Unit 4 Band Theory of Solids

8/3/2019 Unit 4 Band Theory of Solids

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BAND THEORY OF SOLIDS

The structure of much of solid-state theory comes directly from group

theory, but until now there has been no elementary introduction to the band

theory of solids which adopts this approach.

The Kronig Penney Model

Introduction

In the free electron model of a metal the effects of the positive ions,

i.e. the lattice, are neglected. We saw earlier that all energies were allowed

in the free electron case. The free electron theory (V = 0 or constant) in the

form of the Drude theory was reasonably successful in explaining manymetallic properties. However, it was not possible to account for the

differences between metals, insulators and semiconductors using this theory.

In addition, the specific heat capacity of solids was found to be much

smaller than that predicted by the free electron theory.

+ + + +

V(x)

x

Figure 1 Realistic Potential profile for a 1D crystal

In order to account for the differences between electronic materials,

the influence of the crystal lattice on the electrons must be taken into

account when applying Schrodinger's equation to the crystal. The results,

even for a very simple model of the lattice potential, are surprising. We find

that a whole new theory of solids emerges, the Band Theory of Solids. In


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this theory, entities such as bandgaps, positive and negative electron masses,

positive and negative electrons, i.e. holes and electrons, emerge as natural

consequences of the fact that the crystal lattice gives rise to a periodic

potential energy profile within the crystal.

The Nature of the Crystal Potential Field

X-ray diffraction has shown that crystals consist of well defined

periodic arrays of atoms. Let us consider a one dimensional crystal with

atoms which consist of positively charged cores surrounded by loosely

bound outer electrons (the outer electrons are loosely bound due to the

screening effects of the inner electron shells). Since the atoms are in a

periodic arrangement, it is only reasonable to assume that the potential field

arising from the atom cores will also be periodic. The outer electrons, i.e. the

conduction electrons, move in this periodic potential. A realistic potential

profile, based upon Coulomb's potential, i.e. V(x) 1/x, might look like

that shown in the diagram below. However, it is extremely difficult to solve

Schrodinger's equation for a realistic periodic potential. Kronig and Penney

suggested a simplified model consisting of a 1D array of square well

potentials of width a, separated by potential barriers of height V0 and width

b. It is assumed that for any electron, everything else in the crystal can be

represented by this effective potential. This one electron is then consideredto be representative of all other electrons in the system. The value of this

model is that Schrodinger's equation may be solved individually for both

regions. By establishing continuity equations at the boundaries, the form of

overall solutions are arrived at.

-Vo

V(x)

x

Figure 2 Square Well Potential representation of the lattice potential.


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2eV

2

6

b

a

l

Vo

Figure 3 Kronig Penney Model of a 1-D crystal.

Solving Schrodingers equation for the Kronig-Penney model of a 1D

crystal:

2 222

0m

d x

dxE V x x

( )( ( )) ( )+ =

Eqn. 1

d x

dx

mV x E x

2

2 2

20

( )( ( ) ) ( ) =

Region I 0 x a V(x) = 0

Eqn. 2

d x

dx

mEx

2

2 2

20

( )( )+ =

Solution:

Eqn. 3

( ) x Ae Bei x i x= +

I II

0 a l


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2

2

2=

mE

.an energy term

Region II a x l V(x) = V0

Eqn. 4

d x

dx

mV E x

2

2 2 0

20

( )( ) ( ) =

Solution

Eqn. 5

( ) x Ce Dex x= +

2 2 02

= m

V E ( )Now since the barrier is finite, there must be some probability of penetrationby the electrons. Therefore, both the wave function and its first derivative

must be continuous at points such as x = 0, a, l, etc.. This allows us

establish relationships between the constants A, B, C and D:

continuity at:

x = 0

Eqn. 6

1 20 0( ) ( )=

+ = + A B C D

Eqn. 7

1 20 0' '( ) ( )=

= i A i B C D

at x = a

Eqn. 8

1 2( ) ( )a a

Ae Be Ce Di a i a a a

=

+ = +

Eqn. 9

1 2' '( ) ( )a a

i Ae i Be Ce Dei a i a a a

=

=

and at x = l

Eqn. 10


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1 1 20 0( ) ( ) ( )

( )

l e

A B e Ce De

ikl

ikl l l

= =

+ = +

Eqn. 11

1 1 2

0 0' ' '( ) ( ) ( )

( ) ]

l e

i A B e Ce De

ikl

ikl l l

= =

= [

Equations 8-11 have a solution only if the determinant of the coefficients of

A, B, C and D vanish or if,

Eqn. 12

2 2

2

Sinh b Sin a + Cosh b Cos a = Cos kl

for E < V0. For E > Vo, becomes purely imaginary. Equation 12 does not

change significantly if we replace by i :

