Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 1
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
Chapter 1
Exercise Solutions
Ex 1.1
(a) Number of atoms per unit cell
1
1
= 8 +  6 = 4
8
2
4
4
(b) Volume Density = 3 =
a
4.25 10 −8
(
TYU 1.2
(a) Number of atoms per (100) lattice plane
1
= 4 = 1
4
1
1
Surface Density = 2 =
2
a
4.65 10 −8
(
)
3
= 5.21  10 22 cm −3
_______________________________________
Ex 1.2
Intercepts of plane; p=1, q=2, s=2
1 1 1 
Inverse;  , , 
1 2 2 
Multiply by lowest common denominator,
 (211 ) plane
_______________________________________
Ex 1.3
(a) Number of atoms per (100) plane
1
= 1+ 4 = 2
4
2
2
Surface Density = 2 =
a
4.25 10 −8
(
)
−2
= 4.62  10 cm
(b) Number of atoms per (110) lattice plane
1
= 4 = 1
4
Surface Density
1
1
=
=
(a) a 2 4.65 10 −8 2 2
14
( ) (
)
−2
= 3.27  10 cm
(c) Number of atoms per (111) lattice plane
1 1
= 3 =
6 2
1
Lattice plane area = bh
2
14
where b = a 2
( )

h=a 2

)
2
2
2
1
 
− a 2 
2
 
1/ 2
1/ 2
Test Your Understanding Solutions
1 
3

=  2a 2 − a 2  = a
2 
2

Then lattice plane area
 3
1
 = 3 a2
= a 2 a
 2
2
2


Surface Density
1
2
=
= 2.67 10 14 cm −2
2
3
4.65 10 −8
2
_______________________________________
TYU 1.1
TYU 1.3
= 1.11  10 15 cm −2
(b) Number of atoms per (110) plane
1
1
= 2 + 4 = 2
2
4
Surface Density
2
2
=
=
(a) a 2 4.25 10 −8
( ) (
( )
)
2
2
−2
= 7.83  10 cm
_______________________________________
14
Number of atoms per unit cell = 8 
Volume Density = 4  10 22 =
1
=1
8
1
a3
o
 a = 2.92  10 −8 cm = 2.92 A
o
Radius = r = a 2 = 1.46 A
_______________________________________
(
)
o
(a) For (100) planes, distance = a = 4.83 A
(b) For (110) planes, distance
o
a 2 (4.83) 2
=
=
= 3.42 A
2
2
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 1
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
TYU 1.4
(a) 8 corner atoms
(b) 6 face-centered atoms
(c) 4 atoms totally enclosed
_______________________________________
TYU 1.5
Number of atoms in the unit cell
1
1
= 8 + 6 + 4 = 8
8
2
8
8
Volume Density = 3 =
a
5.43 10 −8
(
)
3
= 5 10 22 cm −3
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 1
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Chapter 1
Problem Solutions
1.1
(a) fcc: 8 corner atoms  1 / 8 = 1 atom
6 face atoms  1 / 2 = 3 atoms
Total of 4 atoms per unit cell
(b) bcc: 8 corner atoms  1 / 8 = 1 atom
1 enclosed atom
=1 atom
Total of 2 atoms per unit cell
(c) Diamond: 8 corner atoms  1 / 8 = 1 atom
6 face atoms  1 / 2 = 3 atoms
4 enclosed atoms = 4 atoms
Total of 8 atoms per unit cell
_______________________________________
1.2
(a) Simple cubic lattice: a = 2r
3
Unit cell vol = a 3 = (2r ) = 8r 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

Ratio =
(2) 4 r
3
3

 4r 




 3
(d) Diamond lattice
3




 100 % = 68 %
8
Body diagonal = d = 8r = a 3  a =
 8r 

Unit cell vol = a 3 = 

 3
3
r
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=
= 16 2  r 3
 4 r
4 atoms per cell, so atom vol = (4)
 3
Then
3


(4) 4 r 
 3 
 100 % = 74 %
Ratio =
16 2  r 3
(c) Body-centered cubic lattice
4
d = 4r = a 3  a =
r
3
 4

 r 
Unit cell vol = a = 
 3 




Then
3




3
r
o
a 3 (5.43) 3
=
= 1.176 A
8
8
Center of one silicon atom to center of
Then r =
o
nearest neighbor = 2r = 2.35 A
(b) Number density
8
=
= 5 10 22 cm −3
3
5.43 10 −8
(c) Mass density
N ( At.Wt .) 5 10 22 (28 .09 )
==
=
NA
6.02  10 23
(
)
(
)
  = 2.33 grams/cm 3
_______________________________________
3
3
 4 r 3
2 atoms per cell, so atom vol = (2)
 3
8




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
(
o
(b) a = 2(1.035 ) = 2.07 A
(c) A-atoms: # of atoms = 8 
)
3
Density =
 Density of Ga atoms = 2.22  10 22 cm −3
4 As atoms per unit cell
 Density of As atoms = 2.22  10 22 cm −3
(b) 8 Ge atoms per unit cell
8
Number density =
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
o
= 9.007 A
3
_______________________________________
1.8
(a) 2(1.035 ) 2 = 2(1.035 ) + 2rB
o
rB = 0.4287 A
−8 3
(1.0974 10 )(12.5)
22
6.02 10 23
= 0.228 gm/cm 3
(b) a =
4r
3
o
= 5.196 A
1
# of atoms 8  + 1 = 2
8
2
(5.196 10 )
Mass density =  =
= 4.503 A
2(4r )
(4.5 10 )
−8 3
= 1.4257  10 22 cm −3
o
= 5.515 A
(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
(1.4257 10 )(12.5)
22
6.02 10 23
= 0.296 gm/cm 3
_______________________________________
1.10
From Problem 1.2, percent volume of fcc
atoms is 74%; Therefore after coffee is
ground,
Volume = 0.74 cm 3
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 1
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
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
o
(a) a 3 = 2(2.2 ) + 2(1.8) = 8 A
o
Then a = 4.62 A
Density of A:
1
=
= 1.01 10 22 cm −3
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  10 14 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  10 14 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
Surface Density =
1
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  10 14 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
(c) d =
= 3.048 A
3
_______________________________________
(b) d =
1.19
(a) Simple cubic
(i) (100) plane:
Surface density =
1
1
=
2
a
4.73 10 −8
(
)
2
(ii) (110) plane:
1
a
2
2
= 3.16  10 14 cm −2
(iii) (111) plane:
Area of 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
So h =
(4.73) = 5.793 A
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  10 15 cm −2
_______________________________________
= 4.47  10 14 cm −2
Surface density =
Area of plane
1
= 6.68923 10 −8 5.79304 10 −8
2
= 19 .3755  10 −16 cm 2
1
3
6
Surface density =
19 .3755 10 −16
= 2.58  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  10 14 cm −2
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 1
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(c) (111) plane:
Surface density =
2
( 3 2)(5.43 10 )
−8 2
= 7.83  10 14 cm −2
_______________________________________
1.21
a=
4r
2
=
4(2.37 )
o
= 6.703 A
2
1
1
8 + 6
4
8
2 =
3
(a) #/cm =
a3
6.703  10 −8
(
5 10 17
100 % = 10 −3 %
5 10 22
2 10 15
(b)
100 % = 4 10 −6 %
5 10 22
_______________________________________
)
1.25
(a) Fraction by weight
2  10 16 (10 .82 )

= 1.542  10 − 7
22
5  10 (28 .06 )
(b) Fraction by weight
10 18 (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 =
2
a 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  10 14 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 a o = 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 a o = 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
Exercise Solutions
______________________________________________________________________________________
Chapter 2
Exercise Solutions
Ex. 2.1
(a) E = h =
hc

(b) E n =
(6.625 10 )(310 )
=
−34
10
=
100 10 −8
= 1.9875  10 −17 J
1.9875 10 −17
or E =
= 124 eV
1.6 10 −19
hc 6.625 10 −34 3 10 10
(b) E =
=

4500 10 −8
= 4.417  10 −19 J
4.417 10 −19
or E =
= 2.76 eV
1.6 10 −19
_______________________________________
(
)(
)
 2 2 n 2
2ma 2
(1.054 10 )  n
2(1.67  10 )(12  10 )
−34 2
− 27
2
2
−10 2
= 2.28  10 −23 n 2 J
2.27967 10 −23 n 2
1.6 10 −19
= 1.425  10 −4 n 2 eV
or E n =
Then E1 = 1.425 10 −4 eV
E 2 = 5.70 10 −4 eV
E 3 = 1.28 10 −3 eV
_______________________________________
Ex 2.4
Ex 2.2
(
)(
)(
= 2 9.11 10 −31 12 10 −3 1.6 10 −19
)
1/ 2
= 5.915  10 −26 kg-m/s
=
h 6.625  10 −34
=
= 1.12  10 −8 m
p 5.915  10 − 26
o
or  = 112 A
h 6.625 10 −34
(c) p = =

112 10 −10
= 5.915  10 −26 kg-m/s
(
)
2
1 p 2 1 5.915  10 −26
=
2 m
2 2.2  10 −31
= 7.952  10 −21 J
−21
7.952 10
or E =
= 4.97 10 − 2 eV
1.6 10 −19
_______________________________________
E=
Ex 2.3
(1.054 10 )  n
(9.11 10 )(12 10 )
−34 2
2
)( )
=
2(9.11 10 )(2)(4.555 10 )
−21 1 / 2
−31
1.054  10 −34
or
k 2 = 1.222 10 9 m −1
P = exp − 2k 2 d 
o
(a) d = 10 A = 10 10 −10 m
P = exp − (2) 1.222 10 9 10 10 −10
or
P = 0.0868  8.68 %
 (
)(
)
o
 2 2 n 2
(a) E n =
2ma 2
=
(
2
1
1
m 2 = 9.11 10 −31 10 5
2
2
= 4.555  10 −21 J
Now
2m
(Vo − E ) Set V o = 3E
k2 =
2
Then
1
k2 =
2m(2 E )

E=
(a) p = 2mE
− 31
2
2
−10 2
= 4.179  10 −20 n 2 J
4.179 10 −20 n 2
or E n =
= 0.261n 2 eV
1.6 10 −19
Then
E1 = 0.261 eV, E 2 = 1.045 eV, E 3 = 2.351 eV
(b) d = 100 A = 100 10 −10 m
P = exp − (2) 1.222 10 9 100 10 −10
or
P = 2.43  10 −11  2.43  10 −9 %
 (
)(
)
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 2
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
Ex 2.5
(
2m(VO − E )
(a) k 2 =
=
TYU 2.2
(a) E = (0.8) 1.6 10 −19 = 1.28 10 −19 eV

(
)
(
2 9.11 10 −31 (1.2 − 0.12 ) 1.6 10 −19
(1.054 10 )
t =
)
− 34 2
= 5.3236  10 m −1
Then
 0.12  0.12 
T  16
1 −

1 .2 
 1.2 
9
TYU 2.3
(a) k 2 =
 (
)(
 exp − 2 5.3236 10 9 5 10 −10
T  7.02  10
)
−3
 0.12  0.12 
(b) T  16
1 −

1 .2 
 1.2 
 (
)

1.054 10 −34
=
= 8.23 10 −16 s
E 1.28 10 −19
(b) Same as part (a), t = 8.23  10 −16 s
_______________________________________
2
=
2m(VO − E )
(
2
)
(
)
)
)(
)
(
)
)
2 9.11 10 −31 (0.8 − 0.1) 1.6 10 −19
(1.054 10
− 34 2
= 4.286  10 m −1
 0.1  0.1 
T  16
1 −

 0.8  0.8 
9
)(
 exp − 2 5.3236 10 9 25 10 −10
)
−12
T  3.97  10
_______________________________________
Ex 2.6
 (
 exp − 2 4.2859 10 9 12 10 −10
T  5.97  10
(b) k 2 =
From Example 2.6, we have
−13 .58
−0.0992
En =
=
eV
2 2
n2
(11 .7 ) n
−5
(
)
2 9.11  10 −31 (1.5 − 0.1) 1.6 10 −19
(1.054 10
= 6.061  10 m −1
 0.1  0.1 
T  16
1 −

 1.5  1.5 
− 34 2
9
E1 = −99 .2 meV, E 2 = −24 .8 meV,
E 3 = −11 .0 meV
_______________________________________
 (
)(
 exp − 2 6.061 10 9 12 10 −10
)
−7
T  4.79  10
_______________________________________
Test Your Understanding
TYU 2.1
 1.054 10 −34
(a) p =
=
x
8 10 −10
= 1.318  10 −25 kg-m/s
 d  p 2 
dE
   p
 p =  
(b) E =

dp
 dp  2m 
2p
p p
=
 p =
2m
m
−23
1.2 10
1.318 10 −25
E =
9.11 10 −31
= 1.735  10 −18 J or = 10.85 eV
_______________________________________
(
)(
)
TYU 2.4
T = 5  10 −6
 0.08  0.08 
= 16
1 −
 exp (− 2k 2 a )
0.8 
 0.8 
so that exp (+ 2k 2 a ) = 2.88 10 5
2k 2 a = 12 .571
k2 =
(
)
(
2 9.11 10 −31 (0.8 − 0.08 ) 1.6 10 −19
(1.054 10 )
)
− 34 2
= 4.3467  10 m −1
Then
12.571
a=
= 1.446 10 −9 m
2 4.3467 10 9
9
(
)
o
or a = 14.46 A
_______________________________________
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 
_______________________________________
h
(a) p =
=
2.7
(a) (i)
(
) (
p = 2mE = 2 9.11 10 −31 (1.2) 1.6 10 −19
−25
= 5.915  10
=
kg-m/s
−34
h 6.625  10
=
= 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
= 1.87  10
(
)
Gold: E = 4.90 eV = (4.90 ) 1.6 10 −19 J
So,
6.625  10 −34 3 10 10
=
= 2.54 10 −5 cm
(4.90 ) 1.6 10 −19
or
 = 0.254  m
(
(
)(
)
)
(
)
Cesium: E = 1.90 eV = (1.90 ) 1.6 10 −19 J
So,
6.625  10 −34 3 10 10
=
= 6.54  10 −5 cm
(1.90 ) 1.6 10 −19
or
 = 0.654  m
_______________________________________
(
(
)(
)
)
=
−24
)
kg-m/s
−34
6.625 10
= 3.54 10 −10 m
− 24
1.8704 10
o
or  = 3.54 A
(
)
= 5.915  10
−24
(
(iii) p = 2 9.11 10 −31 (120 ) 1.6 10 −19
=
−34
kg-m/s
6.625 10
= 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
h 6.625  10 −34
= =
= 6.225  10 −9 m
p 1.064  10 − 25
(
2
)
)(
)
−27
o
_______________________________________
2.9
or  = 909 A
_______________________________________
2.11
hc
p
(a) E = h =
Now
p2
h
1  h

 Ee =
E e = e and p e =
e
2m   e
2m




Then
1  h

=
 p 2m   e
hc
2

1  10 h 
 =

2m   p 

2
which yields
100 h
p =
2mc
hc
=
hc
2mc 2
 2mc =
100 h
100
)(
2 9.11  10 −31 3  10 8
100
)
2
= 1.64  10 −15 J = 10.25 keV
_______________________________________
hc

= 1.99  10
2
Set E p = Ee and  p = 10e
(
)
kg-m/s
h 6.625  10 −35
= =
− 9.09  10 −8 m
p 7.288  10 − 27
o
=
)(
)(
= 7.288  10
 = 62.25 A
p
=
(
or
Ep = E =
h
(
Now
E p = h p =
p=
=
(6.625 10 )(3 10 )
−34
8
110 −10
−15
J
Now
E 1.99 10 −15
=
e 1.6 10 −19
V = 1.24  10 4 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 
p =
2p
p p
 p =
2m
m
=
Now p = 2mE
(
) (
= 2 9 10 −31 (16 ) 1.6 10 −19
= 2.147  10
so E =
)
−24
kg-m/s
2.1466 10 −24 8.783 10 −26
(
)(
9 10
= 2.095  10 −19 J
)
−31
2.095 10 −19
= 1.31 eV
1.6 10 −19
(b) (i) p = 8.783 10 −26 kg-m/s
or E =
(
) (
)
= 5.06  10 −23 kg-m/s
(5.06 10 )(8.783 10 )
E =
−26
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 =
(
)
 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
(ii) p = 2 5 10 −28 (16 ) 1.6 10 −19
−23
2.15
(a) Et = 
1.054  10 −34
t =
= 8.23  10 −16 s
(0.8) 1.6 10 −19
(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
+2 1 
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 

+


2m  2 x 2
1 2 x x 
1 2
= j 

2 t
2
2
2.19

−
1  2
1 2


+ V (x ) = j 

2
2m 2 x
2 t
Subtracting these last two equations, we obtain
 2
− 2
2

 1
− 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
2
2
    dx = 1
Note that
=
2
ao
−1
2
 x 
cos 2 
dx = 1
 2 
 − ao

 2
ao
4
ao
2
=
ao
=

 x sin(2nx )  +1 / 2
A  +
=1
4n  −1 / 2
2
 1  1 
1
A  −  −  = 1 = A 2  
2
 4  4 
2
or A = 2
_______________________________________
 − 2x 
 dx
a o 
 exp 
ao
4
 − ao

 2
 − 2 x  ao


 exp 


 a o  ao
2
4

 − 1 
P = (− 1)exp (− 1) − exp  
 2 

which yields
P = 0.239
(c)
ao
P=
−1 / 2
2
2
ao
2
or
+1 / 2
A 2 cos 2 (nx )dx = 1
 
−1
 − 1 = 1 − exp  

 2 
 
2
2
2.18
 − 2 x  ao 4


 exp 


 ao  0
 2
 − x 
 dx
exp 


 a o
 a o 
2

P=
 x sin(x )  + 3
A2  +
=1
2  −1
2
 3  − 1 
A  −   = 1
 2  2 
1
so A 2 =
2
1
or A =
2
_______________________________________
0
  − 2a o
P = (− 1)exp 
  4a o
which yields
P = 0.393
(b)
ao
A
 − 2x 
 dx
a o 
 exp 
or
2.17
+3
4

0
=
2
 2
 − x 
 dx
exp 


 a o
 a o 
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
a/4
2
P =  (x ) dx

a/4
(a)

0
(a) P =
 a 
 
   sin  
2 4
 2 
=  
+
 4  
 a  2

 

 a  

 2   a (1)(a ) 
=   +
4 
 a  8
(b) P =
2




 2  a sin( ) 
=   −
 a   8  8  
 

 a 

or P = 0.25
a/2
(b) P =
a/4

 2x   a / 2
sin


 2  x
 a 
=  +
  
 a  2
4   a / 4

a 

or P = 0.25
+a / 2
(c) P =
2
2  x 
  cos   dx
a
 
 a 
−a / 2





 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
(c) P =
2

 4x   a / 2
sin


 2  x
 a 
=  −
 2  
 a  2
4
  a/4

 a  

 x 

 
sin  

 2  a sin( ) a
 2 
=   +
− −
 a   4  4  8  4  



 

 a 
 a  

2
  a  sin
a/4
  a  cos  a dx
2
 2x 

dx
 a 

 4x   a / 4
sin


2 x
 a 
=   −
 2  
 a  2
4
  0

 a  

or P = 0.409
a/2
2
0
2
2 x 
dx
  cos 
a
 
 2 

 2x   a / 4
sin


2 x
 a 
=   +
  
 a  2
4   0

a 

2
  a  sin




(
)
(
)
sin
2

sin
−
2

2 a
−a

=   −
−
+
 8  
 a   4  8   4 
 
  

 a 
 a  

or P = 1
_______________________________________
2.22
or

8 10 12
= 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
______________________________________________________________________________________
(
k=
)( )
(ii) p = m = 9.11 10 −31 10 4
= 9.11  10 kg-m/s
1
1
E = m 2 = 9.11 10 −31 10 4
2
2
= 4.555  10 −23 J
(
2
2
2
=
= 4.19  10 − 9 m
k
1.5  10 9
(
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
=
2
=
(
)(
)
or  = 7.586  10 rad/s
_______________________________________
13
2.24
(a)
(
)(
p = m = 9.11 10 −31 5 10 4
= 4.555  10
−26
)
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
2
or E n = n 6.686 10 −3 eV
Then
E1 = 6.69 10 −3 eV
)
E 2 = 2.67 10 −2 eV
E 3 = 6.02 10 −2 eV
_______________________________________
2.26
(a) E n =
h 6.625 10 −34
=
= 7.76 10 −9 m
p 8.537 10 −26
2
= 8.097  10 8 m −1

7.76  10 −9
 = k   = 8.097  10 8 9.37  10 4
k=
(
(
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
−25
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 10 15 rad/s
_______________________________________
or  = 41 .9 A
−19
)( )
= 9.11  10
o
(ii)
)
= 2.16  10 rad/s
(b) p = 9.11 10 −31 10 6
)( )
4.555 10
= 2.85 10 − 4 eV
1.6 10 −19
 1.5 10 13
(b) (i)  p = =
= −10 4 m/s
k − 1.5 10 9
or  p = −10 6 cm/s
=
)(
13
−23
or E =
=
(
−27
(
2
2
= 4.32  10 8 m −1
 1.454  10 −8
 = k = 4.32 10 8 5 10 4
o
 = 78.54 A
or
(
2
(
)(
(
)
n (6.018 10 )
=
= n (0.3761 ) eV
= n 2 6.018 10 −20 J
−20
2
or
En
Then
)
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
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
(a) E n =
 2 n 2 2
2ma 2
15  10 −3 =
(
)
2
n 2 1.054  10 −34  2
(
)(
)
2
2 15  10 −3 1.2  10 − 2
(
15 10 −3 = n 2 2.538 10 −62
)
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 =
(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
V ( x ) = 0 for
x
2
2
so in this region
 2 (x ) 2mE
+ 2  (x ) = 0
x 2

The solution is of the form
 (x ) = A cos k x + B sin k x
where
2mE
k=
2
Boundary conditions:
−a
+a
 (x ) = 0 at x =
, x=
2
2
First mode solution:
 1 (x ) = A1 cos k 1 x
where

 2 2
k1 =  E1 =
a
2ma 2
Second mode solution:
 2 (x ) = B 2 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 ) = B 4 sin k 4 x
where
4
16 2  2
k4 =
 E4 =
a
2ma 2
_______________________________________
2.30
The 3-D time-independent wave equation in
cartesian coordinates for V (x, y, z ) = 0 is:
 2 (x, y, z )
x 2
+
 2 (x, y, z )
y 2
+
 2 (x, y, z )
z 2
2mE
 ( x, y , z ) = 0
2
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
______________________________________________________________________________________
YZ
 (x, y )
= Ak y sin k x x  cos k y y
y
2 X
 2Y
2Z
+
XZ
+
XY
x 2
y 2
z 2
2mE
XYZ = 0
2
2mE
Dividing by XYZ and letting k 2 = 2 , we

find
1  2 X 1  2Y 1  2 Z

+ 
+ 
+k2 =0
(1)
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  2Y
1 2Z
 2 = −k y2 and

= −k z2
Y y
Z z 2
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
or
k x2 + k y2 + k z2 = k 2 =
2mE
2
so that
 2 2 2
n x + n 2y + n z2
2
2ma
_______________________________________
E → E nx n y nz =
2.31
 2 (x, y )
(
 2 (x, y )
)
2mE
+ 2  (x, y ) = 0

+
x 2
y 2
Solution is of the form:
 (x, y ) = A sin k x x  sin k y y
(a)
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
 2 (x, y )
= − Ak y2 sin k x x  sin k y y
y 2
Substituting into the original equation, we
find:
2mE
− k x2 − k y2 + 2 = 0
(1)

From the boundary conditions,
o
A sin k x a = 0 , where a = 40 A
So k x =
n x
, n x = 1, 2, 3, ...
a
o
Also A sin k y b = 0 , where b = 20 A
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.
_______________________________________
So k y =
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
or E =
6.155 10 −22
= 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
Combining these two equations, we find
 k − k1 
  B2
A2 =  2
 k 2 + k1 
 2k 2 
  B 2
B1 = 
 k 2 + k1 
The reflection coefficient is
= 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 ) + B 2 exp (− jk 2 x )
where
2m
(E − V O )
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 ( jk 1 x ) + B1 exp (− jk 1 x )
where
2mE
k1 =
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 (− jk 1 x )
 2 (x ) = A2 exp ( jk 2 x ) + B 2 exp (− jk 2 x )
(b)
Boundary conditions:
(1)  1 (x = 0 ) =  2 (x = 0 )
 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 B 2 = −k 1 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*
2.34
 2 ( x ) = A2 exp (− k 2 x )
P=
 (x )
2
2m(Vo − E )
where k 2 =
=
= exp (− 2k 2 x )
A2 A2*
2
(
)
(
2 9.11 10 −31 (3.5 − 2.8) 1.6 10 −19
)
1.054 10 −34
k 2 = 4.286 10 9 m −1
o
(a) For x = 5 A = 5  10 −10 m
P = exp (− 2k 2 x )
 (
)(
= 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
k2 =
k 2 = 4.860  10 9 m −1
−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
  = 1.874  10 5 m/s = 1.874  10 7 cm/s
(
1.2 10
−3
(
= N t 1.6 10
)
−19
(
(
 E 
E
1 −
T  16
 V
V
O
 O 
(a) For m = (0.067 )m o
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


)(
)
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  10 14 m −1
(a)
1
 1 
T  16 1 −  exp − 2 7.274  10 14 10 −14
12
12
 

 (
)(
 (
)
)
1/ 2
= 5.875  10 −7
(b)
T = (10 ) 5.875 10 −7
(
)
= 1.222 exp − 2 7.274 10 14 a
(
)
 1.222

2 7.274  10 14 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
= 1.222 exp − 14 .548 
2m(VO − E )
k2 =
)
or
)(1.874 10 )

 exp (− 2k 2 a )


)
 (
8
2.36
(
 exp − 2 4.124 10 9 15 10 −10
7
N t = 4.002 10 electrons/cm 3
Density of incident electrons,
4.002 10 8
Ni =
= 1.357 10 10 cm −3
0.0295
_______________________________________
)
−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 ( jk 1 x ) + B1 exp (− jk 1 x )
(incident)
(reflected)
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 2
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
where
A1 A1* =
2mE
2
k1 =
2m(VO − E )
2
A1 + B1 = A2 + B 2
d 1 d 2
=

dx
dx
jk 1 A1 − jk 1 B1 = k 2 A2 − k 2 B 2
At x = a :  2 =  3 
A2 exp (k 2 a ) + B 2 exp (− k 2 a )
= A3 exp ( jk 1 a )
d 2 d  3
=

dx
dx
k 2 A2 exp (k 2 a ) − k 2 B 2 exp (− k 2 a )
= jk 1 A3 exp ( jk 1 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 jk 1 k 2 exp (k 2 a ) + exp (− k 2 a ) 
(
)
 exp ( jk 1 a )
We then find
(k
2
2
)
− k12 exp (k 2 a )
− exp (− k 2 a )
+ 4k12 k 22 exp (k 2 a ) + exp (− k 2 a )
2
We have
k2 =
Region III:
 3 (x ) = A3 exp ( jk 1 x ) + B3 exp (− jk 1 x )
(b)
In Region III, the B 3 term represents a
reflected wave. However, once a particle
is transmitted into Region III, there will
not be a reflected wave so that B 3 = 0 .
(c) Boundary conditions:
At x = 0 :  1 =  2 
A1 =
(4k1 k 2 )2
2
Region II:
 2 (x ) = A2 exp (k 2 x ) + B 2 exp (− k 2 x )
where
k2 =
A3 A3*

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 k 1 and
k 2 , we find
(
k 12 + k 22 =
)
2mV O
2
and
 2m(VO − E )   2mE 
k 12 k 22 = 
 2 
2

  
2
 2m 
=  2  (VO − E )(E )
 
2

E
 2m 
=  2  (VO )1 −
V



O


(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



V
V
A1 A1
O 
 O 
_____________________________________
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 ( jk 1 x ) + B1 exp (− jk 1 x )
incident
reflected
where
2mE
k1 =
2
Region II: V = V1
 2 2 (x )
2m(E − V1 )
 2 (x ) = 0 
x
2
 2 (x ) = A2 exp ( jk 2 x ) + B 2 exp (− jk 2 x )
transmitted
reflected
where
2
+
2m(E − V1 )
k2 =
2
Region III: V = V 2
 2 3 (x )
+
2m(E − V2 )
x
2
 3 (x ) = A3 exp ( jk 3 x )
transmitted
where
2
k3 =
 3 (x ) = 0 
2m(E − V2 )
2
There is no reflected wave in Region III.
The transmission coefficient is defined as:
T=
 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 + B 2
 1  2
=

x
x
k 1 A1 − k 1 B1 = k 2 A2 − k 2 B 2
At x = a :  2 =  3 
A2 exp ( jk 2 a ) + B 2 exp (− jk 2 a )
= A3 exp ( jk 3 a )
 2  3
=

x
x
k 2 A2 exp ( jk 2 a ) − k 2 B 2 exp (− jk 2 a )
= 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
4k1 k 3
4k12
T= 3
=
2
k1 (k1 + k 3 )
(k1 + k 3 )2
_______________________________________
2.40
(a) Region I: Since V O  E , we can write
 2 1 (x ) 2m(VO − E )
−
 1 (x ) = 0
x 2
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 (k 1 x )
 2 (x ) = A2 sin (k 2 x ) + B 2 cos(k 2 x )
 3 (x ) = 0
where
k1 =
2m(VO − E )
2
and k 2 =

(b) Boundary conditions
At x = 0 :  1 =  2  B1 = B 2
2mE
2
 1  2
=
 k1 B1 = k 2 A2
x
x
At x = a :  2 =  3 
A2 sin(k 2 a ) + B 2 cos(k 2 a ) = 0
or
B 2 = − A2 tan(k 2 a )
(c)
k 
k1 B1 = k 2 A2  A2 =  1  B1
 k2 
and since B1 = B 2 , then
k 
A2 =  1  B2
 k2 
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 2
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
From B 2 = − A2 tan(k 2 a ) , we can write
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.
_______________________________________
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
 − 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=
2
or
−13 .58
(eV)
n2
n = 1  E1 = −13 .58 eV
En =
 − 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
equation in spherical coordinates reduces to
1   2   2m o
(E − V (r )) = 0
 r
+
r   2
r 2 r 
where
− e2
−2
V (r ) =
=
4 o r m o a o r
For
 100
 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 
 1
=
 
  ao
1




3/ 2
−r 

exp 

 ao 
Then
 100
1
=
r

n = 3  E 3 = −1.51 eV
2.42
We have
3
( )
n = 2  E 2 = −3.395 eV
n = 4  E 4 = −0.849 eV
_______________________________________
4
 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
5/ 2

 −r  r2
 − r 
−

exp 
2r exp 



 ao  ao
 a o 
2m 
2 
+ 2o  E +

mo a o r 
 
 1
 
  ao
−1




 1   1

 
 
    ao




3/ 2
−r 
=0
exp 

 ao 
where
− mo e 4
− 2
(4 o )2 2 2 2mo a o2
Then the above equation becomes
3/ 2
1  1    − r  
−1 
r2 


 
   exp 
2
r
−



2
ao 
  a o    a o   r a o 
2m  −  2
 2 
 = 0
+ 2o 
+
  2m o a o m o a o r 
or
3/ 2
1  1    − r 

   exp 
  a o    a o 
E = E1 =
=

2 
 − 2 1  − 1


+ 2 + 2 +
=0

a
r
a
r
a
a

o 
o
 o
 o

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
Exerc ise Solutions
______________________________________________________________________________________
Chapter 3
Exercise Solutions
Ex 3.3
4 (2m)
h3
3 / 2 2 eV
(a) N =
Ex 3.1
1
m 2
2
E=
3/ 2
1
dE = (2)m  d = m  d
2
(
(
)(
=
Ex 3.2
At ka =  , we have
sin a
−1 = 8
+ cos a
a
From Example 3.2, we have  2 a = 5.141 ,
or E 2 = 7.958 10 −19 J.
At ka = 2 , we see that  3 a = 2 so
or
E3
 a = 2
N = 1.28  10 22 cm −3
4 (2m)
h3
3/ 2
(b) N =
(
(
(
=
=
− 7.958  10
= 3.929  10
J
4 2m p

−19
3/ 2

3/ 2
E − E  dE
)
3/ 2
3
−2
3/ 2

  (E − E )
 3 
(
4 2(0.56 ) 9.11  10 −31
E
E − kT
)
3/ 2
(6.625 10 )
−2
 0.0259 )(1.6 10 )

(− 1)(
3
− 34 3
−19

Or
3/ 2

= 7.92  10 24 m −3
−19
3.929 10
= 2.46 eV
1.6 10 −19
_______________________________________
(
4 2m p
h
)
h3
E − kT
J
−19

)
N = 8.29  10 21 cm −3
_______________________________________
)(4.5 10 )
= 1.189  10
)(
or
−10 2
−18
1eV
= 8.29  10 27 m −3
(2 )2 (1.054 10 −34 )2
= 1.189  10
Then
E = E 3 − E 2
2 eV
= 1.06286  10
2
 2 3 / 2 − 13 / 2 1.6  10 −19
3
N=
−18
2
  E3/ 2
3
56
E
2 9.11  10
3/ 2
or
2ma 2
E =
− 34 3
Ex 3.4
− 31
0
3/ 2
−19
(2 )2  2
=
=
−31
= 1.28  10 28 m −3
or  = 1.76  10 −4 cm/s
_______________________________________
2
2 eV
(
)
(6.625 10 )
2
 2(1.6  10 )
3
)
)( )
= 1.76  10 −6 m/s
2mE 3
2
  E3/ 2
3
4 2 9.11  10
E 10 −12 1.6  10 −19
=
m
9.11  10 −31 10 5
 =
E  dE
0
4 (2m)
=
h3
So
or

or
N = 7.92  10 18 cm −3
_______________________________________
Ex 3.5
gi !
(10 )(9)(8!) = 45
10!
=
=
N i ! (g i − N i )! 8! (10 − 8)! (8!)(2)(1)
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 3
By D. A. Neamen
Exerc ise Solutions
______________________________________________________________________________________
Ex 3.6
(a)
 − (E − E F ) 
f F (E )  exp 

kT


 − (E c + kT 4 − E F ) 
= exp 

kT


 − (0.30 + 0.0259 4 ) 
= exp 

0.0259


= 7.26  10 −6
 − (E − E F ) 
f F (E )  exp 

kT


(b)
Then
 − (E − E F ) 
exp 
 = 0.02
kT


 1 
E − E F = kT ln 
 = 3.9kT
 0.02 
_______________________________________
Test Your Understanding Solutions
 − (0.30 + 0.0259 ) 
= exp 

0.0259


TYU 3.1
At ka = 2 , we see that  3 a = 2 , so
2mE 3
2
−6
= 3.43  10
_______________________________________
or
E3 =
Ex 3.7
 − (E − E F ) 
f F (E ) = exp 

kT


=
 − (0.30 + 0.025 ) 
8  10 − 6 = exp 

kT


 + 0.325 
5
exp 
 = 1.25  10
 kT 
0.325
= ln 1.25  10 5 = 11 .736
kT
0.325
 T 
kT =
= 0.02769 = (0.0259 )

11 .736
 300 
(
)
1
 E − EF 
1 + exp 

 kT 
 − (E − E F )  
 E − EF
exp 
1 + exp 

kT


 kT
2ma 2
(
(2 )2 (1.054 10 −34 )2
)(
2 9.11  10 −31 4.5  10 −10
)
2
2   4 a  3 . Then from
sin  4 a
+ cos  4 a
4a
we find, by trial and error,
 4 a = 7.870 . Then
+1 = 8
E4 =
T = 321 K
_______________________________________
1
 E − EF 
1 + exp 

 kT 
(2 )2  2
= 1.189  10 −18 J
At the other point,  4 a is in the range
so
Ex 3.8
 − (E − E F ) 
exp 
−
kT


 a = 2
=
(7.870 )2  2
2ma 2
(7.870 )2 (1.054 10 −34 )2
(
)(
2 9.11  10 −31 4.5  10 −10
)
2
= 1.8649  10 −18 J
Then
E g = E 4 − E3
= 1.8649  10 −18 − 1.189  10 −18
= 0.02
= 6.762  10 −19 J
or

  − 1 = 0.02

6.762 10 −19
= 4.23 eV
1.6 10 −19
_______________________________________
Eg =
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 3
By D. A. Neamen
Exerc ise Solutions
______________________________________________________________________________________
TYU 3.2
From Example 3.2, for ka =  ,   1 a = 
We have
and E1 = 2.972 10 −19 J.
E=
(2.529 )
2ma
2

2
=
−31
or
m p
mo
−10 2
− 31
1
 E − EF 
1 + exp 

 kT 
 − (E F − E ) 
 exp 

kT


1 − f F (E ) = 1 −
Then
E = 2.972  10 −19 − 1.9258  10 −19
= 1.046  10 −19 J
or
1.046 10 −19
= 0.654 eV
1.6 10 −19
_______________________________________
E =
TYU 3.3
We have E − E c = C1 k 2
 − (0.35 + (0.0259 2)) 
(a) 1 − f F (E ) = exp 

0.0259


= 8.20  10 −7
 − (0.35 + 3(0.0259 2 )) 
(b) 1 − f F (E ) = exp 

0.0259


= 3.02  10 −7
_______________________________________
(E c + 0.32 − E c )(1.6 10 −19 )



= C1 

−10
10

10


2
so that C1 = 5.1876 10 −39
We have
2
m
2
m =

=
2C1
m o 2 m o C1
TYU 3.6
 400 
We find kT = (0.0259 )
 = 0.034533 eV
 300 
(a)
(1.054 10 )
=
2(9.11 10 )(5.1876 10 )
−34 2
(b)
m
= 1.175
mo
_______________________________________
TYU 3.4
We have E − E = −C 2 k 2
(E − 0.875 − E )(1.6 10 −19 )



= −C 2 

−10
 12 10 
 − (E c + kT 4 − E F ) 
f F (E ) = exp 

kT


 − (0.30 + 0.034533 4 ) 
= exp 

0.034533


= 1.31  10 −4
−39
or
so that C 2 = 2.0426 10 −38
= 0.2985
TYU3.5
= 1.9258  10 −19 J
−31
− 38
_______________________________________
(1.054 10 )
2(9.11  10 )(4.5  10 )
(2.529 )
2
2m o C 2
−34 2
−34 2
2
mo
=
(1.054 10 )
=
2(9.11 10 )(2.0426 10 )
For 0  a   and ka = 0 , we have
sin a
+1 = 8
+ cos a
a
By trial and error, a = 2.529 rad.
Then
2mE
 a = 2.529
2
2
m p
2
 − (0.30 + 0.034533 ) 
f F (E ) = exp 

0.034533


= 6.21  10 −5
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 3
By D. A. Neamen
Exerc ise Solutions
______________________________________________________________________________________
TYU 3.7
 400 
We find kT = (0.0259 )
 = 0.034533
 300 
 − (0.35 + (0.034533 2)) 
(a) 1 − f F (E ) = exp 

0.034533


= 2.41  10 −5
 − (0.35 + 3(0.034533 2)) 
(b) 1 − f F (E ) = exp 

0.034533


= 8.85  10 −6
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 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 ) = V O . 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
2mV O 
 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 )
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
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
dx 2
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 )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 
− ( + k ) B exp − j ( + k )x
+ 2 jk  j ( − k )A exp  j ( − k )x 
2
− j ( + k )B exp − j ( + k )x 
(
)
− 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 
− j ( + k )B exp − j ( + k )a 
(
)
(
)
(
)
(
)
+ −  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.
_______________________________________
3.4
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 )
= 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 k a = 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 = k a = n , n = 0, 1, 2, ...  sin y = 0
So that, in general,
d (a ) d
dx
=0=
=
dy
d (ka) dk
And
2mE
=
2
So
−1 / 2
d 1  2mE 
 2m  dE
=  2 
 2 
dk 2   
   dk
This implies that
d
dE
n
=0=
for k =
dk
dk
a
_______________________________________
3.8
(a)  1 a = 
2mo E1
2
a = 
 2 2
E1 =
2m o a 2
=
(
( )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
2mo 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 − E 1
= 1.0198  10 −18 − 3.4114  10 −19
= 6.7868  10 −19 J
6.7868 10 −19
= 4.24 eV
1.6 10 −19
(b)  3 a = 2
or E =
2mo E3
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 =
 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
1.6 10 −19
_______________________________________
or E =
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 3
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
3.9
(a) At ka =  ,  1 a = 
2mo E1
E1 =
2mo E1
a = 
2

3.10
(a)  1 a = 
2
( )2 (1.054 10 −34 )2
(
)(
2 9.11  10 −31 4.2  10 −10
E1 =
)
2
(
)(
2
)
2
E2 =
= 2.5172  10 −19 J
E = E 1 − E o
= 3.4114  10
−19
− 2.5172  10
−19
= 8.942  10 −20 J
−20
8.942 10
= 0.559 eV
1.6 10 −19
(b) At ka = 2 ,  3 a = 2
or E =
2mo E3
2
E3 =
 a = 2
(1.054 10 )
2(9.11  10 )(4.2  10 )
(2 )
−34 2
2
− 31
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
= 3.4474  10
−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 − E 1
= 7.830  10 −19 − 3.4114  10 −19
= 4.4186  10 −19 J
4.4186 10 −19
= 2.76 eV
1.6 10 −19
(b)  3 a = 2
or E =
2mo E3
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
= 1.3646  10 −18 J
At ka =  . From Problem 3.5,
 2 a = 1.729
2mo E 2
(
2 9.11  10 −31 4.2  10 −10
2mo E 2
(0.859 )2 (1.054 10 −34 )2
2 9.11  10 −31 4.2  10 −10
( )2 (1.054 10 −34 )2
= 3.4114  10 −19 J
From Problem 3.6,  2 a = 1.515
= 3.4114  10 −19 J
At ka = 0 , By trial and error,
 o a = 0.859 
Eo =
a = 
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
= 3.50 eV
1.6 10 −19
_____________________________________
or E =
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 3
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
3.11
(a) At ka =  ,  1 a = 
2mo 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 = E 1 − E o
)(4.2 10 )
−19
−10 2
J
= 1.6084  10 −19 J
1.6084 10 −19
= 1.005 eV
1.6 10 −19
(b) At ka = 2 ,  3 a = 2
or E =
2
E3 =
(
(2 )2 (1.054 10 −34 )2
)(
−1
)
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 d 2E 

m *p =  2 

dk 2 

We have that
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
1.6 10 −19
_______________________________________
or E =
−1
 1 d 2E 
m =  2  2 
  dk 
We have
d 2E
d 2E
(
)
(curve B)
curve
A

dk 2
dk 2
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
2mo E 2
3.13
The effective mass is given by
*
= 3.4114  10 −19 − 1.8030  10 −19
2mo E3

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 )
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
2
d E
Points B,C:
0
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
− 7.406 10 −32
 mo
9.11 10 −31
m  = −0.0813 mo
_______________________________________
1
(
)
)
1.054 10 −34
2
Now m =
=
2C1
2 1.25 10 −38
= 4.44  10
(
2
−31
kg
−31
4
.
4437

10
or m  =
 mo
9.11 10 −31
m  = 0.488 mo
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 10 10
Now m  =
2
(
)
)
1.054 10 −34
2
=
2C1
2 1.25 10 −37
= 4.44  10
(
2
−32
kg
−32
4
.
4437

10
or m  =
 mo
9.11 10 −31
m  = 0.0488 mo
_______________________________________
3.17
For A: E − E = −C 2 k 2
(
)
(
 C 2 = 6.25 10 −39
−  2 − 1.054 10 −34
m =
=
2C 2
2 6.25 10 −39
(
2
− 8.8873 10 −31
 mo
9.11 10 −31
m  = − − 0.976 mo
or m  =
(
)
.
2
_______________________________________
3.20
and
d 2E
dk
(
− (0.3) 1.6 10 −19 = −C 2 0.08 10 10
 C 2 = 7.5 10 −38
2

E = E O − E1 cos (k − k O )
Then
dE
= (− E1 )(−  ) sin (k − k O )
dk
= + E1 sin (k − k O )
)
)
= −8.8873  10 −31 kg
For B: E − E = −C 2 k 2
)
)
3.19
(c) Curve A: Effective mass is a constant
Curve B: Effective mass is positive
around k = 0 , and is negative
around k = 
− (0.025 ) 1.6 10 −19 = −C 2 0.08 10 10
(
)
(
)
 C1 = 1.25 10 −37
3.18
(a) (i) E = h
E (1.42 ) 1.6 10 −19
or  = =
h
6.625 10 −34
= 3.429  10 14 Hz
hc c
3 10 10
(ii)  =
= =
E  3.429 10 14
= 8.75  10 −5 cm = 875 nm
E (1.12 ) 1.6 10 −19
(b) (i)  = =
h
6.625 10 −34
= 2.705  10 14 Hz
c
3 10 10
(ii)  = =
 2.705 10 14
= 1.109  10 −4 cm = 1109 nm
_______________________________________
(
For B: E = C i k 2
(
(
2
or m  =
10 2

)
)
= −7.406  10 −32 kg
(
)
= C (0.08 10 )
 C1 = 1.25 10 −38
(
−  2 − 1.054 10 −34
=
2C 2
2 7.5 10 −38
)
2
2
= E1 2 cos (k − k O )
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
YZ
2
2
k = ko
2
E1 2
_______________________________________
m* =


(0.082 m ) (1.64m )
2

(a) mdn
= 4 2 / 3 (mt ) ml
1/ 3
2
= 42 / 3
1/ 3
o
o

mdn
= 0.56 mo
(b)
3
2
1
2
1
=
+
=
+

m cn mt ml 0.082 m o 1.64 m o
=
24.39 0.6098
+
mo
mo

mcn
= 0.12 mo
_______________________________________
3.22

+ (m ) 
= (0.45m ) + (0.082 m ) 

(a) mdp
= (mhh )
3/ 2 2/3
3/ 2
lh
3/ 2 2/ 3
3/ 2
o
= 0.30187 + 0.02348 
2/3
m

dp
o
 mo
= 0.473 m 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
= 0.34 m 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
+
2mE
 XYZ = 0
2
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
+
or
3.21
2 X
 2Y
2Z
+
XZ
+
XY
x 2
y 2
z 2
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  2Y
1 2Z
 2 = −k y2 and

= −k z2
Y y
Z z 2
From the boundary conditions, we find
k y a = n y  and k z a = n z 
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 2
 
n x + n 2y + n z2  
2m
a
_______________________________________
E = E nx n y nz =
(
)
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
2mE
k=

Taking the differential, we obtain
k2 =
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
3/ 2
4 (2m)
g (E ) =
 E
h3
_______________________________________
dk =
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

dk =

1 1 2mn
 
 dE
 2
E
Then
2mn
2a 1
 
 dE
 2
E
Divide by the "volume" a, so
g T (E )dE =
g (E ) =
2mn
1


E
So
g (E ) =
(
2(0.067 ) 9.11  10 −31
1

1.054  10 −34 ( )
E
(
)
1.055  10 18
g (E ) =
m −3 J −1
)
E
_______________________________________
3.26
(a) Silicon, m n = 1.08mo
(
4 2mn
g c (E ) =
(
4 2m n
gc =
3/ 2
E − Ec
)
3 / 2 Ec + 2 kT

h3
=
=
=
h3
)
(
4 2m n
(
h
h

)
3/ 2
3
2
3/ 2
  (E − E c )
3
)
 3/ 2
n
3
4 2m
E − E c  dE
Ec
(
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 )
3/ 2
−21
= 6.0  10 25 m −3
g c = 6.0 10 19 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 c = 7.953 10 55 2 5.5253 10 −21
)
3/ 2
= 9.239  10 25 m −3
or
g c = 9.24 10 19 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
) (
(
)
(i)At T = 300 K, kT = 4.144  10 −21 J
) (
(
3/ 2
g c = 1.2288 10 54 2 4.144 10 −21
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 10 17 cm −3
or g  = 3.27 10 19 cm −3
(ii) At T = 400 K, k T = 5.5253  10 −21 J
) (
(
g c = 1.2288 10 54 2 5.5253 10 −21
)
)
3/ 2
(ii)At T = 400 K, k T = 5.5253  10 −21 J
) (
(
3/ 2
g = 2.3564 10 55 3 5.5253 10 −21
)
3/ 2
= 5.029  10 25 m −3
= 1.427  10 24 m −3
g c = 1.43 10 18 cm −3
_______________________________________
or g  = 5.03 10 19 cm −3
_______________________________________
3.27
(a) Silicon, m p = 0.56 m o
3.28
g  (E ) =
g =
=
=
=
(
4 2m p
(
)
3/ 2
h3
(
4 2m p
h3
4 2m
h

3/ 2
3/ 2
)
 3/ 2
p
3
(

(
=
E

E − E  dE
−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
E − 3 kT

2
3/ 2
 (3kT )
3
(i)At T = 300 K, kT = 4.144  10
) (
(
−21
J
g = 2.969 10 55 3 4.144 10 −21
E = E c + 0.2 eV;
= 2.134  10 46 m −3 J −1
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
(
4 2m p

h3
)
3/ 2
E − E
(
4 2(0.56 ) 9.11  10 −31
(6.625 10 )
)
3/ 2
− 34 3
E − E
g = 0
E = E − 0.1 eV; g  = 5.634 10 45 m −3 J −1
)
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
E = E − 0.4 eV;
= 1.127  10 46 m −3 J −1
_______________________________________
(b) GaAs, m p = 0.48 m o
(
4 2(0.48 ) 9.11  10 −31
(6.625 10 )
= (2.3564 10 )(3kT )
)
− 34 3
55
E − Ec
= 4.4541 10 55 E − E
or g  = 6.34 10 19 cm −3

3/ 2
gc = 0
For E = E ;
) (
)
− 34 3
(b) g  =
= 6.337  10 25 m −3
g =
(6.625 10 )
3/ 2
or g  = 4.12 10 19 cm −3
(
4 2(1.08 ) 9.11  10 −31
g c = 1.509 10 46 m −3 J −1
=
= 4.116  10 25 m −3
g = 2.969 10 55 3 5.5253 10 −21
E − Ec
E = E c + 0.1 eV;
)
(ii)At T = 400 K, k T = 5.5253  10 −21 J
3/ 2
h3
For E = E c ;
3/ 2
55
)
= 1.1929 10 56 E − Ec
E −3kT
)
4 2mn
E − E
h3
4 2m p
(
)
(
(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
gi!
10!
(a) Wi =
=
N i ! (g i − N i )! (7!)(10 − 7 )!
(b)
(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 = k T , f (E ) =
f (E ) = 0.269
(a)
 − (E − E F ) 
f F  exp 

kT


 − 0.30 
−6
E = E c ; f F = exp 
 = 9.32  10
0
.
0259


Ec +
kT
 − (0.30 + 0.0259 2) 
; f F = exp 

0.0259
2


= 5.66  10 −6
 − (0.30 + 0.0259 ) 
E c + k T ; f F = exp 

0.0259


= 3.43  10 −6
3kT
 − (0.30 + 3(0.0259 2)) 
Ec +
; f F = exp 

0.0259
2


= 2.08  10 −6
 − (0.30 + 2(0.0259 )) 
E c + 2k T ; f F = exp 

0.0259


(b) 1 − f F
1

1 + exp (1)
(b) E − E F = 5k T , f (E ) =
f (E ) = 6.69 10
3.34
1

1 + exp (5)
−3
(c) E − E F = 10 k T , f (E ) =
1

1 + exp (10 )
f (E ) = 4.54 10 −5
_______________________________________
 − (E F − E ) 
 exp 

kT


 − 0.25 
−5
E = E ; 1 − f F = exp 
 = 6.43  10
0
.
0259


E −
kT
 − (0.25 + 0.0259 2 ) 
; 1 − f F = exp 

0.0259
2


= 3.90  10 −5
 − (0.25 + 0.0259 ) 
E − k T ; 1 − f F = exp 

0.0259


3.33
1
1 − f (E ) = 1 −
 E − EF
1 + exp 
 kT
= 1.26  10 −6
1
= 1−
 E − EF 
1 + exp 

 kT 
= 2.36  10 −5



or
1
1 − f (E ) =
E −E
1 + exp  F

 kT 
(a) E F − E = k T , 1 − f (E ) = 0.269
(b) E F − E = 5k T , 1 − f (E ) = 6.69 10 −3
(c) E F − E = 10 k T , 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 − 2 k T ;
 − (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 c + E
= E midgap
2
_______________________________________
=
 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
= 9.40 eV
1.6 10 −19
For n = 7 , Empty state
or E 6 =
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
= 12.8 eV
1.6 10 −19
Therefore 9.40  E F  12 .8 eV
_______________________________________
or E 7 =
(
E5 =
2 2
 
n x + n 2y + n z2  
2m
a
(
)
2
2
2
2
−10 2
− 31
3.761 10 −19
= 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
E13 =
(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
= 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
or E13 =
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
E 1 = 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
or E 5 =
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
−34 2
= 3.761  10 −19 J
Or E F =
3.36
(1.054 10 ) ( ) (2 + 2 + 1 )
2(9.11  10 )(12  10 )
or
1
 E 
1 + exp 

 kT 
so f 1 (E1 ) = 1 − f 2 (E 2 )
Q.E.D.
_______________________________________
1 − f 2 (E 2 ) =
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 3
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
3.39
(a) At energy E 1 , 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 ,
f ( E1 ) =
1
 E − EF
1 + exp  1
 kT



=



= 0.01
1
1 + exp (4.6 )
which yields
f (E1 ) = 0.00990  0.01
_______________________________________
3.40
(a)
 − (E − E F ) 
f F = exp 

= 9.32  10
kT
 − (5.80 − 5.50 ) 
 = exp 

0.0259



−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 
=
= 50
or exp 

 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)
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
_______________________________________
f (E ) =
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
 − 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
 − (E F − E 2 ) 
1 − f (E ) = exp 

kT


 − 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
f (E ) = 8.45 10 −13
At E = E 2 ,
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
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 ,
E  E F  exp (−  ) = 0 
 −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 = E midgap ,
f (E ) =
1
 E − EF
1 + exp 
 kT



=
1
 Eg 

1 + exp 

2
kT


Si: E g = 1.12 eV,
f (E ) =
or
1
 1.12 
1 + exp 

 2(0.0259 ) 
f (E ) = 4.07 10 −10
or
1
 0.66 
1 + exp 

 2(0.0259 ) 
f (E ) = 2.93 10 −6
GaAs: E g = 1.42 eV
f (E ) =
or
(a)
 − (E − E F ) 
f F = exp 

kT


 − 0.60 
10 −8 = exp 

 kT 
0.60
= ln 10 +8
or
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
Exercise Solutions
______________________________________________________________________________________
Chapter 4
Exercise Solutions
Ex 4.1
ni = 3.29 10 9 cm −3
For T = 250 K,
 − (E − E F ) 
f F  exp 

kT


 − (E c + kT − E F ) 
= exp 

kT


(
or
 − (E c − E F ) 
n o = N c exp 

kT


(
= 4.7  10
)
ni = 7.13 10 3 cm −3
(b)
 − 0.25 
exp 

 0.0259 
no = 3.02 10 13 cm −3
_______________________________________
Ex 4.2
(
(a) N  = 1.04 10
19
)
 250 


 300 
n i (400 ) 3.288  10 9
=
= 4.61  10 5
n i (250 ) 7.135  10 3
_______________________________________
Ex 4.4
(a) GaAs
E Fi − E midgap =
3/ 2
 − ( E F − E ) 
p o = N  exp 

kT


)
p o = 2.919 10 13 cm −3
2.919 10 13
(b) Ratio =
= 4.54 10 −3
15
6.43 10
_______________________________________
Ex 4.3
(a) For T = 400 K,
)(
)
3


− 1.42
 exp 

 (0.0259 )(400 300 ) 
= 1.081  10
or
19
(b) Ge
3
(0.0259 ) ln  0.37 
4
 0.55 
 −7.70 meV
_______________________________________
E Fi − E midgap =
 − 0.27 
= 7.9115  10 18 exp 

 0.021583 
(




=
 250 
kT = (0.0259 )
 = 0.021583 eV
 300 
 400 
ni2 = 4.7 10 17 7 10 18 

 300 
 m p
3
kT ln  
 mn
4

3
(0.0259 ) ln  0.48 
4
 0.067 
 +38.25 meV
= 7.9115  10 18 cm −3
(
3
= 5.09  10 7
−5
17
)


− 1.42
 exp 

(
)(
)
0
.
0259
250
300


 − (0.25 + 0.0259 ) 
= exp 

0.0259


f F = 2.36 10
)(
 250 
ni2 = 4.7 10 17 7 10 18 

 300 
Ex 4.5
 − ( E F − E ) 
p o = N  exp 

kT


 − 0.215 
= 1.04  10 19 exp 

 0.0259 
(
)
= 2.58  10 15 cm −3
We find E c − E F = 1.12 − 0.215 = 0.905 eV
 − (E c − E F ) 
n o = N c exp 

kT


(
)
 − 0.905 
= 2.8  10 19 exp 

 0.0259 
= 1.87  10 4 cm −3
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 4
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
Ex 4.6
no =
2

pa
=
po + pa
N c F1 / 2 ( F )
2
1.5 10 20 =

(2.8 10 )F ( )
19
1/ 2
F
So F1 / 2 ( F ) = 4.748
E − Ec
From Figure 4.10,  F  3.2 = F
kT
 E F − E c = 3.2(0.0259 ) = 0.08288 eV
_______________________________________
(
)
 250 
N c = 2.8 10 19 

 300 
3/ 2
= 2.13 10 19 cm −3
 250 
kT = (0.0259 )
 = 0.021583 eV
 300 
nd
1
=
no + nd
N
 − (E c − E d ) 
1 + c exp 

2N d
kT


1
=
(7.912 10 ) exp  − 0.045 
 0.021583 
4(10 )


16
= 3.91  10 −2
(b) T = 200 K
N
 200 
= (1.04 10 )

300
(2.13 10 ) exp  − 0.045 
 0.02158 
2(10 )


16
= 7.50  10 −3
(b) T = 200 K
)
 200 
N c = 2.8 10 19 

 300 

= 1.524 10 19 cm −3
 200 
kT = (0.0259 )
 = 0.017267 eV
 300 
nd
1
=
19
no + nd
1.524  10
 − 0.045 
1+
exp 

16
2 10
 0.017267 
(
( )
)
= 1.75  10 −2
(c) Fraction increases as temperature
decreases.
_______________________________________
Ex 4.8
(a) T = 250 K
= 5.661 10 18 cm −3

 200 
kT = (0.0259 )
 = 0.017267 eV
 300 
pa
1
=
18
po + pa
5.661  10
 − 0.045 
1+
exp 

4 10 16
 0.017267 
)
( )
= 8.736  10 −2
_______________________________________
Ex 4.9
From Example 4.3,
ni (250 ) = 7.0 10 7 cm −3
ni (400 ) = 2.38 10 12 cm −3
Then
(a) T = 250 K
no =
=
3/ 2
3/ 2
19
19
1+
(
1+
(
Ex 4.7
(a) T = 250 K
(
1
18
Nd − Na
 N − Na
+  d
2
2

2

 + n i2

7  10 15 − 3  10 15
2
 7 10 15 − 3 10 15
+ 
2

n o = 410 15 cm −3
(
2

 + 7 10 7


(
)
2
)
2
n2
7 10 7
po = i =
= 1.225 cm −3
no
4 10 15
(b) T = 400 K
7  10 15 − 3  10 15
no =
2
 7 10 15 − 3 10 15
+ 
2

2

 + 2.38 10 12


(
)
2
n o  410 15 cm −3
)
 250 
N  = 1.04 10 19 

 300 
3/ 2
= 7.912 10 18 cm −3
 250 
kT = (0.0259 )
 = 0.021583 eV
 300 
(2.38 10 )
12 2
= 1.416  10 9 cm −3
4  10 15
_______________________________________
po =
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 4
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
Ex 4.10
For T = 250 K,
po =
(
)(
)
 250 
ni2 = 1.04 10 19 6 10 18 

 300 
3
 4 10 16 − 8 10 15
+ 
2



− 0.66
 exp 

(
)(
)
0
.
0259
250
300


p o = 3.2 10 16 cm −3
= 1.894  10 24
no =
or ni = 1.376 10 12 cm −3
For T = 350 K,
(
)(
)
 350 
ni2 = 1.04 10 19 6 10 18 

 300 
3
2
(
 2 10 14
 
 2
2
(
)
)
2
= 9.47 10 cm
9

 + 1.80 10 14


(
14 2
−3
)
2
−3
= 1.059  10 cm
3.059  10 14
_______________________________________
(a)
po =
2
Ex 4.13
2
(1.80 10 )
=
Ex 4.11
)
ni2
1.5 10 10
=
= 7.03 10 3 cm −3
16
po
3.2 10




 2.8 10 19 

= (0.0259 ) ln 
15 
 3 10 
E c − E F = 0.2368 eV
_______________________________________
n o = 3.059 10 14 cm −3
po
2
N
E c − E F = kT ln  c
 no

 + 1.376 10 12


n2
1.376 10 12
po = i =
no
2 10 14
(b) T = 350 K
2  10 14
no =
2
)
n o = 3 10 15 cm −3
or ni = 1.80 10 14 cm −3
(a) T = 250 K
2  10 14
no =
2
n o  210 14 cm −3
(
Ex 4.12
no = N d − N a = 8 10 15 − 5 10 15
28
 2 10 14
 
 2
(
2

 + 1.5 10 10


(b) p o = no = ni = 1.5 10 10 cm −3
_______________________________________


− 0.66
 exp 

 (0.0259 ) 350 300 
= 3.236  10
4  10 16 − 8  10 15
2
14
Na − Nd
2
 N − Nd
+  a
2

2

 + n i2

N 
E c − E F = kT ln  c 
 no 
We have E c − E F = (E c − E d ) + (E d − E F )
= 0.05 + 3kT = 0.05 + 3(0.0259 )
= 0.1277 eV
N 
So 0.1277 = (0.0259 ) ln  c 
 no 
Nc
= exp (4.9305 ) = 138 .45
Or
no
Nc
2.8 10 19
=
138 .45
138 .45
17
no = 2.02 10 cm −3
_______________________________________
Then no =
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 4
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
Test Your Understanding Solutions
n i (450 ) 1.7224  10 13
=
= 2.26  10 8
4
n i (200 ) 7.6334  10
_______________________________________
TYU 4.1
(c)
 − (E c − E F ) 
n o = N c exp 

kT


 − 0.22 
= 2.8  10 19 exp 

 0.0259 
TYU 4.4
(a) n
−3
= 5.73  10 cm
Now
E F − E = 1.12 − 0.22 = 0.90 eV
15
2
i
 200 
= (4.7 10 )(7.0 10 )

300
17

= 1.874
 n i = 1.37 cm −3
 − 0.90 
= 1.04  10 19 exp 

 0.0259 
= 8.43  10 3 cm −3
_______________________________________
(


− 1.42
 exp 

(
)(
)
0
.
0259
450
300


ni (450 ) 3.853  10 10
=
= 2.81 10 10
ni (200 )
1.369
_______________________________________
(c)
Now
E c − E F = 1.42 − 0.30 = 1.12 eV
 − 1.12 
n o = 4.7  10 17 exp 

 0.0259 
TYU 4.5
= 0.0779 cm −3
_______________________________________
(
)(1.04 10 )
 200 


 300 
 ni = 2.15 10 10 cm −3
(
)
= 2.967  10 26
 ni = 1.72 10 13 cm −3
)
3


− 0.66
 exp 

 (0.0259 )(450 300 ) 
3


− 1.12
 exp 

 (0.0259 )(450 300 ) 
)(
 450 
(b) ni2 = 1.04 10 19 6.0 10 18 

 300 
 ni = 7.63 10 4 cm −3
)(
3
= 4.639  10 20
= 5.827  10 9
(
)


− 0.66
 exp 

 (0.0259 )(200 300 ) 
3


− 1.12
 exp 

 (0.0259 )(200 300 ) 
 450 
(b) ni2 = 2.8 10 19 1.04 10 19 

 300 
)(
 200 
(a) ni2 = 1.04 10 19 6.0 10 18 

 300 
TYU 4.3
(
3
 ni = 3.85 10 10 cm −3
= 6.53  10 13 cm −3
(a) n = 2.8 10
)
= 1.485  10 21
 − 0.30 
p o = 7.0  10 18 exp 

 0.0259 
19
)(
 450 
(b) ni2 = 4.7 10 17 7.0 10 18 

 300 
TYU 4.2
19



− 1.42
 exp 

(
)(
)
0
.
0259
200
300


 − ( E F − E ) 
p o = N  exp 

kT


2
i
3
18
= 8.820  10 30
 ni = 2.97 10 15 cm −3
ni (450 ) 2.9699 10 15
=
= 1.38  10 5
10
ni (200 ) 2.1539  10
_______________________________________
(c)
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 4
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
TYU 4.6
 m p
3
= kT ln  
 mn
4

E Fi − E midgap
(b) p o =




3
(0.0259 ) 200  ln  0.56 
4
 300   1.08 
 −8.505 meV
3
 400   0.56 
(b) E Fi − E midgap = (0.0259 )
 ln 

4
 300   1.08 
 −17.01 meV
_______________________________________
po =
 m 
m o e 4
− 

 mo 
E=
2
2 2 (4 )
(
2
−12

 1
m ce
2
= 3
+
mo
m
m
2
 1
= 0.120
(



−1
2
TYU 4.9
pa
=
po + pa
1
N
 − (E a − E ) 
1+
exp 

4N a
kT


1
=
19
1.04  10
 − 0.045 
1+
exp 

4 10 17
 0.0259 
(
)(
)
4
−12
2
= −1.0184  10 J
or E = −6.37 meV
Also
r1 (1)(16 )
=
= 133 .3
ao
0.12
_______________________________________
TYU 4.8
2

N c F1 / 2 ( F )
For E c = E F ,  F = 0 and F1 / 2 (0 )  0.65
no =
2

(2.8 10 )(0.65)
19
no = 2.05 10 19 cm −3
3/ 2
For T = 100 K, N c = 5.389 10 18 cm −3
) 4 (16.0)(8.85 10 )
− 34 2
)
 T 
N c = 2.8 10 19 

 300 
−21
(a) n o =
19
TYU 4.10
We have
2 
 1
= 3
+

 1.64 0.082 
− (0.12 ) 9.11  10 −31 1.6  10 −19
2 1.054  10
(1.04 10 )(0.65)
pa
= 0.179
po + pa
_______________________________________
−19 4
(b) Ge:
Conductivity effective mass
E=

or
)(1.6 10 )
) 4 (13.1)(8.85 10 )
−31
= −8.4823  10 −22 J
or E = −5.30 meV
Also
r1
 m  (1)(13 .1)
=r  o  =
= 195 .5
ao
0.067
m 
(
2
( )
− (0.067 ) 9.11  10
2 1.054  10 −34
N  F1 / 2 ( F )
p o = 7.63 10 18 cm −3
_______________________________________
TYU 4.7
(a) GaAs:
(

For E = E F ,  F = 0 and F1 / 2 (0 )  0.65
(a) E Fi − E midgap =
=
2
−1
T = 200 K, N c = 1.524 10 19 cm −3
T = 300 K, N c = 2.8 10 19 cm −3
T = 400 K, N c = 4.311 10 19 cm −3
 T 
kT = (0.0259 )

 300 
For T = 100 K, kT = 0.008633 eV
T = 200 K, kT = 0.01727 eV
T = 300 K, kT = 0.0259 eV
T = 400 K, kT = 0.03453 eV
Fraction of ionized impurity atoms
nd
= 1−
no + nd
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 4
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
(a) For T = 100 K,
nd
1
=
no + nd
N
 − (E c − E d ) 
1 + c exp 

2N d
kT


1
=
5.389  10 18
 − 0.045 
1+
exp 

15
2 10
 0.008633 
= 0.0638
Now, percentage ionized atoms
= (1 − 0.0638 ) 100 = 93 .62 %
(b) For T = 200 K,
nd
1
=
no + nd
N
 − (E c − E d ) 
1 + c exp 

2N d
kT


( )
1
=
1.524  10 19
 − 0.045 
exp 

15
2 10
 0.01727 
= 0.001774
Now, percentage ionized atoms
= (1 − 0.001774 ) 100 = 99 .82 %
(c) For T = 300 K,
nd
1
=
no + nd
Nc
 − (E c − E d ) 
1+
exp 

2N d
kT


1+
( )
1
2.8  10 19
 − 0.045 
1+
exp 

15
2 10
 0.0259 
= 0.000406
Now, percentage ionized atoms
= (1 − 0.000406 ) 100 = 99 .96 %
(d) For T = 400 K,
nd
1
=
no + nd
Nc
 − (E c − E d ) 
1+
exp 

2N d
kT


=
( )
1
=
4.311  10
 − 0.045 
exp 

2 10 15
 0.03453 
= 0.000171
Now, percentage ionized atoms
= (1 − 0.000171 ) 100 = 99 .98 %
_______________________________________
1+
19
( )
TYU 4.11
p o = N a − N d = 2 10 16 − 5 10 15
= 1.5  10 16 cm −3
Then
no =
(
)
ni2
1.8 10 6
=
po
1.5 10 16
2
= 2.16  10 −4 cm −3
_______________________________________
TYU 4.12
2
N
N 
(b) n = d +  d  + n i2
2
 2 
Then
(5 10 )
14 2
1.110 15 = 5 10 14 +
which yields
ni2 = 1.110 29
Now
 − Eg
ni2 = N c N  exp 
 kT
1.110
29
(
= 2.8 10
19
+ ni2




