P.E.S. College of Engineering

Electrical Conductivity in Metals & Semiconductors Dr. S. Gowda Professor of Physics

DEPARTMENT OF PHYSICS

Course Title : Engineering Physics

Course Code : P18PH12/22

Unit - III

ELECTRICAL CONDUCTIVITY and SEMICONDUCTORS

Notes

Academic Year : 2019

P.E.S. College of Engineering Mandya – 571 401, Karnataka

(An Autonomous Institution affiliated to VTU, Belagavi)

1


Electrical Conductivity in Metals

Syllabus: Electrical Conductivity in Metals: Free electron concept. Classical free-electron theory – Assumptions. Failure of classical free-electron theory. Quantum free-electron theory – Assumptions. Fermi-Dirac Statistics. Fermi-energy – Fermi factor, Fermi Velocity, Fermi Temperature. Calculation of Fermi Energy at T = 0 K and T > 0 K. Density of states (with derivation). Merits of quantum free-electron theory. Problems.

Introduction: Materials can be classified into three types based on the conductivity of heat and electricity. They are;

1. Conductors (Example : Metals – Copper, Aluminum, Silver, Gold) 2. Semiconductors (Example : Germanium, Silicon) 3. Insulators ( Example : Wood, Mica, Glass)

Electron Theory of Metals The electron theory of metals explains the following :

• Structural, electrical and thermal properties of materials. • Elasticity, cohesive force and binding in solids. • Behaviour of conductors, semiconductors, insulators etc.,

In solids, electrons in the outermost orbit of atoms are called valance electrons, which determine the properties of the materials. The electron theory is applicable to all solids (both metals and non-metals). This theory explains the electrical, thermal and magnetic properties of solids There are three stages for the development of electron theory of metals; namely, Classical free electron theory, Quantum free electron theory and Zone theory.

1. A classical free electron theory is a macroscopic theory proposed by Paul Drude in 1900. After the discovery of electron by JJ Thomson, this theory was elaborated by Lorentz in 1909. Hence this theory is also known as Drude & Lorentz. According to this theory metals contains free electrons which are responsible for the electrical conductivity in metals and obeys the laws of classical mechanics (Maxwell-Boltzmann distribution).

2. Quantum free electron theory is a microscopic theory developed by Sommerfeld in 1928. According to this theory, the free electrons move with a constant potential obeys quantum laws (Fermi-Dirac statistics).

3. Zone theory was developed by Bloch in 1928. According to this theory, free electrons move in periodic potential provided by lattice. This theory is also known as band theory of solids.

Classical Free Electron Theory (Drude-Lorentz Theory): Free Electron Concept: All metal atoms consists of valance electrons and they are responsible for physical properties of metals such as electrical & thermal conductivities, thermoelectricity, thermionic & Photoelectric effect ete.,

2


According to Drude-Lorentz theory, when a large number of atoms arranged in three dimensional lattice points to form a metal, the boundaries of the neighbouring atoms slightly overlap with each other as shown in figure. Due to this overlapping, the valance electrons of all the atoms are free to move within the metal lattice. These electrons are called free electrons. These free electrons are move randomly in all directions through the conductor with average speed of the order of 106 m/s. This is similar to the motion of gas molecules confined in a vessel. Since the free electrons are responsible for electrical and thermal conduction in metals, they are also called as conduction electrons. All metals contain free electrons which act just as a gas molecules moving in every direction throughout the lattice. The average velocity due to the thermal energy is zero since the electrons are going in every direction. There is a way of affecting this free motion of electrons, which is by use of an electric field. This process is known as electrical conduction and theory is called Drude-Lorentz theory. The assumptions of the Drude-Lorentz classical theory of free-electrons are the following.

Assumptions of Classical free-electron theory:

1. All metals contain large number of free electrons which move freely through the positive ionic core of the metals. Since these free electrons causes conduction in metal under the application of electrical field, they are called as conduction electrons.

2. The free electrons are treated as equivalent to gas molecules; the laws of classical kinetic theory of gases can be applied to them. Therefore these electrons have mean free path (λ), mean collision time (T), average speed (v).

3. In the absence of the electric field, the kinetic energy associated with an electron at a temperature T is given by ½ mvth

2 = 3/2 kT = ½ m c 2

Where vth is the thermal velocity of the electron and which is equal to root mean square velocity c .

4. Since the motion of the electrons is random, the net current is zero in the absence of electric field. But when an electric field is applied, current is produced due to the drift velocity of the electrons.

5. The electric field (or Potential) due to positive ionic cores is considered to be uniform throughout the metal and hence neglected. But the force of attraction between the electrons & lattice ions and the force of repulsion between the electrons themselves are considered to be negligible.

3


Drift velocity, mean free path, Mean collision time, Relaxation time, conductivity, Mobility and resistivity.

1) Drift velocity (Vd) : It is the average velocity acquired by the electrons in a direction opposite to the direction of the applied electric field. Drift velocity is given by 𝑣𝑣𝑑𝑑 = eE

mτ

Where E is the strength of the applied electric field and τ is the relaxation time. 2) Mean Free Path (λ): The average distance travelled by the conduction electrons between

two successive collisions with lattice ions is known as mean free path. i.e., λ = vτ, where τ is the mean collision time & v = vd+vth

v is the combined velocity drift and thermal velocities. Since vd =10-4 m/s is very small compared to thermal velocity vth = 106 m/s, hence λ = vth τ

3) Mean Collision Time (τc) : It is the average time that elapses between two successive collisions of an electron with lattice points.

If v is the total velocity of the electrons due to combined effect of thermal & drift velocities, then τc = λ/v

4) Relaxation Time (τr) : When an electric field is applied, the free electrons drift slowly in a direction opposite to that of the applied electric field with an average velocity v′av

avav vv ′= in the presence of electric field.

If the electric field is turned off suddenly, the average velocity avv reduces exponentially to zero.

Therefore, rtavav evv τ−′= , If t = τr , then aveav vv ′= 1

Therefore, the relaxation time (τr) is time during which the average velocity avv decreases

to ( e1 ) times its value of time when the field is turned off. The relaxation time is a measure

of the rate at which relaxation takes place. For metals the values of τc & τr are equal. Electrical conductivity in metals : Electrical conductivity of a metal is the ability of the

metal to allow electrons to flow through it. The expression for electrical conductivity of a metal is given by

τσm

ne2

=

Where n is the number of free electrons/unit volume & τ is the relaxation time. Mobility of electrons : Mobility is defined as the magnitude of the drift velocity acquired by

the electrons in unit electric field. The expression for the mobility is

τµme

Evd ==

Electrical resistivity : It is the property of the metal and defined as the reciprocal of electrical conductivity.

i.e., τσ

ρ 21

nem

==

4


In metals, the resistivity is due to the scattering of conduction electrons. Scattering of electrons may be due to lattice vibrations or due to impurities.

a) Scattering due to lattice vibrations (Phonons) – When the temperature of metal is increased, due to vibrations of lattice ions, the scattering of electrons may take place. The resistivity due to this type of scattering is called ‘ideal resistivity’ denoted by ρph. This is temperature dependent.

b) Scattering due to impurities – The impurities in metals may also scatter electrons and the resistivity due to this is denoted by ρi. This is temperature independent.

Thus in any metal, the total resistivity is

iph ρρρ +=

This is known as Mattthienssen’s rule

Success of Classical Free Electron Theory: ..

The successes of this theory are:- 1. It verifies Ohm’s law i.e., V = IR 2. It derives Wiedemann-Frenz i.e., T

E

T ∝σσ

3. It explains Electrical (σE) and Thermal conductivity (σT) of metals 4. It explains optical properties of metals.

Failures/Drawbacks of Classical Free-Electron Theory:

1. Specific Heat (CV) : The specific heat of a gas at constant volume is given by RCV 2

3= , where R is the universal gas constant But experimentally it was observed that the specific heat of a metal by its conduction electrons is given by

RTCV410−=

Thus, the experimental value of Cv is very much lesser than the expected value of Cv. According to classical free electron theory Cv is independent of temperature, but the experimental value of Cv is directly proportional to temperature. Hence classical free electron theory fails to explain Cv.

