Chapter3 introduction to the quantum theory of solids

Microelectronics I
Chapter 3: Introduction to the
Quantum Theory of Solids

Microelectronics I : Introduction to the Quantum Theory of Solids
Chapter 3 (part 1)
1. Formation of allowed and forbidden energy band
k-space diagram
(Energy-wave number diagram)
Qualitative and quantitative discussion
Kronig-Penney model
(Energy-wave number diagram)
2. Electrical conduction in solids
Drift current, electron effective mass, concept of hole
Energy band model

Isolated single atom (ex; Si)
electron
energy
Quantized energy level
(quantum state)
1s
2s
2p
3s
3p
+ n=1
n=2
n=3
Crystal (~1020 atom)
electron
energy + + …. = ?
x 1020
1s
2s
2p
3s
3p
1s
2s
2p
3s
3p
1s
2s
2p
3s
3p

Si Crystal
Tetrahedral structure
Diamond structure
Tetrahedral structure
energy
Valence band
conduction band
Energy gap, Eg=1.1 eV
Formation of energy band and energy gap

What happen if 2 identical atoms approach each other ?
r
atom 2atom 1
energy
1s
Isolated atom
z
x x
1s
z
y
x
Distance from center
Probabilitydensity
y
x x
1s 1s
Wave function of two atom electron overlap
interaction

r
atom 2atom 1
When the atoms are far apart
(r=∞), electron from different
atoms can occupy same
energy level.
E1s,atom 1 =E1s, atom 2
As the atoms approach each
other, energy level splits
energy
1s
other, energy level splits
E1s,atom 1 ≠E1s, atom 2
ra
energy
interaction between two overlap wave function
Consistent with Pauli exclusion principle
a ; equilibrium interatomic distance

Regular periodic arrangement of atom (crystal)
ex: 1020 atoms
Total number of quantum states
do not change when forming a
system (crystal)energy
1s
1020 energy levels
a
energy
“energy band”dense allowed energy levels

energy
1020 energy state
1 eV
Consider
1020 energy state
Energy states are equidistant
Energy states are separated by 1/1020 eV = 10-20 eV
(Almost) continuous energy states within energy band

Distance from center
Probabilitydensity
energy
2s
1s
atom 2atom 1
1s
2s
r
atom 2atom 1
energy
1s
a
2s
“there is no energy level”
forbidden band →
energy gap, Eg
As the atoms are brought together,
electron from 2s will interact. Then electron
from 1s.

Si: 1s(2), 2s(2), 2p(6), 3s(2), 3p (2) 14 electrons
Ex;
Tightly bound to
nucleus
Involved in
chemical reactions
energyenergy
3s
3p
energy
Sp3 hybrid orbital
Reform 4 equivalent states 4 equivalent bond (symmetric)

Si Si
Si
Si
Si
energy
+ + + +
energy
filled
empty

Si crystal (1022 atoms/cm3)
filled
empty
energy
conduction band
energy
filled
Valence band
4 x 1022 states/cm3

Forbidden band
→band gap, E
allowed band
Actual band structure “calculated by quantum mechanics”
→band gap, EG
allowed band

Quantitative discussion
Determine the relation between energy of electron(E), wave number (k)
Relation of E and k for free electron
22
Ψ(x,t)= exp ( j(kx-ωt))
E
m
k
E
2
22
h
=
Continuous value of E
K-space diagram
k
E

E-k diagram for electron in quantum well
En=3
m
k
E
n
Lm
E
2
2
22
2
22
h
h
=






=
π
n
L
k 





=
π
E
E-k diagram for electron in crystal? The Kronig-Penney Model
x=Lx=0
En=1
En=2
k
π/L 2π/L

The Kronig-Penney Model
+ + + +
r
e
rV
0
2
4
)(
πε
−
=
Periodic potential
V0
I II I I III II II II
Potential
well
tunneling
Periodic potential
Wave function overlap
-b a
L
Determine a relationship between k, E and V0

Schrodinger equation (E < V0)
Region I 0)(
)( 2
2
2
=+
∂
∂
x
x
x
I
I
ϕα
ϕ
Region II 0)(
)( 2
2
2
=−
∂
∂
x
x
x
II
II
ϕβ
ϕ
2
2 2
h
mE
=α
2
02 )(2
h
EVm −
=β
Potential periodically changes
)()( LxVxV +=
jkx
exUx )()( =ϕ
)()( LxUxU +=
Wave function
amplitude
k; wave number [m-1]
Phase of the wave
Bloch theorem