)(1.04 10 )
19
 T 


 300 
3


− 1.12
 exp 

 (0.0259 )(T 300 ) 
By trial and error,
T  552 K
_______________________________________
TYU 4.13
At T = 550 K,
(
)(
)
 550 
ni2 = 2.8 10 19 1.04 10 19 

 300 
3


− 1.12
 exp 

 (0.0259 )(550 300 ) 
= 1.0236  10 29
or ni = 3.20 10 14 cm −3
2
Nd
N 
+  d  + ni2
2
 2 
Set n o = 1.05 N d
no =
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 4
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
Then
(1.05 − 0.5)N d 2 =  N d 
(0.22913 )N d
 2 
2
+ ni2
= ni
3.20 10
= 1.40 10 15 cm −3
0.22913
_______________________________________
or N d =
14
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 4
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Chapter 4
4.1
(b)
 − Eg
ni2 = N c N  exp 
 kT




(
) = 2.5 10
 − (1.12 )(300 ) 
 T 
= (2.912 10 )
 exp 
(0.0259 )(T ) 
300
ni2 = 5 10 12
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
n i (cm −3 )
200
0.01727
7.68  10 4
400
0.03453
2.38  10 12
600
0.0518
9.74  10 14
T (K)
(b) Germanium
n i (cm −3 )
200
 200 
At T = 200 K, kT = (0.0259 )

 300 
= 0.017267 eV
 400 
At T = 400 K, kT = (0.0259 )

 300 
= 0.034533 eV
2.16  10
10
1.38
400
8.60  10
14
3.28  10
600
3.82  10
16
5.72  10 12
(7.70 10 )
(200 ) (1.40 10 )
n i2 (400 )
(c) GaAs
n i (cm −3 )
n
2
i
10 2
=
2 2
 − Eg 
exp 

 0.034533 
=

3
 − Eg 
 200 
exp 



 300 
 0.017267 
Eg
 Eg

= 8 exp 
−

 0.017267 0.034533 
 400 


 300 
9
_______________________________________
4.2
Plot
_______________________________________
= 3.025  10 17
3


3.025  10 17 = 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



− 1.12
 exp 

 (0.0259 )(T 300 ) 
2.5 10

 = 38.1714


19

23
3
 3.025 10 17
E g (28.9561 ) = ln 
8

or E g = 1.318 eV
 T 
= (2.912 10 )

300
Now
(7.70 10 )
10 2
 400 
= N co N o 

 300 
 − 1.318 
 exp 

 0.034533 
3
38


 − (1.12 )(300 ) 
 exp 

 (0.0259 )(T ) 
By trial and error, T  367.5 K
3
(
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
______________________________________________________________________________________
4.5
(b)
 − 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
n i ( A)
 0.017267 
(b) For T = 300 K,
n i (B )
 − 0.20 
−4
= exp 
 = 4.43  10
n i ( A)
 0.0259 
(c) For T = 400 K,
n i (B )
 − 0.20 
−3
= exp 
 = 3.05  10
n i ( 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 + 4k T and E 2 = E c +
−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 
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
−
 − (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

4.10
E Fi − E midgap
 m *p
3
= kT ln  *
 mn
4

no =




*
n
E Fi − E midgap = −0.0128 eV
m n* = 0.067 mo
= +0.0382 eV
_______________________________________
4.11
E Fi − E midgap =
=
T (K)
N
1
(kT ) ln  
2
 Nc
19

1
(kT ) ln  1.04 1019
2
 2.8 10
kT (eV)





 = −0.4952 (kT )


( E Fi − E midgap )(eV)
− 0.0086
200
0.01727
400
0.03453
− 0.0171
600
0.0518
− 0.0257
_______________________________________
(a) E Fi − E midgap
 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 = k T  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 ) 
n o = K  kT  exp 

kT


_______________________________________
4.14
Let g c (E ) = C1 (E − E c ) for E  E c
Then

4.12




E Fi − Emidgap = −0.0077 eV
Gallium Arsenide: m *p = 0.48 m o ,
1
E − EF

E c 1 + exp 
 kT


 − (E − E F ) 
 K exp 
dE
kT


Ec
Germanium: m *p = 0.37 m o , mn* = 0.55mo
E Fi − Emidgap
(E ) f F (E )dE
c

=K
Silicon: m = 0.56 m o , m = 1.08mo
*
p
g
Ec
no =
g
(E ) f F (E )dE
c
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 = k T  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


 − (E − E c ) 
 dE
kT



 (E − E ) exp 
c
Ec
or
 − (E c − E F ) 
n o = C1 exp 

kT


 (kT )( )exp (−  )(kT )d
0
We find that



 exp (−  )d = exp (−  )(−  − 1) = +1
0
0
So
 − (E c − E F ) 
2
n o = C1 (kT ) exp 

kT


_______________________________________
4.15
r
m
We have 1 =r  o*
ao
m
 o

 
 s
2

0.067
 (13 .6 ) =
(13.6)

(13.1)2

or
E = 0.0053 eV
_______________________________________
4.17


 m*
E = 
 mo



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
N
(a) E c − E F = kT ln  c
 no




 2.8 10 19
= (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  10 19 exp 

 0.0259 
= 6.90  10 3 cm −3
(d) Holes
n 
(e) E F − E Fi = kT ln  o 
 ni 
 7 10 15 

= (0.0259 ) ln 
10 
 1.5 10 
= 0.338 eV
_______________________________________
4.18
 m *  o 
  (13 .6 ) eV
E = 
 
 m o  s 
0.55
(13 .6)  E = 0.029 eV
=
(16 )2
_______________________________________
N
(a) E F − E = kT ln  
 po
4.16
= 1.12 − 0.162 = 0.958 eV
 − 0.958 
(c) n o = 2.8  10 19 exp 

 0.0259 
2
r
m 
We have 1 =r  o* 
ao
m 
For gallium arsenide, r = 13 .1 ,
m = 0.067 mo
*
Then
o
 1 
r1 = (13 .1)
(0.53 ) = 104 A
 0.067 




 1.04 10 19
= (0.0259 ) ln 
16
 2 10
= 0.162 eV
(b) Ec − E F = E g − (E F − E )
(
)
= 2.41  10 3 cm −3




Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 4
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
4.21
p 
(d) E Fi − E F = kT ln  o 
 ni 
 2 10 16 

= (0.0259 ) ln 
10 
 1.5 10 
= 0.365 eV
_______________________________________
4.19




)
 − 0.27637 
(b) p o = 1.04  10 19 exp 

 0.0259 
= 2.414  10 14 cm −3
(c) p-type
_______________________________________
4.20
 375 
(a) kT = (0.0259 )
 = 0.032375 eV
 300 
(
n o = 4.7 10
17
)
 375 


 300 
3/ 2
 − 0.28 
exp 

 0.032375 
= 1.15  10 14 cm −3
E F − E = E g − (Ec − E F ) = 1.42 − 0.28
= 1.14 eV
(
)
 375 
p o = 7 10 18 

 300 
3/ 2
 − 1.14 
exp 

 0.032375 
= 4.99  10 3 cm −3
 4.7 10 17 

(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
 − 1.2046 
p o = 7  10 18 exp 

 0.0259 
(
)
3/ 2
 − 0.28 
exp 

 0.032375 
= 6.86  10 15 cm −3
E F − E = E g − (Ec − E F ) = 1.12 − 0.28
(
= 1.12 − 0.8436
E F − E = 0.2764 eV
(
(
 375 
n o = 2.8 10 19 

 300 
= 0.840 eV
N
(a) E c − E F = kT ln  c
 no




 2.8 10 19
= (0.0259 ) ln 
5
 2 10
= 0.8436 eV
E F − E = E g − (Ec − E F )
 375 
(a) kT = (0.0259 )
 = 0.032375 eV
 300 
)
= 4.42  10 −2 cm −3
_______________________________________
)
 375 
p o = 1.04 10 19 

 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 10 19 

= (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  10 19 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  10 19 exp 

 0.0259 
= 2.10  10 14 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  10 19 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 = n i exp  F

 kT 
(
 400 
kT = (0.0259 )
 = 0.034533 eV
 300 
)
 0.22 
= 1.5  10 10 exp 

 0.0259 
(
= 7.33  10 13 cm −3
 E − EF 
p o = n i exp  Fi

 kT 
(
(
)
= 5.6702  10 24
)
 ni = 2.381 10 12 cm −3
N
(a) E F − E = kT ln  
 po
)
 − 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




= 1.12 − 0.19788 = 0.92212 eV
 − 0.92212 
(c) n o = 2.8  10 19 exp 

 0.0259 
)
= 9.66  10 3 cm −3
(d) Holes
p
(e) E Fi − E F = kT ln  o
 ni
)(
 − 1.12 
 exp 

 0.034533 
= 8.80  10 9 cm −3
 E − EF 
p o = n i exp  Fi

 kT 
(
= 4.3109  10 19 cm −3
(
 0.22 
= 1.8  10 6 exp 

 0.0259 




 1.04 10 19
= (0.0259 ) ln 
15
 5 10
= 0.1979 eV
(b) Ec − E F = E g − (E F − E )
3/ 2
ni2 = 4.3109 10 19 1.601 10 19
−3
= 3.07  10 cm
 E − E Fi 
(b) n o = n i exp  F

 kT 
(
)
 400 
N c = 2.8 10 19 

 300 
)
(
3/ 2
= 1.601  10 19 cm −3
 − 0.22 
= 1.5  10 10 exp 

 0.0259 
6
)
 400 
N  = 1.04 10 19 

 300 




 5 10 15 

= (0.0259 ) ln 
10 
 1.5 10 
= 0.3294 eV
_______________________________________




 1.601 10 19 

= (0.034533 ) ln 
15

 5 10

= 0.2787 eV
(b) E c − E F = 1.12 − 0.27873 = 0.84127 eV
(
)
 − 0.84127 
(c) n o = 4.3109  10 19 exp 

 0.034533 
= 1.134  10 9 cm −3
(d) Holes
p 
(e) E Fi − E F = kT ln  o 
 ni 
 5 10 15 

= (0.034533 ) ln 
12 
 2.381 10 
= 0.2642 eV
_______________________________________
4.26
(a)
(
)
 − 0.25 
p o = 7 10 18 exp 

 0.0259 
= 4.50  10 14 cm −3
E c − E F = 1.42 − 0.25 = 1.17 eV
(
)
 − 1.17 
n o = 4.7  10 17 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
(
N  = 7 10
18
(
)
 400 


 300 
= 8.49  10 9 cm −3
_______________________________________
= 1.078  10 19 cm −3
(
)
 400 
N c = 4.7 10 17 

 300 
= 7.236  10 cm
−3
N
E F − E = kT ln  
 po




17
3/ 2
4.28
(a) n o =
)
4.27
(
p o = 1.04  10
)
no =
=
po =
(
)
 400 


 300 
)
 400 


 300 
17
2

N  F1 / 2 ( F )
2

(1.04 10 )F (  )
19
1/ 2
F
E − E F
kT
E − E F = (3.0 )(0.0259 ) = 0.0777 eV
_______________________________________
We find  F  3.0 =
3/ 2
3/ 2
4.30
E F − E c 4kT
=
=4
kT
kT
Then F1 / 2 ( F )  6.0
(a)  F =
= 4.311  10 19 cm −3
N
E F − E = kT ln  
 po

(4.7 10 )(1.0)
So F1 / 2 ( F ) = 4.26
= 1.601  10 19 cm −3
N c = 2.8 10
2
5 10 19 =
(b) kT = 0.034533 eV
19
19
4.29
n o = 7.23 10 4 cm −3
(
(2.8 10 )(1.0)

)
N  = 1.04 10

= 3.16  10 19 cm −3
2
N c F1 / 2 ( F )
(b) n o =
 − 0.870 
n o = 2.8  10 19 exp 

 0.0259 
19
2
= 5.30  10 17 cm −3
_______________________________________
 − 0.25 
exp 

 0.0259 
= 6.68  10 14 cm −3
E c − E F = 1.12 − 0.25 = 0.870 eV
(
N c F1 / 2 ( F )
E F − E c kT 2
=
= 0.5
kT
kT
Then F1 / 2 ( F )  1.0
= 2.40  10 4 cm −3
_____________________________________
(a)

F =




 − 1.07177 
n o = 7.236  10 17 exp 

 0.034533 
19
2
For E F = E c + kT 2 ,
 1.078 10 19
= (0.034533 ) ln 
14
 4.50 10
= 0.3482 eV
E c − E F = 1.42 − 0.3482 = 1.072 eV
(
)
 − 0.77175 
n o = 4.311  10 19 exp 

 0.034533 
3/ 2




 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
17

p (E ) =
= 3.18  10 18 cm −3
_______________________________________
4.31
For the electron concentration
n(E ) = g c (E ) f F (E )
The Boltzmann approximation applies, so
n(E ) =
(
4 2mn*
h
)
or
 − (E − E F ) 
n (E ) =
(
4 2m n*
h
)
3/ 2
3
 kT
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
3
 − ( E F − E ) 
exp 

kT


E − E
 − (E − E ) 
exp 

kT
kT


Define
x =
E − E
kT
Then
p(x ) = K  x  exp (− x )
E − Ec
 exp 

h
)
 kT
3/ 2
3
(
4 2m *p
3/ 2
E − E
 − (E F − E ) 
 exp 

kT


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 = 3k T eV
we can write
 − 0.045

n o = 2.8  10 19 exp 
− 3
 0.0259

(
)
(
)
= 2.8 10 19 exp (− 4.737 )
or
no = 2.45 10 17 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 = 3k T and E a − E = 0.045 eV
Then
0.045 

p o = 1.04  10 19 exp − 3 −
0.0259 

(
)
(
)
= 1.04 10 19 exp (− 4.737 )
or
p o = 9.12 10 16 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

n o = 4.7  10 17 exp 
− 3
0
.
0259


(
)
(
)
= 4.7 10 17 exp (− 3.224 )
4.35
(a)
)
or
p o = 9.20 10 16 cm −3
_______________________________________
4.33
Plot
_______________________________________
p o = 4 15 − 10 15 = 3 10 15 cm −3
(1.5 10 )
10 2
= 7.5  10 4 cm −3
3  10 15
(b) n o = N d = 310 16 cm −3
10 2
po
= 7.5  10 cm
3
(
)(
)
3
 375 
(d) ni2 = 2.8 10 19 1.04 10 19 

 300 
 − (1.12 )(300 ) 
 exp 

 (0.0259 )(375 ) 
 ni = 7.334 10 11 cm −3
(7.334 10 )
11 2
4  10 15
(
(e) n = 2.8 10
2
i
19
= 1.34  10 8 cm −3
)(1.04 10 )
 450 


 300 
3
 − (1.12 )(300 ) 
 exp 

 (0.0259 )(450 ) 
)
2
= 1.08 10 −3 cm −3
(b) n o = N d = 310 16 cm −3
(1.8 10 )
6 2
= 1.08  10 − 4 cm −3
3  10 16
(c) no = p o = ni = 1.8 10 6 cm −3
po =
(
)(
3
)
 375 
(d) ni2 = 4.7 10 17 7.0 10 18 

 300 
 − (1.42 )(300 ) 
 exp 

 (0.0259 )(375 ) 
 ni = 7.580 10 8 cm −3
p o = N a = 410 15 cm −3
(7.580 10 )
8 2
4  10 15
(
= 1.44  10 2 cm −3
)(
3
)
 450 
(e) ni2 = 4.7 10 17 7.0 10 18 

 300 
 − (1.42 )(300 ) 
 exp 

 (0.0259 )(450 ) 
 ni = 3.853 10 10 cm −3
no = N d = 10 14 cm −3
po
19
(
ni2
1.8 10 6
=
po
3 10 15
(3.853 10 )
=
10 2
p o = N a = 410 15 cm −3
no =
2
13 2
no =
−3
3  10
(c) no = p o = ni = 1.5 10 10 cm −3
16
)
p o = N a − N d = 4 10 15 − 10 15
no =
18
(1.5 10 )
=
(
(1.722 10 )
= 3 10 15 cm −3
= (7 10 )exp (− 4.332 )
no =
2

 + 1.722 10 13


= 2.88 10 12 cm −3
1.029  10 14
_______________________________________
no = 1.87 10 16 cm −3
Assume E a − E = 0.0345 eV
Then
 − 0.0345

p o = 7  10 18 exp 
− 3
0
.
0259


4.34
(a)
 10 14
10 14
+ 
2
 2
= 1.029  10 14 cm −3
po =
or
(
no =
14
= 1.48  10 7 cm −3
10
_______________________________________
4.36
(a) Ge: ni = 2.4 10 13 cm −3
(i) n o =
 ni = 1.722 10 13 cm −3
=
Nd
N
+  d
2
 2
2

 + ni2

 2 10 15
2 10 15
+ 
2
 2
2

 + 2.4 10 13


(
)
2
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 4
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
or
Then
no  N d = 210 15 cm −3
(
f F (E ) =
)
13 2
n2
2.4 10
po = i =
no
2 10 15
(ii) p o  N a − N d = 10 16 − 7 10 15
= 3 10 15 cm −3
n2
2.4 10 13
no = i =
po
3 10 15
)
2
= 1.92  10 11 cm −3
(b) GaAs: ni = 1.8 10 6 cm −3
(i) no  N d = 210 15 cm
(1.8 10 )
6 2
= 1.62  10 −3 cm −3
2  10 15
(ii) p o  N a − N d = 310 15 cm −3
po =
(1.8 10 )
6 2
= 1.08  10 −3 cm −3
3  10 15
(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  10 11 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 10 13 − 110 13
or
p o = 1.5 10 13 cm −3
Then
(
)
ni2
1.5 10 10
=
po
1.5 10 13
Germanium:
no =
po =
2
= 1.5 10 7 cm −3
Na − Nd
 N − Nd
+  a
2
2

 1.5 10 13 
 1.5 10 13
+ 
= 


2
2



or
p o = 3.26 10 13 cm −3
Then
(
2
2

 + ni2


 + 2.4 10 13


(
)
2
)
2
ni2
2.4 10 13
=
= 1.76 10 13 cm −3
p o 3.264 10 13
Gallium Arsenide:
p o = N a − N d = 1.5 10 13 cm −3
and
no =
(
)
2
ni2
1.8 10 6
=
= 0.216 cm −3
po
1.5 10 13
_______________________________________
no =
4.39
(a) N d  N a  n-type
(b) no  N d − N a = 2 10 15 − 1.2 10 15
= 8  10 14 cm −3
po =
(
ni2
1.5 10 10
=
no
8 10 14
)
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 10 15 cm −3
(1.5 10 )
=
10 2
−3
= 5.625  10 cm
4  10 15
_______________________________________
no
4
4.44
Plot
_______________________________________
4.45
4.40
(
)
2
no =
n2
1.5 10 10
no = i =
= 1.125 10 15 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
(
n = 1.04 10
2
i
19
)(6.0 10 )
18
 250 


 300 
3
(1.110


− 0.66
 exp 

 (0.0259 )(250 300 ) 
ni2
n2
1
= i  no2 = ni2
p o 4no
4
2
N
=  a
 2
13 2
so ni = 5.74 10 13 cm −3
N a  N d  p-type
Majority carriers are holes
p o = N a − N d = 3 10 16 − 1.5 10 16
)
2
5 10 16 = N a + 3 10 16 − 1.5 10 16
2

 + 1.8936 10 24

)
N 
7.5735 10 24 − 2.752 10 12 N a +  a 
 2 
N
=  a
 2
) = (4 10 )
n2
1.5 10 10
no = i =
= 1.5 10 4 cm −3
po
1.5 10 16
(b) Boron atoms must be added
p o = N a + N a − N d
2
(
+ ni2
− 4 10 13
(
Na
N 
+  a  + ni2
2
 2 



2
14
= 1.5  10 16 cm −3
Minority carriers are electrons
Then p o = 2.75 10 12 cm −3
N

 2.752 10 12 − a
2


 + ni2


4.9 10 27 = 1.6 10 27 + ni2
4.46
(a)
1
 n o = ni
2
no = 6.88 10 11 cm −3 ,
po =
2
 2 10 14 − 1.2 10 14
+ 
2

po =
 ni = 1.376 10 12 cm −3
So
2  10 14 − 1.2  10 14
2
ni2 3.3 10 27
=
= 3 10 13 cm −3
no 1.110 14
_______________________________________
= 1.8936  10 24
no =
2

 + n i2

So N a = 3.5 10 16 cm −3
2

 + 1.8936 10 24

so that N a = 2.064 10 12 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




 Nd
= (0.0259 ) ln 
10
 1.5  10
p o  n i  n-type
(b) p o =
ni2
n2
 no = i
no
po
no =
(1.5 10 )
10 2
= 1.125  10 16 cm −3
2 10
 electrons are majority carriers
4
p o = 2 10 4 cm −3
 holes are minority carriers
(c) n o = N d − N a
1.125 10 16 = N d − 7 10 15
so N d = 1.825 10 16 cm −3
_______________________________________
4.48
n i (cm −3 )
200
0.01727
2.16  10 10
400
0.03453
8.60  10 14
600
0.0518
3.82  10 16
Na
N
+  a
2
 2
200
1.0  10 15
400
1.49  10
600
3.87  10 16
10 16 cm −3 , E F − E Fi = 0.3473 eV
10 17 cm −3 , E F − E Fi = 0.4070 eV
_______________________________________
4.50
(a) n o =

 + ni2

15
− 0.5 10 15
)
2
(
(E Fi − E F ) (eV)
0.1855
0.01898
0.000674
_______________________________________
4.49
N
(a) E c − E F = kT ln  c
 Nd




 2.8 10 19 

= (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 10 19 1.04 10 19 

 300 
3


− 1.12
 exp 

 (0.0259 )(T 300 ) 
2
15
2
Nd
N
+  d
2
 2
= 0.5 10 15
N a = 10 15 cm −3
p o (cm −3 )
10 15 cm −3 , E F − E Fi = 0.2877 eV
(1.05 10

 + ni2 and

T (K)
For 10 14 cm −3 , E F − E Fi = 0.2280 eV
no = 1.05 N d = 1.05 10 15 cm −3
p 
E Fi − E F = kT ln  o 
 ni 
For Germanium
T (K)
kT (eV)
po =



5.25 10
28
(
= 2.912 10
38
)
 T 


 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 10 19 

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 10 19 

 300 
= 6.696  10 19 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




 6.696 10 19 

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
(
)(
)
n
3
(
n = 2.8 10
N a = 10

 600 


 300 
=
3




3
(0.0259 ) ln (10 )
4
E Fi − Emidgap = +0.0447 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 = n i exp  Fi

kT


( )
 0.4947 
= 10 5 exp 

 0.0259 
Na
N 
+  a  + ni2
2
 2 
= 3.288  10 15 cm −3
 m *p
3
kT ln  *
 mn
4

(b) Impurity atoms to be added so
E midgap − E F = 0.45 eV
2
 3 10 15
3 10 15
+ 
2
 2
cm −3 , E F − E = 0.1697 eV
(a) E Fi − E midgap =
 ni = 9.740 10 14 cm −3
At T = 200 K and T = 400 K,
p o = N a = 310 15 cm −3
At T = 600 K,
=
16
4.53
3
 − 1.12 
 exp 

 0.0518 
po =




N a = 10 17 cm −3 , E F − E = 0.1100 eV
_______________________________________
or
)(1.04 10 )

 7.0 10 18
 = (0.0259 ) ln 

 N
a


N a = 10 15 cm −3 , E F − E = 0.2293 eV
19
19
cm −3 , E Fi − E F = 0.5811 eV
For N a = 10 14 cm −3 , E F − E = 0.2889 eV
 ni = 2.381 10 12 cm −3
At T = 600 K,
19
16
N
E F − E = kT ln  
 Na
 − 1.12 
 exp 

 0.034533 
2
i



(b)
 400 
= (2.8 10 )(1.04 10 )

300


 Na
 = (0.0259 ) ln 
6

 1.8 10

For N a = 10 14 cm −3 , E Fi − E F = 0.4619 eV
N a = 10 17 cm −3 , E Fi − E F = 0.6408 eV
 ni = 7.638 10 4 cm −3
At T = 400 K,
19
N
E Fi − E F = kT ln  a
 ni
N a = 10
 − 1.12 
 exp 

 0.017267 
2
i
4.52
(a)
N a = 10 15 cm −3 , E Fi − E F = 0.5215 eV
At T = 200 K,
 200 
ni2 = 2.8 10 19 1.04 10 19 

 300 
Then, T = 200 K, E Fi − E F = 0.4212 eV
T = 400 K, E Fi − E F = 0.2465 eV
T = 600 K, E Fi − E F = 0.0630 eV
_______________________________________
or
2

 + 9.740 10 14


(
)
2
p o = N a = 1.97 10 13 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  10 15 + 2.8 10 19 exp 

 0.0259 
= 5  10 15 + 6.95  10 15
N
(b) E F − E Fi = kT ln  c
 Nd




 2.8 10 19
= (0.0259 ) ln 
16
 2 10
(c) For part (a);
p o = 210 16 cm −3
or
no =
N d = 1.2 10 16 cm −3
_______________________________________
(
ni2
1.5 10 10
=
po
2 10 16
For part (b):
n o = 210 16 cm −3
 − (E c − E F ) 
N d = N c exp 

kT


(
po =
(
N d = 4.7  10
17
)
 − 0.13346 
exp 

 0.0259 
= 2.718  10 15 cm −3 = N d + 10 15
 N d = 1.718 10 15 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 10 10
=
no
2 10 16
)
2
4.57
 E − E Fi 
n o = n i exp  F

 kT 
(
)
 0.55 
= 1.8  10 6 exp 

 0.0259 
)
 N d = 1.031 10 16 cm −3 Additional
donor atoms
(b) GaAs
 4.7 10 17 

(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
2
= 1.125  10 4 cm −3
_______________________________________
 − 0.1929 
= 2.8  10 19 exp 

 0.0259 
N d = 1.631 10 16 cm −3 = N d + 610 15
)
= 1.125  10 4 cm −3
4.55
(a) Silicon
N 
(i) E c − E F = kT ln  c 
 Nd 
 2.8 10 19 
 = 0.2188 eV
= (0.0259 ) ln 
15 
 6 10 
(ii) E c − E F = 0.2188 − 0.0259 = 0.1929 eV

 = 0.1876 eV


= 3.0  10 15 cm −3
Add additional acceptor impurities
no = N d − N a
3 10 15 = 7 10 15 − N a
 N a = 410 15 cm −3
_______________________________________
4.58
p 
(a) E Fi − E F = kT ln  o 
 ni 
 3 10 15
= (0.0259 ) ln 
10
 1.5 10
n 
(b) E F − E Fi = kT ln  o 
 ni 
 3 10 16
= (0.0259 ) ln 
10
 1.5 10
(c) E F = E Fi

 = 0.3161 eV



 = 0.3758 eV


Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 4
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
4.60
p 
(d) E Fi − E F = kT ln  o 
 ni 
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
N
N 
p o = a +  a  + ni2
2
 2 
5.08  10 15 =
4.59
N
(a) E F − E = kT ln  
 po
(
)
 7.0  10 (375 300 )
 ln 
4  10 15

= 0.2526 eV
 450 
(e) E F − E = (0.0259 )

 300 
(
5  10 15
2




 7.0 10 18 
 = 0.2009 eV
= (0.0259 ) ln 
15 
 3 10 
 7.0 10 18 

(b) E F − E = (0.0259 ) ln 
−4 
 1.08 10 
= 1.360 eV
 7.0 10 18 

(c) E F − E = (0.0259 ) ln 
6 
 1.8 10 
= 0.7508 eV
 375 
(d) 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
15
2

 + ni2


)
= (2.5 10 )
2
− 2.5 10 15
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
−3
N c = 1.2 10 19 
 = 1.633 10 cm
 300 
 350 
19
−3
N  = 1.8 10 19 
 = 2.45 10 cm
 300 
Now
4.064 10 29 = 1.633 10 19 2.45 10 19
(
)(
3/ 2
)
 − Eg 
 exp 

 0.030217 
So
(
)(
)
 1.633 10 19 2.45 10 19 
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 10 15 = 3.5 10 14 cm −3
Replace As atoms  Silicon acts as an
acceptor
N a = (0.95 ) 7 10 15 = 6.65 10 15 cm −3
(b) N a  N d  p-type
(
)
(
)
(c) p o = N a − N d = 6.65 10 15 − 3.5 10 14
= 6.3  10 15 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 10 15 
 = 0.5692 eV
= (0.0259 ) ln 
6 
 1.8 10 
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 5
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
Chapter 5
Exercise Solutions
Ex 5.1
Ex 5.5
J drf  e p po  = e p N a 
(
75 = 1.6 10
−19
dp
dx
d
−x L
= −eD p
10 16 e p
dx
 −1  −x Lp
e
= −eD p 10 16 
 Lp 


J p = −eD p
)(480 )N (120 )

a
which yields N a = 8.14 10 15 cm −3
_______________________________________
( )
Ex 5.2 Using Figure 5.2;
(a) T = 25 C,
(i) N a = 10 16 cm −3 ,   p  410 cm 2 /V-s
=
(ii) N a = 10 18 cm −3 ,   p  130 cm 2 /V-s
(b) N a = 10 cm
14
−3
,
(i) T = 0 C,   p  550 cm 2 /V-s
(ii) T =100 C,   p  300 cm 2 /V-s
_______________________________________
Ex 5.3
(a) For N I = N a + N d = 2.8 10 17 + 8 10 16
= 3.6  10 17 cm −3 ,
  p = 200 cm 2 /V-s
(b)  = e p (N a − N d )
(
)
(
= 1.6 10 −19 (200 ) 2 10 17
)
 = 6.4 (  -cm) −1
(c)  =
1
1
=
= 0.156  -cm
 6.4
_______________________________________
=
V
5
=
= 2500 
I 2  10 −3
RA (2500 ) 10 −6
(b)  =
=
= 2.083  -cm
L
1.2 10 −3
1
1
(c)  = =
= 0.480 (  -cm) −1
 2.083
= e p N a
(a) R =
(
)
0.48
= 3.00  10 18
1.6  10 −19
Using Figure 5.3 and trial and error,
N a  7.3 10 15 cm −3
Then  p N a =
(d)  p  410 cm 2 /V-s
_______________________________________
( )e
+ eD p 10 16
−x Lp
Lp
(1.6 10 )(8)(10 ) e
−19
16
−x Lp
−4
2 10
−x

J p = 64 exp 
 Lp 


(a) For x = 0 ,
J p = 64 A/cm 2
(b) For x = 2  10 −4 cm,
 − 2 10 −4
J p = 64 exp 
−4
 2 10
(c) For x = 10 −3 cm,

 = 23.54 A/cm 2


 − 10 −3 
 = 0.431 A/cm 2
J p = 64 exp 
−4 
2

10


_______________________________________
Ex 5.6
dN d (x )
dx
Ex 5.4

=−
(10 ) e
16
−x L
L
So
 − 10 16  − x L

e


 kT   L 
 x = −

16 − x L
 e  10 e
 kT  1  0.0259
=
  =
−2
 e  L  2  10
or  x = 1.295 V/cm
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 5
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
(b)   e n ( N d − N a )
Ex 5.7
(
Dn
215
n =
=
0
.
0259
kT




 e 
 n = 8301 cm /V-s
_______________________________________
Ex 5.8
From Equation (5.59),
 p (epVxWd )
Ix =
L
(320 ) 1.6 10 −19 10 16 (10 ) 10 −2 8 10 −4
=
0.2
I x = 2.048 10 −4 A
or I x = 0.2048 mA
From Equation (5.53),
I B
2.048 10 −4 5  10 −2
VH = x z =
epd
1.6 10 −19 10 22 8  10 − 6
( )
)( ) (
(
(
)(
)(
)(
)
)(
)
 = 4.8 (  -cm) −1
We find
=
1
 = e n N d =
(1.6 10 )
−19
)
Test Your Understanding Solutions
TYU 5.1
no = N d − N a = 10 15 − 10 14 = 9 10 14 cm
Also
)
2
= 2.5 10 5 cm −3
Now
J drf = e  n no +  p po   e n no 
(
(
)
)
(
1

n
Nd =
1
= 10
0.1
 n N d = 6.25 10 19
Using Figure 5.3 and trial and error,
N d  610 16 cm −3 and
 n  1050 cm 2 /V-s
_______________________________________
TYU 5.4
J diff = eDn
(
1
=
= 0.208 (  -cm)
 4.8
_______________________________________
So
= 8  10 V
or V H = 0.80 mV
_______________________________________
ni2
1.5 10 10
=
no
9 10 14
)
TYU 5.3
−4
po =
(
or
2
(
)
= 1.6 10 −19 (1000 ) 3 10 16
)
= 1.6 10 −19 (1350 ) 9 10 14 (35 )
or
J drf = 6.80 A/cm 2
_______________________________________
 10 15
dn
= −eDn  − 4
dx
 10

−x
 exp 


L 

 n 
We have D n = 25 cm 2 /s
L n = 10 −4 cm = 1  m
Then
−x
J diff = −40 exp 
 A/cm 2
 1 
(a) x = 0 , J diff = −40 A/cm 2
(b) x = 1  m, J diff = −14.7 A/cm 2
(c) x =  , J diff = 0
_______________________________________
TYU 5.5
J diff = −eD p
dp
dx
So
TYU 5.2
(a) For N I = 7 10 16 cm −3 ,
 n  1000 cm 2 /V-s;  p  350 cm 2 /V-s
(
) (0 −0p.010 )
20 = − 1.6 10 −19 (10 )
 p = 1.25 10 17 = 4 10 17 − p
And
p(x = 0.01) = 2.75 10 17 cm −3
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 5
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Chapter 5
5.1
1
1
(a)  =
=
−19
e n N d
1.6 10 (1300 ) 10 15
= 4.808  -cm
(
1
)
( )
1
= 0.208 (  -cm) −1
 4.8077
_______________________________________
(b)  =
5.2

e p
(
−19
)
(
(
)
)
120 = 1.6 10 −19  n N d
=
From Figure 5.3, for N d = 210 17 cm −3 ,
then  n  3800 cm 2 /V-s which gives
 = (1.6 10 −19 )(3800 )(2 10 17 )
 = e p N a
or N a =
1
1.6 10 (220 ) 8 10 16
= 0.355  -cm
(b)  = e n N d
=
=
1.80
(1.6 10 )(380 )
−19
= 2.96  10 16 cm −3
_______________________________________
5.3
(a)  = e n N d
= 121 .6 (  -cm) −1
_______________________________________
5.5
R=
L
A
or  n =
(
)
10 = 1.6 10 −19  n N d
=
From Figure 5.3, for N d = 610 16 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 = 10 16 cm −3
and
 = (1.6 10 −19 )(1050 )(6 10 16 )
0.20 =
1
(1.6 10 )
−19
po =
p Na
From Figure 5.3, for N a = 10 17 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 =
1
(1.6 10 )
−19
p
Na
From Figure 5.3, for N a = 810 16 cm −3 we
find  p  220 cm 2 /V-s which gives
(
ni2
1.8 10 6
=
no
10 16
)
2
= 3.24 10 − 4 cm −3
(b)
J = e n n o 
For GaAs doped at N d = 10 16 cm −3 ,
 n  7500 cm 2 /V-s
Then
J = 1.6 10 −19 (7500 ) 10 16 (10 )
or
J = 120 A/cm 2
(b) (i) p o = N a = 10 16 cm −3
(
)
no =
( )
ni2
= 3.24 10 − 4 cm −3
po
(ii) For GaAs doped at N a = 10 16 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 p o 
(
= 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 10 13 cm −3
(
)
(
)
(d)   e p p o
or
(
)
5.8
L
A
=
(e
L
p Na A
)
For N a = 210 cm
16
−3
, then
 p  400 cm 2 /V-s
R=
(
−19
1.6 10
= 68 .93 
or
(0.075 )
)(400 )(2 10 16 )(8.5 10 −4 )
V
2
=
= 0.0290 A
R 68 .93
I = 29.0 mA
I=
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 =
=
epo
1.6 10 −19 2 10 16
For (a), J =
(
)(
)
= 1.066  10 cm/s
4
9.67 10 −3
= 11.38 A/cm 2
8.5 10 − 4
11.38
d =
1.6 10 −19 2 10 16
For (b), J =
(
)(
)
= 3.55  10 cm/s
_______________________________________
3
5.9
(a) For N d = 210 15 cm −3 , then
 n  8000 cm 2 /V-s
V
5
=
= 200 
I 25  10 − 3
L
R=
(e n N d )A
R=
0.01 = 1.6 10 −19 (480 ) p o
Then
p o = 1.30 10 14 cm −3 = N a − N d
So
N a = 1.30 10 14 + 10 15 = 1.13 10 15 cm −3
Note: For the doping concentrations
obtained, the assumed mobility values are
valid.
_______________________________________
(a) R =
(b) R  L  R = (68 .93 )(3) = 206 .79 
or L = (e n N d )RA
(
)
(
)
(
= 1.6 10 −19 (8000 ) 2 10 15 (200 ) 5 10 −5
= 0.0256 cm
I
(b) J = = eno d
A
I
or  d =
A(eno )
=
25  10 −3
1.6  10 −19 2  10 15
(5 10 )(
−5
= 1.56  10 6 cm/s
(c) I = (en o d )A
(
)(
)(
)(
)(
)
)
)
= 1.6 10 −19 2 10 15 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
______________________________________________________________________________________
5.10
(a)  =
(b) N a = N d = 10 16 cm −3
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
(b)
 d =  n  = (800 )(3)
or
 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
  n  290 cm 2 /V-s
 p  130 cm 2 /V-s
=
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 n o + e p p o e  n +  p n i
)
(
  n  1350 cm 2 /V-s
 p  480 cm 2 /V-s
1
(1350 + 480 ) 1.5 10 10
= 2.28 10  -cm
10
)
)
)
2

or
1
1
=
e n (8) 1.6 10 −19 (1350 )
which gives
no = 5.79 10 14 cm −3
and
no =
(
(
)
)
2
ni2
1.5 10 10
=
= 3.89 10 5 cm −3
no
5.79 10 14
Note: For the doping concentrations obtained
in part (b), the assumed mobility values are
valid.
_______________________________________
po =
(a) N a = N d = 10 14 cm −3
(
−19
ni2
1.8 10 6
=
= 2.49 10 −5 cm −3
po
1.3 10 17
(b) Silicon:
1
 =  e n no
no =
(1.6 10 )
1
(1.6 10 )(290 + 130 )(1.5 10 )
 p  240 cm 2 /V-s
tt =
5
(c) N a = N d = 10 18 cm −3
From Figure 5.3, and using trial and error, we
find
p o  1.3 10 17 cm −3 and
Then
−19
= 2.51 10  -cm
(
 d = 9.5 10 6 cm/s
=
)
5.13
(a) GaAs:
  e p p o  5 = 1.6 10 −19  p p o
For GaAs:  d = 7.5 10 cm/s
Then
d
10 −4
tt =
=
= 1.33  10 −11 s
 d 7.5 10 6
(b) Silicon: For  = 50 kV/cm,
(
(
5
= 9.92 10 5  -cm
_______________________________________
6
=
1
(1250 + 410 ) 1.5 10 10
(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
n i (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 
(
)
( )
 10 19 2
= (0.0259 ) ln 
 3.91 10 9
which gives
E g = 1.122 eV
(
)
2



( )
2
n i (500 K) = 2.27  10 13 cm −3
Then
 i = 1.6 10 −19 2.27 10 13 (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 10 15
)
−1
= 0.2496 (  -cm)
(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 10 15 )
= 0.346 (  -cm) −1
(ii) For T = 400 K ( 127  C),
  n  670 cm 2 /V-s
 = (1.6 10 −19 )(670 )(1.2 10 15 )
= 0.129 (  -cm) −1
_______________________________________
5.17
t
 i = 4.39 10 −6 (  -cm) −1
(ii) Ge:
 i = 1.6 10 −19 2.4 10 13 (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 
2.23  10 − 2 85  10 −8
(iii) GaAs:
200  10 −4
R=
= 9.19  10 12 
2.56  10 −9 85  10 −8
_______________________________________
R=
)
(
(
(ii) Ge:
From Figure 5.3, for N d = 1.2 10 15 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
n i2 (500K) = 10 19
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
=
e n 2 10 16
T
16
) x −
(
)
(
)

x


2(1.111T ) 0



T2
T −

2(1.111T ) 

2
T
(
= 1.6 10
)(750 )(2 10 )(0.55)
(c) I =
V =
L
150 10 − 4
−5
= 1.32  10 A
or I = 13.2  A
(d) Top surface;
 = 1.6 10 −19 (750 ) 2 10 16
(
)
(
(
(
)
)
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  10 19
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 10 15
)
)
 − 1.10 
= 2  10 19 1 10 19 exp 

 0.0259 
2
−1
(
)(
 − Eg 

(a) ni2 = N c N  exp 

 kT 
16
 avg = 1.32 (  -cm) −1
 avg A
(1.32 )(7.5 10 −4 )(10 −4 )
(
5.21
= e n 2 10 16 (0.55 )
−19
(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 = 10 14 cm −3 >> ni  n o = 10 14 cm −3
Then
J =  = e n n o 
(
)
( )
= 1.6 10 −19 (1000 ) 10 14 (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 10 13
) = (5 10 )
2
13 2
+ ni2
and yields
ni2 = 5.25 10 26
3
 − Eg 
 T 

= 2 10 19 110 19 
 exp 

 300 
 kT 
(
)(
)
or
3


− 1.10
 T 
2.625 10 −12 = 
 exp 

(
)(
)
300
0
.
0259
T
300




By trial and error, we find
T = 456 K
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 5
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
5.22
(b) From Figure 5.3,
n-type:  n  1100 cm 2 /V-s
n2
(a)  = e n no + e p p o and no = 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
dpo
p o2
p-type:  p  400 cm 2 /V-s
compensated:  n  1000 cm 2 /V-s
(c) n-type:  = e n n o
(
= 8.8 (  -cm)
p-type:  = e p p o
(
1/ 2
 
p o = n i  n  (Answer to part (b))
 p 


Substituting into the conductivity expression
e n n i2
 =  m in =
1/ 2
ni  n  p
) 
 (
+ e p n i  n  p
) 
1/ 2
which simplifies to
 min = 2eni  n  p
The intrinsic conductivity is defined as
i
 i = eni ( n +  p )  eni =
n +  p
The minimum conductivity can then be
written as
2 i  n  p
 m in =
n +  p
_______________________________________
5.23
(a) n-type: n o = N d = 510 16 cm −3
(
n2
1.5 10 10
po = i =
no
5 10 16
2
10 2
no
−3
= 1.125  10 4 cm −3
2  10
compensated: n o = N d − N a
16
= 5  10
16
− 2  10
16
= 3 10 cm −3
16
(1.5 10 )
=
10 2
po
3  10 16
(
(
)
−1
)
(
= 1.6 10 −19 (1000 ) 3 10 16
)
−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
1

=
1
1
+
1
2
+
1
3
1
1
1
=
+
+
2000 1500 500
= 0.00050 + 0.000667 + 0.0020
1
= 4.5  10 3 cm −3
(1.5 10 )
=
)
= 1.28 (  -cm)
compensated:  = e n n o

p-type: p o = N a = 210 cm
)
or
)
16
(
−1
= 1.6 10 −19 (400 ) 2 10 16
which yields
 (
)
= 1.6 10 −19 (1100 ) 5 10 16
= 7.5 10 3 cm −3
= 0.003167
Then
 = 316 cm 2 /V-s
_______________________________________
5.25
 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
______________________________________________________________________________________
5.26
1

=
1
1
+
1
2
=
(
1
1
+
= 0.006
250 500
Then
 = 167 cm 2 /V-s
_______________________________________
5.27
Plot
_______________________________________
5.28
Plot
_______________________________________
5.29
 5 10 14 − n(0) 
dn

J n = eDn
= eDn 

dx
 0.01 − 0 
 5 10 14 − n(0) 

0.19 = 1.6 10 −19 (25 )

0.010


Then
(0.19 )(0.010 ) = 5 10 14 − n(0)
1.6 10 −19 (25 )
which yields
n(0) = 0.25 10 14 cm −3
_______________________________________
(
)
(
)
5.30
J n = eDn
dn
n
= eDn
dx
x
)
16
15
5.31
dn
n
= eDn
dx
x
 10 15 − 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 10 14 cm −3
(a) J n = eDn
(
)
)
n(x1 ) = 8.91 10 14 cm −3
_______________________________________
5.32
2
dp
d  16 
x 
= −eD p
10 1 +  
dx
dx 
 L  
J p = −eD p
= −eD p 
10 16 
x
 21 + 
L
L


(a) For x = 0 ,
− 1.6 10 −19 (10 ) 10 16 (2)
Jp =
12 10 − 4
= −26.7 A/cm 2
(b) For x = −6  m,
(
) ( )
(
) ( )
6

− 1.6 10 −19 (10 ) 10 16 (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 = eDn
= eDn
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
_______________________________________
(
 10 15 − 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 )

( )
− eDn 10 15 e − x / Ln
Ln
At x = 0 ,
− 1.6 10 −19 (25 ) 10 15
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 10 15
Jp =
= −16 A/cm 2
−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 10 15
(ii) J p =
22.5 10 − 4
= 17.07 A/cm 2
1.6 10 −19 (10 ) 5 10 15 e −1
(b) (i) J p =
50 10 − 4
= 0.589 A/cm 2
1.6 10 −19 (48 ) 5 10 15 e −1
(ii) J p =
22.5 10 − 4
= 6.28 A/cm 2
_______________________________________
(
) (
)
(
) (
)
(
) (
)
(a) J n = eDn
dn
J n = e n n + eDn
dx

(
)
− eDn 2 10 15 e − x / L
L
− 1.6 10 −19 (27 ) 2 10 15 e − x / L
=
15 10 − 4
= −5.76 e − x / L
(b) J p = J Total − J n = −10 − − 5.76 e − 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 ) + eDn
dn(x )
dx
We have  n = 8000 cm 2 /V-s, so that
(
or
) ( )
+ 1.6 10 −19 (25 ) 10 16
 −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 )10 16 exp 
 
 18 

(

dn
d
= eDn
2  10 15 e − x / L
dx
dx
D n = (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  10 15
(
)
 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  10 15
Then
−x
n(x ) = 6.51  10 15 − 3.255  10 15 exp 
 cm −3
d


(b)
At x = 0 , n(0) = 6.51 10 15 − 3.255 10 15
Or
n(0) = 3.26 10 15 cm −3
At x = 50  m,
(
)
 − 50 
n(50 ) = 6.51  10 15 − 3.255  10 15 exp 

 21 .56 
or
n(50 ) = 6.19 10 15 cm −3
(c)
At x = 50  m,
J drf = e n n(50 )
(
)
(
)
= 1.6 10 −19 (8000 ) 6.19 10 15 (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 = n i 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 = eDn
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 10 10 cm −3
Then
− 1.6 10 −19 (25 ) 1.5 10 10 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


(i) At x = 0 , J n = −2.95 10 3 A/cm 2
(ii) At x = 5  m, J n = −23 .7 A/cm 2
_______________________________________
5.39
(a) J n = e n n + eDn
(
)
dn
dx
( )
x

− 80 = 1.6  10 −19 (1000 ) 10 16 1 − 
 L
 − 10 16
+ 1.6 10 −19 (25.9)
 L
−4
−3
where L = 10  10 = 10 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
______________________________________________________________________________________
5.42
(b) For J n = −20 A/cm 2
x 
20 = (1.6 ) − 1 + 41 .44
L 
Then
13 .3
=
V/cm
x

1 − 
 L
_______________________________________

or

− (0.0259 )  − 1 
   N do e − x / L
N do e − x / L  L 
=
0.0259
0.0259
=
L
10  10 − 4
 X = 25 .9 V/cm

)
= −(25.9) 10 10 −4 = −0.0259 V
or  = −25 .9 mV
_______________________________________
5.41
From Example 5.6
(0.0259 ) 10 19 = (0.0259 ) 10 3
x =
10 16 − 10 19 x
1 − 10 3 x
( )
)
(
(
  dx
x
)
)
(
)
or
0
10 −4
( )
(
= −(0.0259 ) 10
3
0
( )
dx
1 − 10 3 x

)
 −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 10 16 exp 
 
  L 
10 −4
V =−
(
)
Now
J drf = e n n
( )
)
(
dN d (x )
dn
= eDn
dx
dx
eDn
−x
=
 N do exp 

(− L )
 L 
We have
 kT 
Dn =  n 
 = (6000 )(0.0259 )
 e 
or
D n = 155 .4 cm 2 /s
Then
− 1.6  10 −19 (155 .4) 5  10 16
−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
(
L
(b)  = −  X dx = −(25 .9 )(L − 0 )
(
0.0259
= 500 V/cm
L
Which yields L = 5.18  10 −5 cm
_______________________________________
So  X =
J diff = eDn
=
0
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) D n = (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
_______________________________________
=
(
)(
)(
)
)(
(1.6 10
−19
(
−5
)(
)
(b) V H = negative  n-type
(c) n =
=
)
=
−I x Bz
edVH
(
)(
)
− 0.5  10 −3 6.5  10 −2
1.6  10 −19 5  10 −5 − 0.825  10 −3
(
)(
)(
)
(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 )
21
−4
n = 4.924  10 21 m −3 = 4.924  10 15 cm −3
(d)
IxL
n =
enVxWd
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 =
enVxWd
−6
(1.6 10
−3
21
or
or
=
(0.5 10 )(10 )
)(6.01 10 )(15 )(10 )(10 )
−3
−19
5.49
(a) V H =  H W = − 16.5 10 −3 5 10 −2
or
V H = −0.825 mV
(a) V H =
(
)(
−3
= 0.03466 m 2 /V-s
or  n = 346 .6 cm 2 /V-s
_______________________________________
5.47
−I x Bz
ned
− 250  10 −6 5  10 −2
=
5  10 21 1.6  10 −19 5  10 −5
)
or n = 6.01  10 15 cm −3
IX L
(c)  n =
enVX Wd
(b) (i)  p =
(
)(
= 6.01  10 m
(ii) D n = (0.0259 )(6200 ) = 160 .6 cm /s
VH =
(
21
2
(a)
(
− I X BZ
− 0.50  10 −3 (0.10 )
=
edVH
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
−I x Bz
edVH
(b) n =
=
(
)(
)
− 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  10 14 cm −3
IxL
(c)  n =
enVxWd
(
)(
)


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
(d)  =
or
1

= e n n
(
)
(
= 1.6  10 − 19 (8182 ) 8.68  10 14
)
 = 0.88 (  -cm)
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 6
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
Chapter 6
Exercise Solutions
Ex 6.4
Ex 6.1
Rn =
n 10 15 e −t / 10
=
 no
10 − 6
Rn =
For t = 0 ,
t = 1  s,
−6
cm −3 s −1
10 15
= 10 21 cm −3 s −1
10 −6
10 15 e −1 / 1
= 3.68 10 20 cm −3 s −1
10 −6
t = 4  s,
Rn =
10 15 e −4 / 1
= 1.83 10 19 cm −3 s −1
10 −6
t = 10  s,
Rn =
10 15 e −10 / 1
= 4.54 10 16 cm −3 s −1
10 −6
_______________________________________
Rn =
Ex 6.2
(a) 10 14 e − t / 50 = 10 14 e −1
 t = 50 ns
(b) 10 14 e −t / 50 = 10 13
 (
)
(a) Ln = Dn no = (25 ) 5 10 −7
1/ 2
= 3.536  10 −3 cm
or L = 35.36  m
n = 10 15 e − x / Ln ( x  0 )
or n = 10 15 e + x / Ln ( x  0 )
(i) n = 10 15 e 0 = 10 15 cm −3
(ii) n = 10 15 e −30 / 35.36 = 4.28  10 14 cm −3
(iii) n = 10 15 e −50 / 35.36 = 2.43  10 14 cm −3
(iv) n = 10 15 e −85 / 35.36 = 9.04  10 13 cm −3
(v) n = 10 15 e −120 / 35.36 = 3.36  10 13 cm −3
_______________________________________
(b)
Ex 6.5
(
)
 − x −  p 0t 2 


exp
1/ 2
4D p t
4D p t


−7
−3
(a)  p  0 t = (400 )(100 ) 10 = 4  10 cm
p(x, t ) =
(
e
−t /  p 0
)
(
)
or = 40  m
(i) x = 20  m.
(
 10 14 
t = 50 10 −9 ln  13  = 1.15 10 −7 s
 10 
or t = 115 ns
_______________________________________
 − − 2 10 −3
0.36788
exp 
−3
−6
3.545 10
 4 10
= 38.18
(ii) x = 40  m,
Ex 6.3
0.36788
exp 0
3.545  10 − 3
= 103 .8
(iii) x = 60  m,
(
)
(
(a) p(t ) = g  po 1 − e
(
= 5  10
21
−t /  po
)(10 )(1 − e
(
(
(i) p(0) = 5 10 1 − e
14
(
)
)
− t /  po
(
−0
− t /  po
)
)= 0
(ii) p 10 −7 = 5 10 14 1 − e −1 / 1
(
)
)
= 3.16  10 14 cm −3
)
(
(iii) p 5 10 −7 = 5 10 14 1 − e −5 / 1
(
)
= 4.966  10 14 cm −3
(iv) p() = 5 10 14 1 − e −
)



2
p =
−7
= 5  10 1 − e
14
p =
)
= 5 10 14 cm −3
(b) p(max ) = 5 10 14 = (0.01)N d
Yes, low-injection condition is met.
_______________________________________
(
)
(
)
 − 2 10 −3
0.36788
p =
exp

−6
3.545  10 −3
 4 10
p = 38 .18
(b) x = 40  m
(i) t = 5  10 −8 s,
 − 2 10 −3
0.60653
exp

−6
2.50663 10 −3
 2 10
= 32.75
p =
2
2






Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 6
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
(ii) t = 10 −7 s,
0.36788
p =
exp 0
3.545  10 − 3
= 103 .8
(iii) t = 2  10 −7 s,
 − − 4 10 −3 2 
0.1353
p =
exp


−6
5.013  10 −3
 8 10

= 3.65
_______________________________________
(
)
Ex 6.6
(a) For N d = 510 15 cm −3 in GaAs, from
Figure 5.3,  n  7500 cm 2 /V-s.
 = e n N d = (1.6 10 −19 )(7500 )(5 10 15 )
= 6 (  -cm) −1
 (13.1) 8.85 10 −14
Then  d = =

6
= 1.93  10 −13 s
or  d = 0.193 ps
(
(b) For N a = 210 cm
16
−3
)
We have
 = e p N a = (1.6  10 −19 )(400 )(2  10 16 )
−1
= 1.28 (  -cm)
 (11.7 ) 8.85 10 −14
Then  d = =

1.28
−13
= 8.09  10 s
or  d = 0.809 ps
_______________________________________
)
Ex 6.7
p o = N a − N d = 10 16 − 3 10 15
2
i
(
)
10 2
n
1.5 10
=
= 3.214 10 4 cm −3
po
7 10 15
(a) In thermal equilibrium,
p 
E Fi − E F = kT ln  o 
 ni 
no =
Ex 6.8
n-type; n o = 10 15 cm −3 , p o = 2.25 10 5 cm −3
 no =  po = 5  10 −7 s
Figure 5.3,  p  400 cm 2 /V-s.
= 7  10 15 cm −3
 7 10 15 + 4 10 14 

= (0.0259 ) ln 

1.5 10 10


= 0.33952 eV
 n + n 

E Fn − E Fi = kT ln  o

 ni 
 3.214 10 4 + 4 10 14 

= (0.0259 ) ln 

1.5 10 10


= 0.26395 eV
_______________________________________
n = p = 10 14 cm −3 ,
in silicon, from
(
(b) Quasi-Fermi levels,
 p + p 

E Fi − E Fp = kT ln  o

 ni 
 7 10 15
= (0.0259 ) ln 
10
 1.5 10
= 0.33808 eV




R=
(
(n

+ n )( p o + p ) − ni2
 po (n o + n + ni ) +  no ( p o + p + ni )
o
)( ) (
)
 (5 10 )(10 + 10 ) + (5 10 )(10 )
 10 15 + 10 14 10 14 − 1.5 10 10
−7
15
2
−7
14
14
or
R = 1.83  10 20 cm −3 s −1
_______________________________________
Ex 6.9
  p0s 
=0
(a) For  p 0 s = 0  p s = p B 
  p0 


From Equation (6.109),
+ x / Lp
− x / Lp
p(x) = g  p0 + Ae
+ Be
As x → , p = g  p 0 = 10 14 cm −3
 A=0
As x → 0, p = 0  B = − g  p 0
(
Then p(x ) = g  p 0 1 − e
(b)
(c)
− x / Lp
)
p(x = 0) = 0
p(0) p(0)
R =
=
 R = 
 p0s
0
Note: p (0 ) = 0 is a result of R =  .
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 6
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
Ex 6.10
TYU 6.4
n-type; Minority carriers = holes
dp
d (p (x ))
J diff = −eD p
= −eD p
dx
dx
−19
15
1.6  10 (10 ) 10
 − 10 
=−
exp 

−4
− (31 .6 10 )
 31 .6 
Then hole diffusion current density
J diff = +0.369 A/cm 2
−x / Lp


se

(a) p (x ) = g  p 0 1 −

D p L p + s 

(i) For s →  ,
(
(
p(x ) = g  p0 1 − e
(ii) For s → 0 ,
p(x) = g  p 0
)
− x / Lp
(
)
(b)
(i) For s →  , p (0 ) = 0
(ii) For s → 0 , p(0) = g  p 0 ,
We have
J diff (electrons) = − J diff (holes)
 p (x ) = constant
_______________________________________
Test Your Understanding Solutions
TYU 6.1
(a) p-type; Minority carriers = electrons
 −t 

(b) n(t ) = n(0) exp 

  no 
Then
 −t 
n(t ) = 10 15 exp 
 cm −3
−6
 5 10 
_______________________________________
(
)(
)

 −t
1 − exp 
−6
 5 10


 −t
or n(t ) = 5 10 14 1 − exp 
−6
 5 10







TYU 6.3
−x

 Lp 


n(x ) = p (x ) = n(0 ) exp 
(10 )(10 −6 ) = 31.6 10 −4 cm
Then
−x

 cm −3
−4
 31 .6  10 
_______________________________________

n(x ) = p(x ) = 10 15 exp 
_______________________________________
TYU 6.5
p =
(
exp − t  po
(4 D t )
)
1/ 2
p
(a)
(b)
(c)
Now
(c) As t →  , n() = 510 14 cm −3
_______________________________________
L p = D p po =
Then electron diffusion current density
J diff = −0.369 A/cm 2
(d)
TYU 6.2
(a) p-type; Minority carriers = electrons

 − t 

(b) n(t ) = g  no 1 − exp 


  no 
= 10 20 5 10 −6
) ( )
exp (− 1 5)
 p = 73.0
(4 )(10 )(10 )
−6
1/ 2
exp (− 5 5)
(4 )(10)(5 10 )
−6
1/ 2
 p = 14.7
exp (− 15 5)
(4 )(10)(15 10 )
−6
1/ 2
exp (− 25 5)
(4 )(10 )(25 10 )
−6
1/ 2
 p = 1.15
 p = 0.120
x =  p  o t = (386 )(10 )t
Then
(a) x = 38.6  m
(b) x = 193  m
(c) x = 579  m
(d) x = 965  m
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 6
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
(c) (i) x −  p  o t
TYU 6.6
Using the results from TYU 6.5, we find
 − x −  p ot 2 
exp − t  p


p =

exp
1/ 2
4D p t
4 D p t


(a) (i) x −  p  o t
(
(
)
)
(
(
= 1.093 10 −2 − (386 )(10 ) 10 −6
= 7.07  10 −3
(

(
)
(
)
)
(
(ii) x −  p  o t
(
= −3.21 10 −3 − (386 )(10 ) 10 −6
= −7.07  10
(
 − − 7.07  10 −3
−6
 4(10 ) 10
(
)
)
2
)



or
p = 20.9
(b) (i) x −  p  o t
(
= 2.64 10 −2 − (386 )(10 ) 5 10 −6
= 7.1 10
)
−3
(
)
 − 7.110 −3 2
−6
 4(10 ) 5 10
p = 14 .7 exp 
(



)
or
p = 11.4
(ii) x −  p  o t
(
= 1.22 10 −2 − (386 )(10 ) 5 10 −6
= −7.1  10 −3
(
)
)
 − − 7.110 −3 2 
p = 14 .7 exp 

−6
 4(10 ) 5 10

or
p = 11.4
(
(
)



(



)
p = 1.06
(ii) x −  p  o t
(
= 5.08 10 −2 − (386 )(10 ) 15 10 −6
= −7.1  10
)
−3
(
)
 − − 7.110 −3 2 

−6
 4(10 ) 15  10 
p = 1.15 exp 
or
(
)
p = 1.06
_______________________________________
−3
p = 73 .0 exp 
2
)
−3
 − 7.1 10 −3 2
p = 1.15 exp 
−6
 4(10 ) 15  10
or
)
)
(
= 7.1 10
)
 − 7.07 10 −3
exp (− 1 5)
p =

exp

−6
 4(10 ) 10
(4 )(10 ) 10 −6 1 / 2
 − 7.07 10 −3 2 
= 73 .0 exp 

−6
 4(10 ) 10

or
p = 20.9
(
= 6.50 10 −2 − (386 )(10 ) 15 10 −6
)
)
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 6
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Chapter 6
6.1
n o = N d = 510 15 cm −3
(
)
2
ni2
1.5 10 10
=
= 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 10 11 cm −3 s −1
−7
2 10
p o + p
4.5 10 4 + 10 14
=
2 10 −7
 p0
(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
_______________________________________
6.4
(a) E = h =
6.2
p o = N a = 210 16 cm −3
(
(a) R  =
)
2
= 1.62 10 − 4 cm −3
n 5 10 14
=
= 10 21 cm −3 s −1
 n 0 5 10 − 7
(b) R p =
 pt =
po
=
 pt
no
 nt
=
(
) (
)
po
2  10 16
 n 0 =
 5  10 − 7
no
1.62  10 − 4
(
(6.625 10 )(3 10 )
−34
)
E = 3.15  10 −19 J; energy of one photon
Now
1 W = 1 J/s  3.17  10 18 photons/s
Volume = (1)(0.1) = 0.1 cm 3
Then
3.17 10 18
g=
0.1
= 3.17  10 19 e-h pairs/cm 3 -s
(b)
n = p = g = 3.17 10 19 10 10 −6
or
n = p = 3.17 10 14 cm −3
_______________________________________
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
 pO
no = N d = 10 16 cm −3
po =
Then
10 16
(
ni2
1.5 10 10
=
no
10 16
2.25  10
 nO
20  10 − 6
which yields
 nO = 8.89 10 +6 s
=
4
)
2
= 2.25 10 4 cm −3
)(
)
6.5
= 6.17  10 s
_______________________________________
13
 nO
8
6300 10 −10
(
no
 n0

=
or
= 5 10 20 cm −3 s −1
_______________________________________
n2
1.8 10 6
no = i =
po
2 10 16
hc
The hole particle current density is
Jp
F p+ =
=  p − D p p
(+ e) p
Now
 • F p+ =  p  • ( p ) − D p  • p
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 •  )
+gp −
p
=
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

n
Multiply Equation (1) by n and Equation
+g−R=
p
=0
For steady-state,
t
Then
0 = − • F p+ + g p − R p
(2) by  p p , and add the two equations.
For a one-dimensional case,
dF p+
= g p − R p = 10 20 − 2  10 19
dx
or
dF p+
= 8 10 19 cm −3 s −1
dx
_______________________________________
6.7
From Equation (6.18),
dFp+
0=−
+ 0 − 2  10 19
dx
or
dFp+
= −2 10 19 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
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 pDn  2

 (n )

 n n +  p p 

  n  p ( p − n)
+ 
  • (n )
  n n +  p p 
 (n )
+ (g − R ) =
t
Define
 n nD p +  p pDn Dn D p (n + p )
D =
=
n n +  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
= −1340 cm 2 /V-s
(b) For holes,  pt =  p 0 = 2  10 −6 s
For electrons,
p
n
=
 nt  p 0
 kT 
(b) D  = D n = 
   n = (0.0259 )(1300 )
 e 
= 33.67 cm 2 /s
=  n 0 = 10 −7 s
(c)  nt
p o = N a = 710 15 cm −3
(
n2
1.5 10 10
no = i =
Na
7 10 15
5.124  10 13
)
2
  nt = 9.12 10 −6 s
_______________________________________
 pt
6.11
3.214 10 4 7 10 15
=
 pt
10 −7
so  pt = 2.18  10 4 s
_______________________________________
6.10
For Ge: ni = 2.4 10 13 cm −3
no =
=
4 10
2
13

 + ni2

 4 10
+ 
 2
= 5.124  10 cm
13
(
13
2

 + 2.4 10 13


(
)
2
−3
)
2
 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 10 13 + 1.124 10 13 )
=
(101)(5.124 10 13 ) + (49.2)(1.124 10 13 )
= 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)
 = e n no + e p po + e( n +  p )(p)
so
ni2
2.4 10 13
=
= 1.124 10 13 cm −3
n o 5.124 10 13
(a) We have:
 n = 3900 cm 2 /V-s, D n = 101 cm 2 /s
po =
 = e n n + e p p
or
2
Nd
N
+  d
2
 2
1.124  10 13
2  10 − 6
=
 nt
= 3.21  10 4 cm −3
no
p
= o
 nt
nn +  p p
(3900 )(1900 )(1.124 10 13 − 5.124 10 13 )
(3900 )(5.124 10 13 ) + (1900 )(1.124 10 13 )
=
From Figure 5.3,  n  1300 cm 2 /V-s
 n  p ( p − n)
(
)
 = e  n +  p (p )
In steady-state, p = g  pO
So that
 = e  n +  p g  pO
(
)(
)
_______________________________________
6.12
(a)
p o = N a = 10 16 cm −3
(
)
2
ni2
1.5 10 10
=
= 2.25 10 4 cm −3
po
10 16
 = 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 10 14
−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
______________________________________________________________________________________
 = (1.6 10
)(380 )(10 )
+ (1.6 10 )(900 + 380 )
)
 (4 10 )(1 − e
) (  -cm)
 = 0.608 + 0.0819 (1 − e
Then
−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 + 0.109 e
1.248 + 0.109 e
I=
(
−t /  p 0
= 2.496 10
)
(
)(
)(
) cm
= (2  10 )(1 − e
= 4  10 21 5  10 −8 1 − e
−t /  p 0
(10 )(10 )
−5
s,
(
p (10 −6 ) = (2  10 14 )1 − e −10
−6
p = (2 10 14 )e
(
− t −10
−6
−8
)
= 1.4  10 16 cm −3
(a) n = p = g  n 0
5 10 14 = 2 10 21 n0
  n 0 = 2.5 10 −7 s
(
(b) n = p = g  n 0 1 − e −t /  n 0
(b) n o = 5 10 15 cm −3
 = e n no + e( n +  p )p
For 0  t  10 −6 s,
R =
)(7500 )(5 10 )
+ (1.6 10 )(7500 + 310 )
)
 (2  10 )(1 − e
15
(
(
 = 6.0 + 0.250 e
) (  -cm)
)
− t −10 −6 /  p 0
(
)
n
5  10
=
1 − e −t /  n 0
 n 0 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
= 6.0 + 0.250 1 − e
(
= 5 10 1 − e
14
14
A
p o = N a − N d = 2 10 16 − 6 10 15
)/  p 0 cm −3
−19
−t /  p 0
6.15
= 2  10 14 cm −3
Then for t  10 −6 s,
+ 2.18 10 −4 e
or I = 2.496 + 0.218e
mA
_______________________________________
−3
/ 510
−3
−t /  p 0
)
−t /  p 0
−t /  p 0
14
For t  10 −6 s,
−t /  p 0
0.05
n = p = g  p0 1 − e
 = (1.6 10
−t /  p 0
14
(ii)  ( ) = 0.690 (  -cm)
_______________________________________
At t = 10
−t /  p 0
−t / 
−1
−6
)
= 4 10 14 e p 0 cm −3
 = 1.6 10 −19 (1300 ) 8 10 15 − 2 10 15
(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  10 14 = 5  10 14 1 − e −t /  n 0
4
)
t =  n0 ln (1.3333 ) = 7.19 10 −8 s
(
)
(
)
1
(ii)   5  10 14 = 5  10 14 1 − e −t /  n 0
2
t =  n 0 ln (2) = 1.73 10 −7 s
6.14
V
L
; R=
A
R
A
I=
V
L
For N I = N d + N a = 8 10 15 + 2 10 15
I=
= 10 16 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  10 14 = 5  10 14 1 − e −t /  n 0
4
t =  n 0 ln (4) = 3.47 10 −7 s
(
)
(
)
(iv) (0.95 ) 5 10 14 = 5 10 14 1 − e −t /  n 0
)
t =  n 0 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  10 14 e − (210 ) / (510 )
1
= 6  10 15 cm −3
po =
(
2
i
n
1.8 10
=
no
6 10 15
(a) Ro =
po
 p0
= 9.16  10 12 cm −3
)
6 2
5.4 10
 4 10 4 =
so  p 0 = 1.35  10
−8
s
(
= 2.7  10 cm
13
(
 p0
(b) (i) n(0) = 5 10 cm
)(1.35 10 )
(
−8
21
−8
s
−t /  p 0
)(
)
)(
= 5  10 20 5  10 −7 1 − e
(
= 2.5  10 1 − e
−t /  p 0
−t /  p 0
At t = 5  10 s,
p = 2.5 10 14 1 − e −1 / 1
(
) cm
(
= 31.08 cm 2 /s

d (n )
d
= eDn
2  10 14 e − x / Ln
dx
dx
−eDn
=
2 10 14 e − x / Ln
Ln
)
)
(
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 10 14 e
(
)
−3
For t  2  10 −6 s,
(
)
− t − 210 −6 /  p O
)
( )(
)
)
(
)
)
− 1.6  10 −19 (31 .08 ) 2  10 14 − x / Ln
e
5.575  10 −3
(
_______________________________________
cm −3
(ii) p 2 10 −6 = 2.454 10 14 cm −3
_______________________________________
6.18
(a) For 0  t  2  10 −6 s
n(t ) = g  n 0 e −t /  n 0
(