2. Mean Free Path (λ) : According to classical free electron theory, the mean free path . τλ c= , where c is the root mean square velocity & τ is the relaxation time of

conduction electrons, we also know that ρ

ττ

ρ 22 nemor

nem

==

By substituting the values of e, m, n & ρ for a metal, τ can be calculated. i.e., τ = 2.47x10-14 s, where c is average velocity of the electron and is equal to 1.15x105 m/s. In this case, λ = 1.15x105x2.47x10-14 = 2.85x10-9 m = 2.85 nm. But the experimental value of λ is found to be 0.285 nm, which is 10 times less than the value obtained from classical free electron theory. Hence classical free electron theory fails to explain λ.

3. Temperature dependence of electrical conductivity : According to classical free electron theory, the electrical conductivity of metals is given by

5


)1(2

→∝= τστσ orm

ne

On the basis of classical free electron theory, the energy of an electron is given by

)2(3

232

21 →∝=∴= Tvor

mkTvkTmv ththth

From definition, mean collision time is inversely proportional to vth .

∴ )3()(11→∝∝∝ Tv

Tor

v thth

ττ

From equations (1) & (3), )4(1→∝

Tσ

But experimentally it has been observed that,

)5(1→∝

Tσ

Hence from equations (4) & (5), it is clear that the prediction of classical free electron theory is not in agreement with the experimental observations. Thus the classical free electron theory fails to explain dependence of T on σ.

4. Dependence of electrical conductivity on electron concentration : As per the classical free electron theory, the electrical conductivity of metals is given by

nor

mne

∝= στσ2

Where ‘n’ is the electron concentration (free electrons) Hence, according to classical free electron theory; bivalent & trivalent metals should posses much higher electrical conductivity than monovalent metals. This is contrary to the experimental observations that the monovalent element metals such as copper & silver are more conducting than Zinc (bivalent) & aluminum (trivalent). Thus the prediction of classical free electron theory that σ ∝ n does not always hold good. Hence classical free electron theory fails to explain dependence of n on σ.

Concentration of Electrons (n) & Electrical Conductivity (σ) of some metals

Metal Valency Electron Concentration n in per m3

Electrical Conductivity σ in Siemen/meter (S/m)

Ag - 47 1 5.85x1028 6.30x107

Cu - 29 1 8.45x1028 5.88x107

Cd - 48 2 9.28x1028 0.15x107

Zn - 30 2 13.10x1028 1.09x107

Ga - 31 3 15.30x1028 0.67x107

Al - 13 3 18.06x1028 3.65x107

Quantum Free Electron Theory (Sommerfeld Theory) :

To overcome the drawbacks of classical free electron theory, Sommerfeld proposed quantum free electron theory. He treated electron as a quantum particle. He retains the vital features of classical free electron theory and included the Pauli Exclusion Principle & Fermi-Dirac statistics. The following are the assumptions of quantum free electron theory.

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1. The free electrons in a metal can have only discrete energy values. Thus the energies are quantized.

2. The electrons obey Pauli’s Exclusion Principle, which states that there cannot be more than two electrons in any energy level.

3. The distribution of electrons in various energy levels obey the Fermi-Dirac quantum statistics.

4. Free electrons have the same potential energy everywhere within the metal, because the potential due to ionic cores is uniform throughout the metal.

5. The force of attraction between electrons & lattice ions and the force of repulsion between electrons can be neglected.

6. Electrons are treated as wave-like particles.

Fermi-Dirac Statistics

There are three statistics are there;

1. Maxwell-Boltzmann Statistics :

kTh

eEf ν

1)( =

- Deals with particles which has no spin - Eg. : Gaseous particles

2. Bose-Einstein Statistics :

1

1)(−

=kTh

eEf ν

- Deals with particles which has integral spin - Known as Bosons - Eg. : Photons. Gluons, 4He atoms,

3. Fermi-Dirac Statistics :

1

1)(+

=kTh

eEf ν

- Deals with particles which has half integral spin - Also known as Fermions - Eg. : Electrons, Protons, Neutrons, Quarks, Neutrinos etc.,

Fermi - level, Fermi - energy and Fermi - factor

As we know that for a metal containing N atoms, there will be N number of energy levels in each band. According to Pauli’s exclusion principle, each energy level can accommodate a maximum of two electrons, one with spin up (+½) and the other with spin down (-½). At absolute zero temperature, two electrons with opposite spins will occupy the lowest available energy level. The next two electrons with opposite spins will occupy the next energy level and so on. Thus, the top most energy level occupied by electrons at absolute zero temperature is called Fermi-energy level. The energy corresponding to that energy level is called Fermi-energy.

7


The energy of the highest occupied level at zero degree absolute is called Fermi energy, and the energy level is referred to as the Fermi level. The Fermi energy is denoted as EF.

All energy levels below Fermi level are completely filled and above which all energy levels are completely empty.

At temperatures above absolute zero, the electrons get thermally excited and move up to higher energy levels. As a result there will be many vacant energy levels below as well as above Fermi energy level. Under thermal equilibrium, the distribution of electrons among various energy levels is given by statistical function f(E). The function f(E) is called Fermi-factor and this gives the probability of occupation of a given energy level under thermal equilibrium. The expression for f(E) is given by

( ) ( ) 11

+= − kTEE Fe

Ef

Where f(E) is called Fermi-Dirac distribution function or Fermi factor, EF is the Fermi energy, k is the Boltzmann constant and T is the temperature of metal under thermal equilibrium.

Note: 1. The Fermi-Dirac distribution function f(E) is used to calculate the probability of an electron occupying a certain energy level. 2. The distribution of electrons among the different energy levels as a function of temperature is known as Fermi-Dirac distribution function.

Variation of Fermi factor with Energy and Temperature

Let us consider the different cases by considering the Fermi factor equation

( ) ( ) 11

+= − kTEE Fe

Ef

Case (i) :- f(E) for E < EF at T = 0 K;

When E < EF & T = 0 K, from the probability function f(E) we have

( ) 110

11

1=

+=

+= ∞−e

Ef

i.e., f(E) = 1 for E < EF at T = 0 K.

This implies that at absolute zero temperature, all the energy levels below EF are 100% occupied which is true from the definition of Fermi energy.

EF En

ergy

T = 0 K Vacant energy levels

Fermi Energy Filled energy levels

8


Case (ii) :- f(E) for E > EF at T = 0 K;

When E > EF & T = 0 K, then f(E) becomes

( ) 011

11

1=

∞=

+∞=

+= ∞e

Ef

i.e., f(E) = 0 for E > EF at T = 0 K.

This implies that at absolute zero temperature, all the energy levels above EF are unoccupied (completely empty) which is true from the definition of Fermi energy.

Case (iii) :- f(E) for E = EF at T = 0 K;

When E = EF & T = 0 K, then f(E) becomes

( ) ateIndetermin1

10

0 =+

=e

Ef

i.e., f(E) = ∞ for E = EF at T = 0 K.

Hence, the occupation of Fermi level at T = 0 K has an undetermined value ranging between zero and unity (0 & 1). The Fermi-Dirac distribution function is discontinuous at E = EF for T = 0 K.

Case (iv) :- f(E) for E = EF at T > 0 K;

When E = EF & T > 0 K, then f(E) becomes

( )21

111

11

0 =+

=+

=e

Ef

i.e., f(E) = ½ for E = EF at T > 0 K.

If E « EF, the probability starts decreasing from 1 and reaches 0.5 (½) at E = EF and for E > EF, it further falls off as shown in figure. In conclusion, the Fermi energy is the most probable or average energy of the electrons in a solid.

The variation of Fermi factor with energy and temperature is as shown in figure given below.

f(E)

E E = EF

0

1.0

0.5

T = 0 K

T > 0 K

Variation of f(E) with E

9


Importance of Fermi Energy Fermi energy level is used to separate the vacant and filled states at 0 K. It is used to know the status of the electrons. Electrons are completely filled below the Fermi energy level and completely empty above

the Fermi level at 0 K. Above 0 K some electrons absorb thermal energy and they jumps to the higher energy

levels. Density of States

The energy levels of electrons in case of single atom are sharp as we know in case of hydrogen atom. In case of solids, the energy levels of electrons will spread out over a range called energy band due to the presence of large number of atoms. Each energy band consists of a number of states at each energy level that are available to be occupied by the electrons; the concept of density states of a system is introduced. The density of states g(E) is defined as the number of energy states available per unit volume per unit energy centered at E. The number of states per unit volume between the energy levels E and E+dE is denoted by g(E)dE.

that metalVolume of taldE in a me and E between E availablergy statesNo. of enedEEg +

=∴ )(

From one-dimensional potential well of width L, we know that the allowed energies for an electron is given by

)1(

8 2

22

→=mL

hnEn

Where, n = 1, 2, 3. . . . . . . are positive integers Since the free electrons in a solid experience a three dimensional potential well, equation (1) can takes the form

( ) )2(

8222

2

2

→++= zyx nnnmLhE

Where nx, ny & nz are non-zero positive integers. Each set of (nx, ny, nz) indicates the permitted energy value.

