Density of states fermi energi.pdf

7/29/2019 Density of states fermi energi.pdf

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ECE 3080 - Dr. Alan DoolittleGeorgia Tech

Lecture 4

Density of States and Fermi Energy Concepts

Reading:

(Contd) Notes and Anderson2 sections 2.8-2.13


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How do electrons and holes populate the bands?

Density of States Concept

In lower level courses, we state that Quantum Mechanics tells us that the

number of available states in a cubic cm per unit of energy, the density of states,

is given by:

eVcm

StatesofNumber

unit

EEEEmm

Eg

EEEEmm

Eg

v

vpp

v

c

cnn

c

3

32

**

32

**

,)(2

)(

,)(2

)(


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Density of States Concept

Thus, the number of states per cubic centimeter between

energy E and E+dE is

otherwise

and

and

0

,EEif)dE(Eg

,EEif)dE(Eg

vv

cc

But where does it come from?


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Where Does the Density of States Concept come from?

Approach:

1. Find the smallest volume of k-space that can hold an electron. This will turn out tobe related to the largest volume of real space that can confine the electron.

2. Next assume that the average energy of the free electrons (free to move), the fermi

energy Ef, corresponds to a wave number kf. This value of kf, defines a volume in

k-space for which all the electrons must be within.

3. We then simply take the ratio of the total volume needed to account for the average

energy of the system to the smallest volume able to hold an electron ant that tells

us the number of electrons.

4. From this we can differentiate to get the density of states distribution.


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How do electrons and holes populate the bands?Derivation of Density of States Concept

First a needed tool: Consider an electron trapped in an energy well with infinite potential barriers.Recall that the reflection coefficient for infinite potential was 1 so the electron can not penetrate the

barrier.

After Neudeck and Pierret Figure 2.4a

2

222

n

22

2

2

2

2

2

22

22

2Eandsin)(

3...,21,nfora

nk0kaAsin0(a)

0B0)0(

:ConditionsBoundary

2or E

22kwhere

cossin)(:SolutionGeneral

0

02

2

man

axnAx

m

kmE

kxBkxAx

kx

Exm

EVm

nn


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What does it mean?

After Neudeck and Pierret Figure 2.4c,d,e

2

222

n2

Eandsin)(ma

n

a

xnAx nn

A standing wave results from

the requirement that there be a

node at the barrier edges (i.e.

BCs: (0)=(a)=0 ) . The

wavelength determines theenergy. Many different possible

states can be occupied by the

electron, each with different

energies and wavelengths.


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What does it mean?

After Neudeck and Pierret Figure 2.5

2

222

n2

Eandsin)(ma

n

a

xnAx nn

Recall, a free particle has E ~k2.

Instead of being continuous in k2, E is

discrete in n2! I.e. the energy values

(and thus, wavelengths/k) of a

confined electron are quantized (takeon only certain values). Note that as

the dimension of the energy well

increases, the spacing between

discrete energy levels (and discrete k

values) reduces. In the infinite

crystal, a continuum same as our free

particle solution is obtained.

Solution for much larger a. Note:

offset vertically for clarity.


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We can use this idea of a set of states in a confined space ( 1D well region) to derive thenumber of states in a given volume (volume of our crystal).

Consider the surfaces of a volume of semiconductor to be infinite potential barriers (i.e.

the electron can not leave the crystal). Thus, the electron is contained in a 3D box.

After Neudeck and Pierret Figure 4.1 and 4.2

m

kmE

kzyx

Exm

EVm

2or E

22kwhere

cz0b,y0a,x0...for

0

02

2

22

2

2

2

2

2

2

2

2

2

22

22


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Using separation of variables...

2222

2

2

22

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

2

0)(

)(

1,0

)(

)(

1,0

)(

)(

1

0)(

)(

1)(

)(

1)(

)(

1

get...we(2)bydividingand(1)into(2)Inserting

)()()(),,((2)

0(1)

zyx

zz

z

y

y

y

xx

x

z

z

y

y

x

x

zyx

kkkkwhere

k

z

z

z

k

y

y

y

k

x

x

x

kz

z

zy

y

yx

x

x

zyxzyx

kzyx

Since k is a constant for a given energy, each of the three terms on the left side must

individually be equal to a constant.

