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


Vunit cell  5.43 A  5.43  10 8 cm  1.60  10  22 cm3 ,
3
3
and one unit cell contains 8 silicon atoms. The atomic density of silicon is
N Si 
8 silicon atoms
 5.00  10 22 (silicon atoms) cm  3 .
Vunit cell
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
Mass1 Si atom 
28.1  g/mole 
 4.67  10 23  g/atom  .
23
6.02  10 atoms/mole 
Therefore, the density of silicon ( Si ) in g/cm3 is
ρ Si  N Si  Mass1 Si atom  2.33 g / cm 3 .
Fermi Function
1.2 (a) Assume E = E f in Equation (1.7.1), f(E) becomes ½. Hence, the probability is ½.
(b) Set E = E c + kT and E f = E c in Equation (1.7.1):
f(E) 
1
1 e
 Ec  kT  Ec /kT

1
 0.27 .
1  e1
The probability of finding electrons in states at E c + kT is 0.27.
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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-E f 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
f(E  E c  kT)  1  f(E  E c  3kT)
1
1
E c  kT  E f /kT  1 
Ec  3kT  E f /kT
1 e
1 e
where
1
1
1 e
E
Ec  3kT  E f /kT
 3 kT  E
/kT
E
 3 kT  E
/kT
f
f
1 e c
1
e c

E  3 kT  E f /kT 
E  3kT  E f /kT
1 e c
1 e c
1
.

 E c  3 kT  E f /kT
1 e
Now, the equation becomes
1
1 e
E c  kT  E f /kT 
1 e

1

 E c  3 kT  E f /kT
.
This is true if and only if
Ec  kT  E f  Ec  3kT  E f  .
Solving the equation above, we find
E f  Ec  2kT .
1.3 (a) Assume E = E f 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
f(E  Ec )  1  f(E  Ev )
1
1
Ec  E f /kT  1 
Ev  E f /kT
1 e
1 e
where
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1
E
1
1 e
Ec  E f /kT
E
/kT
E
E
/kT
1 e v f
1
e v f
1


.
Ev  E f /kT
Ev  E f /kT 
 E v  E f /kT
1 e
1 e
1 e
Now, the equation becomes
1
1 e
Ec  E f /kT 
1 e

1

 E v  E f /kT
.
This is true if and only if
Ec  E f  Ev  E f  .
Solving the equation above, we find
Ef 
Ec  Ev
.
2
(c) The plot of the Fermi-Dirac distribution and the Maxwell-Boltzmann distribution
is shown below.
Probability
1
0.9
Fermi-Dirac
Distribution
0.8
0.7
0.6
Maxwell-Boltzmann
Distribution
0.5
0.4
0.3
0.2
0.1
0
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
(E-E f )/kT
The Boltzmann distribution considerably overestimates the Fermi distribution for
small (E-E f )/kT. If we set (E-E f )/kT = A in Equations (1.7.1) and (1.7.2), we
have
 1 
e  A  1.10
.
A
1  e 
Solving for A, we find
eA 
1  eA
1.10
 e A  10.11 
A  ln 10.11  2.31 .
Therefore, the Boltzmann approximation is accurate to within 10% for (E-E f )/kT
 2.31.
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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
E
/kT
N ( N 2 )10 km
e 10 km
(E
E
)/kT
  ESea Level /kT  e 10 km Sea Level
N ( N 2 ) Sea Level e
where
E10 km  E Sea Level  altitude  mass of N 2 molecule  acceleration of gravity
 10 6 cm  28  1.66  10 24 g  980 cm  s 2  4.56  10 14 erg.
The ratio is
N ( N 2 )10 km
N ( N 2 ) Sea Level
 e ( 4 .5610
14
erg)/( 1.3810 16 erg  K 1 273 K)
 e 1.21  0.30 .
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 N 2 concentration.
(b) We know that
N (O2 )10 km
N (O2 ) Sea Level
 0.25 ,
N ( N 2 )10 km
N ( N 2 ) Sea Level
 0.30 , and
N ( N 2 ) Sea Level
N (O2 ) Sea Level
 4.
Then,
N ( N 2 )10 km
N (O2 )10 km

N ( N 2 )10 km
N ( N 2 ) Sea Level
 0.30  4 

N ( N 2 ) Sea Level
N (O2 ) Sea Level

N (O2 ) Sea Level
N (O2 )10 km
1
 4.8 .
0.25
It is more N 2 -rich than at sea level.
1.5
1  f E f  E   1 
1
E f  E  E f /kT
1 e
E  E  E f /kT
e f

