dx dx 2 p  C 3 x  C 4 For 3L  x   L , g   0 so that 2

d 2 p  2

d p   C 5 and dx

 0 so that

dx p  C 5 x  C 6

The boundary conditions are: (1) p  0 at x  3L (2) p  0 at x  3L (3) p continuous at x  L (4) p continuous at x   L

d p  continuous at x  L dx d p  (6) continuous at x   L dx Applying the boundary conditions, we find G p  o 5L2  x 2 for  L  x   L 2D p

(5)



Go L 3L  x  for 3L  x  L Dp _______________________________________

p 

6.27 V 8   20 V/cm L 0 .4 d 0.25 p    0 t 0 20  32  10  6

0 



 390.6 cm /V-s  p  0 2 t 2 Dp  16t 0



2



6.28 (a)

  x2 Assume that f x, t   4 Dt 1 / 2 exp  4Dt is the solution to the differential equation   2 f  f D 2    x  t To prove: we can write   x2  f 1 / 2   2 x    4 Dt    exp  x  4 Dt   4 Dt  and 2 f x

2

 4 Dt 





D p  10.42 cm /s 2

  x2   2 x   exp   4 Dt   4 Dt



   

       

Also 2 f 1 / 2   x  4 Dt   t  4D

  1    x2   2  exp   t   4Dt       x2  1 / 2   1    4 D   t 3 / 2 exp   2   4Dt  2 f Substituting the expressions for and x 2 f into the differential equation, we find t 0 = 0. Q.E.D. (b) Consider    x2  dx exp  4 Dt   



390.6202 9.35 10 6 2 16 32 10 6

2

1 / 2

  x2  2    exp  4 Dt   4 Dt



G L p  o 3L  x  for L  x  3L Dp



This value is very close to 0.0259 for T  300 K. _______________________________________

Let u  x 2 , then du  2 x  dx or du du dx   2x 2 u Let a  Now

1 4 Dt

Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 

  x2 exp  4 Dt 





2

2 0



 a

1 u



   x2 dx  2 exp   4 Dt   0



exp au du 



 0

1 u

6.31 (a) p-type

 dx   exp au du

 4Dt

or

Then 

4D t x  dx  1 exp  4 Dt 4  D t 4  D t    _______________________________________



2

1

p  E Fi  E F  kT ln  o   ni   5 1015    0.0259 ln  10   1.5 10 

6.29 Plot _______________________________________

E Fi  E F  0.3294 eV (b) n  p  51014 cm 3 and

no 



 10 cm 15



2

 4.5 10 4 cm 3

 n  n   E Fn  E Fi  kT ln  o   ni   4.5 10 4  5 1014    0.0259 ln   1.5 1010  

   

 4 1016    0.0259 ln  10   1.5 10   0.383225 eV (b) n  p  g  p 0  2  10 21 5  10 7



Then

6.30

n (a) E F  E Fi  kT ln  o  ni



ni2 1.5 1010  po 5 1015

or



3

 n  n   E Fn  E Fi  kT ln  o   ni   4 1016  1015    0.0259 ln  10   1.5 10   0.383865 eV  p  p   E Fi  E Fp  kT ln  o   ni   1015    0.0259 ln  10   1.5 10   0.28768 eV (c) E Fn  E F  0.383865  0.383225  0.000640 eV  0.640 meV or _______________________________________

E Fn  E Fi  0.2697 eV and  p  p   E Fi  E Fp  kT ln  o   ni 

 5 1015  5 1014    0.0259 ln   1.5 1010   or

E Fi  E Fp  0.3318 eV _______________________________________ 6.32 (a) For n-type, E Fn  E F  E Fn  E Fi   E F  E Fi 

 n  n  n   kT ln  o  kT ln  o  n  ni   i  n  n    kT ln  o   no   5 1015  n   So 0.00102  0.0259 ln  15   5 10   0.00102  5 1015  n  5 1015 exp   0.0259  Which yields n  21014 cm 3

   

Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

 n  n   (b) E Fn  E Fi  kT ln  o   ni   5 1015  2 1014    0.0259 ln   1.5 1010    0.33038 eV  p  (c) E Fi  E Fp  kT ln    ni   2 1014    0.0259 ln  10   1.5 10   0.2460 eV _______________________________________  n  (a) E Fn  E Fi  kT ln    ni   E  E Fi  or n  ni exp  Fn   kT 



 0.270   1.5 1010 exp    0.0259   5.05 1014 cm 3  p  p   (b) E Fi  E Fp  kT ln  o   ni   6 1015  5.05 1014    0.0259 ln   1.5 1010    0.33618 eV (c) (i) E F  E Fp  E Fi  E Fp  E Fi  E F 



n n  (a) (i) E Fn  E Fi  kT ln  o   ni   1.02 1016  0.0259 ln  6  1.8 10  0.58166 eV

   

 p  (ii) E Fi  E Fp  kT ln    ni   0.02 1016    0.0259 ln  6   1.8 10   0.47982 eV  1.11016   (b) (i) E Fn  E Fi  0.0259 ln  6   1.8 10   0.58361 eV

6.33



6.34



 p  p  p    kT ln  o   kT ln  o  n   ni   i   p o  p    kT ln    po  (ii) E F  E Fp  6 1015  5.05 1014    0.0259 ln   6 1015   3  2.093 10 eV or  2.093 meV _______________________________________

 0.11016   (ii) E Fi  E Fp  0.0259 ln  6   1.8 10   0.52151 eV _______________________________________ 6.35 Quasi-Fermi level for minority carrier electrons:  n  n   E Fn  E Fi  kT ln  o   ni 





2

n2 1.8 10 6 no  i   3.24 10  4 cm 3 po 1016 We have  x  n  1014    50  Then  3.24 10 4  1014 x 50  E Fn  E Fi  kT ln   1.8 10 6   We find

 



x (  m)

0 1 2 10 20 50

( E Fn  E Fi ) (eV) -0.581 +0.361 +0.379 +0.420 +0.438 +0.462



Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Quasi-Fermi level for holes: we have  p  p   E Fi  E Fp  kT ln  o   ni 