J n = −0.1784 e − x / Ln A/cm 2
Holes diffuse at same rate as minority carrier
electrons, so
J p = +0.1784 e − x / Ln A/cm 2
−3
−6
−7
p = 2.5  10 14 1 − e − (210 ) / (510 )
14

(
cm −3
(ii) p 5 10 −7 = 1.58 10 14 cm −3
(b) (i) For 0  t  2  10
1/ 2
(a) n(x ) = p(x ) = 2 10 14 e − x / Ln cm −3
(b) J n = eDn
(
)
= 5.575  10 −3 cm
)
− t −510 −7 /  p O
−6
(
Ln = Dn n 0 = (31.08 ) 10 −6
−3
= 1.58  10 14 cm −3
p(t ) = 1.58 10 14 e
)
6.19
p-type; minority carriers - electrons
 kT 
Dn = 
  n = (0.0259 )(1200 )
 e 
)
−7
For t  5  10 −7 s
)
−3
(iii) n() = 510 14 cm −3
_______________________________________
6.17
(a) (i)For 0  t  5  10 −7 s
14
+ 9.16  10 12
(ii) n 2 10 −6 = 9.16 10 12 cm −3
_______________________________________
p(t ) = g  p 0 1 − e
)
= 4.908 10 14 1 − e −t /  n 0 + 9.16 10 12 cm −3
−3
(c)  =  p 0 = 1.35  10
(
(
n = (5 10 14 − 9.16 10 12 )1 − e −t /  n 0
−4
14
(b) p = g  p 0 = 2  10
(
For t  2  10 −6 s
= 5.4 10 − 4 cm −3
6.20
(a) p-type; p pO = 10 14 cm −3
and
)
= 10 21 5 10 −7 e −t /  n 0
= 5  10 14 e − t /  n 0 cm −3
(
)
2
n i2
1.5  10 10
=
= 2.25  10 6 cm −3
p pO
10 14
(b) Excess minority carrier concentration
n = n p − n pO
n pO =
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
2
dx
Ln
The general solution is of the form
−x
+x
 + B exp 

n = A exp 

L 
L
 n 
 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 )
−
1/ 2
s2 −
−3
= 5  10 cm
d (n )
d
J n = eDn
= eDn
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 ,
Define
J n (0 ) = −0.4 A/cm 2

J p (0) = +0.4 A/cm 2
n(Ln ) = (5 10
J n (Ln ) = −0.4e
−1
)e
=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




n(0) = 5 10 14 cm −3
14
=0
Dividing by D p , we have
 ( )
where Ln = Dn n0 = (25 ) 10
−6
 pO
or
Dp s 2 −  po s −
n(x ) = 5 10 14 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 = 3 L n
n(3Ln ) = (5 10 14 )e −3 = 2.49 10 13 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
D n  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 − − 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  → 
1
and  =
Ln
(b)
For  o = 12 V/cm, then
L =
(kT e) = 0.0259
o
12
= 21 .6 10 − 4 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 )
Dn  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
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
 kT 
where D n =  n 

 e 
so
D n = (1200 )(0.0259 ) = 31 .1 cm /s
and
2
Ln =
L n = 39 .4  m
And
n = G o T (no recombination)
_______________________________________

Ln = Dn nO
or
(31.1)(5 10 −7 ) = 39.4 10 −4 cm
Dp
d 2 (p )
dx
2
+ g  = 0 or
d 2 (p )
dx
2
=−
g
Dp
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 )
(5)
continuous at x = L
dx
d (p )
(6)
continuous at x = −L
dx
Applying the boundary conditions, we find
G
p = o 5 L2 − x 2 for −L  x  +L
2D p
(
)
G L
p = o (3L − x ) for L  x  3L
Dp
p =
G o L
(3L + x ) for −3L  x  −L
Dp
_______________________________________
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
=
(
D p = 10.42 cm /s
2
6.28
(a)
 − x2
exp 
 4 Dt
is the solution to the differential equation
  2 f  f
D 2  =
 x  t
Assume that f (x, t ) = (4 Dt )
−1 / 2
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 )
2
−1 / 2
 − x2
 − 2 x 
 exp 

 4 Dt 
 4 Dt




 − x2
 −2 
+
 exp 
 4 Dt 
 4 Dt
Also
2
f
−1 / 2  − x
= (4 Dt ) 
t
 4D




 − 1 
 − x2 
 2  exp 

 t 
 4 Dt 



 − x2 
−1 / 2  − 1 

+ (4 D )  t −3 / 2 exp 

 2 
 4 Dt 
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
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D t
4D t
−x 
dx =
exp 
=1

4
Dt
4

D
t
4

D
t


−
_______________________________________

2
1
6.29
Plot
_______________________________________
E Fi − E F = 0.3294 eV
(b)
n = p = 510 14 cm −3
and
no =
n
(a) E F − E Fi = kT ln  o
 ni
(
= 10 cm
)(
)
2
= 4.5 10 4 cm −3
 n + n 

E Fn − E Fi = kT ln  o

 ni 
 4.5 10 4 + 5 10 14
= (0.0259 ) ln 
1.5 10 10





 4 10 16 

= (0.0259 ) ln 
10 
 1.5 10 
= 0.383225 eV
(b) n = p = g  p 0 = 2  10 21 5  10 −7
(
ni2
1.5 10 10
=
po
5 10 15
Then
6.30
15




or
Then
+
p 
E Fi − E F = kT ln  o 
 ni 
 5 10 15
= (0.0259 ) ln 
10
 1.5 10




or
)
−3
 n + n 

E Fn − E Fi = kT ln  o

 ni 
 4 10 16 + 10 15 

= (0.0259 ) ln 
10

 1.5 10

= 0.383865 eV
 p + p 

E Fi − E Fp = kT ln  o

 ni 
 10 15 

 (0.0259 ) ln 
10 
 1.5 10 
= 0.28768 eV
(c) E Fn − E F = 0.383865 − 0.383225
= 0.000640 eV
or
= 0.640 meV
_______________________________________
E Fn − E Fi = 0.2697 eV
and
 p + p 

E Fi − E Fp = kT ln  o

 ni 
 5 10 15 + 5 10 14
= (0.0259 ) ln 
1.5 10 10

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 10 15 + n 

So 0.00102 = (0.0259 ) ln 
15

 5 10

 0.00102 
5  10 15 + n = 5  10 15 exp 

 0.0259 
Which yields n  2  10 14 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 10 15 + 2 10 14 

= (0.0259 ) ln 

1.5 10 10


= 0.33038 eV
 p 
(c) E Fi − E Fp  kT ln  
 ni 
 2 10 14 

= (0.0259 ) ln 
10 
 1.5 10 
= 0.2460 eV
_______________________________________
6.33
 n 
(a) E Fn − E Fi  kT ln  
 ni 
 E − E Fi 
or n = ni exp  Fn

kT


(
)
 0.270 
= 1.5  10 10 exp 

 0.0259 
= 5.05  10 14 cm −3
 p + p 

(b) E Fi − E Fp = kT ln  o

 ni 
 6 10 15 + 5.05 10 14 

= (0.0259 ) ln 

1.5 10 10


= 0.33618 eV
(c) (i) E F − E Fp = E Fi − E Fp − (E Fi − E F )
(
)
 p + p 
p 
 − kT ln  o 
= kT ln  o

n 
 ni 
 i 
 p o + p 

= kT ln 

 po 
(ii) E F − E Fp
 6 10 15 + 5.05 10 14 

= (0.0259 ) ln 

6 10 15


−3
= 2.093  10 eV
or
= 2.093 meV
_______________________________________
6.34
n +n

(a) (i) E Fn − E Fi = kT ln  o

 ni 
( )
 (1.02 ) 10 16 

= (0.0259 ) ln 
6 
 1.8 10 
= 0.58166 eV
 p 
(ii) E Fi − E Fp  kT ln  
 ni 
 0.02 10 16 

= (0.0259 ) ln 
6 
 1.8 10 
= 0.47982 eV
 1.110 16 

(b) (i) E Fn − E Fi = (0.0259 ) ln 
6 
 1.8 10 
= 0.58361 eV
 0.110 16 

(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
10 16
We have
 x 
n = 10 14  
 50 
Then
 3.24 10 −4 + 10 14 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 = 10 16 cm −3 and n = p .
We find
x (  m)
p = 10 11 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
(
 n + n 

(b) E Fn − E Fi = kT ln  o

 ni 
 2 10 16 + n 

= (0.0259 ) ln 
10

 1.5 10

n = 10 11 cm −3 , E Fn − E Fi = 0.365273 eV
)
 p + p 
p 
 − kT ln  o 
= kT ln  o

n 
n
i


 i 
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10 12 cm −3
(b)
E Fn − E Fi
0.10877
0.16841
0.22805
0.28768
10 13
10 12
0.365274
10 13
0.365286
14
0.365402
10
15
10
0.366536
_______________________________________
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  = n i . For n = p = 0
R=
− 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  )
 p 
 kT ln  
 ni 
 5 10
= (0.0259 ) ln 
10
 1.5 10
12




or
E Fn − E Fi = 0.1505 eV
_______________________________________
6.37
Plot
_______________________________________
where g o = R o since these are the thermal
equilibrium generation and recombination
rates.
If g  = 0 , then g − R = − R  and
R =
−ni
 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 (n o + 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(n o + p o )
 pO (n o + ni ) +  nO ( p o + ni )
(a) For n-type; n o  p O , n o  n i
Then
R
1
=
= 10 + 7 s −1
n  pO
(b) For intrinsic, n o = p o = n i
Then
2n i
R
=
n  pO (2n i ) +  nO (2n i )
At x = + , p = g  pO so that B = 0 ,
Then
−x

 Lp 


p = g  pO + A exp 
We have
d (p )
Dp
= s(p )
dx x = 0
x =0
We can write
d (p )
−A
and (p ) = g  pO + A
=
x =0
dx x =0 L p
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  n o , p o  n i
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 




)
(10 )(10 −7 ) = 10 −3 cm
L p = D p pO =
Now
p = 10 21 10 −7
( )(
)

 − 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 = 10 14 1 −
 − x 



exp
 L p 
10 4 + s


s

(i) For s = 0 ,
p = 10 14 cm −3
(ii) For s = 2000 cm/s,

 − x 

p = 10 14 1 − 0.167 exp 
 L p 



(iii) For s =  ,

 − x 

p = 10 14 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) = 10 14 cm −3
(ii) For s = 2000 cm/s,
n =
p(0) = 0.833 10 14 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 = 10 15 cm −3
or
n(0) = g  nO = 10 15 cm −3
For x  0
(
)(
)
d (n) n
d (n) n
−
=0
− 2 =0
 nO
dx 2
dx 2
Ln
The solution is of the form
−x
+x
 + B exp 

n = A exp 

L 
 Ln 
 n 
2
W − x 

 Ln 
W 
sinh 
 Ln 
n(0) sinh
2
Dn
At x = 0 ,
n = n(0 ) = A + B
At x = W ,
 −W 
 +W 
 + B exp 

n = 0 = A 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 
  + (W − x ) 
 − (W − x )  
 exp 
 − exp 

  L n

 Ln
 
which can be written as
where
n(0) = 10 15 cm −3 and L n = 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
B=
D p + sW
s
At x = 0 , the flux of excess holes is
d (p )
10 19 = − D p
= −D p A
dx x =0
so that
− 10 19
A=
= −10 18 cm −4
10
and
10 18
(10 + sW ) = 10 18  10 + W 
B=
s
 s

The solution is now
10 

p = 10 18 W − x + 
s 

(
)
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 6
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(a) For s =  ,
p = 10 18 20 10 −4 − x cm −3
Then
d (p )
J p = −eD p
dx
(
(
)
) (
= − 1.6 10 −19 (10 ) − 10 18
)
or
J p = 1.6 A/cm 2
(b) For s = 2  10 3 cm/s,
p = 10 18 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
2 Dn
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 = 10 16 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 10 13
= 0.05848 (  -cm)
Let W = 20  m
(
)(
Then A = Wd = 20 10 −4 4 10 −4
= 80  10
So R = 10 6 =
−8
cm
)
−1
)
2
L
(0.05848 ) 80 10 −8
(
)
−2
Which yields L = 4.68  10 cm
(b) Silicon:
R = 10 6  , p = 510 13 cm −3
For N d = 10 16 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 10 13
= 0.01368 (  -cm)
Let W = 20  m
(
)(
Then A = Wd = 20 10 −4 4 10 −4
= 80  10
So R = 10 =
6
−8
cm
)
−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
Exercise Solutions
______________________________________________________________________________________
Chapter 7
Exercise Solutions
(
Ex 7.1
N N
(a) Vbi = Vt ln  a 2 d
 ni
(




)( )
)
 5 10 15 10 17 
(i) Vbi = (0.0259 ) ln 

2
 1.5 10 10

= 0.736 V
 2  10 16 2 10 15 
(ii) Vbi = (0.0259 ) ln 

2
 1.5  10 10

= 0.671 V
(b)
 5 10 15 10 17 
(i) Vbi = (0.0259 ) ln 

2
 1.8 10 6

= 1.20 V
 2  10 16 2 10 15 
(ii) Vbi = (0.0259 ) ln 

2


1.8 10 6
= 1.14 V
_______________________________________
(
(
(
(
)(
(
)
)
(
)(
(
)
)
(




)(
 5 10 16 5 10 15
= (0.0259 ) ln 
2
 1.5 10 10
= 0.7184 V
(
)
)
(
 Na

N
 d


1


 N + N 
d 
 a

 5 10 15
 
16
 5 10
= 4.52  10 −5 cm
Now
 m ax =
=
eN d x n
s
(1.6 10 )(5 10 )(4.11 10 )
(11.7 )(8.85 10 )
−19
−6
16
−14
= 3.18  10 4 V/cm
_______________________________________
Ex 7.3
N N
(a) Vbi = Vt ln  a 2 d
 ni




(
(
1/ 2
 5 10 15
 
16
 5 10
)
)

1

 5 10 15 + 5 10 16

 x n = 4.11 10 −6 cm
Now
W = x n + x p = 4.11  10 −6 + 4.11  10 −5
)(
(
)
)
)



1


 N + N 
d 
 a

1/ 2
 2(11.7 ) 8.85 10 −14 (0.7184 + 4)
=
1.6 10 −19

 2(11.7 ) 8.85 10 −14 (0.7184 )
=
1.6 10 −19

(
1/ 2
 x p = 4.11  10 −5 cm

 2  (V + V R )  N a
x n =  s bi
N
e

 d








1

 5 10 15 + 5 10 16

 5 10 15 5 10 16
= (0.0259 ) ln 
2
 1.5 10 10
= 0.718 V
Then

2 V
x n =  s bi

 e
(
 5 10 16
 
15
 5 10
)( )
)
Ex 7.2
N N
Vbi = Vt ln  a 2 d
 ni
)
)
 2(11.7 ) 8.85 10 −14 (0.7184 )
xp = 
1.6 10 −19






1/ 2

1

 5 10 15 + 5 10 16






1/ 2





1/ 2
= 1.054  10 −5 cm
or x n = 0.1054  m

 2  (V + V R )  N d
x p =  s bi
N
e

 a

(

1

 N + N
d
 a
)
 2(11.7 ) 8.85 10 −14 (0.7184 + 4)
=
1.6 10 −19

 5  10 16
 
15
 5  10

1


 5  10 15 + 5  10 16 


= 1.054  10 −4 cm
1/ 2
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 7
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
or x p = 1.054  m

 2  (V + V R )  N a + N d
W =  s bi
 N N
e

a
d


(
)





(
Now 7.2  10 4
1/ 2
(
)(
 

 
(
1/ 2
)






1

 5 10 15 + 5 10 16

(
)






1

 5 10 15 + 5 10 16

= 1.432  10 −4 cm
or x p = 1.432  m
(
N N
(a) Vbi = Vt ln  a 2 d
 ni
(
(
(
)(
 

 
(
)
= 1.576  10 −4 cm
or W = 1.576  m
_______________________________________
1/ 2
(
15
)
(
)
16

(
)
(5 10 )(2 10 ) 
(5 10 + 2 10 )
15
15
1/ 2
16
 1.6 10 −19 (13.1) 8.85 10 −14
(c) C  = 
2(1.162 + 8)

16
1/ 2
16
C  = 6.36  10 −9 F/cm 2
_______________________________________
Ex 7.6 For a one-sided junction
 e s N a 
C = 

 2(Vbi + V R ) 
Ex 7.4
(




)
(5 10 )(2 10 ) 

(5 10 + 2 10 )
1/ 2
)
)
 1.6 10 −19 (13.1) 8.85 10 −14
=
2(1.162 + 4)

15
)
 5  10 15 + 5  10 16

15
16
 5  10 5  10
)(


e s N a N d
(b) C  = 

(
)(
)
2
V
+
V
N
+
N
R
a
d 
 bi
1/ 2
 2(11 .7 ) 8.85 10 (0.7184 + 8)
W =
1.6 10 −19

1/ 2
)
C = A  C  = 10 −5 C 
)(
 5 10 3 10
= (0.0259 ) ln 
2
 1.8 10 6
= 1.173 V
 m ax




C  = 8.48  10 −9 F/cm 2
−14
(

 
 2 10 16 5 10 15 
= (0.0259 ) ln 

2


1.8 10 6
= 1.16 V
1/ 2
 2(11.7 ) 8.85 10 −14 (0.7184 + 8)
xp = 
1.6 10 −19

N N
Vbi = Vt ln  a 2 d
 ni
) 
Then V R = 3.21 V
_______________________________________
)
= 1.432  10 −5 cm
or x n = 0.1432  m
 5 10 16
 
15
 5 10
)(
5.184 10 9 = 1.1829 10 9 (Vbi + V R )
Vbi + V R = 1.173 + V R = 4.382
 2(11.7 ) 8.85 10 (0.7184 + 8)
xn = 
1.6 10 10 −19

 5 10
 
16
 5 10
)
Ex 7.5
−14
15
)
(
(
= 1.159  10 −4 cm
or W = 1.159  m
(b) Vbi = 0.718 V
2
 5  10 15 3  10 16

15
16
 5  10 + 3  10
 2(11.7 ) 8.85 10 −14 (0.7184 + 4)
=
1.6 10 −19

 5  10 15 + 5  10 16

15
16
 5  10 5  10
(
 2 1.6  10 −19 (Vbi + V R )
=
 (13 .1) 8.85  10 −14
)
15
(
)

 2e(Vbi + V R )  N a N d
=
N +N
s

d
 a

16
)

0.105  10
)(
−12

 1.6 10
= 10 −5 


(
−19
)(11.7)(8.85 10 )N
−14
2(3 + 0.765 )
(0.105 10 ) = (10 ) (2.20 10 )N
−12 2





1/ 2
a
−5 2
So N a = 5.01 10 cm
15
−3





−32
a
1/ 2
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 7
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
N N
We have Vbi = Vt ln  a 2 d
 ni
= 3.240  10 −5 cm
or W = 0.3240  m




=
(1.5 10 )
10 2
5.01 10
=
 0.765 
exp 

 0.0259 
15
−3
N d = 3.02 10 cm
_______________________________________
17
(

 2  (V )  N
x n =  s bi  a
e

 Nd

(


1


 N + N 
d 
 a

(
1/ 2
(
)
)

)

1

 4  10 15 + 3  10 16

(





1/ 2
)
 2(11.7 ) 8.85 10 −14 (0.6994 )
xp = 
1.6 10 −19

)
 3  10 16
 
15
 4 10
1/ 2
= 3.085  10 −5 cm
or x n = 0.3085  m






1

 4  10 15 + 3  10 16

= 4.469  10 −5 cm
or x p = 0.4469  m
(
1/ 2
)
 2(11.7 ) 8.85 10 −14 (0.6994 )
W =
1.6 10 −19


1

 N + N
d
 a
)





1/ 2
 4  10 15 + 3  10 16

15
16
 4  10 3  10
(
 2(11.7 ) 8.85 10 (0.7722 )
=
1.6 10 −19

−14
 10 16
 
17
 2 10
)(
 4 10 15 3 10 16
(b) Vbi = (0.0259 ) ln 
2
 1.5 10 10
= 0.699 V
 2(11.7 ) 8.85 10 −14 (0.6994 )
xn = 
1.6 10 −19

= 5.96  10 −6 cm
or x n = 0.0596  m
 2  1017 
1



 

16
17
16 

10
2

10
+
10






(
−4
16
−14
 4 10 15
 
16
 3  10
 2(11.7 ) 8.85 10 −14 (0.7722 )
=
1.6 10 −19


 2  (V )  N
x p =  s bi  d
e

 Na

−19
= 4.77  10 4 V/cm
)( )
) 
 2 10 17 10 16
(a) Vbi = (0.0259 ) ln 
2
 1.5 10 10
= 0.772 V
(
(1.6 10 )(10 )(0.3085 10 )
(11.7 )(8.85 10 )
(
Test Your Understanding Solutions
TYU 7.1
eN d x n
s
 m ax =
V 
n2
Then N d = i exp  bi 
Na
 Vt 





1/ 2
= 1.54  10 −6 cm
or x p = 0.0154  m
(
)




(1.6 10 )(3 10 )(5.96 10 )
(11.7 )(8.85 10 )
−19
−6
16
−14
= 2.76  10 4 V/cm
_______________________________________
(
)(
 5 10 16 5 10 15
Vbi = (0.0259 ) ln 
2
 1.8 10 6
= 1.186 V
 2(11.7 ) 8.85 10 −14 (0.7722 )
=
1.6 10 −19

(
 m ax =
TYU 7.2
1/ 2
 2  10 17 + 10 16

17
16
 2 10 10
)
1/ 2
= 5.064  10 −5 cm
or W = 0.5064  m

1

 2 10 17 + 10 16

 2  (V )  N + N d
W =  s bi  a
e

 Na Nd
)(
 

 
 

 
)( )
1/ 2
(

 2  (V )  N
x n =  s bi  a
e

 Nd

)
)



1


 N + N 
d 
 a

1/ 2
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 7
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
(
(
)
 5 10 15
 
16
 5 10






1

 5 10 15 + 5 10 16

1/ 2
= 5.590  10 −6 cm
or x n = 0.05590  m
(


1


 N + N 
d 
 a

1/ 2






1

 5 10 15 + 5 10 16

1/ 2
)




1/ 2
)(
 

 
1/ 2
 5 10 16
 
15
 5 10

1

 5 10 15 + 5 10 16

 5 10 15
 
16
 5 10
−14
= 3.86  10 4 V/cm
_______________________________________
)
)


1


 N + N 
d 
 a

)

1

 5 10 15 + 5 10 16






1/ 2
Also
W = x n + x p = 1.90  10 −4 cm
2(0.718 + 12 )
1.90  10 − 4
= 1.34  10 5 V/cm
_______________________________________
 m ax =
1/ 2
1/ 2
or x p = 1.73  10 −5 cm
Now






 2(11.7 ) 8.85 10 −14 (0.718 + 12 )
xp = 
1.6 10 −19

−6
16

 2  (V + V R )  N a
x n =  s bi
N
e

 d

or
)
(
(1.6 10 )(5 10 )(5.59 10 )
(13.1)(8.85 10 )
(
2(Vbi + V R ) 2(0.718 + 8)
=
W
1.575  10 − 4
or x n = 1.73 10 −4 cm
)(
1/ 2
 2(11.7 ) 8.85 10 −14 (0.718 + 12 )
xn = 
1.6 10 −19

)
 5 10 16 5 10 15
(a) Vbi = (0.0259 ) ln 
2
 1.5 10 10
= 0.718 V
Now for V R = 8 V,





)

1

 5 10 15 + 5 10 16

(
eN d x n
s
(
(
(b) For V R = 12 V
= 6.149  10 −5 cm
or W = 0.6149  m
−19
1/ 2
= 1.11  10 5 V/cm
 5  10 15 + 5  10 16

15
16
 5  10 5  10
(
)





Now
W = x n + x p  W = 1.58  10 −4 cm
Also
 2(13.1) 8.85 10 −14 (1.186 )
W =
1.6 10 −19

TYU 7.3

1

 5 10 16 + 5 10 15

 x p = 1.432  10 −5 cm
 m ax =
 2  (V )  N + N d
W =  s bi  a
e

 Na Nd
(
(
)
 2(11.7 ) 8.85 10 −14 (0.718 + 8)
xp = 
1.6 10 −19

)
= 5.590  10 −5 cm
or x p = 0.5590  m
 m ax =
 5 10 16
 
15
 5 10
 5 10 15
 
16
 5 10
 2(13.1) 8.85 10 −14 (1.186 )
xp = 
1.6 10 −19

 m ax =
(
 x n = 1.432 10 −4 cm

 2  (V )  N
x p =  s bi  d
e

 Na

 5 10 16
 
15
 5 10
)
 2(11.7 ) 8.85 10 −14 (0.718 + 8)
xn = 
1.6 10 −19

 2(13.1) 8.85 10 −14 (1.186 )
xn = 
1.6 10 −19

Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 7
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
TYU 7.4
(
)(
 3 10 16 8 10 15
Vbi = (0.0259 ) ln 
2
 1.5 10 10
= 0.717 V
(

 Na Nd
 e s

C = A


 2(Vbi + V R )  N a + N d
)(
)
(
)





)

1/ 2
 1.6 10 −19 (11.7 ) 8.85 10 −14
= 5 10 −5 
2(Vbi + V R )

(
(
)(
) 
 3 10 16 8 10 15

16
15
 3 10 + 8 10
)
1/ 2

 
or
 5.232  10 −16 
C = 5  10 −5 

 Vbi + V R 
(a) For V R = 2 V,
(
)
(
C = 5  10
−5
1/ 2
 5.232  10 −16 


 0.717 + 2 
)
1/ 2
= 6.94  10 −13 F = 0.694 pF
(b) For V R = 5 V,
 5.232  10 −16 
C = 5  10 −5 

 0.717 + 5 
(
)
1/ 2
= 4.78  10 −13 F = 0.478 pF
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 7
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Chapter 7
7.1
N N
Vbi = Vt ln  a 2 d
 ni
(a)
(b) N d = 510 16 cm −3 , N a = 510 16 cm −3
Si: Vbi = 0.778 V




(
)(
)
 2  10 15 2 10 15 
(i) Vbi = (0.0259 ) ln 

2
 1.5 10 10

= 0.611 V
 2  10 15 2 10 16 
(ii) Vbi = (0.0259 ) ln 

2
 1.5  10 10

= 0.671 V
 2  10 15 2 10 17 
(iii) Vbi = (0.0259 ) ln 

2
 1.5 10 10

= 0.731 V
(b)
 2  10 17 2  10 15 
(i) Vbi = (0.0259 ) ln 

2
 1.5 10 10

= 0.731 V
 2  10 17 2  10 16 
(ii) Vbi = (0.0259 ) ln 

2
 1.5  10 10

= 0.790 V
 2  10 17 2  10 17 
(iii) Vbi = (0.0259 ) ln 

2
 1.5 10 10

= 0.850 V
_______________________________________
(
(
(
(
(
(
(
(
(
(
(
)(
)(
)(
)(
)(
)
)
)
)
)
)
)
)
)
)
)
7.2
Si: ni = 1.5 10 10 cm −3
Ge: ni = 2.4 10 13 cm −3
GaAs: ni = 1.8 10 6 cm −3
N N
Vbi = Vt ln  a 2 d
 ni

 and Vt = 0.0259 V


(a) N d = 10 14 cm −3 , N a = 10 17 cm −3 '
Then
Si: Vbi = 0.635 V
Ge: Vbi = 0.253 V
GaAs: V bi = 1.10 V
Ge: V bi = 0.396 V
GaAs: V bi = 1.25 V
(c) N d = 10 17 cm −3 , N a = 10 17 cm −3
Si: V bi = 0.814 V
Ge: V bi = 0.432 V
GaAs: V bi = 1.28 V
_______________________________________
7.3
(a) Silicon ( T = 300 K)
 Na Nd
Vbi = (0.0259 ) ln 
 1.5  10 10
(
For N a = N d = 10 cm
14
−3
= 10
;
= 10
16
;
= 10
;
(b) GaAs ( T = 300 K)
 Na Nd
Vbi = (0.0259 ) ln 
 1.8 10 6
17
(
For N a = N d = 10 cm
= 10



; Vbi = 0.4561 V
15
14
)
2
−3
= 0.5754 V
= 0.6946 V
= 0.8139 V
)
2



; Vbi = 0.9237 V
= 1.043 V
= 10
;
= 1.162 V
17
= 10
;
= 1.282 V
(c) Silicon (400 K), kT = 0.034533
ni = 2.38 10 12 cm −3
15
;
16
For N a = N d = 10 14 cm −3 ; Vbi = 0.2582 V
= 10 15
= 0.4172 V
= 10
;
= 0.5762 V
17
= 10
;
= 0.7353 V
9
GaAs(400 K), ni = 3.29 10 cm −3
;
16
For N a = N d = 10 14 cm −3 ; Vbi = 0.7129 V
= 10 15
= 0.8719 V
= 10
;
= 1.031 V
17
= 10
;
= 1.190 V
_______________________________________
16
;
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 7
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
7.4
(a) n-side
or
x p = 0.0213  10 −4 cm = 0.0213  m
N
E F − E Fi = kT ln  d
 ni




 5 10 15
= (0.0259 ) ln 
10
 1.5 10
We have
 m ax =




=
or
E F − E Fi = 0.3294 eV
p-side
N 
E Fi − E F = kT ln  a 
 ni 
7.5
(a) n-side




N
E F − E Fi = kT ln  d
 ni




 2 10 16
= (0.0259 ) ln 
10
 1.5 10
)
 2 10 16
= (0.0259 ) ln 
10
 1.5 10
)

Vbi = 0.7363 V
(d)


1


 N + N 

d 
 a
1/ 2
)
 2(11.7 ) 8.85 10 −14 (0.736 )
=
1.6 10 −19


1
 17
 10 + 5  10 15




1/ 2
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 
(
)(
 2 10 16 2 10 16
= (0.0259 ) ln 
2
 1.5 10 10
(
)
)
or
V bi = 0.7305 V
(d)
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

)

1
 17
 10 + 5  10 15





or
or
 5  10 15
 
17
 10




E F − E Fi = 0.3653 eV
p-side
N 
E Fi − E F = kT ln  a 
 ni 
( )(
(
(
−14
or
 10 17 5 10 15
= (0.0259 ) ln 
2
 1.5 10 10
 10 17
 
15
 5  10
−4
15
_______________________________________
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 
(
−19
 m ax = 3.29 10 4 V/cm
or
 Na

N
 d
(1.6 10 )(5 10 )(0.426 10 )
(11.7 )(8.85 10 )
or
 10 17
= (0.0259 ) ln 
10
 1.5 10
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 10 16
 
16
 2 10
For 300 K;

1

 2 10 16 + 2  10 16




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
_______________________________________
(
0.75 x n = 0.25 x p 
)
xn
)
)
=3
xp
Nd
=
=3
Na
xn
So N d = 3 N a
)
(
(
or N d = 1.98  10 16 cm −3
 E − EF
N a = n i exp  Fi
 kT
xp
(



 0.365 
= 1.5  10 10 exp 

 0.0259 
(
(
or 3N a2 = 1.5  10 10



)
2
)
 0.710 
exp 

 0.0259 
which yields N a = 7.766 10 15 cm −3
N d = 2.33 10 16 cm −3
)
 0.330 
= 1.5  10 10 exp 

 0.0259 
or N a = 5.12  10 15 cm −3
(c)
 5.12 10 15 1.98 10 16
Vbi = (0.0259 ) ln 
2

1.5  10 10

2 V
x n =  s bi

 e
(
 Na

N
 d


1


 N + N 
d 
 a

 2(11.7 ) 8.85 10 −14 (0.710 )
=
1.6 10 −19

= 0.695 V
_______________________________________
1
 1 
  
15
3
   4 7.766  10
)(
(
)

1/ 2
)
)
(
)
 Na Nd 
(a) Vbi = (0.0259 ) ln 

2
 1.5  10 10 
 3N a2

0.710 = (0.0259 ) ln 

2
 1.5 10 10 
7.6
 E − E Fi
(b) N d = n i exp  F
 kT
)(
x n = 0.25W = 0.25 x n + x p
xn N d = x p N a 
 m ax = 4.75 10 4 V/cm
)
)
(
−4
16
)(
(
(
or
 m ax =
(
 2  10 15 4 10 16 
Vbi = (0.0259 ) ln 

2


1.8 10 6
= 1.157 V
For 400 K;
 2 10 15 4 10 16 
Vbi = (0.034533 ) ln 

2
 3.28  10 9

= 1.023 V
_______________________________________
(
 

 
1/ 2
)
−6
7.7
200 K; kT = 0.017267 ; n i = 1.38 cm −3
ni = 1.8 10 6 cm −3
300 K; kT = 0.0259 ;
400 K; kT = 0.034533 ; ni = 3.28 10 9 cm −3
For 200 K;
 2  10 15 4  10 16 
Vbi = (0.017267 ) ln 

(1.38 )2


(
= 1.257 V
)(
)
 x n = 9.93 10 cm
or x n = 0.0993  m
(
)
 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
______________________________________________________________________________________
(
 m ax
=
)
 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
 10 16
  15
 10
−4
16
−14
= 3.58  10 4 V/cm
(b) From part (a), we can write
2
 1.180 
3 N a2 = 1.8  10 6 exp 

 0.0259 
(

2 V
x p =  s bi

 e
(
)
 2(13.1) 8.85 10 (1.180 )
xn = 
1.6 10 −19

1
 1 
  
15
 3   4 8.127  10
(
 

 
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
 m ax = 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  10 17 4  10 16 
(a) Vbi = (0.0259 ) ln 

2
 1.5  10 10

= 0.80813 V
(b) V bi increases as temperature decreases
At T = 300 K, we can write
ni2 = 1.5 10 10
 10 16 10 15 

(a) Vbi = (0.0259 ) ln 
2
 1.5  10 10 
or
Vbi = 0.635 V
(b)
 Na

N
 d

1
 16
 10 + 10 15

x p = 0.08644  10 −4 cm = 0.08644  m
 m ax =
= 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

 10 15
  16
 10
 2(13.1) 8.85 10 −14 (1.180 )
xp = 
1.6 10 −19

 m ax =
 Nd

N
 a
 2(11 .7 ) 8.85 10 −14 (0.6350 )
=
1.6 10 −19

N d = 2.438 10 16 cm −3
= 1.324  10 −5 cm
or x n = 0.1324  m
1/ 2
or
which yields N a = 8.127 10 15 cm −3
(



x n = 0.8644 10 −4 cm = 0.8644  m
Now
)
−14

1
 16
 10 + 10 15

(
)
)
2
(
)(
(
)(
)
 − 1.12 
= K 2.8  10 19 1.04  10 19 exp 

 0.0259 
 K = 4.659
At T = 287 K, kT = 0.024778 eV
1/ 2
)
 287 
ni2 = K 2.8 10 19 1.04 10 19 

 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 = 10 16 cm −3 ,
)
 2 10 17 4 10 16 
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




 10 16
= (0.0259 ) ln 
10
 1.5 10
0.82494 − 0.80813
 100 % = 2.08 %
0.80813
 2%
_______________________________________
=
7.11
N N
Vbi = Vt ln  a 2 d
 ni
(
)(
)
(
)
 − 1.12 
= K (2.8  10 )(1.04  10 ) exp 

0.0259
2
19
19

 K = 4.659
At T = 300 K,
(
)(

 4 10 16 2 10 15
Vbi = (0.0259 ) ln 
2
 1.5 10 10
= 0.68886 V
For V bi = 0.550 V, T  300 K
(
n = (4.659 ) 2.8 10
19
(
)
)
For N d = 10 15 cm −3
 10 15
E F − E Fi = (0.0259 ) ln 
10
 1.5 10




(
)(
7.13
N N
(a) Vbi = Vt ln  a 2 d
 ni




( )( )
(
)
V bi = 0.456 V
(b)
 2(11.7 ) 8.85 10 −14 (0.456 )
xn = 
1.6 10 −19

(
)(1.04 10 )
19
E F − E Fi = 0.2877 eV
Then
Vbi = 0.34732 − 0.28768
or
Vbi = 0.0596 V
_______________________________________
or

 380 


 300 
3
 − 1.12 
 exp 

 0.032807 
= 4.112  10 24
Then
E F − E Fi = 0.3473 eV
 10 12 10 16 
= (0.0259 ) ln 

2
 1.5  10 10 
At T = 380 K, kT = 0.032807 eV
Also
2
i
or
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,
ni2 = 1.5 10 10




)
 4 10 16 2 10 15 
Vbi = (0.032807 ) ln 

24
 4.112 10

= 0.5506 V  0.550 V
_______________________________________
)
 10 12
  16
 10

1
 12
 10 + 10 16




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

(
)
 10 16
  12
 10
or
x p = 2.43  10 −3 cm

1
 12
 10 + 10 16




1/ 2
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 7
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(d)
 m ax
7.15
eN d x n
=
s
=
 m ax
(1.6 10 )(10 )(2.43 10 )
(11 .7 )(8.85 10 )
−19
−7
16
or
 m ax = 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




)
)
(c) N d = 810 17 cm −3 , L D = 0.004577  m
Now
(a) Vbi = 0.7427 V
(b) Vbi = 0.8286 V
(c) Vbi = 0.9216 V
Also
 2(11 .7 ) 8.85  10 −14 (Vbi )
xn = 
1.6  10 −19

(
)
(iii)
= 10 ;
(iv)
= 10 17 ;
16
= 0.6946 V
= 0.7543 V
= 0.8139 V
(i) For N a = 10 17 ,
N d = 10 14 ;  m ax = 0.443  10 4 V/cm
(ii)
= 10 15 ;
= 1.46  10 4 V/cm
(iii)
= 10 16 ;
= 4.60  10 4 V/cm
= 11 .2  10 4 V/cm
= 10 ;
(iv)
(b)
(i) For N a = 10 14 , N d = 10 14 ; Vbi = 0.4561 V
)

1

 8 10 17 + N

d
(ii)
= 10 15 ;
(iii)
= 10 16 ;
(iv)
= 10 17 ;
= 0.5157 V
= 0.5754 V
= 0.6350 V
(i) For N a = 10 14 ,
N d = 10 14 ;  m ax = 0.265  10 4 V/cm




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)
1/ 2
17
(b) N d = 2.2 10 16 cm −3 , L D = 0.02760  m
 8 10 17
 
 Nd
)
= 10 15 ;

(a) N d = 810 14 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 = 10 17 , N d = 10 14 ; Vbi = 0.6350 V
(
−14
  kT 

L D =  2s

 e Nd 

 2eVbi
=

 s
(ii)
= 10 15 ;
= 0.381  10 4 V/cm
(iii)
= 10 16 ;
= 0.420  10 4 V/cm
(iv)
= 10 17 ;
= 0.443  10 4 V/cm
(c)
 m ax increases as the doping increases,
and the electric field extends further into
the low-doped side of the pn junction.
_______________________________________
7.16
(
)( )
) 
 5 10 16 10 15
(a) Vbi = (0.0259 ) ln 
2
 1.5  10 10
= 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 10 16
 
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 )
= 1.43  10 4 V/cm
−4
0.9452  10
(ii)For V R = 5 V,
2(0.6767 + 5)
= 4.15  10 4 V/cm
2.738  10 − 4
_______________________________________
 m ax =
)(
 2  10 17 4  10 16
(a) Vbi = (0.0259 ) ln 
2
 1.5  10 10
= 0.8081 V
(b)
(

 2  (V + V R )  N a
x n =  s bi
N
e

 d

)
2(Vbi + V R ) 2(0.8081 + 2.5)
=
W
0.3584  10 − 4
= 1.85  10 5 V/cm
(c)  m ax =
)
)
)(
(
(



1


 N + N 
d 
 a






1/ 2
(
)(
) 
N N
(a) Vbi = Vt ln  a 2 d
 ni
1/ 2





1/ 2
)
1/ 2

 
= 5.78  10 −12 F
or C = 5.78 pF
_______________________________________
7.18
= 0.2987  10 −4 cm
or x n = 0.2987  m

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 10 17 + 4 10 16


 2  (V + V R )  N d
x p =  s bi
N
e

 a



e s N a N d
(d) C = A

 2(Vbi + V R )(N a + N d ) 
 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
1/ 2
Also W = x n + x p = 0.3584  m
 m ax =
17
)(
 

 
= 0.3584  10 cm
or W = 0.3584  m
2(Vbi + V R )
W
(i)For V R = 0 ,
(
1/ 2
−4
 m ax =
(
)





 2  10 17 + 4  10 16

17
16
 2  10 4  10
= 2.738  10 cm
or W = 2.738  m
7.17
1/ 2
 2(11.7 ) 8.85 10 −14 (0.8081 + 2.5)
=
1.6 10 −19

−4
(c)






1

 2 10 17 + 4 10 16

= 5.97  10 −6 cm
or x p = 0.0597  m
−5
= 9.452  10 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





 80 N 2 
= V t ln  2 d 
 ni 
We find
V 
80 N d2 = ni2 exp  bi 
 Vt 
(
= 1.5  10 10
)
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 10 15 cm −3
N a = 2.147 10 cm −3
17
(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
15
 80  2.147  10 + 2.684  10



(3 10 )
(
(
(
)
)




1/ 2
(
)
(4 10 )(4 10 )
15
17

4 10 + 4 10 17 
15
or
9 10 10 = 1.224 10 9 (Vbi + V R )
so that
(Vbi + V R ) = 73.53 V
which yields
V R = 72 .8 V
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
)
 m ax
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  10 15 4 10 17 
(a) Vbi = (0.0259 ) ln 

2
 1.5 10 10

or
V bi = 0.766 V
Now
1/ 2
= 2.83  10 −6 cm
or x p = 0.0283  m
(c)  m ax
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 ) 
 2.147 10 17 2.684 10 15

17
15
 2.147 10 + 2.684 10
(
1/ 2



C  = 4.52  10 −9 F/cm 2
_______________________________________
7.19
(a) Vbi (3 N 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 
N N 


= Vt ln (3) + ln  d 2 a   − Vt ln  d 2 a 

 ni  
 ni 


)(
 4 10 16 4 10 17
(b) Vbi = (0.0259 ) ln 
2
 1.5 10 10
or
V bi = 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 + V R ) = 8.008 V
(
)
)
)

)
(4 10 )(4 10 )
16
17

4 10 + 4 10 17 
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 17 4 10 17 
(c) Vbi = (0.0259 ) ln 

2
 1.5 10 10

or
V bi = 0.886 V
We have
 2 1.6  10 −19 (Vbi + V R )
2
3  10 5 = 
−14
 (11 .7 ) 8.85  10
(
(
)
(
)
 1  5.7543 
=


 3.13  5.8139 
)
or
( A)
= 0.316
(B )
(c)
(
)
(4 10 )(4 10 )

17
17

4 10 + 4 10 17 
17
C j ( A)
so that
(Vbi + V R ) = 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
We find




 10 18 10 16 
VbiB = (0.0259 ) ln 
 = 0.8139 V
2
 1.5 10 10 
We find
W ( A)  5.7543  10 18 + 10 15 

= 

W (B )  5.8139  10 18 + 10 16 
or
W ( A)
= 3.13
W (B )




1/ 2
1/ 2
 VbiB + V R  N a + N dB


 V + V  N + N
R 
a
dA
 biA
 10 15
=  16
 10
 5.8139  10 18 + 10 16


 5.7543  10 18 + 10 15


C j ( A)








1/ 2
1/ 2
= 0.319
_______________________________________
7.22
(a) We have
C j (0)
( )( )
(
)
 10 16
  15
 10


s N a N dB


 2(VbiB + V R )(N a + N dB ) 
1/ 2
 10 18 10 15 
VbiA = (0.0259 ) ln 
 = 0.7543 V
2
 1.5 10 10 
( )( )
(
)
1/ 2
 N
=  dA
 N dB
C j (B )
or
W ( A)  (VbiA + V R ) (N a + N dA )  N dB
=


W (B )  (VbiB + V R ) (N a + N dB )  N dA
=


s N a N dA


 2(VbiA + V R )(N a + N dA ) 
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 + x n
(
)




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 10 16 5 10 15
Vbi = (0.0259 ) ln 
2

1.8  10 6
= 1.162 V
1
C 
Vbi + V R
(
)
 2  10 15 4 10 16
(a) Vbi = (0.0259 ) ln 
2
 1.5  10 10
= 0.6889 V
(
)
(
)
)

)(
)

)
(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  10 17 5  10 15
Vbi = (0.0259 ) ln 
2
 1.5 10 10
= 0.7543 V
(
)
)

)(
)
1/ 2
(
 1.6 10 −19 (11.7 ) 8.85 10 −14
= 8 10 − 4 
2(0.7543 + 10 )

(
1/ 2
1/ 2
(


e s N a N d
(a) C = AC  = A

 2(Vbi + V R )(N a + N d ) 
)


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  10 15 4 10 16
(b) Vbi = (0.0259 ) ln 
2

1.8 10 6
= 1.157 V
(
)
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 10 16 cm −3
and
N d = 3.07 10 15 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
______________________________________________________________________________________
(
(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
L=
(4 N )

= 3.306  10 −3 H = 3.306 mH
2
d
1
(
)(
2 3.306 10 −3 12.14 10 −12
0.6  10 −12 = 2.724  10 − 20
)
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 10 15 cm −3 ,
V bi = 1.10 V
(b) From part (a),
0.6  10 −12 = 2.724  10 − 20
7.26
 2e(Vbi + V R )N d 
 m ax  

s


Let Vbi  0.75 V
(a)
N a = 1.19 10 16 cm −3 ,
Vbi = 1.135 V
_______________________________________
5 2
−19
−14
 N d = 1.88 10 cm
( )
(b) 10
d



7.28
)
 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 ) x n + x p
(0.8)x p = (0.2)xn
xn = 4x p
1/ 2
)

1
 14
 10 + 5  10 15

or
x p = 5.32  10 −6 cm
 N a = 4N d
Also



(
(
(


1


 N + N 
d 
 a

 Nd

N
 a
 10 14
 
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) V bi = V t ln  a 2 d
 n i
(
 5 10 15 10 14
(a) Vbi = (0.0259 ) ln 
2
 1.5  10 10
or
Vbi = 0.5574 V
(b)
−3
5 2
(
Nd
Vbi + 5
By trial and error,
N d = 2.976 10 15 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 10 15 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  10 15
 
14
 10
7.30

1
 14
 10 + 5  10 15




1/ 2
(
(50 10 )
−4 2
=
(
(
)V
)(10 )
2(11 .7 ) 8.85  10
(1.6 10
 10 14
 16
 10

)
−19
−14
R
14








= 0.50  10 −4 cm = 0.50  m
(c)
 m ax 
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
16
d
(
= 1.8  10 6
)
2

 = 1.20

 1.20 
exp 

 0.0259 
)
 N d = 5.36 10 15 cm −3
so
x n = 50 10

)
(
1/ 2
which yields
V R = 193 V
(b)
xp Nd
N
=
 x n = x p  a
xn
Na
 Nd
−4
)
)
 (11.7) 8.85 10 −14 2 10 15
1/ 2
7.29
An n + p junction with N a = 10 14 cm −3 ,
(a) A one-sided junction and assume
V R  Vbi . Then
)(
(
)
 5  10 15 
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  10 17 2  10 15
(a) Vbi = (0.0259 ) ln 
2
 1.5 10 10
= 0.7305 V
2V R
2(193 .15 )
=
W
50 .5  10 − 4
or
 m ax = 7.65  10 4 V/cm
_______________________________________


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 10 16 )(5.36 10 15 )

(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 10 16 )(5.36 10 15 )
(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
n-region, 0  x  x O
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
)
d1  (x ) eN dO
=
=
dx
s
2 s
At x = −2  m  − x O ,  = 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
eN dO x
+ C3
s
We have  2 = 0 at x = x n
2 =
eN dO x n
s
so that for x O  x  x n , we have
 C3 = −
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 
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
 = dx = d (x + x O )dx
s


 x2



 2 + xO  x  + C 2


Let  = 0 at x = − x O , then
=
eN d
s
 x O2

eN d x O2
2

 + C2  C2 =
−
x
O
 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 1.6  10 −19 (Vbi + V R )
4  10 5 = 
 (11 .7 ) 8.85  10 −14
or
 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
_______________________________________
(
)
 (2  10 )(2  10 ) 


16
(a) V B =
s 
2eN B
2
crit
or
(
)(
)
 2
(11.7) 8.85 10 −14 4 10 5
N B = s crit =
2eVB
2 1.6 10 −19 (40 )
Then
(
N B = N a = 1.294 10 cm
16
(b) N B =
)
2
1/ 2

16
16 
 2  10 + 2  10  
 Vbi + V B = 51 .77 V
So V B = 51 .04 V
(b)
 5 10 15 5 10 15 
Vbi = (0.0259 ) ln 

2
 1.5 10 10

= 0.6587 V
Then
 2 1.6  10 −19 (Vbi + V R )
4  10 5 = 
 (11 .7 ) 8.85  10 −14
(
7.35
16
(
−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 10 16 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
=
2eVB
2 1.6 10 −19 (80 )
2
crit
−14
(
)
5 2
= 6.47  10 cm −3
_______________________________________
15
N d = 510 15 cm −3 and V B  100 V, then,
neglecting V bi we have
2 V 
xn   s B 
 eN d 
7.37
(a) For N d = 10 16 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 = 10 15 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

 m ax = 
N +N
s

d
 a
V
=
V

=

Set m ax
R
B
crit and
(
)(




1/ 2
)
 2  10 16 2 10 16 
Vbi = (0.0259 ) ln 

2
 1.5 10 10

= 0.7305 V
(
)
or
( )( )
(
)
 10 18 10 18 
Vbi = (0.0259 ) ln 
 = 0.933 V
2
 1.5  10 10 
Now
eN d x n
 m ax =
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
V bi + 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
)

0=−
3

eaxO3
ea  − x O
+ x O3  + C 2  C 2 =

2 s  3
3 s

Then
(
−19
 x3
 eaxO3
2

+
−
x

x
O
 3
 3
s


_______________________________________
 (x ) = −
)
)(10 )
15
ea
2 s
7.44
We have that
(
)V
)(10 )
2(11 .7 ) 8.85  10
(
At x = + x O and x = − x O ,  = 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

1/ 2
which yields
V R = 4.35 10 3 V
(150 10 )

Set  = 0 at x = − x O , then
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
−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 10 18
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10 15 cm −3 << N d
(
1/ 2
)(
 3 10 17 5 10 15
Then Vbi = (0.0259 ) ln 
2
 1.5 10 10
= 0.7648 V
Now
 1.6 10 −19
C j = 0.45 10 −12 = A
 2(0.7648 + 5)
(
)
(
(
)(
0.45 10 −12 = A 8.476 10 −9
)
 A = 5.31  10 −5 cm 2
)(
)
(
)(
 (11.7) 8.85 10 −14 5 10 15
Cj =
1/ 2
)
 1.6 10 −19
(b) C j = 5.309 10 −5 
 2(Vbi + V R )
(

)
 (11.7) 8.85 10 −14 5 10 15
(
)
)
1/ 2
1.0805  10 −12
Vbi + V R
(i) For V R = 2.5 V, C j = 0.598 pF
(ii) For V R = 0 ,
C j = 1.24 pF
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 8
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
Chapter 8
Exercise Solutions
Ex 8.1
(
)
(
10 2
ni2
1.5 10 10
=
Na
5 10 16
n po =
p no =
2
i
n
1.5 10
=
Nd
2 10 16
V
n p − x p = n po exp  a
 Vt
(
n po =
)
(
2
p no =
= 1.125 10 4 cm −3




(
= 3.57  10 14 cm −3
(
)
(
)
(
)
)
J p (x n ) =
= 8.92  10 14 cm −3
We have that
n p − x p  N a and p n (x n )  N d
(
2
J n − x p = 1.20 A/cm 2
 0.650 
= 1.125  10 exp 

 0.0259 
4
)
)
(
)




(
ni2
1.8 10 6
=
Nd
2 10 16
2
= 1.62  10 −4 cm −3
eDn n po   eVa  
Jn − xp =
 − 1
exp 
Ln   kT  
1.6 10 −19 (210 ) 4.05 10 −4
=
4.583 10 −3
  1.05  
 exp 
 − 1
  0.0259  
 0.650 
= 4.5  10 3 exp 

 0.0259 
V
p n (x n ) = p no exp  a
 Vt
)
= 4.05  10 −4 cm −3
= 4.5 10 3 cm −3
)
(
ni2
1.8 10 6
=
Na
8 10 15
=
)
eD p p no   eVa  
− 1
exp
L p   kT  
(1.6 10 )(8)(1.62 10 )
−19
−4
6.325 10 − 4
  1.05  
 exp 
 − 1
  0.0259  
so low injection applies.
_______________________________________
J p (x n ) = 0.1325 A/cm 2
Ex 8.2
 1
J s = eni2 
 N a
(
= 1.6 10
Dn
 no
1
Nd
)(1.8 10 )
The total current density is:
J T = J n − x p + J p (x n )
Dp 

 po 

(
210
1
+
10 − 7 2  10 16
−18

8
−8 
5  10 
J s = 3.30 10 A/cm
_______________________________________
2
Ex 8.4
In the n-region, for N d = 210 16 cm −3 ,
 n  6000 cm 2 /V-s
or
Ln = Dn no
(210 )(10 −7 ) = 4.583 10 −3 cm
L p = D p po
=
J T = 1.33 A/cm 2
_______________________________________
J = e n N d  n
Ex 8.3
We find
=
)
= 1.20 + 0.1325
6 2
−19
 1

15
 8  10
+
J
1.3325
=
−19
e n N d
1.6 10 (6000 ) 2 10 16
= 0.0694 V/cm
In the p-region, for N a = 810 15 cm −3 ,
n =
(
)
(
)
 p  320 cm 2 /V-cm
(8)(5 10 −8 ) = 6.325 10 −4 cm
J = e p N a  p
or
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 8
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
p =
J
e p N a
=
1.3325
(1.6 10 )(320 )8 10
−19
15
J gen =
= 3.25 V/cm
_______________________________________
=
(
= 1.6 10
 1

15
 2  10
−19
1
Nd
)(1.8 10 )
I Sn = A 
or J s = 1.677 10
A/cm
(
9.8 

5  10 −8 
(
)
or W = 2.141  10 −4 cm
)(
25
5 10 −7
I Sn = 1.273 10 −13 A
I Sp = A 
eni2
Nd
Dp
 po
) (1.6 10 )(1.5 10 )
10 2
−19
8 10 16
Then
I S = I Sn + I Sp = 1.318  10 −13 A
V
(a) I D = I S exp  a
 Vt





)
)
= 2.2  10 −4 A
(
)
 0.610 
(b) I D = 1.318 10 −13 exp 

 0.0259 
1/ 2
 

 




(

 2  10 15 + 8  10 16

15
16
 2  10 8  10
(
2 10
15
 0.550 
= 1.318  10 −13 exp 

 0.0259 
= 2.23  10 −3 A
 2(13.1) 8.85 10 (1.174 + 5)
=
1.6 10 −19

−14
10 2
−19
or
)
)

 2  (V + V R )  N a + N d
W =  s bi
 N N
e

a
d


) (1.6 10 )(1.5 10 )
(
I Sp = 4.5  10 −15 A
 2 10 15 8 10 16
(b) Vbi = (0.0259 ) ln 
2
 1.8 10 6
= 1.174 V
(
Dn
 no
or
2
)(
eni2
Na
= 10 −3 
(
Dp 

 p0 

207
1
+
−8
5  10
8  10 16
=
= 10 −3 
6 2
−17
−8
Ex 8.7
Ex 8.6
 n0
−4
6
6.166 10 −10
= 3.68  10 7
−17
Js
1.677 10
_______________________________________
and V a1 = 1.050 V.
Then
1.42 − V a 2 1.42 − 1.050
=
310
300
which yields
V a 2 = 1.0377 V
so V = 1.0377 −1.050 = −0.0123 V
or V = −12.3 mV per 10 C increase in
temperature.
_______________________________________
+
−19
J gen
(c)
Let T 2 = 310 K, T1 = 300 K, E g = 1.42 eV,
Dn
(1.6 10 )(1.8 10 )(2.141 10 )
2(5  10 )
J gen = 6.166  10 −10 A/cm 2
Ex 8.5
From Example 8.5, we have
E g − eVa 2 E g − eVa1
=
kT2
kT1
 1
(a) J s = eni2 
 N a
eniW
2 0
Now
1/ 2
(a) rd =
Vt
0.0259
=
= 118 
I D 2.2 10 − 4
(b) rd =
0.0259
= 11 .6 
2.23  10 −3
10
10 −7
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 8
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
We find
TYU 8.2
I pO = I Sp
V
exp  a
 Vt
V 

 , I nO = I Sn exp  a 

V 

 t 
(a) I n = A 
Then
(a) I pO = 7.511  10 −6 A;
eni2
Na
Dn
 no
I nO = 2.125 10 A
5  10 16
(b) I p = A 
We find

 I pO pO + I nO nO




−7
(
)( )
+ (2.155 10 )(5 10 )
1
7.617 10 −5 10 −7
2(0.0259 )
−7
or C d = 2.09 10 −8 F = 20.9 nF
_______________________________________
Test Your Understanding Solutions
n po =
(
ni2
1.8 10 6
=
Na
5 10 16
(
ni2
1.8 10 6
=
Nd
5 10 15
For V a (max ) ,
p no =
)
2
)
= 6.48 10 −5 cm −3
2
= 6.48 10 − 4 cm −3
)




10 2
−19
1 10 16

10
 0.625 
 exp 

10 −7
 0.0259 
or I p = 1.09  10 −3 A = 1.09 mA
(c) I Total = I n + I p
= 1.538  10 −4 + 1.087  10 −3
= 1.24  10 −3 A
or I Total = 1.24 mA
_______________________________________
TYU 8.3
From TYU 8.2, I n = 0.154 mA
Now
eD p p no
V 
I p = A
 exp  a 
Wn
 Vt 
We find
p no
(
n2
1.5 10 10
= i =
Nd
10 16
Then
(
)
2
= 2.25 10 4 cm −3
) (1.6 10 )(10 )(2.25 10 )
I p = 10 −3 
V 
p n (x n ) = p no exp  a 
 Vt 
so that
 p (x ) 
 (0.1)N d 
Va = Vt ln  n n  = Vt ln 

 p no 
 p no 
 (0.1) 5 10 15 
= (0.0259 ) ln 
−4 
 6.48 10 
or V a (max ) = 1.067 V
_______________________________________
(
 po
V
 exp  a
 Vt
) (1.6 10 )(1.5 10 )
(
(
)( )
+ (2.125 10 )(5 10 )
−3
Dp
= 10 −3 
or C d = 2.07 10 −9 F = 2.07 nF
TYU 8.1
eni2
Nd
1
7.511 10 −6 10 −7
2(0.0259 )
−4
(b) C d =
25
 0.625 
 exp 

5 10 −7
 0.0259 
or I n = 1.54 10 −4 A = 0.154 mA
I nO = 2.155 10 −3 A
(a) C d =
10 2
−19

(b) I pO = 7.617  10 −5 A;
So




) (1.6 10 )(1.5 10 )
(
= 10 −3 
−4
 1
C d = 
 2Vt
V
 exp  a
 Vt
−19
2 10
4
−4
 0.625 
 exp 

 0.0259 
or I p = 5.44 mA
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 8
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
TYU 8.4
V
(a) J  J s exp  a
 Vt
 1
J s = en 
 N a
I Sp = A 




Dn
2
i
1
+
Nd
 no
(
)(
)
 1

15
 2  10
10
10 − 7



= 9.83  10 −5 A
= 2.137  10 −4 A/cm 2
 1.045 
(b) I D = 5.338  10 − 21 exp 

 0.0259 
)(
= 1.78  10 −3 A
)
 2 10 15 8 10 16 
(b) Vbi = (0.0259 ) ln 

2
 1.5 10 10

= 0.7068 V
We find
 2(11.7 ) 8.85 10 −14 (0.7068 − 0.35 )
W =
1.6 10 −19

)
)
 2  10 15 + 8  10 16

15
16
 2  10 8  10
(
)(
)
 

 
1/ 2
−5
= 4.865  10 cm
Then
V
en W
J rec = i exp  a
2 o
 2Vt
=
−19
(
)(
)
J rec = 5.020 10 A/cm
)
2
Vt
0.0259
=
= 264 
I D 9.828 10 −5
(b) rd =
0.0259
= 14 .6 
1.779  10 −3
We have
V 
V 
I pO = I Sp exp  a  , I nO = I Sn exp  a 
V
 t 
 Vt 
We find
(a) I pO = 1.181  10 −6 A; I nO = 9.71 10 −5 A
Now
 1
C d = 
 2Vt
So
(a) C d =

 I pO pO + I nO nO




(
)( )
+ (9.71 10 )(5 10 )
1
1.181 10 −6 10 −7
2(0.0259 )
−5
−4
J rec 5.020 10
=
= 2.35
J
2.137 10 − 4
_______________________________________
(c)
−7
= 9.40  10 −10 F = 0.940 nF
1
(b) C d =
2.137 10 −5 10 −7
2(0.0259 )
(
(
)( )
)(5 10 )
+ 1.757 10 −3
TYU 8.5
I Sn = A 
(
(a) rd =
I nO = 1.757 10 −3 A
 0.35 
 exp 

 2(0.0259 ) 
−4
Now
(b) I pO = 2.137  10 −5 A;




1.5  10 10 4.865  10 −5
2 10 − 7
(1.6 10 )(
eni2
Na
Dn
 no
−19
2 10 15
or I Sn = 5.274 10 −21 A
−7
= 1.70  10 −8 F = 17.0 nF
_______________________________________
) (1.6 10 )(1.8 10 )
= 10 −3 
9.8
10 −7
 0.970 
(a) I D = 5.338  10 − 21 exp 

 0.0259 
)
(
8 10
16
So I S = I Sn + I Sp = 5.338  10 −21 A
 0.35 
Then J  2.891  10 −10 exp 

 0.0259 
(
6 2
−19
or I Sp = 6.415  10 −23 A
J s = 2.891 10 −10 A/cm 2
(
 po
) (1.6 10 )(1.8 10 )
(
Dp 

 po 

25
1
+
−7
10
8  10 16
(
Dp
= 10 −3 
2
= 1.6 10 −19 1.5 10 10
eni2
Nd
6 2
207
5 10 −7
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 8
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
TYU 8.6
From Figure 5.3, for N d = 810 16 cm −3 ,
(b) erf
  n  900 cm 2 /V-s
t2
 pO
+
In the n-region,
  e n N d
(
= 1.6 10
−19
)(900 )(8 10 )
16
= 11.52 (  -cm)
Then
Rn =
l
A
=
−1
(0.01) = 0.868 
(11.52 )(10 −3 )
In the p-region,
  e p N a
(
)
(
= 1.6 10 −19 (480 ) 2 10 15
)
−1
= 0.1536 (  -cm)
Then
(0.01) = 65.1 
l
Rp =
=
A (0.1536 ) 10 −3
The total resistance is
R = Rn + R p = 66 
(
)
_______________________________________
TYU 8.7
(a) erf
ts
 pO
=
IF
IF + IR
Now
IR 
VR 2
= = 0.5 mA
RR 4
So
erf
ts
 pO
=
1.75
= 0.778
1.75 + 0.5
From Appendix G,
ts
 pO
 0.864
So that
2
t s = (0.864 ) 10 −7 = 0.746 10 −7 s
(
)
 t 2  pO
)
I
= 1 + (0.1) R
 IF



 0.5 
= 1 + (0.1)
 = 1.0286
 1.75 
For N a = 210 15 cm −3 ,
  p  480 cm 2 /V-s
(
exp − t 2  pO
By trial and error
t2
 1.25
 pO
(
)
 t 2 = (1.25 ) 10 −7 = 1.25 10 −7 s
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 8
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Chapter 8
8.1
or
In forward bias
 eV 
I f  I S exp 

 kT 
Then
 eV 
I S exp  1 
I f1
 e 

 kT 
=
= exp 
(V1 − V 2 )
I f2
 eV 
 kT 

I S exp  2 
 kT 
or
 kT   I f 1 
V1 − V 2 = 
 ln
 e   I f 2 
(a)
I f1
For
= 10 , then
I f2
V1 −V 2 = (0.0259 ) ln (10 )
or
V1 −V 2 = 59 .6 mV  60 mV
(b)
I f1
For
= 100 , then
If2
V1 −V 2 = 119 .3 mV  120 mV
_______________________________________
n po =
p no
(
10 2
(
)
ni2
1.5 10
=
Na
8 10 15
n2
1.5 10 10
= i =
Nd
2 10 15
)
= 1.88  10 14 cm −3
(
(
)
 − 0.55 
p n (x n ) = 1.125  10 5 exp 

 0.0259 
0
 − 0.55 
n p − x p = 2.8125  10 4 exp 

 0.0259 
0
_______________________________________
(
) (
)
(
)
(
)
ni2
1.8 10 6
=
Na
4 10 16
n2
1.8  10 6
p no = i =
Nd
10 16
(a) V a = 0.90 V,
(
2
= 8.110 −5 cm −3
2
= 3.24  10 −4 cm −3
)
= 4.0  10 11 cm −3
(
= 1.125 10 cm
5
) (
)
 0.90 
n p − x p = 8.1 10 −5 exp 

 0.0259 
2
−3
= 10 .0  10 10 cm −3
(b) V a = 1.10 V
(
)
 1.10 
p n (x n ) = 3.24 10 − 4 exp 

 0.0259 
= 9.03  10 14 cm −3
(
)
 0.45 
n p − x p = 2.8125  10 4 exp 

 0.0259 
) (
)
 1.10 
n p − x p = 8.1 10 −5 exp 

 0.0259 
)
= 3.95  10 12 cm −3
) (
)
= 4.69  10 13 cm −3
(c) V a = −0.55 V
= 2.8125 10 4 cm −3
 0.45 
p n (x n ) = 1.125  10 5 exp 

 0.0259 
(
) (
 0.55 
n p − x p = 2.8125  10 4 exp 

 0.0259 
)
(
)
 0.90 
p n (x n ) = 3.24 10 − 4 exp 

 0.0259 
V 
p n (x n ) = p no exp  a 
 Vt 
V 
n p − x p = n po exp  a 
 Vt 
(a) V a = 0.45 V,
(
(
 0.55 
p n (x n ) = 1.125  10 5 exp 

 0.0259 
n po =
or
)
(b) V a = 0.55 V,
8.3
V1 −V 2 = (0.0259 ) ln (100 )
8.2
(
n p − x p = 9.88  10 11 cm −3
(c)
= 2.26  10 14 cm −3
p n (x n )  0
(
)
np − xp  0
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 8
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
8.4
(a) n po =
(
2
i
(b) J p (x n ) =
)
10 2
n
1.5 10
=
Na
5 10 16
=
= 4.5  10 3 cm −3
p no
(
n2
1.5 10 10
= i =
Nd
5 10 15
)
2
eD p p no
Lp
10
(
)
)
2
(
ni2
1.5 10 10
=
Nd
3 10 16
)
=
=
eDn n po
Ln
2
i
en
Na
(c) I = I n + I p = 1.85 + 4.52 = 6.37 mA
_______________________________________
8.6
2
=
)
V
exp  a
 Vt
(10 )(1.6 10 )(1.5 10 )
−4
−19
10
10 2
16
25
10 −6
6 2
5 10 16
)
(




)
 0.5 
= 1.8  10 −15 exp 

 0.0259 
or
(




205
5 10 −8
 1.10 
 exp 

 0.0259 
= 1.849 A/cm 2
I n = AJ n (− x p ) = 10 −3 (1.849 ) A
V
I D  I S exp  a
 Vt
I D = 4.36 10 −7 A
(b) For V a = −0.5 V,




V
 exp  a
 Vt
(1.6 10 )(1.8 10 )
or I n = 1.85 mA
 nO
or
(
 no
(
Dn
1
Na
I S = 1.8 10 −15 A
(a) For V a = 0.5 V,



Dn
−19
)
or I p = 4.52 mA
)
8.5
(
9.80
10 −8
 1.10 
 exp 

 0.0259 
16
(
I S = Aeni2 
 (0.1) 7 10 15 
= (0.0259 ) ln 
4 
 3.214 10 
= 0.6165 V
(ii) p-region - lower doped side
_______________________________________
(a) J n − x p =
6 2
For an n + p silicon diode
= 7.5  10 3 cm −3
 (0.1)N a
(i) V a = Vt ln 
 n po