E  E  E f /kT
1 e f
1

 E f  E  E f /kT
1 e
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the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying,
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

1 e

1

 E f  E  E f /kT
1
E f  E  E f /kT
1 e
 f E f  E 
1.6 (a)
150K
1.0
f(E)
0.5
300K
0.0
Ef
E
(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 E f is not within a few kTs from E c or E v . 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
Distribution (E)  E  Ec  e


1 / 2  E  E f /kT
,
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
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the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying,
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

d
E  Ec 1/ 2 e  E  E f /kT  1 1 e  E  E f /kT  E  Ec   1 e  E  E f /kT  0
2 E  Ec
dE
 kT 
The exponential terms cancel out. Solving the remaining equation yields
1
E  Ec 1/ 2  1 E  Ec 1/ 2
kT
2
E  Ec   kT

2
 E  Ec 
kT
.
2
So, the number of carriers in the conduction band peaks at E c +kT/2.
Similarly, in the valence band, the carrier distribution as a function of energy is
proportional to
Distribution (E)  Ev  E  e


1 / 2  E f  E /kT
,
where e-(Ef-E)/kT is Equation (1.7.2). Taking the derivative and setting it to zero, we
obtain


d
Ev  E 1/ 2 e E f  E /kT  1 Ev  E 1/ 2 e  E f  E /kT  Ev  E 1/ 2  1 e E f  E /kT  0 .
dE
2
 kT 
Again, the exponential terms cancel out, and solving the remaining equation yields
1
Ev  E 1/ 2  1 Ev  E 1/ 2
kT
2

Ev  E   kT
2
 E  Ev 
kT
.
2
Therefore, the number of carriers in the valence band peaks at E v -kT/2.
1.8 Since it is given that the semiconductor is non-degenerate (not heavily doped), E f is not
within a few kTs from E c or E v . 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
n
C.B.

Dc(E) f(E) dE   A E  Ec e
Ec


 E  E f /kT
dE .
In order to simplify the integration, we make the following substitutions:
E  Ec
 x  E  kT x  Ec ,
kT
1
dE  dx  dE  kT dx, and x : from 0 to  .
kT
Now the equation becomes
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
n   A kTx e


 kTx  E c  E f / kT
0
kT dx  A kT 
3/ 2
e


 E c  E f / kT


x e  x dx
0
where


0
3


x e  x dx    x 2 e  x dx  3 / 2  
1
2
0
. (Gamma function)
Hence, the electron concentration in the conduction band is
n

2
A kT 
3/ 2
e


 E c  E f / kT
.
Similarly, the hole concentration is given by
p   Dv(E)1-f(E)dE   B Ev  E e
Ev


 E f  E /kT
dE .
-
V.B.
Again, we make the following substitutions to simplify the integration:
Ev  E
1
 x  E   kT x  Ev , 
dE  dx  dE  kT dx, and x : from  to 0 .
kT
kT
Now the equation becomes
0
p    B kTx e


 E f  kTx  E v / kT

kT dx  B kT 
3/ 2
e


 E f  E v / kT


0
x e  x dx
where


0
3

x e  x dx    x 2 e  x dx  3 / 2  
1
0

2
. (Gamma function)
Therefore, the hole concentration in the conduction band is
p

2
B kT 
3/ 2
e


 E f  E v / kT
.
(b) The word “Intrinsic” implies that the electron concentration and the hole concentration
are equal. Therefore,
n p

2
A kT 
3/ 2
e   Ec  Ei  / kT 

2
B kT 
3/ 2
e   Ei  Ev  / kT .
This simplifies to
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A e   E c  Ei  / kT  B e   Ei  Ev  / kT .
Solving for E i yields
Ec  Ev kT  1  Ec  Ev

ln  
 0.009 eV ; k  8.62  10 5 eV K 1 , T  300 K .
2
2  2
2
Ei 
Hence, the intrinsic Fermi level (E i ) is located at 0.009 eV below the mid-bandgap of the
semiconductor.
1.9 The unit step functions set the integration limits. D c (E) is zero for E < E c , and D v (E) is zero
for E > E v . Since it is given that the semiconductor is non-degenerate (not heavily doped), E f
is not within a few kTs from E c or E v . 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

n
C.B.
Dc(E) f(E) dE   A E  Ec e


 E  E f /kT
Ec
dE .
In order to simplify the integration, we make the following substitutions:
E  Ec
 x  E  kT x  Ec ,
kT
1
dE  dx  dE  kT dx, and x : from 0 to  .
kT
Now the equation becomes

n   A kTx e


 kTx  E c  E f / kT
0
kT dx  A kT  e
2


 E c  E f / kT


0
x e  x dx
where


0
x e  x dx  1 .
Hence, the electron concentration in the conduction band is
n  A kT  e
2


