Modern Semiconductor Devices for Integrated Circuits: Answers to Problems (1st Edition) by Chenming Hu

Chapter 1
Visualization of the Silicon Crystal
1.1 (a) Please refer to Figure 1-2. The 8 corner atoms are shared by 8 unit cells and
therefore contribute 1 atom. Similarly, the 6 face atoms are each shared by 2 unit
cells and contribute 3 atoms. And, 4 atoms are located inside the unit cell.
Hence, there are total 8 silicon atoms in each unit cell.
(b) The volume of the unit cell is
    3
22
3
8
3
10
60
1
10
43
5
43
5 cm
.
cm
.
A
.
V cell
unit






 ,
and one unit cell contains 8 silicon atoms. The atomic density of silicon is
3
22
10
00
5
8 


 cm
atoms)
(silicon
.
V
atoms
silicon
N
cell
unit
Si .
Hence, there are 5.001022
silicon atoms in one cubic centimeter.
(c) In order to find the density of silicon, we need to calculate how heavy an
individual silicon atom is
 
 
 
g/atom
.
atoms/mole
.
g/mole
.
Mass atom
Si
23
23
1 10
67
4
10
02
6
1
28 



 .
Therefore, the density of silicon (Si) in g/cm3
is
3
atom
Si
1
Si
Si cm
/
g
2.33
Mass
N
ρ 

 .
Fermi Function
1.2 (a) Assume E = Ef in Equation (1.7.1), f(E) becomes ½. Hence, the probability is ½.
(b) Set E = Ec + kT and Ef = Ec in Equation (1.7.1):
 
  27
0
1
1
1
1
1
.
e
e
f(E) /kT
E
kT
E c
c




 

.
The probability of finding electrons in states at Ec + kT is 0.27.
s
m
t
b
9
8
@
g
m
a
i
l
.
c
o
m
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* For Problem 1.2 Part (b), we cannot use approximations such as Equations (1.7.2)
or (1.7.3) since E-Ef is neither much larger than kT nor much smaller than -kT.
(c) f(E) is the probability of a state being filled at E, and 1-f(E) is the probability of a
state being empty at E. Using Equation (1.7.1), we can rewrite the problem as
kT)
E
f(E
kT)
E
f(E c
c 3
1 





   /kT
E
kT
E
/kT
E
kT
E f
c
f
c
e
e








3
1
1
1
1
1
where
 
 
 
 
 /kT
E
kT
E
/kT
E
kT
E
/kT
E
kT
E
/kT
E
kT
E
/kT
E
kT
E f
c
f
c
f
c
f
c
f
c
e
e
e
e
e

















 3
3
3
3
3
1
1
1
1
1
1
1
 /kT
E
kT
E f
c
e




 3
1
1
.
Now, the equation becomes
   /kT
E
kT
E
/kT
E
kT
E f
c
f
c
e
e








3
1
1
1
1
.
This is true if and only if
 
f
c
f
c E
kT
E
E
kT
E 




 3 .
Solving the equation above, we find
kT
E
E c
f 2

 .
1.3 (a) Assume E = Ef and T > 0K in Equation (1.7.1). f(E) becomes ½. Hence, the
probability is ½.
(b) f(E) is the probability of a state being filled at E, and 1-f(E) is the probability of a
state being empty at E. Using Equation (1.7.1), we can rewrite the problem as
)
E
f(E
)
E
f(E v
c 


 1
   /kT
E
E
/kT
E
E f
v
f
c
e
e





 1
1
1
1
1
where

 
 
 
 
   /kT
E
E
/kT
E
E
/kT
E
E
/kT
E
E
/kT
E
E
/kT
E
E f
v
f
v
f
v
f
v
f
v
f
c
e
e
e
e
e
e

















1
1
1
1
1
1
1
1
1 .
Now, the equation becomes
   /kT
E
E
/kT
E
E f
v
f
c
e
e





 1
1
1
1
.
This is true if and only if
 
f
v
f
c E
E
E
E 


 .
Solving the equation above, we find
2
v
c
f
E
E
E

 .
(c) The plot of the Fermi-Dirac distribution and the Maxwell-Boltzmann distribution
is shown below.
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5 5
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
The Boltzmann distribution considerably overestimates the Fermi distribution for
small (E-Ef)/kT. If we set (E-Ef)/kT = A in Equations (1.7.1) and (1.7.2), we
have









A
A
e
.
e
1
1
10
1 .
Solving for A, we find
  31
2
11
10
ln
11
10
10
1
1
.
.
A
.
e
.
e
e A
A
A






 .
Therefore, the Boltzmann approximation is accurate to within 10% for (E-Ef)/kT
2.31.

