lecture 2_sp_15!04!2015 [compatibility mode]

Si crystal (1022 atoms/cm3)

filled

empty

energy

conduction band

Energy gap, Eg=1.1 eV

energy

Recap

filledValence band

4 x 1022 states/cm3

1

Forbidden band

band gap, E

allowed band

Actual band structure calculated by quantum mechanics

Recap

band gap, EG

allowed band

2

Quantitative discussion

Determine the relation between energy of electron(E), wave number (k)

Relation of E and k for free electron

22

(x,t)= exp ( j(kx-t))

E

Recap

m

kE2

22h

=

Continuous value of E

K-space diagram

k

E

3

E-k diagram for electron in quantum well

E

En=3m

kE

nLm

E

2

222

222

h

h

=

=

pi

nL

k

=

pi

E

Recap

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

x=Lx=0

En=1

En=2

k/L 2/L

4

The Kronig-Penney Model

+ + + +

r

erV

0

2

4)(

pi

=

Periodic potential

Recap

V0

I II I I III II II II

Potential

welltunneling

Periodic potential

Wave function overlap

-b a

L

Determine a relationship between k, E and V0 5

Schrodinger equation (E < V0)

Region I 0)()( 222

=+

x

x

xI

I

Region II 0)()( 222

=

xx

xII

II

22 2

h

mE=

202 )(2h

EVm =

Recap

Potential periodically changes

)()( LxVxV +=jkxexUx )()( =

)()( LxUxU +=

Wave function

amplitude

k; wave number [m-1]

Phase of the wave

Bloch theorem

6

Boundary condition

)()()0()0(bUaU

UU

III

III

=

=Continuous wave function

Recap

)()()0()0(

''

''

bUaUUU

III

III

=

=

Continuous first derivative

7

From Schrodinger equation, Bloch theorem and boundary condition

)cos()cos()cosh()sin()sinh(2

22

kLabab =+

B 0, V0 Approximation for graphic solution

Recap

)cos()cos()sin(20 kaaaabamV

=+

h

)cos()cos()sin(' kaaa

aP =+

20'

h

bamVP =

Gives relation between k, E(from ) and V0

8

)cos()sin()( ' aa

aPaf

+=

Left side

Recap

)cos()( kaaf =Right side

Value must be

between -1 and 1

Allowed value of a9

mE

mE

2

2

22

22

h

h

=

=

Recap

Plot E-k

Discontinuity of E

10

Right side

Shift 2

)2cos()2cos()cos()( pipi nkankakaaf =+==Recap

Shift 2

11

Allowed energy band

Forbidden energy band

From the Kronig-Penney Model (1 dimensional periodic potential function)

Recap

Allowed energy band

Allowed energy band



First Brillouin zone 12

energy

conduction band-

Electrical condition in solids

1. Energy band and the bond model

Valence band

Energy gap, Eg=1.1 eV

+

Breaking of covalent bond

Generation of positive and negative charge

13

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

14

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

=

=n

iieJ

1

Drift current density, [A/cm3]

n; no. of electron per unit volume in the conduction band 15

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 force16

From relation of E and k

mdkEd

m

kE

2

2

2

22

2h

h

=

=

Mass of electron, mMass of electron, m

=

2

2

2

dkEd

mh

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*

17

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

18

4. Concept of hole

Electron fills the empty state

Positive charge empty the state

Hole

19

Microelectronics I : Introduction to the Quantum Theory of Solids

When electric field is applied,

hole

electron

I

Hole moves in same direction as an applied field

20

Metals, Insulators and semiconductor

Conductivity,

( s/m)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

band21

2. Metal

e

full

Partially fillede

No energy gap

Many electron for

conduction

e

3. Semiconductor

e

Almost full

Almost emptyConduction

band

Valence

band

Eg; on the order of 1 eV

Conduction band; electron

Valence band; hole

T> 0K

22

Extension to three dimensions

[110]

1 dimensional model (kronig-Penney Model)

1 potential pattern

[100]

direction

[110]

directionDifferent direction

Different potential patterns

E-k diagram is given by a function of the direction in the crystal23

E-k diagram of Si

Energy gap; Conduction band minimum

valence band maximum

Eg= 1 eV

Indirect bandgap; Indirect bandgap;

Maximum valence band and minimum

conduction band do not occur at the same k

Not suitable for optical device application

(laser)

24


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)

25

How do electrons and holes populate the bands?

Probability of Occupation (Fermi Function) Concept

Now that we know the number of available states at each energy,

then how do the electrons occupy these states?

We need to know how the electrons are distributed in energy.

Again, Quantum Mechanics tells us that the electrons follow the

Fermi-distribution function.Fermi-distribution function.Ef Fermi energy (average energy in the crystal)k Boltzmann constant (k=8.61710-5eV/K)T Temperature in Kelvin (K)

f(E) is the probability that a state at energy E is occupied. 1-f(E) is the probability that a state at energy E is unoccupied.

kTEE feEf /)(1

1)(

++++====

Fermi function applies only under equilibrium conditions, however, is

universal in the sense that it applies with all materials-insulators,

semiconductors, and metals. 27

Fermi-Dirac distribution: Consider T 0 K

For E > EF :

For E < EF :

0)(exp11)( F =++=> EEf

1)(exp11)( F =

+=< EEf

28

E

EF

0 1 f(E)


When E-EF>>kT

EEEf

FF

exp

1)(Maxwell-Boltzmann approximation

Boltzmann approximation

kT

Fexp Maxwell-Boltzmann approximation

Approximation is valid in this range

29

If E = EF then f(EF) =

If then

Thus the following approximation is valid:

i.e., most states at energies 3kT above EF are empty.

kTEE 3F + 1exp F >>

kTEE

=

kTEEEf )(exp)( F

Fermi-Dirac distribution: Consider T > 0 K

30

If then

Thus the following approximation is valid:

So, 1 f(E) = Probability that a state is empty, decays to zero.

