Example 10.1
Given that at
T
= 300
o
K, the electron concentration in silicon is 1.52 x 10
10
electrons /cm
3
and
E
g
= 1.1 eV at 300
o
K.
(a) Find the constant
A
of Equation (10.1).
(b) Use MATLAB to plot the electron concentration versus temperature.
Solution
From Equation (10.1), we have
152 10 300
10 1 5 1 1 300 8 62 10
5
.()
.[./*.* )]
xA e=
−
−
We use MATLAB to solve for
A
.
The width of energy gap with temperature
is given as [1].
ET x
T
T
g
() . .
=−
+
−
117 4 37 10
636
4
2
(10.6)
Using Equations (10.1) and (10.6), we can calculate the electron concentration
at various temperatures.
MATLAB Script
%
% Calculation of the constant A
diary ex10_1.dat
k = 8.62e-5;
na = 1.52e10; ta = 300;
ega = 1.1;
ka = -ega/(k*ta);
t32a = ta.^1.5;
A = na/(t32a*exp(ka));
fprintf('constant A is %10.5e \n', A)
% Electron Concentration vs. temperature
for i = 1:10
t(i) = 273 + 10*(i-1);
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eg(i) = 1.17 - 4.37e-4*(t(i)*t(i))/(t(i) + 636);
t32(i) = t(i).^1.5;
ni(i) = A*t32(i)*exp(-eg(i)/(k*t(i)));
end
semilogy(t,ni)
title('Electron Concentration vs. Temperature')
xlabel('Temperature, K')
ylabel('Electron Concentration, cm-3')
Result for part (a)
constant A is 8.70225e+024
Figure 10.2 shows the plot of the electron concentration versus temperature.
Figure 10.2 Electron Concentration versus Temperature
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10.2 EXTRINSIC SEMICONDUCTOR
10.2.1 Electron and hole concentrations
Extrinsic semiconductors are formed by adding specific amounts of impurity
atoms to the silicon crystal. An n-type semiconductor is formed by doping the
silicon crystal with elements of group V of the periodic table (antimony, arse-
nic, and phosphorus). The impurity atom is called a donor. The majority car-
riers are electrons and the minority carriers are holes. A p-type semiconductor
is formed by doping the silicon crystal with elements of group III of the peri-
odic table (aluminum, boron, gallium, and indium). The impurity atoms are
called acceptor atoms. The majority carriers are holes and minority carriers
are electrons.
In a semiconductor material (intrinsic or extrinsic), the law of mass action
states that
pn
= constant (10.7)
where
p
is the hole concentration
n
is the electron concentration.
For intrinsic semiconductors,
pnn
i
==
(10.8)
and Equation (10.5) becomes
pn n
i
=
2
(10.9)
and
n
i
is given by Equation (10.1).
The law of mass action enables us to calculate the majority and minority car-
rier density in an extrinsic semiconductor material. The charge neutrality
condition of a semiconductor implies that
pN nN
DA
+=+
(10.10)
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where
N
D
is the donor concentration
N
A
is the acceptor concentration
p
is the hole concentration
n
is the electron concentration.
In an n-type semiconductor, the donor concentration is greater than the intrin-
sic electron concentration, i.e.,
N
D
is typically 10
17
cm
-3
and
n
i
= 1.5 x
10
10
cm
-3
in Si at room temperature. Thus, the majority and minority concen-
trations are given by
nN
nD
≅
(10.11)
p
n
N
i
D
≅
2
(10.12)
In a p-type semiconductor, the acceptor concentration
N
A
is greater than the
intrinsic hole concentration
pn
ii
=
. Thus, the majority and minority con-
centrations are given by
pN
pA
≅
(10.13)
n
n
N
i
A
≅
2
(10.14)
The following example gives the minority carrier as a function of doping con-
centration.
Example 10.2
For an n-type semiconductor at 300
o
K, if the doping concentration is varied
from 10
13
to 10
18
atoms/cm
3
, determine the minority carriers in the doped
semiconductors.
Solution
From Equation (10.11) and (10.12),
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Electron concentration =
N
D
and
Hole concentration =
n
N
i
D
2
where
n
i
= 1.5 2 x 10
10
electrons/cm
3
The MATLAB program is as follows:
% hole concentration in a n-type semiconductor
nd = logspace(13,18);
n = nd;
ni = 1.52e10;
ni_sq = ni*ni;
p = ni_sq./nd;
semilogx(nd,p,'b')
title('Hole concentration')
xlabel('Doping concentration, cm-3')
ylabel('Hole concentration, cm-3')
Figure 10.3 shows the hole concentration versus doping.
Figure 10.3 Hole Concentration in N-type Semiconductor (Si)
© 1999 CRC Press LLC
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10.2.2 Fermi level
The Fermi level,
E
F
, is a chemical energy of a material. It is used to describe
the energy level of the electronic state at which an electron has the probability
of 0.5 occupying that state. It is given as
EEEKT
m
m
FCV
n
p
=+−
1
2
4
3
()ln()
*
*
(10.15)
where
E
C
= energy in the conduction band
E
V
= energy in the valence band
and
k, T, m
n
*
and
m
p
*
were defined in Section 10.1.