From equation (1), by taking 22222

2

0 &8 zyx nnnRmLhE ++== equation (2) becomes

)3(20 →= REE

The equation 2222zyx nnnR ++= represents a sphere of radius R formed by the points (nx, ny, nz) with

nx, ny & nz as the three mutually perpendicular axes. Since nx, ny & nz can take only positive integers, the above equation represents th

81 of the sphere called Octant as shown in figures.

nx nz

ny ny

nx

nz

10


The number of allowed energy values N(E)dE available within the sphere of radius R & R+dR is obtained by finding energy difference between E and E+dE. It is equal to the product of th

81

of the volume of the sphere between the two shells of radius R and R+dR and the number of points per unit volume (one).

Thus, ( ) 14 281 xdRRxdEEN π=

or ( ) )4(221 →= dRRdEEN π

Since, each such energy values can accommodate two electrons according to Pauli’s Exclusion Principle, the number of allowed energy states between E and E + dE is

( ) )5(2 2221 →== dRRdRRxdEEZ ππ

In the above equation R is abstract quantity and hence it cannot be measured. So that we wish Express the entire right hand side in terms of energies. From the equation 2

0REE = we get,

00

0 22&

EdERdRorRdREdE

EER ===

)6(21

2 3000

2 →===∴EdEE

EdE

EERRdRdRR

By substituting the value of R2dR in equation (5) we get,

( ) )7(82 2

2

030

→==mLhEBut

E

dEEdEEZ π

( )( )

( ) )8(82

822

21

21

23

2

2

32/3

2

2

2

3

8

→

=

==∴

dEELhmdEEZ

dEEhmLdEEdEEZ

mLh

π

ππ

In the above equations L3 represents the volume of the solid & hence the density of states is the

number of energy states per unit volume. ( )

)10(28)(

)9(28

882

)( OR

)9(82

)(

)()(

21

21

21

21

21

3

2/3

3

2/3

3

2/3

2/3

2

3

→=

→=

=

→

=

=∴

dEEh

mdEEg

dEEhm

dEEh

mxxdEEg

dEEhmdEEgHence

LVolumedEEZdEEg

π

π

π

π

The above equations (9) and (10) are represents the expression for density of energy states.

11


A plot of g(E) versus E is as shown in figure below.

Variation of Fermi energy with temperature

To find the variation of Fermi energy with temperature, let us consider the following.

The number of electrons per unit volume of the solid n is nothing but the product of density of states available and the probability of occupation of electrons among various energy levels up to the Fermi level EF, that is

( ) ( ) )1(0

→= ∫FE

dEE.gEfn

Case (i):- Calculation of Fermi energy at T = 0 K,

We know that, for T = 0 K, we have f(E) = 1 and the Fermi energy for such case is taken as

0FE . Hence the equation (1) for n becomes;

( ) )2(0

0→= ∫

FEdEEgn

By substituting the value of ( ) dEEhmdEEg 2

12/3

28

2

=

π

in equation (2) we get,

23

0

23

0

0 210 2

1

2/3

2

2/3

2

0

2/3

20

2/3

2

833

282

82

82

FF

EE

EhmEx

hmn

dEEhmdEE

hmn FF

=

=

=

= ∫∫

ππ

ππ

or 2/32

83

23

0

=

mhnEF π

∴ ( ) 3/23/23/223/22 3

83

80Bnn

mhn

mhEF =

=

=

ππ

Where 3/22 3

8

=

πmhB is a constant and its value is 5.85x10-38J

T = 0 K

T > 0 K

E = EF

12


Case (ii):- Fermi energy at T > 0 K,

We know that, for T > 0 K, we have f(E) ≠ 1 and the Fermi energy for such case is taken as EF which is given by ;

−=

22

0

0 121

FFF E

kTEE π

For smaller temperatures, the second term of the above equation is very small & hence it vanishes giving,

0FF EE =

implying that at ordinary temperatures EF and0FE essentially the same.

Fermi Temperature (TF) : - Fermi temperature is the temperature at which the average thermal energy of the free electron in a solid is equal to the Fermi energy at 0 K.

But the thermal energy possessed by electrons is given by the product kT.

Thus, when T = TF, kTF = EFo is satisfied

But all practical purposes, EFo = EF k

ET = E kT FFFF =∴∴

This is the expression for Fermi temperature.

For metals, we know that EF will be of the order of few eV, say EF = 5 eV, then

KTF 58000

10x38.110x602.1x5

23

19

== −

−

Fermi velocity (vF) : - The energy of the electrons in metals at Fermi level is EF. The velocity of the electrons which occupy the Fermi level is called Fermi velocity vF.

2/1

2F

2

21

=

=

mEvor

mvE

FF

F

This is the expression for Fermi velocity.

Effective Mass (m*):

When a metal is subjected to an external electric field, a free electron in the metal experiences a force and moves with an average drift velocity in it. The force experienced by the electron is due to the combination of applied external filed and that of a periodic potential due to lattice ions. Because of such combined effect, the electron responds as if it possesses a mass known as

13


effective mass (m*) which is different from its true mass (m). The expression for effective

mass is ( )

22

22

dkEd

hπ

The effective mass of an electron in vacuum is same as the true mass for an electron. It is notice that m* is different for different metals (solids).

Merits or Success of Quantum free electron theory:

The quantum free electron theory solves the flaws of the classical free electron theory which are discussed below.

1. Specific heat of free electrons: According to quantum free electron theory, the electrons occupying energy levels close to EF can absorb heat energy. Such electrons constitute a very small percentage of the total number of free electrons. Hence the specific heat of free electrons is given by

RTE

kCF

V2

=

Since the value of EF ranges from 1 to 10 eV, by taking a typical value of EF = 5 eV, we get

4102 −≈FEk

RTCV410−=∴

which is in agrees well with the experimental results.

2. Temperature dependence of resistivity or conductivity in metals: According to quantum free electron theory, the expressions for electrical conductivity & resistivity of a metal are given by

)1(2

2

→==∗

∗ λρλσ

nevmor

vmne F

F

In the above expression only the mean free path λ is the temperature dependent quantity. In classical theory, the collision was seen as a particle bouncing off another. In the quantum understanding, an electron is viewed as a wave travelling through the medium. If r represents the amplitude of the oscillation of the lattice ions can be considered to present a circular cross section of area πr2 that blocks the path of the electron waves. Hence electron waves are scattered more effectively results in a reduction of mean free path (λ). Thus λ is inversely proportional to the area of cross section.

i.e., )2(12 →∝

rπλ

But the area of cross section is directly proportional to the absolute temperature.

14


TT

T

Trei

∝∝

→∝∴

∝

ρσ

λ

λ

or 1

get we(1)equation in of values thengsubstitutiBy

(3) 1

.,. 2

This is exactly same as the experimental prediction. Thus quantum free electron theory properly explains the dependence of σ on T.

3. Dependence of electrical conductivity on electron concentration : According to

quantum free electro theory, the electrical conductivity in metals is given by

= ∗

Fvmne λσ

2

From the above equation it is clear that the electrical conductivity depends on both the

electron concentration n and Fv

λ .

If we compare the cases of copper and aluminium, the value of n for aluminium is 2.13 times higher than that of copper. But the value of λ/vF for copper is about 3.73 times higher than that of aluminium. Thus the conductivity of copper is more than that of aluminium.

Similarities between Classical and Quantum free electron theory:

The following assumptions apply to both the theories:

1. The valence electrons are treated as though they constitute an ideal gas. 2. The valence electrons can move freely throughout the body of the solid. 3. The mutual repulsion between the electrons, and the force of attraction between the

electrons and ions are considered insignificant.

