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)

electronenergy

Quantized energy level (quantum state)

1s

2s2p

3s3p

+ n=1

n=2

n=3

Crystal (~1020 atom)

electronenergy + + …. = ?

x 1020

1s

2s2p

3s3p

1s

2s2p

3s3p

1s

2s2p

3s3p


Si Crystal

Tetrahedral structureDiamond 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

yx

Distance from center

Pro

babi

lity

dens

ity

yx 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 atomsTotal 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

Pro

babi

lity

dens

ity

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

3s3p

energy

Sp3 hybrid orbital

Reform 4 equivalent states � 4 equivalent bond (symmetric)


Si SiSi

SiSi

energy

+ + + +

energy

filled

empty


Si crystal (1022 atoms/cm3)

filled

empty

energy

conduction band


energy

filledValence 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

kE

2

22h

=

Continuous value of E

K-space diagram

k

E


E-k diagram for electron in quantum well

En=3m

kE

nLm

E

2

2

22

2

22

h

h

=

=

π

nL

k

=

π

E

E-k diagram for electron in crystal? The Kronig-Penney Model

x=Lx=0

En=1

En=2

kπ/L 2π/L


The Kronig-Penney Model

+ + + +r

erV

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

xI

I ϕαϕ

Region II 0)()( 2

2

2

=−∂

∂x

x

xII

II ϕβϕ

2

2 2

h

mE=α

2

02 )(2

h

EVm −=β

�Potential periodically changes

)()( LxVxV +=

jkxexUx )()( =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 =⋅+⋅−

αβαβαβ

αβ

B � 0, V0 � ∞ Approximation for graphic solution

)cos()cos()sin(

2

0 kaaa

abamV=+

α

α

α

h

)cos()cos()sin(' kaa

a

aP =+ α

α

α2

0'

h

bamVP =

Gives relation between k, E(from α) and V0


)cos()sin(

)( ' aa

aPaf α

α

αα +=

Left side

)cos()( kaaf =α

Right side

Value must be between -1 and 1

Allowed value of αa


mE

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

iieJ

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

kE

2

2

2

22

2

h

h

=

=

Mass of electron, mMass of electron, m

=

2

2

2

dk

Edm

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

EFree 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

10310-8

Conductivity; no of charged particle (electron @ hole)

1. Insulatorcarrier

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 fillede

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, Eg2. Effective mass, m*

Q. 1;

Eg=1.42 eV

Calculate the wavelength and Calculate 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

AB

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)

Eh

mEg

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

dEEh

meV

×=

××

××=

=

−

−

−

∫

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> 0KConduction 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


CEEh

mEg −=

3

2/3)2(4)(

π

Energy

Ec


1. Hole as carrier

Empty state

e

eConduction band

Valence band

Ec

Ev

freely freely moves

hole in valence band contribute to conduction

Determine the DOS in the valence band


EEh

mEg 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+

−= ∫ dEEEh

mg

kTEc

Cnπ

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

kTh

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

mg

Ev

v

pπMp; 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

kTh

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

−=

EEEfF

1)(

−+

=

kT

EEEf

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

fCC

2. At Ec+kT

)3.0(0259.01

1

−−+

+

=eVEE

fCC

61032.9

0259.0

3.01

1

)3.0(1

−×=

+

=

−−+

kT

eVEE CC

61043.3

0259.0

3259.01

1

)3.0(0259.01

−×=

+

=

−−++

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

EEEf

F

F

exp1

1)( electron

Hole?

The probability that states are being empty is given by

−+

−=−

kT

EEEf

F

F

exp1

11)(1


Approximation when calculating fF

−+

=

kT

EEEf

F

F

exp1

1)(

When E-EF>>kT

−

≈EE

EfF

F

exp

1)(

Maxwell-Boltzmann approximation

kT

Fexp Maxwell-Boltzmann approximation

Approximation is valid in this range

Chapter3 introduction to the quantum theory of solids

Engineering

Transcript of Chapter3 introduction to the quantum theory of solids