LECTURE 6
BASICS OF SEMICONDUCTOR
PHYSICS
DENSITY OF STATES
• Current is due to the flow of charge carriers
• To know the number of electrons and holes in semiconductor
is important in order to calculate the current flow
• Number of electrons and holes as a function of available energy levels
• Based on Pauli’s exclusion principle, 1 electron per energy level
• Electrons and holes concentration are calculated based on the
calculation of density of allowable energy levels or density of states
• DENSITY OF STATES IS CALCULATED BY SOLVING FREE ELECTRON
IN 3-DIMENSIONAL INFINITE POTENTIAL WELL USING SCHRODINGER
WAVE EQUATION.
DENSITY OF STATES PER UNIT VOLUME OF THE CRYSTAL
g (E) =
3/2
4
π
(2m)
________ √ E
h3
……… (1)
As the energy of this free electron becomes small, the number of
available states decreases.
Example
Consider the density of states for a free electron given by Equation 1.
Calculate the number of states per unit volume with energies between
0 and 1 eV.
N =
∫
4π
________
g (E) dE =
h3
1 eV
0
(2m)3/2
1 eV
∫√E
0
dE
3/2
4
π
(2m)
3/2
.
2/3
.
E
________
N=
h3
-31)]3/2
4
π
[
2
(9.11
x
10
N = __________________ . 2/3 . (1.6 X 10-19) 3/2
(6.625 x10-34)3
= 4.5 x 1027 states / m3
= 4.5 x 1021 states / cm3
STATISTICAL MECHANICS
THREE DISTRIBUTION LAWS DETERMINING THE DISTRIBUTION
OF PARTICLES AMONG AVAILABLE ENERGY STATES
• MAXWELL-BOLTZMANN PROBABILITY FUNCTION
• PARTICLES ARE CONSIDERED AS TO BE DISTINGUISHABLE
• GAS MOLECULES IN A CONTAINER AT LOW PRESSURE
• BOSE-EINSTEIN FUNCTION
• PARTICLES ARE INDISTISTINGUISHABLE
• BLACK BODY RADIATION
• FERMI-DIRAC FUNCTION
• PARTICLES ARE INDISTINGUISHABLE
• ELECTRONS IN CRYSTAL
FERMI-DIRAC PROBABILITY FUNCTION
• GIVES THE PROBABILITY THAT A QUANTUM STATE AT ENERGY
E WILL BE OCCUPIED BY AN ELECTRON
1
_________________
F (E) =
1 + exp (E – EF) / kT
….. (2)
k – Boltzman constant = 1.38 x 10-23 J/K, T in K
Where EF is Fermi Energy, and defined as the energy where the
probability of a state being occupied is 0.5 or 50%.
1
_________________
F (EF) =
=
1 + exp (EF – EF) / kT
1
1
_________
_________
=
1+1
1 + exp (0)
DISTRIBUTION FUNCTION VS ENERGY
LET T = 0K AND E < EF
1
_________________
=
1 + exp (E – EF) / kT
1
_________________
=1
1 + exp (-∞)
F (E < EF) = 1
LET T = 0K AND E > EF
1
_________________
=
1 + exp (E – EF) / kT
F (E > EF) = 0
1
_________________
=0
1 + exp (+∞)
Consider a system of 13 electrons with discrete energy levels as below;
At T = 0 K
E5
E4
E3
E2
E1
• The electrons will be in the lowest energy states, so the probability of
a quantum states being occupied in energy levels E1 through E4 is
unity.
• The probability of quantum states to be occupied in energy level E5 is zero
• In this case, Fermi energy must be above E4 but less than E5
Consider a system of 13 electrons with discrete energy levels as below;
At T > 0 K, E = EF
E5
E4
E3
E2
E1
• Electrons gain certain amount of thermal energy so that some can jump to
higher energy levels.
• As the temperature change, the distribution of electrons vs energy changes
1
1
1
_________
_________
_________________
= 0.5
=
F (EF) =
=
1+1
1 + exp (0)
1 + exp (EF – EF) / kT
Example
Let T = 300K. Determine the probability that an energy level 3kT above
the Fermi energy is occupied by an electron.
1
_________________
F (E) =
1 + exp (E – EF) / kT
1
_________________
=
1 + exp (3kT / kT)
=
1
_________
1 + 20.09
= 4.74 %
SUMMARY
• CONSIDERING A GENERAL CRYSTAL AND APPLYING
THE CONCEPT OF QUANTUM MECHANICS TO
DETERMINE THE CHARACTERISTICS OF ELECTRON
IN SINGLE-CRYSTAL LATTICE.