6.38

 p  (a) E Fi  E Fp  kT ln    ni   p   0.0259 ln   10  1.5 10 

We have p o  1016 cm 3 and n  p . We find x (  m)

p  1011 cm 3 , E Fi  E Fp  0.04914 eV

( E Fi  E Fp ) (eV)

10 12

0 +0.58115 50 +0.58140 _______________________________________

10 14 10 15

6.36 (a) We can write

p  E Fi  E F  kT ln  o   ni  and  p  p   E Fi  E Fp  kT ln  o   ni  so that E Fi  E Fp  E Fi  E F   E F  E Fp



10

0.10877 0.16841 0.22805 0.28768

13

 n  n   (b) E Fn  E Fi  kT ln  o   ni   2 1016  n    0.0259 ln  10   1.5 10  n  1011 cm 3 , E Fn  E Fi  0.365273 eV



 p  p  p    kT ln  o   kT ln  o  n   ni   i 

10 12

0.365274

10

13

0.365286

10

14

0.365402

15

10 0.366536 _______________________________________

or

 p  p    0.01kT E F  E Fp  kT ln  o   po  Then p o  p  exp0.01  1.010 po or p  0.010  low injection, so that po

p  51012 cm 3 (b)

 p  E Fn  E Fi  kT ln    ni   5 10    0.0259 ln  10   1.5 10  12

or E Fn  E Fi  0.1505 eV _______________________________________

6.37 Plot _______________________________________

6.39 (a)

R





C n C p N t np  ni2



C n n  n   C p  p  p 

np  n  2 i

 pO n  n    nO  p  p 

Let n   p   ni . For n  p  0

 ni2  ni   pO ni   nO ni  pO   nO (b) We had defined the net generation rate as g  R  g o  g   R o  R   R

where g o  Ro since these are the thermal equilibrium generation and recombination rates. If g   0 , then g  R   R  and

R 

n i  pO   nO

so that g  R  

ni  pO   nO

Thus a negative recombination rate implies a net positive generation rate. _______________________________________

Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 6.40 We have that C n C p N t np  ni2 R C n n  n   C p  p  p 





np  n  2 i



 pO n  ni    nO  p  ni  If n  n o  n and p  p o  n , then

no  n p o  n  ni2  pO no  n  ni    nO  p o  n  ni  2 no p o  nno  p o   n  ni2   pO no  n  ni    nO  p o  n  ni  2 If n  n i , we can neglect n : also R

no p o  ni2 Then

R

nno  p o   pO no  ni    nO  p o  ni 

(a) For n-type; n o  p O , n o  ni Then R 1   10  7 s 1 n  pO (b) For intrinsic, n o  p o  ni Then 2ni R   n  pO 2ni    nO 2ni 

At x   , p  g  pO so that B  0 , Then x  p  g  pO  A exp  Lp    We have d p  Dp  sp  dx x 0 x 0 We can write d p  A and p   g  pO  A  dx x 0 L p x 0 Then  AD p Lp

R



1

 pO   nO



The excess concentration is then    x  s  p  g  pO 1   exp  L p  Dp Lp  s    where



(c) For p-type; p o  no , p o  ni Then R 1 1    2  10  6 s 1 n  nO 5  10  7 _______________________________________ 6.41 (a) From Equation (6.56) d 2 p  p Dp  g 0 2  pO dx Solution is of the form x     B exp  x  p  g  pO  A exp  Lp   Lp     



L p  D p pO  Now p  10 21 10 7

 

1010 7   10 3 cm



   x  s    1   exp 3  L p   10 10  s  



1  10 7  5 10 7

R  1.67  10  6 s 1 n



Solving for A , we find  sg  pO A Dp s Lp

or

n



 s g  pO  A



or 

p  1014 1 

 (i) For s  0 ,

  x     exp  L p  10 4  s   s

p  1014 cm 3 (ii) For s  2000 cm/s,    x   p  1014 1  0.167 exp  L p     (iii) For s   ,    x   p  1014 1  exp  L p    

Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (b) (i) For s  0 ,

p0  1014 cm 3

n 

(ii) For s  2000 cm/s, p0  0.833 1014 cm 3 (iii) For s   , p0  0 _______________________________________ 6.42

Ln  Dn nO 

255 10 7 

 35.4 10 4 cm (a) At x  0 , g  nO  2 10 21 5 10 7  1015 cm 3 or n0  g  nO  1015 cm 3 For x  0 d 2 n  n d 2 n  n Dn   0   2 0  nO dx 2 dx 2 Ln The solution is of the form x x   B exp  n  A exp  L  L  n   n  At x  0 , n  n0  A  B At x  W ,



W  x    Ln 

n0 sinh 





 W   W    B exp    L  L  n   n  Solving these two equations, we find  n0 exp 2W Ln  A 1  exp 2W Ln  and n0 B 1  exp 2W Ln  Substituting into the general solution, we find n0 n    W    W    exp  exp   L    Ln   n 

n  0  A exp

   W  x     W  x     exp    exp     L n   Ln   which can be written as

W  sinh    Ln 

where n0  1015 cm 3 and Ln  35.4  m (b) If  nO   , we have

d 2 n  0 dx 2 so the solution is of the form n  Cx  D Applying the boundary conditions, we find x  n  n01    W _______________________________________ 6.43 For  pO   , we have

d 2 p 

0 dx 2 So the solution is of the form p  Ax  B At x  W d p   Dp  sp  dx x W x W or  D p A  s AW  B which yields A D p  sW  B s At x  0 , the flux of excess holes is d p  1019   D p  D p A dx x 0 so that  1019 A  1018 cm 4 10 and 1018 10  sW   1018  10  W  B s  s  The solution is now 10   p  1018 W  x   s  

Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 6 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (a) For s   ,

p  1018 20 10 4  x  cm 3

Then J p  eD p



d p  dx

 

  1.6 10 19 10  1018



or

J p  1.6 A/cm 2 (b) For s  210 3 cm/s,

p  1018 70 10 4  x  cm 3

Also J p  1.6 A/cm 2 _______________________________________ 6.44 For W  x  0 d 2 n  Dn  Go  0 dx 2 so that G d n    o x  C1 dx Dn and G n   o x 2  C1 x  C 2 2 Dn For 0  x  W , d 2 n  0 dx 2 so that n  C 3 x  C 4 The boundary conditions are (1) s  0 at x  W so that d n 0 dx x  W (2) s   at x  W so that nW   0 (3) n continuous at x  0 d n  (4) continuous at x  0 dx Applying the boundary conditions, we find G W G W 2 C1  C 3   o and C 2  C 4   o Dn Dn