Eqn. 13

2

Sin b Sin a + Cos b Cos a = Cos kl 2

2

(Cosh x = Cos ix; Sinh x = 1/i Sin ix )

We could plot the left hand side of the last two equations versus E/V 0, (see

Bar-Lev p69). However, to obtain a more convenient form Kronig and

Penney considered the case where the potential barrier becomes a delta

function, that is, the case where V0 is infinitely large, over an infinitesmal

distance b, but the product V0b remains finite and the same, i.e. as V0, b

0, V0b = constant. Now 2

0V and also goes to infinity as V0. Therefore:Lim ( 2

2 2

0)V

What happens the product b as V0 goes to infinity?

b becomes infinitesmal as V0 becomes infinite.

however, since is only proportional to V0 it does not go to

infinity as fast as b goes to zero and so,

the product b goes to zero as V0 goes to infinity


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b 0 as V0 Therefore:

Sin b b

Cos b 1

as V0

Also,as b 0

a l

since l = (a + b).

Equation 13 then reduces to

Eqn. 14

2

Sin l + Cos l = Cos klb

2

or

Eqn. 15

2

2

abSin l

l+ Cos l = Cos kl

If we define

Eqn. 16

Pa b

b

V

=

l i m

2

020

which is a measure of the barrier strength or stopping power, then

Eqn. 17

PSin l

l+ Cos l = Cos kl

or

Eqn. 18

P Sinc l + Cos l = Cos kl

This is the dispersion relationship for electrons in a periodic 1D crystal.


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MathCAD Representation

The Energy Dispersion Relation for the Kronig-Penney Model

When Schrodinger's equation is applied to the K-P model as outlined

above, solutions are only possible which satisfy the following dispersion

relationship between energy (alpha term) and momentum (k term):

PSin l

l+ Cos l = Cos kl

where l=a+b, the lattice spacing;

. .,..4 r a d .4 .

5 0r a d ..4 r a d

D ( ), P .Ps i n ( )

c o s ( )

1 0 5 0 5 1 02

1

0

1

2

3

4

55

2

1

1

D ( ), 0

D ( ), 3

D ( ), 1 0

.4 .4

0

Figure 4 Dispersion relationship for electrons in a periodic

lattice. Barrier stopping power, P, has values 0, 3 and 10.


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2

2

2=

mE

P

mabVo

= 2

Since the dispersion relation will have the same value if Cos(kl) is replaced

by Cos(k+n2 /l)l, the energy curves are unchanged if shifted in k space by

n2 /l. We can therefore represent all the information in the region - /l


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Reduced Zone

0.00E+00

5.00E-01

1.00E+00

1.50E+00

2.00E+00

2.50E+00

-4 -3 -2 -1 0 1 2 3 4k space

Energein

eV


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Effective Mass and Holes

What happens when an electron is accelerated in a lattice with a periodic

potential

ph

k= =

F ma q= =

aq

m=

a = dvdtg

vd

dk

dE

dkg= =

1

aE

= =dv

dt

d

dk

dk

dtg

212


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E

k

vg

k

m*

k

EC

ET

0 /l/2l-/l -/2l

Figure Sketches of the group velocity and effective

mass for the valence band of a semiconductor.


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dE = q dx = q v dt =q dE

dkdtg

dk

dt

qE=

aE

dk=

q d2

2

2

m*d E

dk

2

2

= 2

When an electron is accelerated in a crystal, its response is no longer

determined by the electron rest mass. Instead the electron behaves as though

it has an effective mass given by above equ. The effective mass of the

electron is determined by the curvature of the E versus k diagram. This is

illustrated in the diagram below. Near the bottom of the band, the effective

mass is positive and here electrons behave as normal except that the mass is

no longer constant or equal to the rest mass. Near the top of a given band wesee that the mass of the electron may be negative. In simple terms, this

means that the carriers here move in the opposite direction to the applied

force. The simplest interpretation of this is to assume that we are now

dealing with positive electrons or holes as they are almost universally

referred to.


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Density of States in SemiconductorsOnce the band-structure has been determined, the next task is to

determine the number of modes or states in the band or more importantly,

the density of states per unit energy within a given band. Once we have this

we can then use appropriate statistics to determine how the bands are filled.Since we are interested in the electronic properties of solids, we will be

interested in working out how the bands are filled with electrons. Our

interest will be confined to semiconductors with either an almost empty

conduction band (n-type) or an almost full valence band (p-type). Therefore

we need only consider the bottom of the conduction band or the top of the

valence band and so we may use the parabolic approximation to simplify the

analysis.