So, most states will be filled.

kT (at 300 K) = 0.025eV, Eg(Si) = 1.1eV, so 3kT is very small in

comparison.

kTEE 3F 1expF


Probability of Occupation (Fermi function) Concept

If E Ef +3kT Consequently, above Ef +3kT the Fermi function or filled-state probability

decays exponentially to zero with increasing energy.

kTEEkTEE ff eEfe /)(/)( )(and1 31


Example

The probability that a state is filled at the

conduction band edge (E ) is precisely conduction band edge (Ec) is precisely equal to the probability that a state is

empty at the valence band edge (Ev). Where is the Fermi energy locate?

33

Solution

The Fermi function, f(E), specifies the probability of electron occupying states at a given energy E.The probability that a state is empty (not filled) at a given energy E is equal to 1- f(E).

(((( )))) (((( ))))VC EfEf ====1(((( )))) (((( ))))VC EfEf ====1(((( )))) (((( )))) kTEEC FCeEf /1

1++++

==== (((( )))) (((( )))) (((( )))) kTEEkTEEV VFFV eeEf // 11

1111

++++====

++++====

kTEE

kTEE FVFC

====

2VC

FEEE ++++====

34

The density of electrons (or holes) occupying

the states in energy between E and E + dE is:


Probability of Occupation Concept

dEEfEgc )()(Electrons/cm3 in the conduction band

between E and E + dE (if E E ).

0 Otherwise

dEEfEgc )()(between E and E + dE (if E Ec).

Holes/cm3 in the conduction band

between E and E + dE (if E Ev).dEEfEgv )()(

35

Current flow in semiconductor Number of carriers (electron @ hole)

How to count number of carriers,n?

If we know

1. No. of energy states Density of states (DOS)

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 function36

Density of states (DOS)

EhmEg 3

2/3)2(4)( pi=

A function of energy

As energy decreases available quantum states decreases

Derivation; refer text book

37

Solution

Calculate the density of states (for electron) per unit volume with energies between 0 and 1 eV

Q.

12/3

1

0

)2(4

)(

m

dEEgN

eV

eV

=

pi

321

2/319334

2/331

1

03

2/3

/105.4

)106.1(32

)10625.6()1011.92(4

)2(4

cmstates

dEEhm

eV

=

=

=

pi

pi

38


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 39

CEEhmEg = 3

2/3)2(4)( pi

Energy

Ec

40

1. Hole as carrier

Empty

statee

eConduction

band

Valence

band

Ec

Ev

freely freely

moves

hole in valence band contribute to conduction

Determine the DOS in the valence band

41

EEhmEg v = 3

2/3)2(4)( pi

Energy

Ev

42

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

mgkTEc

Cnpi

Mn; mass of electron

319

2/319334

2/331

2/33

2/3

3

1012.2

)106.10259.0(32

)10625.6()1011.908.12(4

)(32)2(4

=

=

=

cm

kThm

h

n

EcC

pi

pi

Mn; mass of electron

43

Q2;

Determine the total number of energy states in Si between Ev and Ev-kT at T=300K

Solution;

3

2/3)2(4= dEEEh

mg

Ev

v

ppiMp; mass of hole

318

2/319334

2/331

2/33

2/3

3

1092.7

)106.10259.0(32

)10625.6()1011.956.02(4

)(32)2(4

=

=

=

cm

kThm

h

p

kTEvv

pi

pi

Mp; mass of hole

44


Probability of Occupation Concept

45

Metals vs. Semiconductors

Allowed electronic-energy states g(E)

The Fermi level Ef is at an intermediate energy between that of the conduction band edge and that

of the valence band edge.

Fermi level Ef immersed in the continuum of allowed states.

Ef

Ef

Metal Semiconductor

46

Typical band structures of Semiconductor

E

Ec+

CB

E g(E) (EEc)1/2E

Forelectrons

[1f(E)]E

n E ( E )

A re a

Ec

ndEEn E ======== )(


Ev

Ec

0

EF

VB

g(E)fE)

EF

For holes

Energy band

diagram Density of states

Fermi-Dirac

probability

function

probability of

occupancy of a

state

n E ( E ) o r p E ( E )

p E ( E )

A re a = p

Ec

Ev

g(E) X f(E)Energy density of electrons in the CB

number of electrons per unit energy

per unit volume, The area under

nE(E) vs. E is the electron concentration.

number of states

per unit energy per

unit volume

47

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(11

+

=

eVEEf

CC

2. At Ec+kT

)3.0(0259.011

+

+

=

eVEEf

CC

61032.90259.0

3.01

1

)3.0(1

=

+

=

+

kTeVEE CC

61043.30259.03259.01

1

)3.0(0259.01

=

+

=

++

kTeVEE CC

Electron needs higher energy to be at higher energy states. The probability of

electron at Ec+kT lower than at Ec

48

+

=

kTEE

EfF

F

exp1

1)(electron

Hole?

The probability that states are being empty is given by

+

=

kTEE

EfF

F

exp1

11)(1

49

conduction band

energy gap

unfilled

Valence band

Filled

50

lecture 2_sp_15!04!2015 [compatibility mode]

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