Then for W  x  0 G n  o  x 2  2Wx  2W 2 2Dn and for 0  x  W G W n  o W  x  Dn _______________________________________





6.45 Plot _______________________________________ 6.48 (a) GaAs: V 2 R   10 6  I 2  10  6 L and   e  n   p p R  A





p  g  p 0  10 21 5  10 8   5  10 13 cm 3

For N d  1016 cm 3 , from Figure 5.3,

 n  7000 cm 2 /V-s,  p  310 cm 2 /V-s







  1.6 10 19 7000  310 5 1013

 0.05848 (  -cm) Let W  20  m





Then A  Wd  20 10 4 4 10 4 8

 80 10 cm

So R  10 6 



1



2

L 0.05848 80 10 8





2

Which yields L  4.68 10 cm (b) Silicon: R  10 6  , p  51013 cm 3 For N d  1016 cm 3 , from Figure 5.3,

 n  1300 cm 2 /V-s,  p  410 cm 2 /V-s







  1.6 10 19 1300  410 5 1013

 0.01368 (  -cm) Let W  20  m





Then A  Wd  20 10 4 4 10 4 8

 80 10 cm

So R  10 6 



1



2

L 0.01368 80 10 8



2



Which yields L  1.09 10 cm _______________________________________

Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 7 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

Chapter 7 (b) N d  51016 cm 3 , N a  51016 cm 3 Si: Vbi  0.778 V

7.1 N N Vbi  Vt ln  a 2 d  ni

(a)

   

Ge: Vbi  0.396 V







 2 1015 2  1015  (i) Vbi  0.0259  ln   2  1.5  1010   0.611 V  2 1015 2  1016  (ii) Vbi  0.0259  ln   2  1.5  1010   0.671 V  2 1015 2  1017  (iii) Vbi  0.0259  ln   2  1.5 1010   0.731 V (b)  2 1017 2  1015   (i) Vbi  0.0259  ln  2  1.5 1010   0.731 V  2 1017 2  1016  (ii) Vbi  0.0259  ln   10 2  1.5  10 











 0.790 V



































7.3 (a) Silicon ( T  300 K)  Na Nd Vbi  0.0259  ln  10  1.5  10







Si: ni  1.5 1010 cm 3 Ge: ni  2.4 1013 cm 3 GaAs: ni  1.8 10 6 cm 3   and Vt  0.0259 V  

(a) N d  1014 cm 3 , N a  1017 cm 3 ' Then Si: Vbi  0.635 V Ge: Vbi  0.253 V GaAs: Vbi  1.10 V

Ge: Vbi  0.432 V GaAs: Vbi  1.28 V _______________________________________

7.2

N N Vbi  Vt ln  a 2 d  ni

(c) N d  1017 cm 3 , N a  1017 cm 3 Si: Vbi  0.814 V



 2 1017 2  1017   (iii) Vbi  0.0259  ln  2  1.5  1010   0.850 V _______________________________________



GaAs: Vbi  1.25 V





2

  

3

For N a  N d  10 cm ; Vbi  0.4561 V 14

 1015

;

 10

;

16

 10 ; (b) GaAs ( T  300 K)  Na Nd Vbi  0.0259  ln   1.8  10 6 17



 0.5754 V  0.6946 V  0.8139 V



2

  

3

For N a  N d  10 cm ; Vbi  0.9237 V 14

 1.043 V  10 ;  1.162 V 17  10 ;  1.282 V (c) Silicon (400 K), kT  0.034533 ni  2.38 1012 cm 3  1015

;

16

For N a  N d  1014 cm 3 ; Vbi  0.2582 V

 0.4172 V  10  0.5762 V ; 17  10  0.7353 V ; 9 GaAs(400 K), ni  3.29 10 cm 3  1015

;

16

For N a  N d  1014 cm 3 ; Vbi  0.7129 V

 0.8719 V  10 ;  1.031 V 17  10 ;  1.190 V _______________________________________  1015 16

;

Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 7 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ or

7.4 (a) n-side

x p  0.0213  10 4 cm  0.0213  m

N E F  E Fi  kT ln  d  ni

     5 1015    0.0259 ln  10   1.5 10 

We have  max 



or E F  E Fi  0.3294 eV p-side N  E Fi  E F  kT ln  a   ni   1017    0.0259 ln  10   1.5 10  or E Fi  E F  0.4070 eV (b) Vbi  0.3294  0.4070 or Vbi  0.7364 V (c) N N  Vbi  Vt ln  a 2 d   ni 

  

 10 17 5  10 15  0.0259  ln  2  1.5  10 10





N E F  E Fi  kT ln  d  ni

     2 1016    0.0259 ln  10   1.5 10 

or



1/ 2



 1  17  10  5  1015 

  

1/ 2

E F  E Fi  0.3653 eV p-side N  E Fi  E F  kT ln  a   ni   2 1016    0.0259 ln  10   1.5 10  or E Fi  E F  0.3653 eV (b) Vbi  0.3653  0.3653 or Vbi  0.7306 V (c) N N  Vbi  Vt ln  a 2 d   ni 





Vbi  0.7305 V (d)



 1  17  10  5  1015 





or

x n  0.426 10 4 cm  0.426  m Now  211.7  8.85 10 14 0.736 xp   1.6 10 19   5  1015   17  10



 2 1016 2 1016  0.0259  ln  2  1.5 1010

or



14

7.5 (a) n-side

 211.7  8.85 10 14 0.736  1.6 10 19   1017   15  5  10

4

15

_______________________________________



  1    N  N   d   a

19

 max  3.29  10 4 V/cm

Vbi  0.7363 V (d)

 Na  N  d

1.6 10 5 10 0.426 10  11.78.85 10 

or

or

2 V x n   s bi  e

eN d x n s

  