In an intrinsic semiconductor (Si and Ge)
m
n
*
and
m
p
*
are of the same order
of magnitude and typically,
E
F
>>
kT
.
Equation (10.15) simplifies to
EE EE
Fi CV
=≅ +
1
2
()
(10.16)
Equation (10.16) shows that the Fermi energy occurs near the center of the en-
ergy gap in an intrinsic semiconductor. In addition, the Fermi energy can be
thought of as the average energy of mobile carriers in a semiconductor mate-
rial.
In an n-type semiconductor, there is a shift of the Fermi level towards the edge
of the conduction band. The upward shift is dependent on how much the
doped electron density has exceeded the intrinsic value. The relevant equation
is
[]
nne
i
EEkT
Fi
=
−
()/
(10.17)
where
n
is the total electron carrier density
n
i
is the intrinsic electron carrier density
E
F
is the doped Fermi level
E
i
is the intrinsic Fermi level.
© 1999 CRC Press LLC
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In the case of a p-type semiconductor, there is a downward shift in the Fermi
level. The total hole density will be given by
[]
pne
i
EE kT
iF
=
−
()/
(10.18)
Figure 10.4 shows the energy band diagram of intrinsic and extrinsic semicon-
ductors.
E
C
E
I
= E
F
E
V
E
C
E
F
E
I
E
V
E
C
E
F
E
I
E
V
(a) (b) (c )
Figure 10.4 Energy-band Diagram of (a) Intrinsic, (b) N-type, and
(c ) P-type Semiconductors.
10.2.3 Current density and mobility
Two mechanisms account for the movement of carriers in a semiconductor ma-
terial: drift and diffusion. Drift current is caused by the application of an elec-
tric field, whereas diffusion current is obtained when there is a net flow of car-
riers from a region of high concentration to a region of low concentration. The
total drift current density in an extrinsic semiconductor material is
Jqn p
np
=+
()
µµ
Ε
(10.19)
where
J
is current density
n
is mobile electron density
p
is hole density,
µ
n
is mobility of an electron
µ
p
is mobility of a hole
q
is the electron charge
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Ε
is the electric field.
The total conductivity is
σµµ
=+qn p
np
()
Ε
(10.20)
Assuming that there is a diffusion of holes from an area of high concentration
to that of low concentration, then the current density of holes in the x-
direction is
JqD
dp
dx
pp
=−
A/cm
2
(10.21)
where
q
is the electronic charge
D
p
is the hole diffusion constant
p
is the hole concentration.
Equation (10.21) also assumes that, although the hole concentration varies
along the x-direction, it is constant in the y and z-directions. Similarly, the
electron current density,
J
n
, for diffusion of electrons is
JqD
dn
dx
nn
=
A / cm
2
(10.22)
where
D
n
is the electron diffusion constant
n
is the electron concentration.
For silicon,
D
p
= 13 cm
2
/s , and
D
n
= 200 cm
2
/s . The diffusion and mo-
bility constants are related, under steady-state conditions, by the Einstein rela-
tion
D
D
kT
q
n
n
p
p
µµ
==
(10.23)
The following two examples show the effects of doping concentration on mo-
bility and resistivity.
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Example 10.3
From measured data, an empirical relationship between electron (
µ
n
) and hole
(
µ
p
) mobilities versus doping concentration at 300
o
K is given as [2]
µ
nD
D
D
N
xN
xN
()
.
.
.
.
=
+
+
51 10 92
375 10
18 0 91
15 0 91
(10.24)
µ
pn A
A
A
N
xN
xN
()
.
.
.
=
+
+
2 9 10 47 7
586 10
15 0 76
12 0 76
(10.25)
where
N
D
and
N
A
are donor and acceptor concentration per cm
3
,
respectively.
Plot the
µ
n
(
N
D
) and
µ
p
(
N
A
) for the doping concentrations from 10
14
to
10
20
cm
-3
.
Solution
MATLAB Script
% nc - is doping concentration
%
nc = logspace(14,20);
un = (5.1e18 + 92*nc.^0.91)./(3.75e15 + nc.^0.91);
up = (2.90e15 + 47.7*nc.^0.76)./(5.86e12 + nc.^0.76);
semilogx(nc,un,'w',nc,up,'w')
text(8.0e16,1000,'Electron Mobility')
text(5.0e14,560,'Hole Mobility')
title('Mobility versus Doping')
xlabel('Doping Concentration in cm-3')
ylabel('Bulk Mobility (cm2/v.s)')
Figure 10.5 shows the plot of mobility versus doping concentration.
© 1999 CRC Press LLC
© 1999 CRC Press LLC
Figure 10.5 Mobility versus Doping Concentration
Example 10.4
At the temperature of 300
o
K, the resistivity of silicon doped by phosphorus is
given as [ 3]
ρ
n
D
DD
xN
xN xN
=
+
+
−−
375 10
147 10 815 10
15 0 91
17 1 91 1
.
.
.
(10.26)
A similar relation for silicon doped with boron is given as [ 4]
ρ
p
A
AA
xN
xN N
=
+
+
−−
586 10
7 63 10 4 64 10
12 0 76
18 1 76 4
.
*
.
.
(10.27)
where
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