Engineering Physics
Band Structure of Solid State Physics
Solids Srijit Bhattacharjee
❖ Free electron model: In solids loosely bound valence
electrons roam around the solid without being attached
to a particular nucleus.
❖ One can study such state as a free electron model.
❖ Three dimensional free electron gas. It amounts to solve
free electron in a three dimensional box.
❖ Three dimensional free electron F or single electron
~2 k 2
E=
~2 (kx2 + ky2 + kz2 ) 2m
E= k=
n⇡
; n = ±1, ±2, · · ·
2m l
ni ⇡
ki = ; i = x, y, z
li
K-space
s
8
W avef unction : kx ,ky ,kz = sin(kx x) sin(ky y) sin(kz z)
l x ly l z
⇡3 ⇡3
In K-space each single particle state occupies a volume v = l x l y lz = V
If there are N atoms in the sample and they produce q free electrons then
at 0K electrons occupy states up to radius KF (3-d cubical structure is approx-
imated as 3-d sphere)
Each volume v can be occupied by 2 electrons by opposite spins:
N q ⇡3 14 3
= ⇡kF
2 V 83
2
~2 2 ~2 3N q⇡
Fermi energy: EF = k
2m F = 2m ( V )2/3
Fermi energy defines a volume in momentum space whose surface—the Fermi
surface—separates occupied electronic states within the volume from empty ones
without it.
Potential Barrier
Potential:
V0 if 0 < x < a
V (x) =
0 otherwise
Solutions:
1 = Aeikx + Be ikx
, region I, x 0
k2 x k2 x
2 = Fe + Ge , region II, 0 < x < a
ikx
3 = Ce , region III, x a
Here A is incident wave in region I, B is transmitted wave from barrier at x = 0, C is transmitted
wave in region III (x > a). F and G are the wave transmitted and reflected at x = a respectively in
region II. p
2mE 2m(V0 E)
k2 = ; k2 = ,
~ 2 ~
and we are dealing with the case when E < V0 .
Boundary conditions:
0 0 0
1 , 1 , 2 , 2 and 3 , 3 are continuous at x = 0 and x = a. These will give rise four equations of 5
unknowns A, B, C, F, G. Solve these and write B, F, C, G in terms of A.
The transmission co-efficient is given by
2 3 1
⇤ k2 a k2 a 2
C C (e e ) 5
T = ⇤
= 41 + ⇣ ⇣ ⌘⌘
A A E
16 V0 1 E
V0
For large k2 a or for k2 a >> 1 one can write :
✓ ◆
E E
T w 16 1 e 2k2 a
V0 V0
There exists a finite probability for the particle to tunnel through the barrier!
The reflection co-efficient is defined as :
B⇤B
R=
A⇤ A
Check T + R = 1.
Nearly Free Electron Model
A one-dimensional linear periodic array of atoms with the electron potential energy due to the ion cores inside the crystal
Periodicity of lattice is a+b
Solution: (x) = uk (x)eikx with
uk (x = a + b) = uk (x)
Bloch’s theorem
Eq. 2 2mE
P ↵ =
↵a
sin ↵a + cos ↵a = cos ka ~2
Solution of this problem indicates only a band of energy values are allowed
Two terminal cases
1. P ! 0 i.e. the barrier is absent and the electron moves freely. In this
case it is easy to see
cos ↵a = cos ka
2 2 ~2 k 2
k =↵ =) E = 2m
2.P ! 1 i.e. the area of barrier very high.
The width of one energy band decreases as P increases.
This is the limit when ↵a decreases and tends to zero. In this case the
allowed region becomes infinitely narrow.
In this case sin ↵a = 0
=) ↵ = ±n⇡
n2 ⇡ 2 ~ 2
i.e. En = 2ma2
Energy Dispersion Relations
~2 2
E= k
2m
k
Brillouin Zone
Conductor, Semiconductor, Insulator
Density of States
Number of available states per differential energy-interval per unit volume of
the solid as a function of either the energy or momentum of the single-electron states in the solid
For 3-d free electron gas the number of states in a shell of thickness dk :
(1/8 ⇥ 4⇡k 2 dk)
2
⇡ 3 /V
V 2
= 2 k dk
⇡
k2
g(k)dk = 2 dk
⇡
Using energy dispersion relation we get:
p
2m3 1/2
g(E)dE = 2 3 E dE
⇡ ~
Effective mass of electron in a crystal
If an electron is moving in a crystal lattice, due to the presence of Coulomb
interactions it will have an e↵ective mass.