• TO APPLY THOSE CONCEPT TO A SEMICONDUCTOR
MATERIALS IN THERMAL EQUILIBRIUM
• DENSITY OF STATES IN CONDUCTION AND VALENCE
BANDS TO BE APPLIED WITH FERMI-DIRAC FUNCTION
TO DETERMINE THE CONCENTRATIONS OF ELECTRONS
AND HOLES IN THE RESPECTIVE BANDS.
ELECTRONS AND HOLES DISTRIBUTION IN EQUILIBRIUM
THE ELECTRONS DISTRIBUTION IN CONDUCTION BAND IS GIVEN
BY;
n (E) = gc (E) F (E)
Where
….. (3)
gc (E) is density of states in the conduction band
F (E) Fermi-Dirac distribution function
Total electron concentration per unit volume in the conduction band
can be calculated by integrating Equation (3) over entire conduction
band energy.
THE HOLE DISTRIBUTION IN VALENCE BAND IS GIVEN
BY;
p (E) = gv (E) [ 1- F (E)]
Where
….. (4)
gv (E) is density of states in the valence band
[1-F (E)] the probability of states not occupied by electron
Total hole concentration per unit volume in the valence band
can be calculated by integrating Equation (4) over entire valence
band energy.
INTRINSIC SEMICONDUCTOR
• pure s/c material with no impurity atoms and no lattice defect
• At T = 0 K, all states in valence band are filled with electrons, states
in conduction band are empty.
• Fermi energy is between Ec and Ev
• As the temperature increase above 0 K, few electrons will get enough
energy to jump to conduction band and creating holes in the valence
band.
• In intrinsic material, electrons and holes are created in pairs by thermal
energy. So the number of electrons in conduction band is equal to the
number of holes in the valence band
gc
gv
Electron concentration at thermal equilibrium given by;
ni =
∞
∫E
gc (E) F (E) dE
c
ni =
Nc exp
[
-(Ec – EFi )
_________
kT
]
……….. (5)
Where Nc effective density of states in the conduction band
Nc at T = 300K
2.8 x 1019 cm-3
Si
4.7 x 1017 cm-3
GaAs
Ge
1.04 x 1019 cm-3
Holes concentration at thermal equilibrium given by;
pi =
pi =
Ev
∫- ∞ g
v
(E) [1 - F (E)] dE
Nv exp
[
-(EFI – Ev )
_________
kT
]
……….. (6)
Where Nv effective density of states in the valence band
Nv at T = 300K
1.04 x 1019 cm-3
Si
7.0 x 1018 cm-3
GaAs
Ge
6.0 x 1018 cm-3
In intrinsic semiconductor, electron concentration is equal to the
hole concentration, thus
ni = pi and
nipi = ni2 (MASS ACTION LAW) – the product of n p is always
a constant for a given semiconductor material at given temperature
If we take the product of Equation 5 and 6,
-(Ec – EFi )
_________
2
ni = Nc Nv exp
. exp
kT
-(Ec – Ev )
= Nc Nv exp _________
kT
[
[
-Eg
______
= Nc Nv exp
kT
]
]
[
-(EFI – Ev )
_________
kT
]
….. (7)
Nc and Nv vary at T3/2
Example
Calculate the probability that a state in the conduction band is occupied by
an electron and calculate the thermal equilibrium electron concentration
in silicon at T=300K. Assume Fermi Energy is 0.25eV below the conduction
band. The value of Nc for silicon at T=300K is Nc = 2.8 x 1019 cm-3
The probability that an energy state at E = Ec is occupied by electron is
given by Fermi-Dirac probability function
1
_________________
F (Ec) =
1 + exp (Ec – EF) / kT
[-_________
(Ec – EF) ]
exp
kT
For electrons in the conduction band, E > Ec. If (Ec – EF) >> kT, then
(E – EF) >> kT. So the Fermi function can be reduced to Boltzman
Approximation
[- (Ec – EF) ]
F (Ec) = exp _______
kT
kT = 1.38 x 10-23 J/K . 300K = 4.14 x 10-21 J
kT = 4.14 x 10-21 / 1.6 x 10-19 eV = 0.0259 eV
[- 0.25 ]
F (Ec) = exp _______ = 6.43 x 10-5
0.0259
The electron concentration is given by
ni =
Nc exp
-(Ec – EFi )
_________
kT
[
]
= (2.8 x 1019) exp (-0.25 / 0.0259)
= 1.8 x 1015 cm-3
Example
Calculate the thermal equilibrium hole concentration in intrinsic silicon
at T = 400K. Assume Fermi level is 0.27 eV above the valence band. The
value of Nv for silicon at T=300K is Nv = 1.04 x 1019cm-3
At T = 400K
(
)
kT = 0.0259 400/300 = 0.03453 eV
(
)
Nv = (1.04 x 1019) 400/300
3/2
= 1.60 x 1019 cm-3
pi = (1.6 x 1019) exp ( -0.27/0.03453) = 6.43 x 1015 cm-3