Boundary condition
)()(
)0()0(
bUaU
UU
III
III
−=
= Continuous wave function
)()(
)0()0(
''
''
bUaU
UU
III
III
−=
=
Continuous first derivative

From Schrodinger equation, Bloch theorem and boundary condition
)cos()cos()cosh()sin()sinh(
2
22
kLabab =⋅+⋅
−
αβαβ
αβ
αβ
B 0, V0 ∞ Approximation for graphic solution
)cos()cos(
)sin(
2
0
kaa
a
abamV
=+





α
α
α
h
)cos()cos(
)sin('
kaa
a
a
P =+ α
α
α
2
0'
h
bamV
P =
Gives relation between k, E(from α) and V0

)cos(
)sin(
)( '
a
a
a
Paf α
α
α
α +=
Left side
)cos()( kaaf =α
Right side
Value must be
between -1 and 1
Allowed value of αa

m
E
mE
2
2
22
2
2
h
h
α
α
=
=
Plot E-k
Discontinuity of E

)2cos()2cos()cos()( ππα nkankakaaf ==+==
Right side
Shift 2πShift 2π

Allowed energy band
Forbidden energy band
From the Kronig-Penney Model (1 dimensional periodic potential function)
Allowed energy band
Allowed energy band
First Brillouin zone

energy
conduction band
-
Electrical condition in solids
1. Energy band and the bond model
Valence band
+
Breaking of covalent bond
Generation of positive and negative charge

E versus k energy band
conduction band
T = 0 K T > 0 K
When no external force is applied, electron and “empty state” distributions are
symmetrical with k
Valence band

2. Drift current
Current; diffusion current and drift current
When Electric field is applied
E E
dE = F dx = F v dt
“Electron moves to higher empty state”
k k
ENo external force
∑=
υ−=
n
i
ieJ
1
Drift current density, [A/cm3]
n; no. of electron per unit volume in the conduction band

3. Electron effective mass
Fext + Fint = ma
Electron moves differently in the free space and in the crystal (periodical potential)
External forces
(e.g; Electrical field)
Internal forces
(e.g; potential)+ = mass acceleration
Internal forces
Fext = m*a
External forces
(e.g; Electrical field)
Internal forces
(e.g; potential)
= Effective mass acceleration
Effect of internal force

From relation of E and k
mdk
Ed
m
k
E
2
2
2
22
2
h
h
=
=
Mass of electron, mMass of electron, m






=
2
2
2
dk
Ed
m
h
Curvature of E versus k curve
E versus k curve Considering effect of internal force (periodic potential)
m from eq. above is effective mass, m*

E versus k curve
E
Free electron
Electron in crystal A
Electron in crystal B
k
Curvature of E-k depends on the medium that electron moves in
Effective mass changes
m*A m*Bm> >
Ex; m*Si=0.916m0, m*GaAs=0.065m0 m0; in free space

4. Concept of hole
Electron fills the empty state
Positive charge empty the state
“Hole”

When electric field is applied,
hole
electron
I
Hole moves in same direction as an applied field

Metals, Insulators and semiconductor
Conductivity,
σ (S/cm)
MetalSemiconductorInsulator
103
10-8
Conductivity; no of charged particle (electron @ hole)
1. Insulator
carrier
1. Insulator
e
Big energy gap, Eg
empty
full
No charged particle can contribute to
a drift current
Eg; 3.5-6 eV
Conduction
band
Valence
band

2. Metal
e
full
Partially filled
e
No energy gap
Many electron for
conduction
e
3. Semiconductor
e
Almost full
Almost empty
Conduction
band
Valence
band
Eg; on the order of 1 eV
Conduction band; electron
Valence band; hole
T> 0K

from E-k curve , 1. Energy gap, Eg
2. Effective mass, m*
Q. 1;
Eg=1.42 eV
Calculate the wavelength andCalculate the wavelength and
energy of photon released when
electron move from conduction band
to valence band? What is the color
of the light?

Q. 2;
E (eV)
k(Å-1)
0.1
0.7
0.07
A
B
Effective mass of the two electrons?

Extension to three dimensions
[110]
1 dimensional model (kronig-Penney Model)
1 potential pattern
[100]
direction
[110]
direction
Different direction
Different potential patterns
E-k diagram is given by a function of the direction in the crystal

E-k diagram of Si
Energy gap; Conduction band minimum –
valence band maximum
Eg= 1 eV
Indirect bandgap;
Maximum valence band and minimum
conduction band do not occur at the same k
Not suitable for optical device application
(laser)

E-k diagram of GaAs
Eg= 1.4 eV
Direct band gap
suitable for optical device application
(laser)(laser)
Smaller effective mass than Si.
(curvature of the curve)