Differences between Classical and Quantum free electron theory:

Sl. No. Classical free electron theory Quantum free electron theory

1 The energy levels of free electrons are continuous.

The energy values of the free electrons are discontinuous. i.e., the energy levels are discrete.

2 The free electrons may possess same energy.

Free electrons obey Pauli’s exclusion principle. i.e., no two electrons can have same energy.

3 The distribution of electrons in various energy levels obey the Maxwell-Boltzmann statistics.

The distribution of electrons in various energy levels obey the Fermi-Dirac statistics.

15


Problems :

1. The free electron density of aluminium is 18.10x1028 m-3. Calculate its Fermi energy at 0 K. Planck’s constant and mass of free electron are 6.626x10-34 Js and 9.1x10-31 kg.

Solns. n = 18.10x1028 m-3 Planck’s constant, h = 6.626x10-34 Js Mass of an electron, m = 9.11x10-31 kg

2. Calculate the density of states for copper at the Fermi level for T = 0 K. Given that, electron density of copper is 8.5x1028 electrons /m3.

Solns. n = 8.5x1028 m-3 Planck’s constant, h = 6.626x10-34 Js Mass of an electron, m = 9.1x10-31 kg

3. Find the probability of an electron occupying an energy level 0.02 eV above the Fermi level at 200 K and 400 K in a material.

Solns. E - EF = 0.02 eV = 0.02x1.602x10-19 J = 3.204 x10-21 J T1 = 200 K & T2 = 400 K

4. Show that the sum of the probability of occupancy of an energy state at ∆E above the Fermi level and that at ∆E below the Fermi level is unity.

eVxxJx

xxxx

xnm

hEF

68.1110602.1108689.1108689.1

1010.1831011.98

)1060626(38

19

1818

3/228

31

2343/22

0

===

=

=

−

−−

−

−

ππ

3/222/3

23

8,8

2)( 2

1

=

=

ππ n

mhEwheredEE

hmdEEg FF

( )2/1

2/3

234

312/3

2 05.710626.6101.98

28

2)( 2

1

xxxxE

hmEg F

=

=

−

−ππ

( ) eVJxxxxxxn

mhEF 05.7101293.1105.83

101.9810626.63

818

3/228

31

2343/22

==

=

= −

−

−

ππ

( ) ( ) ( ) 24.01188.3

11

1

1

11

11594.1

2001038.110204.323

2111=

+=

+=

+

==+

=

−−

− ee

eEf

xxxkTEETat F

( ) ( ) ( ) 36.017855.1

11

1

1

11

15797.0

4001038.110204.323

2122=

+=

+=

+

==+

=

−−

− ee

eEf

xxxkTEETat F

16


Solns. ( )

( )( ) ( ) 1 ..

)(at level Fermibelow the ocuupation ofy probabilit theis

& )(energy at level Fermi theabobe ocuupation ofy probabilit theis Let

=+

∆−=

∆+=

bE

aE

FbE

FaE

EfEfTS

EEEEf

EEEEf

( ) ( )

( ) ( )

( ) ( )

( ) ( ) ( ) ( ) 11

11

11 ΔEEor E ΔE)(E Elevel, Fermibelow theenergy For

11 ΔEEor E ΔE)(E Elevel, Fermi theaboveenergy For

11know that We

FF

FF

++

+=+∴

+=∴−=−−=

+=∴=−+=

+=

∆−∆

∆−

∆

−

kTEkTEbE

aE

kTEbE

kTEaE

kTEE

eeEfEf

eEf

eEf

eEf

F

( ) ( )

( ) ( )

( ) ( ) 1

11

111

111

11

11

1

Hence., ,1 ,Put

11

=+∴

=++

=+

++

=++

=+

++

=+

=∴=

+

∆−∆

bE

aE

xx

x

bE

aE

kTEkTE

EfEf

xx

xx

xxxEfEf

xexe

5. Calculate the probability of an electron occupying an energy level of 0.05 eV at 500 K above and below the Fermi level.

( )

( ) ( )

( )

( ) ( ) 17615.02385.0

7615.03132.11

13132.01

11

1

1 K., 500 T At,

1

1 ΔEEEor ΔE)(EE level, Fermi thebelowenergy For

2385.01927.41

11927.31

11

1

1 K., 500 T At,

Jx100.05x1.602 eV 05.0EEor ΔE)(EE level, Fermi theaboveenergy For

16087.1

5001038.1x100.05x1.602

FF

16087.1

5001038.1x100.05x1.602

-19FF

23

19-

23

19-

=+=+∴

==+

=+

=

+

==∴

+=∴−=−−=

==+

=+

=

+

==∴

==−+=

−

−

∆−

−

−

bE

aE

xx

bE

kTEbE

xx

aE

EfEf

ee

Ef

eEf

ee

Ef

6. Find the temperature at which there is 1 % probability that a state with 0.5 eV energy above the Fermi energy is occupied.

( ) ( ) ? & ?

K 500 T eV, 05.0EESoln.

F

==

==−bE

aE EfEf

17


( )

( ) ( )

KT

ee

ee

or

eEfTKW

Ef

TT

TxTx

kTEE F

12634.5955804Tor 595.4)99ln(5804

991100or 10001.011

1

1

1

101.0

1

1 ..

?T 0.01 % 1

J100.5x1.602xeV 5.0EESoln.

58045804

5804

1038.1100.5x1.602x

19-F

23

19-

====

=−===+∴

+

=

+

=

+=

===

==−

−

−

7. The Fermi level in potassium is 2.1 eV. What are the energies for which the probability of occupancy at 300 K are 0.99, 0.01 and 0.5?

( )( )

( ) 0.5Effor ? E Find, &0.01Effor ? E Find,

0.99Effor ? E Find, eV 2.1E potassiumFor

Soln.

33

22

11

F

======

=

( ) ( )( )

eVx

xxEf

sllyeV

xxx

Ef

sllyeV

xxx

Ef

EfEfkTEE

Efe

eEfTKW

F

kTEEkTEE

F

F

1.201.2)0(x02584.01.2

15.0

1ln10602.1

3001038.11.21)(

1lnkTEE

bygiven is )f(Efor E 2187.21187.01.2)5950.4x(02584.01.2

101.01ln

10602.13001038.11.21

)(1lnkTEE

bygiven is )f(Efor E 9813.11187.01.2)5950.4(x02584.01.2

199.01ln

10602.13001038.11.21

)(1lnkTEE

bygiven is )f(Efor E

1)(

1lnkTEEor 1)(

1ln

get, wesides,both on logarithm natural Taking

1)(

1or 1

1 ..

19

23

3F3

33

19

23

2F2

22

19

23

1F1

11

F

=+=+=

−+=

−+=

=+=+=

−+=

−+=

=−=−+=

−+=

−+=

∴

−+=

−=

−

−=+

=

−

−

−

−

−

−

−−

8. Calculate the Fermi energy in eV for a metal at 0 K, whose density is 10500 kgm-3, atomic weight is 107.9 and it has one conduction electron per atom.

18


Solns. Density of metal, ρ = 10500 kgm-3 Atomic weight of metal, wt. = 107.9 Fermi energy, EF = ? We know that the concentration of electrons in metal, n is given by

( )

eVxxJx

xxxx

xnm

hE

mx

F

51.510602.1108173.8108173.8

10816.531011.98

)1060626(38

10861.5107.9

1 x 60022x10 x 05001(wt.)ht atomicweig

ratomlectronspeno.offreee x )scostant(NAvogadro' x )density(n

19

1818

3/228

31

2343/2

3/22

32826

A

0

===

=

=

==

=

−

−−

−

−

−

ππ

ρ

REVIEW QUESTIONS

Objective / Short Answer Questions: 1. _ _ _ _ _ _ _ _ _ _ electrons are free or conduction electrons.

2. The classical free electron theory was proposed by _ _ _ _ _ _ _ _ _ _.

3. The free electrons in classical free electron theory are treated as _ _ _ _ _ _ _ _ _ _ _ _ _.

4. In classical free electron theory,_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ distribution law is applicable.

5. According to classical free electron theory, the expression for specific heat at constant

volume is _ _ _ _ _ _ _ _ _ _ _.

6. According to classical free electron theory, electrical conductivity varies inversely as _ _ _ _

_ _ _ _ _ _ _ temperature.