1/ 2

2 V xn   s bi  e

 Na  N  d

  1    N  N  d   a 

1/ 2



Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 7 By D. A. Neamen Problem Solutions ______________________________________________________________________________________





 211.7 8.85 10 14 0.7305  1.6 10 19   2  1016   16  2  10

For 300 K;

 1   2  1016  2  1016 

  

1/ 2

x n  0.154 10 4 cm  0.154  m By symmetry x p  0.154  10 4 cm  0.154  m Now eN d x n s



7.8

1.6 10 2 10 0.1537 10  11.78.85 10  19

14

or _______________________________________







    0.330  exp   0.0259 









200 K; kT  0.017267 ; ni  1.38 cm 3

ni  1.8 10 6 cm 3

400 K; kT  0.034533 ; ni  3.28 10 9 cm 3 For 200 K;  2  1015 4  1016  Vbi  0.017267  ln   1.382  



 1.257 V



xp Nd  3 Na xn





2



 0.710  exp   0.0259 

N d  2.33 1016 cm 3

7.7 300 K; kT  0.0259 ;

3

which yields N a  7.766 1015 cm 3

or N a  5.12 1015 cm 3 (c)  5.12  1015 1.98  1016  Vbi  0.0259  ln   2   1.5  1010  0.695 V _______________________________________





So N d  3N a

or 3N a2  1.5 1010

 E  EF N a  ni exp Fi  kT



xn





or N d  1.98 1016 cm 3

 1.5 10

xp



 E  E Fi  (b) N d  ni exp F   kT   0.365   1.5 1010 exp   0.0259 

10

0.75xn  0.25x p 



 Na Nd  (a) Vbi  0.0259  ln   2  1.5  1010   3N a2  0.710  0.0259  ln   2 10  1.5 10 

7.6





xn  0.25W  0.25 xn  x p

xn N d  x p N a 

 max  4.75  10 4 V/cm









4

16







or

 max 



 2 1015 4  1016  Vbi  0.0259  ln   2   1.8  10 6  1.157 V For 400 K;  2 1015 4 1016  Vbi  0.034533 ln   2  3.28 10 9   1.023 V _______________________________________



 2  V x n   s bi  e



 Na  N  d

  1    N  N  d   a

1/ 2



 211.7 8.85 10 14 0.710  1.6 10 19  1  1     3 4 7 . 766  10 15  



 x n  9.93 10 6 cm or x n  0.0993  m





    

1/ 2



 211.7 8.85 10 14 0.710 xp   1.6 10 19  1  3     15  1   4 7.766  10



 2.979 10 5 cm or x p  0.2979  m



    

1/ 2

Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 7 By D. A. Neamen Problem Solutions ______________________________________________________________________________________



 max 





 211.7 8.85 10 14 0.6350  1.6 10 19 

Now eN d x n s

1.6 10 2.33 10 0.0993 10  11.78.85 10  19

 1016   15  10

4

16

14

 2 V x p   s bi   e

which yields N a  8.127 1015 cm 3





 213.1 8.85 10 1.180 xn   1.6 10 19  1  1     15  3   4 8.127  10



 1.324 10 5 cm or x n  0.1324  m

    

1/ 2





  

1/ 2

    

eN d x n s



1.6 10 10 0.8644 10  11.78.85 10  19



1/ 2

4

15

14

or  max  1.34  10 4 V/cm

_______________________________________

eN d x n s

7.10

1.6 10 2.438 10 0.1324 10   13.18.85 10  4

16

14

 4.45  10 4 V/cm _______________________________________



    

  1    N  N  d   a







 2 1017 4  1016   (a) Vbi  0.0259  ln  2  1.5  1010   0.80813 V (b) V bi increases as temperature decreases At T  300 K, we can write

ni2  1.5 1010

 1016 1015   (a) Vbi  0.0259  ln  2  1.5  1010  or Vbi  0.635 V (b)  Na  N  d

 1  16  10  1015 

x p  0.08644  10 4 cm  0.08644  m

 max 

 3.973 10 5 cm or x p  0.3973  m

 2  V x n   s bi  e



or



1  3     15  1   4 8.127  10

7.9

1/ 2

(c)



19

  1    N  N  d   a 

 1015   16  10

 213.1 8.85 10 14 1.180 xp   1.6 10 19 

 max 

 Nd  N  a

 211.7 8.85 10 14 0.6350  1.6 10 19 

N d  2.438 1016 cm 3



1/ 2

x n  0.8644 10 4 cm  0.8644  m Now



14

  

or

 3.58  10 4 V/cm (b) From part (a), we can write 2  1.180  3N a2  1.8 10 6 exp   0.0259 



 1  16  10  1015 







2













  1.12   K 2.8 1019 1.04 1019 exp   0.0259   K  4.659 At T  287 K, kT  0.024778 eV 1/ 2

 287  ni2  K 2.8 1019 1.04 1019    300 

3

  1.12   exp   0.024778 





 4.659 2.5496 10 38 2.3404 10 20 So n  2.780 10 2 i

19



Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 7 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ Then





7.12 (b) For N d  1016 cm 3 ,



 2 1017 4 1016  Vbi  0.024778 ln   19  2.780 10   0.82494 V We find Vbi 287   Vbi 300   100% Vbi 300 

N E F  E Fi  kT ln  d  ni

     1016    0.0259 ln  10   1.5 10 

0.82494  0.80813  100%  2.08% 0.80813  2% _______________________________________ 

or E F  E Fi  0.3473 eV

For N d  1015 cm 3

 1015   E F  E Fi  0.0259 ln  10   1.5 10 

7.11 N N Vbi  Vt ln  a 2 d  ni

or

   







16 15  T   4  10 2  10  0.550  0.0259   ln   2 ni  300    Using the procedure from Problem 7.10, we can write, for T  300 K,

    1.12   K 2.8 10 1.04 10 exp  0.0259

ni2  1.5 1010

2

19

19



 K  4.659 At T  300 K,







 4  1016 2 1015 Vbi  0.0259  ln  2  1.5  1010  0.68886 V For Vbi  0.550 V,  T  300 K At T  380 K, kT  0.032807 eV Also



n  4.659 2.8 10 2 i

19











N N (a) Vbi  Vt ln  a 2 d  ni

   