Probability
Fermi-Dirac
Distribution
Maxwell-Boltzmann
Distribution
(E-Ef)/kT

1.4 (a) Please refer to the example in Sec. 1.7.2. The ratio of the nitrogen concentration
at 10 km above sea level to the nitrogen concentration at sea level is given by
)/kT
E
(E
/kT
E
/kT
E
Level
Sea
km Level
Sea
km
Level
Sea
km
e
e
e
N
N
N
N 




 10
10
)
(
)
(
2
10
2
where
gravity
of
on
accelerati
molecule
N
of
mass
altitude
E
E Level
Sea
km 


 2
10
.
10
56
4
980
10
66
1
28
10 14
2
24
6
erg
.
s
cm
g
.
cm 










The ratio is
30
0
)
(
)
( 21
1
273
10
38
1
10
56
4
2
10
2 1
16
14
.
e
e
N
N
N
N .
K)
K
erg
.
erg)/(
.
(
Level
Sea
km


 




 


.
Since nitrogen is lighter than oxygen, the potential energy difference for nitrogen
is smaller, and consequently the exponential term for nitrogen is larger than 0.25
for oxygen. Therefore, the nitrogen concentration at 10 km is more than 25% of
the sea level N2 concentration.
(b) We know that
25
.
0
)
(
)
(
2
10
2

Level
Sea
km
O
N
O
N
, 30
.
0
)
(
)
(
2
10
2

Level
Sea
km
N
N
N
N
, and 4
)
(
)
(
2
2

Level
Sea
Level
Sea
O
N
N
N
.
Then,
km
Level
Sea
Level
Sea
Level
Sea
Level
Sea
km
km
km
O
N
O
N
O
N
N
N
N
N
N
N
O
N
N
N
10
2
2
2
2
2
10
2
10
2
10
2
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(



8
.
4
25
.
0
1
4
30
.
0 


 .
It is more N2-rich than at sea level.
1.5
   /kT
E
E
E
f f
f
e
E
E
f 








1
1
1
1
 
 /kT
E
E
E
/kT
E
E
E
f
f
f
f
e
e








1
 /kT
E
E
E f
f
e






1
1

 /kT
E
E
E f
f
e






1
1
 /kT
E
E
E f
f
e





1
1
 
E
E
f f 


1.6 (a)
0.0
0.5
1.0
150K
f(E)
Ef E
300K
(b) At 0K, the probability of a state below the Fermi level being filled is 1 and a state
above the Fermi level being filled is 0. So a total of 7 states are filled which
means there are 14 electrons (since 2 electrons can occupy each state) in the
system.
Density of States
1.7 Since the semiconductor is assumed to be, We are asked to use Equations (1.7.2) and
(1.7.4) to approximate the Fermi distribution. (This means that the doping
concentration is low and Ef is not within a few kTs from Ec or Ev. A lightly doped
semiconductor is known as a non-degenerate semiconductor.) The carrier distribution
as a function of energy in the conduction band is proportional to
   /kT
E
E
/
c
f
e
E
E
(E)
on
Distributi




2
1
,
where e-(E-Ef)/kT
is from Equation (1.7.2). Taking the derivative with respect to E and
setting it to zero, we obtain

   
     
0
1
1
2
1
2
1


















 /kT
E
E
c
/kT
E
E
c
/kT
E
E
/
c
f
f
f
e
kT
E
E
e
E
E
e
E
E
dE
d
The exponential terms cancel out. Solving the remaining equation yields
     
2
2
1
2
1 2
1
2
1 kT
E
E
kT
E
E
E
E
kT
E
E c
c
/
c
/
c 









.
So, the number of carriers in the conduction band peaks at Ec+kT/2.
Similarly, in the valence band, the carrier distribution as a function of energy is
proportional to
   /kT
E
E
/
v
f
e
E
E
(E)
on
Distributi




2
1
,
where e-(Ef-E)/kT
is Equation (1.7.2). Taking the derivative and setting it to zero, we
obtain
   