= 4.521 A/cm 2
I p = AJ p (x n ) = 10 −3 (4.521 ) A
= 3.214  10 4 cm −3
p no =
V
 exp  a
 Vt
−19
 p (x ) 
or V a = Vt ln  n n 
 p no 
 (0.1) 5 10 15 
= (0.0259 ) ln 

4
 4.5 10

= 0.599 V
(ii) n-region - lower doped side
(b) n po
 p0




(1.6 10 )(1.8 10 )
=
V 
(i) p n (x n ) = p no exp  a 
 Vt 
(
Dp
eni2
Nd
= 4.5  10 4 cm −3
n2
1.5 10 10
= i =
Na
7 10 15
V
exp  a
 Vt
)
  − 0.5  
I D = 1.8 10 −15 exp 
 − 1
  0.0259  
or
I D  − I S = −1.8 10 −15 A
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 8
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
8.7
8.9
 1
J s = eni2 
 N a
Dn
+
 no
(
1
Nd
)(
= 1.6 10 −19 2.4 10 13
 1

15
 4  10
)
Dp 

 p0 

2
90
1
+
−6
2  10
2  10 17
48
2  10 − 6



J s = 1.568 10 −4 A/cm 2
V
(a) I = AJ s exp  a
 Vt
(




)(
)
 0.25 
= 10 − 4 1.568  10 − 4 exp 

 0.0259 
= 2.44  10 −4 A
or I = 0.244 mA
(
)(
(b) I = − I s = − AJ s = − 10 −4 1.568 10 −4
)
−8
= −1.568  10 A
_______________________________________
Dn
 no
(
+
1
Nd
)(
= 1.6 10 −19 1.5 10 10
 1

17
 5  10
)
Dp 

 p0 

10
8  10 −8
J s = 5.145 10 −11 A/cm 2
(
)(
I s = AJ s = 2 10 −4 5.145 10 −11
= 1.029  10
V
(b) I = I s exp  a
 Vt
−14



)
A
)
 0.45 
(i) I = 1.029  10 −14 exp 

 0.0259 
= 3.61  10 −7 A
(
)
 0.55 
(ii) I = 1.029  10 −14 exp 

 0.0259 
= 1.72  10 −5 A
(
 I s = 6.305 10 −15 A = 6.305  10 −12 mA
I s 6.305 10 −12
=
A
2 10 − 4
= 3.153  10 −8 mA/cm 2
V 
Case 2: I = I s exp  a 
 Vt 
 0.70 
= 2  10 −12 exp 

 0.0259 
or I = 1.093 mA
I
2 10 −12
Js = s =
A 110 −3
= 2  10 −9 mA/cm 2
V 
Case 3: I = AJ s exp  a 
 Vt 
 I 
So Va = Vt ln 

 AJ s 


0.80
= (0.0259 ) ln  − 4
−7 
 10 10 
V a = 0.6502 V
Js =
(




(




 0.65 
0.50  10 −3 = I s exp 

 0.0259 
2
25
1
+
−7
10
8  10 15
8.10
V
Case 1: I = I s exp  a
 Vt
8.8
 1
(a) J s = eni2 
 N a
We have
 V  
I = I S exp   − 1
  Vt  
or we can write this as
V 
I
+ 1 = exp  
IS
 Vt 
so that
 I

V = Vt ln  + 1
 IS

In reverse bias, I is negative, so at
I
= −0.90 , we have
IS
V = (0.0259 ) ln (1 − 0.90 )
or
V = −59.6 mV
_______________________________________
)
 0.65 
(iii) I = 1.029  10 −14 exp 

 0.0259 
= 8.16  10 −4 A
_______________________________________
)
(
)(
)
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 8
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Then
(
)(
)
I s = AJ s = 10 −4 10 −7 = 10 −11 mA
I
V
exp  a
 Vt
Case 4: I s =
=




1.20
 0.72 
exp 

 0.0259 
I s = 1.014 10 −12 mA
I s 1.014  10 −12
=
Js
2  10 −8
A=
= 5.07  10 −5 cm 2
_______________________________________
8.11
eDn n po
(a)
Jn
Ln
=
J n + J p eDn n po eD p p no
+
Ln
Lp
Dn
 no
=
Dn
 no


n i2
Na
or
1+
D p no  N a

Dn po  N d
D p no  N a

Dn po  N d
2
i





1
=
 0.90 − 1

Dn po  1

− 1

D p no  0.90 
Na
=
Nd
(25 )(10 −7 ) (0.1111 )
(10 )(5  10 −7 )
=
1
25

Na
5 10 − 7
Dp n
n
+

Na
 po N d
2
i
1
0.90 =
8.12
The cross-sectional area is
I 10 10 −3
A= =
= 5 10 − 4 cm 2
J
20
We have
V 
 0.65 
J  J S exp  D   20 = J S exp 

 0.0259 
 Vt 
which yields
J S = 2.522 10 −10 A/cm 2
We can write
 1
Dp 
Dn
1

J S = eni2 
+
 N a  nO N d  pO 
We want
Dn
1

N a  nO
= 0.10
Dp
Dn
1
1

+

N a  nO N d
 pO
Nd
Na
= 12.73
= 0.07857 or
Na
Nd
(b) From part (a),
Na
=
Nd
Dn po  1

− 1

D p no  0.20 
=
(25 )(10 −7 ) (4)
(10 )(5  10 −7 )
Na
Nd
= 2.828 or
= 0.354
Nd
Na
_______________________________________
1
Na
25
1
10
+

−7
N
5 10
5 10 − 7
d
7.071  10 3
= 0.10
Na
3
3
7.071  10 +
4.472  10
Nd
which yields
Na
= 14.23
Nd
Now
=
(
(
)
)(
J S = 2.522 10 −10 = 1.6 10 −19 1.5 10 10
)
2

1
25
1
10 


+


−7
Nd
5  10
5  10 − 7 
 (14 .23 )N d
We find
N d = 7.09 10 14 cm −3
and
N a = 1.01 10 16 cm −3
_______________________________________
8.13
Plot
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 8
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
8.14
(a)
8.15
(a) p-side;
eDn n po
N
E Fi − E F = kT ln  a
 ni




 5 10 15
= (0.0259 ) ln 
10
 1.5 10
Jn
Ln
=
J n + J p eDn n po eD p p no
+
Ln
Lp
Dn

 no
=
Dn
 no

n i2
Na
or
D p n i2
n i2
+

Na
 po N d
1
=
D p no  N a

D n po  N d
1+




We have
Dp  p

1
1
=
=
and no =
Dn
 n 2.4
 po 0.1
so
Jn
=
Jn + J p
1
1 1  Na
1+


2.4 0.1  N d




or
Jn
=
Jn + J p
e n N d
L
e n N d + n  e p N a
Lp
We have
 n = e n N d and  p = e p N a
Also
Ln
=
Lp
E Fi − E F = 0.329 eV
Also on the n-side;
N
E F − E Fi = kT ln  d
 ni




 10 17
= (0.0259 ) ln 
10
 1.5 10
D n no
=
D p po
Then
(
2 .4
= 4.90
0.1
)
n  p
Jn
=
Jn + J p
 n  p + 4.90
(
)
_______________________________________




or
E F − E Fi = 0.407 eV
(b) We can find
D n = (1250 )(0.0259 ) = 32 .4 cm 2 /s
D p = (320 )(0.0259 ) = 8.29 cm 2 /s
Now
 1
Dp 
Dn
1

J S = eni2 
+
 N a  nO N d  pO 
(
1
N 
1 + (2.04 ) a 
 Nd 
(b) Using Einstein's relation, we can write
e n ni2

Jn
Ln N a
=
Jn + J p
e n n i2 e p ni2

+

Ln N a
Lp Nd
=




)(
= 1.6 10 −19 1.5 10 10
 1

15
 5  10
)
2
8.29 

10 − 7 
32 .4
1
+
10 − 6 10 17
or
J S = 4.426 10 −11 A/cm 2
Then
I S = AJ S = 10 −4 4.426 10 −11
or
I S = 4.426 10 −15 A
We find
V 
I = I S exp  D 
 Vt 
 0.5 
= 4.426  10 −15 exp 

 0.0259 
or
I = 1.07  10 −6 A = 1.07  A
(c) The hole current is
(
(
I p = eni2 A 
)(
)
)
1
Nd
Dp
 po
V
exp  D
 Vt




Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 8
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(
)(
= 1.6  10 −19 1.5  10 10
) (10 )
2
−4
V 
(e) I p (x n ) = I sp exp  a 
 Vt 
1 

17
 10 
V 
8.29
exp  D 
−7
10
 Vt 

(
or
V 
I p = 3.278 10 −16 exp  D  (A)
 Vt 
Then
I p J p 3.278  10 −16
=
=
= 0.0741
I
J S 4.426  10 −15
_______________________________________
I Total
(a) I sp
(
= 1.6 10
−19
= 1.820  10 −4 A
Now
 − (1 2 )L p
1


I p  x n + L p  = I p (x n ) exp 

2
Lp



(

Dp n
 = eA


 po N d

)(5 10 )
(
= 1.6 10
−19
(
)
1.5 10 10
10

8 10 −8 1.5 10 16
2

Dn n
 = eA



Na
no

)(5 10 )
= 9.710  10 −5 A
_______________________________________
(
1.5 10 10
25

2 10 −7
5 10 16
I sn = 4.025 10 −15 A
(
)(
)
2
)
 5 10 16 1.5  10 16 
(c) Vbi = (0.0259 ) ln 

2


1.5  10 10
= 0.746826 V
V a = (0.8)Vbi = (0.8)(0.746826 ) = 0.59746 V
(
)
V  n
V 
p n (x n ) = p no exp  a  =
exp  a 
 Vt  N d
 Vt 
2
i
=
(1.5 10 )
10 2
1.5 10 16
 0.59746 
exp 

 0.0259 
= 1.56  10 14 cm −3
(
(d) I n − x p
)
V 
= I n (x n ) = I sn exp  a 
 Vt 
 0.59746 
= 4.025  10 −15 exp 

 0.0259 
(
Then
1
1




I n  x n + L p  = I Total − I p  x n + L p 
2
2




= 1.820  10 −4 − 8.4896  10 −5
2
i
−4
)
= 8.4896  10 −5 A
I sp = 1.342  10 −14 A
 eDn n po
(b) I sn = A
 Ln




 −1 
= 1.3997  10 − 4 exp  
 2 
2
i
−4
= 1.3997  10 −4 A
= In + I p
= 4.1981  10 −5 + 1.3997  10 −4
8.16
 eD p p no
= A
 Lp

)
 0.59746 
= 1.342  10 −14 exp 

 0.0259 
8.17
(a) The excess hole concentration is given by
p n = p n − p no
 V
= p no exp  a
  Vt
We find
p no =
(
ni2
1.5 10 10
=
Nd
10 16
and
)
2
= 2.25 10 4 cm −3
(8)(0.01 10 −6 )
L p = D p pO =
= 2.828  10 −4 cm = 2.828  m
Then
p n = (2.25 10 4 )exp 


 0.610  
 − 1
 0.0259  
−x


 exp 

−4
 2.828  10 
)
= 4.1981  10 −5 A
−x
 

 − 1  exp 

 Lp 
 


or
p n = (3.81 10 14 )exp 
−x

 cm −3
−4
 2.828  10 

Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 8
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(b) We have
(b) Problem 8.8
d (p n )
J p = −eD p
dx
=
(
eD p 3.808 10 14
2.828 10
−4
) exp 
−x



−4
2
.
828

10


At x = 3  10 −4 cm,
1.6  10 −19 (8) 3.808  10 14
 −3 
J p (3) =
exp 

−4
2.828  10
 2.828 
or
J p (3) = 0.5966 A/cm 2
(
) (
J no =
 (0.1)N d 
p 
or V a = Vt ln  n  = V t ln  2

p
 n i N d 
 no 
 (0.1)N d2 
= Vt ln 

2
 n i

)
(c) We have
eDn n po
V 
J no =
exp  a 
Ln
 Vt 
We can determine that
n po = 4.5  10 3 cm −3 and L n = 10 .72  m
Then
V 
p n = p no exp  a 
 Vt 
(1.6 10 )(23)(4.5 10 )
−19
3
10.72 10 − 4
 0.610 
 exp 

 0.0259 
(
)
 (0.1) 8  10 15 2 
= (0.0259 ) ln 

2
 1.5  10 10

= 0.623 V
_______________________________________
(
)
8.19
The excess electron concentration is given by
n p = n p − n po
 V  
−x

= n po exp  a  − 1  exp 

  Vt  
 Ln 
The total number of excess electrons is


N p = A n p dx
or
0
J no = 0.2615 A/cm
We can also find
J po = 1.724 A/cm 2
2
We may note that

−x
−x
dx = − L n exp 

exp 

 L  0 = Ln
 Ln 
 n 
0
Then
 V  
N p = AL n n po exp  a  − 1
  Vt  
We find that
D n = 25 cm 2 /s and L n = 50 .0  m
Also

Then at x = 3  m,
J n (3) = J no + J po − J p (3)
= 0.2615 +1.724 − 0.5966
or
J n (3) = 1.39 A/cm 2
_______________________________________
8.18
n po =
(a) Problem 8.7
V 
n p = n po exp  a 
 Vt 
 np 


 = Vt ln  (0.1)N a 
or V a = Vt ln 
2
 n po 
 n i N a 


 (0.1)N a2 
= Vt ln 

2
 n i

)
2
= 2.81 10 4 cm −3
Then
N p = 10 −3 50 .0  10 −4 2.8125  10 4
(
)(
)(
 V
 exp  a
  Vt
)
 
 − 1

 
or
(
)
 (0.1) 4 10 15 2 
= (0.0259 ) ln 

2
 2.4 10 13

= 0.205 V
(
(
ni2
1.5 10 10
=
Na
8 10 15
)
 V  
N p = (0.1406 )exp  a  − 1
  Vt  
Then, we find the total number of excess
electrons in the p-region to be:
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 8
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
 E g 2 − 0.59 

10 3 = exp 

 0.0259 
Then
E g 2 = 0.59 + (0.0259 ) ln 10 3
(a) V a = 0.3 V, N p = 1.51  10 4
(b) V a = 0.4 V, N p = 7.17  10 5
(c) V a = 0.5 V, N p = 3.40  10 7
( )
Similarly, the total number of excess holes in
the n-region is found to be
 V  
Pn = AL p p no exp  a  − 1
  Vt  
We find that
D p = 10.0 cm 2 /s and L p = 10.0  m
Also
p no
(
n2
1.5 10 10
= i =
Nd
10 16
)
2
= 2.25 10 4 cm −3
Then
(
Pn = 2.25  10
−2
  Va
exp 
  Vt
)
E g 2 = 0.769 eV
_______________________________________
8.21
(a) We have
 1
Dp
Dn
1
I S = Aeni2 
+
 N a  nO N d  pO
which can be written in the form
I S = C ni2



3
 − Eg 
 T 

= C N cO N O 
 exp 

 300 
 kT 
 
 − 1

 
or
So
(a) V a = 0.3 V, Pn = 2.41 10 3
(b) V a = 0.4 V, Pn = 1.15 10 5
(c) V a = 0.5 V, Pn = 5.45 10 6
_______________________________________
8.20
 − Eg 
V 
 eV
  exp  a
I  ni2 exp  a   exp 

V
kT
 kT
 t 


Then
 eVa − E g 

I  exp 

 kT 
so
 eVa1 − E g1 

exp 

kT
I1


=
I2
 eVa 2 − E g 2 

exp 

kT


or
 eVa1 − eVa 2 − E g1 + E g 2 
I1

= exp 

I2
kT


or



We then have
 0.255 − 0.32 − 0.525 + E g 2
10 10 −3
= exp 
−6
0.0259
10 10

or
 − Eg 

I S = CT 3 exp 

 kT 
(b) Taking the ratio
 − Eg

3 exp 
I S 2  T2 
 kT2
=   
I S 1  T1 
 − Eg
exp 
 kT1








3


 1
1 
  exp + E g 

−


 kT1 kT2 
1
= 38.61
For T1 = 300 K, k T1 = 0.0259 ,
kT1
T
=  2
 T1
For T 2 = 400 K, kT2 = 0.03453 ,
1
= 28.96
kT2
(i) Germanium: E g = 0.66 eV
3
I S 2  400 
=
 exp (0.66 )(38.61 − 28.96 )
I S1  300 
I
or S 2 = 1383
I S1
(ii) Silicon: E g = 1.12 eV
I S2




I S1
3
 400 
 exp (1.12 )(38 .61 − 28 .96 )
= 
 300 
I S2
= 1.17 10 5
I S1
_______________________________________
or
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 8
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
8.22
Plot
_______________________________________
 n2
=  i
 Nd
8.23
First case:
If
V 
= exp  a 
Is
 Vt 
Va
0.50
=
= 0.05049 V
or Vt =
If
ln 2  10 4
ln
Is
(
 (0.1)N d2 
or V a = Vt ln 

2
 n i

(
)(
or ni2 = 8.2519 10 27
Now
 − Eg
ni2 = N c N  exp 
 kT







3
L p = D p po =
(10 )(10
) = 10
or L p = 10  m;  Wn  L p
(a) J p (x n ) =
eD p p no
Wn
V
exp  a
 Vt




−3
10 2
−4
15
(
V
exp  a
 Vt





V
 exp  a
V

 t





AeDn n po
Ln
D n  n i2

 no  N a
)(
= 10 −3 1.6 10 −19
)

 − (1.12 )(300 ) 
 T 
2.8337 10 −11 = 
 exp 

300


 (0.0259 )(T ) 
By trial and error,
T  502 K
The reverse-bias current is limiting factor.
_______________________________________
−7




−19
19


− 1.12
 exp 

 (0.0259 )(T 300 ) 
8.24

V
 exp  a
V

 t

−3
In =
= Ae
3
)
I p = 4.565  10 −3 A
10 

10 − 7 

)
 0.5516 
 exp 

 0.0259 
)( )
19




(10 )(1.6 10 )(10 )(1.5 10 )
(0.7 10 )(2 10 )
=
T 
(8.2519 10 ) = (2.8 10 )(1.04 10 ) 300

27
AeD p  n i2

W n  N d
(ii) I p =
25
1
+
−7
5  10
2  1017
(
(
1.2 10 −6 = 5 10 −4 1.6 10 −19 ni2
 1

15
 4  10

V
 exp  a
V

 t

 (0.1) 2 10 15 2 
= (0.0259 ) ln 

2
 1.5  10 10

V a = 0.5516 V
)
 T 
Now 0.05049 = (0.0259 )

 300 
T = 584.8 K
Second case:
 1 D
Dp
1
n
I s = Aeni2 
+
 N a  no N d  po
V 
(i) p n (x n ) = (0.1)N d = p no exp  a 
 Vt 
(
)
2
1.5 10 10
25

−7
5 10
2 10 17
 0.5516 
 exp 

 0.0259 
I n = 2.26 10 −6 A
I = In + I p
= 2.26  10 −6 + 4.565  10 −3
= 4.567  10 −3 A
or I = 4.567 mA
V
(b) (i) n p − x p = (0.1)N a = n po exp  a
 Vt
(
)
 n2
=  i
 Na
cm
 (0.1)N a2 
or V a = Vt ln 

2
 n i

(

V
 exp  a
V

 t

)
 (0.1) 2 10 15 2 
= (0.0259 ) ln 

2
 1.5  10 10

V a = 0.5516 V
(




)




Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 8
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
AeD p  n i2

W n  N d
(ii) I p =
=

V
 exp  a
V

 t

Then, from the first boundary condition, we
obtain
 V  
p no exp  a  − 1
  Vt  




(10 )(1.6 10 )(10 )(1.5 10 )
(0.7 10 )(2 10 )
−3
10 2
−19
−4
17
 0.5516 
exp 

 0.0259 
I p = 4.565  10 −5 A
I n = Ae
(
D n  n i2

 no  N a
)(
= 10 −3 1.6 10 −19
)

V
 exp  a
V

 t

(




)
2
1.5 10 10
25

−7
5 10
2 10 15
 0.5516 
 exp 

 0.0259 
I n = 2.2597 10 −4 A
I = In + I p
= 2.2597  10 −4 + 4.565  10 −5
= 2.716  10 −4 A
or I = 0.2716 mA
_______________________________________
8.25
(a) We can write for the n-region
d 2 (p n ) p n
− 2 =0
dx 2
Lp
The general solution is of the form
+x


 + B exp  − x 
p n = A exp 
 Lp 
 Lp 




The boundary condition at x = x n gives

 Va  
 − 1

 Vt  
p n (x n ) = p no exp 

 + xn 


 + B exp  − x n 
= A exp 
 Lp 
 Lp 




and the boundary condition at x = x n + W n
gives
p n ( x n + W n ) = 0
 x + Wn 


 + B exp  − (x n + W n ) 
= A exp  n
 Lp 


Lp




From this equation, we have
 − 2( x n + W n ) 
A = − B exp 

Lp


 − (x n + 2W n ) 
 − xn
= − B exp 
 + B exp 
Lp


 Lp
 − x 
 − 2Wn
= B exp  n  1 − exp 
 L p 
 Lp



We then obtain
 V  
p no exp  a  − 1
  Vt  
B=
 − x 
 − 2W n 

exp  n  1 − exp 
 L p 
 L p 




which can be written as








 x + Wn 
 V  
p no exp  a  − 1  exp  n

  Vt  
 L p 
B=
W 
 − Wn 

exp  n  − exp 
 Lp 
 Lp 




We can also find
 − (x n + Wn ) 
 V  
− p no exp  a  − 1  exp 

Lp
  Vt  


A=
W 
 − Wn 

exp  n  − exp 
 Lp 
 Lp 




The solution can now be written as
 V  
p no exp  a  − 1
  Vt  
p n =
W 
2 sinh n 
 Lp 



 − (x n + Wn − x )  
  x + Wn − x 

 exp  n
 − exp 

L
L



 
p
p
 

or finally
 x + Wn − x 

sinh n


L
  Va  
p


p n = p no exp   − 1 
V


  t  
W
sinh n 
 Lp 


Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 8
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(b)
d (p n )
J p = −eD p
dx
x = xn
 V  
− eD p p no exp  a  − 1
  Vt  
=
W 
sinh n 
 Lp 


 −1 


 cosh x n + W n − x 

 Lp 

 x = xn
Lp




Then
W   V  
coth n   exp  a  − 1
 Lp  
Lp

   Vt  
_______________________________________
Jp =
eD p p no
8.26
V 
I D  ni2 exp  D 
 Vt 
For the temperature range 300  T  320 K,
neglect the change in N c and N  .
Then
 − Eg 
 eV 
  exp  D 
I D  exp 

 kT 
 kT 
(
)
 − E g − eVD 
 exp 

kT


Taking the ratio of currents, but maintaining
I D a constant, we have
(
)
(
)
 − E g − eV D1 
exp 

kT1


1=
 − E g − eV D 2 
exp 

kT2


We then have
E g − eVD1 E g − eVD 2
=
kT1
kT2
We have
T = 300 K , V D1 = 0.60 V and
k T1 = 0.0259 eV,
kT1
= 0.0259 V
e
T = 310 K ,
kT2 = 0.02676 eV,
kT2
= 0.02676 V
e
T = 320 K ,
kT3 = 0.02763 eV,
kT3
= 0.02763 V
e
For T = 310 K ,
1.12 − 0.60 1.12 − V D 2
=
0.0259
0.02676
which yields
V D 2 = 0.5827 V
For T = 320 K ,
1.12 − 0.60 1.12 − V D 3
=
0.0259
0.02763
which yields
V D 3 = 0.5653 V
_______________________________________
8.27
(a) We can write
 eV 
I D = C  ni2  exp  a 
 kT 
where C is a constant, independent of
temperature.
As a first approximation, neglect the
variation of N c and N  with temperature
over the range of interest. We can then write
 − Eg 
V 
  exp  a 
I D = C1  exp 
V 

 t 
 kT 
 − E g − eVa 
= C1  exp 

kT


where C 1 is another constant, independent of
temperature. We find
E g − eVa
C 
= ln  1 
kT
 ID 
or
 kT   C1 

Va = E g − 
  ln 
 e   ID 
_______________________________________
(
)
8.28
 1
(a) I s = Aeni2 
 N a
(
)(
Dn
 n0
)(
= 10 −4 1.6 10 −19 1.5 10 10
 1

16
 4  10
We find
Aeni W
2 0
)
1
Nd
Dp 

 p0 

2
25
1
+
−7
10
4  10 16
I s = 2.323 10 −15 A
(b) I gen =
+
10
10 − 7



Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 8
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Aeni W
I s = I gen =
 4  10 16 4 10 16 
2 0
Vbi = (0.0259 ) ln 

−4
10 2
10 1.6  10 −19 4.734  10 14 6.109  10 −5
 1.5 10

=
2 10 − 7
= 0.7665 V
and
Then
1/ 2
I s + I gen = 2.314  10 −6 A

 2 s (Vbi + V R )  N a + N d 

W =
 N N 
or I s = I gen = 2.314  A
e

a
d 



(b)
From Problem 8.28
 2(11.7 ) 8.85 10 −14 (0.7665 + 5)
=
I s = 2.323 10 −15 A
1.6 10 −19

I gen = 7.331  10 −11 A
1/ 2
16
16 
 4  10 + 4  10  

V 
V 
16
16  
So I = I s exp  a  = I gen exp  a 
 4  10 4  10  
V
 t 
 2Vt 
W = 6.109  10 −5 cm
V 
Then
2.323 10 −15 exp  a 
 Vt 
10 −4 1.6  10 −19 1.5  10 10 6.109  10 −5
I gen =
−7
V 
2 10
= 7.331 10 −11 exp  a 
= 7.331  10 −11 A
 2Vt 
(
)(
(
(
I gen
)
(
)(
)(
)(
(
)(
)
(
)
 1 D
D p  AeniW
1
n
=
Aeni2 
+
2 0
 N a  n0 N d  p 0 
 1
25
1
10 
ni 
+

16
10 − 7 4  10 16 10 − 7 
 4  10
W
6.109  10 −5
=
=
2 0
2 10 − 7
(

 = 3.1558 10 4


(
)
8.30
 kT 
Dn = 
   n = (0.0259 )(5500 )
 e 
)
Then
ni2 = 2.2407 10 29
)
Va = 2Vt ln 3.1558 10 4
= 0.5366 V
_______________________________________
3.0545 10 2
3.9528 10 −13 + 2.50 10 −13
= 4.734  10 14 cm −3
= 142 .5 cm 2 /s
D p = (0.0259 )(220 ) = 5.70 cm 2 /s
(a)
)(1.04 10 )
19
 T 


 300 
3
 − (1.12 )(300 )
 T 
7.6947 10 = 
 exp 

300


 (0.0259 )(T ) 
By trial and error,
T  567 K
We have
3
−10
)
)
)
V
exp  a
 2Vt
so ni =
(
)(
V 
exp  a 
−11
 Vt  7.331 10
=
 V  2.323 10 −15
exp  a 
 2Vt 
8.29
(a) Set I S = I gen ,
= 2.8 10
(
(
=
19
)(
)
7.331  10 −11
= 3.16  10 4
Is
2.323  10 −15
_______________________________________
(c)
)(
)
(
(
)
 1
(i) I s = Aeni2 
 N a
(
Dn
 n0
)(
+
1
Nd
)(
Dp 

 p0 

= 2 10 −4 1.6 10 −19 1.8 10 6
 1

16
 7  10
142 .5
1
+
2  10 −8 7  10 16
I s = 1.50 10 −22 A
)
2
5.70 

2  10 −8 
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 8
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
8.31
Using results from Problem 8.30, we find
V a = 0.4 V, I d = 7.64 10 −16 A,
V 
(ii) I D = I s exp  a 
 Vt 
(
)
 0.6 
= 1.50  10 − 22 exp 

 0.0259 
I rec = 1.35 10 −10 A, I T  1.35 10 −10 A
V a = 0.6 V, I d = 1.73 10 −12 A
= 1.726  10 −12 A
(
(iii) I D = 1.50  10
− 22
)
I rec = 6.44 10 −9 A, I T  6.44 10 −9 A
 0.8 
exp 

 0.0259 
V a = 0.8 V, I d = 3.90 10 −9 A
I rec = 3.06 10 −7 A, I T = 3.10 10 −7 A
= 3.896  10 −9 A
(
)
V a = 1.0 V, I d = 8.80 10 −6 A
A
V a = 1.2 V. I d = 1.99 10 −2 A
 1.0 
(iv) I D = 1.50  10 − 22 exp 

 0.0259 
(b) I gen =
= 8.795  10
Aeni W
−6
2 0
(
)(
I rec = 1.45 10 −5 A, I T = 2.33 10 −5 A
I rec = 6.90 10 −4 A, I T = 2.06 10 −2 A
_______________________________________
)
 7  10 16 7 10 16 
Vbi = (0.0259 ) ln 

2


1.8  10 6
= 1.263 V
 2(13.1) 8.85 10 −14 (1.263 + 3)
W =
1.6 10 −19

(
(
)
)
 7  10 16 + 7  10 16

16
16
 7  10 7  10
(
= 4.201  10
8.32
Plot
_______________________________________
−5
)(
 

 
8.33
Plot
_______________________________________
1/ 2
)
8.34
We have that
cm
(i)Then
I gen =
(2 10 )(1.6 10 )(1.8 10 )(4.201 10 )
2(2  10 )
−4
−19
−8
= 6.049  10
−14
A
V
(ii) I rec = I ro exp  a
 2Vt
(
(iii) I rec
6




)

0.6 

= 6  10 −14 exp 
(
2
0
.0259 ) 

= 6.436  10 −9 A

0.8 

= 6  10 −14 exp 
 2(0.0259 ) 
(iv) I rec
(
)
= 3.058  10 −7 A
 1.0


= 6  10 −14 exp 
(
)
2
0
.
0259


(
)
−5
= 1.453  10 A
_______________________________________
−5
R=
np − ni2
 pO (n + n ) +  nO ( p + p )
Let  pO =  nO   O and n  = p  = n i
We can write
 E − E Fi 
n = ni exp  Fn

kT


and
 E Fi − E Fp 

p = ni exp 

kT


We also have
(E Fn − E Fi ) + E Fi − E Fp = eVa
(
)
so that
E Fi − E Fp = eVa − (E Fn − E Fi )
(
Then
)
 eV − (E Fn − E Fi ) 
p = n i exp  a

kT


 eVa 
 − (E Fn − E Fi ) 
= n i exp 
  exp 

kT
 kT 


Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 8
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Define
eV
 E − E Fi 

 a = a and  =  Fn
kT
kT


Then the recombination rate can be written as
ni e ni e a  e − − ni2
R=
 O ni e + ni + ni e a  e − + ni
or
n i e a − 1
R=
 O 2 + e + e a  e −
To find the maximum recombination rate, set
dR
=0
d

)(
(
(

=
or
0=
)
)
(


)  d 2 + e
d
n i e a − 1
O
(
)  (−1)2 + e
n i e a − 1
O

+ ea  e −

+ e a  e −


 e −e
which simplifies to
− ni e a − 1
0=

(
) e − e
2 + e + e
O
a


 e −


−1
e
e
=e
a
 =

=
(
or
Rmax =

)
(

)
ni e a − 1
O 2+e
(
(e

0
In this case, G = g  = 410 19 cm −3 s −1 and is
a constant through the space charge region.
Then
J gen = eg W
a 2
)
ni e a − 1
+ e a
2




(
)(
)
 5 10 15 5 10 15 
= (0.0259 ) ln 

2
 1.5 10 10

(
)
V bi = 0.659 V
Also
 2  (V + V R )  N a + N d

W =  s bi
 N N
e

a
d

(

)




1/ 2
 2(11 .7 ) 8.85 10 −14 (0.659 + 10 )
=
1.6 10 −19

a
2
Then the maximum recombination rate
becomes
n i e a − 1
Rmax =
 O 2 + e a 2 + e a  e − a 2

W
J gen = eGdx
or
 e −
The denominator is not zero, so we have
0 = e  − e  a  e −
or
2
8.35
We have

2
a
Q.E.D.
_______________________________________
N N
Vbi = Vt ln  a 2 d
 ni
−2
−
ni
 eV 
exp  a 
2 O
 2kT 
We find

a
R m ax =
 5  10 15 + 5  10 15
 
15
15
 5  10 5  10
(
)(
)
or
W = 2.35  10 −4 cm
Then
J gen = 1.6  10 −19 4  10 19 2.35  10 −4
(
)(
)(




1/ 2
)
or
)
2 O  a 2 + 1
which can be written as
  eV  
ni exp  a  − 1
  kT  
Rm ax =
  eV  
2 O exp  a  + 1
  2kT  
If V a  (k T e ) , then we can neglect the (-1)
term in the numerator and the (+1) term in the
denominator, so we finally have
J gen = 1.5  10 −3 A/cm 2
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 8
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
8.36
8.38
 1
J S = eni2 
 N a
(
Dn
 nO
+
)(
= 1.6  10 −19 1.5  10
1
Nd
Dp 

 pO 

)
 1

16
 3  10
10 2
+
1
10 18
(a) C d =
(
18
10 − 7
6
10 −7



or
J S = 1.638 10 −11 A/cm 2
Now
V 
J D = J S exp  D 
 Vt 
We want
J = 0 = JG − JD
or
V 
0 = 25 10 −3 − 1.638 10 −11 exp  D 
 Vt 
which can be written as
V 
25 10 −3
exp  D  =
= 1.526 10 9
−11
V
1
.
638

10
 t 
We find
V D = Vt ln 1.526 10 9
or
V D = 0.548 V
_______________________________________
(
Q
, For I D = 1.2 mA
V
)
)(
)
)(
)
Q = C d  V = 1.158 10 −8 50 10 −3
−10
= 5.79  10 C
(b) For I D = 0.12 mA
(
Q = C d  V = 1.158 10 −9 50 10 −3
−11
= 5.79  10 C
_______________________________________
8.39
For a p + n diode
I DQ pO
I DQ
gd =
, Cd =
2Vt
Vt
Now
10 −3
gd =
= 3.86 10 − 2 S
0.0259
and
10 −3 10 −7
Cd =
= 1.93 10 −9 F
2(0.0259 )
We have
g − jC d
1
1
Z= =
= d2
Y g d + jC d
g d +  2 C d2
(
)(
)
where  = 2 f
We obtain
f = 10 kHz ,
Z = 25.9 − j 0.0814
f = 100 kHz , Z = 25.9 − j 0.814
8.37
(a) rd =
Cd =
f = 1 MHz ,
Vt
0.0259
=
= 21 .6 
I DQ 1.2  10 −3
I DQ 0
2Vt
=
(1.2 10 )(0.5 10 )
−3
−6
2(0.0259 )
−8
= 1.16  10 F
or C d = 11 .6 nF
0.0259
= 216 
0.12  10 −3
0.12 10 −3 0.5 10 −6
Cd =
2(0.0259 )
)(
f = 10 MHz , Z = 2.38 − j 7.49
_______________________________________
8.40
Reverse bias


e s N a N d
C j = AC  = A  

 2(Vbi + V R )(N a + N d ) 
(b) rd =
(
Z = 23 .6 − j 7.41
)
= 1.16  10 −9 F
or C d = 1.16 nF
_______________________________________
(
)(
1/ 2
)
 5 10 17 8 10 15 
Vbi = (0.0259 ) ln 

2
 1.5 10 10

= 0.790 V
 1.6 10 −19 (11.7 ) 8.85 10 −14
C j = 2 10 − 4 
2(Vbi + V R )

(
(
)(
)
(
)
(
)(
) 
 5  10 17 8  10 15

17
15
 5  10 + 8  10

 
)
1/ 2
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 8
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Cj =
5.1078  10 −12
Vbi + V r
V R (V)
 pO = 2(0.0259 )(2.5  10 −6 )
or
 pO = 1.3  10 −7 s
F
C j (pF)
10
5
3
1
0
−0.20
−0.40
At 1 mA,
C d = 2.5 10 −6 10 −3
or
C d = 2.5 10 −9 F
_______________________________________
1.555
2.123
2.624
3.818
5.747
6.650
8.179
(
Then
I po p 0 I po 8 10 −8
Cd =
=
= 1.544 10 −6 I po
2Vt
2(0.0259 )
I po = Ae
(
= 2 10
−4
) (
D p  ni2

 p 0  N d
)(1.6 10 )
(
I po = 1.006 10
V a (V)
0.20
−14
C d (F)
3.51  10
)

V
  exp  a
V

 t

−19
)




(
C j (F)
+ 6.650  10
)
2




= C Total (F)
−12
 6.650  10 −12
0.40
7.92  10 −14 + 8.179  10 −12
= 8.258  10 −12
0.60
1.79 10
−10
+ ...
 1.79  10 −10
_______________________________________
8.41
For a p + n diode, I pO  I nO , then
 1 
 I pO pO
C d = 

 2Vt 
Now
 pO
= 2.5  10 − 6 F/A
2Vt
Then
(
)
(i) C d =
or I po =
or I po
1.5 10 10
10

8 10 −8
8 10 15
V
 exp  a
 Vt
V 
exp  a  A
 Vt 
+
−17
)
8.42
(a) N a  N d  I po  I no
Forward bias
For N a  N d  I po  I no
(
)(
I po p 0
2Vt
2Vt (C d )
 p0
=
(
2(0.0259 ) 10 −9
10 −7
)
= 5.18  10 −4 A
= 0.518 mA
(ii) I po = Ae
Dp

 p0
V
ni2
exp  a
Nd
 Vt
(




)(
0.518  10 −3 = 5 10 − 4 1.6  10 −19

(1.5 10 )
10 2
8  10
15
)
10
10 − 7
V
exp  a
 Vt
 0.518 10 −3 

Va = (0.0259 ) ln 
−14 
 2.25 10 
= 0.618 V
V
0.0259
= 50 
(iii) rd = t =
I D 0.518 10 −3
(b)
2V (C ) 2(0.0259 ) 0.25 10 −9
(i) I po = t d =
 p0
10 −7
(




)
= 1.295  10 −4 A
or I po = 0.1295 mA
 0.1295 10 −3 

(ii) Va = (0.0259 ) ln 
−14 
 2.25 10

= 0.5821 V
0.0259
= 200 
(iii) rd =
0.1295  10 −3
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 8
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(b) Set R = 0
(i) For I D = 1 mA,
8.43
(a) p-region:
pL
L
L
Rp =
=
=
A
 p A e p N a A
(
 10 −3
V = (0.0259 ) ln  −10
 10
or V = 0.417 V
)
so
Rp =
or
0.2
(1.6 10 )(480 )(10 )(10 )
−19
16
−2
(ii) For I D = 10 mA,
 10 −2 
V = (0.0259 ) ln  −10 
 10 
or V = 0.477 V
_______________________________________
R p = 26 
n-region:
Rn =
n L
A
=
L
nA
=
L
(e n N d )A
so
Rn =
0.10
(1350 ) 10 15 10 −2
(1.6 10 )
−19
( )(
8.45
)
(a) rd =
or
I
V a = Vt ln  D
 Is
R = 72.3 
(b)
V = IR  0.1 = I (72 .3)
which yields
I = 1.38 mA
_______________________________________
=
 n L(n )
A(n )
(0.2)(10
2 10
+
−2
−5
 p L( p )
A( p )
) + (0.1)(10 )
−2
2 10
1
1 dI D
=
= I S 
rd dVa
 Vt

V
 exp  a

V

 t




or
 10 −3
V = 10 −3 (150 ) + (0.0259 ) ln  −10
 10
or V = 0.567 V
(ii) For I D = 10 mA,




 10 −2
V = 10 − 2 (150 ) + (0.0259 ) ln  −10
 10
or V = 1.98 V




)
Vt 0.0259
=
= 4.3167 10 − 4 A
rd
60
 4.3167 10 −4 

Va = (0.0259 ) ln 
−12

 5 10

= 0.4733 V
_______________________________________
(a)
I 
V = I D R + Vt ln  D 
 IS 
(a) (i) For I D = 1 mA,
)




8.46
R = 150 
We can write
(
(b) I D =
−5
or
(




 8.09375 10 −4
= (0.0259 ) ln 
−12
 5 10
V a = 0.4896 V
or
R=
Vt
V
0.0259
 ID = t =
ID
rd
32
or I D = 8.09375 10 −4 A
R n = 46 .3 
The total resistance is
R = R p + Rn = 26 + 46.3
8.44




1
10 −13
 0.020 
=
 exp 

rd 0.0259
 0.0259 
which yields
rd = 1.2 10 11 
(b)
1
10 −13
 − 0.02 
=
 exp 

rd 0.0259
 0.0259 
which yields
rd = 5.6 10 11 
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 8
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
8.47
8.49
C j = 18 pF at V R = 0
I
(a) If R = 0.2
IF
Then we have
ts
erf
 pO
C j = 4.2 pF at V R = 10 V
IF
=
=
IF + IR
1
1
=
I R 1 + 0.2
1+
IF
We have
 nO =  pO = 10 −7 s , I F = 2 mA
and
IR 
or
ts
erf
 pO
So
= 0.833
 I 
 2
t s   pO ln 1 + F  = 10 − 7 ln 1 + 
I
 1
R 

(
We find
ts
 pO
(b) If
= 0.978 
ts
 pO
= 0.956
IR
= 1.0 , then
IF
erf
ts
 pO
=
V R 10
=
= 1 mA
R 10
1
= 0.50
1+1
)
or
t s = 1.110 −7 s
Also
18 + 4.2
C avg =
= 11 .1 pF
2
The time constant is
 S = RC avg = 10 4 11 .1 10 −12
( )(
)
= 1.11  10 −7 s
Now, the turn-off time is
t off = t s +  S = (1.1 + 1.11) 10 −7
which yields
ts
= 0.228
 pO
_______________________________________
8.48
or
t off = 2.21  10 −7 s
_______________________________________
(a) erf
ts
p
IF
=
IF + IR
8.50
erf 0.3 = erf (0.5477 )
 erf (0.55 ) = 0.56332
1
Then 0.56332 =
I
1+ R
IF
IR
1
=
− 1 = 0.775
I F 0.56332
(b) erf
t2
 p0
+
 −t
exp  2
  p0

 t
 2
  p0

By trial and error,
t2




I 
= 1 + (0.1) R 
 IF 




= 1+ (0.1)(0.775 )
= 1.0775
 0.80
 p0
_______________________________________
(
(
)
 5 10 19 2 
Vbi = (0.0259 ) ln 
 = 1.136 V
2
 1.5 10 10 
We find
)
 2  (V − V a )  N a + N d

W =  s bi
 N N
e

a
d

(
)




1/ 2
 2(11.7 ) 8.85 10 −14 (1.136 − 0.40 )
=
1.6 10 −19

 5  10 19 + 5  10 19

2

5  10 19

(
)




1/ 2
which yields
o
W = 6.17  10 −7 cm = 61.7 A
_______________________________________
8.51
Sketch
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 8
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
8.53
From Figure 7.15, N d  910 15 cm −3
Let N a = 510 17 cm −3
V
p n (x n ) = (0.1)N d = 9 10 14 = p no exp  a
 Vt
p no =
(
ni2
1.5 10 10
=
Nd
9 10 15
)




2
= 2.5 10 4 cm −3
 9 10 14
Then Va = (0.0259 ) ln 
4
 2.5 10
I
50  10 −3
Is =
=
V 
 0.6295 

exp  a  exp 
 0.0259 
 Vt 

 = 0.6295 V


= 1.389  10 −12 A
Is 
AeD p p no
Lp
1.389  10
=
Aeni2
Nd
=
Dp
 p0
−12
(
)(
A 1.6 10 −19 1.5 10 10
9 10 15
(
)
2
1.389 10 −12 = A 2.828 10 −11
−2
)
10
2 10 −7
or A = 4.91  10 cm
_______________________________________
2
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 9
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
Chapter 9
Exercise Solutions
Ex 9.3
Ex 9.1
 B 0 =  m −  = 4.55 − 4.07 = 0.48 V
N 
 2  (V + V R ) 
x n =  s bi

eN d


 4.7 10 17 

 n = Vt ln  c  = (0.0259 ) ln 
15 
 Nd 
 5 10 
= 0.1177 V
Vbi =  B 0 −  n = 0.48 − 0.1177 = 0.3623 V
2 V 
x n =  s bi 
 eN d 
(
)(
)
(
(
 m ax =
)
)
(
−19
−5
16
−14
(
2
(13.1) 8.85 10 −14 2.335 10 12
)(
1/ 2
(1.6 10 )(10 )(8.309 10 )
(11.7 )(8.85 10 )
= 1.284  10 5 V/cm
2
1

2
e s
 1 
 
 C 
V R
−3
)(
x n = 8.309 10 −5 cm
2
(
e
4 s
 1.6  10 −19 6.42  10 4 
=
−14 
 4 (11 .7 ) 8.85  10

 = 0.0281 V
(b) V R = 5 V,
 1 
 
8.5 10 12
 C 

= 2.335 10 12
V R
3 + 0.64
)
or N d = 4.62 10 cm
_______________________________________
18
−5
16
−14
Then  =
Ex 9.2
From Figure 9.3,
Vbi  0.64 V
(1.6 10 )
−19
= 6.42  10 4 V/cm
−5
= 2.24  10 4 V/cm
_______________________________________
−19
(1.6 10 )(10 )(4.155 10 )
(11.7 )(8.85 10 )
=
−14
=
eN d x n
s
 m ax =
(1.6 10 )(5 10 )(3.24 10 )
=
(13.1)(8.85 10 )
Then N d =
1/ 2
 x n = 4.155 10 −5 cm
1/ 2
)
15
1/ 2
)( )
)(V + V )
= 1.294 10 −9 bi
R
(a) V R = 1 V, V bi = 0.334 V
= 3.24  10 −5 cm
eN d x n
 m ax =
s
−19
)
−14

 2(11 .7 ) 8.85 10 (Vbi + V R ) 

=

−19
16


1
.
6

10
10


1/ 2
 2(13 .1) 8.85  10 −14 (0.3623 ) 
=

1.6  10 −19 5  10 15


(
(
(
1/ 2
)(
)
1/ 2
 1.6  10 −19 1.284  10 5 
 = 

−14
 4 (11 .7 ) 8.85  10

 = 0.0397 V
_______________________________________
(
)
Ex 9.4
A =
4 emn k 2
h3
Assume mn = mo , then
A =
(
)(
)(
)
4 1.6  10 −19 9.11  10 −31 1.38  10 −23
(6.625 10
)
2
− 34 3
= 1.20  10 6 A/K 2 -m 2
 A  = 120 A/K 2 -cm 2
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 9
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
Ex 9.5
V 
I  AJ s exp  a 
 Vt 
 I 

so that V a = Vt ln 

 AJ s 
For the pn junction:


10 10 −6

Va = (0.0259 ) ln  − 4
−11 
 10 3.66 10

= 0.5628 V
For the Schottky junction:


10 10 −6

Va = (0.0259 ) ln  − 4
−5 
 10 5.98 10 
= 0.1922 V
_______________________________________
(
)(
(
)
)(
)
p
N 
eVbi = E + kT ln  Po  n 
 p no N P 
 10 15 6 10 18

= 0.70 + (0.0259 ) ln 
11
18 
 5.76 10 7 10 
or V bi = 0.889 V
_______________________________________
(
(
 I S = 4.66 10 −12 A
_______________________________________
Ex 9.7
We have
 2.8 10 19 
 = 0.237 V
= (0.0259 ) ln 
15 
 3 10 
Vbi =  Bo −  n = 0.49 − 0.237 = 0.253 V
 2  (V + V R ) 
(c) x n =  s bi

eN d


(
)
)(
1/ 2
)
o
or x n = 128 .7 A
_______________________________________
Ex 9.8
From Example 9.8, E = 0.70 eV.
We find
Now
(
)(
1/ 2
)
(1.6 10 )(3 10 )(1.505 10 )
(11.7 )(8.85 10 )
−19
−4
15
−14
or  m ax = 6.98  10 4 V/cm
= 1.287  10 −6 cm
p no =
)
or x n = 1.505 10 −4 cm
Then
eN d x n
 m ax =
s
1/ 2
 2(13 .1) 8.85  10 −14 (0.80 ) 
=

−19
7  10 18
 1.6  10

(
(
1/ 2
 2(11 .7 ) 8.85  10 −14 (0.253 + 5) 
=

1.6  10 −19 3  10 15


=
2 V 
x n =  s bi 
 eN d 
)
N 
(b)  n = Vt ln  c 
 Nd 
(
)
)(
TYU 9.1
(a)  Bo = 4.5 − 4.01 = 0.49 V
Then
 − 0.3 
I S = 5 10 − 7 exp 

 0.0259 
)
Test Your Understanding Solutions
Ex 9.6
 V − 0.3 

I ST exp  a

 − 0.3 
I
 Vt 

= 1 = ST exp 
IS
Vt 
 Va 

I S exp  
 Vt 
( )(
ni2
2.4 10 13
=
Nd
10 15
)
2
= 5.76 10 11 cm −3
 e s N d 
(d) C  = 

 2(Vbi + V R ) 
(
)
1/ 2
(
)(
)
 1.6  10 −19 (11 .7 ) 8.85  10 −14 3  10 15 
=

2(0.253 + 5)


−9
2
or C  = 6.88  10 F/cm
_______________________________________
TYU 9.2
(a)  Bo = 5.12 − 4.07 = 1.05 V
 4.7 10 17 
 = 0.131 V
(b)  n = (0.0259 ) ln 
15 
 3 10 
Vbi = 1.05 − 0.131 = 0.919 V
1/ 2
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 9
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
(
)
 2(13 .1) 8.85  10 −14 (0.919 + 5) 
(c) x n = 

1.6  10 −19 3  10 15


(
)(
1/ 2
)
−4
or x n = 1.69 10 cm
Now
1.6  10 −19 3  10 15 1.69 10 −4
 m ax =
(13.1) 8.85 10 −14
(
)(
or  m ax = 7  10 V/cm
)(
(
)
)
4
(
)
(
 1.6 10 −19 (13.1) 8.85 10 −14
(d) C  = 
2(0.919 + 5)

(
 310 15
)
)
1/ 2
or C  = 6.86  10 −9 F/cm 2
_______________________________________
TYU 9.3
 =
e
4 s
(
)(
)
)
 1.6  10 −19 6.98  10 4 
=
−14 
 4 (11 .7 ) 8.85  10

or  = 0.0293 V
xm =
(
1/ 2
e
16 s 


1.6  10 −19
=
−14
4 
6.98  10 
16 (11 .7 ) 8.85  10
(
)(
1/ 2
)
o
or x m = 2.10 10 −7 cm = 21.0 A
_______________________________________
TYU 9.4
V 
I = I S exp  a 
 Vt 
Then
 I 
V a = Vt ln  
 IS 
(a) For the pn junction diode
 100 10 −6
Va = (0.0259 ) ln 
−14
 10
For the Schottky diode
 100 10 −6
Va = (0.0259 ) ln 
−9
 10

 = 0.596 V



 = 0.298 V


(b) For the pn junction diode
 10 −3 
Va = (0.0259 ) ln  −14  = 0.656 V
 10 
For the Schottky diode
 10 −3 
Va = (0.0259 ) ln  −9  = 0.358 V
 10 
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 9
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Chapter 9
9.1
9.2
(a) Vbi =  B 0 −  n
(a) We have
N
e n = eVt ln  c
 Nd




 2.8 10 19
= (0.0259 ) ln 
16
 10
N 

 = 0.206 eV


(c)
 BO =  m −  = 4.28 − 4.01
or
 BO = 0.27 V
and
Vbi =  BO −  n = 0.27 − 0.206
or
V bi = 0.064 V
Also
2 V 
x d =  s bi 
 eN d 
)
V bi increases,  B 0 remains constant
 2(11 .7 ) 8.85  10 −14 (0.064 ) 
=

1.6  10 −19 10 16


(
1/ 2
)( )
or
x d = 9.110 −6 cm
Then
eN d x d
 m ax =
s
 2.8 10 19 

(c)  n = (0.0259 ) ln 
15

 10

V
= 0.2652
Vbi = 0.65 − 0.2652 = 0.3848 V
V bi decreases,  B 0 remains constant
_______________________________________
9.3
(a)  B 0 =  m −  = 5.1 − 4.01 = 1.09 V
(b) Vbi =  B 0 −  n
(1.6 10 )(10 )(9.1`10 )
=
(11.7 )(8.85 10 )
−19
 2.8 10 19 

= (0.0259 ) ln 
15 
 5 10 
=0.2235 V
Vbi = 0.65 − 0.2235 = 0.4265 V
 2.8 10 19 

(b)  n = (0.0259 ) ln 
16

 10

= 0.2056 V
Vbi = 0.65 − 0.2056 = 0.4444 V
1/ 2
(
 n = Vt ln  c 
 Nd 
−6
16
−14
or
 m ax = 1.41  10 4 V/cm
(d)
Using the figure,  Bn = 0.55 V
So
Vbi =  Bn −  n = 0.55 − 0.206
or
V bi = 0.344 V
We then find
x n = 2.11 10 −5 cm
and
 m ax = 3.26 10 4 V/cm
_______________________________________
N 
 n = Vt ln  c 
 Nd 
 2.8 10 19 
 = 0.2056 V
= (0.0259 ) ln 
16

 10

Vbi = 1.09 − 0.2056 = 0.8844 V
 2  (V + V R ) 
(c) x n =  s bi

eN d


1/ 2
(
(
)
)( )
 2(11 .7 ) 8.85  10 −14 (0.8844 + 1) 
(i) x n = 

1.6  10 −19 10 16


= 4.939  10 −5 cm
or x n = 0.4939  m
 m ax =
=
eN d x n
s
(1.6 10 )(10 )(4.939 10 )
(11.7 )(8.85 10 )
−19
−5
16
= 7.63  10 4 V/cm
−14
1/ 2
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 9
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(
(
)
)( )
 2(11 .7 ) 8.85  10 −14 (0.8844 + 5) 
(ii) x n = 

1.6  10 −19 10 16


(
)(
1/ 2
)
= 1.292  10 cm
or x n = 1.292  m
= 8.727  10 cm
or x n = 0.8728  m
(1.6 10 )(10 )(8.727 10 )
(11.7 )(8.85 10 )
−19
)
−4
−5
 m ax =
(
 2(13 .1) 8.85  10 −14 (0.7623 + 5) 
(ii) x n = 

1.6  10 −19 5  10 15


1/ 2
−5
16
 m ax =
−14
(1.6 10 )(5 10 )(1.292 10 )
(13.1)(8.85 10 )
−19
−4
15
−14
= 1.35  10 5 V/cm
_______________________________________
= 8.92  10 4 V/cm
_______________________________________
9.4
(a)  B 0 =  m −  = 5.1 − 4.07 = 1.03 V
9.6
1/ 2
 4.7 10 17 
 = 0.1177 V
(b)  n = (0.0259 ) ln 
15 
 5 10 
(c) Vbi = 1.03 − 0.1177 = 0.9123 V
(d)
(
)
 2(13 .1) 8.85  10 −14 (0.9123 + 1) 
(i) x n = 

1.6  10 −19 5  10 15


(
)(
)
(i)
= 7.445  10 cm
or x n = 0.7445  m
(1.6 10 )(5 10 )(7.445 10 )
(13.1)(8.85 10 )
−19
−5
15
−14
= 5.14  10 V/cm
4
(
)
 2(13 .1) 8.85  10 −14 (0.9123 + 5) 
(ii) x n = 

1.6  10 −19 5  10 15


(
 2.8 10 19 
 = 0.265 V
15

 10

Vbi = 0.88 − 0.265 = 0.615 V
 n = (0.0259 ) ln 
1/ 2
−5
 m ax =
 e s N d 
(a) C  = 

 2(Vbi + V R ) 
We have  B 0 = 0.88 V
)(
1/ 2
)
)(
(
= 7.16  10 −13 F
or C = 0.716 pF
(ii)
(
C = 10
−4
−4
= 1.309  10 cm
or x n = 1.309  m
 m ax =
(1.6 10 )(5 10 )(1.309 10 )
(13.1)(8.85 10 )
−19
−4
15
−14
= 9.03  10 4 V/cm
_______________________________________
 2(13 .1) 8.85  10 −14 (0.7623 + 1) 
(i) x n = 

1.6  10 −19 5  10 15


(
)(
 m ax =
1/ 2
)
= 7.147  10 −5 cm
or x n = 0.7147  m
(1.6 10 )(5 10 )(7.147 10 )
(13.1)(8.85 10 )
−19
= 4.93  10 4 V/cm
−5
15
−14
)( )
1/ 2
)(
)
(
)( )
1/ 2
 1.6  10 −19 (11 .7 ) 8.85  10 −14 10 15 


2(0.615 + 5)


= 3.84  10 −13 F
or C = 0.384 pF
 2.8 10 19 
 = 0.206 V
(b)  n = (0.0259 ) ln 
16

 10

Vbi = 0.88 − 0.206 = 0.674 V
(i)
)(
(
(b)  n = 0.1177 V
(c) Vbi = 0.88 − 0.1177 = 0.7623 V
(d)
)
(
)
(
)( )
1/ 2
)
(
)( )
1/ 2
 1.6  10 −19 (11 .7 ) 8.85  10 −14 10 16 
C = 10 − 4 

2(0.674 + 1)


9.5
(
)
 1.6  10 −19 (11 .7 ) 8.85  10 −14 10 15 
C = 10 − 4 

2(0.615 + 1)


= 2.22  10 −12 F
or C = 2.22 pF
(ii)
(
C = 10
−4
)(
 1.6  10 −19 (11 .7 ) 8.85  10 −14 10 16 


2(0.6745 + 5)


= 1.21  10 −12 F
or C = 1.21 pF
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 9
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
9.7
 m ax =
(a) From the figure, Vbi = 0.90 V
(b) We find
(
)
(
)(
 4.7 10 17
= (0.0259 ) ln 
16
 1.04 10
−4
15
−14
(b)
(i)  =
(
e
4 s
)(
)
)
 1.6  10 −19 4.66  10 4 
=
−14 
 4 (11 .7 ) 8.85  10

= 0.0239 V
)
xm =
(
e
16 s 
(
(
)
)(
1/ 2
)
or
x m = 2.57  10 −7 cm




(
 n = 0.0986 V
(d)
 Bn = Vbi +  n = 0.90 + 0.0986
or
 Bn = 0.9986 V
_______________________________________
1/ 2


1.6  10 −19
=
−14
4 
4.66  10 
16 (11 .7 ) 8.85  10
)(
)
)
 1.6  10 −19 9.14  10 4 
(ii)  = 
−14 
 4 (11 .7 ) 8.85  10

= 0.0335 V
or
9.8
−19
= 9.14  10 4 V/cm
2
 1 
 
3 10 15 − 0
 C 
=
= 1.034 10 15
V R
2 − (− 0.90 )
and
2
1.034 10 15 =
e s N d
We can then write
2
Nd =
1.6 10 −19 (13.1) 8.85 10 −14 1.034 10 15
or
N d = 1.04 10 16 cm −3
(c)
N 
 n = Vt ln  c 
 Nd 
(1.6 10 )(5 10 )(1.183 10 )
(11.7 )(8.85 10 )
(
1/ 2


1.6  10 −19
xm = 
−14
4 
9.14  10 
16 (11 .7 ) 8.85  10
(
)(
)
−7
= 1.83  10 cm
_______________________________________
9.9
We have
From Figure 9.5,  BO  0.63 V
 2.8 10 
 = 0.224 V
(a)  n = (0.0259 ) ln 
15 
 5 10 
Vbi =  B 0 −  n = 0.63 − 0.224 = 0.406 V
19
(
)
 2(11 .7 ) 8.85  10 −14 (0.406 + 1) 
(i) x n = 

1.6  10 −19 5  10 15


(
)(
1/ 2
)
= 6.033  10 −5 cm
or x n = 0.6033  m
 m ax =
(1.6 10 )(5 10 )(6.033 10 )
(11.7 )(8.85 10 )
−19
−5
15
−14
= 4.66  10 4 V/cm
(
)
 2(11 .7 ) 8.85  10 (0.406 + 5) 
(ii) x n = 

1.6  10 −19 5  10 15


(
−14
)(
−4
= 1.183  10 cm
or x n = 1.183  m
)
1/ 2
1/ 2
−  (x ) =
−e
− x
16 s x
e (x ) =
e2
+ ex
16 s x
or
Now
d (e (x ))
− e2
=0=
+ e
dx
16 s x 2
Solving for x, we find
e
x = xm =
16 s 
Substituting this value of x = x m into the
equation for the potential, we find
e
e
 =
+
16

s 
e
16 s
16 s 
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 9
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
which yields
e
 =
4 s
(b) We have
E g − eO − e Bn
(
=
_______________________________________
 4.7 10 17 
 = 0.0997 V
(a)  n = (0.0259 ) ln 
16

 10

Vbi =  B 0 −  n = 0.88 − 0.0997  0.780 V
)
 2(13 .1) 8.85  10 −14 (0.780 ) 
xn = 

1.6  10 −19 10 16


(
=
1/ 2
)( )
−14
(1.6 10 )
4 (13 .1)(8.85  10 )
(0.044 ) (4 )(13.1)(8.85 10 )
=
−19
−14
−14
2
1.6 10 −19
= 1.763  10 5 V/cm
Now
(1.6 10 )(10 )x
(13.1)(8.85 10 )
−19
 = 1.763 10 5 =
16
n
−14
 x n = 1.277 10 −4 cm
And
(
x n2 = 1.277 10 −4
=
1/ 2
)
2
(
)
(1.6 10 )(10 )
2(13 .1) 8.85  10 −14 (0.780 + V R )
−19
16
 V R = 10 .5 V
_______________________________________
9.11
Plot
_______________________________________
9.12
(a)  BO =  m −  = 5.2 − 4.07
or
 BO = 1.13 V
Bn
13

or
−8

0.83 −  Bn
= 0.038  Bn − 0.10 − 0.221(1.13 −  Bn )
e
4 s
=
−14
( )

(8.85 10 ) 5.2 − (4.07 +  )
−
 10 
(25  10 )
e
e 
−5
16
= 4.64  10 4 V/cm
(b)  = (0.05 )(0.88 ) = 0.044 V
(0.044 )2
2(1.6 10 )(13.1)(8.85 10 )
−19




−14
(1.6 10 )(10 )(3.362 10 )
(13.1)(8.85 10 )
=
1
 10 13
e
 e
 10 16 ( Bn − 0.10 )
= 3.362  10 cm
or x n = 0.3362  m
−19
i
 m − ( +  Bn )
eDit 
which becomes
e(1.43 − 0.60 −  Bn )
−5
 m ax =
2e s N d ( Bn −  n )
−
9.10
From Figure 9.5,  BO  0.88 V
(
1
eDit
)
We find
 Bn = 0.858 V
(c)
If  m = 4.5 V, then
 BO =  m −  = 4.5 − 4.07
or
 BO = 0.43 V
From part (b), we have
0.83 −  Bn
= 0.038  Bn − 0.10 − 0.2214.5 − (4.07 +  Bn )
We then find
 Bn = 0.733 V
With interface states, the barrier height is less
sensitive to the metal work function.
_______________________________________
9.13
We have that
E g − eO − e Bn
(
=
1
eDit
)
2e s N d ( Bn −  n )
−
i
 m − ( +  Bn )
eDit 
Let eDit = D it (cm −2 eV −1 )
Then we can write
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 9
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
e(1.12 − 0.230 − 0.60 )
=
(
)
(
1
2 1.6 10 −19 (11.7 ) 8.85 10 −14
Dit
(
 5 10
−
16
)
(
)(0.60 − 0.164 )
(8.85 10 ) 4.75 − (4.01 + 0.60)
D  (20 10 )
We then find
Dit = 4.97 10 11 cm −2 eV −1
_______________________________________
9.14
 2.8 10 19 
 = 0.224 V
(a)  n = (0.0259 ) ln 
15 
 5 10 
(b) Vbi =  Bn −  n = 0.89 − 0.224 = 0.666 V
 − e Bn 

(c) J sT = A T 2 exp 
 kT 
 − 0.89 
2
= (120 )(300 ) exp 

 0.0259 
J sT = 1.29 10 −8 A/cm 2
 J 
5

 = (0.0259 ) ln 
Va = Vt ln 

−8

J
 1.29 10 
 sT 
V a = 0.512 V
_______________________________________
9.15
(a)  B 0  0.63 V
 − 0.63 
2
J sT = (120 )(300 ) exp 

 0.0259 
= 2.948  10 −4 A/cm 2
)
I sT = 10 −4 2.948 10 −4 = 2.948 10 −8 A
 I
(i) Va = Vt ln 
 I sT
 V
(i) I = I sT exp  a
  V t
 
 − 1

 
 10  10 −6

V a = (0.030217 ) ln 
+ 1
1.296  10 − 6

V
= 0.0654
 100 10 −6

(ii) Va = (0.030217 ) ln 
+ 1
−6
1.296 10

= 0.1317 V


10 −3

(iii) Va  (0.030217 ) ln 
−6 
 1.296 10 
= 0.201 V
_______________________________________
9.16
(a)  Bn  0.88 V
(d)
)(
)
= 1.296  10 −6 A
−14
−8
 350 
kT = (0.0259 )
 = 0.030217 eV
 300 
 − 0.63 
2
I sT = 10 − 4 (120 )(350 ) exp 

 0.030217 
1/ 2
it
(
(b)




 10 10 −6 

= (0.0259 ) ln 
−8 
 2.948 10 
= 0.151 V
 100 10 −6 

(ii) Va = (0.0259 ) ln 
−8 
 2.948 10 
= 0.211 V


10 −3

(iii) Va = (0.0259 ) ln 
−8 
 2.948 10 
= 0.270 V
 − 0.88 
2
(b) J sT = (1.12 )(300 ) exp 

 0.0259 
= 1.768  10 −10 A/cm 2
10


(c) V a = (0.0259 ) ln 

−10
1
.
768

10


= 0.641 V
(d) V a = Vt ln (2 ) = (0.0259 ) ln (2 )
= 0.0180 V
_______________________________________
9.17
Plot
_______________________________________
9.18
From the figure,  Bn = 0.68 V
 −
J ST = A*T 2 exp  Bn
 Vt

  
  exp 


V 

 t 
  
 − 0.68 
2

= (120 )(300 ) exp 
  exp 

0
.
0259


 Vt 
or
  

J ST = 4.277 10 −5 exp 

 Vt 
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 9
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
We have
 =
Also
e
4 s
Now

 = 0.2056 V


and
Vbi =  Bn −  n = 0.68 − 0.2056 = 0.4744 V
(a) We find for V R = 2 V,
 2  (V + V R ) 
x d =  s bi

eN d


(
1/ 2
)
1/ 2
)( )
x d = 0.566 10 −4 cm = 0.566  m
Then
eN d x d
 m ax =
s
(1.6 10 )(10 )(0.566 10 )
=
(11.7 )(8.85 10 )
−19
−14
or
)
 1.6  10 −19 8.745  10 4 
 = 

−14
 4 (11 .7 ) 8.85  10

)
(
)
1/ 2
 = 0.03803 V
Then
)
or
J ST 2 = 1.86 10 −4 A/cm 2
Finally,
I R 2 = 1.86 10 −8 A
_______________________________________
9.19
We have that

J
−
s →m

=  x dn
Ec
g c (E ) =
)
 0.0328 
J ST 1 = 4.277  10 −5 exp 

 0.0259 
dn =
or
For A = 10 −4 cm 2 , we find
I R1 = 1.52 10 −8 A
(b) For V R = 4 V, then
)
)( )
 2(11 .7 ) 8.85  10 (4.4744 ) 
xd = 

1.6  10 −19 10 16


1/ 2
(
4 2mn*
)
3/ 2
3/ 2
E − Ec
h3
1 * 2
m n = E − E c
2
We can then write
E − Ec = 
or
x d = 0.761 10 −4 cm = 0.761  m
)
 − (E − E F ) 
 exp 
 dE
kT


If the energy above E c is kinetic energy, then
J ST 1 = 1.52 10 −4 A/cm 2
−14
(
4 2mn*
E − Ec
h3
and assuming the Boltzmann approximation
 − (E − E F ) 
f F (E ) = exp 

kT


Then
 = 0.0328 V
Then
(
)
or
1/ 2
or
(
)(
The incremental electron concentration is
dn = g c (E ) f F (E )dE
where
 m ax = 8.745  10 4 V/cm
(
(
 1.6  10 −19 1.176  10 5 
 = 

−14
 4 (11 .7 ) 8.85  10

−4
16
)(
and
(
or
(
−14
 0.03803 
J ST 2 = 4.277  10 −5 exp 

 0.0259 
 2(11 .7 ) 8.85  10 −14 (2.4744 ) 
=

1.6  10 −19 10 16


(
−4
16
 m ax = 1.176  10 5 V/cm
 2.8 10 19
= (0.0259 ) ln 
16
 10
Now
−19
or
N 
 n = Vt ln  c 
 Nd 
(
(1.6 10 )(10 )(0.761 10 )
(11.7 )(8.85 10 )
 m ax =
and
mn*
2
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 9
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
1 *
m n  2d = m n*d
2
We can also write
E − E F = (E − E c ) + (E c − E F )
m n* y2
 2kT 
=    y =  * 
2kT
 mn 
dE =
=
1/ 2
so that
3

− e n
 exp 

 kT

 m* 
J s−→ m = 2 n 
 h 



 − m n* 2
 exp 
 2k T
3
m 
− e n 
 exp 
J s−→ m = 2


 kT 
 h 

 − m n* x2
  x exp 
 2kT
 Ox


 − m n* y2
 exp 
 2kT
−


d x



d
 y

 − m n* z2
 exp 
 2k T
−




 − e(Vbi − V a ) 
2
 exp 
   exp −  d
kT

 0
(
)
 (
)


