Engineering Physics

(203192102/203192151)
Irfan Malek, Assistant Professor,
ASH Department, PIET, PARUL UNIVERSITY.
 CHAPTER-1
Electronic Materials
Classification of solids based on band structure
LOWEST BAND WITH SIGNIFICANTLY UNOCCUPIED STATES
(CONDUCTION BAND)

BAND OCCUPIED BY OUTERMOST ELECTRONS

(VALENCE BAND)
How Resistivity changes with temperature?
Free electron theory (Drude Theory – 1900)

Ion core

Free electron
Postulates of Free electron theory

• Free electrons  molecules of a gas

Move freely in the space between ions

• Electrons obey Maxwell-Boltzmann statistics

• When no electric field  electrons move in random directions

Making elastic collisions with no loss of energy

• The interaction of free electrons with ion cores is negligible.

• The free electrons find uniform electric field of positive ions

Advantages of Free electron theory

It explains …….
• Electrical & thermal conductivity of metals

• Optical properties of metals

• Ohm’s law and Wiedemann-franz law (ratio of thermal to electrical

conductivity is proportional to temperature)
Drawbacks of Free electron theory

It fails to explain ………..

• Heat capacity of metals

• Superconductivity of metals

• Electrical conductivity of semiconductors or insulators

• Ferromagnetism

• Quantum phenomena like photo- electric effect, Compton effect,

black body radiation
Fermi energy level

The uppermost filled

energy level at 0 K

OR

The highest occupied

energy level at 0 K
 Derivation of Fermi energy
• The wave function of an electron
in a semiconductor material of
length L, width L and height L is
given by
1
 ( x, y , z ) 
ik y y
e ikx x e e ikz z
V
…………… (1)

Where, V is the volume of

semiconductor

V  L3
 Concept of Fermi energy
• Since electron is confined in the material and can not come out of
the material, we have following boundary conditions :

 ( x, y, z )  ( x  L, y, z ) …………… (2)

 ( x, y, z )  ( x, y  L, z ) …………… (3)

 ( x, y, z )  ( x, y, z  L) …………… (4)
Concept•ofFrom
Fermi
eq. (1)energy
and (2)
1 1
e ik z z  e ik x ( x  L ) e
ik y y ik y y
e ik x x e e ik z z
V V

 eik x L  1
 cos(k x L)  i sin(k x L)  1

 k x L  nx 2 Where, n x  0, 1, 2, 3,......... ...

2
 k x  nx …………… (5)
L

• From eq. (1) and (3)

2
 k y  ny …………… (6)
L
Concept of Fermi energy
• From eq. (1) and (4)

2
 k z  nz …………… (7)
L
• Now the wave vector in 3-
dimension is given by
 ^ ^ ^
K  kx i  ky j  kz k

 2 ^ 2 ^ 2 ^
K  nx i  ny j  nz k
L L L

• The points corresponding to

this K-vector are in K-space
• All these points are
Concept of Fermi energy
separated by
2
L
• As shown in the
diagram, volume
occupied by single
K-point is
 2  8 3
3

   3
 L  L
• Therefore, no. of K-
points per unit
volume of K-space
L3 V
 3
8 3
8
• Suppose there are N no. of electrons within volume V which occupy all energy
Concept of Fermi energy
states in K-space which are inside the sphere of radius K F at 0 K and all the
states outside the sphere are un occupied.
• The surface of this sphere will have states of same energy which is the
maximum possible energy of electron at 0 K, known as fermi energy. Hence this
surface is known as Fermi surface.
4
• Volume of fermi sphere is  KF3
3
• The total no. of K-points within
this sphere

= (volume of sphere) X (no. of K-

points per unit volume)
4 V
  KF3  3
3 8
• Concept of Fermi
Since every point energy
corresponds to two possible states (for spin up electron and
spin down electron) , The total no. of states within fermi sphere,