     
   
0
1
2
1 2
1
2
1
2
1


















 /kT
E
E
/
v
/kT
E
E
/
v
/kT
E
E
/
v
f
f
f
e
kT
E
E
e
E
E
e
E
E
dE
d
.
Again, the exponential terms cancel out, and solving the remaining equation yields
     
2
2
1
2
1 2
1
2
1 kT
E
E
kT
E
E
E
E
kT
E
E v
v
/
v
/
v 









.
Therefore, the number of carriers in the valence band peaks at Ev-kT/2.
1.8 Since it is given that the semiconductor is non-degenerate (not heavily doped), Ef is not
within a few kTs from Ec or Ev. We can use Equations (1.7.2) and (1.7.4) to approximate the
Fermi-Dirac distribution.
(a) The electron concentration in the conduction band is given by
 
dE
e
E
E
A
dE
f(E)
(E)
D
n
/kT
E
E
E
c
C.B.
c
f
c




 

 .
In order to simplify the integration, we make the following substitutions:









to
from
x
and
dx
kT
dE
dx
dE
kT
E
x
kT
E
x
kT
E
E
c
c
0
:
,
1
, .
Now the equation becomes

 
   
dx
e
x
e
kT
A
dx
kT
e
kTx
A
n x
kT
E
E
kT
E
E
kTx f
c
f
c 








 

0
/
2
/
3
/
0
where
   
2
2
/
3
0
1
2
3
0




 
 



 dx
e
x
dx
e
x x
x
. (Gamma function)
Hence, the electron concentration in the conduction band is
    kT
E
E f
c
e
kT
A
n
/
2
/
3
2




.
Similarly, the hole concentration is given by
   
dE
e
E
E
B
dE
-f(E)
(E)
D
p
/kT
E
E
E
-
v
V.B.
v
f
v 



 

 1 .
Again, we make the following substitutions to simplify the integration:
0
:
,
1
, to
from
x
and
dx
kT
dE
dx
dE
kT
E
x
kT
E
x
kT
E
E
v
v












.
 
   
dx
e
x
e
kT
B
dx
kT
e
kTx
B
p x
kT
E
E
kT
E
kTx
E v
f
v
f 






 
 


0
/
2
/
3
/
0
where
   
2
2
/
3
0
1
2
3
0




 
 



 dx
e
x
dx
e
x x
x
. (Gamma function)
Therefore, the hole concentration in the conduction band is
    kT
E
E v
f
e
kT
B
p
/
2
/
3
2




.
(b) The word “Intrinsic” implies that the electron concentration and the hole concentration
are equal. Therefore,
   
    kT
E
E
kT
E
E v
i
i
c
e
kT
B
e
kT
A
p
n /
2
/
3
/
2
/
3
2
2









.
This simplifies to

    kT
E
E
kT
E
E v
i
i
c
e
B
e
A /
/ 



 .
Solving for Ei yields
K
T
K
eV
k
eV
E
E
kT
E
E
E v
c
v
c
i 300
,
10
62
.
8
;
009
.
0
2
2
1
ln
2
2
1
5














 

.
Hence, the intrinsic Fermi level (Ei) is located at 0.009 eV below the mid-bandgap of the
semiconductor.
1.9 The unit step functions set the integration limits. Dc(E) is zero for E < Ec, and Dv(E) is zero
for E > Ev. Since it is given that the semiconductor is non-degenerate (not heavily doped), Ef
is not within a few kTs from Ec or Ev. We can use Equations (1.7.2) and (1.7.4) to
approximate the Fermi-Dirac distribution.
(a) The electron concentration in the conduction band is given by
   
dE
e
E
E
A
dE
f(E)
(E)
D
n
/kT
E
E
E
c
C.B.
c
f
c




 

 .









to
from
x
and
dx
kT
dE
dx
dE
kT
E
x
kT
E
x
kT
E
E
c
c
0
:
,
1
, .
 
   
dx
e
x
e
kT
A
dx
kT
e
kTx
A
n x
kT
E
E
kT
E
E
kTx f
c
f
c 








 

0
/
2
/
0
where
1
0



 dx
e
x x
.
Hence, the electron concentration in the conduction band is
    kT
E
E f
c
e
kT
A
n
/
2 