Consider a box shaped crystal of sides Lx, Ly, Lz with interatomic

spacings lx, ly, lz respectively. There are Nx=Lx/lx, Ny=Ly/ly, and Nz=Lz/lzcorresponding values of kx, ky, kz. A single level, therefore, requires a

section of Brillouin zone in the kx direction of width

kx

= =

22

l

x x

x

N L

and the volume in k space taken by a single level is kx ky kz, since twostates with opposing spin can have the same energy, a single state requires

only half this volume. Therefore a single state volume is:

12

32

2 k k k

L L Lx y z x y z=

( )

The number of allowed states N in k space in a spherical shell of radius k

and thickness dk is:

N' = =

42

k dk

(2 )2L L L

kL L L dk

2

3

x y z

x y z

The density of states per unit volume is:


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dkk

=dN2

For the bottom of the conduction band, E Ek

2m

C

2

n

* + 2

, implying:

km

E - E2n

*

2 C=2

( )

and also,

dk =m

kdE

n

*

2

Therefore, the density of states in the conduction band is,

dN =4

hm (E E ) dE

= N (E)dE

3 n

*

C

C

12

( )23

2

where the density of states function is given by

N (E)=

4

h m (E E )C 3 n*

C

12

( )2

32

.

Similarly for the valence band, the density of states function has the form,

dN =4

hm (E E ) dE

= N (E)dE

3 p

*

V

V

12

( )23

2


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Occupation of allowed states.

For collections of particles, e.g. atoms, molecules, electrons, a statistical

treatment which describes the average rather than detailed properties of a

typical component of the complete assembly of particles is most useful since

the behaviour of the group of particles can then be deduced directly. The

type of statistics used depends on -

the type of particle present (neutral, charged, mass, etc..)

the possible interactions between them.

Classical particles obey Maxwell-Boltzmann statistics where there is no

restriction on the energy of the particles, e.g. ideal gas,

dN = N FMB(E) dE

where N is the number of neutral molecules per cubic meter.FMB is the distribution function and gives the fraction of thetotal number of molecules per unit volume in the energy

range dE. In the high energy region,

F (E) eMB

E

k TB

EC

EV

En

E

k Ep

NC(E)

NV(E)

Conduction Band

Valence Band

Figure 5 Density of states in the

conduction and valence band with

parabolic band structure.


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Particles which obey the exclusion principle, e.g. electrons, interact quantum

mechanically in such a way that the occupancy of a particular state is

restricted by the Pauli exclusion principle. For such particles, Fermi-Dirac

statistics apply. The Fermi-Dirac distribution function has the form,

F

e

FD E E

k T

F

B

=

+

1

1

This gives for any ensemble obeying the exclusion principle, the probability

that a particular state E is occupied.

For high energy states, i.e. E>>EF, FFD FMB. At high energies, the

number of electrons distributed over many available states is small and there

are many more energy levels than electrons to occupy them. Under these

conditions, there is little chance of two or more electrons occupying thesame state and whether the exclusion principle is included in the statistics or

not becomes irrelevant to the form of the distribution.

F ( ),E T ..4

E

( ).k T

3

2

e

E

.k T

0 1 2 3 4 5 60

2 1 02 0

4 1 02 0

F ( ),E 2 0 0

F ( ),E 3 0 0

F ( ),E 4 0 0

E

.k T

Figure Maxwell-Boltzmann distribution for a classicalensemble of particles.


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At very low temperatures there is a small but finite probability that

electrons will occupy available states for which E>EF but the probability

rapidly decreases with increasing energy. As the temperature is increased,

this tail of the probability function becomes more pronounced and the

probability of occupancy of higher energy states is correspondingly

increased.

Note that the probability that an electron occupies the Fermi Energy

Level, EF, (referred to simply as the Fermi level) is always independent of

the actual temperature. Also, the probability function is symmetric about the

Fermi level.

Properties of Semiconductors

Extrinsic Semiconductors

Intrinsic Semiconductors

Doping

F ( ),E T1

1 e

E E F

.k T

0 0 . 5 1 1 . 5 20

0 . 5

1F ( ),E 2

F ( ),E 5 0

F ( ),E 1 0 0

E

E F

Figure :Fermi-Dirac distribution function for electrons at

temperatures 2, 50 and 100K.


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Carrier Density

A Conduction Band

dE

e TkEE

B

F

+

= 1

1)E(E)m2(h4=(E)dE(E)FNN(E)dE 2

12

3

C*n3FDC

The density of electrons in the whole conduction band is,

=Top

C

E

EFDC dEEFENn )()(

Since EF is located at least a few kBT below EC, we can make the following

approximations,

(1) We can replace FFD(E) by FMB(E) since (E-EF/kBT) >> 1

(2) Recall FMB(E) goes to zero as E goes to infinity. Therefore, we can

replace the top limit of the integral by without changing the

result.

