ece3080-l-4-density of states fermi...

ECE 3080 - Dr. Alan DoolittleGeorgia Tech

Lecture 4

Density of States and Fermi Energy Concepts

Reading:

(Cont’d) Notes and Anderson2 sections 2.8-2.13


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

vvpp

v

ccnn

c

3

32

**

32

**

,)(2

)(

,)(2

)(


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

Thus, the number of states per cubic centimeter between energy E’ and E’+dE is

otherwiseandand

0,EE’ if )dE(E’g,EE’ if )dE(E’g

vv

cc

But where does it come from?


How do electrons and holes populate the bands?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 to be 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 fermienergy 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.


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

22

2

2

22

22

2 Eand sin)(

3...,2 1,nfor a

nk 0kaAsin 0(a)

0 B 0)0( :ConditionsBoundary

2or E 22k where

cossin)( :Solution General

0

02

2

man

axnAx

mkmE

kxBkxAx

kx

Exm

EVm

nn



What does it mean?

After Neudeck and Pierret Figure 2.4c,d,e

2

222

n 2 Eand sin)(

man

axnAx nn

A standing wave results from the requirement that there be a node at the barrier edges (i.e. BC’s: (0)=(a)=0 ) . The wavelength determines the

energy. Many different possible “states” can be occupied by the

electron, each with different energies and wavelengths.



What does it mean?

After Neudeck and Pierret Figure 2.5

2

222

n 2 Eand sin)(

man

axnAx 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 (take on 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.



We can use this idea of a set of states in a confined space ( 1D well region) to derive the number 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

mkmE

kzyx

Exm

EVm

2or E 22k where

cz0 b,y0 a,x0...for

0

02

2

22

2

22

2

2

2

2

2

2

22

22



Using separation of variables...

2222

22

22

2

22

2

2

22

2

2

2

2

2

22

2

2

2

2

2

0)()(

1 ,0)(

)(1 ,0)(

)(1

0)()(

1)()(

1)()(

1

get... we(2)by dividing and (1) into (2) Inserting)()()(),,( (2)

0 (1)

zyx

zz

zy

y

yx

x

x

z

z

y

y

x

x

zyx

kkkkwhere

kz

zz

ky

yy

kx

xx

kz

zzy

yyx

xx

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...



Cont’d...

2

2

2

2

2

222

nz ny, nx,

zz

yy

xx

2 Eand

3...,2 1,nfor ,a

nk ,a

nk ,

ank where

sinsinsin),,()()()(),,(

cn

bn

an

m

zkykxkAzyxzyxzyx

zyx

zyx

zyx

0

a

b

c

kx

ky

kzEach solution (i.e. each combination 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 a “fermi-sphere” (see next page).



Cont’d...f

22

f Eenergy electron average for the valuemomentum a defines 2

E mk f

Volume of a single state “cube”: V3

state single

Vcba

Volume of a “fermi-sphere”: 34V 3

sphere-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 energy of the crystal (known as the fermi energy).

“V” is the physical volume of the crystal where as all other volumes used here refer to volume in k-space. Note that: Vsingle-state is the smallest unit in k-space. Vsingle-state is required to “hold” a single electron.



Cont’d...k-space volume of a single state “cube”: V

3

state single

Vcba

k-space volume of a “fermi-sphere”: 34V 3

sphere-fermi

fk

Number of filled states in a fermi-sphere: 2

3

3

3

sin 3

4

34

21

21

212N

f

f

stategle

spherefermi kV

V

kxxx

VV

Correction for allowing 2 electrons per state (+/- spin)

Correction for redundancy in counting identical states resulting

from +/- nx, +/- ny, +/- nz. Specifically, sin(-)=sin(+) so the state would be the same. Same as

counting only the positive octant in fermi-sphere.



Cont’d...

Number of filled states in a fermi-sphere:

31

2

f2

3 3k 3

N

VNkV f

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

density electron theis n where2

3 E 2

3

2

E3

222

f

32

22

22

f mn

mV

N

mk f

23

222

3 233

N

mEVkV f

Thus,



Cont’d...

E

V

mmEV

32

2

21

22

2mm G(E)

223

23

G(E)

VdEdN

per volumeenergy per states of # G(E)

23

222

3 233

N

mEVkV f

Applying to the semiconductor we must recognize m m* and since 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:


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, 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”.

)(~

,tan1

1)( )(

crystaltheinenergyaverageenergyFermiEand

KelvinineTemperaturTtconsBoltzmankwheree

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



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 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)).


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 kT3 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

eEf )(

1

1)(


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



!N ! iii

ii NS

SW

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

i ii

i

ii NS

SWW!N ! i

When we consider more than one level (i.e. all i’s) the number of ways we can arrange the electrons increases as the product of the Wi’s .

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 Ni’s . 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 Stirling’s Approximation to eliminate the factorials.

i

i!Nln! ln!lnln iii NSSW

or

or

Ei

Ei

Ei

Ei+2

Ei+1

Ei



Using Stirling’s Approximation, ln(x!) ~ (xln(x) – x) so that the above becomes,

iii

iiii

ii

NlnNlnlnln terms,like Collecting

NNlnNlnlnln

!Nln! ln!lnln

iiiiii

iiiiiiiii

iii

NSNSSSW

NSNSNSSSSW

NSSW

Now we can maximize W with respect to Ni’s . Note that since d(lnW)=dW/W when dW=0, d(ln[W])=0.

i

ii

ii

i

1ln0ln

1Nln1lnln

1Nln1lnlnln

ii

i

iii

iiii

dNNSWd

dNNSWd

NSNW

dNWd

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

(4)

1lnln used have weWhere xdx

xxd



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

i allfor 01ln that requires which

1ln0lni

ii

i

iii

i

ENS

dNENSWd



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

kT

EEi

ii

Ei

ii

ii

i

Fi

i

eNSEf

eNSEf

ENS

1

1)(

kT1 and

kTE-

have wetors,semiconduc of of case thein1

1)(

01ln

F


How do electrons and holes populate the bands?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

,EE if dEf(E)]-(E)[1g

,EE if dEf(E)(E)g

vv

cc

Electrons/cm3 in the conduction band between Energy E and E+dE

Holes/cm3 in the valence band between Energy E and E+dE


How do electrons and holes populate the bands?Band Occupation


How do electrons and holes populate the bands?Intrinsic Energy (or Intrinsic Level)

…Equal numbers of electrons and holes

Ef is said to equal Ei (intrinsic energy) when…


How do electrons and holes populate the bands?Additional Dopant States

Intrinsic: Equal number of electrons and holes

n-type: more electrons than holes

p-type: more holes than electrons

ece3080-l-4-density of states fermi...

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