 .
Similarly, the hole concentration is given by
     
dE
e
E
E
B
dE
-f(E)
(E)
D
p
/kT
E
E
E
-
v
V.B.
v
f
v 



 

 1 .
Again, we make the following substitutions to simplify the integration:

0
:
,
1
, to
from
x
and
dx
kT
dE
dx
dE
kT
E
x
kT
E
x
kT
E
E
v
v












.
 
   
dx
e
x
e
kT
B
dx
kT
e
kTx
B
p x
kT
E
E
kT
E
kTx
E v
f
v
f 






 
 


0
/
2
/
0
where
1
0



 dx
e
x x
.
Therefore, the hole concentration in the conduction band is
    kT
E
E v
f
e
kT
B
p
/
2 

 .
(b) The word “Intrinsic” implies that the electron concentration and the hole concentration are
equal. Therefore,
   
    kT
E
E
kT
E
E v
i
i
c
e
kT
B
e
kT
A
p
n /
2
/
2 





 .
This simplifies to
    kT
E
E
kT
E
E v
i
i
c
e
B
e
A /
/ 



 .
If we solve for Ei, we obtain
.
300
,
10
62
.
8
;
009
.
0
2
2
1
ln
2
2
1
5
K
T
K
eV
k
eV
E
E
kT
E
E
E v
c
v
c
i 













 

Hence, the intrinsic Fermi level (Ei) is located at 0.009 eV below the mid-bandgap of the
semiconductor.
1.10 (a) The carrier distribution as a function of energy in the conduction band is proportional
to
   /kT
E
E
/
c
f
e
E
E
(E)
on
Distributi




2
1
,
where e-(E-Ef)/kT
is from Equation (1.7.2). Taking the derivative and setting it to zero, we
obtain
   
     
0
1
1
2
1
2
1


















 /kT
E
E
c
/kT
E
E
c
/kT
E
E
/
c
f
f
f
e
kT
E
E
e
E
E
e
E
E
dE
d
.
The exponential terms cancel out. Solving the remaining equation yields

     
2
2
1
2
1 2
1
2
1 kT
E
E
kT
E
E
E
E
kT
E
E c
c
/
c
/
c 









.
Hence, the number of carriers in the conduction band peaks at Ec+kT/2.
(b) The electron concentration in the conduction band is given by
 


C.B.
Band
Conduction
the
of
Top
E
c
c
c
dE
f(E)
(E)
D
dE
f(E)
(E)
D
n .
We assume that the function f(E) falls off rapidly such that
0




Band
Conduction
the
of
Top
E
c
Band
Conduction
the
of
Top
c
c
dE
f(E)
(E)
D
dE
f(E)
(E)
D
.
Now we may change the upper limit of integration from the Top of the Conduction Band
to ∞:
 
dE
e
E
E
A
n
/kT
E
E
E
c
f
c



 
 .
Also, in order to simplify the integration, we make the following substitutions:









to
from
x
and
dx
kT
dE
dx
dE
kT
E
x
kT
E
x
kT
E
E
c
c
0
:
,
1
, .
The equation becomes
 
   
dx
e
x
e
kT
A
dx
kT
e
kTx
A
n x
kT
E
E
kT
E
E
kTx f
c
f
c 








 

0
/
2
/
3
/
0
where
   
2
2
/
3
0
1
2
3
0




 
 



 dx
e
x
dx
e
x x
x
. (Gamma function)
Therefore, the electron concentration in the conduction band is
    kT
E
E f
c
e
kT
A
n
/
2
/
3
2




.
(b) The ratio of the peak electron concentration at E = Ec+(1/2)kT to the electron
concentration at E = Ec+40kT is

   
    













kT
E
kT
E
c
c
kT
E
kT
E
C
c
c
c
f
c
f
c
e
E
kT
E
A
e
E
kT
E
A
kT
E
n
kT
E
n
/
5
.
0
2
/
1
/
40
2
/
1
5
.
0
40
)
2
1
(
)
40
(
     
  16
5
.
39
/
5
.
0
40
2
/
1
10
60
.
5
)
5
.
0
/
40
(
5
.
0
/
40 










 e
e
kT
kT
kT
E
kT
E
E
kT
E f
c
f
c
.
The ratio is very small, and this result justifies our assumption in Part (b).
(c) The kinetic energy of an electron at E is equal to E-EC. The average kinetic energy of
electrons is


electrons
of
number
total
electrons
all
of
energy
kinetic
the
of
sum
E
K .
.
 