7. According to Pauli’s exclusion principle_ _ _ _ _ _ _ _ _ electrons can accommodate in

each energy level.

8. Maxwell-Boltzmann distribution law is applicable for the particle has _ _ _ _ _ _ _ _spin.

9. According to Pauli’s exclusion principle no two electrons can have _ _ _ _ _ _ _ quantum

number.

10. Experimentally specific heat at constant volume is given by _ _ _ _ _ _ _ _ _ _ _ _ _.

11. The quantum free electron theory was proposed by _ _ _ _ _ _ _ _ _ _ _ _ _ _.

12. In quantum free electron theory _ _ _ _ _ _ __ _ _ _ _ _ _ distribution law is applicable.

13. According to quantum free electron theory, the expression for specific heat at constant

volume is _ _ _ _ _ _ _.

14. The value of Fermi function in Fermi level at T > 0 K is _ _ _ _ _ _ _ _ _ _ _ _ _.

19


15. According to quantum free electron theory, electrical conductivity varies _ _ _ _ _ _ _ with

temperature.

16. Fermi-Dirac distribution law is applicable for the particle has _ _ _ _ _ _ _ _ _ _ _ _ _ spin.

17. The average distance traverses between two successive collisions of electrons in metal is

called _ _ _ _ _ _ _ _ _ _ _ _ _ _.

18. The velocity acquired by the electrons in a direction opposite to that of an applied electric

field is called _ _ _ _ _ _ _ _ _ _ _ _ _ _ .

Long Answer Questions: 1. What are the assumptions of classical free electron theory?

2. Define the terms: Fermi-level, Fermi-energy and Fermi-factor.

3. Mention the drawback/Failures of classical free electron theory.

4. What are the Success and Failures of Classical free electron theory?

5. What are the assumptions of quantum free electron theory?

6. Define the terms : Fermi-temperature, Fermi-velocity and density of states.

7. What are the successes of quantum free electron theory? Explain.

8. How quantum free electron theory successfully explain the failures of classical free

electron theory.

9. What are the merits of quantum free electron theory? Explain.

10. Write down the Fermi-Dirac equation for the probability of occupation of an energy level E

by an electron. Show that the probability of its occupancy by an electron is zero if E > EF

and unity if E < EF at temperature 0 K.

11. Define Fermi energy. Explain Fermi Dirac distribution for electrons in a metal at

temperature T = 0 K and T > 0 K.

12. Define density of states? Derive an expression for density of states in metals.

13. Write an expression for the Fermi energy distribution function FD (E) and discuss its

behavior with change in temperature. Plot FD (E) versus E for T = 0 K, and T > 0 K.

14. Derive an expression for carrier concentration in a metal by using the Fermi distribution

function. Explain the variation of Fermi level with temperature.

15. Obtain an expression for Fermi energy at temperature T = 0 K and relate it to Fermi energy

at non zero temperature.

16. Explain the dependence of electrical conductivity on temperature and electron

concentration on the basis of classical free electron theory.

17. Explain the meaning of ‘density of states’. Derive an expression for the number of allowed

states for unit volume of a solid.

20


18. Derive an expression for number of allowed energy states for a unit volume of solid.

19. Elucidate the difference between classical free electron theory and quantum free electron

theory.

20. Write down the difference between classical and quantum free electron theories.

Problems: 1. Calculate the density of states for copper at the Fermi level for T = 0 K.

2. Use Fermi distribution function to obtain the value of f(E) for E - EF = 0.01 eV.

3. Calculate the Fermi energy and Fermi temperature in a metal if the Fermi velocity of

electrons in the metal is 0.86x106 m/s.

4. Find the electron density for a metal with Fermi energy of 3 eV.

5. Find the probability that an energy level at 0.2 eV below Fermi level being occupied at

temperatures 300 K and 1000 K.

6. At what temperature can we expect a 10% probability that electron in silver have an energy

which 1% above the Fermi energy? The Fermi energy of silver is 5.5 eV.

7. The density if zinc is 7.13x103 kg/m3 and its atomic weight is 65.4. Calculate the value of

Fermi energy in zinc.

8. Consider the silver in the metallic state, with the one free electron per atom. Calculate its

Fermi energy. The density of silver is 10.5 gram/cm3 and its atomic weight is 108.

9. Evaluate the Fermi function for an energy kT above the Fermi energy.

* * * * END * * * *

21


SEMICONDUCTORS

Syllabus: Semiconductors - Classification of Semiconductors, Fermi level in intrinsic and extrinsic semiconductors. Expression for electron concentration in conduction band and Mention the expression for hole concentration in valance band of an intrinsic semiconductor. Relation between EF and Eg. Expression for intrinsic carrier concentration and conductivity of an intrinsic semiconductor. Numerical Problems.

Introduction: Band theory of solids: Band theory of solids can explain the conductivity of conductors, insulators and semiconductors on energy gap between the bands.

Energy bands in solids: A solid contains a large number of atoms closely packed together. In an isolated atom the electrons are tightly bound and have discrete, sharp energy level (1s,2s, 2p,3s,3p,3d,….so on). Each of these energy levels can hold a finite number of electrons with different magnetic and spin quantum numbers. A given atom has some of these levels filled and some empty depending on the total number of electrons associated with the atom. For example, an atom of carbon has six electrons out of which 2 electrons is in 1s, 2 electrons in 2s and 2 electrons in 2p levels. However 2p level can hold 6 electrons out of which 4 levels are empty. When two atoms are brought closer, the orbits of outer electrons overlap and hence the electrons are interact. As a result of this, two molecular orbitals are formed, one with lower energy called bonding orbital and the other with higher energy called antibonding orbital. Then the combination of two atomic orbitals results in two molecular orbitals which extend over both atoms. Hence the individual valence electrons are no longer localized to their original atoms. This results in shifting and splitting of each energy level into two energy levels. This splitting of energy levels is depends on the inter atomic distance between the atoms. If more and more atoms are brought closer, more and more closely spaced energy levels are formed. In a solid of N atoms each main energy level splits into a band of N closely spaced energy levels. The levels are so close, the energy separation between the successive levels in an allowed band is of the order of 10-27eV. The width of the energy band depends on the extent of overlapping of orbits. Hence the outer band is very wide. As we go to inner energy levels or orbits the width of the energy bands decreases. The energy bands are spreaded by gaps called ‘forbidden energy gap’. The splitting of energy levels and the energy bands at equilibrium inter atomic distance are shown in figure.

22


Classification of solids into conductors, insulators and semiconductors Based on electrical properties, solids can be classified as conductors, insulators and semiconductors. The order of magnitude of electrical conductivity serves as an indicator for the classification. There are few exceptions like conductors with incomplete inner orbits (like Sn, Mn, Zr, Rh etc.,) showing higher resistivity and narrow band gap semiconductors (like InSb, Bi2Te3 etc.,) with exceptionally high conductivity. Hence conductivity alone should not be the basis for classification of solids. The temperature coefficient of resistivity helps us to classify and distinguish between highly conducting semiconductors and less conducting metals. The positive value of temperature coefficient of resistivity for conductors and negative value for semiconductors can be explained on the basis of band theory.

Conductors: These are the solids which really conduct electricity. There are two types of energy bands in conductors depending on the electronic configuration of atoms.

In alkali metals and other metals having configuration ns1 or ns2np1 etc. having unpaired electrons in the outermost orbit of their atoms and hence the valence band is partially filled. So that, the electrons are easily excited to the higher levels in the same band. As a very large number of vacant levels exists, a large current can flow in conductors. In conductors having paired electrons in the outermost orbit the valence band is completely filled. So they should not conduct electric current. But it is observed that they also conduct electric current. This is because conduction band overlap with the valence band forming a composite band which is also partially filled as in the earlier case. In these conductors the forbidden gap/energy gap Eg = 0.

Insulators: These are solids, which don’t conduct electric current. In insulators valence band is completely filled and conduction band is completely empty. They are separated by a very wide energy gap of the order of 6 eV (> 3 eV).

Since the valence band is completely filled, the electrons can’t move, so they cannot conduct electricity. The conduction band is completely empty having no electrons to move. To excite electrons from valence band to conduction band, we require an electric field of the order of 6x108 V/m. hence an insulator doesn’t conduct electric current under normal condition.