Therefore,

dEeTk

EE

B

F

C

21

23

EC

*

n3)E(E)m2(

h

4=n

Now

= TkEE

Tk

EE

Tk

EE

B

FC

B

C

B

F

eee .

dEeeTk

EE

Tk

EE

B

C

B

FC

C

21

23

EC

*

n3)E(E)m2(

h

4=n

21

21

21

)(

=

Tk

EEkTEE

B

CC

Let x = (E-EC)/kBT

Also dE=kBTdx


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Substituting,

dxeeTk xTk

EE

BB

FC

0

*

n3

21

23

x)m2(h

4=n

202

1 =

dxx

= TkEE

C

Tk

EE

B

B

FC

B

FC

eN

eTk 23

)m2(h

2=n *n3

where NC is a constant and is known as the Effective Density of States in the

conduction band.

B Valence Band

Similarly for holes in the valence band,

= TkEE

V

Tk

EE

B

B

VF

B

VF

eN

eTk 23

)m2(h

2=p *p3

where NV is a constant and is known as the Effective Density of States in thevalence band.

At

300K

Si GaAs

NC 2.8 x 1019cm-3 4.7 x 1017 cm-3

NV 1.04 x 1019 cm-3 7.0 x 1018 cm-3

Fermi Level in Semiconductors

Intrinsic semiconductors

In intrinsic semiconductors, n=p=ni where ni is the intrinsic carrier density.


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= TkEE

V

Tk

EE

CB

ViF

B

iFC

eNeN

2

32

==

n

pTk

EEE

V

C

m

me

N

NB

iFVC

+

+=

*4

3

2n

pBVCF

m

mLn

TkEEE

i

In both Si and Ge, mn* mp

*. Therefore, EFi is practically mid-way between

EC and EV, i.e. at EG/2.

Law of Mass Action

= TkEE

Tk

EE

VCB

ViF

B

iFC

eeNNnp

2

i

Tk

E

VC neNNnpB

G

==

= TkE

B

G

eATnp 3

where A is a constant. The electron-hole product in semiconductors in

thermal equilibrium is a function of T only since mn* and mp* and EG are

relatively independent of T. This is a general result holding for both intrinsic

and extrinsic semiconductors and is known as the law of mass action; the

electron-hole product is a constant at a given temperature.

ni = 1.45 x 1010cm-3(Si)

ni = 1.79 x 106cm-3(GaAs)


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Fermi Level in Extrinsic Semiconductors

Doping

n-typeAddition of Gr V donor Impurity level just below Ec!

p-type Addition of Gr VI acceptor Impurity level just above Ev!

ND Density of donor atoms

ND+ Density of ionised donors

NA Density of acceptor atoms

NA+ Density of ionised acceptors

n Free electron density

p Free hole density

ni Intrinsic electron/hole carrier density.

Extrinsic Semiconductorpn

Charge Neutrality must prevail

p + ND+ = n + NA

-

+

= TkEEE

V

C B

VFC

eN

N

p

n 2

Conduction Band

Valence Band

EV

EC

ED

EA

Pair Production

EFi

Free Electrons

Free Holes


23/25

+

+

+=

p

nLn

Tk

N

NLn

TkEEE B

C

VBVCF 222

+=

p

nLn

TkEE BFF i

2

The Fermi level moves up towards EC when donor impurities are added. The

reverse happens when acceptors are added, i.e. EF moves down towards EV.

Hall Effect

If an electric current flows through a conductor in a magnetic field, the

magnetic field exerts a transverse force on the moving charge carriers which

tends to push them to one side of the conductor. This is most evident in a

thin flat conductor as illustrated. A buildup of charge at the sides of the

conductors will balance this magnetic influence, producing a measurable

voltage between the two sides of the conductor. The presence of this

ndE

pdE

Holes

Electrons

NC

NV

E

EC

EVp

n

F(E)EF


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measurable transverse voltage is called the Hall effect after E. H. Hall who

discovered it in 1879.

Note that the direction of the current I in the diagram is that ofconventional

current, so that the motion of electrons is in the opposite direction. That

further confuses all the "right hand rule" manipulations you have to go

through to get the direction of the forces.

At equilibrium

e EH = e v B

EH = v B ----(1)If Jx is the current density in the X-direction then

Jx = n e v

Or v= Jx/ n e ---------(2)

Where n is the concentration of the current carrier

Therefore

EH = Jx B / n e -------(3)

Hall coefficient RH is defined in terms of density Jx by the relation

EH = RH Jx B --------- (4)RH = 1/n e ----------(5)

RH is hall coefficient

In this case

RH= -1/n e

It is negative because the electric field developed is in the ve of Y-

direction
http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elecur.html#c3http://hyperphysics.phy-astr.gsu.edu/hbase/electric/elecur.html#c3


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For P-type semi conductor

RH= 1/p e

Where p is the positive hole density

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Unit 4 Band Theory of Solids

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