 (
2

)
−
The current is due to all x-directed velocities
that are greater than  Ox and for all y- and
z- directed velocities. Then

2
 2kT 
 − e n
  *   exp 
 kT
 mn 
 exp −  d  exp −  2 d
The differential volume element is
4  2 d = d x d y d z

3


  4  2 d


We can write
 2 =  x2 +  y2 +  z2
*
n

 2kT 
m n* z2
=  2   z =  *   
2kT
 mn 
Substituting the new variables, we have
1 * 2
m n + e n
2
 m*
dn = 2 n
 h
1/ 2
2

d z


We can write
1 * 2
m n Ox = e(Vbi − V a )
2
Make a change of variables:
m n* x2
2(Vbi − V a )
=2 +
2kT
kT
or
e(V − Va )
2kT 
 x2 = *  2 + bi

kT
mn 

Taking the differential, we find
 2kT 
 x d x =  * d
 mn 
We may note that when  x =  Ox ,  = 0 .
We may define other change of variables,
−
_______________________________________
9.20
For the Schottky diode,
V 
0.80 10 −3 = 10 − 4 6 10 −8 exp  a 
 Vt 
 0.80 10 −3 
(a) Va (SB) = (0.0259 ) ln  − 4
−8 
 10 6 10 
= 0.4845 V
Then
V a ( pn) = 0.4845 + 0.285 = 0.7695 V
(
)(
)
(
(
)(
)
)
 0.7695 
(b) 0.80  10 −3 = A pn 10 −11 exp 

 0.0259 
 A pn = 0.998  10 −5  10 −5 cm 2
_______________________________________
9.21
For the pn junction,
I s = 8 10 − 4 8 10 − 13 = 6.4 10 − 16 A
(
)(
)
 150 10 −6
(a) Va = (0.0259 ) ln 
−16
 6.4 10
= 0.678 V
 700 10 −6
(b) Va = (0.0259 ) ln 
−16
 6.4 10
= 0.718 V
 1.2 10 −3
(c) Va = (0.0259 ) ln 
−16
 6.4 10
= 0.732 V












Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 9
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
For the Schottky junction,
I sT = 8 10 −4 6 10 −9 = 4.8 10 −12 A
(
)(
)
 150 10 −6 

(a) Va = (0.0259 ) ln 
−12 
 4.8 10 
= 0.447 V
 700 10 −6 

(b) Va = (0.0259 ) ln 
−12 
 4.8 10 
= 0.487 V
 1.2 10 −3 

(c) Va = (0.0259 ) ln 
−12 
 4.8 10 
= 0.501 V
_______________________________________
9.22
(a) (i) I = 0.80 mA in each diode
(ii)

0.8 10 −3
Va (SB) = (0.0259 ) ln 
−4
−9
 8 10 6 10
= 0.490 V

0.8 10 −3
Va ( pn) = (0.0259 ) ln 
−4
−13
 8 10 8 10
= 0.721 V
(b) Same voltage across each diode
I = 0.8  10 −3 = I SB + I pn
(
(
(
= 8 10
−4
)(6 10 )
−9
)(
)
)(
V
exp  a
 Vt



)(
(
For the pn junction diode,
 1.143 
V a = (0.0259 ) ln 
 = 0.6907 V
−12
 3  10 
For the Schottky diode,
 1.143 
V a = (0.0259 ) ln 
 = 0.4447 V
−8
 4  10 
(b) For the pn junction diode,
3
 − Eg 
 T 

J S  ni2  
 exp 

 300 
 kT 
Then
J S (400 )  400 
=

J S (300 )  300 







)
3
− Eg
Eg 

 exp 
+

 (0.0259 )(400 300 ) 0.0259 
1.12 
 1.12
= 2.37 exp 
−

 0.0259 0.03453 
)
V 
+ 8  10 − 4 8  10 − 13 exp  a 
 Vt 
V 
= 4.8 10 −12 + 6.4 10 −16 exp  a 
 Vt 
(
9.23
(a) For I = 0.8 mA, we find
0.8 10 −3
J=
= 1.143 A/cm 2
7 10 − 4
We have
 J 

Va = Vt ln 

 JS 
or
J S (400 )
= 1.17 10 5
J S (300 )
Now
I = 7 10 −4 1.17 10 5 3 10 −12
(
)(
)(
 0.6907 
 exp 

 0.03453 
)
Then


0.8 10 −3
Va = (0.0259 ) ln 
−12
−16 
 4.8 10 + 6.4 10 
V a = 0.49032 V
(
)
(
)
 0.49032 
I SB = 4.8  10 −12 exp 

 0.0259 
 I SB = 0.7998 mA
 0.49032 
I pn = 6.4  10 −16 exp 

 0.0259 
 I pn  0.107  A
_______________________________________
)
or
I = 120 mA
For the Schottky diode,
 − e BO
J ST  T 2 exp 
 kT
Now
J ST (400 )  400 
=

J ST (300 )  300 



2


−  BO

 exp 
+ BO 
 (0.0259 )(400 300 ) 0.0259 
0.82 
 0.82
= 1.778 exp 
−

 0.0259 0.03453 
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 9
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(
or
J ST (400 )
= 4.856 10 3
J ST (300 )
Then
I = 7 10 −4 4.856 10 3 4 10 −8
(
)(
)(
)
 0.4447 
 exp 

 0.03453 
or
I = 53.3 mA
_______________________________________
9.24
Plot
_______________________________________
9.25
Rc 10 −4
=
= 0.1 
A 10 −3
R
10 −4
(b) R = c = − 4 = 1 
A 10
R
10 −4
(c) R = c = −5 = 10 
A 10
_______________________________________
(a) R =
9.26
Rc 5 10 −5
=
= 5
A
10 −5
(i) V = IR = (1)(5) = 5 mV
(ii) V = IR = (0.1)(5) = 0.5 mV
−5
5 10
= 50 
10 −6
(i) V = IR = (1)(50 ) = 50 mV
(ii) V = IR = (0.1)(50 ) = 5 mV
_______________________________________
(b) R =
9.27
 
Vt exp  Bn 
 Vt 
Rc =
 2
AT
 R A T 2 
or  Bn = Vt ln  c

 Vt

(
9.28
(b) We need  n =  m −  = 4.2 − 4.0 = 0.20 V
And
N 
 n = Vt ln  c 
 Nd 
or
 2.8 10 19 

0.20 = (0.0259 ) ln 

 Nd

which yields
N d = 1.24 10 16 cm −3
(c)
Barrier height = 0.20 V
_______________________________________
9.29
We have that
−eN d
(x n − x )
=
s
Then
eN 
x2 
 = − dx = d  x n  x −  + C 2
s 
2 
Let  = 0 at x = 0  C 2 = 0 , so

(a) R =

x2 
 xn  x −


2 

At x = x n ,  = V bi , so
=
eN d
s
 = Vbi =
)
eN d x n2

s
2
or
xn =
 5  10 −5 (120 )(300 )2 
(a)  Bn = (0.0259 ) ln 

0.0259


= 0.258 V
)
 5  10 −6 (120 )(300 )2 
(b)  Bn = (0.0259 ) ln 

0.0259


V
= 0.198
_______________________________________
2 s Vbi
eN d
Also
V bi =  BO −  n
where
N 
 n = Vt ln  c 
 Nd 
Now for

0.70
 = BO =
= 0.35 V
2
2
we have
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 9
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(1.6 10 )N x (50 10 )
(11.7)(8.85 10 )
(50 10 ) 
−
−19
0.35 =
−8
d
−14
n
−8 2
2
or
(
0.35 = 7.73 10 −14 N d x n − 25 10 −8
We have
(
)

)
 2(11 .7 ) 8.85 10 −14 Vbi 
xn = 

1.6  10 −19 N d


and
Vbi = 0.70 −  n
By trial and error, we find
N d = 3.5 10 18 cm −3
_______________________________________
(
)
1/ 2
9.34
Consider an n-P heterojunction in thermal
equilibrium. Poisson's equation is
d 2
 (x )
d
=−
=−
2

dx
dx
In the n-region,
d n  (x ) eN dn
=
=
dx
n
n
For uniform doping, we have
eN dn x
n =
+ C1
n
The boundary condition is
 n = 0 at x = − x n , so we obatin
C1 =
eN dn x n
n
Then
eN dn
(x + x n )
n
In the P-region,
d p
eN
= − aP
dx
P
which gives
eN x
 P = − aP + C 2
P
We have the boundary condition that
 P = 0 at x = x P , so that
n =
9.30
N 
(b)  BO =  p = Vt ln   
 Na 
 1.04 10 19
= (0.0259 ) ln 
16
 5 10




or
 BO = 0.138 V
_______________________________________
9.31
Sketches
_______________________________________
C2 =
eN aP x P
P
Then
9.32
Sketches
_______________________________________
9.33
Electron affinity rule
Ec = e  n −  p
(
)
For GaAs,  = 4.07 and for AlAs,  = 3.5 .
If we assume a linear extrapolation between
GaAs and AlAs, then for
Al 0.3 Ga 0 .7 As   = 3.90
Then
E c = 4.07 − 3.90 = 0.17 eV
_______________________________________
eN aP
(x P − x )
P
Assuming zero surface charge density at
x = 0 , the electric flux density D is
continuous, so n  n (0 ) = P  P (0 ) , which
yields
N dn x n = N aP x P
We can determine the electric potential as
P =
 n (x ) = −  n dx

 eN x 2 eN dn x n x 
= −  dn +
 + C3
n 
 2 n
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 9
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Now
Vbin =  n (0) −  n (− x n )

eN dn x n2 eN dn x n2 
= C 3 − C 3 −
+

2 n
n 

or
eN dn x n2
2 n
Similarly on the P-side, we find
eN aP x P2
VbiP =
2 P
We have that
eN dn x n2 eN aP x P2
Vbi = Vbin + VbiP =
+
2 n
2 P
We can write
N 
x P = x n  dn 
 N aP 
Substituting and collecting terms, we find
2
 e  N N + e n N dn
 2
Vbi =  P dn aP
  xn
2 n P N aP


Solving for x n , we have
Vbin =


2 n P N aPVbi
xn = 

(
)
eN

N
+

N
n
dn 
 dn P aP
Similarly on the P-side, we have
1/ 2
1/ 2


2 n  P N dnVbi
xP = 

 eN aP ( P N aP + n N dn ) 
The total space charge width is then
W = xn + xP
Substituting and collecting terms, we obtain
 2 n  P Vbi (N aP + N dn ) 
W =

 eN dn N aP ( P N aP + n N dn ) 
_______________________________________
1/ 2
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 10
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
Chapter 10
Exercise Solutions
Ex 10.1
N 
 2 10 15 

 fp = Vt ln  a  = (0.0259 ) ln 
10 
 ni 
 1.5 10 
= 0.3056 V
1/ 2
 4 s  fp 
x dT = 
 eN a
(


)
 4(11 .7 ) 8.85  10 (0.3056 ) 
=

1.6  10 −19 2  10 15


(
−14
)(
Ex 10.4
From Figure 10.16,  ms  +0.28 V
We find
N 
 2 10 16
 fp = Vt ln  a  = (0.0259 ) ln 
10
 ni 
 1.5 10
= 0.3653 V
 4 s  fp 
x dT = 

 eN a 
1/ 2
)
(
x dT = 6.29 10 cm
or x dT = 0.629  m
_______________________________________
)
= 8.629  10 −7 F/cm 2
Then
V FB =  ms −
Qss
C ox
= −1.03 −
(2 10 )(1.6 10 )
10
)(
1/ 2
)
(
)(
)(
= 1.6 10 −19 2 10 16 2.1744 10 −5
= 6.958  10
−8
)
C/cm 2
Then
=
t 
 (max ) − Qss ) ox  +  ms + 2 fp
VTN = ( QSD
 ox 
−8
−19
6.958  10 − 1.6  10
2  10 10 80  10 −8
(
) (
)(
(3.9)(8.85 10
−14
)
)(
)
+ 0.28 + 2(0.3653 )
VTN = 0.1539 + 0.28 + 2(0.3653 )
= 1.16 V
_______________________________________
Ex 10.3
From Figure 10.16,  ms  −1.03 V
(
)
= 2.174  10 cm
 (max ) = eN a x dT
QSD
16
ox (3.9) 8.85 10 −14
=
t ox
40  10 −8
(
−5
Ex 10.2
C ox =
1/ 2
 4(11 .7 ) 8.85  10 −14 (0.3653 ) 
=

1.6  10 −19 2  10 16


−5
N 
 10


 fp = Vt ln  a  = (0.0259 ) ln 
10 
 ni 
 1.5 10 
= 0.347 V

Eg


 ms =  m −    +
+  fp 
2e



= 3.20 − (3.25 + 0.56 + 0.347 )
 ms = −0.957 V
_______________________________________




Ex 10.5
From Figure 10.16,  ms  1.06 V
We find
N 
 2 10 16
 fn = Vt ln  d  = (0.0259 ) ln 
10
 ni 
 1.5 10
= 0.3653 V
−19
8.629 10 −7
V FB = −1.034 V
_______________________________________
x dT
 4 s  fn 
=

 eN d 
1/ 2
(
)
 4(11 .7 ) 8.85  10 −14 (0.3653 ) 
=

1.6  10 −19 2  10 16


(
= 2.174  10 −5 cm
)(
)
1/ 2




Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 10
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
 (max ) = eN d  fn
QSD
(
= 1.6 10
−19
We find

C FB
2.174  10 −7
=
= 0.504
C ox
4.314  10 − 7
_______________________________________
)(2 10 )(2.1744 10 )
−5
16
= 6.958  10 −8 C/cm 2
Now
t 
 (max ) − Qss   ox  +  ms − 2 fn
VTP = − QSD
 ox 
− 6.958  10 −8 − 5  1010 1.6  10 −19 200  10 −8
=
(3.9) 8.85  10 −14
+ 1.06 − 2(0.3653 )
VTP = −0.4495 + 1.06 − 2(0.3653 )
or VTP = −0.12 V
_______________________________________


(
) (
Ex 10.6
)(
(
)
(
ox (3.9) 8.85 10 −14
=
t ox
80 10 −8
C ox =
)(
)
= 4.314  10 −7 F/cm 2
 3 10 16
10
 1.5 10
 fp = (0.0259 ) ln 
x dT
(
)

 = 0.3758 V


 4(11 .7 ) 8.85  10 14 (0.3758 ) 
=

1.6  10 −19 3  10 16


(
)(
C m in =
=

  x dT


80 10
−8
)
2
A/V
or k n = 0.2804 mA/V 2
Now
k W
2
I D = n  (VGS − VT )
2 L
0.2804
(12 )(VGS − 0.4)2
=
2
2
(a) I D = (1.6826 )(0.8 − 0.4) = 0.269 mA
(
ox (3.9) 8.85 10 −14
=
t ox
80 10 −8
)
= 4.314  10 −7 F/cm 2
W n C ox
(VGS − VT )2
I D (sat ) =
2L
Then
)
I D 2 (sat ) − I D1 (sat ) =
W n C ox
(VGS 2 − VGS1 )
2L
0.295  10 −3 − 0.132  10 −3
ox
  V 
t ox +  ox  t s
 s  eN a
(3.9) 8.85 10 −14
 3.9 
+

 11 .7 
= 2.804  10
−4
= 1.7176  10 −2 − 1.1489  10 −2 = 5.687  10 −3
Then
(
=
(
C ox =
= 5.076  10 −8 F/cm 2
We find
C m in 5.076  10 −8
=
= 0.1177
C ox
4.314  10 − 7
 =
C FB
Then
k n =  n C ox = (650 ) 4.314 10 −7
2
(3.9)(8.85 10 −14 )
Now
= 4.314  10 −7 F/cm 2
Ex 10.8
We find
(
)
(c) I D = (1.6826 )(1.6 − 0.4) = 2.423 mA
_______________________________________
)
 3 .9 
−5
80  10 −8 + 
 1.80  10
 11 .7 
(
ox (3.9) 8.85 10 −14
=
t ox
80 10 −8
2
ox

t ox +  ox
 s
C ox =
(b) I D = (1.6826 )(1.2 − 0.4) = 1.077 mA
1/ 2
= 1.80  10 −5 cm
Now
)
Ex 10.7
We find
)
Or
(0.0259 )(11.7)(8.85 10
−14
(1.6 10 )(3 10 )
 = 2.174 10 −7 F/cm 2
C FB
(6) n (4.314 10 −7 )
(1.25 − 1.0)
2(1.5)
5.687  10 −3 =
−19
16
)
 5.687  10 −3


0.25

2

 = 8.628  10 − 7   n


  n  600 cm 2 /V-s
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 10
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
We now find
(6)(600 )(4.314 10 −7 ) (1.0 − V )2
0.132 10 −3 =
T
2(1.5)
1/ 2
 0.132  10 −3 


 5.1768  10 − 4  = 1.0 − VT


VT = 0.495 V
_______________________________________
Test Your Understanding Solutions
TYU 10.1
N 
(a)  fn = Vt ln  d 
 ni 
 8 10 15
= (0.0259 ) ln 
10
 1.5 10
 4 s  fn 
x dT = 

 eN d 
Ex 10.9
N
(a)  fp = Vt ln  a
 ni

 10 16
 = (0.0259 ) ln 

 1.5 10 10


= 0.3473 V

(3.9) 8.85 10 −14
= ox =
t ox
120  10 −8
(
C ox
= 2.876  10
−7
F/cm
C ox
(
(
)( )
2.876 10 −7
(b) VT = 
(i) VT
 2
(
fp
+ VSB − 2 fp


= (0.200 )1.3018 − 0.8334 

= (0.200 ) 2(0.3473 ) + 2 − 2(0.3473 )
= (0.200 )1.6415 − 0.8334 
=
 n (VGS − VT )
2 L2
(420 )(1.5 − 0.4)
(
2 1.2 10
 f T = 5.11 GHz
)
−4 2
(
)

 = 0.3832 V


 4(11 .7 ) 8.85  10 −14 (0.3832 ) 
=

1.6  10 −19 4  10 16


(
)(
1/ 2
)
= 1.575  10 cm
or x dT = 0.1575  m
_______________________________________
N 
or

 3 10 16 

 fp = Vt ln  a  = (0.0259 ) ln 
10 
 ni 
 1.5 10 

VT = 0.162 V
_______________________________________
fT =
)
−5
VT = 0.0937 V
Ex 10.10
1/ 2
TYU 10.2
= (0.200 ) 2(0.3473 ) + 1 − 2(0.3473 )
(ii) VT
)(
 4 10 16
(b)  fn = (0.0259 ) ln 
10
 1.5 10
2
 = 0.200 V 1 / 2
)
= 3.324  10 cm
or x dT = 0.3324  m
)
2 1.6 10 −19 (11.7 ) 8.85 10 −14 10 16
=
(
 4(11 .7 ) 8.85  10 −14 (0.3415 ) 
=

1.6  10 −19 8  10 15


x dT
)
1/ 2
−5
2e s N a
=





 = 0.3415 V


 fp = 0.376 V
We have
 Eg

 ms = −
+  fp  = −(0.560 + 0.376 )
 2e

or
 ms = −0.936 V
_______________________________________
TYU 10.3
From TYU 10.2,
 fp = 0.376 V
= 5.11  10 9 Hz
_______________________________________
We have
 Eg

 ms = 
−  fp  = 0.560 − 0.376
 2e

or
 ms = +0.184 V
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 10
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
TYU10.4
V FB =  ms
TYU 10.6
k n W
2
 (VGS − VT )
2 L
From Ex 10.7, k n = 0.2804 mA/V 2
Then
 0.2804  W 
2
0.100 = 
 (1.0 − 0.4)
 2  L 
W

= 1.98
L
_______________________________________
Q
− ss
C ox
ID =
 3 10 16 
 = 0.3758 V
10 
 1.5 10 
From Equation (10.17)
Eg
1.12
ms =
−  fp =
− 0.3758
2e
2
= 0.1842 V

(3.9) 8.85 10 −14
C ox = ox =
t ox
160  10 −8
 fp = (0.0259 ) ln 
(
)
TYU 10.7
= 2.157  10 −7 F/cm 2
V FB = 0.1842 −
C ox =
(8 10 )(1.6 10 )
−19
10
= 0.125 V
_______________________________________
(

 = 0.3758 V


)
 4(11 .7 ) 8.85  10 (0.3758 ) 
x dT = 

1.6  10 −19 3  10 16


(
−14
)(
(
Then
V SG = 1 V
)
(
)(
)(
= 1.6 10 −19 3 10 16 1.80 10 −5
−8
)
= 8.644  10 C/cm
From Figure 10.16,  ms  −1.13 V
+ 0.65 =
8.644 10
(
−19

 +  ms + 2 fp


10
)
+ 0.65 = 2.2713 10 5 t ox − 1.13 + 0.7516
t ox = 4.52 10 −6 cm
o
)
2
 I D = 0.525 mA
V SG = 2 V  I D = 3.74 mA
_______________________________________
TYU 10.8
− (1.6 10 )(5 10 )(t ox )
(3.9)(8.85 10 −14 )
+ (− 1.13 ) + 2(0.3758 )
−8
)
V SG = 1.5 V  I D = 1.77 mA
2
t
 (max ) − Qss ) ox
VTN = ( QSD
 ox
  p C ox  W 
 (V SG + VT )2
I D = 
 L 
2


 (310 ) 1.569 10 −7 
2
=
  (60 )  (V SG − 0.4)
2


= 1.459 10 −3 (V SG − 0.4) A
1/ 2
−5
= 1.80  10 cm
 (max ) = eN a x dT
QSD
Now
(
TYU 10.5
 fp
)
= 1.569  10 −7 F/cm 2
2.157 10 −7
 3 10 16
= (0.0259 ) ln 
10
 1.5 10
(
ox (3.9) 8.85 10 −14
=
t ox
220 10 −8
or t ox = 45 .2 nm = 452 A
_______________________________________
  p C ox  W 
 (V SG + VT )2
I D = 
 L 
2



(310 ) 1.569 10 −7    W 
200 10 −6 = 
  
2

 L
(
)
 (1.25 − 0.4)
2
W 
   = 11 .4
L
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 10
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
TYU 10.9
(a) C ox =
(
ox (3.9) 8.85 10 −14
=
t ox
120  10 −8
)
= 2.876  10 −7 F/cm 2
=
=
2e s N a
(
C ox
)
(
)( )
2 1.6 10 −19 (11.7 ) 8.85 10 −14 10 15
2.876 10
 = 0.0633 V 1 / 2
−7
N 
 10 15
(b)  fp = Vt ln  a  = (0.0259 ) ln 
10
 ni 
 1.5 10
= 0.2877 V
VT = 
 2
(i) VT
fp
+ VSB − 2 fp






= (0.0633 ) 2(0.2877 ) + 1 − 2(0.2877 )
= (0.0633 )1.2551 − 0.7586 

VT = 0.0314 V
(ii) VT

= (0.0633 ) 2(0.2877 ) + 2 − 2(0.2877 )
= (0.0633 )1.6048 − 0.7586 

VT = 0.0536 V
_______________________________________
TYU 10.10
CM
= 1 + g m RL
C gdT
We find
C ox =
(
ox (3.9) 8.85 10 −14
=
t ox
180  10 −8
)
= 1.9175  10 −7 F/cm 2
W 
g m =    n C ox (VGS − VT )
L
(
)
 24 
−7
=
(420 ) 1.9175  10 (1.5 − 0.4)
 1.2 
= 1.772  10 −3 A/V
Then
CM
= 1 + 1.772  10 −3 100  10 3
C gdT
(
)(
)
= 178
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 10
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Chapter 10
10.1
(a) p-type; inversion
(b) p-type; depletion
(c) p-type; accumulation
(d) n-type; inversion
_______________________________________
10.2
N 
(a) (i)  fp = Vt ln  a 
 ni 
 7 10 15
= (0.0259 ) ln 
10
 1.5 10
= 0.3381 V
 4 s  fp 
x dT = 

 eN a 
)
)(
(
 3 10 16 

(ii)  fp = (0.03022 ) ln 
11 
 1.93 10 
= 0.3613 V
(
)
)(
)
= 1.77  10 cm
or x dT = 0.177  m
_______________________________________
1/ 2
)
10.3
(a)
 4 s  fn 
= eN d 

 eN d 
 (max ) = eN d x dT
Q SD

(




)
(
= 1.80  10 cm
or x dT = 0.180  m
 350 
(b) kT = (0.0259 )
 = 0.03022 V
 300 
 − Eg 

ni2 = N c N  exp 

 kT 
 350 
= 2.8 10 19 1.04 10 19 

 300 
so ni = 1.93 10 11 cm −3
 7 10 15 

(i)  fp = (0.03022 ) ln 
11 
 1.93 10 
= 0.3173 V
(1.25 10 )
= (1.6 10 )(N
−19
d
)(4)(11.7)(8.85 10 −14 )(0.30 )
 N d = 7.86 10 14 cm −3
2nd approximation:
 7.86 10 14
 fn = (0.0259 ) ln 
10
 1.5 10
Then

 = 0.2814 V


(1.25 10 )
= (1.6 10 )(N )(4)(11.7)(8.85 10 )(0.2814 )
−8 2
−19
−14
d
3
 − 1.12 
 exp 

 0.03022 
= 3.71  10 22
1/ 2
−8 2
1/ 2
)
)
)
Then
−5
)(
1/ 2
1st approximation: Let  fn = 0.30 V
 4(11 .7 ) 8.85  10 −14 (0.3758 ) 
x dT = 

1.6  10 −19 3  10 16


)(
1/ 2
−5
= (eN d ) 4 s  fn
 3 10 16
(ii)  fp = (0.0259 ) ln 
10
 1.5 10
= 0.3758 V
(
)
= 3.43  10 cm
or x dT = 0.343  m
= 3.54  10 −5 cm
or x dT = 0.354  m
(
)(
1/ 2
−5
(




 4(11 .7 ) 8.85  10 −14 (0.3381 ) 
=

1.6  10 −19 7  10 15


(
)
 4(11 .7 ) 8.85  10 −14 (0.3613 ) 
x dT = 

1.6  10 −19 3  10 16


1/ 2
(
(
 4(11 .7 ) 8.85  10 −14 (0.3173 ) 
x dT = 

1.6  10 −19 7  10 15


 N d = 8.38 10 14 cm −3
 8.38 10 14 
 = 0.2831 V
(b)  fn = (0.0259 ) ln 
10 
 1.5 10 
 s = 2 fn = 2(0.2831 ) = 0.566 V
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 10
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
10.4
p-type silicon
(a) Aluminum gate

Eg


 ms =  m −    +
+  fp 
2e



We have
N 
 fp = Vt ln  a 
 ni 
 6 10
= (0.0259 ) ln 
10
 1.5 10
15

 = 0.334 V


Then
 ms = 3.20 − (3.25 + 0.56 + 0.334 )
or
 ms = −0.944 V
(b) n + polysilicon gate
 Eg

 ms = −
+  fp  = −(0.56 + 0.334 )
 2e

or
 ms = −0.894 V
(c) p + polysilicon gate
 Eg

 ms = 
−  fp  = (0.56 − 0.334 )
 2e

or
 ms = +0.226 V
_______________________________________
10.5
 4 10 16 
 = 0.3832 V
 fp = (0.0259 ) ln 
10 
 1.5 10 
Eg


 ms =  m −    +
+  fp 
2e


= 3.20 − (3.25 + 0.56 + 0.3832 )
 ms = −0.9932 V
_______________________________________
10.6
(a) N d  210 17 cm −3
(b) Not possible -  ms is always positive.
(c) N d  210 15 cm −3
_______________________________________
10.7
From Problem 10.5,  ms = −0.9932 V
Q
V FB =  ms − ss
C ox
(a) C ox =
(
ox (3.9) 8.85 10 −14
=
t ox
200 10 −8
)
= 1.726  10 −7 F/cm 2
V FB = −0.9932 −
(5 10 )(1.6 10 )
−19
10
1.726 10 −7
= −1.040 V
(3.9) 8.85 10 −14
(b) C ox =
80 10 −8
= 4.314  10 −7 F/cm 2
5 10 10 1.6 10 −19
V FB = −0.9932 −
4.314 10 −7
= −1.012 V
_______________________________________
(
)
(
)(
)
10.8
(a)  ms  −0.42 V
V FB =  ms = −0.42 V
(b)
(3.9) 8.85 10 −14 = 1.726 10 −7 F/cm 2
C ox =
200 10 −8
Q
4  10 10 1.6  10 −19
(i) V FB = − ss = −
C ox
1.726  10 − 7
= −0.0371 V
10 11 1.6 10 −19
(ii) V FB = −
1.726 10 −7
= −0.0927 V
(c) V FB =  ms = −0.42 V
(
)
(
( )(
C ox =
)(
)
)
(3.9)(8.85 10 −14 ) = 2.876 10 −7 F/cm 2
120 10 −8
4 10 10 1.6 10 −19
(i) V FB = −
2.876 10 −7
= −0.0223 V
10 11 1.6 10 −19
(ii) V FB = −
2.876 10 −7
= −0.0556 V
_______________________________________
(
( )(
)(
)
)
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 10
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
10.9
 ms
VTN =
Eg


=  m −    +
+  fp 
2e


 2 10 16
10
 1.5 10
(
10.11
x dT
)
)

 = 0.3653 V


 4(11 .7 ) 8.85  10 −14 (0.3653 ) 
x dT = 

1.6  10 −19 2  10 16


)(
)
)(
)(
)(
= 1.6 10 −19 2 10 16 2.174 10 −5
C/cm
= 2.507  10
−8
)
2
C/cm
(3.9) 8.85 10 −14 = 2.301 10 −7 F/cm 2
C ox =
150 10 −8
 (max ) + Q ss 
 Q SD
VTP = − 
 +  ms − 2 fn
C ox


(
)
(
1/ 2
)
= 2.174  10 cm

(
QSD max ) = eN a x dT
)
2
(3.9)(8.85 10 −14 ) = 2.301 10 −7 F/cm 2
150 10 −8
)(
1/ 2
)(
)
 2.507 10 −8 + 1.6 10 −19 7 10 10 
= −

2.301 10 −7


+  ms − 2(0.3161 )
VTP = −0.7898 +  ms
−5
C ox =
(
(
 2 10 16
10
 1.5 10
= 6.958  10
)
 4(11 .7 ) 8.85  10 −14 (0.3161 ) 
=

1.6  10 −19 3  10 15


= 5.223  10 cm
 (max ) = eN d x dT
QSD
10.10
−8
(

 = 0.3161 V


−5
Q ss
= 1.2  10 10 cm −2
e
_______________________________________
(
 3 10 15
10
 1.5 10
 fn = (0.0259 ) ln 
= 1.6 10 −19 3 10 15 5.223 10 −5
 fp = (0.0259 ) ln 
)
(c) Al gate on p-type:  ms  −0.95 V
VTN = 0.9843 − 0.95 = +0.0343 V
_______________________________________
or
)(
)(
(b) p + poly gate on p-type:  ms  +0.28 V
VTN = 0.9843 + 0.28 = +1.26 V
= 1.92  10 −9 C/cm 2
(
(
(a) n + poly gate on p-type:  ms  −1.12 V
VTN = 0.9843 − 1.12 = −0.136 V
)
(
+  ms + 2 fp
6.958 10 −8 − 7 10 10 1.6 10 −19
2.301 10 −7
+  ms + 2(0.3653 )
= 0.9843 +  ms

 = 0.365 V


Then
 ms = 3.20 − (3.25 + 0.56 + 0.365 )
or
 ms = −0.975 V
Now
Q
V FB =  ms − ss
C ox
or
Q ss = ( ms − V FB )C ox
We have

(3.9) 8.85 10 −14
C ox = ox =
t ox
450 10 −8
or
C ox = 7.67 10 −8 F/cm 2
So now
Qss = − 0.975 − (− 1) 7.67 10 −8
(
C ox
=
where
 fp = (0.0259 ) ln 
 (max ) − Qss
QSD
(a) n + poly gate on n-type:  ms  −0.41 V
VTP = −0.7898 − 0.41 = −1.20 V
(b) p + poly gate on n-type:  ms  +1.0 V
VTP = −0.7898 + 1.0 = +0.210 V
(c) Al gate on n-type:  ms  −0.29 V
VTP = −0.7898 − 0.29 = −1.08 V
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 10
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
10.12
Now
 5 10 15 
 = 0.3294 V
 fp = (0.0259 ) ln 
10 
 1.5 10 
The surface potential is
 s = 2 fp = 2(0.3294 ) = 0.659 V
We have
V FB =  ms
Now
VT =
 4 10 16
10
 1.5 10
 fp = (0.0259 ) ln 
= 0.3832 V
x dT
Q
− ss = −0.90 V
C ox
 (max )
QSD
C ox
(
)(
(
)
)(
)
We also find

(3.9) 8.85 10 −14
C ox = ox =
t ox
400 10 −8
or
C ox = 8.629 10 −8 F/cm 2
Then
3.304 10 −8
VT =
+ 0.659 − 0.90
8.629 10 −8
or
VT = +0.142 V
_______________________________________
10.13
C ox =
)
(
ox (3.9) 8.85 10 −14
=
t ox
220 10 −8
)
= 1.569  10 −7 F/cm 2
(
Qss = 1.6 10
−19
)(
)(4 10 )
10
= 6.4  10 −9 C/cm 2
By trial and error, let N a = 410 16 cm −3 .
)
2
1.008 10 −7 − 6.4 10 −9
1.569 10 −7
− 0.94 + 2(0.3832 )
Then VTN = 0.428 V  0.45 V
_______________________________________
=
 (max ) = 3.304  10 −8 C/cm 2
Q SD
(
)
= 1.008  10 C/cm
 ms  −0.94 V
Then
 (max ) − Qss
QSD
VTN =
+  ms + 2 fp
C ox
1/ 2
x dT = 0.413 10 −4 cm
Then
 (max ) = 1.6  10 −19 5  10 15 0.413  10 −4
Q SD
or
)(
−7
)
)(
)(
1/ 2
= 1.6 10 −19 4 10 16 1.575 10 −5
or
(
(
= 1.575  10 cm
 (max )
QSD
 4(11 .7 ) 8.85  10 −14 (0.3294 ) 
=

1.6  10 −19 5  10 15


(
)
 4(11 .7 ) 8.85  10 −14 (0.3832 ) 
=

1.6  10 −19 4  10 16


−5
+  s + V FB
We obtain
1/ 2
 4 s  fp 
x dT = 

 eN a 
(




10.14
C ox =
(
ox (3.9) 8.85 10 −14
=
t ox
180  10 −8
)
= 1.9175  10 −7 F/cm −3
(
)(
Qss = 1.6 10 −19 4 10 10
−9
= 6.4  10 C/cm
)
2
By trial and error, let N d = 510 16 cm −3
Now
 5 10 16 

 fn = (0.0259 ) ln 
10 
 1.5 10 
= 0.3890 V
(
)
 4(11 .7 ) 8.85  10 −14 (0.3890 ) 
x dT = 

1.6  10 −19 5  10 16


(
)(
1/ 2
)
−5
= 1.419  10 cm

QSD (max )
(
)(
)(
= 1.6 10 −19 5 10 16 1.419 10 −5
−7
= 1.135  10 C/cm
 ms  +1.10 V
Then
−3
)
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 10
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
VTP = −
( Q (max ) + Q )
SD
ss
C ox
(1.135 10
Now
+  ms − 2 fn
V FB =  ms −
)
−7
+ 6.4 10 −9
1.9175 10 −7
+ 1.10 − 2(0.3890 )
Then VTP = −0.303 V, which is within the
specified value.
_______________________________________
=−
10.15
We have C ox = 1.569 10 −7 F/cm 2
= −1.03 −
x dT
)
(
)(
= 1.182  10 −4 cm
 (max )
QSD
(
1.9175 10
(




)
(
)( )(
= 1.6 10 −19 10 15 8.630 10 −5
= 1.381  10
Now
1/ 2
)
)( )
= 8.630  10 −5 cm
 (max )
QSD
(
VTN =
−8
C/cm
 (max )
QSD
C ox
1/ 2
)
2
+ V FB + 2 fp
1.381 10 −8
− 1.08 + 2(0.2877 )
1.9175 10 −7
or VTN = −0.433 V
_______________________________________
=
)(
)(
)
= 9.456  10 C/cm
 ms  −0.33 V
Then
( QSD (max ) + Qss )
VTP = −
+  ms − 2 fn
C ox
2
 9.456 10 −9 + 6.4 10 −9 

= −

1.569 10 −7


− 0.33 − 2(0.2697 )
= 0.970 V
Then VTP = −0.970 V  −0.975 V which
meets the specification.
_______________________________________
10.16
(a)  ms  −1.03 V
C ox =
10
−7
 4(11 .7 ) 8.85  10 −14 (0.2877 ) 
x dT = 

1.6  10 −19 10 15


= 1.6 10 −19 5 10 14 1.182 10 −4
−9
−19
 10 15
(b)  fp = (0.0259 ) ln 
10
 1.5 10
= 0.2877 V
By trial and error, let N d = 510 14 cm −3
Now
 5 10 14 

 fn = (0.0259 ) ln 
10 
 1.5 10 
= 0.2697 V
(
(1.6 10 )(6 10 )
V FB = −1.08 V
Qss = 6.4 10 −9 C/cm 2
 4(11 .7 ) 8.85  10 −14 (0.2697 ) 
=

1.6  10 −19 5  10 14


Qss
C ox
(3.9)(8.85 10 −14 )
180 10 −8
= 1.9175  10 −7 F/cm 2
10.17
(a) We have n-type material under the gate, so
1/ 2
 4 s  fn 
x dT = t C = 

 eN d 
where
 10 15 
 = 0.288 V
 fn = (0.0259 ) ln 
10 
 1.5 10 
Then
(
)
 4(11 .7 ) 8.85  10 −14 (0.288 ) 
x dT = 

1.6  10 −19 10 15


(
1/ 2
)( )
or
x dT = t C = 0.863 10 −4 cm = 0.863  m
(b)
t 
 (max ) + Qss ) ox  +  ms − 2 fn
VT = −( QSD
 ox 
For an n + polysilicon gate,
Eg

 ms = −
−  fn  = −(0.56 − 0.288 )
 2e



or
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 10
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
 ms = −0.272 V
Now
 (max ) = (1.6  10 −19 )(10 15 )(0.863  10 −4 )
Q SD
or
 (max ) = 1.38  10 −8 C/cm 2
Q SD
We have
Qss = 1.6 10 −19 10 10 = 1.6 10 −9 C/cm 2
We now find
− 1.38 10 −8 + 1.6 10 −9
VT =
500 10 −8
(3.9) 8.85 10 −14
− 0.272 − 2(0.288 )
or
VT = −1.07 V
_______________________________________
(
)( )
(
(
)(
)
)
10.18
Eg


(b)  ms =  m −    +
+  fp 
2e


where
 m −   = −0.20 V
and
 10 16 
 = 0.3473 V
 fp = (0.0259 ) ln 
10 
 1.5 10 
Then
 ms = −0.20 − (0.56 + 0.3473 )
or
 ms = −1.107 V
(c) For Q ss = 0
t
 (max )  ox
VTN = QSD
 ox
We find
x dT
(
)
(
or
)(
 (max ) = 4.797  10 −8 C/cm 2
Q SD
Then
−14
10.19
Plot
_______________________________________
10.20
Plot
_______________________________________
10.21
Plot
_______________________________________
C ox =
1/ 2
)( )
)(
−8
10.23
(a) For f = 1 Hz (low freq),
or
(
−8
− 1.107 + 2(0.3473 )
or
VT = +0.00455 V  0 V
_______________________________________
=
x dT = 0.30 10 −4 cm = 0.30  m
Now
 (max ) = 1.6  10 −19 10 16 0.30  10 −4
Q SD
(4.797 10 )(300 10 )
(3.9)(8.85 10 )
10.22
Plot
_______________________________________

 +  ms + 2 fp


 4(11 .7 ) 8.85  10 −14 (0.3473 ) 
=

1.6  10 −19 10 16


VT =
)
(
ox (3.9) 8.85 10 −14
=
t ox
120  10 −8
= 2.876  10 −7 F/cm 2
ox
 =
C FB
  V 
t ox +  ox  t s
 s  eN a
(3.9) 8.85 10 −14
(
 3.9 
120 10 −8 + 

 11.7 
)
(0.0259 )(11.7)(8.85 10 −14 )
(1.6 10 )(10 )
−19
 = 1.346 10 −7 F/cm 2
C FB
ox
C m in =
 
t ox +  ox   x dT
 s 
Now
 10 16
 fp = (0.0259 ) ln 
10
 1.5 10
x dT
)
(
)

 = 0.3473 V


 4(11 .7 ) 8.85  10 −14 (0.3473 ) 
=

1.6  10 −19 10 16


(
−5
= 3.00  10 cm
16
)( )
1/ 2
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 10
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Then
C m in =
(3.9)(8.85 10
−14
)
x dT
(
 3 .9 
−5
120  10 −8 + 
 3.00  10
11
.
7


)
C m in =
C ox = 2.876 10 −7 F/cm 2 (unchanged)
 = 1.346 10 F/cm
C FB
(unchanged)
 = 3.083 10 −8 F/cm 2 (unchanged)
C min
 (max )
QSD
C ox
C/cm
)
2
4.80 10
VTN =
− 1.10 + 2(0.3473 )
2.876 10 −7
VTN = −0.2385 V
_______________________________________
=
(
ox (3.9) 8.85 10 −14
=
t ox
120  10 −8
= 2.876  10 −7 F/cm 2
ox
 =
C FB
  V 
t ox +  ox  t s
 s  eN a
(3.9) 8.85 10 −14
(
 3.9 
120 10 −8 + 

 11.7 
)
= 8.504  10 −9 F/cm 2
C  (inv) = C ox = 2.876 10 −7 F/cm 2
(b) f = 1 MHz (high freq),
 (max )
Q SD
C ox
+ V FB − 2 fn
Now
 (max ) = eN d x dT
QSD
(
)(
)(
= 1.6 10 −19 5 10 14 1.182 10 −4
= 9.456  10
−9
C/cm
)
2
Then
9.456 10 −9
+ 0.95 − 2(0.2697 )
2.876 10 −7
VTP = +0.378 V
_______________________________________
VTP = −
f = 1 Hz (low freq),
C ox =
(
 3 .9 
−4
120  10 −8 + 
 1.182  10
 11 .7 
VTP = −
−8
10.24
(a)
(3.9)(8.85 10 −14 )
 = 8.504 10 −9 F/cm 2
C  (inv) = C min
(c) V FB =  ms  0.95 V
)( )(
= 4.80  10
cm
 = 8.504 10 −9 F/cm 2 (unchanged)
C min
= 1.6 10 −19 10 16 3.00 10 −5
−8
)
 = 4.726 10 −8 F/cm 2 (unchanged)
C FB
+ V FB + 2 fp
Now
 (max ) = eN a x dT
QSD
(
−4
1/ 2
C ox = 2.876 10 −7 F/cm 2 (unchanged)
 = 3.083 10 −8 F/cm 2
C  (inv) = C min
(c) V FB =  ms = −1.10 V
VTN =
)
)(
= 1.182  10
C  (inv) = C ox = 2.876 10 −7 F/cm 2
(b) f = 1 MHz (high freq),
2
(
Then
= 3.083  10 −8 F/cm 2
−7
(
 4(11 .7 ) 8.85  10 −14 (0.2697 ) 
=

1.6  10 −19 5  10 14


)
)
(0.0259 )(11.7)(8.85 10 −14 )
(1.6 10 )(5 10 )
 = 4.726 10 −8 F/cm 2
C FB
ox
C m in =
 
t ox +  ox   x dT
 s 
Now
 5 10 14
 fn = (0.0259 ) ln 
10
 1.5 10
−19
14
10.25
The amount of fixed oxide charge at x is
 ( x )x C/cm 2
By lever action, the effect of this oxide charge
on the flatband voltage is
1  x 
   (x )x
V FB = −
C ox  t ox 
If we add the effect at each point, we must
integrate so that
V FB
1
=−
C ox
to x

0
x (x )
dx
t ox
_______________________________________

 = 0.2697 V


Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 10
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
10.26
(a) We have  ( x ) =
Then
VFB = −
−
=−
1
Cox
1
Cox
1
QSS
2
t
0
x( x )
dx
t ox
t ox
t ox − t
t ox
QSS
t ox
t
VFB = −
SS
SS
ox
ox
ox
or
VFB
ox
0
−8
10
−14
10
−8
t ox
 x  O
xI
F
G
Ht J
Kdx
ox
1
Cox
O

af z
x
t ox
O
t ox
3
t ox
2
x dx
2
0
 O t ox
 I af
3
3
F
t
G
J
Ht K
−b
1.28  10 g
b200  10 g
=
3(3.9)b
8.85  10 g
VFB = −
ox
SS
ox
1
which becomes
1
gb8  10 g
200  10
=−
ox
−19
Cox
−19
−2
z
t ox
1
dx
FQ I t − at − t f = − Q
C H
t K
C
Ft I
= −Q G J
H K
−b
1.6  10 g
b8  10 gb200  10 g
=
(3.9)b
8.85  10 g
1
t ox  O = QSS   O =
or  O = 1.28  10
Now
z
F
IJF I
G
z
HK
b gHK
t ox
b
2 1.6  10

2

=−
0
ox
2
ox
ox
Then
−2
VFB
−8
2
−14
or
VFB = −0.0742 V
or V FB = −0.0494 V
_______________________________________
(b)
We have
10.27
Sketch
_______________________________________
( x ) =
QSS
t ox
b1.6  10 gb8  10 g
−19
=
10
200  10
−8
10.28
Sketch
_______________________________________
= 6.4  10 =  O
−3
Now
VFB = −
1
Cox
z
t ox
0
x( x )

dx = − O
t ox
Cox t ox
or
 O t ox
2
VFB = −
=
2 ox
b
gb200  10 g
2(3.9)b
8.85  10 g
− 6.4  10
−3
−14
or
VFB = −0.0371 V
(c)
Fx I
( x ) =  GJ
Ht K
O
ox
We find
−8
2
z
t ox
xdx
0
10.29
(b)
N N
V FB = −Vbi = −Vt ln  a 2 d
 ni




( )( )
(
)
 10 16 10 16 
= −(0.0259 ) ln 

2
 1.5 10 10 
or
V FB = −0.695 V
(c) Apply V G = −3 V, Vox  3 V
For V G = +3 V,

d
=−
dx
s
n-side:  = eN d
eN
eN x
d
= − d   = − d + C1
dx
s
s
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 10
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
 = 0 at x = − x n , then C1 = −
eN d x n
s
so
eN d
(x + x n ) for − x n  x  0
s
In the oxide,  = 0 , so
=−
d
= 0   = constant. From the
dx
boundary conditions, in the oxide
eN x
=− d n
s
In the p-region,
eN
eN a x

d
=−
=+ a =
+ C2
dx
s
s
s
 = 0 at x = (t ox + x p ) , then
(
) 
eN a
t ox + x p − x
s
eN a x p
eN x
=− d n
At x = t ox ,  = −
s
s
=−
So that N a x p = N d x n
Since N a = N d , then x n = x p
The potential is

 = − dx
For zero bias, we can write
Vn + Vox + V p = Vbi
where Vn , Vox , V p are the voltage drops across
the n-region, the oxide, and the p-region,
respectively. For the oxide:
eN d x n t ox
Vox =   t ox =
s
For the n-region:

eN d  x 2

V n (x ) =
+ x n  x  + C 

s  2

Arbitrarily, set V n = 0 at x = − x n , then
2
n
eN d x
so that
2 s
eN d
( x + x n )2
V n (x ) =
2 s
C =
eN d x n2
At x = 0 , Vn =
which is the voltage
2 s
drop across the n-region. Because of
symmetry, Vn = V p . Then for zero bias, we
have
2V n + V ox = V bi
which can be written as
eN d x n2 eN d x n t ox
+
= Vbi
s
s
or
V 
x n2 + x n t ox − bi s = 0
eN d
Solving for x n , we obtain
2
t ox
t   V
+  ox  + s bi
2
eN d
 2 
If we apply a voltage V G , then replace V bi by
Vbi + VG , so
xn = −
t
 t   (V + VG )
x n = x p = − ox +  ox  + s bi
2
eN d
 2 
We find
500  10 −8
xn = x p = −
2
2
(
)
 500 10 −8 
(11.7) 8.85 10 −14 (3.695 )
 +
+ 

2
1.6 10 −19 10 16


which yields
x n = x p = 4.646  10 −5 cm
2
(
)( )
Now
Vox =
=
eN d x n t ox
s
(1.6 10 )(10 )(4.646 10 )(500 10 )
(11.7 )(8.85 10 )
−19
−5
16
−8
−14
or
V ox = 0.359 V
We also find
eN d x n2
Vn = V p =
2 s
(1.6 10 )(10 )(4.646 10 )
=
2(11.7 )(8.85 10 )
−19
−5 2
16
−14
or
Vn = V p = 1.67 V
_______________________________________
10.30
(a) n-type
(b) We have
200 10 −12
C ox =
= 110 −7 F/cm 2
2 10 −3
Also
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 10
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
C ox =
(
ox

(3.9) 8.85 10 −14
 t ox = ox =
t ox
C ox
1 10 − 7
)
 (max ) − Q ss
 Q SD
 
−
+  ms + 2 fp  
C ox

 
Using the definition of threshold voltage V T ,
we have
Q n = −C ox (VGS − V DS ) − VT 
At saturation
V DS = V DS (sat ) = VGS − VT
which then makes Q n equal to zero at the
or
o
−6
t ox = 3.45 10 cm = 34.5 nm = 345 A
(c)
Q
V FB =  ms − ss
C ox
or
Q
− 0.80 = −0.50 − ss− 7
10
which yields
Qss = 3 10 −8 C/cm 2 = 1.875  10 11 cm −2
(d)
ox
 =
C FB
    kT  s 

t ox +  ox  


 s   e  eN d 
 (
drain terminal.
_______________________________________
10.33
(0.0259 )(11.7)(8.85 10 −14 ) 
−19
16

which yields
 = 7.82 10 −8 F/cm 2
C FB
or
C FB = 156 pF
_______________________________________
10.31
(a) Point 1: Inversion
2: Threshold
3: Depletion
4: Flat-band
5: Accumulation
_______________________________________
10.32
We have
Qn = −Cox (VGS − Vx ) −  ms + 2 fp

(

Now let V x = V DS , so

Q n = −C ox (VGS − V DS )


) 
 


k p W
2

2(V SG + VT )V SD − V SD
2 L
 0.10 
2
=
(15 ) 2(0.8 − 0.4)(0.25 ) − (0.25 )
2




 (max ))
− (Q ss + Q SD
(
10.34
(a) I D =
)
 (max ) + Q ss
 Q SD
+
−  ms + 2 fp
C ox

 (max ) is a
For a p-type substrate, Q SD
negative value, so we can write


) 
(1.6 10 )(2 10 )
k n W
2

2(VGS − VT )V DS − V DS
2 L
 0.18 
2
=
(8) 2(0.8 − 0.4)(0.2) − (0.2)
 2 
= 0.0864 mA
k W
2
(b) I D = n  (VGS − VT )
2 L
 0.18 
2
=
(8)(0.8 − 0.4)
2


= 0.1152 mA
(c) Same as (b), I D = 0.1152 mA
k W
2
(d) I D = n  (VGS − VT )
2 L
 0.18 
2
=
(8)(1.2 − 0.4 )
 2 
= 0.4608 mA
_______________________________________
(a) I D =
= (3.9) 8.85 10 −14  3.45 10 −6
 3.9 
+

 11.7 

Q n = −C ox  (VGS − V DS )

I D = 0.103 mA
k p W
2
 (V SG + VT )
(b) I D =
2 L
 0.10 
2
=
(15 )(0.8 − 0.4 )
 2 
= 0.12 mA


Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 10
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
k p W
2
 (V SG + VT )
2 L
 0.10 
2
=
(15 )(1.2 − 0.4)
 2 
= 0.48 mA
(d) Same as (c), I D = 0.48 mA
_______________________________________
10.37
(c) I D =
C ox =
110 10 −8
 C W (425 ) 3.138 10 −7 (20 )
K n = n ox =
2L
2(1.2)
(
V GS = 0.6 V, V DS (sat ) = 0.15 V,
k W
2
(a) I D = n  (VGS − VT )
2 L
 0.6  W 
2
1 .0 = 
 (1.4 − 0.8)
 2  L 
VGS
W
= 9.26
L
 0.6 
2
(b) I D = 
(9.26 )(1.85 − 0.8)
2


= 3.06 mA
k W
2
2(VGS − VT )V DS − V DS
(c) I D = n 
2 L
 0.6 
2
=
(9.26 ) 2(1.2 − 0.8)(0.15 ) − (0.15 )
 2 
= 0.271 mA
_______________________________________

10.36
(a) Assume biased in saturation region
k p W
2
ID =
 (V SG + VT )
2 L
 0.12 
2
0.10 = 
(20 )(0 + VT )
2


 VT = +0.289 V
Note: V SD = 1.0 V  V SG + VT = 0 + 0.289 V
So the transistor is biased in the saturation
region.
 0.12 
2
(b) I D = 
(20 )(0.4 + 0.289 )
2


= 0.570 mA
 0.12 
(c) I D = 
(20 )2(0.6 + 0.289 )(0.15 )
 2 
− (0.15 )
or
V GS
V GS


2

I D = 0.293 mA
_______________________________________
)
= 1.111  10 −3 A/V 2 =1.111 mA/V 2
(a) VGS = 0 , I D = 0
10.35

(3.9)(8.85 10 −14 ) = 3.138 10 −7 F/cm 2

I D (sat ) = (1.111)(0.6 − 0.45 )2
= 0.025 mA
= 1.2 V, V DS (sat ) = 0.75 V,
I D (sat ) = (1.111)(1.2 − 0.45 )
= 0.625 mA
= 1.8 V, V DS (sat ) = 1.35 V,
I D (sat ) = (1.111)(1.8 − 0.45 )
= 2.025 mA
= 2.4 V, V DS (sat ) = 1.95 V,
2
2
I D (sat ) = (1.111)(2.4 − 0.45 )
= 4.225 mA
(c) I D = 0 for V GS  0.45 V
V GS = 0.6 V,
2

I D = (1.111) 2(0.6 − 0.45 )(0.1) − (0.1)
= 0.0222 mA
VGS = 1.2 V,

2
I D = (1.111) 2(1.2 − 0.45 )(0.1) − (0.1)
= 0.156 mA
V GS = 1.8 V,

2
I D = (1.111) 2(1.8 − 0.45 )(0.1) − (0.1)
= 0.289 mA
V GS = 2.4 V,



2


I D = (1.111) 2(2.4 − 0.45 )(0.1) − (0.1)
= 0.422 mA
_______________________________________
10.38
C ox =
2
(
ox (3.9) 8.85 10 −14
=
t ox
110  10 −8
)
= 3.138  10 −7 F/cm 2
 p C oxW
Kp =
2L
(210 ) 3.138 10 −7 (35 )
=
2(1.2)
(
)
= 9.61  10 −4 A/V 2 =0.961 mA/V 2
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 10
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(a) V SG = 0 , I D = 0
V SG = 0.6 V, V SD (sat ) = 0.25 V
10.40
Sketch
_______________________________________
I D (sat ) = (0.961)(0.6 − 0.35 )
= 0.060 mA
V SG = 1.2 V, V SD (sat ) = 0.85 V
2
I D (sat ) = (0.961)(1.2 − 0.35 )
= 0.694 mA
V SG = 1.8 V, V SD (sat ) = 1.45 V
10.41
Sketch
_______________________________________
2
10.42
We have
V DS (sat ) = VGS − VT = V DS − VT
so that
V DS = V DS (sat ) + VT
I D (sat ) = (0.961)(1.8 − 0.35 )
= 2.02 mA
V SG = 2.4 V, V SD (sat ) = 2.05 V
2
I D (sat ) = (0.961)(2.4 − 0.35 )
= 4.04 mA
(c) I D = 0 for V SG  0.35 V
V SG = 0.6 V
2

I D = (0.961) 2(0.6 − 0.35 )(0.1) − (0.1)
= 0.0384 mA
V SG = 1.2 V

V SG
I D = (0.961) 2(1.2 − 0.35 )(0.1) − (0.1)
= 0.154 mA
= 1 .8 V

2

2
I D = (0.961) 2(1.8 − 0.35 )(0.1) − (0.1)
= 0.269 mA
V SG = 2.4 V

2
2



I D = (0.961) 2(2.4 − 0.35 )(0.1) − (0.1)
= 0.384 mA
_______________________________________
10.39
(a) From Problem 10.37, K n = 1.111 mA/V 2
For VGS = −0.8 V, I D = 0
VGS = 0 , V DS (sat ) = 0.8 V
I D (sat ) = (1.111)(0 + 0.8)
= 0.711 mA
VGS = +0.8 V, V DS (sat ) = 1.6 V
2
I D (sat ) = (1.111)(0.8 + 0.8)
= 2.84 mA
VGS = 1.6 V, V DS (sat ) = 2.4 V
2
I D (sat ) = (1.111)(1.6 + 0.8)
= 6.40 mA
_______________________________________
2
Since V DS  V DS (sat ) , the transistor is always
biased in the saturation region. Then
2
I D = K n (VGS − VT )
where, from Problem 10.37,
K n = 1.111 mA/V 2 and VT = 0.45 V
Then
V DS = V GS
I D (mA)
0
0
1
0.336
2
2.67
3
7.22
4
14.0
5
23.0
_______________________________________
10.43
From Problem 10.38, K p = 0.961 mA/V 2

2
I D = K p 2(V SG + VT )(V SD ) − V SD
gd =
I D
V SD
VSD → 0

= 2 K p (V SG + VT )
For V SG  0.35 V, g d = 0
For V SG  0.35 V,
g d = 2(0.961 )(V SG − 0.35 )
For V SG = 2.4 V,
g d = 2(0.961 )(2.4 − 0.35 )
= 3.94 mA/V
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 10
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
10.44
I D
(a) g m =
VGS
 

2
K n 2(VGS − VT )(V DS ) − V DS
VGS
= K n (2V DS )
=

1.25 = K n (2 )(0.05 )
 K n = 12 .5 mA/V 2

(b) I D = (12.5)2(0.8 − 0.3)(0.05 )− (0.05 )
= 0.594 mA
2
(c) I D = (12.5)(0.8 − 0.3)
= 3.125 mA
_______________________________________
2
)
2
I D (sat ) = 18  A
(d)
V DS  V DS (sat )

−5

)2(3 − 0.8)(1) − (1) 
2
or
I D = 42 .5  A
_______________________________________
10.47
(a) C ox =
(3.9)(8.85 10 −14 )
180 10 −8
= 1.9175  10 −7 F/cm 2
(
(i) k n =  n C ox = (450 ) 1.9175 10 −7
)
−5
= 8.629  10 A/V 2
or k n = 86 .29  A/V 2
I D (sat )  0.033 ,
then
W n C ox
 (3 − 0.2)
2L
W n C ox
= 0.139 10 −3
2L
or
(
)
or
(
VGS = 3 V, we have
or
(
I D (sat ) = 1.25 10 −5 (2 − 0.8)
= 1.25 10
W n C ox
I D (sat ) =
 (VGS − VT )
2L
where

(3.9) 8.85 10 −14
C ox = ox =
t ox
425  10 −8
or
C ox = 8.12 10 −8 F/cm 2
We are given W L = 10 . From the graph, for
0.033 =
so V DS  V DS (sat )
2
I D = K n 2(VGS − VT )V DS − V DS
10.45
We find that VT  0.2 V
Now
(
(b)
2
2
(sat )
I D (sat ) = K n (VGS − VT ) = K nV DS
so
2
2 10 −4 = K n (4)
which yields
K n = 12 .5  A/V 2
(c)
V DS (sat ) = VGS − VT = 2 − 0.8 = 1.2 V
)
1
(10 ) n 8.12 10 −8 = 0.139 10 −3
2
which yields
 n = 342 cm 2 /V-s
_______________________________________
10.46
(a)
V DS (sat ) = VGS − VT
or
4 = VGS − 0.8  VGS = 4.8 V
 k
(ii) I D (sat ) =  n
 2
 W 
2
 (VGS − VT )
 L 
 0.08629  W 
2
0.8 = 
 (2 − 0.4 )
2

 L 
W

= 7.24
L
(b) (i) k p =  p C ox = (210 ) 1.9175  10 −7
(
= 4.027  10
−5
or k p = 40.27  A/V
A/V
)
2
2
 k p  W 
2
(ii) I D (sat ) =   (V SG + VT )
2
L
  
 0.04027  W 
2
0.8 = 
 (2 − 0.4)
2
L

 
W
= 15 .5
L
_______________________________________

Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 10
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
10.48
From Problem 10.37, K n = 1.111 mA/V 2
 

2
K n 2(VGS − VT )(V DS ) − V DS
VGS
= K n (2V DS ) = (1.111 )(2 )(0.1)
(a) g mL =



10.49
From Problem 10.38, K p = 0.961 mA/V 2
(a) g mL =
 

2
K p 2(V SG + VT )(V SD ) − V SD
V SG

= K p (2VSD ) = (0.961)(2)(0.1)
or g mL = 0.192 mA/V


2
K p (V SG + VT )
V SG
(b) g ms =
(
)
= 3.452  10 −4 A/V 2
or K n = 0.3452 mA/V 2
For I D = 0 , VGS = VTO = 0.7713 V
For I D = 0.5 = (0.3452 )(VGS − 0.7713 )
 V GS = 1.975 V
2
(c) (i) For V SB = 0 , VT = VTO = 0.7713 V
(ii) V SB = 1 V,

VT = (0.5594 ) 2(0.389 ) + 1
− 2(0.389 )


= 0.2525 V
VT = 0.7713 + 0.2525 = 1.024 V
= 2K p (VSG + VT ) = 2(0.961)(1.5 − 0.35)
(iii) V SB = 2 V,
or g ms = 2.21 mA/V
_______________________________________

VT = (0.5594 ) 2(0.389 ) + 2
− 2(0.389 )
10.50
C ox
Now C ox
Then
(
(iv) V SB = 4 V,
(3.9)(8.85 10 −14 )
=
2 1.6 10
− 2(0.389 )
16
_______________________________________
2.301 10 −7
(b)  fp
1/ 2
 5 10 16
= (0.0259 ) ln 
10
 1.5 10
(
)

 = 0.3890 V


 4(11 .7 ) 8.85  10 −14 (0.3890 ) 
(i) x dT = 

1.6  10 −19 5  10 16


(
)(
= 1.419  10 −5 cm
 (max )
QSD
(

10 16
10
 1.5 10
 fp = (0.0259 ) ln 

 = 0.3473 V


 2 + V − 2 
= (0.12 ) 2(0.3473 ) + 2.5
VT = 
)
fp
SB
fp
− 2(0.3473 )
)(
= 1.135  10 C/cm
10.51
1/ 2
)(
= 1.6 10 −19 5 10 16 1.419 10 −5
−7
2

= 0.7294 V
VT = 0.7713 + 0.7294 = 1.501 V
)(11.7)(8.85 10 )(5 10 )
−14
 = 0.5594 V

VT = (0.5594 ) 2(0.389 ) + 4
150 10 −8
= 2.301  10 −7 F/cm 2
−19

= 0.4390 V
VT = 0.7713 + 0.4390 = 1.210 V
2e s N a
(a)  =
=
+ V FB + 2 fp
1.135 10 −7
− 0.5 + 2(0.3890 )
2.301 10 −7
= 0.7713 V
 C W
K n = n ox
2L
(
450 ) 2.301 10 −7 (8)
=
2(1.2)

2
K n (VGS − VT )
VGS
= 2 K n (VGS − VT ) = 2(1.111 )(1.5 − 0.45 )
so g ms = 2.33 mA/V
_______________________________________
C ox
=
so g mL = 0.222 mA/V
(b) g ms =
 (max )
QSD
VTO =
)
or
VT = 0.114 V

Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 10
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Now VT = VTO + VT
0.5 = VTO + 0.114
 VTO = 0.386 V
_______________________________________
10.52
(a) C ox =
(3.9)(8.85 10 −14 )
C ox = 8.63 10 −8 F/cm 2
We find
Qss = 1.6 10 −19 5 10 10 = 8 10 −9 C/cm 2
Then
 (max ) − Q ss
Q SD
VT =
+  ms + 2 fp
C ox
(
=
(
)
(
)(
2 1.6 10 −19 (11.7 ) 8.85 10 −14 5 10 15
)
1.726 10 −7
 = 0.2358 V 1 / 2
 2
fn

 = 0.3294 V


+ V BS − 2 fn



 = 0.288 V


(
)
(
or
)( )
or
 (max ) = 1.38  10 −8 C/cm 2
Q SD
Also
C ox =
or
)( )(
(
ox (3.9) 8.85 10 −14
=
t ox
400 10 −8
)

W n C ox
(VGS − VT )
L
W n ox
(VGS − VT )
=
Lt ox
gm =
or
(


0.357 = 0.211 0.576 + VSB − 0.576
10.55
(a)
1/ 2
x dT = 0.863 10 −4 cm
Now
 (max ) = 1.6  10 −19 10 15 0.863  10 −4
Q SD
)( )
−8
10.54
Plot
_______________________________________
1/ 2
 4(11 .7 ) 8.85  10 (0.288 ) 
=

1.6  10 −19 10 15


(

which yields
V SB = 5.43 V
_______________________________________
also
−14
+ V SB − 2 fp
8.63 10

10.53
(a) n + poly-to-p-type   ms = −1.0 V
x dT
)
fp
 2(0.288 ) + VSB − 2(0.288 )
_______________________________________
 4 s  fp 
=

 eN a 
(
 2
2 1.6 10 −19 (11.7 ) 8.85 10 −14 10 15
+ 0.357 =
 V BS = 2.39 V
10 15
10
 1.5 10
C ox
or
− 2(0.3294 )

2e s N a
VT =

− 0.22 = −(0.2358 ) 2(0.3294 ) + VBS
 fp = (0.0259 ) ln 
(b) For NMOS, apply V SB and V T shifts in a
positive direction, so for VT = 0 , we want
VT = +0.357 V.
So
 5 10 15
(b)  fn = (0.0259 ) ln 
10
 1.5 10
VT = −

 − 1.0 + 2(0.288 )


or
VT = −0.357 V
2e s N d
C ox
)
 1.38 10 −8 − 8 10 −9
= 
8.63 10 −8

200 10 −8
= 1.726  10 −7 F/cm 2
=
)(
)
=
or
(10 )(400 )(3.9)(8.85 10 −14 ) (5 − 0.65)
475 10 −8
g m = 1.26 mS
Now
gm
g
1
g m =
 m = 0.8 =
1 + g m rs
gm
1 + g m rs
which yields
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 10
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
rs =
or
Then
C M = 1.035 10 −14
1  1
1  1


− 1 =
− 1


g m  0.8  1.26  0.8 
(
10.56
(a) The ideal cutoff frequency for no overlap
capacitance is,
gm
 (V − V )
fT =
= n GS 2 T
2 C gs
2 L
(400 )(4 − 0.75 )
(
2 2 10
(
(
=
)
(
(3.9)(8.85 10
)(
−14
)
)
500 10
 0.75 10 −4 20 10 −4
(
)(
)
C gsT = 3.797  10 −14 F
We now find
W n C ox
(VGS − VT )
L
20  10 −4 (400 )(3.9) 8.85  10 −14
=
2  10 − 4 500  10 −8
 (4 − 0.75 )
)(
S
gm
2 C gsT + C M
fT =
(
=
=
gm =
g m = 0.8974 10
)
or
C gdT = 1.035  10 −14 F
−3
)(
)
0.8974  10 −3
2 3.797  10 −14 + 1.032  10 −13
(
)
(
)
)
)
We find
g m = WC ox ds
or
or
500 10
−4
+ 0.75 10 −4 20 10 −4
 (2 10
(
−8
(
−8
(
We find
C gdT = C ox 0.75  10 −4 20  10 −4
)
)
)
10.57
(a) For the ideal case

4  10 6
f T = ds =
2 L 2 2  10 − 4
or
f T = 3.18 GHz
(b) With overlap capacitance (using the
values from Problem 10.56),
gm
fT =
2 C gdT + C M
where
C M = C gdT (1 + g m RL )
(
(3.9)(8.85 10
−14
f T = 1.01 GHz
_______________________________________
)
f T = 5.17 GHz
(b) Now
gm
fT =
2 C gsT + C M
C M = 1.032 10 −13 F
Now
C gsT = C ox L + 0.75  10 −4 (W )
or
−4 2
or
Also
)
or
(b) For VGS = 3 V, g m = 0.683 mS
Then
0.683
g m =
= 0.602 mS
1 + (0.683 )(0.198 )
or
g m
0.602
=
= 0.88
g m 0.683
which is a 12% reduction.
_______________________________________
=
)(
 1 + 0.8974 10 −3 10 10 3
rs = 0.198 k 
=
 (
)
)
(20 10 )(3.9)(8.85 10 )(4 10 )
−4
−14
6
500 10 −8
or
g m = 0.5522 10 −3 S
We have
C M = C gdT (1 + g m RL )
(
= 1.035 10 −14
 (
)
)(
 1 + 0.5522 10 −3 10 10 3
or
C M = 6.750 10 −14 F
)
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 10
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Then
fT =
or
0.5522  10 −3
2 3.797  10 −14 + 6.75  10 −14
(
)
f T = 0.833 GHz
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 11
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
Chapter 11
Exercise Solutions
Ex 11.1
I D
ID
Ex 11.4
=
L
=
L − L