 

C.B.
c
C.B.
c
c
dE
f(E)
(E)
D
dE
f(E)
(E)
D
E
E
   
 
dE
e
E
E
A
dE
e
E
E
A
E
E
/kT
E
E
E
c
/kT
E
E
E
c
c
f
c
f
c











 .









to
from
x
and
dx
kT
dE
dx
dE
kT
E
x
kT
E
x
kT
E
E
c
c
0
:
,
1
, .
   
   
   
 
   
  dx
e
x
e
kT
A
dx
e
x
e
kT
A
dx
kT
e
kTx
A
dx
kT
e
kTx
A
x
kT
E
E
x
kT
E
E
kT
E
E
kTx
kT
E
E
kTx
f
c
f
c
f
c
f
c



















 
0
2
/
1
/
2
/
3
0
2
/
3
/
2
/
5
/
0
2
/
1
/
0
2
/
3
where
     
4
3
2
/
5
0
1
2
5
0
2
/
3 



 
 



 dx
e
x
dx
e
x x
x
(Gamma functions)
and
     
2
2
/
3
0
1
2
3
0
2
/
1 



 
 



 dx
e
x
dx
e
x x
x
. (Gamma functions)
Hence, the average kinetic energy is (3/2)kT.
Electron and Hole Concentrations
1.11 (a) We use Equation (1.8.11) to calculate the hole concentration:

  3
15
3
5
2
10
2
2
10
10
/
10
/ 






 cm
cm
n
n
p
n
p
n i
i .
(b) Please refer to Equations (1.9.3a) and (1.9.3b). Since Nd-Na >> ni and all the impurities
are ionized, n = Nd-Na, and p = (ni)2
/(Nd-Na).
(c) Since the Fermi level is located 0.26 eV above Ei and closer to Ec, the sample is n-type.
If we assume that Ei is located at the mid-bandgap (~ 0.55 eV), then Ec-Ef = 0.29 eV.
Ef
Ec
Ei
Ev
1
2
3
1: Ec-Ei =0.55 eV
2: Ec-Ef =0.29 eV
3: Ef-Ei =0.26 eV
Using Equations (1.8.5) and (1.8.11), we find
  3
5
2
3
14
/
10
49
.
2
/
10
01
.
4 








 cm
n
n
p
and
cm
e
N
n i
kT
E
E
c
f
c
.
Therefore, the electron concentration is 4.011014
cm-3
, and the hole concentration is
2.49105
cm-3
.
* There is another way to solve this problem:
  3
5
2
3
14
/
10
55
.
4
/
10
20
.
2 







 cm
n
n
p
and
cm
e
n
n i
kT
E
E
i
i
f
.
(d) If T = 800 K, there is enough thermal energy to free more electrons from silicon-
silicon bonds. Hence, using Equation (1.8.12), we first calculate the intrinsic carrier
density ni at 800 K:
        3
16
2
/
10
56
.
2
800
800 



 cm
e
K
N
K
N
n
kT
E
v
c
i
g
.
where
3
20
3
2
/
3
19
2
/
3
2
10
22
.
1
300
10
8
.
2
2
2
)
800
( 



















 cm
cm
K
T
h
kT
m
K
T
N dn
c

and
.
10
53
.
4
300
10
04
.
1
2
2
)
800
( 3
19
3
2
/
3
19
2
/
3
2




















 cm
cm
K
T
h
kT
m
K
T
N dp
v

Clearly, ni at 800K is much larger than Nd-Na (which is equal to n from the previous
part). Hence the electron concentration is nni, and the hole concentration is
p=(ni)2
/nni. The semiconductor is intrinsic at 800K, and Ef is located very close to the
mid-bandgap.
Nearly Intrinsic Semiconductor