4 V
 2  K F  3
3

3 8
3
VK F

3 2
• And all these states are occupied, that means,

No. of states = No. of electrons

3
VK F
N
3 2
3
N K
  F2
V 3
 Concept
 n
K of Fermi energy
F
3

Where, n 
N
is the number density of electrons.
3 2 V

• Therefore, Fermi radius

1
K F  (3 2 n) 3
…………… (8)

• Therefore, Fermi energy

2
2KF
EF 
2m
2
2
 EF  (3 2 n) 3 …………… (9)
2m
Density of states (for Conductors)
• Important quantity which decides electronic properties
• Defined as number of available states per unit volume per unit energy range

• The relation for Fermi energy is given by,

2
2
EF  (3 2 n) 3
2m
Where n is the number of energy states available per unit volume per unit energy
• In general, above can also be written as,
2
2
E (3 2 n) 3 …………… (1)
2m
Concept
Where Eof Fermi
is energy energy
level below which number of electron states per unit
volume are n.

2
2mE
 (3 2 n) 
3
2

 2mE 
3
 3 2 n   2  2
  

1  2mE  2
3
 n  2  2  ……………… (2)
3   
• Concept oftheFermi
If we increase energy energy
from E to E + dE, then the number of available
states per unit volume will also change from n to n + dn

• That means dE range of energy contains dn states per unit volume.

• Therefore, the ratio

gives the number of
available states per unit
volume per unit energy
range. This is known as
density of states and is
denoted by g(E).
dn
 g (E)  ……. (3)
dE
Concept of Fermi energy
From eq. (2) and (3),

dn 1  2m  32 3 12
g (E)   2  2 
E
dE 3    2

1  2m  2 2
3 1
 g (E)  2  2 
E ……………… (4)
2   

• Equation (4) gives the relation for density of states g(E) as a

function of energy (E)
Energy dependence of Density of states

• This is the plot of density of states

as a function of energy. The density
of states increases in parabolic
manner as the energy increases.
 E-k diagrams
• Gives energy-momentum
relationship for an electron.
• Useful to study band structure.
• Gives an idea of band gap, whether
material has direct bandgap or
indirect bandgap.

• For a free electron, the relation

between energy E and wave vector k is
ℏ2 𝑘 2
𝐸=
2𝑚
 Concept of Fermi energy

Direct lattice Reciprocal lattice

• Relationship between unit vectors of  2  2
reciprocal lattice and unit vectors of real b1  b2 
lattice can be given by a1 a2
• The wave function describing an electron in a periodic lattice at position
Concept
r is given by of Fermi energy
  ik r
 (r )  u (r ) e ………………… (1)

u (r ) is amplitude of wave associated with moving electron.
• The wave function describing an electron in a periodic lattice at position
r + R is given by
    ik r ik R
 (r  R)  u (r  R) e e
  
But due to periodic lattice, amplitude of wave is also periodic. i.e. u(r )  u(r  R)
   ik r ik R
Therefore,  (r  R)  u (r ) e e
   ik R
From eq. (1),  (r  R)  (r ) e ………………… (2)
• For every wave
Concept of Fermi
vector associated energy
with electron, we
can find a
reciprocal lattice
vector
corresponding to
the nearest point
such that we can
connect them by a
small vector given
by
  
k '  k K

𝑘 ′ 𝑖𝑠 𝑠𝑚𝑎𝑙𝑙 𝑣𝑒𝑐𝑡𝑜𝑟 𝑘 𝑖𝑠 𝑤𝑎𝑣𝑒 𝑣𝑒𝑐𝑡𝑜𝑟 𝐾 𝑖𝑠 𝑟𝑒𝑐𝑖𝑝𝑟𝑜𝑐𝑎𝑙 𝑙𝑎𝑡𝑡𝑖𝑐𝑒 𝑣𝑒𝑐𝑡𝑜𝑟

    i ( k '  K ) R
Concept  (energy
of Fermi
From eq. (2), r  R)   (r ) e
   ik ' R iK R
  (r  R)   (r ) e e
 
But, e iK R
1
   ik ' R
  (r  R)   (r ) e ………………… (3)

From eq. (2) and (3), we can say that 𝑘 and 𝑘 ′ are equivalent.
• Also from the diagram we can conclude that single 𝑘 ′ can represent number of 𝑘

• These 𝑘 ′ vectors are within Brillouin zone

• Thus in E-k diagram, on X-axis, we can consider 𝑘 ′ instead of 𝑘 and the entire
diagram can be reduced within the zone boundaries.
 