    

or



Vbi  0.456 V (b)  211.7 8.85 10 14 0.456 xn   1.6 10 19 



1.04 10  19

7.13

 1012 1016   0.0259  ln   2  1.5  1010 

 380     300 



 4 1016 2 1015  Vbi  0.032807  ln   24  4.112 10   0.5506 V  0.550 V _______________________________________



 1012   16  10

3

  1.12   exp   0.032807   4.112 10 24 Then

E F  E Fi  0.2877 eV Then Vbi  0.34732  0.28768 or Vbi  0.0596 V _______________________________________

 1  12  10  1016 

  

1/ 2

or

x n  2.43 10 7 cm (c)  211.7  8.85 10 14 0.456 xp   1.6 10 19 





 1016   12  10

or x p  2.43  10 3 cm

 1   12  10  1016   

1/ 2

Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 7 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (d)  max

7.15 eN d x n  s



 max

1.6 10 10 2.43 10  11.78.85 10  19

7

16

or  max  3.75  10 2 V/cm

_______________________________________ 7.14 Assume silicon, so



1/ 2





 11.7  8.85 10 14 0.0259  1.6 10 19  2  1.6 10 19 N d or



 1.676  10 L D   Nd 

5

   









(iii)

 10 ;

(iv)

 1017 ;

16

N d  1014 ;  max  0.443  10 4 V/cm (ii)

 1015 ;

 1.46 10 4 V/cm

(iii)

 1016 ;

 4.60  10 4 V/cm

 1017 ;  11.2 10 4 V/cm (iv) (b) (i) For N a  1014 , N d  1014 ; Vbi  0.4561 V

(ii)

 1015 ;

(c) N d  81017 cm 3 , L D  0.004577  m Now (a) Vbi  0.7427 V

(iii)

 10 ;

(iv)

 1017 ;

 8  1017    Nd



 1   8 1017  N  d

   

1/ 2

Then (a) x n  1.096  m (b) x n  0.2178  m (c) x n  0.02730  m Now L (a) D  0.1320 xn L (b) D  0.1267 xn LD  0.1677 xn _______________________________________

(c)

16

 0.5157 V  0.5754 V  0.6350 V

(i) For N a  1014 ,

N d  1014 ;  max  0.265  10 4 V/cm

(c) Vbi  0.9216 V Also  211.7  8.85  10 14 Vbi  xn   1.6  10 19 



 0.6946 V  0.7543 V  0.8139 V

(i) For N a  1017 ,

(b) N d  2.2 1016 cm 3 , L D  0.02760  m

(b) Vbi  0.8286 V

1/ 2



 1015 ;



(a) N d  81014 cm 3 , L D  0.1447  m

   

(ii)

1/ 2

1/ 2

 Na Nd  N N d  a

We find 2 1.6 10 19 2e   3.0904 10 7 s 11.7 8.85 10 14 (a) (i) For N a  1017 , N d  1014 ; Vbi  0.6350 V



14

  kT   L D   2s   e Nd 

 2eVbi   s

(ii)

 1015 ;

 0.38110 4 V/cm

(iii)

 1016 ;

 0.420 10 4 V/cm

(iv)

 1017 ;

 0.443  10 4 V/cm

 max increases as the doping increases, and the electric field extends further into the low-doped side of the pn junction. _______________________________________ (c)

7.16



   

 5  1016 1015 (a) Vbi  0.0259  ln  2  1.5  1010  0.6767 V



 2  V  V R   N a  N d  (b) W   s bi  N N e  a d 

   

1/ 2

Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 7 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ (i) For V R  0 ,







 211.7 8.85 10 14 0.6767  W  1.6 10 19   5  10 16  10 15  16 15  5  10 10



 

    

 4 1016   17  2 10

1/ 2



 2  V  V R   N a  N d  W   s bi  N N e  a d 



 211.7 8.85 10 14 0.6767  5 W  1.6 10 19   5  10 16  10 15  16 15  5  10 10



 



    

1/ 2



20.6767  5  4.15  10 4 V/cm 2.738  10  4 _______________________________________  max 



 2 1017 4  1016 (a) Vbi  0.0259  ln  2  1.5  1010  0.8081 V (b)



 2  V  V R   N a  x n   s bi N e   d

(c)  max 



1/ 2





 1.85 10 5 V/cm   e s N a N d (d) C  A   2Vbi  V R N a  N d  











1/ 2

7.18 N N (a) Vbi  Vt ln  a 2 d  ni

1/ 2

 0.2987 10 4 cm or x n  0.2987  m

 1   N  N d  a



 



1/ 2

  

 5.78 10 12 F or C  5.78 pF _______________________________________

  1    N  N  d   a

    



1/ 2

 1.6 10 19 11.7 8.85 10 14  2 10  4  20.8081  2.5 



 1   2 1017  4  1016 

  2  V  V R   N d x p   s bi N e   a 

2Vbi  V R  20.8081  2.5  W 0.3584  10  4

 2  10 17 4  10 16  17 16  2  10  4  10

 211.7 8.85 10 14 0.8081  2.5  1.6 10 19   2 10   16  4 10



    

 0.3584 10 cm or W  0.3584  m Also W  xn  x p  0.3584  m

20.6767   1.43  10 4 V/cm 4 0.9452  10 (ii)For V R  5 V,

17

1/ 2

4

 max 





   

 2  10 17  4  10 16  17 16  2  10 4  10

 2.738 10 cm or W  2.738  m 2Vbi  V R  (c)  max  W (i)For V R  0 ,



1/ 2

 211.7 8.85 10 14 0.8081  2.5  1.6 10 19 

4

7.17

    

 1   2 1017  4  1016 

 5.97 10 6 cm or x p  0.0597  m

 9.452 10 5 cm or W  0.9452  m

(ii) For V R  5 V,



 211.7 8.85 10 14 0.8081  2.5  1.6 10 19 

    

1/ 2

   

 80 N 2   Vt ln  2 d   ni  We find V 80 N d2  ni2 exp bi  Vt



 1.5 1010

   



2

 5.762 10 32

 0.740  exp   0.0259 

Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 7 By D. A. Neamen Problem Solutions ______________________________________________________________________________________  Vt ln 3  0.0259  ln 3  0.02845 V