1.12 Applying Equation (1.8.11) to this problem yields
3
13
3
12
2
2
2
10
41
.
1
10
07
.
7
2
1
2
2
/ 










 cm
n
and
cm
n
p
n
p
p
n
p
n i
i
i .
1.13 (a) B is a group III element. When added to Si (which belongs to Group IV), it acts as an
acceptor producing a large number of holes. Hence, this becomes a P-type Si film.
(b) At T = 300 K, since the intrinsic carrier density is negligible compared to the dopant
concentration, p=Na=41016
cm-3
, and n = (ni=1010
cm-3
)2
/p = 2500 cm-3
.
At T = 600 K,
        3
15
2
/
10
16
.
1
600
600 



 cm
e
K
N
K
N
n
kT
E
v
c
i
g
where
3
19
3
2
/
3
19
2
/
3
2
10
92
.
7
300
10
8
.
2
2
2
)
600
( 



















 cm
cm
K
T
h
kT
m
K
T
N dn
c

and
.
10
94
.
2
300
10
04
.
1
2
2
)
600
( 3
19
3
2
/
3
19
2
/
3
2




















 cm
cm
K
T
h
kT
m
K
T
N dp
v

The intrinsic carrier concentration is no more negligible compared to the dopant
concentration. Thus, we have
  3
16
3
15
16
10
12
.
4
10
16
.
1
10
4 








 cm
cm
n
N
p i
a , and
  3
13
3
16
2
3
15
2
10
27
.
3
10
12
.
4
/
10
16
.
1
/ 







 cm
cm
cm
p
n
n i .
The electron concentration has increased by many orders of magnitude.
(c) At high temperatures, there is enough thermal energy to free more electrons from
silicon-silicon bonds, and consequently, the number of intrinsic carriers increases.
(d) Using Equation 1.8.8, we calculate the position of the Fermi level with respect to Ev.
   
  K
T
eV
T
p
T
N
kT
E
E v
v
f 600
,
34
.
0
/
ln 


 .
At 600 K, the Fermi level is located 0.34 eV above the valence band.
Incomplete Ionization of Dopants and Freeze-out
1.14 From Equation (1.9.1), we know that n + Na
-
= p + Nd
+
. Since Nd
+
is much larger than Na
-
,
all the samples are n-type, and n  Nd
+
- Na
-
= 31015
/cm3
. This value is assumed to be
constant. Using the Equations (1.8.10) and (1.9.3b),

     
kT
E
CT
kT
E
N
N
N
N
n
p g
g
v
c
a
d
i /
exp
/
exp
/ 3
2





 

,
where C is a temperature independent constant. Using the sensitivity of p defined by
p/T,
   
kT
E
CT
kT
E
T
p g
g /
exp
/
3
/ 2






Therefore, the larger the energy gap is the less sensitive to temperature the minority carrier
is. For the definition of the sensitivity of p,
    T
kT
E
p
T
p g /
/
3
/
/ 



The temperature sensitivity of the minority carrier is greater for larger Eg.
1.15 (a) Let us first consider the case of n-type doping. The dopant atoms are located at
energy Ed inside the bandgap, near the conduction band edge. The problem states that
we are considering the situation in which half the impurity atoms are ionized, i.e.
n=Nd/2. In other words, the probability of dopant atoms being ionized is ½, or
conversely, the probability that a state at the donor energy ED is filled is ½.
From Problem 1.2 part (a), we know that if f (ED)=1/2, then ED=Ef. From Equation
1.8.5,
  kT
E
E
c
f
c
e
N
n
/


 .
We also know that Ef=ED and Ec-ED=0.05eV.
3
2
/
3
19
2
/
3
2
300
10
8
.
2
2
2
)
( 















 cm
K
T
h
kT
m
T
N dn
c

.
      kT
E
E
d
c
D
kT
E
E
c
kT
E
E
c
D
c
D
c
f
c
e
N
T
N
N
e
T
N
e
T
N /
/
/ )
(
2
2
)
(
)
( 







 .
This equation can be solved iteratively. Starting with an arbitrary guess of 100K for T,
we find T converges to 84.4 K.
Similarly, for boron
3
2
/
3
19
2
/
3
2
300
10
04
.
1
2
2
)
( 















 cm
K
T
h
kT
m
T
N dp
v

.
      kT
E
E
a
v
a
kT
E
E
v
kT
E
E
v
v
a
v
a
v
f
e
N
T
N
N
e
T
N
e
T
N /
/
/ )
(
2
2
)
(
)
( 







 .
Starting from T =100K, we find T converges to 67.7K.