 E c  E f / kT
.
Similarly, the hole concentration is given by
p   Dv(E)1-f(E)dE   B  Ev  E e
Ev
V.B.
-


 E f  E /kT
dE .
Again, we make the following substitutions to simplify the integration:
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Ev  E
1
 x  E   kT x  Ev , 
dE  dx  dE  kT dx, and x : from  to 0 .
kT
kT
Now the equation becomes
0
p    B kTx e


 E f  kTx  E v / kT

kT dx  B kT  e
2


 E f  E v / kT


0
x e  x dx
where


0
x e  x dx  1 .
Therefore, the hole concentration in the conduction band is
p  B kT  e
2


 E f  E v / kT
.
(b) The word “Intrinsic” implies that the electron concentration and the hole concentration are
equal. Therefore,
n  p  A kT  e   E c  Ei  / kT  B kT  e   Ei  E v  / kT .
2
2
This simplifies to
A e   Ec  Ei  / kT  B e   Ei  Ev  / kT .
If we solve for E i , we obtain
Ei 
Ec  Ev kT  1  Ec  Ev

ln  
 0.009 eV ; k  8.62  10 5 eV K 1 , T  300 K .
2
2  2
2
Hence, the intrinsic Fermi level (E i ) 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
1 / 2  E  E f /kT
Distributi on (E)  E  Ec  e
,
where e-(E-Ef)/kT is from Equation (1.7.2). Taking the derivative and setting it to zero, we
obtain


d
E  Ec 1/ 2 e E  E f /kT  1 1 e E  E f /kT  E  Ec   1 e E  E f /kT  0 .
2 E  Ec
dE
 kT 
The exponential terms cancel out. Solving the remaining equation yields
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the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying,
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1
E  Ec 1/ 2  1 E  Ec 1/ 2
kT
2

E  Ec   kT
2
 E  Ec 
kT
.
2
Hence, the number of carriers in the conduction band peaks at E c +kT/2.
(b) The electron concentration in the conduction band is given by
n
C.B.
Dc(E) f(E) dE  
Top of the Conduction Band
Ec
Dc(E) f(E) dE .
We assume that the function f(E) falls off rapidly such that



Top of the Conduction Band
Top of the Conduction Band
Dc(E) f(E) dE
 0.
Dc(E) f(E) dE
Ec
Now we may change the upper limit of integration from the Top of the Conduction Band
to ∞:

n   A E  Ec e


 E  E f /kT
Ec
dE .
Also, in order to simplify the integration, we make the following substitutions:
E  Ec
 x  E  kT x  Ec ,
kT
1
dE  dx  dE  kT dx, and x : from 0 to  .
kT
The equation becomes

n   A kTx e


 kTx  E c  E f / kT
0
kT dx  A kT 
3/ 2
e


 E c  E f / kT


0
x e  x dx
where


0
3

x e  x dx    x 2 e  x dx  3 / 2  
1
0

2
. (Gamma function)
Therefore, the electron concentration in the conduction band is
n

2
A kT 
3/ 2
e


 E c  E f / kT
.
(b) The ratio of the peak electron concentration at E = E c +(1/2)kT to the electron
concentration at E = E c +40kT is
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
 / kT
f
n( Ec  40kT ) AEc  40kT  EC  e c


1 / 2  E c  0.5 kT  E f  / kT
1


A
E
0
.
5
kT
E
e


c
c
n( Ec  kT )
2
1 / 2  E  40 kT  E f  E c  0.5 kT  E f  / kT
 40kT / 0.5kT  e c
 (40 / 0.5)e  39.5  5.60  1016 .
1 / 2  E  40 kT  E
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-E C . The average kinetic energy of
electrons is
K .E . 
sum of the kinetic energy of all electrons

total number of electrons
 E  E D (E) f(E) dE
 D (E) f(E) dE
E  E  A E  E e 




 A EE e

c
C.B.
C.B.
c
c


 E  E f /kT
Ec
c
c

 E  E f /kT
c
Ec
dE
.
dE
In order to simplify the integration, we make the following substitutions:
E  Ec
 x  E  kT x  Ec ,
kT
1
dE  dx  dE  kT dx, and x : from 0 to  .
kT
Now the equation becomes