This mass is di↵erent from the mass of a free electron
dE
Group velocity of electron: vg = ~dk
If the electron is accelerated due to The presence of an electric field E then:
dE = eEvg dt
dvg 1 d2 E dk
We also have: dt =~ dk2 dt Acceleration (classical)= eE/m
dk/dt = eE/~
⇣ 2
⌘ 1
Comparing we get: m⇤ = ~2 d E
dk2
Semiconductors
Silicon crystal at T=0 K Silicon crystal at T> 0K
Intrinsic semiconductor
N-type semiconductor with donor impurity P-type semiconductor with acceptor impurity
The densities of thermally generated electrons and holes in semiconductors are generally very
small at room temperature.
For extrinsic semiconductors (doped), there will be plenty of free charge carriers
It is easier to break the covalent bond for a doped semiconductor
Band structure for intrinsic semiconductors
In a semiconductor, most of the energy bands will be basically totally
filled, while the higher energy bands are basically totally empty. We
usually concentrate on the highest filled valence band and lowest unfilled
conduction band to study the properties of semiconductors.
The discrete energy levels of Silicon atoms form energy bands
Band diagram in the presence of donors and acceptors
Density of states for valence and conduction bands
p
8⇡mn 2mn (E Ec )
Dc (E) = ,E Ec
h3
p
8⇡mp 2mp (Ev E)
Dv (E) = , E Ev
h3
Thermal Equilibrium and Fermi Distribution
Charge carriers follow Fermi-Dirac statistics to occupy states at thermal equilibrium
❖ At high Energy the FD distribution reduces to Maxwell-
Boltzmann form
(E EF )/kT
f (E) ⇡ e
❖ Hole concentration:
1 f (E)
Fermi level for n-type material. EF moves closer to EC.
Fermi level for a p-type material
Electron and Hole Concentrations
No. of electrons per unit volume (concentration) in conduction band
Z T op of conduction band⇡1
n= f (E)Dc (E)dE
Ec
= Nc e (Ec EF )/kT
✓ ◆3/2
2⇡mn kT
Nc = 2
h2
K=Boltzmann constant Nc =E↵ective density of states
Z Ev
p= (1 f (E))Dv (E)dE
bottom of valence band
(EF Ev )/kT
= Nv e
Fermi level of intrinsic semiconductor
For intrinsic semiconductors, when no dopant present n=p=ni (intrinsic carrier concntn.)
np = Nc Nv e(Ec Ev )/kT
= Nc Nv e Eg /kT
Electron and Hole concentrations in an weakly doped semiconductors are
equal.
The recombination rate becomes equal to the rate of thermal generation of electron–hole pairs
np = n2i
p
Eg /2kT
ni = Nc Nv e
Finding the Fermi level
Where is EF located in the energy band of silicon, at 300K with n = 1017cm–3?
Ec EF = kT. ln(Nc /n)
19 17
= .026 ln(2.8 ⇥ 10 /10 )
= 0.146eV
The EF is located 146 meV below the Ec .
Charge Neutrality Relationship for extrinsic semiconductor
n + Na = p + Nd
"✓ ◆2 #1/2
Nd Na Nd Na
n= + + n2i
2 2
And
"✓ ◆2 #1/2
Na Nd Na Nd
p= + + n2i
2 2
Equilibrium values
Heavily one sided doped scenario
Nd Na ⇡ Nd >> ni ; Na Nd ⇡ Na >> ni
These are applicable in most practical scenarios
If Nd >> Na , Nd >> ni
Majority carriers are electrons n ⇡ Nd
n2i
p⇡
Nd
If Na >> Nd , Na >> ni
Majority carriers are holes p ⇡ Na
n2i
p⇡
Na
Doped semiconductor where ni >> |Nd Na |
• Increasing temperature increases the number of intrinsic carriers
• All semiconductors become intrinsic at sufficiently high temperatures
n ⇡ p ⇡ ni
✓ ◆
Nd
EF Ei = kT ln
ni
for
Nd >> Na , Nd >> ni