 N d  2.684 1015 cm 3 N a  2.147 1017 cm 3 (b)  2  V  V R   N a  x n   s bi N e   d





  1    N  N  d   a

1/ 2



1/ 2

7.20



  1    N  N  d   a 

1/ 2

 211.7  8.85 10 14 0.740  10  1.6 10 19  1  1     17 80 2 . 147  10  2.684  10 15  

  

3 10 









   

1/ 2





4 10 4 10  15

17

 4 10  4 1017  15

9 1010  1.224 10 9 Vbi  V R  so that Vbi  VR   73.53 V which yields V R  72.8 V







 

 2.147  1017 2.684  1015  17 15  2.147  10  2.684  10



1/ 2

  

C   4.52 10 9 F/cm 2 _______________________________________

7.19 (a) Vbi 3N a   Vbi N a   N 3 N   N N  Vt ln  d 2 a   Vt ln  d 2 a  n i   n i

 N N   Vt ln 3  ln  d 2 a   n i 



or

1/ 2

 1.6 10 19 11.7 8.85 10 14  20.740  10 



 2 1.6  10 19 Vbi  V R   14  11.7  8.85  10



 9.38  10 4 V/cm



 max

5 2

  e s N a N d (d) C      2Vbi  V R N a  N d  



 2eVbi  V R   N a N d   N N s d  a 

or

2Vbi  V R   W 20.740  10   2.262  0.028310  4





 4 1015 4  1017  (a) Vbi  0.0259  ln   10 2  1.5 10  or Vbi  0.766 V Now

1/ 2

 2.83 10 6 cm or x p  0.0283  m

(c)  max

C 3 N a   3 N a     3  1.732 C N a   N a  (c) For a larger doping, the space charge width narrows which results in a larger capacitance. _______________________________________ 1/ 2

 2.262 10 4 cm or x n  2.262  m

  2  V  V R   N d x p   s bi N e   a 

1/ 2

So

 211.7  8.85 10 14 0.740  10  1.6 10 19  1  80       17 15  1   2.147  10  2.684  10 

 e s N a  (b) C      2Vbi  V R  

  Nd Na     Vt ln  2    n i 











  

  



 4  1016 4 1017 (b) Vbi  0.0259  ln  2  1.5 1010 or Vbi  0.826 V We have  2 1.6  10 19 Vbi  V R  2 3  10 5   14  11.7  8.85  10

so that Vbi  VR   8.008 V







 



4 10 4 10  16

17

 4 10  4 1017  16

Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 7 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ which yields V R  7.18 V

(b)





2VbiA  V R   A  W  A W B  VbiA  V R    B  2VbiB  V R  W  A VbiB  V R W B 



 4  10 4  10  (c) Vbi  0.0259  ln   10 2  1.5  10  or Vbi  0.886 V We have  2 1.6  10 19 Vbi  V R  2 3  10 5   14  11.7  8.85  10







17



17



 1  5.7543      3.13  5.8139 



or

 A  0.316 B  (c)

  4 10 4 10   17

17

 4 10  4 1017 

  s N a N dA      2 V  V N  N R a dA   biA

1/ 2

  s N a N dB    2VbiB  V R N a  N dB  

1/ 2

17

C j  A

so that Vbi  VR   1.456 V which yields V R  0.570 V _______________________________________

C j B 

7.21 (a)

 2 s VbiA  V R   N a  N dA    N N e a dA  

W  A  W B   2  V  V   N  N s biB R dB  a   N N e  a dB 

   

1/ 2

   

1/ 2

or

 N   dA  N dB

 VbiB  V R  N a  N dB    V  V  N  N R  a dA  biA

 1015   16  10

 5.8139  1018  1016      5.7543  1018  1015    

C j  A C j B 

   

1/ 2

1/ 2

 0.319

_______________________________________

or W  A  VbiA  V R  N a  N dA   N dB    W B   VbiB  V R  N a  N dB   N dA

We find

   

1/ 2

7.22 (a) We have C j 0 

    

 1018 1015  VbiA  0.0259  ln    0.7543 V 2  1.5  1010 

    

 1018 1016  VbiB  0.0259  ln    0.8139 V 2  1.5  1010  We find W  A  5.7543  1018  10 15      W B   5.8139  10 18  1016   10 16   15  10

or



W  A  3.13 W B 

   

1/ 2

C j 10 

or



 s N a N d     2Vbi N a  N d  

  s N a N d    2Vbi  V R N a  N d  

C j 0 

 V  VR  3.13   bi  C j 10   Vbi For V R  10 V, we find

3.132 Vbi

1/ 2

 Vbi  10

or Vbi  1.137 V (b) x p  0.2W  0.2 x p  xn





   

1/ 2

1/ 2

Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 7 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

N  0.25  d xn Na Now N N Vbi  Vt ln  a 2 d  ni so



16

15

   

C





 2  1016 5  1015 Vbi  0.0259  ln  2  1.8 10 6  1.162 V 1 C  Vbi  V R





 2 1015 4  1016 (a) Vbi  0.0259  ln  2  1.5  1010  0.6889 V









 







1/ 2



2 10 4 10   2 10  4 10  15

16

15

1/ 2

16

6.6457 10 12 1.157  V R

(i) For V R  0 , C  6.178 pF (ii) For V R  5 V, C  2.678 pF _______________________________________ 7.25





 2 1017 5 1015 Vbi  0.0259  ln  2  1.5 1010  0.7543 V





 

  e s N a N d (a) C  AC   A   2Vbi  V R N a  N d  







1/ 2



 1.6 10 19 11.7 8.85 10 14  8 10  4  20.7543  10 



  e s N a N d C  AC   A   2Vbi  V R N a  N d  



 1.6  1019 13.1 8.85  1014  5  10 4  21.157  VR  



1.162  V R 2 1.162  0.5 1.162  V R 2 1.50 2  1.662 which yields V R 2  2.58 V _______________________________________







1.50 



0.6889  V R

  e s N a N d C  AC   A   2Vbi  V R N a  N d  

C

1/ 2

16

 2 1015 4  1016 (b) Vbi  0.0259  ln  2  1.8  10 6  1.157 V





1/ 2



6.2806 10 12

(i) For V R  0 , C  7.567 pF (ii) For V R  5 V, C  2.633 pF



Vbi  V R 2 C V R1  So  C V R 2  Vbi  V R1

7.24



15







2 10 4 10    2 10  4 10 

 0.25 N a2  1.137  0.0259  ln   2  1.8  10 6  We can then write  1.137  1.8 10 6 Na  exp   0.25  20.0259  which yields N a  1.23 1016 cm 3 and N d  3.07 1015 cm 3 _______________________________________