Semiconductors: These are the solids, having conductivity less than the conductors and greater than the insulators. The energy band diagram of a semiconductor is similar to that of an

23


insulator but, the valence band and conduction band are separated by a small narrow forbidden energy gap of the order of < 3 eV.

Theory of semiconductors After the investigation of p-n diode and transistor, the importance of semiconductor was realized. Today all electronic circuits are making use of semiconductors. The invention of integrated circuits leads to the manufacture of computers. The fast growth of our knowledge and invention of new electronic devices is due to semiconductors only. Hence from an engineer’s point of view it is necessary to know about the theory of semiconductors. The electrical properties of semiconductors can changed by adding impurities to them. Therefore we have two types of semiconductors; pure and impure. The pure semiconductors are called intrinsic semiconductors and the impure semiconductors are called extrinsic semiconductors. In an intrinsic semiconductors, the – ve and + ve charge carriers (electrons & holes) are equal in number. The carriers contributing to electrical conductivity are generated due to breaking of covalent bonds by thermal excitation process. Pure germanium and silicon are the examples for intrinsic semiconductors. Each atom of these materials are having 4 valence electrons and all of them are participate in covalent bond formation with neighboring four atoms at 0 K. Hence no electrons are free in the material; as a result the material behaves like an insulator. Suppose the semiconductor heated to T K , the covalent bond is broken and electron in it acquire energy more from valence band to conduction band creates a vacant site in the valence band called hole. Thus the breaking of each covalent bond results in the formation of electron-hole pair. In the energy band diagram, the holes are shown in valence band. As the temperature increases the no of holes in the valence band and no of electrons in the conduction band increases, consequently conductivity also increases. Extrinsic semiconductors are obtained by adding very small impurities to pure semiconductors in a controlled manner. This process is called “doping” and the impurity added is called “dopant”. Doped semiconductors are called ‘extrinsic semiconductors’. By virtue of small impurity in an extrinsic semiconductor, it posses high electrical conductivity. There are two types of extrinsic semiconductors are there. They are

1. n - type semiconductors and 2. p - type semiconductors.

i. n - type semiconductors: If a donor impurity (pentavalent atoms) are added to a

tetravalent pure semiconductor such as germanium or silicon the – ve charge carriers are more than the + ve charge carriers. These types of semiconductors are called – ve type or n - type semiconductors.

24


ii. p - type semiconductors: If an acceptor impurity (trivalent atoms) are added to pure semiconductors, the +ve charge carriers are more than the – ve charge carriers. These types of semiconductors are called + ve type or p-type semiconductors.

Further due to thermal excitation, electron-hole pair is formed by the breaking of bonds, which become additional charge carriers. Thus both type charge carriers are observed in extrinsic semiconductors. However, the type of charge carriers which owe their formation due to only the thermal excitation and not due to doping are very small in number and they are called ‘minority carriers’. The other type charge carriers obtained due to both doping as well as thermal excitation are called ‘majority carriers’.

Significance of Fermi level According to Fermi-Dirac statistics, at temperature T = 0 K , all the energy levels below Fermi level are completely filled and the ones above are completely empty. Thus Fermi level (EF) at 0 K acts as a distinguished energy position between the filled and unfilled energy states. If we consider a material at temperature T > 0 K say at room temperature, the electrons just below the Fermi level absorbs energy and finds the place above Fermi level corresponding to the energy absorbed. Similarly in case of semiconductors at temperature T > 0 K, only the electrons in the upper part of the topmost filled energy band (in valence band) are capable of absorbing energy and occupy the vacant energy levels (in conduction band) above them. These electrons can’t stay in the higher energy level for long time and return to the lower energy levels whichever is vacant. The thermal excitation again takes which are vacant. This process of excitation and de-excitation continues as long as the thermal energy supply exists. As the temperature increases, electrons can be elevated to higher vacant energy levels, hence the Fermi energy levels (EF) is also increases. On the contrary, as the temperature decreases, the excitation energy reduces, hence EF also reduces.

Fermi level in intrinsic semiconductor

Energy gap (Eg) is the difference between the bottom of the conduction band (Ec) and top of the valence band (Ev) i.e., Eg = Ec - Ev. For convention, the energy at the top of the valence band is taken as zero (i.e., Ev = 0) for reference. At T = 0 K, all the energy levels in the valence band are completely filled and all energy levels in the conduction band are completely empty. But at room temperature, due to thermal excitation, some of the electrons at the top of the valence band are able to jump the energy gap and occupy some energy levels at the bottom part of the conduction band. These electrons return soon to the vacant energy levels left in the valence band. The electrons in this set of energy levels continue to undergo excitation and de-excitation, and thus becomes conduction electrons. Based on this one can say that, conduction

25


electrons between the energy levels in the bottom part of conduction band and the top portion of the valence band. Due to this distribution, the average energy of the electrons taking part in conduction will be almost equal to (Ec + Ev)/2 = Eg/2 (since Ev = 0 & Ec = Eg). Thus Fermi level lies in the mid part of the forbidden gap for an intrinsic semiconductors. Impurities level and Fermi level in extrinsic semiconductor

a) n - type semiconductors :

In case of n-type semiconductors, the electrons of donor atoms are free for the movement as compared to the resistance of electrons in that material, and hence they possess more energy. As a result the energy level for the electrons will be elevated to a position much higher than the valence band and lie very close to the conduction band as shown in the figure. These levels are called donor levels. The energy difference between the top of the valence band and the donor level is denoted as Ed. At low temperatures, thermal energy will be less and hence electrons can’t move from valence band to conduction band. But the same energy will be enough to transfer the electrons from donor level to the conduction band (since, Eg - Ed is small). These electrons are mainly responsible for conduction in n - type materials. Since these electrons occupy the energy levels very close to the bottom of the conduction band, the electrons get distributed between the energy levels in the bottom part of the conduction band and the donor levels. The difference between these two states will be almost equal to Ec - Ed (or Eg - Ed). Thus the average energy of electrons participating in conduction becomes

21

(Ec + Ed) or21 (Eg + Ed), and hence Fermi level in an n-type material at low temperature will

be located in the forbidden band at the level 21 (Ec + Ed) or

21 (Eg - Ed) just below the bottom of

conduction band. b) p - type semiconductors:

26


In case of p - type semiconducting materials, the acceptor atoms give rise to holes which are relatively free compared to those which are in the conduction band. Hence holes due to acceptor atoms possess higher energy than those in the conduction band. Since the energy for holes increases in the downward direction in the band diagram & they occupy energy levels in the band gap close to the valence band as shown in figure. These energy levels referred as acceptor levels. The energy difference between the acceptor levels & the top of the valence band is denoted by Ea - Ev = Ea (since Ev = 0). Since Ea is small, a small amount of thermal energy is sufficient to transfer the holes from the acceptor levels to the valence band. At low temperature, the holes participate in the conduction undergo excitation and de-excitation between the acceptor level and energy levels in the top portion of the valence band. As a result, the average energy of the holes participating in conduction becomes (Ea + Ev)/2 or Ea/2 (since Ev = 0). Hence, the Fermi level in p-type material at low temperature will be located in the forbidden gap at the level (Ea + Ev)/2 or Ea/2 from the top of the valence band.

Carrier concentration in an intrinsic semiconductor

1) Expression for concentration (density) of electrons. The number of electrons in the conduction band/unit volume of the material is called the electron concentration or electron density. Similarly, the number of holes in the valence band/unit volume of the material is called the hole concentration or hole density. In general the number of charge carriers/unit volume of the material is called carrier concentration or density. The number of electrons per unit volume in the energy range E & E + dE depends on the density of available state and the Fermi-Dirac distribution function. It is given by

( ) ( ) )1(→×= EfdEEgdNe

Where g(E)dE is the number of allowed energy states (density of states) per unit volume in the energy range E & E + dE, and f(E) is the Fermi distribution function represents the probability of occupation of states with energy E. The concentration of electrons in the conduction band (Ne) can be obtained by integrating equation (1) from E = Ec to E = ∞.