1
= 1.35
L
1−
L
(
L
1
= 1−
= 0.259
L
1.35
C ox =
L
0.1807
=
= 0.698  m
0.259
0.259
_______________________________________
Then L =
Ex 11.2
From Figure 11.10,  n  550 cm 2 /V-s
_______________________________________
 fp
x dT
(3.9)(8.85 10 −14 )
120 10 −8
= 2.876  10 −7 F/cm 2
 10 16
= (0.0259 ) ln 
10
 1.5 10
(
)( )
)
W=
(
)
2
eN a  x dT
C ox (VT )
(1.6 10 )(10 ) 2 (0.30 10 )
 
=
(4.314 10 )(0.1)
−19
−4 2
16
−7
W = 5.243  10 −5 cm
or
W = 0.524  m
_______________________________________
N N
Vbi = Vt ln  a 2 d
 ni
(
1/ 2
= 0.30  10 cm

eN x  r j 
2x
VT = − a dT   1 + dT − 1

C ox  L 
rj

 
−19
16
−4
1.6 10
10 0.30 10
=−
2.876 10 −7
 0.25 

 1 + 2(0.3) − 1


0.25
 0.75 

VT = −0.0469 V
_______________________________________




)( ) = 0.902 V
) 
 3 10 16 10 19
= (0.0259 ) ln 
2
 1.5 10 10
)( )
)( )(
(3.9)(8.85 10 −14 )
80 10 −8
= 4.314  10 −7 F/cm 2
−4
(
1/ 2
Ex 11.5

 = 0.3473 V


 4(11 .7 ) 8.85  10 −14 (0.3473 ) 
=

1.6  10 −19 10 16


(
)
= 0.30  10 −4 cm
From Example 11.1, we have
L = 0.1807  m
C ox =
(

 = 0.3473 V


 4(11 .7 ) 8.85  10 −14 (0.3473 ) 
x dT = 

1.6  10 −19 10 16


or
Ex 11.3
10 16
10
 1.5 10
 fp = (0.0259 ) ln 
 2 s V bi 
x dO = 

 eN a 
(
1/ 2
(
)
 2(11 .7 ) 8.85  10 −14 (0.902 ) 
=

1.6  10 −19 3  10 16


)
(
)(
1/ 2
)
−5
= 1.973  10 cm = 0.1973  m
x d = L − x dO = 0.8 − 0.1973 = 0.6027  m
Also
 2  (V + V DS ) 
x d =  s bi

eN a


or
Vbi + V DS =
x d2 eN a
2 s
1/ 2
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 11
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
=
(0.6027 10 ) (1.6 10 )(3 10 )
2(11.7 )(8.85 10 )
−4 2
−19
16
−14
Ex 11.7
(a) N h = 10 18 120 10 −8 = 1.2  10 12 cm −2
Qss
Vbi + V DS = 8.419 V
Then
V DS = 8.419 − 0.902 = 7.52 V
_______________________________________

10 15
10
 1.5 10
4 V 
=  s bi 
 eN a 

 = 0.2877 V


1/ 2
(
)
 4(11 .7 ) 8.85  10 (0.2877 ) 
=

1.6  10 −19 10 15


(
= 8.630  10
( )(
)
= (8 10 )(0.2) = 1.6 10
(3.9)(8.85 10 )
=
11
C ox
Then
cm −2
(
)(
)
Q ss
1.6  10 11 1.6  10 −19
=−
C ox
4.314  10 − 7
= −0.0593 V
(c) Threshold voltage shift decreases when
the oxide thickness decreases.
_______________________________________
120 10 −8
= 2.876  10 −7 F/cm 2
The initial threshold voltage is
eN x
VTO = V FBO + 2 fpO + a dTO
C ox
= +0.95 + 2(0.2877 )
(1.6 10 )(10 )(8.63 10 )
−5
2.876 10 −7
Test Your Understanding Solutions
TYU 11.1
VTO = 1.573 V
Now
VT = VTO + VT
I D1
I D2
0.40 = 1.573 + VT
 VT = −1.173 V
Negative VT  implant donor ions
Now
eDI
VT =
C ox
or
VT C ox (1.173 ) 2.876 10 −7
DI =
=
e
1.6 10 −19
12
−2
= 2.11  10 cm
_______________________________________
(
11
80 10 −8
= 4.314  10 −7 F/cm 2
VT = −
15
)
−14
cm = 0.863  m
−19
)(
(b) N h = 10 18 80 10 −8 = 8 10 11 cm −2
1/ 2
(3.9)(8.85 10 −14 )
=
+
(
Q ss
2.4  10 11 1.6  10 −19
=−
C ox
2.876  10 − 7
= −0.134 V
)( )
−5
cm −2
120 10
= 2.876  10 −7 F/cm 2
VT = −
Qss
−14
11
−8
Then
 fpO = (0.0259 ) ln 
C ox
12
−14
C ox
Ex 11.6
We find
x dTO
( )(
)
= (1.2 10 )(0.2) = 2.4 10
(3.9)(8.85 10 )
=
)
V 
exp  GS1 
 V − VGS 2
 Vt 
=
= exp  GS1
Vt
V 

exp  GS 2 
 Vt 




or
I 
VGS1 − VGS 2 = Vt ln  D1 
 I D2 
Then
VGS1 − VGS 2 = (0.0259 ) ln (10 ) = 0.05964 V
or VGS1 − VGS 2 = 59 .64 mV
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 11
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
TYU 11.2
(a) I D (sat ) =
 n C oxW
(VGS − VT )
2L
(1000 ) 10 −8 10 −3 (V − 0.4)2
=
GS
2 10 − 4
2
( )(
( )
)
= 5.0 10 −5 (VGS − 0.4) A
2
or I D (sat ) = 50(VGS − 0.4)  A
(b) I D (sat ) = WC ox sat (VGS − VT )
2
(
)(
)(
)
= 10 −3 10 −8 5 10 6 (VGS − 0.4)
= 5.0 10 (VGS − 0.4) A
or I D (sat ) = 50 (VGS − 0.4 )  A
_______________________________________
−5
TYU 11.3
L → k L = (0.7 )(1)  L = 0.7  m
W → k W = (0.7 )(10 )  W = 7  m
o
t ox → ktox = (0.7 )(250 )  t ox = 175 A
N a 5 10 15
=
 N a = 7.14 10 15 cm −3
k
0.7
V D → kVD = (0.7 )(3)  V D = 2.1 V
_______________________________________
Na →
TYU 11.4
We have from Example 11.6,
VTO = −0.419 V, C ox = 1.9175 10 −7 F/cm 2
VT = VTO +
eDI
C ox
(1.6 10 )D
−19
(a) + 0.25 = −0.419 +
1.9175 10
 D I = 8.02 10 11 cm −2
I
−7
(1.6 10 )D
−19
(b) − 0.25 = −0.419 +
I
1.9175 10 −7
11
 D I = 2.03 10 cm −2
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Chapter 11
11.1
(a)
11.3
 V
I D = 10 −15 exp  GS
 (2.1)Vt
For VGS = 0.5 V,
I D = 10
−15




 2 10 16
10
 1.5 10
 fp = (0.0259 ) ln 
We find that
I D = 9.83 10 A
For V GS = 0.7 V,
L =
( )
For V GS = 0.9 V,
(b)
Power: P = I T
Then
For VGS = 0.5 V,
For V GS = 0.7 V,
(

)
 0.3653 + 2 − 0.3653 + 0.6
(
L = 2.544 10 −5

)
 0.3653 + 4 − 0.3653 + 0.6
cm = 0.2816  m
− V T = 2 .0 − 0 . 4 = 1 . 6 V
L = 2.816  10
(c) V DS (sat ) = VGS
(
I 
VGS 2 − VGS1 = nVt ln  D 2 
 I D1 
(a) VGS 2 − VGS1 = (0.0259 ) ln (10 )
= 0.0596 V
(b) VGS 2 − VGS1 = (1.5)(0.0259 ) ln (10 )
= 0.0895 V
(c) VGS 2 − VGS1 = (2.1)(0.0259 ) ln (10 )
= 0.125 V
_______________________________________

−5
L = 2.544 10 −5

)
 0.3653 + 2 − 0.3653 + 1.6

L = 3.461  10 −6 cm = 0.0346  m
(d) V DS (sat ) = VGS − VT = 2.0 − 0.4 = 1.6 V
(
L = 2.544 10 −5
I D2
I D1

(b) V DS (sat ) = VGS − VT = 1.0 − 0.4 = 0.6 V
11.2
V 
exp  GS 2 
 (V − VGS1 ) 
 nVt 
=
= exp  GS 2

nVt
 VGS1 




exp 

 nVt 

L = 1.413  10 −5 cm = 0.1413  m
P = 49.2  W
P = 1.94 mW
For V GS = 0.9 V, P = 77 mW
_______________________________________
cm/V 1 / 2
+ V DS −  fp + V DS (sat )
fp
L = 2.544 10 −5
I T = 15 .4 mA
 V DD

2 s
eN a
)(
−5
)
)
(a) V DS (sat ) = VGS − VT = 1.0 − 0.4 = 0.6 V
−8
I D = 1.54 10 A
Then the total current is:
I T = I D 10 6
For VGS = 0.5 V, I T = 9.83  A
For V GS = 0.7 V, I T = 0.388 mA
(
= 2.544  10
−12
I D = 3.88 10 −10 A
For V GS = 0.9 V,
(
2(11 .7 ) 8.85  10 −14
1.6  10 −19 2  10 16
2 s
=
eN a


0.5
exp 

(
)(
)
2
.
1
0
.
0259



 = 0.3653 V



)
 0.3653 + 4 −
0.3653 + 1.6

−5
cm = 0.1749  m
_______________________________________
L = 1.749  10
11.4
 2 10 16
10
 1.5 10
 fp = (0.0259 ) ln 
We find that
2 s
=
eN a
(

 = 0.3653 V


2(11 .7 ) 8.85  10 −14
1.6  10 −19 2  10 16
(
)(
)
)
= 2.544  10 −5 cm/V 1 / 2
V DS (sat ) = VGS − VT = 2.0 − 0.4 = 1.6 V
L =
2 s
eN a

fp
+ V DS −  fp + V DS (sat )

Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(
(a) L = 2.544 10 −5

)
 0.3653 + 3 − 0.3653 + 1.6
L = 1.10  10
−5

cm = 0.110  m

)

L
0.2326
= 0.10 =
L
L
 L = 2.326  m
_______________________________________
Now
(

(3.9) 8.85 10
= ox =
t ox
120  10 −8
−14
)

)
 4(11 .7 ) 8.85  10 (0.3832 ) 
x dT = 

1.6  10 −19 4  10 16


(
)(
)(
1/ 2
)
= 1.6 10 −19 4 10 16 1.575 10 −5
−7
= 1.008  10 C/cm
)
2
So
1.008 10 −7
− 0.5223 + 2(0.3832 )
2.876 10 −7
= 0.595 V
V DS (sat ) = VGS − VT = 1.25 − 0.595 = 0.655 V
VT =
(

cm = 0.0735  m
(ii) L = 1.799 10


−5
)
 0.3832 + 0.655 + 2 − 0.3832 + 0.655

−5
L = 1.303  10 cm = 0.1303  m
(
)
−5

cm = 0.2205  m
L
0.2205
= 0.12 =
L
L
L = 1.84  m
_______________________________________
2.876 10 −7
= 1.575  10 −5 cm
 (max )
QSD
)
(b)
−19
)(
L = 7.35  10
−6
L = 2.205  10
10
(
−  fp + V DS (sat )
 0.3832 + 0.655 + 1 − 0.3832 + 0.655

V FB = −0.5223 V
Now
 (max )
QSD
VT =
+ V FB + 2 fp
C ox
We find
 4 10 16 
 = 0.3832 V
 fp = (0.0259 ) ln 
10 
 1.5 10 
(
−5
 0.3832 + 0.655 + 4 − 0.3832 + 0.655
(4 10 )(1.6 10 )
−14
(
+ V DS (sat ) + V DS
fp
(iii) L = 1.799 10 −5
= 2.876  10 −7 F/cm 2
Q
V FB =  ms − ss
C ox
= −0.5 −

(i) L = 1.799 10
L = 2.326  10 −5 cm = 0.2326  m
C ox
2 s
eN a
(a) L =
 0.3653 + 5 − 0.3653 + 1.6
11.5
)(
= 1.799  10 cm/V 1 / 2
L
0.110
= 0.10 =
L
L
 L = 1.10  m
(
(
)
)
−5
Now
(b) L = 2.544 10 −5
(
2(11 .7 ) 8.85  10 −14
1.6  10 −19 4  10 16
2 s
=
eN a
11.6
 3 10 16
10
 1.5 10
 fp = (0.0259 ) ln 
(

 = 0.3758 V


2(11 .7 ) 8.85  10 −14
1.6  10 −19 3  10 16
2 s
=
eN a
(
= 2.077  10
)(
−5
)
)
cm/V 1 / 2
(a) Ideal,
k W
2
I D = n  (VGS − VT )
2 L
 0.05  15 
2
=

(1.0 − 0.4 )
2
0
.
80



= 0.16875 mA
(i) V DS (sat ) = 1.0 − 0.4 = 0.6 V
(
L = 2.077 10 −5

)
 0.3758 + 2 − 0.3758 + 0.6
−5
= 1.150  10 cm = 0.115  m
 L 
I D = 
 ID
 L − L 

Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
11.7
(a)
0.80


=
(0.16875 )
 0.80 − 0.115 
= 0.19708 mA
(ii) L = 2.077 10 −5
(
)

k n W
2
 (VGS − VT )
2 L
 0.075 
2
=
(10 )(0.8 − 0.35 )
2


= 0.07594 mA = 75 .94  A
(ii) I D = I D (1 + V DS )
= (75 .9375 )1 + (0.02 )(1.5)
= 78 .22  A
(i) I D =
 0.3758 + 4 − 0.3758 + 0.6

−5
= 2.293  10 cm = 0.2293  m
 L 
I D = 
 ID
 L − L 
0.80


=
(0.16875 )
 0.80 − 0.2293 
= 0.23655 mA
1
1
=
 I D (0.02 )(75 .94 )
= 0.658 M  = 658 k 
(iii) ro =
(b)
 I D
ro = 
 V DS




−1
 (0.23655 − 0.19708 ) 10 −3 
=

4−2


 ro = 5.07 10 4  = 50.7 k 
(c) V DS (sat ) = VGS − VT = 2.0 − 0.4 = 1.6 V
(
)
(i) L = 2.077 10 −5

 0.3758 + 2 − 0.3758 + 1.6
= 2.819  10
−6

cm = 0.02819  m
 L 
I D = 
 ID
 L − L 


 sat =
= 1.425  10 −5 cm = 0.1425  m
We find
 L 
I D = 
 ID
 L − L 




−1
 (0.20532 − 0.17491 ) 10 −3 
=

4−2


V DS (sat )
L
L (  m)
3
0.80


=
(0.16875 )
0
.
80
−
0
.
1425


= 0.20532 mA
 I D
ro = 
 V DS
 0.075 
2
(i) I D = 
(10 )(1.25 − 0.35 )
2


= 0.30375 mA
(ii) I D = (0.30375 )1 + (0.02 )(1.5)
= 0.3129 mA
1
(iii) ro =
= 165 k 
(0.02 )(0.30375 )
_______________________________________
11.9
(a) Assume V DS (sat ) = 1 V. Then
−5
 0.3758 + 4 − 0.3758 + 0.6
(b)
11.8
Plot
_______________________________________
0.80


=
(0.16875 )
 0.80 − 0.02819 
= 0.17491 mA
(ii) L = (2.077 10 )
−1
−1
ro = 6.577 10 4  = 65.77 k 
_______________________________________
 sat (V/cm)
3.33  10 3
1
0 .5
1 10 4
0.25
4  10 4
0.13
7.69  10 4
2  10 4
(b)
Assume  n = 500 cm 2 /V-s, we have
 =  n  sat
Then
For L = 3  m,  = 1.67  10 6 cm/s
For L = 1  m,  = 5 10 6 cm/s
For L  0.5  m,   10 7 cm/s
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
11.10
k n =  n C ox =
(a)
 n ox
IDL
1
L
=

ro (L − L )2 V DS
t ox
(425 )(3.9)(8.85 10 −14 )
=
=
−8
110 10
= 1.334  10 −4 A/V 2 = 0.1334 mA/V 2
 k   W 
2
I D =  n  (VGS − VT )
2
L
  
L =

fp
2 s 1
  fp + V DS
eN a 2
L
=
V DS
(
We find
2 s
=
eN a
(
(

11.11
(a)
W n C ox
(VGS − VT )2
2L
 10 
2
=  (500 ) 6.9  10 −8 (VGS − 1)
 2
(

I D (sat ) = 0.173 (VGS − 1)
and
  eff
Let  eff =  O 
 C
)
)
We find
Now
2.077  10 −5
L
=
= 6.290  10 − 6 cm/V
V DS 2 0.3758 + 2.35
C ox =
VGS
t ox
(
ox

(3.9) 8.85 10 −14
 t ox = ox =
t ox
C ox
6.9  10 −8
)
or
0.3758 + 2.35
t ox = 500 A
Then
= 1.660  10 cm = 0.166  m
−1 / 3
 C = 2.5 10 4 V/cm
V DS
− 0.3758 + 0.35




Where  O = 1000 cm 2 /V-s and
Let  eff =
−5
(mA)
(b)

 = 0.3758 V


= V DS (sat ) + V DS = VGS − VT + V DS
= 0.8 − 0.45 + 2 = 2.35 V
)
2
I D (sat ) = 0.173 (VGS − 1) (mA) 1 / 2
 3 10 16
10
 1.5 10
(
)
or
 fp = (0.0259 ) ln 
L = 2.077  10 −5
)
I D (sat ) =
= 2.077  10 −5 cm/V 1 / 2
(

)(
 ro = 5.865 10 4  = 58.65 k 
_______________________________________
−1 / 2
)(
)(
= 1.705  10 −5
)
2(11 .7 ) 8.85  10 −14
1.6  10 −19 3  10 16
−6
−4 2
(




+ V DS −  fp + V DS (sat )
−4
 ro = 1.04 10 5  = 104 k 
(b)
0.1362 10 −3 0.8 10 −4
1
=
 6.290 10 −6
ro
(0.8 − 0.166 )10 −4 2
Now
2 s
eN a
−3
= 9.615  10 −6
 0.1334  20 
2
=

(0.8 − 0.45 )
 2  1.2 
= 0.1362 mA
I D

1
  L 
=
=
 ID 

ro V DS V DS  L − L 


L − L−1
= IDL
V DS
L
−2 
= I D L(− 1)(L − L )  −
 V DS
IDL
L
=

2
(L − L ) V DS
(0.1362 10 )(1.2 10 )  (6.290 10 )
(1.2 − 0.166 )10 

o
)
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
I D (sat )
V GS
 eff
 eff
1
2
3
-4  10 5
-397
6  10 5
347
4
5
8  10
315
0.989
292
1.27
5
10  10 5
0
0.370
0.692
(
11.12
Plot
_______________________________________
(a) C ox =
(3.9)(8.85 10 −14 )
)
(i) I D = (0.410 ) 2(2)(0.5) − (0.5)
= 0.7175 mA
2
(ii) I D = (0.410 ) 2(2)(1.0) − (1.0)
= 1.23 mA
2




(iii) I D = (0.410 ) 2(2)(1.25 ) − (1.25 )
= 1.409 mA
2
(iv) I D = (0.410 ) 2(2)(2) − (2)
= 1.64 mA
(b) I D = WC ox (VGS − VT ) ds

2


−10
ds
−7
ds
(
 0.5 
6
(i) For V DS = 0.5 V,  ds = 
 4  10
1
.
25


)(
= 1.6  10 6 cm/s
)
)(
)
(c) For part (a), V DS (sat ) = 2 V
For part (b), V DS (sat ) = 1.25 V
_______________________________________
11.15
(a) Non-saturation region
1
W 
2
I D =  n C ox   2(VGS − VT )V DS − V DS
2
L
We have

C
C ox = ox  ox
t ox
k
and
W  kW , L  kL
also
V GS  k VGS , V DS  k VDS
So
1  C  kW 
I D =  n  ox 

2  k  kL 

( )(
)
= (3.452 10 ) A
= (3.452 10 ) mA
(
(
I D = 3.452 10 −7 4 10 6
= 1.38 mA
(iv) For V DS = 2 V,  ds = 410 6 cm/s
I D = 1.38 mA
 2(kVGS − VT )kVDS − (kVDS )
= 10 −3 1.726 10 −7 (2) ds
I D = 3.452 10 −7 1.6 10 6
= 0.552 mA
)

= 4.10  10 −4 A/V 2 = 0.410 mA/V 2
For VGS − VT = 2 V, V DS (sat ) = 2 V

= 3.2  10 6 cm/s
11.14
Plot
_______________________________________
200 10 −8
= 1.726  10 −7 F/cm 2
 C W
K n = n ox
2L
(475 ) 1.726 10 −7 (10 )
=
2(1.0)
(
)(
)
I D = 3.452 10 −7 3.2 10 6
= 1.10 mA
(iii) For V DS = 1.25 V,  ds = 410 6 cm/s
(c) The slope of the variable mobility curve is
not constant, but is continually decreasing.
_______________________________________
11.13
(
 1.0 
6
(ii) For V DS = 1.0 V,  ds = 
 4  10
 1.25 
)

2

Then
I D   k ID
In the saturation region,
1  C  kW 
2
I D =  n  ox 
kVGS − VT 
2  k  kL 
Then
I D   k ID
(b)
P = I DV DD  (kI D )(kVDD )  k 2 P
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
11.16
I D (sat ) = WC ox (VGS − VT ) sat
C
 (kW ) ox
 k

(kVGS − VT ) sat

or
I D (sat )   k I D (sat )
_______________________________________
11.17
(a)
k n W
2
  (VGS − VT )
2 L
 0.15  6 
2
=

(3 − 0.45 )
 2  1.2 
2.438 mA
(ii) Scaled device:
V D = VGS = k (3) = (0.65 )(3) = 1.95 V
(i) I D (max ) =
 0.15   0.15 
k n = 
=
 = 0.2308 mA/V 2
 k   0.65 
L = k (1.2 ) = (0.65 )(1.2 ) = 0.78  m
W = k (6 ) = (0.65 )(6 ) = 3.90  m
Then
 0.2308  3.9 
2
I D (max ) = 

(1.95 − 0.45 )
 2  0.78 
= 1.298 mA
(
) = I D (max )V D = (2.438 )(3)
P
max
(b) (i)
= 7.314 mW
(ii) P (max ) = (1.298 )(1.95 )
= 2.531 mW
_______________________________________
11.18
C ox =
(
ox (3.9) 8.85 10 −14
=
t ox
120  10 −8

eN a x dT  r j 
2x

1 + dT − 1

C ox  L 
rj

 
−19
16
1.6 10
5 10 1.419 10 −5
=−
2.876 10 −7
 0.25 

 1 + 2(0.1419 ) − 1


0.25
 0.80 

VT = −0.0569 V
_______________________________________
VT = −
)
(
11.19
C ox =
 fp
 5 10 16
10
 1.5 10
(
(
)
(
= 1.419  10 −5 cm
)(
)
1/ 2
(
= 2.174  10
VT = −
)

 = 0.3653 V


)(
−5
1/ 2
)
cm
2 10 16 2.174 10 −5
(1.6 10 )(
)(
−19
)
4.314 10 −7
 0.30 

 1 + 2(0.2174 ) − 1


0.30
 0.70 

VT = −0.0391 V
VT = VTO + VT
0.35 = VTO − 0.0391
 VTO = 0.389 V
_______________________________________
C ox =
 4(11 .7 ) 8.85  10 −14 (0.3890 ) 
x dT = 

1.6  10 −19 5  10 16


)
(3.9)(8.85 10 −14 )
80 10 −8
= 4.314  10 −7 F/cm 2
 2 10 16
= (0.0259 ) ln 
10
 1.5 10
11.20

 = 0.3890 V


)(
 4(11 .7 ) 8.85  10 −14 (0.3653 ) 
x dT = 

1.6  10 −19 2  10 16


= 2.876  10 −7 F/cm 2
 fp = (0.0259 ) ln 
)(
 fp
(3.9)(8.85 10 −14 )
200 10 −8
= 1.726  10 −7 F/cm 2
 3 10 16
= (0.0259 ) ln 
10
 1.5 10
(
)

 = 0.3758 V


 4(11 .7 ) 8.85  10 −14 (0.3758 ) 
x dT = 

1.6  10 −19 3  10 16


(
= 1.80  10 −5 cm
)(
)
1/ 2
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
VT = −0.15
=−
(1.6 10 )(310 )(1.80 10 )
−19
16
−5
1.726 10 −7
 0.30 

 1 + 2(0.18 ) − 1


0.30
 L 

 0.30 
0.15 = (0.5006 )
(0.4832 )
 L 
 L = 0.484  m
_______________________________________
11.21
We have
L  = L − (a + b )
and from the geometry
2
2
2
= (r j + x dS )
(1) (a + r j ) + x dT
and
2
2
2
= (r j + x dD )
(2) (b + r j ) + x dT
From (1)
(a + r j )2 = (r j + x dS )2 − x dT2
so that
a=
(r
j
+ x dS
)
2
−x
2
dT
threshold equation. Then
 (max )
QB
QSD
VT =
−
C ox
C ox
=
eN a x dT
C ox
 a + b  eN a x dT
1 − 2 L  − C


ox
or
eN a x dT (a + b )

C ox
2L
Then substituting, we obtain

rj 
eN x
2 x dS

VT = − a dT 
+  2 − 1
 1 +
C ox
2 L 
rj


VT = −
− rj
which can be written as
2

 x dS   x dT

 −
a = r j  1 +

r j   r j


The average bulk charge in the trapezoid (per
unit area) is
 L + L 
Q B  L = eN a x dT 

 2 
or
 L + L 
Q B = eN a x dT 

 2L 
We can write
L + L 1 L 1 1
L − (a + b)
= +
= +
2L
2 2L 2 2L
which is
L + L
a+b
= 1−
2L
2L

 (max ) in the
Q
Now, B replaces QSD
2


 − 1




or
2
2


2 x dS  x dS   x dT 

a = rj  1+
+
−
− 1
 rj   rj 
rj

 



Define
x2 − x2
 2 = dS 2 dT
rj
We can then write


2x
a = r j  1 + dS +  2 − 1
rj


Similarly from (2), we will have


2x
b = r j  1 + dD +  2 − 1
rj


where
x2 − x2
 2 = dD 2 dT
rj


2x

+  1 + dD +  2 − 1 
r

 
j

Note that if x dS = x dD = x dT , then  =  = 0
and the expression for VT reduces to that
given in the text.
_______________________________________
11.22
We have L = 0 , so Equation (11.27)
becomes
L + L
L 1

=
2L
2L 2


2x
 rj 

= 1 −  1 + dT − 1 
L
rj
 




or
 1
rj 
2x
 1 + dT − 1 =
L 
rj
 2

Then Equation (11.28) is
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
1
Q B = eN a x dT  
2
Then change in the threshold voltage is
 (max )
QB
QSD
VT =
−
C oc
C ox
or
(1 2)(eN a x dT ) (eN a x dT )
VT =
−
C ox
C ox
which becomes
1 eN x
VT = −  a dT
2 C ox
_______________________________________
VT =
11.25
VT = −
eN a x dT
C ox

rj

L




2x

 1 + dT − 1 
rj

 

N 
e a (kxdT ) 
 krj
 k 
−

 C ox 
 kL


 k 
  x dT

 W
−19
or
−5 2
16
−7
−4
VT = +0.0257 V
_______________________________________
11.27
C ox =
 fp
(3.9)(8.85 10 −14 )
120 10 −8
= 2.876  10 −7 F/cm 2
 10 16
= (0.0259 ) ln 
10
 1.5 10
(

 = 0.3473 V


)
 4(11 .7 ) 8.85  10 −14 (0.3473 ) 
x dT = 

1.6  10 −19 10 16


(
1/ 2
)( )
−5
= 3.0  10 cm
eN a x dT   x dT
VT =

C ox  W
In this case,  = 1
So



(1.6 10 )(10 )(1.0)(3 10 )
0.045 =
(2.876 10 )(W )
−19

 
2kxdT
− 1 
 1+
krj

 



(1.6 10 )(3 10 )(1.80 10 )  2 
 
=
(4.314 10 )(2.2 10 )
11.23
Plot
_______________________________________
11.24
Plot
_______________________________________
eN a x dT
C ox
−5 2
16
−7
 W = 1.11  m
_______________________________________
 V T  k V T
_______________________________________
11.28
Plot
_______________________________________
11.26
11.29
C ox
 fp
(3.9)(8.85 10 −14 )
=
80 10 −8
= 4.314  10 −7 F/cm 2
 3 10 16
= (0.0259 ) ln 
10
 1.5 10
(
)

 = 0.3758 V


 4(11 .7 ) 8.85  10 (0.3758 ) 
x dT = 

1.6  10 −19 3  10 16


(
−5
= 1.80  10 cm
−14
)(
eN a x dT   x dT 


C ox  W 
Assume that  is a constant, then
VT =
N 
e a (kxdT )
  kxdT
 k 
V T 

 C ox 
 kW


 k 
1/ 2
)
or



 V T  k V T
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
11.30
(a)
Now
(i)  ox
V
 G
t ox
6  10 =
6

V CE
10
−8
0.0941
14 .5
10
−7
0.274
13 .5
10
−6
0.454
12 .3
10
−5
0.634
10 .7
10 −4
0.814
8 .6
−3
0.994
2 .7
ID
VG
200  10 −8
 VG = 12 V
12
= 4V
(ii) V G =
3
(b)
VG
(i) 6  10 6 =
80  10 −8
 V G = 4 .8 V
10
_______________________________________
11.33
One Debye length is
 (kT e ) 
LD =  s

 eN a 
4 .8
= 1 .6 V
3
_______________________________________
(ii) V G =
11.31
(a) VG = (8)(3) = 24 =  ox t ox = 6 10 6 (t ox )
(
)
t ox = 4 10 −6 cm
o
or t ox = 40 nm = 400 A
(
)
(b) VG = (12 )(3) = 36 =  ox t ox = 6 10 6 (t ox )
t ox = 6 10 −6 cm
o
or t ox = 60 nm = 600 A
_______________________________________
11.32
Snapback breakdown means M = 1 , where
 IO 
 = (0.18 ) log 10 

−9 
 3 10 
and
1
M=
m
V 
1 −  CE 
 V BO 
Let V BO = 15 V and m = 3 . Now when
M = 1 =
V
1 −  CE
 15
we can write this as
V
1 −  CE
 15
3
(
(



1/ 2
)( )
or
L D = 4.09 10 −6 cm
Six Debye lengths is then
6L D = 0.246 10 −4 cm = 0.246  m
From Example 11.5, we have
x dO = 0.336  m, which is the zero-biased
source-substrate junction width.
At near punch-through, we will have
x dO + 6 L D + x d = L
where x d is the reverse-biased drainsubstrate junction width. Now
0.336 + 0.246 + x d = 1.2
or
x d = 0.618  m
Then, at near punch-through we have
 2  (V + V DS ) 
x d =  s bi

eN a


1/ 2
or
3

 =   VCE = 15  3 1 − 

)
 (11 .7 ) 8.85  10 −14 (0.0259 ) 
=

1.6  10 −19 10 16


Vbi + V DS =

1/ 2
=
x d2 eN a
2 s
(0.618 10 ) (1.6 10 )(10 )
2(11.7 )(8.85 10 )
which yields
Vbi + V DS = 2.95 V
−4 2
−19
−14
16
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
From Example 11.5, we have V bi = 0.874 V,
so
V DS = 2.08 V
which is the near punch-through voltage. The
ideal punch-through voltage was
V DS = 4.9 V
_______________________________________
11.35
With a source-to-substrate voltage of 2 volts,
 2  (V + V SB ) 
x dO =  s bi

eN a


(
1/ 2
)
 2(11 .7 ) 8.85  10 −14 (0.902 + 2 ) 
=

1.6  10 −19 3  10 16


(
)(
1/ 2
)
or
11.34
( )(
(
)
 10 3 10 
Vbi = (0.0259 ) ln 
 = 0.902 V
2
 1.5 10 10

The zero-biased source-substrate junction
width is given by
19
 2 s V bi 
x dO = 

 eN a 
16
)
(
)
)(
(
(
)
1/ 2
)
(
)(
1/ 2
)
11.36
or
C ox =
−6
L D = 2.36 10 cm
so that
6L D = 0.142 10 −4 cm = 0.142  m
Now
x dO + 6 L D + x d = L
(
1/ 2
)
 2(11 .7 ) 8.85  10 −14 (0.902 + 5) 
=

1.6  10 −19 3  10 16


(
)(
(3.9)(8.85 10 −14 ) = 2.876 10 −7 F/cm 2
120 10 −8
eDI
VT =
C ox
Implant acceptor ions for a positive threshold
voltage shift.
VT (C ox ) (0.80 ) 2.876 10 −7
DI =
=
e
1.6 10 −19
12
−2
= 1.438  10 cm
_______________________________________
(
We have for V DS = 5 V,
 2  (V + V DS ) 
x d =  s bi

eN a


)
x d = 0.584 10 −4 cm = 0.584  m
Then
L = x dO + 6 L D + x d
= 0.354 + 0.142 + 0.584
or
L = 1.08  m
_______________________________________
x dO = 0.197 10 cm = 0.197  m
The Debye length is
 (11 .7 ) 8.85  10 −14 (0.0259 ) 
=

1.6  10 −19 3  10 16


)(
1/ 2
or
−4
 (kT e ) 
LD =  s

 eN a 
)
1/ 2
 2(11 .7 ) 8.85  10 −14 (0.902 + 5 + 2 ) 
=

1.6  10 −19 3  10 16


1/ 2
or
(
 2  (V + V DS + V SB ) 
x d =  s bi

eN a


1/ 2
 2(11 .7 ) 8.85  10 −14 (0.902 ) 
=

1.6  10 −19 3  10 16


(
x dO = 0.354 10 −4 cm = 0.354  m
We have 6 L D = 0.142  m from the previous
problem.
Now
1/ 2
)
or
x d = 0.505 10 −4 cm = 0.505  m
Then
L = 0.197 + 0.142 + 0.505
or
L = 0.844  m
_______________________________________
11.37
C ox =
)
(3.9)(8.85 10 −14 )
180 10 −8
= 1.9175  10 −7 F/cm 2
eDI
VT =
C ox
Implant donor ions for a negative threshold
voltage shift.
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
VT C ox
DI =
=
(0.60 )(1.9175 10 −7 )
V FB =  ms −
−19
e
1.6 10
11
−2
= 7.19  10 cm
_______________________________________
= +1.08 −
11.38
(a)  ms  −1.08 V
C ox =
V FB
 fn
(3.9)(8.85 10 −14 )
150 10 −8
= 2.301  10 −7 F/cm 2
Q
=  ms − ss
C ox
= −1.08 −
x dT
(
2.301 10
 6 10 15
10
 1.5 10
)
)(
= 3.797  10 −5 cm
 (max )
QSD
(
)(

 = 0.3341 V


1/ 2
)
)(
)
2
3.645 10
− 1.115 + 2(0.3341 )
2.301 10 −7
VTO = −0.2884 V
(b) For a positive threshold voltage shift, add
acceptor ions.
VT = VT − VTO = 0.50 − (− 0.2884 )
= 0.788 V
Then
(VT )C ox (0.788 ) 2.301 10 −7
DI =
=
e
1.6 10 −19
12
−2
= 1.13  10 cm
_______________________________________
(
C ox =
(3.9)(8.85 10 −14 )
180 10 −8
= 1.9175  10 −7 F/cm 2
)
)(
)(
1/ 2
)
)(
VTO
)
)
= 6.958  10 C/cm
 (max )
QSD
=−
+ V FB − 2 fn
C ox
2
6.958 10 −8
+ 0.9966 − 2(0.3653 )
1.9175 10 −7
VTO = −0.0969 V
(b) For a negative threshold voltage shift,
add donor ions.
VT = VT − VTO = −0.40 − (− 0.0969 )
= −0.3031 V
VT C ox
Then D I =
e
(
0.3031 ) 1.9175 10 −7
=
1.6 10 −19
= 3.63  10 11 cm −2
_______________________________________
=−
(
−8
11.39
(a)  ms  +1.08 V
(
−8
= 3.645  10 C/cm
 (max )
QSD
VTO =
+ V FB + 2 fp
C ox
=
(

 = 0.3653 V


 4(11 .7 ) 8.85  10 −14 (0.3653 ) 
=

1.6  10 −19 2  10 16


(
= 1.6 10 −19 6 10 15 3.797 10 −5
−8
1.9175 10
= +0.9966 V
 2 10 16
= (0.0259 ) ln 
10
 1.5 10
−7
= 1.6 10 −19 2 10 16 2.1744 10 −5
−7
 4(11 .7 ) 8.85  10 −14 (0.3341 ) 
=

1.6  10 −19 6  10 15


(
−19
11
= 2.1744  10 cm
 (max )
QSD
−19
10
 fp = (0.0259 ) ln 
x dT
(10 )(1.6 10 )
−5
(5 10 )(1.6 10 )
= −1.115 V
Qss
C ox
)
11.40
 4 10 15
(a)  fp = (0.0259 ) ln 
10
 1.5 10
(3.9) 8.85 10 −14
C ox =
80 10 −8
= 4.314  10 −7 F/cm 2
(
)
(
)

 = 0.3236 V


 4(11 .7 ) 8.85  10 −14 (0.3236 ) 
x dT = 

1.6  10 −19 4  10 15


(
)(
= 4.576  10 −5 cm
 (max )
QSD
(
)(
)(
1/ 2
)
= 1.6 10 −19 4 10 15 4.576 10 −5
= 2.929  10
−8
C/cm
2
)
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
VTO =
 (max )
QSD
C ox
Then
+ V FB + 2 fp
2.929 10 −8
− 1.25 + 2(0.3236 )
4.314 10 −7
= −0.5349 V
(b) For a positive threshold voltage shift, add
acceptor ions.
VT = VT − VTO = 0.40 − (− 0.5349 )
= 0.9349 V
(VT )C ox
Then D I =
e
(0.9349 ) 4.314 10 −7
=
1.6 10 −19
= 2.52  10 12 cm −2
(c) Add acceptor ions.
VT = VT − VTO = −0.40 − (− 0.5349 )
= 0.1349 V
(0.1349 ) 4.314 10 −7
Then DI =
1.6 10 −19
= 3.64  10 11 cm −2
_______________________________________
=
(
)
(
)
11.41
The total space charge width is greater than
x i , so from Chapter 10
VT =
2e s N a
C ox
 2
fp
+ V SB − 2 fp

Now

10 14
10
 1.5 10
 fp = (0.0259 ) ln 
and
C ox

 = 0.228 V


(3.9)(8.85 10 −14 )
=
2(1.6 10 )(11.7)(8.85 10 )(10 )
−19
−14
14
6.90  10 −8

 2(0.228 ) + VSB − 2(0.228 )
or

1
0.0443
3
0.0987
5
0.1385
_______________________________________
11.42
(a)

10 17
10
 1.5 10
 fn = (0.0259 ) ln 
and
x dT
VT = 0.0834 0.456 + VSB − 0.456


1/ 2
(
)

 = 0.407 V


 4(11 .7 ) 8.85  10 −14 (0.407 ) 
=

1.6  10 −19 10 17


(
= 1.026  10
)(
1/ 2
)
−5
cm
n poly on n-type  ms = −0.32 V
We have
 (max ) = 1.6  10 −19 10 17 1.026  10 −5
Q SD
+
(
)(
)(
)
= 1.64  10 −7 C/cm 2
Now
VTP = − 1.64 10 −7 − 1.6 10 −19 5 10 10


(
80  10
−8
(3.9)(8.85 10 −14 )
)(
)
− 0.32 − 2(0.407 )
or
VTP = −1.53 V (Enhancement PMOS)
(b) For VT = 0 , shift threshold voltage in
positive direction, so implant acceptor ions.
(VT )C ox
eDI
VT =
 DI =
C ox
e
so
(1.53 )(3.9) 8.85 10 −14
DI =
80  10 −8 1.6  10 −19
or
D I = 4.13 10 12 cm −2
_______________________________________
(
500 10 −8
= 6.90  10 −8 F/cm 2
Then
V T
=
VT (V)
V SB (V)
(
)(
)
)
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 11
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
11.43
The areal density of generated holes is
= 8 10 12 10 5 750 10 −8 = 6 10 12 cm −2
The equivalent surface charge trapped is
= (0.10 ) 6 10 12 = 6 10 11 cm −2
Then
Q
VT = − ss
C ox
(
)( )(
(
=−
)
)
(6 10 )(1.6 10 )(750 10 )
(3.9)(8.85 10 )
−19
11
−8
−14
or
VT = −2.09 V
_______________________________________
11.45
We have the areal density of generated holes
as
= ( g )( )(t ox )
where g is the generation rate and  is the
radiation dose. The equivalent charge trapped
is
= xg t ox
where x is the fraction of generated holes
trapped.
Then
 exg 
Q
exg t ox
(t )2
VT = − ss = −
= −
(ox t ox )  ox  ox
C ox
or
VT  − (t ox )
_______________________________________
2
11.44
The areal density of generated holes is
6  10 12 cm −2 . Now

(3.9) 8.85 10 −14
C ox = ox =
t ox
750  10 −8
(
)
= 4.6  10 −8 F/cm 2
Then
(
) (
)
Q ss
6  10 12 (x ) 1.6  10 −19
=−
C ox
4.6  10 −8
where x is the fraction of holes that may be
trapped. For VT = −0.50 V we find
x = 0.024  x = 2.4%
_______________________________________
VT = −
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 12
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
Chapter 12
Exercise Solutions
Ex 12.5
Ex 12.1
1
 N  D  x 
1 +  B  E  B 
 N E  D B  x E 
1
=
15
 5  10  10  0.80 
 
1 + 
18
 20  0.60 
 10

= 0.9967
_______________________________________
=
Ex 12.2
1
1
=
x 
 1.2 

cosh B  cosh
 10 .0 
 LB 
= 0.9928
_______________________________________
T =
1
x 
cosh B 
 LB 
From Example 12.5,
Then
1
0.9980 =
x
cosh B
 10
T =
Ex 12.6
1
=
1+
1
 − V BE
J r0
1+
 exp 
J s0
 2Vt
=




Ex 12.4
=
N
1 +  B
 NE
1
 D E

 D B
 x B 


 x E 
1
0.9950 =
 NB 
1 + 
(1.0 )(1.0 )
18 
 6  10 
 N B = 3.02 10 16 cm −3
_______________________________________
 − V BE
J r0
 exp 
J s0
 2Vt
0.9950 =
1

 − 0.65 
  exp 


 2(0.0259 ) 

 1.2 10
1 + 
−9
 1.804  10
= 0.99976
_______________________________________
−7



 1 
x B = (10 ) cosh−1 

 0.9980 
 x B = 0.633  m
_______________________________________
Ex 12.3
=
L B = 10  m.




1
 10 
 − V BE
1 +  −11   exp 
 10 
 2Vt
−8
 − V BE
exp 
 2Vt




 0.005025
−6
=
  −8  = 5.025 10
10
 

 10 −11 


or
1


V BE = 2Vt ln 

−6
 5.025  10 
= 2(0.0259 )(12 .201 )
V BE = 0.6320 V
_______________________________________
Ex 12.7
1
(a)  =
1+
=
 − V BE
J r0
 exp 
J s0
 2Vt
 5  10 − 9
1 + 
− 11
 2  10
= 0.99392




1

 − 0.55 
  exp 

 2(0.0259 ) 




Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 12
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
 =  T  = (0.9980 )(0.9980 )(0.99392 )
= 0.98995

0.98995
=
= 98 .5
Then  =
1 −  1 − 0.98995
1
(b)  =
−9
 5 10 
 − 0.65 
  exp 

1 + 
−11 
2

10
 2(0.0259 ) 


= 0.99911
 =  T  = (0.9980 )(0.9980 )(0.99911 )
= 0.99512

0.99512
=
=
= 204
1 −  1 − 0.99512
_______________________________________
Ex 12.8
The space charge width extending into the
base region is
p E 0 =
=
(
)

N N
Now Vbi = Vt ln  B 2 C
 ni
(
For VCB
(
(
)(
)
)
1/ 2





1/ 2
)
15
)

1/ 2
−6
= 5.18  10 cm = 0.0518  m
(
10
20
 0.23 
 exp 

 0.0259 
ex B2 0 N B (N C + N B )

2 s
NC
(1.6 10 )(0.80 10 )
=
2(11.7 )(8.85 10 )
(5 10 )(2 10 + 5 10 )

(2 10 )
−4 2
−19
−14
15
16
15

x dB = 9.956 10 −12 (0.6946 + 2)
For VCB = 10 V,
10 2




V pt = 643 V
(b) From Figure 7.15, BV  180 V
_______________________________________
Ex 12.11
(a) From Figure 7.15, BVCBO  125 V
 5 10 2 10
= (0.0259 ) ln 
2
 1.5 10 10
= 0.6946 V
= 2 V,
16
(1.5 10 )
(a) V pt =
1/ 2




 E g
ni2
 exp 
NE
 kT
16
 2 10 15
1


16
16
5  10 + 2  10 15
 5 10
= 9.956 10 −12 (Vbi + VCB )
2
= 1.618  10 4 cm −3
_______________________________________
)
(
)
N E = 10 20 cm −3 ,
 2(11 .7 ) 8.85  10 (Vbi + VCB )
=

1.6  10 −19
−14
(
ni2
1.5 10 10
=
= 2.25 cm −3
NE
10 20
From Figure 12.26, E g = 0.23 eV for
pE0 =
Ex 12.10


1
 2  (V + V BC )  N C

x dB =  s bi



(
)
e
N
N
+
N

B
C 
 B


(
Ex 12.9
Neglecting bandgap narrowing,
)

x dB = 9.956 10 −12 (0.6946 + 10 )
1/ 2
= 1.03  10 −5 cm = 0.103  m
Neglecting the B-E space charge width, we
find the neutral base width to be:
VCB = 2 V,
x B = x B 0 − x dB = 0.70 − 0.0518 = 0.6482  m
VCB = 10 V,
x B = 0.70 − 0.103 = 0.597  m
_______________________________________
(b) BVCEO =
BVCBO
=
125
= 25 V
125
_______________________________________
n

3
Ex 12.12
We find
 I (1 −  R ) + I B  F 
VCE (sat ) = Vt ln  C


  F I B − (1 −  F )I C  R 

(0.5)(1 − 0.05 ) + 0.05
0.992 
= (0.0259 ) ln 


(
)(
)
(
)(
)
0
.
992
0
.
05
−
1
−
0
.
992
0
.
5
0.05 

V CE (sat ) = 0.141 V
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 12
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
Test Your Understanding Solutions
Ex 12.13
1
We have f =
2 r C 
TYU 12.1
1
1
Or C =
=
2 r f 2 2.6 10 3 35 10 6
(
)(
)
C = 1.75 10 −12 F = 1.75 pF
_______________________________________
Ex 12.14
re =
 e = re  C je = (518 )(0.40  10 −12 )
= 2.072  10 −10 s = 207.2 ps
(
x B2
0.5 10 −4
=
2 Dn
2(25 )
)
2
= 5  10 −11 s = 50 ps
x
2.4  10 −4
 d = dc =
s
10 7
= 1  10 −12 s = 1 ps
Now
 ec =  e +  b +  d +  c
= 207.2 + 50 + 24 +1 = 282.2 ps
Then
1
1
fT =
=
2  ec 2 282 .2 10 −12
)
= 5.64  10 Hz = 564 MHz
8
Also
fT

=
= 2.25  10 4 cm −3
V 
(a) n B (0) = n BO exp  BE 
 Vt 
= 3.81  10 14 cm −3
(b) Using Equation (12.15a), we find
x
x 2 22
= 0.10
At x = B , B =
2
LB
10
Then
sinh(0.10 ) = 0.100167
(
 c = rc  C  = (20 )(0.05  10 −12 )
f =
2
sinh(0.20 ) = 0.20134
So
2.25 10 4
n(x = x B / 2) =
(0.20134 )
= 2.4  10 −11 s = 24 ps
(
)
 0.61 
= 2.25  10 4 exp 

 0.0259 
Vt
0.0259
=
= 518 
I E 50 10 −6
b =
n BO
(
n2
1.5 10 10
= i =
NB
10 16
564 10 3
100
= 5.64  10 6 Hz = 5.64 MHz
_______________________________________
)
  0.61  

 exp 
 − 1 (0.100167 ) − (0.100167 )
  0.0259  

or
n(x = x B / 2) = 1.8947 10 14 cm −3
(c) For an ideal linear function
3.8084  10 14
n B (x = x B / 2) =
2
= 1.9042  10 14 cm −3
1.8947
= 0.9950
Ratio =
1.9042
_______________________________________
TYU 12.2
p EO
(
n2
1.5 10 10
= i =
NE
10 18
)
2
V
(a) p E (0) = p EO exp  BE
 Vt
(
= 2.25  10 2 cm −3




)
 0.61 
= 2.25  10 2 exp 

 0.0259 
= 3.808  10 12 cm −3
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 12
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
(b) We have
 x 2
sinh E  = sinh(0.5) = 0.5211
 LE 
and
sinh(1) = 1.1752
Then, using Equation (12.21a),
3.808 10 12 (0.5211 )
p E (x  = x E 2) =
1.1752
= 1.689  10 12 cm −3
_______________________________________
(
)
TYU 12.3

 − x   

p C (x ) = p CO + p C = p CO 1 − exp 


 LC  

Then
 − x  

0.95 = 1 − exp 

 LC 
or
 − x  
 + x  
1
 = 0.05  exp 

exp 
 L  = 0.05

L
 C 
 C 
so
x 
 1 
= ln 
3
LC
 0.05 
_______________________________________
TYU 12.4
We have LB =
LE =
(20 )(10 −7 ) = 1.414 10 −3 cm
(8)(10 −8 ) = 2.828 10 −4 cm
Now
1
=
1+
=
 8  1.414 10 −3
 
−4

 20  2.828  10
 8  1.414  10 −3
 
 20  2.828  10 − 4


1
N 
1 +  B (0.2319 )
 NE 
which yields
NB
= 0.02167
NE
0.9950 =
or N B = 1.08 10 17 cm −3
_______________________________________
−1
or  = 0.99586
Then
 =  T  = (0.99586 )(0.9967 )(0.9967 )
= 0.9893
Now

0.9893
=
=
= 92 .4
1 −  1 − 0.9893
_______________________________________
TYU 12.6
1
x
cosh B
 LB
= 0.9984
T =



=
1
 0.80  10 − 4
cosh
−3
 1.414  10




Then
 = (0.9967 )(0.9984 )(0.9967 ) = 0.9918
Now
0.9918
=
= 121
1 − 0.9918
_______________________________________
TYU 12.7
(a) R =
(S 2)
 xB L
We find
 = e p N B
(
)
( )
= 1.6 10 −19 (400 ) 10 16
= 0.640 (  -cm)
−1
Then
 tanh(0.0707 ) 

 tanh(0.7072 ) 






 tanh(0.07072 ) 

 
 tanh(0.3536 ) 
p EO D E L B tanh(x B L B )

n BO D B L E tanh(x E L E )
1
N
1 +  B
 NE
or
TYU 12.5
  5  10 16
 = 1 + 
  5  10 18
R=
5  10 −4
(0.640 ) 0.80 10 − 4 10 10 − 4
(
)(
= 9.77 10 
3
(
)(
(b) V = IR = 5 10 −6 9.77 10 3
= 0.04883 V
or V = 48.83 mV
)
)
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 12
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
(c)
V (x = 0) 
exp  BE

Vt
J (x = 0)


=
J (x = S 2)
V BE (x = S 2) 
exp 

Vt


 V 
0.04883 
 = exp 
= exp 


V
 0.0259 
 t 
= 6.59
_______________________________________
TYU 12.8
V +VA
ro = CE
IC
or
V +VA
I C = CE
ro
Now
8 + 125
I C1 =
= 0.6650 mA
200
and
2 + 125
I C2 =
= 0.6350 mA
200
Then
I C = 0.6650 − 0.6350 = 0.030 mA
or
I C = 30  A
_______________________________________
TYU 12.9
1
x 
cosh B 
 LB 
For x B = 0.80  m,  T = 0.9984
For x B = 1  m,  T = 0.9975
So
0.9975   T  0.9984
(a)  T =
(b)  =  T (0.9967 )(0.9967 )
So that
0.9909    0.99182
Then

=
 109    121
1−
_______________________________________
TYU 12.10
V pt =
ex B2 N B (N C + N B )

2 s
NC
70 =
(1.6 10 )(0.7 10 )
2(11.7 )(8.85 10 )
(310 )(310

−4 2
−19
−14
16
16
+ NC
)
NC
−3
 N C = 5.81 10 cm
_______________________________________
15
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Chapter 12
12.1
Sketch
_______________________________________
Then

5 10 −3
−13
 1.738 10
 BE = (0.0259 ) ln 




12.2
Sketch
_______________________________________
  BE = 0.6237 V
_______________________________________
12.3
12.5
eDn ABE n B 0
(a) I S =
xB
=
(a)  =
(1.6 10 )(18)(5 10 )(4 10 )
−19
−5
3
−4
0.80 10
= 7.2  10 A
 
(b) I C = I S exp  BE 
 Vt 
 0.58 
(i) I C = 7.2  10 −15 exp 

 0.0259 
IB =
)
IB =
= 3.827  10 −5 A = 38 .27  A
(
)
 0.65 
(ii) I C = 7.2  10 −15 exp 

 0.0259 
= 5.710  10
(
)
A = 0.571 mA
 0.72 
exp 

 0.0259 
12.4
iC =

eDn ABE
 n B 0  exp  BE
xB
 Vt
2 10 −3 =
(1.6 10 )(22)A




−19
(
−4
)
 0.60 
 2  10 4 exp 

 0.0259 
 ABE = 1.975 10 −4 cm 2
(b) 5 10
−3
(1.6 10 )(22 )(1.975 10 )
=
−19
5  10 −3
−4
0.80 10 − 4
 

 2  10 4 exp  BE 
 0.0259 
 

= 1.738  10 −13 exp  BE 
0
.
0259


(
(
)
)
0.571
= 0.008695 mA
65 .67
= 8.695  A
8.519
= 0.1297 mA
65 .67
8.519
IE =
= 8.649 mA
0.9850
0.9940
= 165 .7
(c)  =
1 − 0.9940
(i) For I C = 38 .27  A,
IB =
38 .27
= 0.2310  A
165 .7
38 .27
IE =
= 38 .50  A
0.9940
(ii) For I C = 0.571 mA,
IB =
BE
0.80 10
=
0.571
= 0.5797 mA
0.9850
(iii) For I C = 8.519 mA,
= 8.519  10 −3 A = 8.519 mA
_______________________________________
(a)
IC
IE =
−4
(iii) I C = 7.2  10 −15
(b)
(i) For I C = 38 .27  A,
38.27
= 0.5828  A
 65.67
I
38 .27
IE = C =
= 38 .85  A
 0.9850
(ii) For I C = 0.571 mA,
−15
(

0.9850
=
= 65 .7
1 −  1 − 0.9850
IB =
0.571
= 0.003446 mA
165 .7
= 3.446  A
IE =
0.571
= 0.5744 mA
0.9940
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(iii) For I C = 8.519 mA,
IB =
8.519
= 0.05141 mA
165 .7
= 51 .41  A
8.519
IE =
= 8.570 mA
0.9940
_______________________________________
(b) For VCE = 3 V, I C = 0
(i) VCE = VCC − I C RC
0.2 = 3 − I C (10 ) , I C = 0.28 mA
(ii) For VCB = 0  VCE = V BE = 0.65 V
0.65 = 3 − I C (10 ) , I C = 0.235 mA
_______________________________________
12.9
12.6
I
0.625
(a)  = C =
= 148 .8
I B 0.0042
(a)
148 .8
=
=
= 0.9933
1 +  149 .8
I
0.625
IE = C =
= 0.6292 mA
 0.9933
I
1.254
= 0.9851
(b)  = C =
I E 1.273

0.9851
=
= 66 .0
1 −  1 − 0.9851
I
1.254
IB = C =
= 0.0190 mA

66
= 19.0  A
(ii) For VCB = 0  VCE = V BE = 0.65 V
0.65 = 3 − I C (25 ) , I C = 0.094 mA
(
ni2
1.5 10 10
=
NC
10 15
pC 0 =
)
2
)
2
= 2.25  10 5 cm −3
V
(b) n B (0) = n B 0 exp  BE
 Vt
(




)
 0.640 
= 1.125  10 4 exp 

 0.0259 
= 6.064  10 14 cm −3
V 
p E (0) = p E 0 exp  BE 
 Vt 
I E = (1 +  )I B = (151 )(0.065 ) = 9.815  A
_______________________________________
0.2 = 3 − I C (25 ) , I C = 0.112 mA
(
= 1.125  10 4 cm −3
150
= 0.99338
151
I C =  I B = (150 )(0.065 ) = 9.75  A
12.8
(a) For VCE = 3 V, I C = 0
(i) VCE = VCC − I C RC
2
ni2
1.5 10 10
=
NB
2 10 16
n B0 =
(c)  =
12.7
(c) For i B = 0.05 mA,
i C =  i B = (100 )(0.05 )
or
i C = 5 mA
We have
 CE = VCC − i C R = 10 − (5)(1)
or
 CE = 5 V
_______________________________________
)
= 2.8125  10 2 cm −3

=
(
ni2
1.5 10 10
=
NE
8 10 17
pE0 =
(
)
 0.640 
= 2.8125  10 2 exp 

 0.0259 
= 1.516  10 13 cm −3
_______________________________________
12.10
(a) n E 0 =
(
ni2
1.5 10 10
=
NE
5 10 17
)
= 4.5  10 2 cm −3
p B0 =
(
ni2
1.5 10 10
=
NB
10 16
= 2.25  10 4 cm −3
nC 0 =
(
ni2
1.5 10 10
=
NC
10 15
= 2.25  10 5 cm −3
2
)
2
)
2
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
At x = 0 ,
V 
(b) p B (0) = p B 0 exp  EB 
 Vt 
(
)
 0.615 
= 2.25  10 4 exp 

 0.0259 
d (n B )
dx
=
x =0
  V BE
exp 
 x B    Vt


L B sinh

 LB 
 
 − 1

 
x
 cosh B
 LB
 
 + 1
 
  V BE
exp 
 x B    Vt

L B sinh
 LB 
 
 − 1

 
− n BO
= 4.62  10 14 cm −3
V 
n E (0) = n E 0 exp  EB 
 Vt 
(
At x = x B ,
)
 0.615 
= 4.5  10 2 exp 

 0.0259 
d (n B )
dx
= 9.24  10 12 cm −3
_______________________________________
12.11
(a) n B 0
(
n2
1.5 10 10
= i =
NB
2 10 16
)
2
Now
n B (0) = (0.1)(N B ) = 2 10 15 cm −3
V 
= nB 0 exp  BE 
 Vt 
 2 10 15
Then V BE = (0.0259 ) ln 
4
 1.125 10
= 0.6709 V
(
ni2
1.5 10 10
=
NE
8 10 17




)
2
= 2.8125  10 2 cm −3
p E (0) = p E 0
V
exp  BE
 Vt
(

+


)
 0.6709 
= 2.8125  10 exp 

 0.0259 
2
= 5.0  10 13 cm −3
_______________________________________
12.12
We have
d (n B )
=
dx
  V BE
exp 
 x B    Vt

sinh
 LB 
n BO
x = xB
− n BO
x
+ cosh B
 LB



Taking the ratio
d (n B )
dx x = x B
= 1.125  10 4 cm −3
(b) p E 0 =
=
 
 − 1

 
 −1 
x −x 1
 x 
 cosh B
 −

 
cosh
 LB 
 LB  LB
 L B 
d (n B )
dx
x =0
  V BE  
x 
 − 1 + cosh B 
exp 
L 

  Vt  
 B
=
  V BE  
x 
 − 1  cosh B  + 1
exp 
L 

  Vt  
 B
1

x 
cosh B 
 LB 
(a) For
xB
= 0.1  Ratio = 0.9950
LB
(b) For
xB
= 1.0  Ratio = 0.648
LB
xB
= 10  Ratio = 9.08  10 −5
LB
_______________________________________
(c) For
12.13
In the base of the transistor, we have
d 2 (n B (x )) n B (x )
DB
−
=0
 BO
dx 2
or
d 2 (n B (x )) n B (x )
−
=0
dx 2
L2B
where LB = DB BO
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
The general solution to the differential
equation is of the form
 x 
−x

n B (x ) = A exp   + B exp 
 LB 
 LB 
From the boundary conditions, we have
n B (0) = A + B = n B (0) − n BO
 V
= n BO exp  BE
  Vt
 
 − 1

 
Also
x 
−x 
n B (x B ) = A exp  B  + B exp  B 
L
 B
 LB 
= − n BO
From the first boundary condition, we can
write
 V  
A = n BO exp  BE  − 1 − B
  Vt  
Substituting into the second boundary
condition, we find
 x 
 − x B 

B exp  B  − exp 
  L B 
 L B 
 V  
x 
= n BO exp  BE  − 1  exp  B  + n BO
  Vt  
 LB 
Solving for B, we find
 V  
x 
n BO exp  BE  − 1  exp  B  + n BO
V
  t  
 LB 
B=
x 
2 sinh B 
 LB 
We then find
 V  
 − xB 
 − n BO
− n BO exp  BE  − 1  exp 
  Vt  
 LB 
A=
x 
2 sinh B 
 LB 
_______________________________________
12.14
In the base of the pnp transistor, we have
d 2 (p B (x )) p B (x )
DB
−
=0
 BO
dx 2
or
d 2 (p B (x )) p B (x )
−
=0
dx
L2B
where LB = DB BO
The general solution is of the form
 x 
−x

p B (x ) = A exp   + B exp 
 LB 
 LB 
From the boundary conditions, we can write
p B (0 ) = A + B = p B (0) − p BO
 V
= p BO exp  EB
  Vt
 
 − 1

 
Also
 xB 
 − xB
 + B exp 
L
 B
 LB
p B (x B ) = A exp 



= − p BO
From the first boundary condition equation,
we find
 V  
A = p BO exp  EB  − 1 − B
  Vt  
Substituting into the second boundary
equation, we obtain
 V  
x 
p BO exp  EB  − 1  exp  B  + p BO
  Vt  
 LB 
B=
x 
2 sinh B 
 LB 
and then we obtain
 V  
 − xB 
 − p BO
− p BO exp  EB  − 1  exp 
  Vt  
 LB 
A=
x 
2 sinh B 
 LB 
Substituting the expressions for A and B into
the general solution and collecting terms, we
obtain
  V EB  
p BO
 − 1
p B (x ) =
exp 
 x B    Vt  

sinh
 LB 
x −x
 x 
 − sinh

 sinh B
 LB 
 L B 
_______________________________________
12.15
For the idealized straight line approximation,
the total minority carrier concentration is
given by
  V   x − x 

nB (x ) = nBO exp  BE    B

  Vt   xB 
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
The excess carrier concentration is
n B (x ) = n B (x ) − n BO
so for the idealized case, we can write

  V   x − x  

 − 1
n BO (x ) = n BO exp  BE    B
V
x

  t   B  

1
At x = x B , we have
2

x 
 1   V  

n BO  B  = n BO  exp  BE  − 1
2
2
V



   t  

For the actual case, we have
  V BE  
n BO
x 
 − 1
n B  B  =
exp 
 x B    Vt  
 2 

sinh
 LB 
 x
 sinh B
 2LB
(a) For

 x
 − sinh B

 2LB



xB
= 0.10 , we have
LB
 x 
sinh B  = 0.0500208
 2LB 
and
x 
sinh B  = 0.100167
 LB 
Then
x 
x 
n BO  B  − n B  B 
 2 
 2 
x


n BO  B 
2


=
  V BE
exp 
  Vt

 = 0.5211

=
  V BR
exp 
  Vt




 (0.50 − 0.4434 ) − 1.0 + 0.8868


V 
1
exp  BE  − 1
2
 Vt 
V 
Again assume that exp  BE   1 . Then the
 Vt 
ratio becomes
0.0566
=
= 0.1132  11 .32 %
0.50
_______________________________________
 p B 0 = 510 3 cm −3
p B0 =
ni2
NB
(
n2
1.5 10 10
 NB = i =
p B0
5 10 3
)
2
= 4.5  10 16 cm −3

 −1


V 
If we assume that exp  BE   1 , then we
 Vt 
find that the ratio is
0.00063
=
= 0.00126  0.126 %
0.50
x
(b) For B = 1.0 , we have
LB
 x
sinh B
 2LB
x 
sinh B  = 1.1752
 LB 
Then
x 
x
n BO  B  − n B  B
 2 
 2
x 
n BO  B 
 2 
12.16
(a) p B (x B ) = −5 10 3 = − p B 0

 (0.50 − 0.49937 ) − 1.0 + 0.99875


V
1
exp  BE
2
 Vt
and
 V EB 


 Vt 
 p (0) 
 V EB = Vt ln  B 
 p B0 
 10 15 

= (0.0259 ) ln 
3 
 5 10 
= 0.6740 V
(b) Using the linear approximation,
V
d (p B (x )) eDB p B 0
J = eDB

exp  EB
dx
xB
 Vt
p B (0)  p B 0 exp 
Since x B  L B , J
Then
 J
x =0
x = xB




Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(1.6  10 )(10 )(5  10 ) exp  0.674
−19
J =
0.8  10
3
= 20.0 A/cm 2
(c) Using Equation (12.15a),
− p B0
d (p B (0))
=
dx
x
L B  sinh B
 LB

  V
 exp  EB

  Vt
Now


 0.0259 
−4



 
 x −x
 x
 − 1 cosh B
 L  + cosh L

B


 B
 
d (p B (x ))
=
dx
J = eDB





eDB p B 0
x
L B sinh B
 LB




 x −x
 x 
  V  

 + cosh

 exp  EB  − 1 cosh B

 LB 
 L B 
  Vt  

For x = 0 , sinh(1) = 1.1752 , cosh(1) = 1.5431
cosh(0 ) = 1.0
Then
1.6  10 −19 (10 ) 5  10 3
J
=
x =0
10  10 − 4 (1.1752 )
(
(
) (
)
)
  0.6740  

 exp 
 − 1 (1.5431 ) + (1.0 )
  0.0259  

= 2.1042 A/cm 2
J
=
J
x = xB
(1.6 10 )(10 )(5 10 )
(10 10 )(1.1752 )
−19
3
−4
  0.6740  

 exp 
 − 1 (1.0 ) + (1.5431 )
  0.0259  

= 1.3636 A/cm 2
J
x = xB
(d) For part (b),
J
x = xB
= 1.0
J
x =0
For part (c),
J
x = xB
J
=
where LB = DB BO
The general solution is of the form
 x 
−x

n B (x ) = A exp   + B exp 
 LB 
 LB 
If x B  L B , then also x  L B , so that

n B (x )  A1 +

x
LB


x
 + B1 −

 LB



 x 

= ( A + B ) + ( A − B )
 LB 
which can be written as
 x 
n B (x ) = C + D 
 LB 
The boundary conditions are
 V  
n B (0) = C = n BO exp  BE  − 1
  Vt  
and
x =0
For x = x B ,
12.17
(a) For an npn transistor biased in saturation,
the excess minority carrier electron
concentration in the base is found from
d 2 (n B (x )) n B (x )
DB
−
=0
 BO
dx 2
or
d 2 (n B (x )) n B (x )
−
=0
dx 2
L2B
1.3636
= 0.648
2.1042
x =0
_______________________________________
 V
 xB 
 = n BO exp  BC

  Vt
 LB 
The coefficient D can be written as

L 
  V  
D =  B (n BO )exp  BC  − 1

 xB 
  Vt  
n B (x B ) = C + D
 
 − 1

 
  V  

− exp  BE  − 1 
V
  t   

The excess electron concentration is then
given by

x 
  V   
n B (x ) = n BO exp  BE  − 1  1 − 

  Vt    x B 
 V
+ exp  BC
  Vt
   x
 − 1  


   x B



Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(b) The electron diffusion current density is
d (n B (x ))
J n = eDB
dx

  V
= eDB n BO exp  BE

  Vt
   −1 
 − 1  



   x B 
 V
+ exp  BC
  Vt
   1
 − 1  


   x B



or
 V 
eD B n BO   V BE 
 − exp  BC 
exp 

 V 
xB
  Vt 
 t 
(c) The total excess charge in the base region
is
V
4.8611 10 11 = 5.4662 10 11 − exp  BC
 Vt
(
Jn = −
=
xB


  
x2
 − 1   x −


2x B
  
 V
+ exp  BC
  Vt
   x2
 − 1  


   2 x B

  V BE
exp 

  Vt
xB



 0


 
 − 1

 
12.18
(a) Using the linear approximation, we can
write
 V 
eD B n B 0   V BE 
 − exp  BC 
Jn =
exp 