7.23



 1.6 10 19 11.7 8.85 10 14  5 10  4  20.6889  V R  

Then xp

2 10 5 10    2 10  5 10  17

17

C  4.904 10 12 F 1 1 f  L 2 C 2 f  2 LC

15

15



1/ 2

Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 7 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

L



4.904 10 2 1.25 10  12

2

6







 1.6 10 19 13.1 8.85 10 14  10  4  2Vbi  2 

1

4 N  

 3.306 10 3 H  3.306 mH

2 d

1



2 3.306 10

3

0.6 10 12  2.724 10  20

12.14 10  12

1/ 2

1





2 3.306  10 3 6.704  10 12



1/ 2

 1.069 10 6 Hz  1.069 MHz _______________________________________

N a  6.016 1015 cm 3 , Vbi  1.10 V (b) From part (a), 0.6 10 12  2.724 10  20

7.26  2eVbi  V R N d   max    s   Let Vbi  0.75 V

(a)

N a  1.19 1016 cm 3 , Vbi  1.135 V _______________________________________

5 2

19

14

 N d  1.88 10 cm

 

(b) 10

d

  

7.28

3



 2 1.6  10 19 0.75  10 N d    11.7  8.85 10 14  





2 V x p   s bi  e

3

 N d  3.0110 cm _______________________________________ 15

7.27



x p  0.20W  0.20 xn  x p

0.8x p  0.2xn xn  4x p

1/ 2



 1  14  10  5  1015 

or x p  5.32  10 6 cm

 N a  4N d

Also

  







  1    N  N  d   a 

 Nd  N  a

 1014   15  5  10

 

 4 N d2  0.0259  ln   1.8  10 6

   

 211.7 8.85 10 14 0.5574  1.6 10 19 



N a x p  N d xn  N d 4x p N N (a) Vbi  Vt ln  a 2 d  n i



 5  1015 1014 (a) Vbi  0.0259  ln  2  1.5  1010 or Vbi  0.5574 V (b)

5 2



Nd Vbi  5

By trial and error, N d  2.976 1015 cm 3 ,

1/ 2

2.5 10   21.6  10 0.75  10 N  11.7 8.85 10   16

Nd Vbi  2

By trial and error, N d  1.504 1015 cm 3 ,

 7.94  10 Hz  0.794 MHz (ii) For V R  5 V, C  6.704 pF 5

f 

1/ 2

5 N d 

(b) (i) For V R  1 V, C  12.14 pF

f 





2

2 V x n   s bi  e

  

  e s N a N d C  AC   A   2Vbi  V R N a  N d  

1/ 2

 Na  N  d

 1   N  N d  a

   

1/ 2

  

1/ 2

Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 7 By D. A. Neamen Problem Solutions ______________________________________________________________________________________





 211.7 8.85 10 14 0.5574  1.6 10 19   5  1015   14  10

7.30

 1  14  10  5  1015 

  

1/ 2



50 10 

4 2





1/ 2





C





1/ 2

1.287 10 13 Vbi  V R

(i) For V R  1 V, C  9.783 10 14 F (ii) For V R  3 V, C  6.663 10 14 F (iii) For V R  5 V, C  5.376 10 14 F _______________________________________ 7.31

 



 8  10 16 N d (a) Vbi  0.0259  ln  2  1.8  10 6

8 10 N  1.8 10 

6 2

16

d

1/ 2



   1.20 

 1.20  exp   0.0259 

 N d  5.36 1015 cm 3



V 10 

211.7 8.85 10

1.6 10

19

14

R

14

which yields V R  193 V (b) xp Nd N    x n  x p  a  xn Na  Nd  so  1014  x n  50 10  4  16   10  4  0.50 10 cm  0.50  m (c) 2V 2193.15  max  R  W 50.5  10  4 or  max  7.65  10 4 V/cm







 11.7 8.85 10 14 2 1015

1/ 2

7.29 An n  p junction with N a  1014 cm 3 , (a) A one-sided junction and assume V R  Vbi . Then







 5  1015  1   14    14 15    10  10  5  10  which becomes 9 10 6  1.269 10 7 Vbi  V R  We find V R  70.4 V _______________________________________

or



 1.6 10 19  10 5   2Vbi  V R 

 211.7  8.85  10 14 Vbi  V R  30  10  4   1.6  10 19 

2 V  xp   s R   eN a 



 e s N d  (b) C  AC   A     2Vbi  V R  

or

x n  2.66 10 4 cm (c) For x n  30  m, we have



 2 1017 2  1015 (a) Vbi  0.0259  ln  2  1.5 1010  0.7305 V



_______________________________________

  e s N a N d (b) C  AC   A   2Vbi  V R N a  N d  



1/ 2



 1.6 10 19 1.10 10 12  A  21.20  1.0

13.18.85 10 14 8 1016 5.36 1015  

8 10

16

 5.36  10 15

 A  7.56 10 5 cm 2



1/ 2

 





 1.6 10 19 (c) 0.80 10 12  7.56 10 5   2Vbi  V R 





13.18.85 10 14 8 1016 5.36 1015 

8 10

16

 5.36  10 15

1.0582 10 8 



1/ 2

 

2.1585 10 8 Vbi  V R

 Vbi  V R  4.161  1.20  V R

V R  2.96 V _______________________________________

Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 7 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 7.32 Plot _______________________________________ 7.33 N N (a) Vbi  Vt ln  aO 2 dO  ni (c) p-region eN d  x     aO dx s s or eN x    aO  C1 s We have

   

