band theory of solidsnvnvb

8/13/2019 Band Theory of Solidsnvnvb

1/28

BAND THEORY OF SOLIDS


2/28

Valence Band

band occupied by valenceelectron

may be completely orpartially filled

some of the electrons

may be loosely attached tothe nucleus and leavevalence band. These arethe free electrons.

Conduction Band

free electrons take part inthe conduction process.Thus they are calledconduction electrons

Band occupied by the

conduction electronsMay be empty or partially

filled

Band gap- difference of energy between the bottom of

the conduction band and the top of the valence band

of the electrons in a crystalline solid.


3/28

Depending on band gap, a solid is classified as conductor,

semi conductor or insulator.


4/28

Semiconductors

Narrow bandgap (0.1eV-

1.0eV)

May beintrinsic or

extrinsic


5/28

Intrinsic Semiconductors

Electrical properties based on electronic

structure inherent to the pure material.

Example: Silicon and Germanium

Combination of Group IIIA and VA ( gallium

arsenide) or combination of Group 11B and

VIA(cadmium sulfide)

Conductivity due to electrons and holes

Equal number of electrons and holes


6/28

Intrinsic Semiconductors


7/28

Intrinsic Conductivity

3

3

= +

= ( + ) since

number of electrons/mnumber of holes/m

e h

e h

e h

ne pe

ne n p

n p


8/28

Example

• Determine the conductivity of intrinsic silicon

at 300K given that the electron and hole

mobilities are 0.135m2 /Vs and 0.048m2 /Vs,

respectively and that the concentration of

conducting electrons is 1 x1016 /m3 .


9/28

Electron concentration in the

conduction band

• Assumptions:

1. Electrons in the conduction band behave as

free particles with an effective massm

The number of conduction electrons per cubic

meter whose energies lie between E and

E+dE is


10/28

3/2 1/23

( ) ( )

where ( ) is the density of states at the bottom

of conduction band

4( ) (2 ) ( )

=energy at the bottom of conduction band

C

C

dn N E f E dE

N E

N E m E Eh

E


11/28

3/2 1/2

3

3/2 1/23

3/2 1/2

3

4(2 ) [( ) /{1 exp( )/ }]

For and , 1 in the denominator may be

neglected.

4 (2 ) [( ) /{exp( )/ }]

4(2 ) [( ) exp( )/ ]

C

C

C

c C F B E

C F B

c C F B E

C F B E

n m E E E E k T dE

h E E E E k T

n m E E E E k T dEh

m E E E E k T dEh

3/2 1/2

3

4(2 ) exp( ) [( ) exp( )/ ]

C F c C C B

Em E E E E E E k T dE

h


12/28

3/2 1/2 1/2

3 0

3/2 3/2

3

2 3/2

Let ( )/

4

(2 ) exp( )/ ( )

4(2 ) ( ) [exp( )/ ]

2

2[2 / ] [exp( )/ ]

C B B

x

F C B B B

B F C B

B F C B

E E k T x k Tdx dE

n m E E k T x k T e k Tdx h

n m k T E E k T h

n mk T h E E k T


13/28

Hole Concentration in the Valence

Band

1

3/2 1/2

3

( ) ( ) 1 [1 exp( )/ ]

1 [1 exp( )/ ]exp( )/

4

( ) [2 ] [ ]

energy at top of valence band

F B

F B

F B

h v

v

f hole f E E E k T

E E k T E E k T

N E m E Eh

E


14/28

3/2

2

( )(1 ( ))Following the same procedure:

22 exp( )/

v E

h Bv B

p N E f E dE

m k T p E E k T h


15/28

Intrinsic concentration of charge

carriers

3

3/2

2

3

3

3/2

2

24 ( ) exp( )/

exp( / )

2

where 4 ( )

is the energy gap or band gap

Bh v C B

g B

B

h

g

k T np mm E E k T

h

np AT E k T

k

A mm h

E


16/28

Energy Band Diagram and Fermi Level

for Intrinsic Semiconductor

• Show that the fermi energy is at the center of

the energy gap.

• Find the position of Fermi energy at room

temperature for germanium crystal having

5x1022 atoms/m3

•For an intrinsic semiconductor having band gap

of 0.7eV, calculate the hole density and

electron density at room temperature.


17/28

Extrinsic Semiconductor

• Doping- process of adding impurity to make

an insulator a conductor

Two types of extrinsic semiconductor

a) N type

b) P type


18/28

N- type

• Donor Impurity- has one more valence

electron than the atom it replaces

• Ex: Phosphorus atom in silicon

• There is one extra electron that will not

participate in the covalent bonding

• All donor levels will be fully activated and

donor atoms


19/28

N-type


20/28

3/2

2

2

2 exp( )/

density of donor atoms

exp( )/

exp( )/

ln

B

d F C B

D

C F C B

C F C B

D

C F C B

D

mk T

n N E E k T h

N

N E E k T

N E E k T

N

N E E k T N

ED

EF


21/28

P-type

• Introduction of acceptor impurity-one

electron less than the atom it replaces

• Introduces an incomplete bond resulting to a

hole in the v alence band


22/28

P-type


23/28

• Show that

ln V F V B A

N E E k T

N


24/28

HALL EFFECT

X x X x x X X

X x x x X X

X X X X q X X

X X x x x x x

X x x x x x x

B (in)

VD

I

H D V Ew v Bw

w

F

This voltage is known as Hall

Voltage.


25/28

Example

• A strip of aluminum of width =1.5cm and

thickness t= 250m is placed in a uniform

magnetic field of 0.55T oriented perpendicular

to the plane of the strip. When a current of25A is established in the strip, a voltage of

1.64V is measured across the width of the

strip. What is the density of charge carriers inaluminum and how many charge carriers are

provided by each atom?


26/28

Solution

29 3

3

3 3

2.10 10 carriers/m

number of atoms of aluminum/m

2.72 10 kg/m 26.98kg/mole

No. of charge carriers/atom =3.5

H D

A

jBw iBV v Bw

nq qnt

n x N

N M

x M


27/28

Hall Coefficient

specimen thickness

1R

H H

e H

R IBV t

t

ne


28/28

Example

• Some metal alloy is known to have electrical

conductivity and electron mobility values of

1.2x107 (m)-1 and 0.0050m2 /Vs,

respectively. Through a specimen of this alloythat is 35mm thick is passed a current of

40A. What magnetic field would you need to

be imposed to yield a Hall voltage of -3.5x107 V?

band theory of solidsnvnvb

Documents