 A kTx 
 A kTx 
0

0
3/ 2
1/ 2
e
e


 kTx  E c  E f / kT


 kTx  E c  E f / kT
kT dx
kT dx
A kT 
e
A kT 
e
5/ 2

3/ 2



0
/ kT 
 E c  E f / kT
 Ec  E f


 x 
 x 
0
3/ 2 x
1/ 2
e dx
e  x dx
where
5

3 
3/ 2 x
1  x




x
e
dx

x
0
0 2 e dx  5 / 2 

4
(Gamma functions)
and
3


1  x
1/ 2  x




x
e
dx

x
0
0 2 e dx  3 / 2 

2
. (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:
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 
2
n  p  ni2  p  ni2 / n  1010 / 10 5 cm 3  1015 cm 3 .
(b) Please refer to Equations (1.9.3a) and (1.9.3b). Since N d -N a >> n i and all the impurities
are ionized, n = N d -N a , and p = (n i )2/(N d -N a ).
(c) Since the Fermi level is located 0.26 eV above E i and closer to E c , the sample is n-type.
If we assume that E i is located at the mid-bandgap (~ 0.55 eV), then E c -E f = 0.29 eV.
1
Ec
Ef
Ei
2
3
1: E c -E i =0.55 eV
2: E c -E f =0.29 eV
3: E f -E i =0.26 eV
Ev
Using Equations (1.8.5) and (1.8.11), we find
n  Nc e


 E c  E f / kT
 4.01  1014 cm  3 and p  ni2 / n  2.49  105 cm  3 .
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:
n  ni e
E f  Ei  / kT
 2.20  1014 cm  3 and p  ni2 / n  4.55  105 cm  3 .
(d) If T = 800 K, there is enough thermal energy to free more electrons from siliconsilicon bonds. Hence, using Equation (1.8.12), we first calculate the intrinsic carrier
density n i at 800 K:
ni  N c 800 K N v 800 K  e
 
 E g /  2 kT 
 2.56  1016 cm  3 .
where
 2 mdn kT 
N c (T  800 K )  2 
2

 h
3/ 2
 T 
 2.8  10  

 300 K 
3/ 2
19
cm  3  1.22  1020 cm  3
and
 2 mdp kT 
N v (T  800 K )  2

2
 h

3/ 2
 T 
 1.04  10  

 300 K 
19
3/ 2
cm  3  4.53  1019 cm  3 .
Clearly, n i at 800K is much larger than N d -N a (which is equal to n from the previous
part). Hence the electron concentration is nn i , and the hole concentration is
p=(n i )2/nn i . The semiconductor is intrinsic at 800K, and E f is located very close to the
mid-bandgap.
Nearly Intrinsic Semiconductor
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1.12 Applying Equation (1.8.11) to this problem yields
ni2 / p  n  2 p  2 p 2  ni2  p 
1
ni  7.07  1012 cm  3 and n  1.41  1013 cm  3 .
2
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=N a =41016 cm-3, and n = (n i =1010cm-3)2/p = 2500 cm-3.
At T = 600 K,
ni  N c 600 K N v 600 K  e
 
 E g /  2 kT 
 1.16  1015 cm  3
where
 2 mdn kT 
N c (T  600 K )  2 
2

 h
3/ 2
 T 
 2.8  10  

 300 K 
3/ 2
19
cm  3  7.92  1019 cm  3
and
 2 mdp kT 
N v (T  600 K )  2

2
 h

3/ 2
 T 
 1.04  10  

 300 K 
19
3/ 2
cm  3  2.94  1019 cm  3 .
The intrinsic carrier concentration is no more negligible compared to the dopant
concentration. Thus, we have


p  N a  ni  4  1016  1.16  1015 cm 3  4.12  1016 cm 3 , and


2
n  ni2 / p  1.16  1015 cm 3 / 4.12  1016 cm 3  3.27  1013 cm3 .
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 E v .
E f  Ev  kT lnN v T  / pT   0.34 eV , T  600 K .
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 + N a - = p + N d +. Since N d + is much larger than N a -,
all the samples are n-type, and n  N d + - N a - = 31015 /cm3. This value is assumed to be
constant. Using the Equations (1.8.10) and (1.9.3b),
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