• Concept
Consider oneof Fermi energy
dimensional reciprocal lattice points on the k- axis
2
• Separation between any two points is
a  
 to 
• The Brillouin zone of point kept at origin will be from a a
 
• Thus infinite no. of k vectors are transformed into finite no. of k ' vectors
within the Brillouin zone boundaries   to  
a a
  
Transformation of k into k'
Concept of Fermi energy
    
k K k '  k K
 3 2  
to  to 
a a a a a
 3 2  
 to    to 
a a a a a

3 5 4  
to  to 
a a a a a
3 5 4  
 to    to 
a a a a a
   3
• Concept oftable
First row in the Fermi energy
means that the range of k can be shifted from a
to
a
  2
TO  to  by subtracting
a a a

  3
• Second row in the table means that the range of kcan be shifted from  to 
  a a
TO  to  by adding 2
a a
a
Concept of Fermi energy
Kronig-Penney Model

• Establishes the fact that band gap exists in semiconductors

• Describes motion of electron in one dimensional periodic potential

• The potentials of
adjacent atoms overlap
and gives rise to periodic
potential
 This periodic potential can be approximated to square well having two regions:
Concept of Fermi energy
• Region I (potential V = 0) of length a,
• Region II (potential V = V0 ) of length b,
• Periodicity is (a+b)
 Solving Schrodinger equation in both the regions, we get
Concept of Fermi energy
( 2   2 )
 sin(a) sin(b)  cos(a) cos(b)  cos k (a  b) ……. (1)
2
2m ( E  V0 )
where,  2  2mE
2
and 2 
 2
2mV0 E
Now we define,  0  and 
2 V0
Therefore,   0 
For E  V0 we get,    0   1
Substituting in eq. 1
1  2
sin( 0 a  ) sin( 0b   1)  cos( 0 a  ) cos( 0b   1)  cos k (a  b)
2  (  1)
……. (2)
 For E  V ofwe
Concept get, energy
Fermi
0  i 1   0

Substituting in eq. 1

1  2
sin( 0 a  ) sinh( 0b 1   )  cos( 0 a  ) cosh( 0b 1   )  cos k (a  b)
2  (  1)
……. (3)

• From eq. (2) and (3), we can conclude that only those values of energy  will be
allowed for which the value of L.H.S comes between -1 and +1 because R.H.S is
cos k ( a  b)
 2 2
a  b  a  b 
0

3 3
  a  b  a  b 
a  b  a  b 

Eq.(20)
Eq. 3 Eq. 2
Eq.(18)

 2 3
k k k
There is bandgap at: a  b  a  b  a  b 
 E

BAND GAP

BAND GAP

BAND GAP

3 2 1 -1 -2 -3
 2
AFTER SHIFTING BY
a  b

3 2   2 3  
   
a  b  a  b  a  b  a  b  a  b  a  b  a  b  a  b 
Formation of energy bands in solids
• Why bands are formed in solids?
• Qualitative explanation of how the band gap arises

• In single isolated atom,

electrons occupy discrete
energy levels as per their
energy.
•Concept of Fermi
Splitting of single energy energy
level into two when atoms containing
single electrons are very close
Concept of Fermi energy
• If there are N no. of such atoms very close to each other, then energy level n=1
will split into N levels.
• Similarly, if electrons are filled
Concept of Fermi energy
up to 3rd orbit in an atom and
when such atoms are brought
close to each other, then first
n=3 level will split because of
interaction of outermost
electrons, then if the
interatomic distance further
decreases, electrons of level
n=2 will start interaction and
thus n=2 level will split.