 V 
x B   Vt 
 t 
n B0
(
= 4.5  10 cm
3
Then
125 =
)
2
   − x  
  exp 




   LC 
 V
= p C 0 exp  BC
  Vt
  − x   

 (− LC )exp 




  LC  0

V
= p C 0 LC exp  BC
 Vt
We find
pC 0 =




(
ni2
1.5 10 10
=
NC
10 15
)
2
= 2.25  10 5 cm −3
Then
Qp
(
)(
)
 0.643 
= 2.25  10 5 35  10 − 4 exp 

e
 0.0259 
−3
= 4.77  10 13 cm −2
_______________________________________
(1.6 10 )(25)(4.5 10 )
−19
= 9.56  10 10 cm −2
e
(d) In the collector,
 V
p C (x )  p C 0 exp  BC
  Vt
Now

Qp
= p C (x )dx 
e
0

  V  

+ exp  BC  − 1 
  Vt   

_______________________________________
n2
1.5 10 10
= i =
NB
5 10 16

Qn




which yields
− en BO x B
2
−4
3
= (0.1575 ) 5.466 10 11 + 6.052 10 10
0
QnB =
(4.5 10 )(0.7 10 )




2
  0.70 
 0.643 
 exp 
 + exp 

 0.0259 
  0.0259 
QnB = −e n B (x )dx

  V
= −en BO exp  BE

  Vt
)
V BC = (0.0259 ) ln 6.051 10 10
= 0.6430 V
(b) VCE (sat ) = V BE − V BC = 0.70 − 0.6430
= 0.057 V
(c) We have
Qn
V
n x  V 
= B 0 B exp  BE  + exp  BC
e
2   Vt 
 Vt




3
0.7 10 − 4
  0.70 
 V BC
 exp 
 − exp 
  0.0259 
 Vt




12.19
(b)
n BO =
and
(
ni2
1.5 10 10
=
NB
10 17
)
2
= 2.25 10 3 cm −3
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(
ni2
1.5 10 10
=
NC
7 10 15
At x = x B ,
p CO =
)
2
= 3.21 10 4 cm −3
V 
n B (x B ) = n BO exp  BC 
 Vt 
(
12.20
Low-injection limit is reached when
p C (0) = (0.10 )N C = (0.10 ) 5 10 14
or
p C (0) = 5  10 13 cm −3
We have
(
)
 0.565 
= 2.25  10 3 exp 

 0.0259 
or n B (x B ) = 6.7 10 12 cm −3
At x  = 0 ,
V
p C (0) = p CO exp  BC
 Vt
(




( )(
(
or x p1 = 1.23  10
)
)
)

1

 7 10 15 + 10 17




1/ 2
cm
From the B-E space charge region
 10 19 10 17 
Vbi 2 = (0.0259 ) ln 

2
 1.5 10 10 
= 0.933 V
Then
 2(11 .7 ) 8.85 10 −14 (0.933 + 2)
x p2 = 
1.6 10 −19

( )( )
(
)
(
 10 19
  17
 10
)

1
 19
 10 + 10 17




1/ 2
or
x p 2 = 1.94  10 −5 cm
Now
x B = x BO − x p1 − x p 2
= 1.20 − 0.0123 − 0.194
or
= 4.5 10 5 cm −3
or
or p C (0) = 9.56 10 13 cm −3
(c) From the B-C space charge region,
 10 17 7  10 15 
Vbi1 = (0.0259 ) ln 

2
 1.5  10 10

= 0.745 V
Then
 2(11.7 ) 8.85 10 −14 (0.745 − 0.565 )
x p1 = 
1.6 10 −19

−6
2
V 
p C (0) = p CO exp  CB 
 Vt 
)
 7 10 15
 
17
 10
)
Also
 0.565 
= 3.21  10 4 exp 

 0.0259 
(
(
ni2
1.5 10 10
=
NC
5 10 14
p CO =
)
x B = 0.994  m
_______________________________________
 p (0) 
VCB = Vt ln  C 
 p CO 
 5 10 13 

= (0.0259 ) ln 
5 
 4.5 10 
or
V CB = 0.48 V
_______________________________________
12.21
(a)
(i)  =
1+
(ii)  T =
(iii)  =
1
=
I pE
I nE
1
= 0.99305
0.0035
1+
0.50
I nC 0.495
=
= 0.990
I nE
0.50
I nE + I pE
I nE + I R + I pE
0.50 + 0.0035
= 0.990167
0.50 + 0.005 + 0.0035
(iv)  =  T  = (0.99305 )(0.990 )(0.990167 )
= 0.97345

0.97345
=
= 36 .7
(v)  =
1 −  1 − 0.97345

120
(b) For  = 120   =
=
1 +  121
 = 0.991736
Then  =  T =  = 0.997238
=
I nC
I
= nC
I nE 0.50
= 0.4986 mA
 T = 0.997238 =
 I nC
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
 = 0.997238 =
1+
1
=
I pE
I nE
1+
1
I pE
(b)  =
0.50
 I pE = 0.00138 mA = 1.38  A
=
I nE + I R + I pE
0.50 + 0.00138
0.50 + I R + 0.00138
 I R = 0.00139 mA = 1.39  A
_______________________________________
0.997238 =
12.22
(a) Using Equation (12.37)
eDB p B 0 ABE
I nC =
LB
   V EB  

 − 1
 exp 


   Vt  

1

+





xB 
 sinh x B 


tanh
L 
 L 

 B
 B 

p B0
(
n2
1.5 10 10
= i =
NB
10 16
)
2
J nE =
(10 )(5 10 −7 )
= 2.236  10 −3 cm
We find
 0.70 10 −4
x 
sinh B  = sinh
−3
 LB 
 2.236 10
) (
eD B n BO
LB
We find that
n BO =




)(
   V BE  

 − 1
 exp 


   Vt  

1
+


 xB  
 tanh x B 

sinh
L 


 B
 LB  

(
ni2
1.5 10 10
=
NB
5 10 16
and
= 0.03131
 0.70 10 −4 
 xB 

 = tanh
tanh
 2.236 10 −3 
 LB 


= 0.03130
Then
1.6 10 −19 (10 ) 2.25 10 4 5 10 −4
I nC =
2.236 10 −3
   0.550  

 − 1
 exp 

1 
   0.0259  

+

0.03131
0.03130 





−4
I nC = 4.29 10 A = 0.429 mA
(
16
12.23
(a) We have
= 2.25  10 4 cm −3
L B = D B B 0 =
1
= 0.95969
 10
 15  0.7 


1+ 
 

17 
 5 10  10  0.5 
1
1
T =
=
x 
 0.70 10 − 4 

cosh B  cosh
−3 
 LB 
 2.236 10 
= 0.99951
 =  T  = (0.95969 )(0.99951 )(0.995 )
= 0.95442

0.95442
=
=
= 20 .94
1 −  1 − 0.95442
Then
I C =  I B = (20 .94 )(0.80 ) = 16 .75  A
(c) I C = I E = (0.95442 )(125 ) = 119 .3  A
_______________________________________
=
I nE + I pE
Now
1
N B DE x B
1+


N E DB x E
L B = D B BO =
)
2
= 4.5 10 3 cm −3
(15)(5 10 −8 )
= 8.660  10 −4 cm
Then
J nE =
)
(1.6 10 )(15)(4.5 10 )
−19
3
−4
8.66 10


 0.60 

 exp 

1


 0.0259 

+

0
.
70
0
.
70



 tanh
 sinh


 8.66 
 8.66  
or
J nE = 1.779 A/cm 2
We also have
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
J pE =
eDE p EO
LE
  V BE
exp 
  Vt
(
)
Now
p EO
n2
1.5 10 10
= i =
NE
10 18
and
L E = D E EO =
 
1
 − 1 

x
 
tanh E
 LE



or
2
= 2.25 10 2 cm −3
(8)(10 −8 )
= 2.828  10 −4 cm
Then
J pE =
(1.6 10 )(8)(2.25 10 )
−19
2
2.828 10 − 4
  0.60  
 exp 
 − 1 
  0.0259  
1
 0.8 
tanh

 2.828 
or
J pE = 0.04251 A/cm 2
We can find
eDB n BO
J nC =
LB
   V BE  

 − 1
 exp 


   Vt  

1

+

 xB  
 sinh x B 


tanh
L 
 L 

 B
 B 

=
(1.6 10 )(15)(4.5 10 )
−19
−4
   0.60  

 − 1
 exp 

1
   0.0259  


+

 0.7  
 sinh 0.7 
tanh


 8.66 
 8.66  

or
2
The recombination current density is
V 
J R = J ro exp  BE 
 2Vt 
 0.60 
= 3  10 −8 exp 

 2(0.0259 ) 
or
J R = 3.218 10 −3 A/cm 2
(
)
 = 0.9767
We also find
J
1.773
 T = nC =
J nE 1.779
or
 T = 0.9966
Also
J nE + J pE
=
J nE + J R + J pE
=
1.779 + 0.04251
1.779 + 0.003218 + 0.04251
or
 = 0.9982
Then
 =  T  = (0.9767 )(0.9966 )(0.9982 )
or
 = 0.9716
Now

0.9716
=
=
1 −  1 − 0.9716
or
 = 34.2
_______________________________________
3
8.66 10
J nC = 1.773 A/cm
(b) Using the calculated current densities, we
find
J nE
1.779
 =
=
J nE + J pE 1.779 + 0.04251
12.24
(a) We have
=
N D x
1
 1− B  E  B
N B DE x B
N E DB x E
1+


N E DB x E
or
  1− K 
NB
NE
(i) Now
2 N BO
K
NE
 (B )
=
N
 ( A)
1 − BO  K
NE
1−
 2 N BO
 N

 1 −
 K 1 + BO  K 
NE
NE



2 N BO
N BO
 1−
K +
K
Ne
NE
or finally
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
N
D x
 (B )
= 1 − BO  E  B
 ( A)
N E DB x E
(c) Neglect any change in space charge width.
1
=
 − V BE 
J

1 + rO  exp 

J sO
 2Vt 
(ii)
We have
  1−
N B DE x B
x


= 1− K  B
N E DB x E
xE
 1−
Then
 (B )
=
 ( A)

x 
x
 1 − K   BO 1 + K   BO
2
x
xE
E 

x
x
 1 − K   BO + K   BO
2x E
xE



x BO
2x E
J sO  n BO =
or finally
 T (C )
3 x 
 1 +  BO 
 T ( A)
8  LB 
2
K
J sOA
ni2
NB
2 N BO K N BO K
N K
 (B )
 1−
+
= 1 − BO
 ( A)
C
C
C
 (B )
= 1−
 ( A)
 −V
J rO exp  BE
 2Vt
 eDB n BO

 xB







(ii) We find
 (C )
 1+
 ( A)
 −V
J rO exp  BE
 2Vt




 eDB n BO 


 xB 
(d) Device C has the largest  . The emitter
injection efficiency, base transport, and
recombination factors all increase.
_______________________________________
 1  x  2  1  x  2 
 1 −  BO  1 +  BO  
 8  L B   2  L B  



2
J sOB
+
Then finally
 1 x 2 
BO
1 −

 8  L B  


=
 1 x 2 
BO
1 −

 2  L B  


1  x BO 
1  x BO 

 + 

8  LB 
2  LB 
K
Now
so
N
D x
 (C )
= 1 + BO  E  BO
 ( A)
N E DB 2x E
(b) (i) We find
 T (B )
=1
 T ( A)
(ii)
2
 1  (x

BO 2 )  
1 − 




 T (C )  2  L B  
=
 T ( A)
 1 x 2 
1 −  BO 
 2  L B  


K
J sOB 
K 
K 
1 +

 1 −



K
 J sOB  J sOA 
1−
J sOA
1−
 1−
or
 1−

K
 = 1−

J sO

(i)
x BO
2x E
 (C )
=
x
 ( A)
1 − K   BO
xE
1− K 
= 1+ K  
 − V BE
J rO
 exp 
J sO
 2Vt
12.25
(a) We have
=
2
1
1
=
N B DE x B
N
1+


1+ K  B
N E DB x E
NE
or
NB
NE
Then
  1− K 
(i)
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
NB
2 N EO
 (B )
=
N
 ( A)
1− K  B
N EO
1− K 

N B 
N
1 + K  B
 1 − K 


2 N EO 
N EO

NB
N
 1− K 
+K B
2 N EO
N EO
= 1+ K 
or
NB
2 N EO
N B DE x B
 (B )
= 1+


 ( A)
2 N EO D B x E
 1−
(ii) Now
1
=
1+ K 
xB
xE
 1− K 
1− K 
J sO 
so
xB
(x EO 2)
= 1− K 

x
1 + K   B

x EO

x
+ K  B
x EO
xB
x EO
or finally
N D
x
 (C )
= 1− B  E  B
 ( A)
N E D B x EO
(b) We have
1  xB
2  LB
 T = 1 − 
(i)
 T (B )
=1
 T ( A)
(ii)
or
x
1− K  B
x EO

2x
 1 − K   B
x EO

x
 1 − 2K   B
x EO
 T (C )
=1
 T ( A)



2
K
K
+
J sOB J sOA
Now
xB
xE
Then
 (C )
=
 ( A)




(c) Neglect any change in space charge width.
1
=
 − V BE 
J

1 + rO exp 

J sO
 2Vt 
1
K
=
 1−
K
J sO
1+
J sO
(i)
K
1−
J sOB 
 (B )
K 
K 
1 +

=
 1 −
 J

K
 ( A)
J
sOB 
sOA 

1−
J sOA




1
N E xE
 (B )
= 1 − K (2 N EO ) + K (N EO )
 ( A)
 (B )
= 1 − K   N EO
 ( A)
Recombination factor decreases
(ii) We have
 (C )
x 
= 1 − K  EO  + K (x EO )
 ( A)
 2 
or
 (C )
1
= 1 + K   x EO
 ( A)
2
Recombination factor increases
_______________________________________
12.26
(b)
n BO =
(
ni2
1.5 10 10
=
NB
10 17
)
2
= 2.25 10 3 cm −3
Then
V
n B (0) = n BO exp  BC
 Vt
(




)
 0.6 
= 2.25  10 3 exp 

 0.0259 
= 2.59  10 13 cm −3
Now
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
J nC =
eDB n B (0)
xB
=
(b) For D E = D B , L E = L B , x E = x B , we
have
1
1
=
=
1 + ( p EO n BO ) 1 + (N B N E )
(1.6 10 )(20 )(2.59 10 )
−19
13
10
−4
or
J nC = 0.828 A/cm
Assuming a long collector
V
eDC p nO
J pC =
exp  BC
LC
 Vt
where
p cO =
(
ni2
1.5 10 10
=
NC
10 16
and
LC = DC CO =




)
2
= 2.25 10 4 cm −3
J pC =
(15)(2 10 −7 )
(1.6 10 )(15)(2.25 10 )
−19
4
1.732 10 −3
 0 .6 
 exp 

 0.0259 
or
J pC = 0.359 A/cm 2
The collector current is
I C = J nC + J pC  A
(
)
(
= (0.828 + 0.359 ) 10 −3
)
1− 
NB NE


0.01
0.10
1.0
10.0
0.990
0.909
0.50
0.0909
99
9.99
1.0
0.10
If N B N E  0.01 , the emitter injection
efficiency is probably not the limiting factor.
If, however, N B N E  0.01 , then the current
gain is small and the emitter injection
efficiency is probably the limiting factor.
_______________________________________
12.28
We have
or
I C = 1.19 mA
The emitter current is
I E = J nC  A = (0.828 ) 10 −3
or
I E = 0.828 mA
_______________________________________
(

(c) For x B L B  0.10 , the value of  is
unreasonably large, which means that the
base transport factor is not the limiting factor.
For x B L B  1.0 , the value of  is very
small, which means that the base transport
factor will probably be the limiting factor.
= 1.732  10 −3 cm
Then
=
and
2
)
12.27
(a)
J sO =
Now
n BO =
eDB n BO
L B tanh(x B L B )
(
ni2
1.5 10 10
=
NB
10 17
and
L B = D B BO =
)
2
= 2.25 10 3 cm −3
(25 )(10 −7 )
= 15 .8  10 −4 cm
T
1
T =
and  =
cosh(x B L B )
1−T
x B LB
T

0.01
0.10
1.0
10.0
0.99995
0.995
0.648
0.0000908
19,999
199
1.84
0
Then
J sO =
(1.6 10 )(25 )(2.25 10 )
(15.8 10 ) tanh(0.7 15.8)
−19
−4
or
J sO = 1.287 10 −10 A/cm 2
Now
3
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
1
=
1+
 − V BE
J rO
exp 
J sO
 2Vt
0.99656 =




 N E = 4.61 10 18 cm −3
_______________________________________
1
=
1+
 − V BE 
2 10 −9
exp 

1.287 10 −10
 2(0.0259 ) 
12.30
(a) We have J rO = 5 10 −8 A/cm 2
We find
or
(a)
1
=
 − V BE 
1 + (15 .54 ) exp 

 0.0518 
n BO =
L B = D B BO =

=


0.7535
0.99316
0.999855
3.06
145
6,902
J sO =

1+ 
 =  T 
=
J sO =
L B = D B B 0 =
 − V BE
J rO
exp 
J sO
 2Vt




= 0.55 V.
 = 0.995
=
1
 0.80 10 − 4
cosh
−3
 2.145 10





0.993377
=
= 0.99656
 T  (0.99930 )(0.9975 )
1
N B DE x B
1+


N E DB x E
1
For T = 300 K and V BE
Then



B
1.139  10 −11
tanh(x B L B )
1+
(23)(2 10 −7 )
=
3
−4
=
= 2.145  10 −3 cm
1
x
cosh B
 LB
= 0.99930
−19
We have
150
= 0.993377
151
Let x B = 0.80  m
(1.6 10 )(25 )(4.5 10 )
(15.8 10 ) tanh(x L )
B
  T = 0.995867
 =
(25 )(10 −7 )
or
0.993377 = ( T )(0.9975 )
T =
= 4.5 10 3 cm −3
eDB n BO
L B tanh(x B L B )
=
(c) If V BE  0.4 V, the recombination factor is
likely the limiting factor in the current gain.
_______________________________________
=
2
Then
0.20
0.40
0.60
12.29
)
= 15 .8  10 −4 cm
1− 
Now
V BE
=
(
ni2
1.5 10 10
=
NB
5 10 16
and
and
(b)
Now
1
 2  10  8  0.80 
 
1 + 


 N E  23  0.35 
16
1
 5  10 −8 
x 
− 0.55 
  tanh B   exp 
1 + 

−11 
L 
0
1
.
139

10
 .0518 
 B


which yields
xB
= 0.0468
LB
or
x B = (0.0468 )(15 .8) = 0.739  m
(b) For T = 400 K and J rO = 5 10 −8 A/cm 2 ,
− Eg


exp 

n BO (400 )  400 
 (0.0259 )(400 300 ) 
=
 
n BO (300 )  300 
 − Eg 
exp 

 0.0259 
For E g = 1.12 eV,
3
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
n BO (400 )
= 1.175 10 5
n BO (300 )
or
(
(b)
(i) ro =
)(
n BO (400 ) = 1.175 10 5 4.5 10 3
)
= 5.29  10 8 cm −3
Then
J sO =
(1.6 10 )(25 )(5.29 10 )
(15.8 10 ) tanh(0.739 15.8)
−19
8
−4
or
J sO = 2.865 10 −5 A/cm 2
Finally
1
=
−8


5 10
− 0.55
1+
 exp 

−5
2.865 10
 2(0.0259 )(400 300 ) 
or
 = 0.9999994
_______________________________________
12.31
Plot
_______________________________________
12.32
Plot
_______________________________________
12.33
Plot
_______________________________________
12.34
Plot
_______________________________________
12.35
(a) I C =
(V + V )
1
(VCE + V A )  ro = CE A
ro
IC
2 + 120
= 101 .67 k 
1.2
1
1
=
= 0.00984 (k  ) −1
(ii) g o =
ro 101 .67
(i) ro =
= 9.84 10 −6 ()
(iii) I C =
−1
4 + 120
= 1.22 mA
101 .667
VCE + V A 2 + 160
=
= 648 k 
IC
0.25
(ii) g o =
1
1
=
= 0.00154 (k  ) −1
ro 648
= 1.54 10 −6 ()
−1
4 + 160
= 0.253 mA
648
_______________________________________
(iii) I C =
12.36
V EC
V EC 5 − 2
 I C =
=
I C
ro
180
I C = 0.01667 mA = 16 .67  A
_______________________________________
ro =
12.37
x dB


1
 2  (V + VCB )  N C

=  s bi



(
)
e
N
N
+
N

B
C 
 B


(
1/ 2
)
 2(11 .7 ) 8.85  10 −14 (Vbi + VCB )
=

1.6  10 −19
(
 2 10 15
1


16
15
2 10 + 2  10 16
 2 10
= 5.8832 10 −11
Now
N N
Vbi = Vt ln  B 2 C
 ni
(
)(V
bi




(

+ VCB )
)(





1/ 2
)
1/ 2
)
 2 10 15 2 10 16 
= (0.0259 ) ln 

2
 1.5 10 10

= 0.6709 V
(i) For VCB = 4 V, x dB = 0.1658  m
(ii) For VCB = 8 V, x dB = 0.2259  m
(
)
(iii) For VCB = 12 V, x dB = 0.2730  m
Neglecting the B-E space charge width,
(i) For VCB = 4 V,
x B = 0.85 − 0.1658 = 0.6842  m
(ii) For VCB = 8 V,
x B = 0.85 − 0.2259 = 0.6241  m
(iii) For VCB = 12 V,
x B = 0.85 − 0.2730 = 0.5770  m
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
x B = x BO − x p
Now
V 
eDB n B 0
JC =
exp  BE 
xB
 Vt 
where
n B0
(
n2
1.5 10 10
= i =
NB
2 10 16
Now
)
2
= 1.125  10 4 cm −3
so
JC =
(1.6 10 )(25 )(1.125 10 ) exp  0.650
−19
4


 0.0259 
xB
3.5686  10
=
xB
−3
A/cm
(
(
 5 10 15
 
16
 3 10
(ii)For VCB = 8 V, J C = 57 .18 A/cm 2
J C
JC
=
(b)
VCE VCE + V A
(
n2
1.5 10 10
= i =
NB
3 10 16
)
2
= 7.5 10 3 cm −3
and
V
n B (0) = n BO exp  BE
 Vt
(




)
 0 .7 
= 7.5  10 3 exp 

 0.0259 
or
n B (0) = 4.10 10 15 cm −3
We have
dnB eDB n B (0)
J = eDB
=
dx
xB
=
(1.6 10 )(20 )(4.10 10 )
−19
15
xB
or
1.312  10 −2
A/cm 2
xB
Neglecting the space charge width at the B-E
junction, we have
J=



1




 N B + N C 
or
(

1

 5 10 15 + 3 10 16

)

x p = 6.163 10 −11 (Vbi + VCB )
1/ 2
For VCB = 5 V, x p = 0.1875  m
61.85 − 52.16
52.16
=
12 − 4
4 + 0.650 + V A
 V A = 38 .4 V
_______________________________________
n BO
)
)
)
1/ 2
 2(11 .7 ) 8.85  10 −14 (Vbi + VCB )
=
1.6  10 −19

2
(iii)For VCB = 12 V, J C = 61 .85 A/cm 2
(
 2  (V + VCB )  N C

x p =  s bi
e

 NB
(i)For VCB = 4 V, J C = 52 .16 A/cm 2
12.38
We find
)(
 3 10 16 5 10 15
Vbi = (0.0259 ) ln 
2
 1.5 10 10
or
Vbi = 0.705 V
Also
For VCB = 10 V, x p = 0.2569  m
(a) For x BO = 1.0  m
For VCB = 5 V,
x B = 1.0 − 0.1875 = 0.8125  m
Then
1.312 10 −2
J=
= 161 .5 A/cm 2
0.8125 10 − 4
For VCB = 10 V,
x B = 1.0 − 0.2569 = 0.7431  m
and
1.312 10 −2
J=
= 176 .6 A/cm 2
−4
0.7431 10
We can write
J
(VCE + V A )
J=
VCE
where
J
J
176 .6 − 161 .5
=
=
VCE VCB
10 − 5
= 3.02 A/cm 2 /V
Then
161 .5 = (3.02 )(5.7 + V A )
which yields
V A = 47 .8 V



1/ 2
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(b) For x BO = 0.80  m
For VCB = 5 V,
x B = 0.80 − 0.1875 = 0.6125  m
Then
1.312 10 −2
J=
= 214 .2 A/cm 2
0.6125 10 − 4
For VCB = 10 V,
x B = 0.80 − 0.2569 = 0.5431  m
and
1.312 10 −2
J=
= 241 .6 A/cm 2
0.5431 10 − 4
Now
J
J
241 .6 − 214 .2
=
=
VCE VCB
10 − 5
= 5.48 A/cm 2 /V
We can write
J
(VCE + V A )
J=
VCE
or
12.39
(a)
x dB


1
 2  (V + V BC )  N C

=  s bi



e

 N B (N B + N C )  


(
)
 2(11 .7 ) 8.85  10 −14 (Vbi + V BC )
=

1.6  10 −19
10 15
1
  16  15
10 + 10 16
10
(
(
)

= 1.1766 10 −10 (Vbi + VBC )
Now
N N 
Vbi = Vt ln  B 2 C 
 ni 





)
( )( )
(
)
 10 15 10 16 
= (0.0259 ) ln 

2
 1.5  10 10 
= 0.6350 V
For V BC = 1 V, x dB = 0.1387  m
For V BC = 5 V, x dB = 0.2575  m
Then x dB = 0.2575 − 0.1387 = 0.1188  m
(c) For x BO = 0.60  m
For VCB = 5 V,
(b) I C =
x B = 0.60 − 0.2569 = 0.3431  m
and
1.312 10 −2
J=
= 382 .4 A/cm 2
−4
0.3431 10
Now
J
J
382 .4 − 318 .1
=
=
VCE VCB
10 − 5
= 12.86 A/cm 2 /V
We can write
J
(VCE + V A )
J=
VCE
or
318 .1 = (12 .86 )(5.7 + V A )
which yields
V A = 19 .0 V
_______________________________________
1/ 2
1/ 2
214 .2 = (5.48 )(5.7 + V A )
which yields
V A = 33 .4 V
x B = 0.60 − 0.1875 = 0.4125  m
Then
1.312 10 −2
J=
= 318 .1 A/cm 2
0.4125 10 − 4
For VCB = 10 V,
1/ 2
V 
eDB p B 0 ABE
exp  EB 
xB
 Vt 
We find
p B0 =
(
ni2
1.5 10 10
=
NB
10 16
)
2
= 2.25  10 4 cm −3
Then
IC =
(1.6 10 )(10 )(2.25 10 )(10 )
−19
4
−4
xB
 0.625 
 exp 

 0.0259 
=
1.0874  10 −7
A
xB
For V BC = 1 V, I C =
1.0874  10 −7
(0.70 − 0.1387 )10 − 4
= 1.937  10 −3 A = 1.937 mA
For V BC = 5 V, I C =
1.0874  10 −7
(0.70 − 0.2575 )10 − 4
= 2.456  10 −3 A = 2.456 mA
Then
I C = 2.456 − 1.937 = 0.519 mA
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
I C
IC
=
V BC V EC + V A
(c)
12.41
(a) We have
=
0.519  10 −3
1.937  10 −3
=
5 −1
1 + 0.625 + V A
V A = 13 .3 V
For x B = x E , L B = L E , D B = D E , we obtain
V + V A 1.625 + 13.3
(d) ro = EC
=
IC
1.937 10 −3
 =
= 7.705 10 3  = 7.705 k 
_______________________________________
12.40
Let x B = x E , L B = L E , D B = D E
Then the emitter injection efficiency is
1
1
=
=
2
p EO
n
N
1+
1 + iE  2B
n BO
N E niB
12.42
(a)

0.980
0.909
0.8999
8.99
10 19
0.990
0.980
49 .3
10 20
0.9990
0.989
90 .2
(b) Taking into account bandgap narrowing,
we find


10 17

0
0 .5
0.495
0.98
18
25
0.792
0.784
3.63
10 19
80
0.820
0.812
4.32
230
0.122
0.121
0.14
10
10
20
10 19
 0.080 
exp 

 0.0259 
( )(Length) = (S 2)
Area
 (x B L )
(S 2)
=
(e p N B )(x B L )
(i) R =
0.495
E g (meV)
NB
which yields
N B = 1.83 10 15 cm −3
(b) Neglecting bandgap narrowing, we would
have
1
1
=
 0.996 =
NB
NB
1+
1 + 19
NE
10
which yields
N B = 4.02 10 16 cm −3
_______________________________________
0 .5
NE
1
1+
18
10
For N E = 10 19 cm −3 , we have E g = 80 meV
0.996 =
For no bandgap narrowing, n iE2 = n i2 .
With bandgap narrowing,
 E g 

niE2 = ni2 exp 

 kT 
Then
1
 =
 E g 
N

1 + B exp 

NE
 kT 
(a) No bandgap narrowing, so E g = 0
10 17
1
 E g 
NB

1+
exp 

NE
 kT 
Then
where n iB2 = n i2 .
 =  T  =  (0.995 )2 . We find
NE


1
p EO D E L B tanh(x B L B )
1+

n BO D B L E tanh(x E L E )
_______________________________________
=
1
(1.6 10 )(250 )(2 10 )
−19

16
5  10 −4
(0.65  10 )(25  10 )
−4
−4
R = 3.846 10 3  = 3.846 k 
(
)(
(ii) V = (I B 2)R = 5 10 −6 3.846 10 3
= 0.01923 V
)
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
V (x = S 2 ) 
exp  BE

n B (x = S 2 )
 0.0259 
=
(iii)
n B (x = 0)
V (x = 0 ) 
exp  BE

 0.0259 
Now
R=
545 .8 =
 0.60 
exp 

 0.0259 
=
 0.60 + V 
exp 

 0.0259 
(b)
1
(1.6 10 )(250 )(2 10 )
16
1.5  10 −4
0.65  10 − 4 25  10 − 4
(
)(
)
= 1.154 10  = 1.154 k 
3
(
)(
16
(S 2)
(0.65 10 )(25 10 )
−4
−4
12.44
(a)
Then
n B (x = S 2)
= 0.476
n B (x = 0)

−19
 S = 1.42  10 −4 cm = 1.42  m
_______________________________________
 − 0.01923 
= exp 

 0.0259 
−19
1
(1.6 10 )(250 )(2 10 )

 − V 
= exp 

 0.0259 
(i) R =
(S 2)
1

e p N B x B L
(ii) V = (I B 2)R = 5 10 −6 1.154 10 3
= 0.005769 V
 − V 
n B (x = S 2)
(iii)
= exp 

n B (x = 0)
 Vt 
)
 − 0.005769 
= exp 

 0.0259 
 − ax 

N B = N B (0) exp 
 xB 
where
 N (0) 
a = ln  B
0
 N B (x B ) 
and is a constant. In thermal equilibrium
dN B
J p = e p N B  − eD p
=0
dx
so that
D p 1 dN B  kT  1 dN B
=


=  

 p N B dx  e  N B dx
which becomes
−a
 − ax 
 kT  1
  exp 

=
 N B (0)  

e
N
x

 B
 B 
 xB 
 kT   − a  1
 
=
 NB
  
 e   xB  N B
Then
n B (x = S 2)
= 0.80
n B (x = 0)
_______________________________________
or
12.43
(b) The electric field is in the negative
x-direction which will aid the flow of
minority carrier electrons across the base.
 − V 
n B (x = S 2)
= 0.90 = exp 

n B (x = 0)
 Vt 
Then
(c)
 1 
 1 
V = Vt ln 
 = (0.0259 ) ln 

 0.90 
 0.90 
V = 0.002729 V = (I B 2)R
(
)
= 5 10 −6 R
R = 545 .8 
 a  kT 

 = −

 x B  e 
which is a constant.
dn
dx
Assuming no recombination in the base, J n
will be a constant across the base. Then
J
dn   n 
dn   
  n = n =
+ 
+ n 

dx  Dn 
eDn dx
 Vt 
J n = e n n + eDn
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
 kT 
where V t = 

 e 
The homogeneous solution to the differential
equation is found from
dn H
+ An H = 0
dx

where A =
Vt
The solution is of the form
n H = n H (0 ) exp (− Ax )
The particular solution is found from
nP  A = B
Jn
eDn
The particular solution is then
 Jn 




J
B  eDn  J nVt
nP = =
=
= n
A
eDn  e n 

 
V 
 t
The total solution is then
J
n = n + n H (0) exp (− Ax )
e n 
and
V 
V
ni2
n(0) = n pO exp  BE  =
exp  BE
 Vt  N B (0)
 Vt
12.46
We want BVCEO = 60 V
Then
BVCBO
BVCBO
BVCEO =
 60 =
3
n 
50
which yields
BVCBO = 221 V
For this breakdown voltage, we need
N C  1.5 10 15 cm −3
The depletion width into the collector at this
voltage is
xC = x n
 2  (V + V BC )  N B 

1



=  s bi




e

 N C  N B + N C 
We find
 1.5 10 15 10 16 
Vbi = (0.0259 ) ln 
 = 0.646 V
2
 1.5 10 10

and
V BC  BVCEO = 60 V
so that
 2(11.7 ) 8.85 10 −14 (0.646 + 60 )
xC = 
1.6 10 −19

1/ 2
where B =
(
)( )
)
(
(
)
 10 16
 
15
 1.5  10





1
 16
 10 + 1.5 10 15




1/ 2
or
Then
V
n
exp  BE
N B (0)
 Vt

J
− n
 e 
n

_______________________________________
n H (0) =
2
i
12.45
(a) For N C = 210 15 cm −3 ,
BV BC 0  180 V

0.9930
=
= 141 .86
1 −  1 − 0.9930
BVBC 0
180
BVEC 0 =
=
= 34.5 V
3
n 
141 .86
x C = 6.75 10 −4 cm = 6.75  m
_______________________________________
12.47
(a) For N C = 810 15 cm −3 ,
BVCB 0  64 V
(b) V pt =
(b)  =
−3
(c) For N B = 510 cm ,
BV EB  19 V
_______________________________________
16
=
ex B2 0 N B (N C + N B )

2 s
NC
(1.6 10 )(0.50 10 )
2(11.7 )(8.85 10 )
(5 10 )(8 10 + 5 10 )

−4 2
−19
−14
16
15
16
8 10 15
V pt = 70.0 V
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
12.48
(a) V pt
(a) For V CE (sat ) = 0.30 V, we find
ex 2 N (N + N B )
= B0  B C
2 s
NC
=
(1.6 10 )(0.65 10 )
2(11.7 )(8.85 10 )
(2 10 )(5 10 + 2 10 )

−4 2
−19
15
16
5 15
V pt = 32.6 V

 2eV pt
 m ax  

 s
(
 N B NC

N +N
C
 B
)





(
(
(c) For V CE (sat ) = 0.10 V, we find
I B = 0.105 mA = 105  A
_______________________________________
1/ 2
 2 1.6 10 −19 (32 .6)
=
−14
 (11 .7 ) 8.85 10
)
)(
)
 5 10 15 2 10 16  



15
16
 5 10 + 2 10  

1/ 2
 m ax = 2.01  10 5 V/cm
_______________________________________
ex B2 0 N B (N C + N B )

2 s
NC
(1.6 10 )(x )
2(11.7 )(8.85 10 )
(5 10 )(3 10

−19
15 =
12.51
For an npn transistor biased in the active
mode, we have V BC  0 , so that
V 
exp  BC   0 . Now
 Vt 
I E + I B + I C = 0  I B = − (I C + I E )
Then we have


 V  


I B = − F I ES exp  BE  − 1 + I CS 
V


  t  



 V

− −  R I CS − I ES exp  BE

  Vt

2
B0
−14
16
15
+ 5 10 16
)
3 10
−5
 x B 0 = 1.483 10 cm = 0.1483  m
_______________________________________
12.50
We have
We find
I B = 0.01014 mA = 10 .14  A
(b) For V CE (sat ) = 0.20 V, we find
I B = 0.0119 mA = 11.9  A
(b) From Chapter 7,
V pt =
 0.8 + I B 
(4.95 )
= 
 0.99 I B − 0.01 
−14
16
12.49
 0.30 
exp 
= 1.0726  10 5

 0.0259 
15
 I (1 −  R ) + I B  F 
VCE (sat ) = Vt  ln  C


  F I B − I C (1 −  F )  R 
We can write
(1)(1 − 0.2) + I B  0.99 
V (sat ) 
exp  CE
=



 0.0259  (0.99 )I B − (1)(1 − 0.99 )  0.20 
or
V (sat )  0.8 + I B 
(4.95 )
exp  CE
 = 
 0.0259   0.99 I B − 0.01 
 
 − 1 


  

or
 V
I B = (1 −  F )I ES exp  BE
  Vt
 
 − 1

 
− (1 −  R )I CS
_______________________________________
12.52
We can write
 V
I ES exp  BE
  Vt
 
 − 1

 
 V  
=  R I CS exp  BC  − 1 − I E
  Vt  
Substituting, we find


 V  


I C =  F  R I CS exp  BC  − 1 − I E 


  Vt  


 V
− I CS exp  BC
  Vt
 
 − 1

 
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
From the definition of currents, we have
I E = − I C for the case of I B = 0 . Then
 V
I C =  F  R I CS exp  BC
  Vt
 
 − 1 +  F I C

 
 V  
− I CS exp  BC  − 1
  Vt  
When a C-E voltage is applied, then the B-C
V 
becomes reverse biased, so exp  BC   0 .
 Vt 
Then
I C = − F  R I CS +  F I C + I CS
Finally, we find
I (1 −  F  R )
I C = I CEO = CS
1− F
_______________________________________
 
 − 1

 
 V
− I CS exp  BC
  Vt
 
 − 1

 
For V BE = 0.2 V,
(
)
  0.20  
I C = (0.992 ) 5 10 −14 exp 
 − 1
  0.0259  
  V
 
− 10 −13 exp  BC  − 1
0
.
0259
 
 
−10
= 1.1197  10
  V
 
− 10 −13 exp  BC  − 1
0
.
0259
 
 
For VCB = −V BC = −0.5 V
(
)
(
)
I C = 1.1197 10 −10 − 2.4214 10 −5
= −2.421  10 −5 A = −24.21  A
For VCB = −V BC = −0.25 V
I C = 1.1197 10 −10 − 1.5561 10 −9
= −1.44  10 −9 A
For V CB = −V BC  0 V
I C = 2.5277 10 −7
(
− 10
= −2.396  10 −5 A  −24  A
For VCB = −V BC = −0.25 V
I C = 2.5277 10 −7 − 1.5561 10 −9
= 2.51  10 −7 A = +0.251  A
For V CB = −V BC  0 V
I C = 2.5277 10 −7 A = 0.2528  A
(c) For V BE = 0.6 V,
I C = 5.7063 10 −4
  V
 
− 10 −13 exp  BC  − 1
  0.0259  
= −V BC = −0.5 V
(
For VCB
)
I C = 5.7063 10 −4 − 2.4214 10 −5
I C = 5.7063 10 −4 A = 0.5706 mA
_______________________________________
12.54
 I (1 −  R ) + I B  F 
V CE (sat ) = Vt ln  C


  F I B − (1 −  F )I C  R 

(5)(1 − 0.15 ) + I B
 0.975 
= (0.0259 ) ln 


 (0.975 )I B − (1 − 0.975 )(5)  0.150 


4.25 + I B
(6.5)
= (0.0259 ) ln 
 (0.975 )I B − 0.125

I B = 0.15 A, VCE (sat ) = 0.187 V
I B = 0.25 A, VCE (sat ) = 0.143 V
I B = 0.50 A, VCE (sat ) = 0.115 V
I B = 1.0 A, VCE (sat ) = 0.0956 V
_______________________________________
12.55
(a) (i) re =
Vt
0.0259
=
= 0.1036 k 
IE
0.25
 e = reC je = (103 .6)(0.35  10 −12 )
I C = 1.1197 10 −10 A
(b) For V BE = 0.4 V,
−13
I C = 2.5277 10 −7 − 2.4214 10 −5
= 5.464  10 −4 A = 0.5464 mA
For VCB = −V BC  −0.25 V
12.53
 V
(a) I C =  F I ES exp  BE
  Vt
For VCB = −V BC = −0.5 V
  V BC  
 − 1
exp 
  0.0259  
)
= 3.626  10 −11 s = 36.26 ps
(
x2
0.65 10 −4
(ii)  b = B =
2 Dn
2(25 )
)
2
= 8.45  10 −11 s = 84.5 ps
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 12
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(iii)  d =
x dc
s
=
2.2  10 −4
10 7
= 2.2  10 −11 s = 22 ps
(iv)  c = rc C  + C s
(
)
= (18 )(0.020 + 0.020 )10 −12
(b)  ec
= 7.2  10 −13 s = 0.72 ps
=  e + b + d + c
= 36.26 + 84.5 + 22 + 0.72 = 143.48 ps
1
1
(c) f T =
=
2  ec 2 143 .48 10 −12
(
)
= 1.109  10 Hz = 1.109 GHz
f
1.109 10 9
f = T =

125
9
(d)
= 8.87  10 6 Hz = 8.87 MHz
_______________________________________
12.56
(
)
2
x2
0.5 10 −4
b = B =
= 6.25 10 −11 s
2DB
2(20 )
We have
 b = 0.2 ec
so that
 ec = 3.125 10 −10 s
Then
1
1
fT =
=
2  ec 2 3.125 10 −10
or
f T = 5.09 10 8 Hz = 509 MHz
_______________________________________
(
)
12.57
We have
 ec =  e +  b +  d +  c
We are given
 b = 100 ps and  e = 25 ps
We find
x
1.2 10 −4
d = d =
= 1.2 10 −11 s
s
10 7
or
 d = 12 ps
Also
 c = rc C c = (10 ) 0.110 −12 = 10 −12 s
or
 c = 1 ps
Then
 ec = 25 + 100 + 12 + 1 = 138 ps
We obtain
1
1
fT =
=
2  ec 2 138 10 −12
(
)
(
)
= 1.15  10 Hz
9
or
f T = 1.15 GHz
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 13
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
Chapter 13
Exercise Solutions
Ex 13.1
N N
Vbi = Vt ln  a 2 d
 ni
(
V pO =




)( )
)
 2  10 16 10 18 
= (0.0259 ) ln 
 = 0.832 V
2
 1.5  10 10

V pO = Vbi − V p = 0.832 − (− 2.50 )
(
= 3.332 V
Now
 2 s V pO 
a=

 eN d 
(
−14
)(
)
)
(1.6 10 )(0.4 10 ) (10 )
2(11.7 )(8.85 10 )
−4 2
−19
16
( )( )
(
)
_______________________________________
I P1 =
=
 n (eN d )2 Wa 3 
6 s
 

 
Or
I D1 (sat ) = 22 .13  A
_______________________________________
g ms (max ) =
=
3I P1
V PO

Vbi
1 −
V PO




3(0.23735 ) 
0.8139
1 −
1.236 
1.236



g ms (max ) = 0.1086 mA/V
_______________________________________
Ex 13.5


 L 
(900 )(1.6 10 −19 )(10 16 )2
6(11.7 )(8.85 10 −14 )
(
0.8139
1.236
Then
−14
 10 18 10 16 
Vbi = (0.0259 ) ln 
 = 0.8139 V
2
 1.5 10 10 
V p = V pO − Vbi = 1.236 − 0.8139 = 0.422 V
Ex 13.3



 

 2 Vbi
1 −
 3 V pO

Ex 13.4
From Ex 13.3, Vbi = 0.8139 V,
V pO = 1.236 V, I P1 = 0.23735 mA
ea 2 N a
2 s
= 1.236 V
16
−14
 2
 1 −
 3
= 0.02213 mA
1/ 2
Ex 13.2
=
−4 2
−19

 0.8139 
= (0.23735 )1 − 3

 1.236 

= 4.64  10 −5 cm = 0.464  m
_______________________________________
V pO =
(1.6 10 )(0.40 10 ) (10 )
2(11.7 )(8.85 10 )
= 1.236 V

V

I D1 (sat ) = I P1 1 − 3 bi
 V pO



1/ 2
 2(11 .7 ) 8.85  10 (3.332 ) 
=

1.6  10 −19 2  10 16


(
=
ea 2 N d
2 s
)(
V pO =
)
 50 10 −4 0.40 10 −4 3 


5 10 − 4


= 2.37  10 −4 A = 0.237 mA
 10 18 10 16 
Vbi = (0.0259 ) ln 
 = 0.8139 V
2
 1.5 10 10 
( )( )
(
)
=
ea 2 N d
2 s
(1.6 10 )(0.40 10 ) (5 10 )
2(13.1)(8.85 10 )
−19
−4 2
15
−14
= 0.5520 V
N 
 n = Vt ln  c 
 Nd 
 4.7 10 17
= (0.0259 ) ln 
15
 5 10

 = 0.1177 V


Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 13
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
Vbi =  Bn −  n = 0.85 − 0.1177 = 0.7323 V
VT = Vbi − V pO = 0.7323 − 0.5520
= 0.180 V
_______________________________________
Ex 13.6
N 
 4.7 10
 = (0.0259 ) ln 

 10 16


 n = Vt ln  c
 Nd
= 0.0997 V
17




Vbi = 0.89 − 0.0997 = 0.790 V
V pO = Vbi − VT = 0.790 − 0.25 = 0.540 V
Then
(
)( )
)
a 2 1.6  10 −19 10 16
2(13 .1) 8.85  10 −14
 a = 0.280  m
_______________________________________
0.540 =
Ex 13.7
kn =
(
2aL
(7000 )(13 .1) 8.85 10 −14 25 10 −4
=
2 0.40  10 − 4 0.8  10 − 4
(
)(
)(
)
= 3.17  10 −3 A/V 2 = 3.17 mA/V 2
)
I D1 (sat ) = k n (VGS − VT )
From exercise problem Ex 13.5,
VT = 0.180 V
Then
2
I D1 (sat ) = (3.17 )(0.50 − 0.180 )
= 0.325 mA
_______________________________________
2
Ex 13.8
 2  (V (2 ) − V DS (sat )) 
L(2 ) =  s DS

eN d


(
1/ 2
)
 2(11 .7 ) 8.85  10 −14 (2.5) 
=

1.6  10 −19 10 16


(
1/ 2
)( )
−4
= 0.5688  10 cm = 0.5688  m
 2  (V (1) − V DS (sat )) 
L(1) =  s DS

eN d


(
)
1/ 2
 2(11 .7 ) 8.85  10 −14 (2.0 ) 
=

1.6  10 −19 10 16


(
 10 
= (4.0)
 = 4.11709 mA
 9.7156 


L
I D 1 (1) = I D1 

 L − (1 2)L(1)
 10 
= (4.0)
 = 4.10442 mA
 9.7456 
Now
rds =
1/ 2
)( )
−4
= 0.5088  10 cm = 0.5088  m
V DS (2) − V DS (1)
I D 1 (2) − I D 1 (1)
2.5 − 2.0
= 39 .46 k 
4.11709 − 4.10442
_______________________________________
=
Ex 13.9
fT =
 n s W
(


L
I D 1 (2) = I D1 

 L − (1 2)L(2) 
e n N d a 2
2 s L2
(1.6 10 )(1000 )(5 10 )(0.50 10 )
=
2 (11 .7 )(8.85  10 )(2  10 )
−19
−4 2
15
−4 2
−14
f T = 7.69 10 9 Hz = 7.69 GHz
_______________________________________
Test Your Understanding Solutions
TYU 13.1
(
)(
 5 10 18 5 10 15
Vbi = (0.0259 ) ln 
2
 1.8 10 6
= 1.305 V

2 V
x n =  s bi

 e
(
 Na

N
 d
(
)
)


1


 N + N 
d 
 a


1/ 2
)
 2(13.1) 8.85 10 −14 (1.305 )
=
1.6 10 −19

 5 10 18 
1

 
15 
18
15
 5 10  5 10 + 5 10
which yields
x n = 6.147 10 −5 cm = 0.6147  m
We want
0.6147
a=
= 0.512  m
1.2





1/ 2
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 13
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
So that
V pO
TYU 13.3
ea 2 N d
=
2 s
fT =
(1.6 10 )(0.5123 10 ) (5 10 )
2(13.1)(8.85 10 )
−4 2
−19
=
15
−14
e p N a a 2
2 s L2
(1.6 10 )(400 )(2 10 )(0.50 10 )
=
2 (11 .7 )(8.85  10 )(4  10 )
−19
−4 2
16
−14
−4 2
= 0.905 V
Then
V p = Vbi − V pO = 1.305 − 0.905
f T = 3.07 10 9 Hz = 3.07 GHz
_______________________________________
 V p  +0.40 V
TYU 13.4
_______________________________________
TYU 13.2
I P1 =
(


 L 
)(
)
−4 3
5 10
−4





or I p1 = 6.593  10 −4 A = 0.6593 mA
Now
ea 2 N a
2 s
(1.6 10 )(0.50 10 ) (2 10 )
2(11.7 )(8.85 10 )
−4 2
−19
16
−14
or V pO = 3.863 V
Also
(
)(
)
 5  10 18 2  10 16 
Vbi = (0.0259 ) ln 

2
 1.5  10 10

= 0.8736 V
Now

 V  2 Vbi

I D1 (sat ) = I P1 1 − 3 bi 1 −
 V pO  3 V pO




(
)
−4 2
f T = 1.07 10 11 Hz = 107 GHz
_______________________________________
2
(
)
(40 10 )(0.50 10 )

−4 2
15
−14
−4
=
2 s L2
−19
 (400 ) 1.6  10 −19 2  10 16
=
6(11 .7 ) 8.85  10 −14

V pO =
e n N d a 2
(1.6 10 )(6500 )(3 10 )(0.50 10 )
=
2 (13 .1)(8.85  10 )(10 )
 p (eN a )2 Wa 3 
6 s
fT =



 


 0.8736 
= (0.6593 )1 − 3

 3.863 

  2  0.8736  
 1 −  

  3  3.863  
or I D1 (sat ) = 0.3538 mA
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 13
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Chapter 13
13.1
Sketch
_______________________________________
(c) VDS (sat ) = V pO − (Vbi − VGS )
(i) V DS (sat ) = 3.312 − (1.328 − 0 )
= 1.984 V
(ii) V DS (sat ) = 3.312 − (1.328 − (− 1.0 ))
= 0.984 V
_______________________________________
13.2
Sketch
_______________________________________
13.3
(a) V pO =
(i) V pO =
13.4
ea 2 N d
2 s
(1.6 10 )(0.40 10 ) (3 10 )
2(13.1)(8.85 10 )
−4 2
−19
16
−14
= 3.312 V
(
)(
 3 10 16 2 10 18
(ii) Vbi = (0.0259 ) ln 
2
 1.8 10 6
= 1.328 V
V p = Vbi − V pO = 1.328 − 3.312
(
)
(i) V pO =
)
(
)
(
)(
(
)
1/ 2
)(
(
)
)(
)
h2 = 4.57 10 −5 cm = 0.457  m
h2  a  a − h2 = 0
(
)
)

)
1/ 2
(
)(
1/ 2
)
−5
h2 = 2.42 10 cm = 0.242  m
a − h2 = 0.40 − 0.242 = 0.158  m
1/ 2
)
 2(13 .1) 8.85  10 −14 (1.328 + 2.5 − (− 0.5)) 
=

1.6  10 −19 3  10 16


(
)(
 2(11 .7 ) 8.85  10 −14 (0.860 + 0 − (− 0.5)) 
=

1.6  10 −19 3  10 16


(ii) h 2
(
)
 2(11 .7 ) 8.85  10 −14 (0.860 + 0.5 − (− 0.5)) 
=

1.6  10 −19 3  10 16


h2 = 3.35 10 −5 cm = 0.335  m
a − h2 = 0.40 − 0.335 = 0.065  m
(iii) h 2
(
 2  (V + V DS − VGS ) 
(b) h2 =  s bi

eN d


(i) h 2
(
)
 2(13 .1) 8.85  10 −14 (1.328 + 0.5 − (− 0.5)) 
=

1.6  10 −19 3  10 16


(
16
−14
= −2.849 V
1/ 2
h2 = 2.97 10 −5 cm = 0.297  m
a − h2 = 0.40 − 0.297 = 0.103  m
(ii) h 2
−4 2
−19
 3 10 16 2 10 18
(ii) Vbi = (0.0259 ) ln 
2
 1.5 10 10
= 0.860 V
V p = Vbi − V pO = 0.860 − 3.709

 2(13 .1) 8.85  10 −14 (1.328 + 0 − (− 0.5)) 
=

1.6  10 −19 3  10 16


(1.6 10 )(0.40 10 ) (3 10 )
2(11.7 )(8.85 10 )
= 3.709 V
= −1.984 V
 2  (V + V DS − VGS ) 
(b) h2 =  s bi

eN d


(i) h 2
ea 2 N d
2 s
(a) V pO =
1/ 2
(
)(
1/ 2
)
h2 = 2.83 10 −5 cm = 0.283  m
a − h2 = 0.40 − 0.283 = 0.117  m
(iii) h 2
(
)
 2(11 .7 ) 8.85  10 −14 (0.860 + 2.5 − (− 0.5)) 
=

1.6  10 −19 3  10 16


(
−5
)(
)
h2 = 4.08 10 cm = 0.408  m
h2  a  a − h2 = 0
1/ 2
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 13
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(c) VDS (sat ) = V pO − (Vbi − VGS )
or
(i) V DS (sat ) = 3.705 − (0.860 − 0 )
= 2.845 V
(ii) V DS (sat ) = 3.705 − (0.860 − (− 1.0 ))
= 1.845 V
_______________________________________
(a) V pO =
2 s V pO
Na =
ea N a
 Na =
2 s
(
)
(
ea 2
2(13.1) 8.85 10 −14 (2.75 )
−4 2
= 9.433  10 15 cm −3
(
−4 2
)( )
) 
(
−14
GS
−19
GS
 2  (V + V SD + VGS ) 
(d) h2 =  s bi

eN a


 2  (V + V SD + VGS ) 
h2 =  s bi

eN a


−4 2
−14
(0.50 10 )
 2(13 .1)(8.85  10 )(1.28 + 0 + V ) 
=
(1.6 10 )(9.433 10 ) 

2.5 10 = (1.5363 10 )(1.28 + V )
SD
−19
GS
15
−9
GS
 VGS = 0.347 V
 2  (V + V SD + VGS ) 
(d) h2 =  s bi

eN a


)
)
(0.65 10 ) = (1.5363 10 )(1.28 + V )
−4 2
−9
SD
 V SD = 1.47 V
_______________________________________
13.6
(a) N a =
=
2 s V pO
ea 2
2(11.7 ) 8.85 10 −14 (2.75 )
(
)
(1.6 10 )(0.65 10 )
−19
SD
 V SD = 1.94 V
_______________________________________
N N
(a) Vbi = Vt ln  a 2 d
 ni
 2(13 .1) 8.85  10 (1.28 + V SD ) 
=

1.6  10 −19 9.433  10 15


)(
−9
13.7
1/ 2
−4 2
(
15
−4 2
−14
−19
1/ 2
(0.65 10 )
 2(11 .7 )(8.85  10 )(0.8095 + V ) 
=
(1.6 10 )(8.425 10 ) 

(0.65 10 ) = (1.536 10 )(0.8095 + V )
1/ 2
−4 2
(
−9
 VGS = 0.8178 V
(c) a − h2 = 0.15 = 0.65 − h2
h2 = 0.50  m
−14
15
−9
= 1.47 V
(0.65 10 )
1/ 2
(0.50 10 )
 2(11 .7 )(8.85  10 )(0.8095 + 0 + V ) 
=

(1.6 10 )(8.425 10 )


2.5 10 = (1.536 10 )(0.8095 + V )
 9.433 10 15 10 18
(b) Vbi = (0.0259 ) ln 
2

1.8 10 6
= 1.280 V
V p = V pO − Vbi = 2.75 − 1.280
−9
(
 2  (V + V SD + VGS ) 
h2 =  s bi

eN a


(1.6 10 )(0.65 10 )
−19
)( )
) 
 8.425 10 15 10 18
(b) Vbi = (0.0259 ) ln 
2

1.5 10 10
= 0.8095 V
V p = V pO − Vbi = 2.75 − 0.8095
= 1.9405 V
(c) a − h2 = 0.15 = 0.65 − h2
 h 2 = 0.50  m
13.5
2
N a = 8.425  10 15 cm −3
−4 2




(
)(
 2 10 16 3 10 18
= (0.0259 ) ln 
2
 1.5 10 10
= 0.860 V
V p = V pO − Vbi
(
)
)

3.0 = V pO − 0.860  V pO = 3.86 V
 2 s V pO 
Now a = 

 eN a 
(
1/ 2
)
 2(11 .7 ) 8.85  10 −14 (3.86 ) 
=

−19
2  10 16
 1.6  10

(
−5
)(
)
 5.0  10 cm = 0.50  m
1/ 2
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 13
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(b) V pO = 3.86 V
13.10
(c) VSD (sat ) = V pO − (Vbi + VGS )
(i) V SD (sat ) = 3.86 − 0.86 = 3.0 V
(ii) V SD (sat ) = 3.86 − (0.86 + 1.5) = 1.5 V
_______________________________________
 2 s V pO 
a=

 eN a 




(
)(
 2 10 16 3 10 18
= (0.0259 ) ln 
2
 1.8 10 6
= 1.328 V
V p = V pO − Vbi
(
)
 2 s V pO 
a=

 eN a 
(
)
)(
1/ 2
13.11
(a)
)
I P1 =
=
)(
 4  10 16 4 10 18
Vbi = (0.0259 ) ln 
2
 1.5  10 10
= 0.886 V
We find
5 = V pO − 0.886  V pO = 5.886 V
)
)
 2(11 .7 ) 8.85  10 (5.886 ) 
=

1.6  10 −19 4  10 16


(
= 4.36  10
)(
−5
)

6 s L
(1000 )(1.6 10 −19 )(10 16 )2
6(11.7 )(8.85 10 −14 )

I P1 = 1.03 mA
(b)
ea 2 N d
V PO =
2 s
(400 10 )(0.5 10 )
−4
−4 3
20  10 − 4
(
1/ 2
)
cm = 0.436  m
(b) (i) V pO = 5.886 V
(ii) V p = Vbi − V pO = 0.886 − 5.886 = −5.0 V
_______________________________________
)(
(
) ( )
) 
 1.6 10 −19 0.5 10 −4 2 10 16
=
2(11.7 ) 8.85 10 −14

or
V PO = 1.93 V
Also
1/ 2
(
 n (eN d )2 Wa 3
or
13.9
(a) VDS (sat ) = V pO − (Vbi − VGS )
−14
)
(ii) V p = V pO − Vbi = 5.764 −1.264
(i) V SD (sat ) = 4.328 − (1.328 + 0 ) = 3.0 V
(ii) V SD (sat ) = 4.328 − (1.328 + 1.5) = 1.5 V
_______________________________________
 2 s V pO 
a=

 eN a 
)(
1/ 2
(b)
(i) V pO = 5.764 V
(c) VSD (sat ) = V pO − (Vbi + VGS )
(
)
= 1.293  10 cm = 1.293  m
(b) V pO = 4.328 V
(
(
−4
= 5.60  10 −5 cm = 0.560  m
Now
1/ 2
= 4.5 V
_______________________________________
 2(13 .1) 8.85  10 −14 (4.328 ) 
=

1.6  10 −19 2  10 16


(
(

1/ 2
(
 2(13 .1) 8.85  10 −14 (5.764 ) 
=

1.6  10 −19 5  10 15


)
3.0 = V pO −1.328  V pO = 4.328 V
)( )
) 
3.5 = V pO − (1.264 + 1.0)  V pO = 5.764 V
13.8
N N
(a) Vbi = Vt ln  a 2 d
 ni
(
 5 10 15 10 18
(a) Vbi = (0.0259 ) ln 
2
 1.8 10 6
= 1.264 V
VSD (sat ) = V pO − (Vbi + VGS )
( )( )
(
)
 10 19 10 16 
Vbi = (0.0259 ) ln 
 = 0.874 V
2
 1.5  10 10 
Now
V DS (sat ) = V PO − (Vbi − VGS )
= 1.93 − 0.874 + VGS
or
V DS (sat ) = 1.056 + VGS
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 13
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
We have
V P = Vbi − V PO = 0.874 − 1.93 = −1.056 V
Then
(i) For VGS = 0 , V DS (sat ) = 1.06 V
1
V P = −0.264 V,
4
V DS (sat ) = 0.792 V
(ii) For VGS =
(iii) For VGS =
(iv) For V P =
1
V P = −0.528 V,
2
V DS (sat ) = 0.528 V
3
V P = −0.792 V,
4
V DS (sat ) = 0.264 V
 2 Vbi − VGS
 1 −
 3
V PO








 2 0.874 − VGS 

 1 −
 3

1.93


(i) For VGS = 0 , I D1 (sat ) = 0.258 mA
(ii) For VGS = −0.264 V,
I D1 (sat ) = 0.141 mA
(iii) For VGS = −0.528 V,
I D1 (sat ) = 0.0608 mA
(iv) For VGS = −0.792 V,
I D1 (sat ) = 0.0148 mA
_______________________________________
13.12
  V −V
GS
g d = GO1 1 −  bi
  V PO

where
G O1 =
(




1/ 2
3I P1 3 1.03  10 −3
=
V PO
1.93




)
or
GO1 = 1.60 10 −3 S = 1.60 mS
Then
13.13
n-channel JFET - GaAs
(a)
e N Wa
GO1 = n d
L
1.6 10 −19 (8000 ) 2 10 16
=
10 10 − 4
 30 10 −4 0.35 10 −4
or
GO1 = 2.69 10 −3 S
(b)
V DS (sat ) = V PO − (Vbi − VGS )
We have
ea 2 N d
V PO =
2 s
)
(
)
(





 0.874 − VGS
= (1.03 )1 − 3
1.93


g d (mS)
0
0.453
0.523
-0.264
0.590
0.371
-0.528
0.726
0.237
-0.792
0.863
0.114
-1.056
1.0
0
_______________________________________
(
(c)

 V − VGS
I D1 (sat ) = I P1 1 − 3 bi

 V PO
(Vbi − VGS ) / V PO
V GS
=
)(
)
(1.6 10 )(0.35 10 ) (2 10 )
2(13.1)(8.85 10 )
−4 2
−19
16
−14
or
V PO = 1.69 V
We find
(
)(
)
 5  10 18 2  10 16 
Vbi = (0.0259 ) ln 

2


1.8 10 6
or
V bi = 1.34 V
Then
V P = Vbi − V PO = 1.34 − 1.69 = −0.35 V
We then obtain
V DS (sat ) = 1.69 − (1.34 − VGS ) = 0.35 + VGS
(
)
For VGS = 0 , V DS (sat ) = 0.35 V
For VGS =
1
V P = −0.175 V,
2
V DS (sat ) = 0.175 V
(c)

 V − VGS
I D1 (sat ) = I P1 1 − 3 bi

 V PO




 2 Vbi − VGS
 1 −
 3
V PO





Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 13
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
where
 n (eN d )2 Wa 3
I P1 =
=
6 s L
(8000 )(1.6 10 −19 )(2 10 16 )2
6(13 .1)(8.85 10 −14 )

(30 10 )(0.35 10 )
−4 3
−4
10  10 − 4
or
I P1 = 1.515 mA
Then

 1.34 − VGS 
I D1 (sat ) = (1.515 )1 − 3

 1.69 

 2 1.34 − VGS 
 (mA)
 1 −
 3

1
.
69


= 0 , I D1 (sat ) = 0.0506 mA
For VGS
and
For VGS = −0.175 V,
I D1 (sat ) = 0.0124 mA
_______________________________________
13.14
g mS =
3I P1
V PO

1 − Vbi − VGS

V PO





3(1.03 ) 
0.874
1−

1.93 
1.93




or
g mS (max ) = 0.524 mS
For W = 400  m, we have
g mS (max ) =
or
_______________________________________
13.15
The maximum transconductance occurs for
VGS = 0 , so we have

1 − Vbi

V PO

(1.6 10 )(0.5 10 ) (1.5 10 )
2(13.1)(8.85 10 )
or
V PO = 2.59 V
Now
Vbi =  Bn −  n
where
N
 n = Vt ln  c
 Nd
−19
−4 2
16
−14

 4.7 10 17
 = (0.0259 ) ln 

 1.5 10 16






 n = 0.0892 V
so that
g mS (max ) = 13 .1 mS/cm = 1.31 mS/mm
3I P1
V PO
13.16
n-channel MESFET - GaAs
(a)
ea 2 N d
V PO =
2 s
or
0.524
400  10 − 4
(a) g mS (max ) =
 10 
g mS (max ) = (0.2947 )  = 1.47 mS
 2
_______________________________________
=
We have
I P1 = 1.03 mA, V PO = 1.93 V, V bi = 0.874 V
The maximum transconductance occurs when
VGS = 0 . Then
g mS (max ) =
which can be written as

Vbi 
g mS (max ) = GO1 1 −

V PO 

We found
G O1 = 2.69 mS, V bi = 1.34 V, V PO = 1.69 V
Then

1.34 
g mS (max ) = (2.69 )1 −

1.69 

or
g mS (max ) = 0.295 mS
This is for a channel length of L = 10  m.
(b) If the channel length is reduced to
L = 2  m, then




Vbi = 0.90 − 0.0892 = 0.811 V
Then
VT = Vbi − V PO = 0.811 − 2.59
or
VT = −1.78 V
(b )If VT  0 for an n-channel device, the
device is a depletion mode MESFET.
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 13
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(b) VT = Vbi − V pO
13.17
n-channel MESFET - GaAs
(a) We want VT = +0.10 V
Then
VT = Vbi − V PO =  Bn −  n − V PO
so
 N  ea 2 N d
VT = 0.10 = 0.89 − Vt ln  c  −
2 s
 Nd 
which can be written as
17


(0.0259 ) ln  4.7 10 
 Nd

(1.6 10 )(0.35 10 ) N
+
2(13 .1)(8.85 10 )
We find
N 
Then
VT = 0.788 − 1.5 = −0.712 V
 2  (V + V DS − VGS ) 
(c) h2 =  s bi

eN d


d
−14
= 0.89 − 0.10
(
or

(0.0259 ) ln  4.7 10

Nd
17
(




)
1/ 2
1/ 2
a − h2 = 0.330 − 0.1677 = 0.1623  m
(ii)
(

)
h 2 = 7.246  10 − 10 (0.788 + 1.0 − 0.4)
1/ 2
= 3.171  10 −5 cm = 0.3171  m
a − h2 = 0.330 − 0.3171 = 0.0129  m
(iii)
N c (400 )  400 
=
 = 1.54
N c (300 )  300 
Then
N c (400 ) = 4.7 10 17 (1.54 )
3/ 2
(
(
h2  a  a − h2 = 0
_______________________________________
Also
 400 
Vt = (0.0259 )
 = 0.03453
 300 
Then
 7.24 10 17
VT = 0.89 − (0.03453 ) ln 
15
 8.110
(
)(
13.19
(a) V pO =




− 8.453 10 −17 8.110 15
13.18
1/ 2
 2(13 .1) 8.85  10 −14 (1.5) 
=

19
16
 1.6  10 2  10

1/ 2
)
= 3.30  10 −5 cm = 0.330  m
ea 2 N d
2 s
(1.6 10 )(0.50 10 ) (5 10 )
=
2(13.1)(8.85 10 )
−19
)
which becomes
VT = +0.050 V
_______________________________________
)(
1/ 2
= 5.64  10 −5 cm = 0.564  m
= 7.24  10 17 cm −3
)

)
h2 = 7.246 10 −10 (0.788 + 4.0 − 0.4)
)
(
)
)(
)
)(0.788 + 0 − 0.4)
= 1.677  10 −5 cm = 0.1677  m
By trial and error,
N d = 8.110 15 cm −3
(b) At T = 400 K
 2 s V pO 
(a) a = 

 eN d 
(
(i) h2 = 7.246 10 −10
+ 8.453 10 −17 N d = 0.79
(
(
1/ 2
 2(13 .1) 8.85 10 −14 (Vbi + V DS − VGS ) 
=

1.6 10 −19 2 10 16


−4 2
−19
 4.7 10 17 

 n = Vt ln  c  = (0.0259 ) ln 
16 
 Nd 
 2 10 
= 0.0818 V
Vbi =  Bn −  n = 0.87 − 0.0818 = 0.788 V
−4 2
−14
= 0.8626 V
We find
 4.7 10 17 

15 
 5 10 
 n = (0.0259 ) ln 
= 0.1177 V
Vbi =  Bn −  n = 0.87 − 0.1177
= 0.7523 V
VT = Vbi − V pO = 0.7523 − 0.8626
= −0.1103 V
15
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 13
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
 4.7 10 17 
 = 0.0713 V
(b)  n = (0.0259 ) ln 
16 
 3 10 
Vbi =  Bn −  n = 0.87 − 0.0713
= 0.7987 V
VT = Vbi − V pO
and
Vbi =  Bn −  n = 0.82 − 0.206 = 0.614 V
With V DS = 0 and V GS = 0.35 V, we find
a − h = 0.075  10 −4
 2  (V − VGS ) 
= a −  s bi

eN d


or V pO = Vbi − VT = 0.7987 − (− 0.1103 )
= 0.909 V
so that
a = 0.075  10 −4
Then
 2 s V pO 
a=

 eN d 
(
1/ 2
(
)
)(
1/ 2
)
= 2.095  10 −5 cm = 0.2095  m
_______________________________________
13.20
VT = Vbi − V PO =  Bn −  n − V PO
We want VT = 0.5 V, so
0.5 = 0.85 −  n − V PO
Now
 4.7 10 17
 n = (0.0259 ) ln 
 Nd
or
(1.6 10 )(0.26 10 ) (10 )
= 0.614 −
2(11.7 )(8.85 10 )
−4 2
16
−14
= (Vbi − VT ) − (Vbi − VGS )
or
V DS (sat ) = VGS − VT = 0.35 − 0.092
which yields
V DS (sat ) = 0.258 V
_______________________________________
−4 2
d
−14
(
a = 0.26  10 −4 cm = 0.26  m
Now
ea 2 N d
VT = Vbi − V PO = 0.614 −
2 s
or
We obtain
VT = 0.092 V
(b)
V DS (sat ) = V PO − (Vbi − VGS )