p  ni2 / N d  N a  N c N v exp E g / kT   CT 3 exp E g / kT  ,
where C is a temperature independent constant. Using the sensitivity of p defined by
p/T,
p / T  3  E g / kT  CT 2 exp E g / kT 
Therefore, the larger the energy gap is the less sensitive to temperature the minority carrier
is. For the definition of the sensitivity of p,
p / T  / p  3  Eg / kT / T
The temperature sensitivity of the minority carrier is greater for larger E g .
1.15 (a) Let us first consider the case of n-type doping. The dopant atoms are located at
energy E d 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=N d /2. In other words, the probability of dopant atoms being ionized is ½, or
conversely, the probability that a state at the donor energy E D is filled is ½.
From Problem 1.2 part (a), we know that if f (E D )=1/2, then E D =E f . From Equation
1.8.5,
n  Nce


 Ec  E f / kT
.
We also know that E f =E D and E c -E D =0.05eV.
 2 mdn kT 
N c (T )  2
2

 h
N c (T )e


 E c  E f / kT
3/ 2
3/ 2
 T 
3
 2.8  10  
 cm .
K
300


N
2 N c (T )
 N c (T )e   Ec  E D  / kT  D 
 e Ec  E D  / kT .
Nd
2
19
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
 2 mdp kT 
N v (T )  2

2

 h
N v (T )e


 E f  E v / kT
3/ 2
3/ 2
 T 
3
 1.04  10  
 cm .
300
K


N
2 N v (T )
 N v (T )e   E a  Ev  / kT  a 
 e E a  Ev  / kT .
Na
2
19
Starting from T =100K, we find T converges to 67.7K.
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(b) We want to find T where n i is 10N d . This can be written as
ni  N c T N v T  e
 
 E g /  2 kT 
 T 
 1.71  10  

 300 K 
19
3/ 2
e
 
 E g /  2 kT 
 10 N d
where
 2 mdn kT 
N c (T )  2
2

 h
3/ 2
 2 mdp kT 
N v (T )  2

2

 h
3/ 2
 T 
 2.8  1019  

 300 K 
3/ 2
 T 
 1.04  10  

 300 K 
cm  3 and
3/ 2
19
cm 3 .
We need to solve the equation iteratively, as in part (a) for n i =10N d =1017cm-3. Starting
from T=300K, we get T=777 K for n i =10N d .
For n i =10N a , we simply replace N d in the equation above with N a . Starting from T
=300K, we find T=635 K.
(c) If we assume full ionization of impurities at T = 300 K,
For arsenic: n  N d  1016 cm 3  ni , p 
2
ni
 2.1  104 cm  3
Nd
2
n
For boron: p  N a  10 cm  ni , n  i  2.1  105 cm  3
Na
15
3
(d) Please refer to the example in Section 2.8. For arsenic,
E f  E v  kT ln
N v 1.04  1019 cm 3

 0.88 eV .
p
2.1  10 4 cm 3
For boron,
E f  Ev  kT ln
N v 1.04  1019 cm 3

 0.24 eV .
p
1015 cm  3
(e) In case of arsenic + boron,


2
n2
1010 cm 3
n  N d  N a  9  10 cm , and p  i 
 1.11  10 4 cm  3 , and
n 9  1015 cm  3
 1.04  1019 cm 3 
 Nv 
  0.90 eV .




E f  Ev  kT ln   0.026 eV ln
4
3 

1
.
11
10
cm
 p 


15
3
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1.16 (a) If we assume full ionization of impurities, the electron concentration is n  N d =
1017cm-3. The hole concentration is p=(n i )2/n=(1010 cm-3)2/1017cm-3=103 cm-3.
The Fermi level position, with respect to E c , is


Ec  E f  kT lnN c / n   0.026 ln 2.8  1019 cm 3 / 1017 cm 3  0.15 eV .
E f is located 0.15 eV below E c .
(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.
ED  E f  Ec  E f   Ec  ED   0.1eV , and
nD  N d
1
1 e
E D  E f  / kT

1017 cm 3 
15
3
0.1 eV / 0.026 eV   2.09  10 cm  2% of N d .
1 e
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  N d =
1019cm-3. The hole concentration is p=(n i )2/n = (1010cm-3)2/1019 cm-3 = 10 cm-3.
The Fermi level position, with respect to E c , is


Ec  E f  kT lnN c / n   0.026  ln 2.8  1019 cm 3 / 1019 cm 3  0.027 eV .
It is located 0.027 eV below E c .
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.
E D  E f  E c  E f   E c  E D   0.023 eV , and
nD  N d
1
1 e
E D  E f  / kT