• If the interatomic distance

further decreases, then
electrons of level n=1 will
also start interacting and the
level n=1 will also split at very
small interatomic distance.
 Qualitative explanation of how bandgap arises in Si

• Si has total 14
electrons which
will be arranged as
shown in the
diagram

1s 2 2s 2 2 p 6 3s 2 3 p 2
• If N no. of Silicon atoms are brought very close to each other such that all the
Concept of Fermi energy
levels are split to form bands, then

1s band will contain 2N electrons

2s band will contain 2N electrons
2p band will contain 6N electrons
3s band will contain 2N electrons
3p band will contain 2N electrons
• Concept
As shown in of Fermi energy
the diagram,
3s band contains 2N states in which
2N electrons are filled up.
3p band contains 6N states in which
2N electrons are filled up. Hence 4N
states are empty.
• When 3s band and 3p band overlap
with each other,
2N states of 3p band combine with
2N states of 3s band.

• Therefore, now lower band

containing 4N states and upper
band containing 4N states are
separated by some bandgap.
Effective mass of electron

• Mass of electron under the influence of periodic potential in a

material is called effective mass

• When electron is moving through a crystal, it experiences internal forces due

to positive ions and other electrons present. Hence, the original mass of
electron seems to change and it is called effective mass of electron.

• Effective mass can be positive or it can be negative also

 Effective mass of electron
• For free electron,
2k 2
E
2m
dE  2 2k  2 k
  
dk 2m m

d 2E 2
 2

dk m

1 d 2E 1
 2  …………………. (1)
 dk 2 m
• Concept
The energyofof Fermi
electronenergy
near the
bottom of the conduction band can
be approximated by parabola
• E-k diagram for Conduction band
• Therefore we can write,

E  EC  c1 k 2 ………..…. (2)
Where EC is the energy at the
bottom of the conduction band

For k  0 , E  EC
For k  0 , E  EC

• Therefore c1 must be positive

• Concept
Differentiating
ofeq.(2)
Fermi energy

dE d 2E 1 d 2E 1
 2c1 k  2
 2c1  2 2
 2 2c1 ……….……. (3)
dk dk  dk 

• From eq.(1) and (3)

1 1 2
 2 2c1  m ………………. (4)
m  2c1

• Since c1 is positive, mass of electron is positive in conduction band.

• The energy ofmass
Effective electron
ofnear the top of in valence band
electron
the valence band can be approximated
by parabola
• E-k diagram for Valence band
• Therefore we can write,

EV  E  c2 k 2 ………..…. (5)

Where EV is the energy at the top of

the valence band

• For k  0 , E  EV

• For k  0 , E  EV

• Therefore c2 must be positive

 Concept of Fermi energy
• Differentiating eq.(5)

dE d 2E 1 d 2 E 1
  2c2 k   2  2c2  2 2
 2 2c2 …………. (6)
dk dk  dk 

• From eq.(1) and (6)

1 1 2
 2 2c2  m ……….…. (7)
m  2c2

• Since c2 is positive, mass of electron is negative in valence band.

Direct and Indirect band gap materials

• In real three dimensional

materials, the spacing between
atoms may not be same in all
directions

• This gives rise to different band

structure in different materials

• Therefore, along different

directions, the curvature of E-k
diagram will be different.
Direct band gap materials
• In such materials, the top of the
valence band and the bottom of the
conduction band occur at same value
of k
• In direct band gap materials, when
electron makes transition from
conduction band to valence band, the
value of k does not change. Hence.
The momentum of electron is
conserved.
• Direct band gap materials are quite
useful in optoelectronic devices like
LASER, LED, etc.
Indirect band gap materials
• In such materials, the top of the valence
band and the bottom of the conduction
band occur at different values of k
• In Indirect band gap materials, when
electron makes transition from
conduction band to valence band, the
value of k changes. Hence, The
momentum of electron also changes.
Therefore, momentum is not conserved.
• Therefore, electron must interact with
lattice in order to adjust its momentum
while going from one band to another
band.
Examples of direct and indirect band gap materials

• Examples of direct bandgap materials include amorphous

silicon and some III-V materials such as InAs, GaAs, etc

• Indirect bandgap materials include crystalline silicon and

Ge.