(1.6 10 )(0.25 10 ) N
2(13 .1)(8.85  10 )
1/ 2
or
VT
ea 2 N d
2 s
−19
=
(
−19
and
V PO =
)
)( )
 2(11 .7 ) 8.85  10 −14 (0.614 − 0.35 ) 
+

1.6  10 −19 10 16


 2(13 .1) 8.85  10 −14 (0.909 ) 
=

1.6  10 −19 3  10 16


(
1/ 2
)
13.22
V PO = 4.31 10 −17 N d
Then
 4.7 10 17
0.5 = 0.85 − (0.0259 ) ln 
 Nd
(




)
− 4.31 10 −17 N d
By trial and error
N d = 5.45 10 15 cm −3
_______________________________________
13.21
n-channel MESFET - silicon
(a) For a gold contact,  Bn = 0.82 V.
We find
 2.8 10 19 
 = 0.206 V
 n = (0.0259 ) ln 
16

 10

 4.7 10 17 

(a)  n = (0.0259 ) ln 
16 
 2 10 
= 0.0818 V
(i) Vbi =  Bn −  n = 0.90 − 0.0818
= 0.818 V
(ii) V pO =
=
ea 2 N d
2 s
(1.6 10 )(0.65 10 ) (2 10 )
2(13.1)(8.85 10 )
−19
−4 2
−14
= 5.83 V
(iii) VT = Vbi − V pO = 0.818 − 5.83
= −5.012 V
16
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 13
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(b) VDS (sat ) = V pO − (Vbi − VGS )
(i) V DS (sat ) = 5.83 − (0.818 − (− 1.0 ))
= 4.01 V
(ii) V DS (sat ) = 5.83 − (0.818 − (− 2.0 ))
= 3.01 V
(
)
(iii) V DS sat = 5.83 − (0.818 − (− 3.0 ))
= 2.01 V
_______________________________________
13.23
(a) k n =
 n s W
2aL
(6500 )(13.1)(8.85 10 −14 )(12 10 −4 )
=
(
)(
2 0.25  10 − 4 1.5  10 − 4
−3
)
= 1.206  10 A/V = 1.206 mA/V 2
2
(b) I D1 (sat ) = k n (VGS − VT )
2
(i) I D1 (sat ) = (1.206 )(0.25 − 0.15 )
= 0.01206 mA = 12 .06  A
2
(ii) I D1 (sat ) = (1.206 )(0.45 − 0.15 )
= 0.1085 mA
(c) V DS (sat ) = VGS − VT

I D

2
=
k n (VGS − VT )
VGS VGS
= 2k n (VGS − VT )
1.25 = 2k n (0.45 − 0.15 )
(a) g ms =
 k n = 2.083 mA/V
kn =
2.083  10 −3 =

( )(
(
 10 18 3 10 16
Vbi = (0.0259 ) ln 
2
 1.5 10 10
= 0.8424 V
V pO =
)
)

ea 2 N d
2 s
(1.6 10 )(0.50 10 ) (3 10 )
=
2(11.7 )(8.85 10 )
−4 2
−19
16
−14
= 5.795 V
(a) VDS (sat ) = V pO − (Vbi − VGS )
(
)
1/ 2
 2(11 .7 ) 8.85  10 −14 (10 − 4.953 ) 
=

1.6  10 −19 3  10 16


(
)(
1/ 2
)
−5
L = 4.666  10 cm
Now
(1 2)L = 0.90
L
= 1−
L
L
L
4.666 10 −5
L=
=
2(0.10 )
2(0.10 )
L = 2.333  10 −4 cm = 2.333  m
(b) VDS (sat ) = V pO − (Vbi − VGS )
2
(6500 )(13.1)(8.85 10 −14 )W
(
2 0.25  10
−4
)(1.5 10 )
−4
−3
 W = 2.073  10 cm = 20 .73  m
(b) I D1 (sat ) = k n (VGS − VT )
13.27
 2  (V − V DS (sat )) 
L =  s DS

eN d


 n s W
2aL
13.26
Plot
_______________________________________
= 5.795 − 0.8424 = 4.953 V
2
(i) V DS (sat ) = 0.25 − 0.15 = 0.10 V
(ii) V DS (sat ) = 0.45 − 0.15 = 0.30 V
_______________________________________
13.24
13.25
Plot
_______________________________________
2
(i) I D1 (sat ) = (2.083 )(0.25 − 0.15 )
= 0.02083 mA = 20 .83  A
= 5.795 − (0.8424 + 3)
= 1.953 V
(
2
)
= 5.892  10
)(
−5
1/ 2
)
cm
Then
2
(ii) I D1 (sat ) = (2.083 )(0.45 − 0.15 )
= 0.1875 mA
_______________________________________
(
 2(11 .7 ) 8.85  10 −14 (10 − 1.953 ) 
L = 

1.6  10 −19 3  10 16


L=
L
5.892 10 −5
=
2(0.10 )
2(0.10 )
= 2.946  10 −4 cm = 2.946  m
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 13
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
13.28
We have that


L

I D 1 = I D1 
 L − (1 2)L 
Assuming that we are in the saturation region,
then I D 1 = I D 1 (sat ) and I D1 = I D1 (sat ) .
We can write
1
I D 1 (sat ) = I D1 (sat ) 
1 L
1− 
2 L
If L  L , then
 1 L 
I D 1 (sat )  I D1 (sat )1 + 

 2 L 
We have that
 2  (V − V DS (sat )) 
L =  s DS

eN d


1/ 2
 2  V  V (sat ) 

=  s DS 1 − DS
V DS 
 eN d 
which can be written as
1/ 2
 V DS (sat ) 
1 −


V DS 

The parameter  is not independent of V DS .
Define
V DS
x
V DS (sat )
and consider the function
1 1
f = 1 − 
x x
which is directly proportional to  . Then
x
f (x )
1.5
1.75
2.0
2.25
2.50
2.75
3.0
h2 = h sat
)
 2  (V + V DS − VGS ) 
=  s bi

eN d


and
(
)(
 5  10 18 4  10 16
Vbi = (0.0259 ) ln 
2
 1.5  10 10
or
V bi = 0.8915 V
For VGS = 0 , we obtain
(
1/ 2
1  2 s

2 L  eN d V DS
( )(
(
(
)
)
1/ 2
)

 2(11 .7 ) 8.85  10 −14 (0.8915 + 2 ) 
h sat = 

1.6  10 −19 4  10 16


 2 s  V DS (sat ) 
1 −

L = V DS 

V DS 
 eN d V DS 
If we write
I D 1 (sat ) = I D1 (sat )(1 + V DS )
then by comparing equations, we have
=
13.29
(a) Saturation occurs when  = 1 10 4 V/cm.
As a first approximation, let
V
 = DS
L
Then
V DS =   L = 10 4 2 10 −4 = 2 V
(b) We have that
1/ 2
0.222
0.245
0.250
0.247
0.240
0.231
0.222
So that  is nearly a constant.
_______________________________________
)(
1/ 2
)
or
hsat = 0.306 10 −4 cm = 0.306  m
(c) We then find
I D1 (sat ) = eN d  sat (a − h sat )W
(
)(
)( )
 (0.50 − 0.306 )(10 )(30 10 )
= 1.6 10 −19 4 10 16 10 7
−4
or
−4
I D1 (sat ) = 3.72 mA
(d) For VGS = 0 , we have

V
I D1 (sat ) = I P1 1 − 3 bi

 V PO
Now
I P1 =
=




 n (eN d )2 Wa 3
6 s L
(1000 )(1.6 10 −19 )(4 10 16 )2
6(11.7 )(8.85 10 −14 )

or
 2 Vbi
 1−
 3 V
PO

I P1 = 12 .36 mA
(30 10 )(0.5 10 )
(2 10 )
−4 3
−4
−4
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 13
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Also
V PO =
=
13.31
(a)
ea 2 N d
2 s
 =  n  = (8000 )(5 10 3 ) = 4 10 7 cm/s
(1.6 10 )(0.5 10 ) (4 10 )
2(11 .7 )(8.85 10 )
−4 2
−19
16
Then
−14
td =
or
V PO = 7.726 V
Then
or

 0.8915 
I D1 (sat ) = (12.36 )1 − 3

 7.726 

 2
 1 −
 3

0.8915
7.726




or
I D1 (sat ) = 9.05 mA
_______________________________________
13.30
(a) If L = 1  m, then saturation will occur
when
V DS =   L = 10 4 110 −4 = 1 V
We find
( )(
)
 2  (V + V DS − VGS ) 
h2 = h sat =  s bi

eN d


We have Vbi = 0.8915 V and for VGS = 0 , we
obtain
(
)
 2(11 .7 ) 8.85  10 (0.8915 + 1) 
h sat = 

1.6  10 −19 4  10 16


(
)(
or
−4
hsat = 0.247 10 cm = 0.247  m
Then
I D1 (sat ) = eN d  sat (a − h sat )W
(
)(
)( )
 (0.50 − 0.247 )(10 )(30 10 )
= 1.6 10 −19 4 10 16 10 7
−4
−4
or
I D1 (sat ) = 4.86 mA
If velocity saturation did not occur, then from
the previous problem, we would have
2
I D1 (sat ) = (9.05 )  = 18 .1 mA
1
(b) If velocity saturation occurs, then the
relation I D1 (sat )  (1 L ) does not apply.
_______________________________________
2 10 −4
= 5 10 −12 s
4 10 7
(b) Assume  =  sat = 10 7 cm/s
Then
L
2 10 −4
td =
=
= 2  10 −11 s
7
 sat
10
or
t d = 20 ps
_______________________________________
13.32
(a)
 =  n  = (1000 )(10 4 ) = 10 7 cm/s
Then
td =
or
L

=
2 10 −4
= 2 10 −11 s
10 7
t d = 20 ps
(b) For  =  sat = 10 7 cm/s
td =
1/ 2
)

=
t d = 5 ps
1/ 2
−14
L
L
=
 sat
2 10 −4
= 2  10 −11 s
10 7
or
t d = 20 ps
_______________________________________
13.33
The reverse-bias current is dominated by the
generation current. We have
V P = Vbi − V PO
We find
 5 10 18 3 10 16 
Vbi = (0.0259 ) ln 

2
 1.5 10 10

or
V bi = 0.884 V
(
Also V PO =
(
ea 2 N d
2 s
)(
(
)(
)
)(
)
 1.6 10 −19 0.3 10 −4 2 3 10 16
=
2(11 .7 ) 8.85 10 −14

(
)
)

Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 13
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
or
13.34
(a) The ideal transconductance for VGS = 0 is
V PO = 2.086 V
Then
V P = 0.884 − 2.086 = −1.20 V
Let V GS = −1.20 V
Now
 2  (V + V DS − VGS ) 
x n =  s bi

eN d


(
 2(11 .7 ) 8.85 10 −14
=
1.6 10 −19

(

or
)
1/ 2
(
)
(3 10 )
1/ 2


16

)
x n = 4.314 10 −10 (2.084 + VDS )
(a) For V DS = 0 , x n = 0.30  m
(b) For V DS = 1 V, x n = 0.365  m
(c) For V DS = 5 V, x n = 0.553  m
1/ 2
=
(
)
(

 30 10 − 4


(
)
)(
)
+ (x n ) 0.6 10 −4 30 10 −4
or
(
Vol = 10.8 10 −12 + x n 18 10 −8
(a) For V DS = 0 , Vol = 1.62  10
)
−11
cm 3
(b) For V DS = 1 V, Vol = 1.737  10 −11 cm 3
(c) For V DS = 5 V, Vol = 2.075  10 −11 cm 3
The generation current at the drain is
 n 
I DG = e i   Vol
 2 O 
 1.5 10 10 
= 1.6 10 −19 
 Vol
−8 
 2 5 10 
or
(
)
(
)
(
(
)
)(
(1.6 10 )(0.3 10 ) (7 10 )
2(13 .1)(8.85 10 )
−19
−4 2
16
−14
or
The depletion region volume at the drain is
L
Vol = (a ) (W ) + (x n )(2a )(W )
2
 2.4 10 −4
= 0.3 10 − 4 
2

)
(
(0.884 + V DS − (− 1.20 )) 
(

Vbi 
g mS = GO1 1 −

V PO 

where
e N Wa
GO1 = n d
L
1.6 10 −19 (4500 ) 7 10 16
=
1.5 10 − 4
 5 10 −4 0.3 10 −4
or
G O1 = 5.04 mS
We find
ea 2 N d
V PO =
2 s
)
I DG = 2.4 10 −2  Vol
(a) For V DS = 0 , I DG = 0.39 pA
(b) For V DS = 1 V, I DG = 0.42 pA
(c) For V DS = 5 V, I DG = 0.50 pA
_______________________________________
V PO = 4.347 V
We have
 4.7 10 17 
 = 0.049 V
16 
 7 10 
 n = (0.0259 ) ln 
so that
Vbi =  Bn −  n = 0.89 − 0.049 = 0.841 V
Then

0.841 
g mS = (5.04 )1 −

4.347 

or
g mS = 2.82 mS
(b) With a source resistance
gm
g
1
g m =
 m =
1 + g m rs
g m 1 + g m rs
For
g m
1
= 0.80 =
gm
1 + (2.823 )rs
we obtain
rs = 88 .6 
(c)
L L
L
rs =
=
=
A A (e n n )A
)
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 13
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
so
(
)
(
L = (88.56 ) 1.6 10 −19 (4500 ) 7 10 16
(
)(
)
 0.3 10 −4 5 10 −4
)
or
13.36
(a) For constant mobility
e N a 2
fT = n d 2
2 s L
L = 0.67  10 −4 cm = 0.67  m
_______________________________________
13.35
Considering the capacitance charging time,
we have
gm
fT =
2 C G
where
 WL
CG = s
a
=
(13.1)(8.85 10 −14 )(5 10 −4 )(1.5 10 −4 )
0.3 10
=
(
13.37
C G = 2.9 10 F
We must use g m , so we obtain
=
11
−15
Hz
We can also write
1
1
fT =
C =
2  C
2 f T
so
1
C =
= 1.285 10 −12 s
2 1.238 10 11
The channel transit time is
1.5 10 −4
tt =
= 1.5 10 −11 s
10 7
The total time constant is
 = 1.5  10 −11 + 1.285  10 −12
)
= 1.629  10 −11 s
Taking into account the channel transit time
and the capacitance charging time, we find
1
1
fT =
=
2  2 1.629 10 −11
or
f T = 9.77 10 9 Hz = 9.77 GHz
_______________________________________
(
)
)
f T = 1.33 10 Hz = 13.3 GHz
_______________________________________
fT =
(2.82 10 )(0.80 ) = 1.238 10
2 (2.9  10 )
−4 2
−14
10
−4
−3
−4 2
16
(b) For saturation velocity model

10 7
f T = sat =
2 L 2 1.2  10 − 4
−15
(
−19
f T = 4.12 10 11 Hz = 412 GHz
or
fT =
(1.6 10 )(7500 )(4 10 )(0.30 10 )
2 (13 .1)(8.85  10 )(1.2  10 )
e n N d a 2
2 s L2
(1.6 10 )(1000 )(2 10 )(0.40 10 )
2 (11 .7 )(8.85 10 )L
−19
−4 2
16
−14
2
786 .975
L2
786 .975
fT =
2
3 10 − 4
fT =
(a)
(
)
= 8.74  10 9 Hz = 8.74 GHz
786 .975
fT =
2
1.5 10 − 4
(b)
(
)
= 3.50  10 Hz = 35.0 GHz
_______________________________________
10
13.38
fT =
e p N a a 2
2 s L2
or
 e p N a a 2 
L=

 2 s f T 
(
)
1/ 2
(
(
)(
 1.6 10 −19 (420 ) 2 10 16 0.40 10 − 4
=
2 (11.7) 8.85 10 −14 f T

18.18
L=
fT
)
)
2



1/ 2
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 13
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(a) L =
 g mS 

 = 5.02 S/cm = 502 mS/mm
 W 
(b) At V g = 0 , we obtain
18.18
5 10
9
= 2.57  10 −4 cm = 2.57  m
(b) L =
N
I D (sat )
=
− V − V 
(d + d ) off O s
W
(
18.18
12 10
9
= 1.66  10 −4 cm = 1.66  m
_______________________________________
13.39
(a)
V off
I D (sat )
= 5.37 A/cm = 537 mA/mm
W
_______________________________________
E c
= B −
− VP2
e
VP2 =
=
13.41
eN d d d2
2 N
(1.6 10 )(3 10 )(350 10 )
2(12.2)(8.85 10 )
−19
−8 2
18
−14
or
Voff = −2.07 V
(b)
N
V g − Voff
e(d + d )
For V g = 0 , we have
(
nS =
nS =
)
(12.2)(8.85 10 −14 ) (2.07 )
(1.6 10 −19 )(350 + 80 )(10 −8 )
or
n S = 3.25 10 12 cm −2
_______________________________________
13.40
(a) We have
I D (sat ) =
N W
V − V − V 
(d + d ) g off O s
(
)
We find
 g mS 
  I D (sat ) N  s

 =

=
 W  V g  W  (d + d )
=
E c
− VP2
e
We want Voff = −0.3 V, so
V off =  B −
−0.30 = 0.85 − 0.22 − V P 2
or
V P 2 = 0.93 V
We have
eN d d d2
VP2 =
2 N
or
2 N V P 2
d d2 =
eN d
V P 2 = 2.72 V
Then
Voff = 0.89 − 0.24 − 2.72
or
(12.2)(8.85  10 − 12 )
(2.07 − 1)(2  10 7 )
−8
(350 + 80 )(10 )
or
where
or
=
)
(12 .2)(8.85 10 −14 )(2 10 7 )
(350 + 80 )(10 −8 )
(
)
2(12 .2) 8.85 10 −14 (0.93 )
1.6  10 −19 2 10 18
We then obtain
=
(
)(
)
o
d d = 2.51 10 −6 cm = 251 A
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 14
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
Chapter 14
Exercise Solutions
Ex 14.1
For silicon,  = 0.8  m    10 3 cm −1
 = 0.6  m    4.5  10 3 cm −1
Let d = 5  m = 5  10 −4 cm
(a) For  = 0.8  m,
(190 )(10 −7 )
Ln = Dn n 0 =
= 4.36  10 −3 cm
(10 )(10 −8 )
L p = D p p 0 =
I  (d )
= exp (− d )
I 0
= 3.16  10 −4 cm
 ( )(
= exp − 10 5 10
= 0.607 = 60.7%
(b) For  = 0.6  m,
I  (d )
= exp (− d )
I 0
3
(
−4
Now
)
 Dn
Dp
J S = eni2 
+
 Ln N a L p N d

(
)
)(
Ex 14.2
For  = 1  m in silicon,   10 2 cm −1
Now
1.24 1.24
E = h =
=
= 1.24 eV

1.0
(a) I  (d ) = I  0 exp (− d )
 ( )(5 10 )
= (0.10 ) exp − 10
= 0.0951 W/cm 2
I (d )
10 2 (0.0951 )
g = 
=
h
1.6  10 −19 (1.24 )
( )
)
= 4.79  10 19 cm −3 s −1
(b) I  (d ) = I  0 exp (− d )
 ( )(
= (0.10 ) exp − 10 2 20 10 −4
)
2


190
10

+
−3
17
−4
16 
3.16 10 2 10 
 4.36 10 10
(
)( ) (
−18
)(
J S = 1.046 10 A/cm
We find
 J 
Voc = Vt ln 1 + L 
 JS 
)
= 0.0819 W/cm
I  (d )
(10 2 )(0.0819 )
g =
=
h
(1.6 10 −19 )(1.24 )
2
= 4.13  10 19 cm −3 s −1
_______________________________________
N N
(b) Vbi = Vt ln  a 2 d
 ni
)
2

20 10 −3
= (0.0259 ) ln 1 +
−18
 1.046 10
= 0.971 V
−4
2
)(




= 1.6 10 −19 1.8 10 6
= exp − 4.5 10 3 5 10 −4
= 0.105 = 10.5%
_______________________________________
(
Ex 14.3
(a) We find




( )(
(




)
 10 17 2  10 16 
= (0.0259 ) ln 

2
 1.8  10 6

= 1.24 V
So
Voc 0.971
=
= 0.783
Vbi 1.240
_______________________________________
)
Ex 14.4
The photocurrent is given by
I L = eGL p  n +  p A
(
(
= 1.6 10 −19
)
)(10 )(10 )
21
−6
(
)
 (1000 + 400 ) 10 −7 (10 )
−7
I L = 2.24 10 A = 0.224  A
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 14
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
Ex 14.5
We find
N N
Vbi = Vt ln  a 2 d
 ni
Ex 14.7
(a) For x = 0.15 , E g  1.60 eV




1.24
= 0.775  m
1.60
(b) For x = 0.30 , E g  1.76 eV
=
( )( )
(
)
 10 15 10 15 
= (0.0259 ) ln 

2
 1.5  10 10 
= 0.575 V
 2  (V + V R )  N a + N d

W =  s bi
 N N
e

a
d

(
)
1.24
= 0.705  m
1.76
_______________________________________
=




1/ 2
Ex 14.8
For GaAs, n 2 = 3.8
For GaP, n 2 = 3.2
 2(11 .7 ) 8.85 10 −14 (0.575 + 5)
=
1.6 10 −19

 10 15 + 10 15
  15
15
 10 10
 

 
1/ 2
( )( )
= 3.80  10 −4 cm
Then
J L = e W + Ln + L p GL
(
)
From Example 14.5, L n = 35 .4  m
(
L p = 10.0  m
)
(
)( )
J L = 1.6 10 −19 (3.80 + 35.4 + 10.0) 10 −4 10 21
J L = 0.787 A/cm
Now
J L1 = eWG L
(
2
)(
)( )
Then for GaAs 0 .6 P 0 .4 ,
n 2 = (3.8 − 3.2 )(0.6 ) + 3.2 = 3.56
Then
2
2
 n − n1 
 3.56 − 1.0 
 = 
 =  2
 = 0.315
 3.56 + 1.0 
 n 2 + n1 
_______________________________________
Ex 14.9
For GaAs 0 .6 P 0 .4 , n 2 = 3.56 (See Exercise
Ex 14.8)
Then
n 
 1.0 
 c = sin −1  1  = sin −1 
 = 16.3
n
 3.56 
 2
_______________________________________
= 1.6 10 −19 3.80 10 −4 10 21
= 0.0608 A/cm 2
Then
J L1 0.0608
=
= 0.0773
JL
0.787
_______________________________________
(
)(10 )
 1 − exp − (10 )(20 10 )
= 1.6 10
17
2
−4
J L = 2.90 10 −3 A/cm 2 = 2.90 mA/cm 2
(b) For  = 10 cm
J L = 1.6 10 −19 10 17
4
(
TYU 14.1
I  = I o exp (−  x )
(a) For  = 1  m,   10 2 cm −1
(i) x = 5  m
Ex 14.6
(a) For  = 10 2 cm −1
J L = e o 1 − exp (− W )
−19
Test Your Understanding Solutions
−1
 ( )(
I  = (0.1) exp − 10 2 5 10 −4
= 0.0951 W/cm
(ii) x = 20  m
)
2
 ( )(
I  = (0.1) exp − 10 2 20 10 −4
)
= 0.0819 W/cm
(b) For  = 0.6  m,   4  10 3 cm −1
(i) x = 5  m
2
)( )
 1 − exp − (10 )(20 10 )
4
−4
J L = 1.6 10 −2 A/cm 2 = 16.0 mA/cm 2
_______________________________________
(
)(
I  = (0.1) exp − 4 10 3 5 10 −4
= 0.0135 W/cm
2
)
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 14
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
(ii) x = 20  m
(
)(
I = (0.1) exp − 4 10 20 10
3
−4
)
= 3.35  10 −5 W/cm 2
_______________________________________
TYU 14.2
From Example 14.3,
Now
 J
VOC = Vt ln 1 + L
 JS
J S = 3.6 10 −11 A/cm 2




We have
 J 
0.60 = (0.0259 ) ln 1 + L 
 JS 
which yields
JL
= 1.15 10 10
JS
or
J L = 0.414 A/cm 2
_______________________________________
TYU 14.3
From Example 14.3, J S = 3.6 10 −11 A/cm 2
We have J L = 15 10 −3 A/cm 2
→ J L = 150 10 −3 A/cm 2
Now
 J 
VOC = Vt ln 1 + L 
 JS 
 150 10 −3 

= (0.0259 ) ln 1 +
−11 
 3.6 10 
or V OC = 0.574 V
_______________________________________
TYU 14.4
 Vm
1 +
 V
t


V
 exp  m

V

 t
 V
I m = I L − I S exp  m
  Vt
(
)
I S = 3.6 10 −11 (1) = 3.6 10 −11 A
 
 − 1

 
)
  0.521  
= 0.414 − 3.6 10 −11 exp 
 − 1
  0.0259  
or I m = 0.394 A
So
Pm = I mV m = (0.394 )(0.521 ) = 0.205 W
_______________________________________
TYU 14.5
Use results from Example 14.5;
We have V R = 4.5 V
Then
 2(11.7 ) 8.85 10 −14 (0.695 + 4.5)
W =
1.6 10 −19

(
)
 10 16 + 10 16
  16
16
 10 10
 

 
1/ 2
( )( )
or W = 1.16  m
Now
V
0. 5
IL = R =
= 0.10 mA
R
5
Then
0.10 10 −3
JL =
= 0.10 A/cm 2
10 −3
We have
J L = e W + Ln + L p GL
(
)
and
0.10 = 1.6 10 −19 (1.16 + 35.4 + 10 ) 10 −4 G L
which yields
G L = 1.34 10 20 cm −3 s −1
(
)
(
)
For  = 1  m,  = 10 2 cm −1
1.24
= 1.24 eV
1
 I
(G )(h )
GL =
 I = L
h

E = h =

J
 = 1+ L

JS

So
 Vm 
 Vm 
10
1 +



 V  exp  V  = 1.15 10
t 

 t 
By trial and error,
V m = 0.521 V
We have I L = (0.414 )(1) = 0.414 A
(
Then
or
I =
(1.34 10 )(1.24 )(1.6 10 )
−19
20
2
10
= 0.266 W/cm 2
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 14
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Chapter 14
14.1
 m ax
(b)  =
1.24
=
m
Eg
1.24
= 1.11  m
1.12
1.24
= 1.88  m
(b) Ge:  m ax =
0.66
1.24
= 0.873  m
(c) GaAs:  m ax =
1.42
1.24
= 0.919  m
(d) InP:  m ax =
1.35
_______________________________________
(a) Si:  m ax =
14.2
(a) For  = 480 nm,
1.24 1.24
E=
=
= 2.58 eV

0.480
For  = 725 nm,
1.24
E=
= 1.71 eV
0.725
(b) For E = 0.87 eV,
1.24 1.24
=
=
= 1.43  m
E
0.87
For E = 1.32 eV,
1.24
=
= 0.939  m
1.32
For E = 1.90 eV,
1.24
=
= 0.653  m
1.90
_______________________________________
14.3
(i) From Figure 14.4,   2.6  10 4 cm −1
I (d )
(ii) 
= exp (− d )
I 0
(
)
)(
= exp − 2.6 10 4 0.80 10 −4
= 0.125
Fraction absorbed = 1 − 0.125 = 0.875
_______________________________________
14.4
g =
 I (x )
h
For h = 1.3 eV,  =
1.24
= 0.95  m
1.3
For silicon:   3 10 2 cm −1
Then for I (x ) = 10 −2 W/cm 2 , we obtain
g =
(3 10 )(10 )
(1.6 10 )(1.3)
−2
2
−19
or
g  = 1.44 10 19 cm −3 s −1
The excess concentration is
n = g  = 1.44 10 19 10 −6
or
n = 1.44  10 13 cm −3
_______________________________________
(
)(
14.5
(a) p = g  p 0  g  =
)
p
 p0
5 10 15
= 2.5 10 22 cm −3 s −1
2 10 −7
For h = 1.65 eV,
1.24
 =
= 0.752  m
1.65
From Figure 14.4,   9  10 3 cm −1
(g )(h )
I 0 =

2.5 10 22 1.6 10 −19 (1.65 )
=
9 10 3
= 0.733 W/cm 2
g =
1.24
= 0.752  m
(a)  =
1.65
(i) From Figure 14.4,   9  10 3 cm −1
I (d )
= exp (− d )
(ii) 
I 0
(
1.24
= 0.653  m
1.90
)(
)
= exp − 9 10 3 1.2 10 −4
= 0.340
Fraction absorbed = 1 − 0.34 = 0.66
(
)(
)
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 14
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(b)
I  (d )
= 0.1 = exp (− d )
I 0
(
where L p = D p p
) 
0.1 = exp − 9 10 3 (d )
d=
1
 1 
ln 

3
9  10
 0 .1 
= 2.56  10 −4 cm = 2.56  m
_______________________________________
14.6
1.24 1.24
=
=
= 0.886  m
E
1.40
From Figure 14.4,   4.5  10 2 cm −1
I (d )
= 0.1 = exp (− d )
(a) 
I 0
d=
1
 1 
 1 
ln 
ln 
=

  0.1  4.5  10 2  0.1 
1
= 5.12  10 −3 cm = 51.2  m
(b) d =
1
 1 
ln 

2
4.5  10
 0.3 
= 2.68  10 −3 cm = 26.8  m
_______________________________________
14.7
GaAs:
For x = 1  m = 10 −4 cm, we have 50%
absorbed or 50% transmitted, then
I (x )
= 0.50 = exp (−  x )
IO
We can write
1  1   1 
 =    ln 
 =  − 4   ln (2)
 x   0.5   10 
or
 = 0.69  10 4 cm −1
This value corresponds to
 = 0.75  m , E = 1.65 eV
_______________________________________
14.8
The ambipolar transport equation for minority
carrier holes in steady state is
d 2 (p n )
p
Dp
+ GL − n = 0
2
p
dx
or
d 2 (p n ) p n
G
− 2 =− L
2
Dp
dx
Lp
The photon flux in the semiconductor is
 ( x ) =  O exp (−  x )
and the generation rate is
G L =   (x ) =   O exp (−  x )
so the differential equation becomes
d 2 (p n ) p n
 O
− 2 =−
exp (−  x )
Dp
dx 2
Lp
The general solution is of the form
−x


 + B exp  + x 
p n (x ) = A exp 
 Lp 
 Lp 




  O p
− 2 2
 exp (−  x )
 L p −1
As x →  , p n = 0 so that B = 0 . Then
 − x    O p
−
 exp (−  x )
 L p   2 L2 − 1
p


At x = 0 , we have
d (p n )
Dp
= sp n
dx x = 0
x =0
so we can write
  O p
p n
= A− 2 2
x =0
 L p −1
p n (x ) = A exp 
and
d (p n )
dx
=−
x =0
2
A   O p
+ 2 2
L p  L p −1
Then we have
AD p  2  O p D p
s  O  p
−
+
= sA − 2 2
2 2
Lp
 L p −1
 L p −1
Solving for A, we find
  O p  s +  D p 
A= 2 2


 L p − 1  s + D p L p 
The solution can now be written as
  O p
p n (x ) = 2 2
 L p −1
(
)
 s + Dp

−x
 − exp (−  x )

 exp 
 Lp 
 s + D p L p



_______________________________________
(
)
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 14
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
14.9
We have
d 2 n p
n p
Dn
+ GL −
=0
2
n
dx
or
d 2 n p n p
G
− 2 =− L
Dn
dx 2
Ln
The solution is then

 x 
 x 
n p (x ) = G L n 1 − cosh  + B sinh 

 L n 
 Ln 
where B was just given.
_______________________________________
( )
( )
14.10
The general solution can be written in the
form
 x 
 x 
n p (x ) = A cosh  + B sinh  + G L n
 Ln 
 Ln 
For s =  at x = 0 means n p (0) = 0 . Then
0 = A + G L n  A = −G L  n
At x = W ,
d n p
− Dn
= s o n p
dx x =W
x =W
L p = D p p 0 =
= 2.236  10 −3 cm
 Dn
Dp
+
Now J S = eni2 
 Ln N a L p N d

(
(
dx
=−
x =W
so we can write
W
G L n D n
sinh
Ln
 Ln
−10
(
A/cm
)( )
2
I S = AJ S = (5) 1.79 10 −10

 + G L n


W 
G L n
sinh 
Ln
 Ln 
+
2
)( ) (
J S = 1.790 10
W
n p (W ) = −G L n cosh
 Ln
( )
)






25
10

+
−3
16
−3
15 
2.236 10 10 
 5 10 10
Now
and
d n p
)(
= 1.6 10 −19 1.5 10 10
( )




W
+ B sinh
 Ln
(25)(10 −6 ) = 5 10 −3 cm
(10 )(5 10 −7 )
Ln = Dn n 0 =
where Ln = Dn n
= 8.950  10 −10 A
(a) I L = eG L AW
We find
 10 16 10 15 
Vbi = (0.0259 ) ln 
 = 0.6350 V
2
 1.5 10 10 
( )( )
(
)

2 V
W =  s bi

 e
W 
B
cosh 
Ln
 Ln 
 BDn
W
−

 L cosh L
n

 n
W 
= s o − G L  n cosh 
 Ln 
(
 Na + Nd

 N N
a
d

)





1/ 2
 2(11.7 ) 8.85 10 −14 (0.635 )
=
1.6 10 −19






W 
 + G L n 
+ B sinh


 Ln 
Solving for B, we obtain


W 
 W   
G L  Ln sinh  + s o n cosh  − 1 

 Ln 
 Ln   

B=
W 
W 
Dn
cosh  + s o sinh 
Ln
 Ln 
 Ln 
)
 10 16 + 10 15
  16
15
 10 10
 

 
1/ 2
( )( )
W = 9.508  10 −5 cm
Then
I L = 1.6 10 −19 5 10 21 (5) 9.508 10 −5
= 0.380 A = 380 mA
(
)(
) (
)
 I 
(b) Voc = Vt ln 1 + L 
 IS 
0.380 

= (0.0259 ) ln 1 +

−10
 8.95  10 
V oc = 0.5145 V
Voc 0.5145
=
= 0.810
Vbi
0.635
_______________________________________
(c)
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 14
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
14.11
From Problem 14.10, I S = 8.95 10 −10 A
 I 
(a) Voc = Vt ln 1 + L 
 IS 

120 10 −3
= (0.0259 ) ln 1 +
−10
 8.95 10
= 0.4847 V
 V  
(b) I = I L − I S exp   − 1
  Vt  




 V
(ii) 1 + m
 Vt

I
 = 1+ L

IS

10 10 −3
= 1+
8.95 10 −10
= 1.117  10 7
By trial and error, V m  0.351 V
Now
 V  
I m = I L − I S exp  m  − 1
  Vt  
= 10  10 −3
100  10 −3 = 120  10 −3
 V
− 8.95  10 −10 exp 
  Vt
V = 0.4383 V
 Vm 
V 
I
 exp  m  = 1 + L
(c) 1 +



IS
 Vt 
 Vt 
 
 − 1

 
120 10 −3
8.95 10 −10
= 1.341  10 8
By trial and error, V m  0.412 V
Now
 V  
I m = I L − I S exp  m  − 1
  Vt  
= 1+
= 120  10 −3
(
)
  0.412  
− 8.95 10 −10 exp 
 − 1
  0.0259  
 I m = 112 .75 10 −3 A = 112.75 mA
Pm = I mV m = (112 .75 )(0.412 )
= 46.5 mW
V
0.412
(d) Vm = I m R L  R L = m =
I m 0.11275
R L = 3.65 
_______________________________________
14.12
From Problem 14.10, I S = 8.95 10 −10 A
(a)
 I 
(i) Voc = Vt ln 1 + L 
 IS 

10 10 −3
= (0.0259 ) ln 1 +
−10
 8.95 10
= 0.420 V

V
 exp  m

V

 t




(
)
  0.351  
− 8.95 10 −10 exp 
 − 1
  0.0259  
I m = 9.31 10 −3 A = 9.31 mA
Then
Pm = I mV m = (9.31)(0.351 ) = 3.27 mW
(b)

100 10 −3 

(i) Voc = (0.0259 ) ln 1 +
−10 
 8.95 10 
= 0.480 V
 V 
V 
I
(ii) 1 + m  exp  m  = 1 + L
IS
 Vt 
 Vt 
100 10 −3
= 1+
8.95 10 −10
= 1.117  10 8
By trial and error, V m  0.407 V
Now
 V  
I m = I L − I S exp  m  − 1
  Vt  
= 100  10 −3
(
)
  0.407  
− 8.95 10 −10 exp 
 − 1
  0.0259  
I m = 9.40 10 −2 A = 94.0 mA
Then
Pm = I mV m = (94 .0 )(0.407 ) = 38 .3 mW
Pm 2 38.3
=
= 11.7
Pm1 3.27
_______________________________________
(c)
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 14
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
14.13
VOC
 J 
= Vt ln 1 + L 
 JS 
 30 10 −3
= (0.0259 ) ln 1 +
JS

 V  
I = 50  10 −3 − 4.579  10 −12 exp   − 1
  Vt  
We see that when I = 0 , V = V OC = 0.599 V.
We find
(




Dn
n
(
+
)(
J S = 1.6 10 −19 1.8 10 6
 1

 N a
Dp 

p 

1
Nd
)
2
225
1
+ 19
−8
5  10
10
7
5  10 −8



or
 6.708 10 4

J S = 5.184 10 −7 
+ 1.183 10 −15 
N
a


Then
J S (A/cm 2 )
VOC (V)
N a (cm −3 )
(
)
10 15
3.477  10 −17
0.891
10
16
3.478  10
−18
0.950
10
17
3.484  10
−19
1.01
3.539  10 −20
1.07
10 18
_______________________________________
14.14
(a)
I L = J L  A = 25 10 −3 (2) = 50 10 −3 A
We have
 1 D
Dp 
1
n

J S = eni2 
+
Nd  p 
 N a  n

or
(
(
)
)(
J S = 1.6 10 −19 1.5 10 10
)
2
 1
18
1

+ 19
16
−6
3

10
5

10
10

which becomes
J S = 2.289 10 −12 A/cm 2
or
I S = 4.579 10 −12 A
We have
 V  
I = I L − I S exp   − 1
  Vt  
or

6
−7 
5  10 
I (mA)
50
50
50
50
49.98
49.84
48.89
42.36
33.46
14.19
V (V)
0
0.1
0.2
0.3
0.4
0.45
0.50
0.55
0.57
0.59
where
 1
J S = eni2 
 N a
which becomes
)
(b) The voltage at the maximum power point
is found from
 Vm 
 Vm 
I
 = 1+ L
1 +
  exp 

V
V
IS
t 

 t 
50 10 −3
4.58 10 −12
= 1.092  10 10
= 1+
By trial and error,
V m = 0.520 V
At this point, we find
I m = 47 .6 mA
so the maximum power is
Pm = I mV m = (47 .6 )(0.520 )
or
Pm = 24 .8 mW
(c) We have
V V
0.520
V = IR  R = = m =
I
I m 47.6 10 −3
or
R = 10.9 
_______________________________________
14.15
 180 10 −3
(a) Voc = (0.0259 ) ln 1 +
2 10 −9

= 0.474 V
 V 
V 
I
(b) 1 + m  exp  m  = 1 + L
IS
 Vt 
 Vt 
= 1+




180 10 −3
= 9  10 7
−9
2 10
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 14
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(e) Then I = (6 )(0.09463 ) = 0.5678 A
By trial and error, V m  0.402 V
V 
I m  I L − I S exp  m 
 Vt 
(
)
 0.402 
= 180  10 −3 − 2  10 −9 exp 

 0.0259 
−1
= 1.69  10 A = 169 mA
Pm = I mV m = (169 )(0.402 ) = 67 .9 mW
Vm 0.402
=
= 2.379 
I m 0.169
(d) R L → (1.5)(2.379 ) = 3.568 
Now
V 
V
I=
= I L − I S exp  
RL
 Vt 
V
 V 
= 180  10 −3 − 2  10 −9 exp 

3.568
 0.0259 
By trial and error, V  0.444 V
V
0.444
=
= 0.1244 A
Then I =
R L 3.568
P = IV = (124 .4 )(0.444 ) = 55 .2 mW
_______________________________________
(c) R L =
(
)
14.16
 100 10
(a) Voc = (0.0259 ) ln 1 +
10 −10

= 0.5367 V
 V 
V 
I
(b) 1 + m  exp  m  = 1 + L
IS
 Vt 
 Vt 
= 1+
−3




100 10 −3
10 −10
= 10 9
By trial and error, V m  0.461 V
Then
 0.461 
I m = 100  10 −3 − 10 −10 exp 

 0.0259 
(
)
= 9.463  10 −2 A = 94.63 mA
Pm = I mV m = (94 .63 )(0.461 ) = 43 .62 mW
10
= 21 .7 → n = 22 cells
0.461
(d) Now V = (22 )(0.461 ) = 10 .14 V
P = IV
5.2 = I (10 .14 )  I = 0.5128 A
(c) n =
Then n  =
V
10 .14
=
= 17 .86 
I 0.5678
_______________________________________
So R L =
0.5128
= 5.42 → n  = 6
0.09463
14.17
Let x = 0 correspond to the edge of the space
charge region in the p-type material. Then in
the p-region
d 2 n p
n p
Dn
+ GL −
=0
2
n
dx
or
d 2 n p n p
G
− 2 =− L
2
Dn
dx
Ln
( )
( )
where
G L =   (x ) =   O exp (−  x )
Then we have
d 2 n p n p
 O
− 2 =−
exp (−  x )
2
Dn
dx
Ln
( )
The general solution is of the form
−x
+x
 + B exp 

n p (x ) = A exp 

L 
 Ln 
 n 
  O n
exp (−  x )
 2 L2n − 1
As x →  , n p = 0 so that B = 0 . Then
−
 − x    O n
− 2 2
 exp (−  x )

 Ln   Ln − 1
 
We also have n p (0) = 0 = A − 2 O2 n ,
 Ln − 1
which yields
 
A = 2 O2 n
 Ln − 1
We then obtain

    − x 
 − exp (−  x )
n p (x ) = 2 O2 n exp 

L
 L n − 1   n 

where  O is the incident flux at x = 0 .
_______________________________________
n p (x ) = A exp 
14.18
For 90% absorption, we have
(x )
= exp (−  x ) = 0.10
O
Then
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 14
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
exp (+  x ) =
14.20
n-type, so holes are the minority carrier
(a)
p = G L p = 10 21 10 −8
1
= 10
0. 1
or
(
1
x =    ln (10 )
 
p = n = 10 13 cm −3
(b)
 = e(p )  n +  p
14.19
(a) n o = N d = 510 15 cm −3
I = e n n o A
(
)
(
= 1.6 10 −19 (1200 ) 5 10 15
(
(
)
)
p
14
= 2.56  10 −2 (  -cm) −1
(d) I L = ( )A
(
)(
)
3


= 2.56  10 − 2 5 10 − 4 

−4
 120  10 
−3
−19
3.2  10 −3
10 21 5  10 − 4 120  10 − 4
)(
)(10 )(8000 + 250 )
13
(
or
)(
0.66  10 −3
10 21 10 − 4 100  10 − 4
(1.6 10 )(
−19
)
)(
)(
)(
)
)
ph = 4.125
_______________________________________
14.21
 ( x ) =  O exp (−  x )
The electron-hole generation rate is
g  =   (x ) =   O exp (−  x )
and the excess carrier concentration is
p =  p (x)
Now
 = e(p )  n +  p
(
)
and
J L = ( )
The photocurrent is now found from
= 3.2  10 A = 3.2 mA
IL
(e)  ph =
eG L AL
(1.6 10 )(
)
 = 1.32  10 −2 (  -cm) −1
(c)
( )AV
I L = J L  A = ( )A =
L
−2
1.32 10 10 −4 (5)
=
100 10 − 4
or
I L = 0.66 mA
(d)
IL
 ph =
eG L AL
=
( )( )
 = e(p )( +  )
= (1.6 10 )(10 )(1200 + 400 )
−19
−19
or
3


 5  10 − 4 

−4
 120  10 
I = 0.12 A = 120 mA
(b) p = GL p 0 = 10 21 10 −7 = 10 14 cm −3
n
(
= 1.6 10
and for h = 2.0 eV,   10 5 cm −1 .
Then
 1 
x =  5   ln (10 ) = 0.23  10 − 4 cm
 10 
or
x = 0.23  m
_______________________________________
=
)
or
For h = 1.7 eV ,   10 4 cm −1
Then
 1 
x =  4   ln (10 ) = 2.3  10 − 4 cm
 10 
or
x = 2.3  m
(c)
)(
IL =
ph = 3.33
W
xO
0
0
 ( )  dA =  dy  ( )  dx
(
xO
) 
= We  n +  p  p  dx
_______________________________________
0
Then
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 14
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(
)
I L = We  n +  p   O p exp (−  x )dx
(

(
= 1.6 10
0
)
= We  n +  p   O p
which becomes
I L = We n +  p O p 1 − exp (−  xO )
)
Now
I L = 50 10 −4 1.6 10 −19 (1200 + 450 )(50 )
(
)(
( )(2 10
 10
16
−7
)
)1 − exp (− (5 10 )(10 ))
or
I L = 0.131  A
_______________________________________
14.22
( )(
(
 10 2 10
Vbi = (0.0259 ) ln 
2
 1.5 10 10
= 0.6530 V
16
15
)
)
= 10 −3 cm
(
)





1/ 2
)
 10 16 + 2  10 15
  16
15
 10 2  10
( )(
(
 

 
1/ 2
)
Then
W = 2.095  10 −4 cm
(a) I L1 = eWG L A
)(
)( )(
= 1.6 10 −19 2.095 10 −4 10 21 10 −3
= 3.352  10
−5
A = 33 .52  A
(b) In n-region,
p = G L p 0 = 10 21 10 −7 = 10 14 cm −3
(
)(
)
In p-region,
n = G L n0 = 10 21 5 10 −7
( )(
Dp
or
d 2 (p n )
p
+ GL − n = 0
2

p
dx
d 2 (p n ) p n
G
− 2 =− L
2
Dp
dx 
Lp
which yields
G L L2p
p np =
= G L p
Dp
 2(11.7 ) 8.85 10 −14 (0.653 + 5)
=
1.6 10 −19

(
 (2.095 + 35.36 + 10.0)10 −4
The general solution is found to be
 − x 


 + B exp  + x  
p nh (x ) = A exp 
 Lp 
 Lp 




The particular solution is found from
p np
G
− 2 =− L
Dp
Lp
(10 )(10 −7 )

 2  (V + V R )  N a + N d
W =  s bi
 N N
e

a
d


−3
positive in the negative x direction. The
homogenerous solution is found from
d 2 (p nh ) p nh
− 2 =0
dx  2
Lp

= 3.536  10 −3 cm
L p = D p p 0 =
)(10 )(10 )
where L p = D p p and where x  is
(25)(5 10 −7 )
Ln = Dn n 0 =
)
21
14.23
In the n-region under steady state and for
 = 0 , we have
−4
4
−19
I L = 7.593 10 −4 A = 0.7593 mA
_______________________________________
 1
 xO
 − exp (−  x )
 
 0
(
(
(c) I L = eGL A W + Ln + L p
xO
)
= 5 10 14 cm −3
)
The total solution is the sum of the
homogeneous and particular solutions, so we
have
 − x 


 + B exp  x   + G L p
p n (x ) = A exp 
 Lp 
 Lp 




One boundary condition is that p n remains
finite as x  →  which means that B = 0 .
Then at x  = 0 , p n (0 ) = 0 = p n (0 ) + p nO , so
that p n (0 ) = − p nO .
We find that
A = − p nO + GL p
(
)
The solution is then written as
 − x 

 Lp 


p n (x ) = G L p − (G L p + p nO ) exp 
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 14
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
The diffusion current density is found as
d (p n (x ))
J p = −eD p
dx
x = 0
But
d (p n )
d (p n )
=−
dx
dx 
since x and x  are in opposite directions.
So
d (p n )
J p = +eD p
dx  x= 0
(
= eD p − GL p + p nO
14.25
(a) G L 0 =
I  0
(10 3 )(0.080 )
=
h
(1.6 10 −19 )(1.5)
= 3.33  10 20 cm −3 s −1
Then
G L (x ) = G L 0 exp (− x )
) ( ) 
(
= 3.33 10 20 exp − 10 3 (x )
(b) J L = e o 1 − exp (− W )
eG L 0
1 − exp (− W )

(1.6 10 −19 )(3.333 10 20 )
=
(10 3 )
)
=
 −1 


 exp  − x  

 Lp 
 L p  x= 0




 ( )(

 1 − exp − 10 3 100 10 −4
Finally
J p = eGL L p +
eD p p nO
−2
)
J L = 5.33 10 A/cm = 53.3 mA/cm 2
_______________________________________
Lp
2
_______________________________________
14.26
(a) J L = eWG L
14.24
(a) J L = e o 1 − exp (− W )
(
)(5 10 )
 1 − exp − (10 )(2 10 )
Diode A: J L = 1.6 10
−19
17
−4
4
J L = 6.92 10 −2 A/cm 2
(
)(5 10 )
 1 − exp − (10 )(10 10 )
Diode B: J L = 1.6 10
−19
17
−4
4
J L  8.0 10 −2 A/cm 2
(
)(
)
 1 − exp − (10 )(80 10 )
Diode C: J L = 1.6 10 −19 5 10 17
−4
4
J L = 8.0 10 −2 A/cm 2
(b) J L = e o 1 − exp (− W )
(
)(
)
 1 − exp − (5 10 )(2 10 )
Diode A: J L = 1.6 10 −19 5 10 17
2
−4
J L = 7.613 10 −3 A/cm 2
(
)(
)
 1 − exp − (5 10 )(10 10 )
Diode B: J L = 1.6 10 −19 5 10 17
2
−4
J L = 3.148 10 −2 A/cm 2
(
)(
)
 1 − exp − (5 10 )(80 10 )
Diode C: J L = 1.6 10 −19 5 10 17
2
−4
J L = 7.853 10 −2 A/cm 2
_______________________________________
(
)(
)( )
= 1.6 10 −19 20 10 −4 10 21
= 0.32 A/cm
(b) J L = e o 1 − exp (− W )
2
eG L 0
1 − exp (− W )

(1.6 10 −19 )(10 21 )
=
(10 3 )
=
 ( )(

 1 − exp − 10 3 20 10 −4
)
J L = 0.138 A/cm
_______________________________________
2
14.27
The minimum  occurs when  = 1  m
which gives  = 10 2 cm −1 . We want
(x )
= exp (−  x ) = 0.10
O
which can be written as
1
exp (+  x ) =
= 10
0.10
Then
1
1
x = ln (10 ) = 2 ln (10 ) = 2.30  10 − 2 cm

10
or
x = 230  m
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 14
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
14.28
For Al x Ga 1− x As system, a direct bandgap for
0  x  0.45 , we have
E g = 1.424 + 1.247 x
At x = 0.45 , E g = 1.985 eV, so for the direct
bandgap
1.424  E g  1.985 eV
which yields
0.625    0.871  m
_______________________________________
14.29
(a) From Figure 14.24, E g  1.64 eV
=
1.24 1.24
=
= 0.756  m
Eg
1.64
(b) From Figure 14.24, E g  1.78 eV
=
1.24 1.24
=
= 0.697  m
Eg
1.78
_______________________________________
1.24
1.24
=
= 1.85 eV

0.670
From Figure 14.23, x  0.35
_______________________________________
14.31
Eg =
where T1 = 1 − R1 and where R1 is the
reflection coefficient (Fresnel loss), and the
factor T 2 is the fraction of photons that do not
experience total internal reflection. We have
 n − n1 

R1 =  2
 n 2 + n1 
so that
2
2
 n − n1 

T1 = 1 − R1 = 1 −  2
 n 2 + n1 
which reduces to
4n1 n 2
T1 =
(n1 + n 2 )2
Now consider the solid angle from the source
point. The surface area described by the solid
angle is  p 2 . The factor T1 is given by
 p2
4 R 2
From the geometry, we have
  p 2

sin C  =
 p = 2 R sin C
R
 2 
 2
Then the area is
 
A =  p 2 = 4 R 2  sin 2  C 
T1 =
14.30
Eg =
14.33
We can write the external quantum efficiency
as
 ext = T1T2
1.24
1.24
=
= 1.85 eV

0.670
From Figure 14.24, x  0.38
_______________________________________
14.32
(a) For GaAs, n 2 = 3.66 and for air, n1 = 1.0 .
The critical angle is
n 
 1 
 C = sin −1  1  = sin −1 
 = 15.86
 3.66 
 n2 
The fraction of photons that will not
experience total internal reflection is
2 C 2(15 .86 )
=
 8.81 %
360
360
(b)Fresnel loss:
2
2
 n − n1 
 3.66 − 1 
 = 
R =  2
 = 0.3258
n
+
n
 3.66 + 1 
1 
 2
The fraction of photons emitted is then
(0.0881 )(1 − 0.3258 ) = 0.0594  5.94 %
_______________________________________



 2 
Now
 p2
 
= sin 2  C 
4 R 2
 2 
From a trig identity, we have
  1
sin 2  C  = (1 − cos C )
T1 =
 2 
2
Then
1
(1 − cos C )
2
The external quantum efficiency is now
4n1 n 2
1
 ext = T1T2 =
 (1 − cos C )
2
(n1 + n 2 ) 2
T1 =
or
 ext =
2n1 n 2
(n1 + n 2 )2
(1 − cos C )
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 14
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
14.34
For an optical cavity, we have

N  = L
2
If  changes slightly, then N changes slightly
also. We can write
N 1 1 (N 1 + 1) 2
=
2
2
Rearranging terms, we find
N 1 1 (N 1 + 1) 2 N 1 1 N 1  2  2
−
=
−
−
=0
2
2
2
2
2
If we define  = 1 −  2 , then we have
N1

 = 2
2
2
We can approximate  2 =  , then
N1
2L
= L  N1 =
2

Then
1 2L


 =
2 
2
which yields
 =
2
2L
_______________________________________
14.35
For GaAs:
h = 1.42 eV   =
Then
 =
2
2L
=
1.24
= 0.873  m
1.42
(0.873 10 )
2(75 10 )
−4 2
−4
= 5.08 10 − 7 cm
or
 = 5.08 10 −3  m
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 15
By D. A. Neamen
Exercise Solutions
______________________________________________________________________________________
Chapter 15
Exercise Solutions
Test Your Understanding Solutions
Ex 15.1
TYU 15.1
(a) Collector Region,
V
(a) VCEQ = CC = 30 V
2
PT = I CQ VCEQ
30 = I CQ (30 )  I CQ = 1 A
I C (max ) = 2I CQ = 2 A
RL =
VCC
60
=
= 30 
I C (max ) 2
V
(b) VCEQ = CC = 20 V
2
PT = I CQ VCEQ
30 = I CQ (20 )  I CQ = 1.5 A
I C (max ) = 2I CQ = 3 A
RL =
VCC
40
=
= 13.3 
I C (max ) 3
VCC
= 10 V
2
1

5
P(max ) =   I C , m ax   VCEQ =  (10 )
2

2
= 25 W
I C (max ) = 5 A
(c) VCEQ =
VCC
20
=
= 4
I C (max ) 5
_______________________________________
RL =
Ex 15.2
(a) BV DSS = V DD = 24 V
I D, m ax =
V DD 24
=
=2A
RL
12
1
 1
  2  24 
PT =   I D , m ax   V DD  =   
2
 2
  2  2 
= 12 W
(b) BV DSS = V DD = 40 V
I D, m ax =
V DD 40
=
=5A
RL
8
1
 1
  5  40 
PT =   I D , m ax   V DD  =   
2
 2
  2  2 
= 50 W
_______________________________________

 2  (V + V R )  N a
x n =  s bi
N
e

 d

Neglecting V bi ,
(

1

 N + N
d
 a





1/ 2
)
 2(11.7 ) 8.85 10 −14 (200 )
xn = 
1.6 10 −19

 10 16
  14
 10

1
 16
 10 + 10 14






1/ 2
or x n = 50 .6  m
(b) Base Region,
 2(11.7 ) 8.85 10 −14 (200 )
xp = 
1.6 10 −19

(
 10 14
  16
 10
or x p = 0.506  m
)

1
 16
 10 + 10 14






1/ 2
_______________________________________
TYU 15.2
VCE = 2VCC − I C R E
= 20 − I C (0.2 )
So
0 = 20 − I C (max )(0.2 )
 I C (max ) = 100 mA
Maximum power at the center of the load line
 2V  I (max )   20  0.10 
Pm ax =  CC  C
 =  

2
 2 
  2  2 
Pm ax = 0.5 W
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 15
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
Chapter 15
15.1 See diagrams in Figure 8.29
_______________________________________
15.2
V
0.60 − 0.15
=
= −25 
I (2 − 20 ) 10 −3
_______________________________________
R=
15.6
s
10 7
= 5  10 9 Hz
2 L 2 10  10 − 4
= 5 GHz
_______________________________________
f =
=
15.7
(a) n po =
15.3
fr =
1
2 R m in C j
R m in
−1
Rp
1
=
2 (10 ) 2 10 −9
(
)
10
−1
1
= 2.39  10 Hz = 23.9 MHz
_______________________________________
7
15.4
(a) no L = 10 12 cm −2
10 12
= 10 −3 cm = 10  m
10 15
L
10 −3
=
= 6.667  10 −11 s
(ii)  =
 d 1.5 10 7
(i) L =
1
 6.667  10 −11
= 1.5  10 10 Hz = 15 GHz
(iii) f =
1
=
(b)
10 12
= 10 − 4 cm = 1  m
10 16
L
10 −4
=
= 6.667  10 −12 s
(ii)  =
 d 1.5 10 7
(i) L =
1
1
 6.667  10 −12
= 1.5  10 11 Hz = 150 GHz
_______________________________________
(iii) f =
=
15.5
V
9
=
= 6  10 3 V/cm
L 15  10 − 4
(b)  d  1.5 10 7 cm/s
(a)  =
d
1.5 10 7
= 110 10 Hz
L 15 10 − 4
= 10 GHz
_______________________________________
(c)
f =
=
(
)
(
ni2
1.5 10 10
=
NB
8 10 15
)
2
= 2.8125  10 4 cm −3
V 
(i) n p (0)  n po exp  BE 
 Vt 
 n p (0 ) 

 V BE = Vt ln 
 n po 




10 14

= (0.0259 ) ln 
4 
 2.8125 10 
= 0.5696 V
(ii) Neglecting any recombination in the base
eDB n po A
V 
IC 
exp  BE 
xB
 Vt 
1.6 10 −19 (20 ) 2.8125 10 4 (0.4)
=
2 10 − 4
 0.5696 
 exp 

 0.0259 
I C = 0.640 A
(
) (
)
(b) n p (0) = (0.1)N B = 8  10 14 cm −3
 8 10 14 

(i) V BE  (0.0259 ) ln 
4 
 2.8125 10 
= 0.6234 V
1.6 10 −19 (20 ) 2.8125 10 4 (0.4)
(ii) I C =
2 10 − 4
 0.6234 
 exp 

 0.0259 
(
) (
)
I C = 5.12 A
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 15
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
15.8
(a) From Figure 7.15, BV BC  450 V
(b) V pt =
15.10
(a) BVCEO =
N (N + N B )
ex
 B C
2 s
NC
2
B
(1.6 10 )(2 10 )
=
2(11.7 )(8.85 10 )
(8 10 )(6 10

−4 2
−19
−14
15
14
+ 8 10 15
)
6 10 14
V pt = 354 .4 V
(c) From Figure 7.15, BV BE  65 V
_______________________________________
15.9
From the junction breakdown curve, for
BVCBO = 1000 V, we need the collector doping
concentration to be N C  210 14 cm −3 .
Depletion width into the base (neglect V bi ).
2 V
x p =  s BC
e

(
(
 NC

 NB


1




 N B + N C 
1/ 2
)
 2(11.7 ) 8.85 10 −14 (1000 )
=
1.6 10 −19

)
 2 10 14
 
15
 5 10

1

 5 10 15 + 2  10 14




1/ 2



1/ 2
or
x p = 3.16  10 −4 cm = 3.16  m
(Minimum base width)
Depletion width into the collector
2 V
x n =  s BC
e

(
(
 NB

N
 C


1


 N + N 
C 
 B

1/ 2
)
 2(11.7 ) 8.85 10 −14 (1000 )
=
1.6 10 −19

 5 10 15
 
14
 2 10
)

1

 5 10 15 + 2  10 14

BVCBO
n

300
(i) BVCEO =
3
(ii) BVCEO =
3
10
300
50
= 139 V
= 81.4 V
(b)
125
(i) BVCEO =
3
(ii) BVCEO =
3
10
125
= 58.0 V
= 33.9 V
50
_______________________________________
15.11
(a) We have
 eff =  A  B +  A +  B
so
180 = 25  B + 25 +  B
or
155 = 26  B
which yields
 B = 5.96
(b) We have
 B i EA = i CB
or
 1+  A 
  iCA = iCB
 B 
 A 
so
(5.96 ) 1 + 25   iCA = 20
 25 
which yields
i CA = 3.23 A
_______________________________________
15.12
or
x n = 78 .9 10 −4 cm = 78.9  m
(Minimum collector width)
_______________________________________
(
1

(b) PT =   I C , m ax  VCEQ
2

)
1

30 =   I C , m ax (60 )
2


 I C ,max = 1.0 A
RL =
VCEQ
I CQ
=
60
= 120 
0.5
VCE , max = 120 V
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 15
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
(
1

(c) PT =   I C , m ax  VCEQ
2

)
2
30 =  VCEQ
2
 VCEQ = 30 V
RL =
VCEQ
I CQ
=
30
= 30 
1
VCE ,max = 2VCEQ = 2(30 ) = 60 V
(d) Same as part (b)
_______________________________________
15.13
V 
(a) PT = VCEQ I CQ =  CC   I CQ
 2 
10 = 6I CQ  I CQ = 1.667 A
RL =
VCEQ
I CQ
=
6
= 3.60 
1.667
(b) I C ,max = 2I CQ = 2(1.667 ) = 3.333 A
_______________________________________
15.14
If V CC = 25 V, then
V
25
I C (max ) = CC =
= 0.25 A  I C , rated
R L 100
The power
P = I C VCE = I C (VCC − I C R L )
Now, to find the maximum power point
dP
= 0 = VCC − 2 I C R L = 25 − I C (2)(100 )
dI C
which yields
I C = 0.125 A
So
P (max ) = (0.125 )25 − (0.125 )(100 )
or
P (max ) = 1.56 W  PT
So maximum VCC is V CC = 25 V
_______________________________________
15.15
V DS
ID
Power dissipated in the transistor
V2
P = I DV DS = DS
Ron
Now Ron =
We have
200 − V DS
ID =
100
so we can write
V2
 200 − V DS 
P = 
 V DS = DS
Ron
 100 
For T = 25 C, R on = 2  .
Then
V2
 200 − V DS 

  V DS = DS
2
 100 
which yields
V DS = 3.92 V
The power is
 200 − 3.92 
P=
(3.92 ) = 7.69 W
 100

We then have
V DS (V)
R on (  )
T (C)
P (W)
25
2.0
3.92
7.69
50
2.33
4.56
8.91
75
2.67
5.19
10.1
100
3.0
5.83
11.3
_______________________________________
15.16
(a) We have, for three devices in parallel,
V V V
+ +
= 5  V (1.51) = 5
1.8 2 2.2
or
V = 3.311 V
V
Then, I = , so that
R
I 1 = 1.839 A
I 2 = 1.656 A
I 3 = 1.505 A
Now, P = IV , so
P1 = 6.09 W
P2 = 5.48 W
P3 = 4.98 W
(b) Now
1
1 
 1
V
+
+
 = 5  V = 3.882 V
 1.8 3.6 2.2 
Then
I 1 = 2.157 A, P1 = 8.37 W
I 2 = 1.078 A, P2 = 4.19 W
I 3 = 1.765 A, P3 = 6.85 W
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 15
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
15.17
(a) Let the n-drift region doping
concentration be N d = 10 14 cm −3 .
15.18
(b) In the saturation region,
2
2
I D = K n (VGS − VT ) = (0.20 )(VGS − 2)
V DS = V DD − I D R L = 60 − I D (10 )
( )( )
(
)
 10 14 10 15 
Vbi = (0.0259 ) ln 

2
 1.5  10 10 
= 0.516 V
For the base region,

 2  (V + V R )  N d
x p =  s bi
N
e

 a

(
)
For VGS = 4 V, I D = 0.8 A, V DS = 52 V
P = I D V DS = (0.8)(52 ) = 41 .6 W


1


 N + N 
d 
 a

1/ 2
For VGS = 8 V, transistor biased in the
nonsaturation region.
60 − V DS
2
= (0.20 ) 2(8 − 2 )V DS − V DS
10
 2(11.7 ) 8.85 10 −14 (0.516 + 200 )
=
1.6 10 −19

 10 14
  15
 10

1


 14
 10 + 10 15 




1/ 2
)
 2(11.7 ) 8.85 10 −14 (0.516 + 200 )
xn = 
1.6 10 −19


1


 14
 10 + 10 15 



15.19
1/ 2
(
)

1

 N + N
d
 a





1/ 2





1/ 2
 2(11.7 ) 8.85 10 −14 (0.516 + 80 )
=
1.6 10 −19

 10 14
  15
 10

1
 14
 10 + 10 15

= channel length
 2(11.7 ) 8.85 10 −14 (0.516 + 80 )
xn = 
1.6 10 −19

)
 10 15
  14
 10

1
 14
 10 + 10 15

 V   I D , m ax 

(b) P =  DD   

 2   2 
V
V
R L = DD = 10  I D , m ax = DD
I D , m ax
10
Then
x p = 3.08  10 −4 cm = 3.08  m
(
1
 1

(a) P =   V DD   I D , m ax 
2
2



 60  I 
45 =   D   I D , m ax = 3 A
 2  2 
V
60
R L = DD =
= 20 
I D , m ax
3
x n = 4.86 10 −3 cm = 48.6  m
= drift region width
(b) Assume N d = 10 14 cm −3
V bi = 0.516 V

 2  (V + V R )  N d
x p =  s bi
N
e

 a

2
We obtain 2.0V DS
− 25V DS + 60 = 0
 V DS = 3.24 V, I D = 5.676 A
For VGS = 6 V, P  PT so transistor may
be damaged.
_____________________________________
= channel length
 10 15
  14
 10

 P = (3.24 )(5.676 ) = 18 .39 W
x p = 4.86  10 −4 cm = 4.86  m
(
For VGS = 6 V, I D = 3.2 A, V DS = 28 V
P = (3.2 )(28 ) = 89 .6 W





1/ 2
x n = 3.08 10 −3 cm = 30.8  m
= drift region width
_______________________________________
V  V 
P =  DD    DD 
 2   20 
Or
V2
45 = DD
40
 V DD = 42 .4 V
_______________________________________
Semiconductor Physics and Devices: Basic Principles, 4th edition
Chapter 15
By D. A. Neamen
Problem Solutions
______________________________________________________________________________________
15.20
We have  1 +  2 = 1 . Now
1 =
1
1 + 1
and  2 =
2
1+  2
so


1 +  2 = 1 + 2 = 1
1 + 1 1 +  2
which can be written as
 (1 +  2 ) +  2 (1 +  1 )
1= 1
(1 +  1 )(1 +  2 )
or
(1 +  1 )(1 +  2 ) =  1 (1 +  2 ) +  2 (1 +  1 )
Expanding, we find
1 + 1 +  2 + 1 2
= 1 + 1 2 +  2 + 1  2
which yields
1 2 = 1
_______________________________________
15.21
The reverse-biased p-well to substrate junction
corresponds to the J 2 junction in an SCR. The
photocurrent generated in this junction will be
similar to the avalanche generated current in an
SCR, which can trigger the device.
_______________________________________
15.22
Case 1: Terminal 1(+), terminal 2(-), and I G
negative: this triggering was discussed in the
text.
Case 2: Terminal 1(+), terminal 2(-), and I G
positive: the gate current enters the P2 region
directly so that J3 becomes forward biased.
Electrons are injected from N2 and diffuse into
N1, lowering the potential of N1. The junction
J2 becomes more forward biased, and the
increased current triggers the SCR so that
P2N1P1N4 turns on.
Case 3: Terminal 1(-), terminal 2(+), and I G
positive: the gate current enters the P2 region
directly so that the J3 junction becomes more
forward biased. More electrons are injected
from N2 into N1 so that J1 also becomes more
forward biased. The increased current triggers
the P1N1P2N2 device into its conducting state.
Case 4: Terminal 1(-), terminal 2(+), and I G
negative: in this case, the J4 junction becomes
forward biased. Electrons are injected from N3
and diffuse into N1. The potential of N1 is
lowered which increases the forward biased
potential of J1. This increased current then
triggers the P1N1P2N2 device into its
conducting state.
_______________________________________