(b) We want to find T where ni is 10Nd. This can be written as
           
d
kT
E
kT
E
v
c
i N
e
K
T
e
T
N
T
N
n g
g
10
300
10
71
.
1
2
/
2
/
3
19
2
/













where
3
2
/
3
19
2
/
3
2
300
10
8
.
2
2
2
)
( 















 cm
K
T
h
kT
m
T
N dn
c

and
3
2
/
3
19
2
/
3
2
300
10
04
.
1
2
2
)
( 















 cm
K
T
h
kT
m
T
N dp
v

.
We need to solve the equation iteratively, as in part (a) for ni=10Nd=1017
cm-3
. Starting
from T=300K, we get T=777 K for ni=10Nd.
For ni=10Na, we simply replace Nd in the equation above with Na. Starting from T

=300K, we find T=635 K.
(c) If we assume full ionization of impurities at T = 300 K,
For arsenic: n ,
i
d n
cm
N 

 3
16
10 3
4
2
10
1
.
2 


 cm
N
n
p
d
i
For boron: p ,
i
a n
cm
N 

 3
15
10  3
5
2
10
1
.
2 


 cm
N
n
n
a
i
(d) Please refer to the example in Section 2.8. For arsenic,
eV
cm
cm
p
N
kT
E
E v
v
f 88
.
0
10
1
.
2
10
04
.
1
ln 3
4
3
19





 

.
For boron,
eV
cm
cm
p
N
kT
E
E v
v
f 24
.
0
10
10
04
.
1
ln 3
15
3
19




 

.
(e) In case of arsenic + boron,
3
15
10
9 



 cm
N
N
n a
d , and
  3
4
3
15
2
3
10
2
10
11
.
1
10
9
10 






 cm
cm
cm
n
n
p i
, and
  eV
cm
cm
eV
p
N
kT
E
E v
v
f 90
.
0
10
11
.
1
10
04
.
1
ln
026
.
0
ln 3
4
3
19





















 

.

1.16 (a) If we assume full ionization of impurities, the electron concentration is n  Nd =
1017
cm-3
. The hole concentration is p=(ni)2
/n=(1010
cm-3
)2
/1017
cm-3
=103
cm-3
.
The Fermi level position, with respect to Ec, is
    eV
cm
cm
n
N
kT
E
E c
f
c 15
.
0
10
/
10
8
.
2
ln
026
.
0
/
ln 3
17
3
19




 

.
Ef is located 0.15 eV below Ec.
(b) In order to check the full ionization assumption with the calculated Fermi level, we
need to find the percentage of donors occupied by electrons.
    eV
E
E
E
E
E
E D
c
f
c
f
D 1
.
0





 , and
 
 
  d
eV
eV
kT
E
E
d
D N
of
cm
e
cm
e
N
n f
D
%
2
10
09
.
2
1
10
1
1 3
15
026
.
0
/
1
.
0
3
17
/






 


.
Since only 2% of dopants are not ionized, it is fine to assume that the impurities are
fully ionized.
(c) We assume full ionization of impurities, the electron concentration is n  Nd =
1019
cm-3
/n = (1010
cm-3
)2
/1019
cm-3
= 10 cm-3
.
    eV
cm
cm
n
N
kT
E
E c
f
c 027
.
0
10
/
10
8
.
2
ln
026
.
0
/
ln 3
19
3
19





 

.
It is located 0.027 eV below Ec.
Again, we need to find the percentage of donors occupied by electrons in order to
check the full ionization assumption with the calculated Fermi level.
    eV
E
E
E
E
E
E D
c
f
c
f
D 023
.
0






 , and
 
 
  d
eV
eV
kT
E
E
d
D N
of
cm
e
cm
e
N
n f
D
%
71
10
08
.
7
1
10
1
1 3
18
026
.
0
/
023
.
0
3
19
/






 



.
Since 71% of dopants are not ionized, the full ionization assumption is not correct.
(d) For T=30 K, we need to use Equation (1.10.2) to find the electron concentration since
the temperature is extremely low. First, we calculate Nc and Nv at T=30K:
3
17
3
2
/
3
19
2
/
3
2
10
85
.
8
300
10
8
.
2
2
2
)
30
( 



















 cm
cm
K
T
h
kT
m
K
T
N dn
c

and
.
10
29
.
3
300
10
04
.
1
2
2
)
30
( 3
17
3
2
/
3
19
2
/
3
2




















 cm
cm
K
T
h
kT
m
K
T
N
dp
v

The electron concentration is

      3
8
2
/
10
43
.
8
2
30 




 cm
e
N
K
N
n kT
E
E
d
c D
c
.
And, the hole concentration is
0
/
2

 n
n
p i
where
        3
75
2
/
10
32
.
2
2
30
30 




 cm
e
K
N
K
N
n
kT
E
v
c
i
g
.
Since ni is extremely small, we can assume that all the electrons are contributed by
ionized dopants. Hence,
 