1019 cm 3 
 7.08  1018 cm  3  71% of N d .
1  e 0.023 eV / 0.026 eV 
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 N c and N v at T=30K:
 2 m dn kT 
N c (T  30 K )  2 

h2


3/ 2
 T 
 2.8  10  

 300 K 
3/ 2
19
cm 3  8.85  1017 cm 3
and
 2 mdp kT 
N v (T  30 K )  2 

h2


3/ 2
 T 
 1.04  10  

 300 K 
19
3/ 2
cm 3  3.29  1017 cm 3 .
The electron concentration is
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n
N c 30 K N d  Ec  ED  / 2 kT 
e
 8.43  10 8 cm 3 .
2
And, the hole concentration is
p  ni2 / n  0
where
ni 
N c 30 K N v 30 K  E g  / 2 kT 
e
 2.32  10 75 cm 3 .
2
Since n i is extremely small, we can assume that all the electrons are contributed by
ionized dopants. Hence,
1
8.43  108 cm 3


8
3
n  N d 1 

8
.
43

10
cm

 8.43  10  9 .

E D  E f  / kT
17
3
10 cm
 1 e

The full ionization assumption is not correct since only 8.4310-7% of N d is ionized.
To locate the Fermi level,
1


n 

  1  0.048 eV .
ED  E f  kT ln 1 
  Nd 



E c -E f = 0.05-0.048 = 0.002 eV. Therefore, the Fermi level is positioned 0.002 eV
below E c , between E c and E D .
1.17 (a) We assume full ionization of impurities, the electron concentration is n  N d =
1016cm-3. The hole concentration is p=(n i )2/n = (1010cm-3)2/1016cm-3 = 104 cm-3.
The Fermi level position, with respect to E c , is


Ec  E f  kT lnN c / n   0.026 ln 2.8  1019 cm 3 / 1016 cm 3  0.21eV .
It is located 0.21 eV below E c .
We need to find the percentage of donors occupied by electrons in order to check the
full ionization assumption with the calculated Fermi level.
ED  E f  Ec  E f   Ec  ED   0.16 eV , and
nD  N d
1
1 e
E D  E f  / kT


1016 cm 3
13
3
1

0.16 eV / 0.026 eV   2.12  10 cm  2.12  10 % of N d .
1 e
Since only 0.21% of dopants are not ionized, the full ionization assumption is correct.
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from
the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying,
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(b) We assume full ionization of impurities, the electron concentration is n  N d =
1018cm-3. The hole concentration is p=(n i )2/n = (1010cm-3)2/1018cm-3 = 102 cm-3.
The Fermi level position with respect to E c is


Ec  E f  kT lnN c / n  0.026 ln 2.8  1019 cm 3 / 1018 cm 3  0.087 eV .
It is located 0.087 eV below E c .
We need to find the percentage of donors occupied by electrons in order to check the
full ionization assumption with the calculated Fermi level.
ED  E f  Ec  E f   Ec  ED   0.037 eV , and
nD  N d
1
1 e
E D  E f  / kT



1018 cm 3
17
3
0.037 eV / 0.026 eV   1.94  10 cm  19% of N d .
1 e
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  N d =
1019cm-3. The hole concentration is p=(n i )2/n = (1010cm-3)2/1019cm-3 = 10 cm-3.
The Fermi level position, with respect to E c , is


Ec  E f  kT lnN c / n  0.026 ln 2.8  1019 cm 3 / 1019 cm 3  0.027 eV .
It is located 0.027 eV below E c .
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.
ED  E f  Ec  E f   Ec  ED   0.023 eV , and
nD  N d
1
1 e
E D  E f  / kT



1019 cm3
 7.08  1018 cm 3  71% of N d .
1  e 0.023 eV / 0.026 eV 
Since 71% of dopants are not ionized, the full ionization assumption is not correct.
Since N d is not fully ionized and N d (ionized) << N d (not-ionized),
n  N d 1  f ED   N d e
E D  E f  / kT
 N ce


 E c  E f / kT
.
Solving the equation above for E f yields
Ef 
ED  Ec   kT ln N d  .
2
2
N 
 c
© 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This publication is protected by Copyright and written permission should be obtained from
the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying,
recording, or likewise. For information regarding permission(s), write to: Rights and Permissions Department, Pearson Education, Inc., Upper Saddle River, NJ 07458.