( ) ( ) )2(→×==∴ ∫∫

∞∞

EfdEEgNdNcc E

eE

e

According to Fermi-Dirac statistics, the Fermi distribution function is given by,

( )1

1

+

=

−

kTEE F

eEf

At room temperature kT = 0.026 eV is very small

For energy E > EC,

−

kTEE F

e >> 1 Hence

−

−

=+ kTEE

kTEE FF

ee 1

( ) )3(→=∴

−

−e kTEE F

Ef According to free electron theory of metals, the density of states per unit volume is given by,

27


( ) ( )

)4(28212

3

3 →= dEEh

mdEEg eπ In the present case electron is not free, it is moving in a periodic potential. So, me in the above equation should be replaced by me

* known as effective mass of the electron in the conduction band. In the conduction band of the semiconductors, the energy of electrons starts from Ec, hence the energy of the electron E is replaced by E - Ec. ∴equation (4) becomes,

( ) ( ) ( ) )5(2821

23

3 →−=∗

dEEEh

mdEEg ceπ

Substitute the values of f(E) & g(E)dE from equations (3) & (5) in equation (2), we get

( ) ( ) ( ) ( ) )6(282821

23

21

23

33 →×−=×−=

−

−∞∗

−

−∞ ∗

∫∫ dEeEEh

mdEeEEh

mN kTEE

Ec

ekTEE

cE

ee

F

c

F

c

ππ

Let us multiply the right side of the above equation by ( )( )

−

kTE

kTE cc

ekTkT

23

23

which is equal to unity.

After rearranging we get

( ) ( )kTdEe

kTEEkT

hmN kT

EkTE

kTE

kTE

E

cee

ccF

c

−++−∞∗

×

−

= ∫21

23

3

23

28 π

( ) ( )kTdEee

kTEEkT

hmN kT

EE

kTEE

E

cee

Fcc

c

−−

−−

∞∗

×

−

= ∫21

23

3

23

28 π

OR ( )

)7(28 21

3

23

→

−

=

−

−∞

−

−∗

∫ kTdEe

kTEEe

hkTmN kT

EE

E

ckTEE

ee

c

c

Fcπ

Let xkT

EE c =−

∴ kTdEdx = & when E = Ec, x = 0 & E = ∞, x = ∞

∴x varies from 0 to ∞, hence equation(7) becomes

( ) )8(28

03

23

21

→= ∫∞

−

−

−∗

dxexeh

kTmN xkTEE

ee

Fcπ

But ∫∞

− =0 2

21 πdxex x

, therefore equation (8) becomes

( ) )9(2

28 23

2

*223

23

→

=××=

−

−

−

−∗

=

−−

× kTEE

ckT

EEe

e

FcFc

eNeh

kTmNkT

FEcE

eh

kTemπππ

Where 23

222

=

∗

hkTmN e

cπ

Therefore equation (9) represents the expression for electron concentration in conduction band of intrinsic semiconductor. Where Nc is the temperature-dependent material constant known as effective density of states in the conduction band.

28


In silicon at T = 300 K, Nc = 2.8x1025/m3. Since Ec - Ev = Eg & if Ev = 0, then Ec = Eg. Therefore equation (9) becomes,

)10(2223

2 →=×

=

−

−∗

kTEE

ckT

EE

ee

gFgF

eNeh

kTmN π

This is also the expression for electron concentration. Expression for hole concentration: The number of holes per unit volume of a semiconductor in the energy range E & E + dE in the valence band is given by

( ) ( ) )1(→×= dEEgEfdN hhh

Where ( ) dEEgh is the density of states available for holes in the energy range E & E + dE in

the valence band and ( )Efh is the probability of finding the holes in the energy state E.

To evaluate Nh, let us consider the probability of having a hole at some energy E in the valence band as just the same as the probability of absence of an electron at the same energy E. As per the rule of probability, The probability of absence = 1 – probability of presence. Therefore Hole probability = 1 – electron probability

Hence, the probability that a hole is available with an energy E is ( ) ( )EfEfh −= 1

( )11

11

1

11

+

=

+

−+=

+

−=

−

−

−

−

−

kTEE

kTEE

kTEE

kTEE

kTEEh F

F

F

F

F

e

e

e

e

eEf

Or ( )

−

− +

=+

=kT

EE

kTEE

h F

F ee

Ef1

111

1

Since kT is smaller, 1>>

−

kTEEF

e , then

−

−

≈+ kTEE

kTEE FF

ee 1

( ) ( ) ( ) )2(→== −−− FF EEEEh eeEf

The density of states for holes in valence band for the energy range E & E + dE is given by,

( ) ( ) ( ) )3(2821

23

3 →−=∗

dEEEh

mdEEg Vh

hπ

Where mh* is the effective mass of the hole in valence band and E = Ev - E is the energy of hole

in the valence band. By substituting the values of fh(E) and gh(E)dE in equation (1), we get

( ) ( ) )4(2821

23

3 →×−=

−∗

dEeEEh

mdN kTEE

Vh

h

Fπ

29


The concentration of holes in the valence band can be obtained by integrating equation (4) from E = - ∞ to E = Ev

( ) ( )∫∫∞−

−∗

∞−

×−==∴V FV E

kTEE

Vh

E

hh dEeEEh

mdNN 21

23

3

28 π

Multiplying the right hand side of the above equation by ( )( )

−

kTE

kTE VV

ekTkT

23

23

& by rearranging we

get,

( ) ( )kTdEe

kTEEkT

hmN kT

EkTE

kTE

kTEE

Vhh

VVFV

−+−

∞−

∗

×

−

= ∫21

23

3

23

28 π

( ) )5(28 21

3

23

→

−

×= ∫∞−

−

−

−∗

kTdE

kTEEee

hkTmN V

EkT

EEkT

EEh

h

V VFVπ

Let xkT

EEV =

−

−=∴

kTdEdx

or dx

kTdE

−=

Also for E = Ev, x = 0 & E = - ∞, x = ∞

( ) ( ) ( ) ( ) )6(2828

03

23

0

3

23

21

21

→=−=∴ ∫∫∞

−

−∗

∞

−

−∗

dxexeh

kTmdxexeh

kTmN xkTEE

hxkTEE

hh

FVFV ππ

But ( )20

21 π

=∫∞

− dxex x therefore equation (6) becomes,

( ) )7(222

28 23

23

23

→×=×

=××=

−−

−−∗

−∗kT

EE

VkT

EEhkT

EEh

h

VFVFFV

eNeh

kTmeh

kTmN πππ

Where23

2

22

=

∗

hkTm

N hV

π is the effective density of states in the valence band for holes.

Equation (7) represents the expressions for hole concentration or density. Since Ev = 0, equation (7) becomes

)8(2223

2 →=×

=

−

−∗kTE

VkTE

hh

FF

eNeh

kTmN π

This is also the expression for hole concentration. Relation between Fermi energy and energy gap for an intrinsic semiconductor (or Fermi energy in intrinsic semiconductor) For an intrinsic semiconductor, the number of electrons per unit volume in conduction band is equal to the number of holes per unit volume in valence band. ∴ Ne = Nh

30


By substituting the values of Ne and Nh we get

−∗

−

−∗

×

=×

kTEE

hkTEE

eFVFc

eh

kTmeh

kTm 23

23

22

2222 ππ

( ) ( )

−

∗

−

−∗ =× kT

EE

hkT

EE

e

FVFc

emem 23

23

23

23

)(2

OR

=

=

∗

∗

+−

∗

∗

+−+−

e

hkTEEE

e

hkTEEEE

mme

mme

VCF

FVFc

Taking logarithm on both sides, we get

=+−

+−

∗

∗

∗

∗

=

e

hVCF

e

hVCF

mmEEE

mm

kTEEE

ekT

e

log23

log23

)(2 OR

)(2

=

+

− ∗

∗

e

hVCF m

mEEEe

kT log43

2

+

+= ∗

∗

e

hVCF m

mkTEEEe

log43

2

At T = 0 K,

+

=2

VCF

EEE This is the expression for Fermi energy.

As we know that, Ev = 0 & Ec = Eg, then

2g

F

EE =

Thus the Fermi level is in the middle of the band gap for an intrinsic semiconductor.