So this is just 3 equivalent 1D solutions which we have already done...


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Contd...

2

2

2

2

2

222

nzny,nx,

zz

y

yx

x

2Eand

3...,21,nfor,a

nk,

a

nk,

a

nkwhere

sinsinsin),,(

)()()(),,(

c

n

b

n

a

n

m

zkykxkAzyx

zyxzyx

zyx

zyx

zyx

0

a

b

c

kx

ky

kzEach solution (i.e. eachcombination of nx, ny, nz)

results in a volume of k-space. If we add up all

possible combinations, we

would have an infinite

solution. Thus, we will only

consider states contained in afermi-sphere (see next

page).


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Contd...

f

22

f Eenergyelectronaveragefor thevaluemomentumadefines2

E m

kf

Volume of a single state cube: V3

statesingle

Vcba

Volume of a fermi-sphere: 3

4V 3sphere-fermi

fk

A Fermi-Sphere is

defined by the

number of states in

k-space necessary to

hold all the electrons

needed to add up to

the average energyof the crystal

(known as the fermi

energy).

V is the physical

volume of thecrystal where as all

other volumes used

here refer to volume

in k-space. Note

that: Vsingle-state is thesmallest unit in k-

space. Vsingle-state is

required to hold a

single electron.


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Contd...k-space volume of a single state cube: V

3

statesingle

Vcba

k-space volume of a fermi-sphere: 3

4V 3sphere-fermi

fk

Number of filled states

in a fermi-sphere: 2

3

3

3

sin 3

4

3

4

2

1

2

1

2

12N

f

f

stategle

spherefermi kV

V

k

xxxV

V

Correction for

allowing 2 electrons

per state (+/- spin)

Correction for redundancy in

counting identical states resulting

from +/- nx, +/- ny, +/- nz.

Specifically, sin(-)=sin(+) so thestate would be the same. Same ascounting only the positive octant in

fermi-sphere.


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Contd...

Number of filled states

in a fermi-sphere:

31

2

f2

33

k3

N

V

NkV f

Efvaries in Si from 0 to ~1.1 eV as n varies from 0 to ~5e21cm-3

densityelectrontheisnwhere

2

3E

2

3

2

E3

222

f

3

2

22

22

fm

n

m

V

N

m

kf

23

222

32

33

N

mEVkV f

Thus,


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Contd...

E

V

mmEV

32

2

21

22

2mmG(E)

22

3

2

3

G(E)

VdE

dN

per volumeenergyperstatesof#G(E)

23

222

32

33

N

mEVkV f

Applying to the semiconductor we must recognize m m* andsince we have only considered kinetic energy (not the potential

energy) we have E E-Ec

cEE 32** 2mm

G(E)

Finally, we can define the density of states function:


15/27



Probability of Occupation (Fermi Function) Concept

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

)(~

,tan

1

1)(

)(

crystaltheinenergyaverageenergyFermiEand

KelvinineTemperaturTtconsBoltzmankwhere

e

Ef

F

kTEE F

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


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At T=0K, occupancy is digital: No occupation of states above EF and

complete occupation of states below EF

At T>0K, occupation probability is reduced with increasing energy.

f(E=EF) = 1/2 regardless of temperature.


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At T=0K, occupancy is digital: No occupation of states above EF and

complete occupation of states below EF

At T>0K, occupation probability is reduced with increasing energy.

f(E=EF) = 1/2 regardless of temperature.

At higher temperatures, higher energy states can be occupied, leaving more

lower energy states unoccupied (1-f(E)).


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0.00

0.20

0.40

0.60

0.80

1.00

1.20

0 0.2 0.4 0.6 0.8 1 1.2

f(E

)

E [eV]

3 kT3 kT

3 kT3 kT

+/-3 kT

Ef=0.55 eV



For E > (Ef+3kT):

f(E) ~ e-(E-Ef)/kT~0

For E < (Ef-3kT):

f(E) ~ 1-e-(E-Ef)/kT~1

T=10 K,

kT=0.00086 eV

T=300K,

kT=0.0259

T=450K,

kT=0.039

kTEE F

e

Ef)(

1

1)(


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But where did we get the fermi distribution function?Probability of Occupation (Fermi Function) Concept

After Neudeck and Pierret Figure 4.5

Consider a system of N total electrons spread between S allowable states. At each energy, Ei,we have Si available states with Ni of these Si states filled.