Current flow in semiconductor ∝ Number of carriers (electron @ hole)
How to count number of carriers,n?
If we know
1. No. of energy states
Assumption; Pauli exclusion principle
1. No. of energy states
2. Occupied energy states
Density of states (DOS)
The probability that energy states is
occupied
“Fermi-Dirac distribution function”
n = DOS x “Fermi-Dirac distribution function”

Density of states (DOS)
E
h
m
Eg 3
2/3
)2(4
)(
π
=
A function of energy
As energy decreases available quantum states decreases
Derivation; refer text book

Solution
Calculate the density of states per unit volume with energies between 0 and 1 eV
Q.
12/3
1
0
)2(4
)(
m
dEEgN
eV
eV
= ∫
π
321
2/319
334
2/331
1
0
3
2/3
/105.4
)106.1(
3
2
)10625.6(
)1011.92(4
)2(4
cmstates
dEE
h
m
eV
×=
×
×
××
=
=
−
−
−
∫
π
π

Extension to semiconductor
Our concern; no of carrier that contribute to conduction (flow of current)
Free electron or hole
1. Electron as carrier
e
T> 0K
Conduction
band
Can freely moves
e
e band
Valence
band
Ec
Ev
Electron in conduction band contribute to conduction
Determine the DOS in the conduction band

CEE
h
m
Eg −= 3
2/3
)2(4
)(
π
Energy
Ec

1. Hole as carrier
Empty
state
e
e
Conduction
band
Valence
band
Ec
Ev
freelyfreely
moves
hole in valence band contribute to conduction
Determine the DOS in the valence band

EE
h
m
Eg v −= 3
2/3
)2(4
)(
π
Energy
Ev

Q1;
Determine the total number of energy states in Si between Ec and Ec+kT at
T=300K
Solution;
3
2/3
)2(4
+
−= ∫ dEEE
h
m
g
kTEc
C
nπ
Mn; mass of electron
319
2/319
334
2/331
2/3
3
2/3
3
1012.2
)106.10259.0(
3
2
)10625.6(
)1011.908.12(4
)(
3
2)2(4
−
−
−
−
×=
××





×
×××
=






=
∫
cm
kT
h
m
h
n
Ec
C
π
π
Mn; mass of electron

Q2;
Determine the total number of energy states in Si between Ev and Ev-kT at
T=300K
Solution;
3
2/3
)2(4
−= ∫ dEEE
h
m
g
Ev
v
pπ
Mp; mass of hole
318
2/319
334
2/331
2/3
3
2/3
3
1092.7
)106.10259.0(
3
2
)10625.6(
)1011.956.02(4
)(
3
2)2(4
−
−
−
−
−
×=
××





×
×××
=






=
∫
cm
kT
h
m
h
p
kTEv
v
π
π
Mp; mass of hole

The probability that energy states is occupied
“Fermi-Dirac distribution function”
Statistical behavior of a large number of electrons
Distribution function
 −
=
EE
EfF
1
)(





 −
+
=
kT
EE
Ef
F
F
exp1
)(
EF; Fermi energy
Fermi energy;
Energy of the highest occupied quantum state

For temperature above 0 K, some electrons jump to higher energy level.
So some energy states above EF will be occupied by electrons and some
energy states below EF will be empty

Q;
Assume that EF is 0.30 eV below Ec. Determine the probability of a states being
occupied by an electron at Ec and at Ec+kT (T=300K)
Solution;
1. At Ec
)3.0(
1
1



 −−
+
=
eVEE
f
CC
2. At Ec+kT
)3.0(0259.0
1
1



 −−+
+
=
eVEE
f
CC
6
1032.9
0259.0
3.0
1
1
)3.0(
1
−
×=






+
=





 −−
+
kT
eVEE CC
6
1043.3
0259.0
3259.0
1
1
)3.0(0259.0
1
−
×=






+
=





 −−+
+
kT
eVEE CC
Electron needs higher energy to be at higher energy states. The probability
of electron at Ec+kT lower than at Ec






 −
+
=
kT
EE
Ef
F
F
exp1
1
)( electron
Hole?
The probability that states are being empty is given by





 −
+
−=−
kT
EE
Ef
F
F
exp1
1
1)(1

Approximation when calculating fF





 −
+
=
kT
EE
Ef
F
F
exp1
1
)(
When E-EF>>kT



 −
≈
EE
Ef
F
F
exp
1
)(
Maxwell-Boltzmann approximation





kT
F
exp Maxwell-Boltzmann approximation
Approximation is valid in this range

Chapter3 introduction to the quantum theory of solids

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