9
3
17
3
8
3
8
/
10
43
.
8
10
10
43
.
8
10
43
.
8
1
1
1 



















cm
cm
cm
e
N
n kT
E
E
d f
D
.
The full ionization assumption is not correct since only 8.4310-7
% of Nd is ionized.
To locate the Fermi level,
eV
N
n
kT
E
E
d
f
D 048
.
0
1
1
ln
1























.
Ec-Ef = 0.05-0.048 = 0.002 eV. Therefore, the Fermi level is positioned 0.002 eV
below Ec, between Ec and ED.
1.17 (a) We assume full ionization of impurities, the electron concentration is n  Nd =
1016
cm-3
/n = (1010
cm-3
)2
/1016
cm-3
= 104
cm-3
.
    eV
cm
cm
n
N
kT
E
E c
f
c 21
.
0
10
/
10
8
.
2
ln
026
.
0
/
ln 3
16
3
19




 

.
We need to find the percentage of donors occupied by electrons in order to check the
full ionization assumption with the calculated Fermi level.
    eV
E
E
E
E
E
E D
c
f
c
f
D 16
.
0





 , and
 
 
  .
%
10
12
.
2
10
12
.
2
1
10
1
1 1
3
13
026
.
0
/
16
.
0
3
16
/ d
eV
eV
kT
E
E
d
D N
of
cm
e
cm
e
N
n f
D












Since only 0.21% of dopants are not ionized, the full ionization assumption is correct.

(b) We assume full ionization of impurities, the electron concentration is n  Nd =
1018
cm-3
/n = (1010
cm-3
)2
/1018
cm-3
= 102
cm-3
.
The Fermi level position with respect to Ec is
    eV
cm
cm
n
N
kT
E
E c
f
c 087
.
0
10
/
10
8
.
2
ln
026
.
0
/
ln 3
18
3
19




 

.
We need to find the percentage of donors occupied by electrons in order to check the
full ionization assumption with the calculated Fermi level.
    eV
E
E
E
E
E
E D
c
f
c
f
D 037
.
0





 , and
 
 
  d
eV
eV
kT
E
E
d
D N
of
cm
e
cm
e
N
n f
D
%
19
10
94
.
1
1
10
1
1 3
17
026
.
0
/
037
.
0
3
18
/






 


.
Since 19% of dopants are not ionized, the full ionization assumption is not accurate but
acceptable.
(c) We assume full ionization of impurities, the electron concentration is n  Nd =
1019
cm-3
/n = (1010
cm-3
)2
/1019
cm-3
= 10 cm-3
.
    eV
cm
cm
n
N
kT
E
E c
f
c 027
.
0
10
/
10
8
.
2
ln
026
.
0
/
ln 3
19
3
19




 

.
Again, we need to find the percentage of donors occupied by electrons in order to
check the full ionization assumption with the calculated Fermi level.
    eV
E
E
E
E
E
E D
c
f
c
f
D 023
.
0






 , and
 
 
  d
eV
eV
kT
E
E
d
D N
of
cm
e
cm
e
N
n f
D
%
71
10
08
.
7
1
10
1
1 3
18
026
.
0
/
023
.
0
3
19
/






 



.
Since 71% of dopants are not ionized, the full ionization assumption is not correct.
Since Nd is not fully ionized and Nd(ionized) << Nd(not-ionized),
 
      kT
E
E
c
kT
E
E
d
D
d
f
c
f
D
e
N
e
N
E
f
N
n
/
/
1






 .
Solving the equation above for Ef yields
 











c
d
c
D
f
N
N
kT
E
E
E ln
2
2
.

Modern Semiconductor Devices for Integrated Circuits: Answers to Problems (1st Edition) by Chenming Hu

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Modern Semiconductor Devices for Integrated Circuits: Answers to Problems (1st Edition) by Chenming Hu