LECTURE 6BASICS 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) = 4 π (2m)3/2 ________ √ E

h3

As the energy of this free electron becomes small, the number of available states decreases.

……… (1)

Example

Consider the density of states for a free electron given by Equation 1.Calculate the number of states per unit volume with energies between0 and 1 eV.

N = ∫0

1 eV

g (E) dE =4 π (2m)3/2 ________ √ E

h3∫

0

1 eV

dE

N = 4 π (2m)3/2 ________h3

. 2/3 . E3/2

N = 4 π [ 2 (9.11 x 10-31)]3/2

(6.625 x10-34)3. 2/3 . (1.6 X 10-19) 3/2__________________

= 4.5 x 1027 states / m3 = 4.5 x 1021 states / cm3

STATISTICAL MECHANICS

THREE DISTRIBUTION LAWS DETERMINING THE DISTRIBUTIONOF 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

F (E) = 1_________________

1 + exp (E – EF) / kT

• GIVES THE PROBABILITY THAT A QUANTUM STATE AT ENERGY E WILL BE OCCUPIED BY AN ELECTRON

Where EF is Fermi Energy, and defined as the energy where theprobability of a state being occupied is 0.5 or 50%.

F (EF) = 1_________________

1 + exp (EF – EF) / kT1_________

1 + exp (0)= =

1_________1 + 1

k – Boltzman constant = 1.38 x 10-23 J/K, T in K

….. (2)

DISTRIBUTION FUNCTION VS ENERGY

LET T = 0K AND E < EF

1_________________1 + exp (E – EF) / kT =

1_________________1 + exp (-∞)

= 1

F (E < EF) = 1

LET T = 0K AND E > EF

1_________________1 + exp (E – EF) / kT =

1_________________1 + exp (+∞)

= 0

F (E > EF) = 0

Consider a system of 13 electrons with discrete energy levels as below;

E1

E2

E3

E4

E5

At T = 0 K

• 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;

E1

E2

E3

E4

E5

At T > 0 K, E = EF

• 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

F (EF) = 1_________________

1 + exp (EF – EF) / kT1_________

1 + exp (0)= =

1_________1 + 1

= 0.5

Example

Let T = 300K. Determine the probability that an energy level 3kT abovethe Fermi energy is occupied by an electron.

1_________________1 + exp (E – EF) / kT

F (E) =

= 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 GIVENBY;

n (E) = gc (E) F (E) ….. (3)

Where g

c (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 conductionband energy.

THE HOLE DISTRIBUTION IN VALENCE BAND IS GIVENBY;

p (E) = gv (E) [ 1- F (E)] ….. (4)

Where g

v (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 valenceband 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

gv

gc

Electron concentration at thermal equilibrium given by;

ni = ∫Ec

∞gc (E) F (E) dE

ni = Nc exp [ -(Ec – EFi ) ]_________kT

……….. (5)

Where Nc effective density of states in the conduction band

SiGaAsGe

Nc

2.8 x 1019 cm-3

4.7 x 1017 cm-3

1.04 x 1019 cm-3

at T = 300K

Holes concentration at thermal equilibrium given by;

pi = ∫Ev

- ∞gv (E) [1 - F (E)] dE

pi = Nv exp [ -(EFI – Ev ) ]_________kT

……….. (6)

Where Nv effective density of states in the valence band

SiGaAsGe

Nv

1.04 x 1019 cm-3

7.0 x 1018 cm-3

6.0 x 1018 cm-3

at T = 300K

In intrinsic semiconductor, electron concentration is equal to thehole 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,

ni2 = Nc Nv exp [ -(Ec – EFi ) ]_________

kT. exp [ -(EFI – Ev ) ]_________

kT

= Nc Nv exp [ -(Ec – Ev )_________kT

]

= Nc Nv exp -Eg

kT______ ….. (7)

Nc and Nv vary at T3/2

Example

Calculate the probability that a state in the conduction band is occupied byan electron and calculate the thermal equilibrium electron concentrationin silicon at T=300K. Assume Fermi Energy is 0.25eV below the conductionband. 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 isgiven by Fermi-Dirac probability function

F (Ec) = 1_________________

1 + exp (Ec – EF) / 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

exp [- (Ec – EF) ]_________

kT

exp [- (Ec – EF) ]_______

kTF (Ec) =

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

exp [- 0.25 ]_______0.0259

F (Ec) = = 6.43 x 10-5

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 siliconat T = 400K. Assume Fermi level is 0.27 eV above the valence band. Thevalue 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