• Some III-V materials are indirect bandgap as well, for

example Alsb
Applications of direct and indirect band gap materials
• Transition of electrons from conduction band to valence band takes place
vertically in direct band gap semiconductors in which momentum of
electron is conserved and there is no need for electron to interact with the
lattice. Hence they possess high efficiency for optoelectronic devices.
• Transition of electrons from conduction band to valence band takes place
obliquely in indirect band gap semiconductors in which momentum of
electron is not conserved and there is a need for electron to interact with
the lattice. Hence they possess less efficiency for optoelectronic devices.
• For sources of light like LED or LASER, direct band gap semiconductors
must be used.
• For detectors of light like photo diodes, photo transistors , solar cells,
indirect band gap semiconductors can be used.
Density of states in semiconductors

• Density of states is a continuous function of energy in case of

conductors

• In case of semiconductors, it is not continuous as there is a band gap

• Variation of density of states as a function of energy is different in

valence band and in conduction band
Density of states in semiconductors
• Density of states for a free electron is given by
1  2m  32 12
g (E)    E ……………. (1)
2 2   2 

• In eq. (1), E is the energy of a free electron given by

2 2
E  k ……………. (2)
2m

• When electron is in conduction band, the equation of its energy is given by,
2 2
E  EC  k ……………. (3)
2m
• Applications
Therefore in eq. (1)of
wedirect
can replace E  EC
E byindirect
and band gap materials
1  2m  2
3
E  E C  2
1
 g C (E)  2  2  ……………. (4)
2   
• Eq. (4) gives density of states in conduction band.
• When electron is in valence band, the equation of its energy is given by,
2 2
EV  E  k ……………. (5)
2m
• Therefore in eq. (1) we can replace E by EV  E

1  2m  2
3
EV  E 2
1
 gV ( E )  2  2  ……………(6)
2   

• Eq. (6) gives density of states in valence band.

Energy dependence of density of states

1  2m  2
3

 E  EC 2
1
gC ( E )  
2 2   2 

1  2m  2
3
EV  E 2
1
gV ( E )  2  2 
2   
Occupation probability
• Occupation probability means the probability for a given state to be
occupied by electron.

It is given by,
N (E)
f F (E) 
g (E)
Where,

N(E) is the number of states occupied per unit volume per unit energy.
g(E) is the number of states available per unit volume per unit energy.
 Example of Occupation probability

• For example, as shown in the

diagram, N(E) = 12 and g(E) = 21

N (E)
 f F (E)   0.57
g (E)

• Thus, there are 57% chances that

a given state will be occupied by
the electron.
Fermi function (Occupation probability function)

• All the electrons in a semiconductor are indistinguishable particles

and they occupy different quantum states.

• Therefore the distribution of electrons in various energy states can be

determined from Fermi-Dirac distribution law which gives occupation
probability for electron states.

• Fermi-Dirac occupation probability is given by

1
f F (E) 
 E  EF 
1  exp  
 KT 
 Temperature dependence of Fermi function

• At 0 K temperature:

• For E < EF
1
f F (E)   1
1  exp  

• Thus, at 0 K temperature, all the

states below fermi level are 100%
occupied.
 Temperature dependence of Fermi function

• At 0 K temperature:

• For E > E F
1 1
f F (E)    0
1  exp   

• Thus, at 0 K temperature, all

the states above fermi level
are 100% empty.
 Temperature dependence of Fermi function

• At any temperature

When E = E F
1 1 1
f F (E)   
1  exp0 1  1 2

• Therefore, at any temperature,

there is 50% probability for the
fermi level to be occupied.
 Temperature
Concept dependence
of Fermi energy of Fermi function

• At T > 0 K :

• Due to thermal excitation, some

electrons will go to higher levels

• Thus, occupation probability will

decrease below fermi level and
increase above fermi level.
 Temperature dependence of Fermi function

• Here 𝑇1 < 𝑇2 < 𝑇3

• The diagram shows that as the

temperature increases more and
more, occupation probability
decreases further and further
below fermi level.
• Similarly, as the temperature
increases more and more,
occupation probability increases
further and further above fermi
level.
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