Intrinsic carrier concentration In an intrinsic semiconductor there will be continuous production of electron-hole pairs due to thermal energy. At the same time there will be decrease in the number of electrons and holes due to electron-hole combination. As long as temperature remains constant the electron-hole concentration will also remain constant. Thus at temperature T K, Ne = Nh = Ni. Where Ni is called intrinsic charge carrier concentration. Therefore the product of concentrations is Ne x Nh = Ni

2. On substituting for Ne & Nh in the above equation, we get

−++−

−

−

−

×=××=kT

FEVEFEcEkT

FEVEkT

FEcE

eNNeeNNN VcVci2

31


−

−

−

×=×=kT

gE

kTVEcE

eNNeNNN VcVci2

Where Ec - Ev = Eg

−

×=kTgE

eNNN Vci

2

On substituting for Nc & Nv, we get

−∗∗

×

×

= kT

E

hei

g

eh

kTmh

kTmN 222

21

23

23

2222 ππ

( ) )1(22 22

432

3

→

=

−

∗∗ kTE

hei

g

emmhkTN π

)2(223

→=

−

kTE

i

g

eCTN Where ( )432

3

2

22 ∗∗

= he mm

hkC π

Equations (1) & (2) represents charge carrier concentrations in an intrinsic semiconductor. Conductivity in intrinsic semiconductor: Conductivity will take place in an intrinsic semiconductor due to the movement of electrons and holes (charge carriers) under the application of electric field. The average velocity acquired by the charge carriers in the presence of electric field E, is called drift velocity Vd. The drift velocity, Vd α E or Vd = µE → (1)

Where µ is called mobility of charge carriers. It is defined as the velocity acquired by a carrier

per unit electric field. i.e. )2(EV= d →µ

The net transfer of charges through the cross-section of the specimen gives the electric current in it. The total electric current (I) in the semiconductor specimen is the sum of the current due to drifting of electrons in the conduction band (Ie) and the current due to drifting of holes in the valence band(Ih).

)3(→+=∴ he III

To find an expression for electrical conductivity, first we shall consider the flow of electrons in the semiconductor. Let ‘A’ be the area of cross-section, Ne is the number of electrons/unit volume and ‘e’ is the magnitude of charge of an electron, then the flow of electron/second in the conduction band gives the electron current Ie.

)4(→=∴ deee eAvNI

∴ Current density of electrons is given by,

deee

e evNAIJ ==

)5(→= EeNJ eee µ Since Ev ede µ=

But ohm’s law gives the relation for Je as )6(→= EJ ee σ

32


On comparing equations (5) & (6), we get )7(→= eee eN µσ

Is the electronic conductivity is the conduction band. Similarly, the conductivity due to holes in the valence band is given by

)8(→= hhh eN µσ

Where, Nh and μh represents the concentration & mobility of holes respectively. The total conductivity of the semiconductor is given by,

hheehe eNeN µµσσσ +=+=

or ( ) )9(→+= hhee NNe µµσ

In an intrinsic semiconductor, ihe NNN == is the intrinsic carrier density, so that equation (9) becomes

( ) )10(→+= heii eN µµσ

Using the expression for Ni from intrinsic semiconductor

i.e., ( )

−

∗∗ ×××

= kT

E

hei

g

eTmmh

kN 22

23

432

3

22 πin equation (10), we get

( ) ( )

−

∗∗ ×+××××

= kT

E

hehei

g

eeTmmh

k 22

23

432

3

22 µµπσ

or )11(2 →=

−

kTE

oi

g

eσσ

where ( ) ( )heheo eTmmh

k µµπσ +××××

= ∗∗ 2

3432

3

222

Equation (10) represents the expression for electrical conductivity in intrinsic semiconductor.

Problems : 1. The mobility of electrons and holes in a sample of intrinsic germanium at 300K are 0.36 m2/Vs and 0.14 m2/Vs respectively. If the resistivity of the specimen is 2.2 Ω.m, compute the intrinsic concentration. we know that,

( )heii eN µµσ +=

Or ( )heiii en µµσ

ρ+

==11

( ) ( )14.036.010602.12.211

19 +××=

+=∴ −

heii e

nµµρ

318 /1067.5 mni ×=

2. The following data are given for intrinsic germanium at 300 K. ni = 2.4x1019/m3, μe = 0.39 m2/Vs, μh = 0.19 m2/Vs. calculate the resistivity of the sample.

Solutions, µe = 0.36 m2/Vs, µh = 0.14 m2/Vs ρi = 2.2 Ωm ni = ?

33


we know that,

( )heii en µµ

ρ+

=1

ii σ

ρ 1=

( ) mi Ω=+×××

= − 448.019.039.010602.1104.2

11919ρ

3. Find the conductivity of intrinsic silicon at 300 K. It is given that ni at 300 K in silicon is 1.5x1016/m3 and mobilities of electrons and holes in silicon are 0.13 m2/Vs and 0.05 m2/Vs respectively. Solution, we know that, ni = 1.5x1016/m3 σi = ni e (µe+µh) μe = 0.13 m2/Vs σi = 1.5x1016x1.602x1019 (0.13+0.05) μh = 0.05 m2/Vs σi = 4.325x10-4 /Ωm

4. For an intrinsic semiconductor with gap width Eg = 0.8 eV, calculate the concentration of intrinsic charge carriers at 300 K assuming that mn

* = mp*= me (rest mass of electrons).

we know that,

−

= kT

E

ei

g

ehkTmN 2

2

23

22 π ep mm == ∗∗em

−

×

=∴ kT

E

ei

g

eThkmN 2

223

23

22 π

( ) [ ]

×××××

−

−

−−−

×

×

××××= 3001038.12

10602.18.0

234

312323

19

23

23

exp30010626.6

101.91038.122 πiN

m/10 3.0410 1.22 5196.15 102.4 = N 18-721i ×=××××

5. The effective mass of the electron in silicon is 0.31 me, where me is the free electron mass. Find the electron concentration for silicon at 300 K assuming that the Fermi level lies exactly in the middle of the energy gap, given that the energy gap for silicon is 1.1 eV.

we know that,

−

−∗

×

= kT

EEe

e

Fc

eh

kTmN23

2

22 π

( )kTE

ekT

EE

ee

g

gg

eh

kTmeh

kTmN 22

2

2

23

23

2222−∗

+−

∗

×

=×

=∴

ππ

( )

×××××

−

−

−−−

−

×

×

×××××= 3001038.12

10602.11.1

234

233123

1923

10626.63001038.1101.922 eNe

π

3151024 1007.210784.4103255.4 mNe ×=×××= −

6. Calculate the position of the Fermi level for pure silicon at 300 K, if the electron

Solutions, ni = 2.4x1019/m3 μe = 0.39 m2/Vs μh = 0.19 m2/Vs ρi = ?

Solutions, me

* = 0.31me T = 300K if Ev = 0, Ec = Eg & also Eg=1.1eV=1.1x1.602x10-19J here EF = Eg/2 & me

*= 0.31me

EF = Eg/2, Ne=?

Solutions, Eg = 0.8 eV T=300 K Ni = ?

34


concentration is 2x1015m-3. Given that for silicon, the energy gap is 1.1eV, and the effective mass of electron is 0.31me, where me is the free electron mass.

we know that,

−∗

×

= kT

EE

ee

gF

eh

kTmN23

2

22 π

( )

−

−

−−

×

×

××××××=× kT

EE gF

e23

234

233115

10626.63001038.1101.931.022102 π

−

××=× kTEE gF

e2415 103255.4102

or 10

24

15

10624.4103255.4

102 −

−

×=×

×=kT

EE gF

e

or ( ) 5.2110624.4ln 10 −=×=

− −

kTEE gF

( )5.21−×=−∴ kTEE gF

or ( )5.21−×+= kTEE gF

( )5.213001038.110602.11.1 2319 ×××−××= −−FE

1919 1089.0107622.1 −− ×−×=FE

JEF1910872.0 −×=

or eVEF 544.010602.110872.0

19

19

=××

= −

−

The position of Fermi level is 0.544 eV. Some important questions

1. Explain the significance of Fermi level in an intrinsic semiconductor. 2. Explain the significance of Fermi levels in extrinsic semiconductor. 3. Explain the significance of Fermi level in n-type extrinsic semiconductor. 4. Explain the significance of Fermi level in p-type extrinsic semiconductor. 5. Derive the expression for electron concentration in an intrinsic semiconductor. 6. Derive the expression for hole concentration in an intrinsic semiconductor. 7. Derive the expression for electrical conductivity of an intrinsic semiconductor.

* * * End * * *

Solutions, 315102 −×= mNe

JEg1910602.11.1 −××=

ee mm 31.0=∗

T=300 K EF=?

35

P.E.S. College of Engineering

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