  0 at x   x p  C1  



eN aO x  xp s

eN dO x eN dO  x    x n  O  2 s s  2  _______________________________________ 1 

7.34

d 2 x   x  dx    2 s dx dx For 2  x  1  m,  x   eN d So eN d x d eN d    C1 dx s s (a)

eN aO x p s

Then for  x p  x  0 

Then for 0  x  x O we have



n-region, 0  x  x O

d 1  x  eN dO   dx s 2 s

At x  2  m   xO ,   0 So eN d x O C1  s Then eN d x  x O   s

At x  0 , 0  x  1  , so 0  

or eN dO x 1   C2 2 s

n-region, x O  x  x n

d 2  x  eN dO   dx s s

or 2 

eN dO x  C3 s

We have  2  0 at x  x n  C3  

eN dO x n s

so that for x O  x  x n , we have 2  

eN dO x n  x  s

We also have  2  1 at x  x O Then eN dO x O eN  C 2   dO x n  x O  2 s s which gives eN  x  C 2   dO  x n  O  s  2 



eN d  1  210  4 s

1.6 10 5 10  110  11.78.85 10  19

15

4

14

or

0  7.726 10 4 V/cm (c) Magnitude of potential difference is eN d x  x O dx   dx  s





 x2     2  xO  x   C 2   Let   0 at x   xO , then 

eN d s

 x O2  eN d x O2 2    x  C  C  O  2 2  2 2 s   Then we can write eN d x  x O 2   2 s At x  1  m 0

eN d s

1 

1.6 10 5 10  1  210  211.7 8.85 10  19

15

4 2

14

or

1  3.863 V Potential difference across the intrinsic region

Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 7 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

 2  0  d  7.726  10 4 2  10 4 

Then

 2  15.45 V By symmetry, the potential difference across the p-region space-charge region is also 3.863 V. The total reverse-bias voltage is then V R  23.863  15.45  23.2 V _______________________________________ 7.35 (a) V B 

s  2eN B

2 crit

or



 

 2 11.7 8.85 10 14 4 10 5 N B  s crit  2eV B 2 1.6 10 19 40 Then



N B  N a  1.294 10 cm 16

(b) N B 





 2 1.6  10 19 Vbi  V R  4  10 5    11.7  8.85  10 14

or



2

   2  10 2  10     16

16

1/ 2

 16 16   2  10  2  10    Vbi  V B  51.77 V

So V B  51.04 V (b)





 5  1015 5  1015 Vbi  0.0259  ln  2  1.5  1010  0.6587 V Then  2 1.6  10 19 Vbi  V R  4  10 5    11.7  8.85  10 14



3









   5  10 5  10     15

11.78.85 10 14 4 10 5 2 21.6 10 19 20



15

1/ 2

 15 15   5  10  5  10    Vbi  V R  207.1

Or N B  N a  2.59 1016 cm 3 _______________________________________

So V R  206 V _______________________________________

7.36

7.39 For a silicon p  n junction with

Na 



 

s  11.7 8.85 10 4 10  2eV B 2 1.6 10 19 80 2 crit

14





5 2

 6.47 10 cm 3 _______________________________________ 15

N d  51015 cm 3 and V B  100 V, then, neglecting V bi we have  2 s V B  xn     eN d 

7.37 (a) For N d  1016 cm 3 , from Figure 7.15, V B  75 V



1/ 2



 211.7  8.85  10 14 100     19 5  10 15   1.6  10





1/ 2



(b) For N d  1015 cm 3 , V B  450 V _______________________________________

x n min   5.09 10 4 cm  5.09  m _______________________________________

7.38 (a) From Equation (7.36),

7.40 We find

 2eVbi  V R   N a N d   max   N N s  d  a Set  max   crit and V R  V B





   

1/ 2



 2 1016 2  1016  Vbi  0.0259  ln   10 2  1.5 10   0.7305 V





or

    

 1018 1018  Vbi  0.0259  ln    0.933 V 2  1.5 1010  Now eN d x n  max  s so

Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 7 By D. A. Neamen Problem Solutions ______________________________________________________________________________________

1.6 10 10 x 11.78.85 10  19

10 6 

18

n

14

which yields x n  6.47 10 6 cm Now  2  V  V R   N a  1   x n   s bi    e   N d  N a  N d Then  211.7 8.85 10 14 2 6.47 10 6   1.6 10 19 







   

1/ 2



 10 18  1   Vbi  V R  18  18  18   10  10  10  which yields Vbi  V R  6.468 V or V R  5.54 V _______________________________________

7.41 Assume silicon: For an n  p junction  2  V  V R   x p   s bi  eN a   Assume Vbi  V R

(a) For x p  75  m

75 10 

4 2



4 2



1.6 10





3  eax O3 ea   x O  x O3   C 2  C 2   2 s  3 3 s 

Then



19

 eax O ea  x 3   xO2  x     3 2 s  3 s  _______________________________________ 3

 x   

 10  15

7.44 We have that



V 10 

211.7 8.85 10





0

which yields V R  4.35 10 3 V

150 10 



1/ 2

211.7 8.85 10 14 V R

(b) For x p  150  m

7.43 (a) For the linearly graded junction  x   eax Then d  x  eax   dx s s Now eax ea x 2  dx    C1 s s 2 At x   x O and x   xO ,   0 So 2 2 ea  x O  ea  x O  0  C  C   1 1 s  2  s  2  Then ea  x 2  x O2 2 s (b)  ea  x 3 2  x    dx     xO  x  C 2 2 s  3  Set   0 at x   xO , then

19

14

R

1.6 10 which yields V R  1.74 10 4 V Note: From Figure 7.15, the breakdown voltage is approximately 300 V. So, in each case, breakdown is reached first. _______________________________________ 15

7.42 Impurity gradien 2 1018 a  10 22 cm 4 4 2 10 From Figure 7.15, V B  15 V _______________________________________

 ea 2s  C    12Vbi  V R   Then

1/ 3

7.2 10   a 1.6  10 11.7 8.85  10   9 3

19

14

2

  

120.7  3.5  which yields a  1.110 20 cm 4 _______________________________________

Semiconductor Physics and Devices: Basic Principles, 4th edition Chapter 7 By D. A. Neamen Problem Solutions ______________________________________________________________________________________ 7.45  e s N a  (a) C j  AC   A     2Vbi  V R  

Let N a  51015 cm 3