We assume the electrons are indistinguishable1.

1A simple test as to whether a particle

is indistinguishable (statistically

invariant) is when two particles are

interchanged, did the electronic

configuration change?

Constraints for electrons:

(1) Each allowed state can accommodate at most, only one electron (neglecting

spin for the moment)

(2) N=Ni=constant; the total number of electrons is fixed(3) Etotal=EiNi; the total system energy is fixed


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!N! iiii

iNS

SW

How many ways, Wi, can we arrange at each energy, Ei, Ni indistinguishable electrons intothe Si available states.

i ii

i

i

i

NS

SWW

!N! i

When we consider more than one level (i.e. all is) the number of ways we can arrange theelectrons increases as the product of the Wis .

If all possible distributions are equally likely, then the probability of obtaining a specific distribution is

proportional to the number of ways that distribution can be constructed (in statistics, this is the

distribution with the most (complexions). For example, interchanging the blue and red electron would

result in two different ways (complexions) of obtaining the same distribution. The most probable

distribution is the one that has the most variations that repeat that distribution. To find that maximum,

we want to maximize W with respect to Nis . Thus, we will find dW/dNi = 0. However, to eliminate the

factorials, we will first take the natural log of the above...

... then we will take d(ln[W])/dNi=0. Before we do that, we can use Stirlings Approximation to

eliminate the factorials.

i

i!Nln!ln!lnln iii NSSW

or

or

Ei

Ei

Ei

Ei+2

Ei+1

Ei


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Using Stirlings Approximation, ln(x!) ~ (xln(x) x) so that the above becomes,

i

ii

i

iii

i

i

NlnNlnlnln

terms,likeCollecting

NNlnNlnlnln

!Nln!ln!lnln

iiiiii

iiiiiiiii

iii

NSNSSSW

NSNSNSSSSW

NSSW

Now we can maximize W with respect to Nis . Note that since d(lnW)=dW/W when dW=0,

d(ln[W])=0.

i

i

i

i

i

i

1ln0ln

1Nln1lnln

1Nln1lnlnln

i

i

i

iii

ii

ii

dN

N

SWd

dNNSWd

NSN

W

dN

Wd

To find the maximum, we set the derivative equal to 0...

(4)

1ln

lnusedhaveweWhere x

dx

xxd


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From our original constraints, (2) and (3), we get...

ii

ii

0

0

iitotalii

ii

NdEENE

NdNN

Using the method of undetermined multipliers (Lagrange multiplier method) we multiply the

above constraints by constants and and add to equation 4 to get ...

i

i

0

0

ii

i

NdE

Nd

iallfor01lnthatrequireswhich

1ln0lni

i

i

i

ii

i

i

EN

S

dNENSWd


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This final relationship can be solved for the ratio of filled states, Ni per states available Si ...

kT

EE

i

ii

E

i

ii

i

i

i

Fi

i

eN

SEf

eN

SEf

EN

S

1

1)(

kT

1and

kT

E-

havewetors,semiconducofofcasethein

1

1)(

01ln

F


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Probability of Occupation

We now have the density of states describing the density of

available states versus energy and the probability of a state

being occupied or empty. Thus, the density of electrons (or

holes) occupying the states in energy between E and E+dE

is:

otherwise

and

and

0

,EEifdEf(E)]-(E)[1g

,EEifdEf(E)(E)g

vv

cc

Electrons/cm3

in the conductionband between Energy E and E+dE

Holes/cm3 in the valence band

between Energy E and E+dE


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Band Occupation


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Intrinsic Energy (or Intrinsic Level)

Equal

numbers ofelectrons and

holes

Efis said to equal

Ei(intrinsic

energy) when


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Additional Dopant States

Intrinsic:

Equal number

of electrons

and holes

n-type: more

electrons than

holes

p-type: moreholes than

electrons

Density of states fermi energi.pdf

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