1

Chapter 2-3. States and carrier distributions

Determine the carrier distribution with respect to energy in different bands.

Determine the quantitative information of carrier concentrations in different bands.

You should have read pages 40 – 49 For next class, read pages 49 – 57

So far we have concentrated on carrier properties of qualitative nature. We also need to:

Two concepts will be introduced to determine this:

• Density of states

• Fermi-Dirac distribution and Fermi-Level

2

Density of states

There are 4 states per atom or 4 51022 / cm3 states in each of conduction and valence bands of Si.

The distribution of these states in the bands are not uniform, but follows a distribution function given by the following equations.

3

Dependence of DOS near band edges

4

More on density of states (DOS)

gc(E) dE represents the # of conduction band states/cm3 lying in the energy range between E and E + dE

gv(E) dE represents the # of valence band states/cm3 lying in the energy range between E and E + dE

More states are available available away from the band edges, similar to a seating arrangement in a football field

Units for gc(E) and gv (E): # per unit volume per unit energy, i.e., # / (cm3 eV)

Energy bands are drawn with respect to electron energies

5

Some examples

Calculate the density of states per cm3 that lie between the energies EC and EC+1eV.

To calculate this, you have to integrate the equation between EC and EC+1eV.

If you integrate from Ec, bottom to Ec, top, what should you get?

321327

2/319334

2/331

12/33

2/3*

1

3

2/3*1

32

**

/108.5/108.5

]0)106.1[(3

2

)/10625.6(

)1011.918.1(28

|)(3

2)(28

)()(28)(2

cmstatesxmstatesxD

JxsJx

KgxxD

EEh

mD

dEEEh

mdE

EEmmD

eVEEC

n

eVE

E

Cn

eVE

E

Cnn

C

C

C

C

C

C

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Fermi-Dirac distribution and the Fermi-level

Density of states tells us how many states exist at a given energy E. The Fermi function f(E) specifies how many of the existing states at the energy E will be filled with electrons. The function f(E) specifies, under equilibrium conditions, the probability that an available state at an energy E will be occupied by an electron. It is a probability distribution function.

EF = Fermi energy or Fermi level

k = Boltzmann constant = 1.38 1023 J/K = 8.6 105 eV/K

T = absolute temperature in K

7

Distribution function for gas molecules

Example: Gas molecules follow a different distributionfunction: The Maxwell-Boltzmann distribution

kT

)EE(

kT

E

kT

E

n

n12

1

2

e

e

e

1

2

Let us look at the Fermi-Dirac distribution more closely.

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Fermi-Dirac distribution: Consider T 0 K

For E > EF :

For E < EF :

0)(exp1

1)( F

EEf

1)(exp1

1)( F

EEf

E

EF

0 1 f(E)

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If E = EF then f(EF) = ½

If then

Thus the following approximation is valid:

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

If then

Thus the following approximation is valid:

So, 1f(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 1exp F

kT

EE

kT

EEEf

)(exp)( F

kTEE 3F 1exp F

kT

EE

kT

EEEf Fexp1)(

Fermi-Dirac distribution: Consider T > 0 K

10

Temperature dependence of Fermi-Dirac distribution

11

Exercise 2.3

12

Example

Assume that the density of states is the same in the conduction band (CB) and valence band (VB). Then, the probability that a state is filled at the conduction band edge (EC) is equal to the probability that a state is empty at the valence band edge.

Where is the Fermi level located?

)(1)( VC EfEf

kT

EE

kT

EE

kT

EE VFFVFC

e1

1

e1

11

e1

1

VFFC EEEE 2

VCF

EEE

This corresponds to intrinsic material, where the # of electrons at EC = # of holes (empty states) at EV. Note that the probability within the band gap is finite, but there are no states available, so electrons cannot be found there.

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Equilibrium distribution of carriers

Distribution of carriers = DOS probability of occupancy = g(E) f(E)

(where DOS = Density of states)

Total number of electrons in CB (conduction band) =

top

Cd)()(C0

E

EEEfEgn

Total number of holes in VB (valence band) =

V

Bottomd)(1)(V0

E

EEEfEgp

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Fermi-level positioning and carrier distributions

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Visualization of carrier distribution

One way to convey the carrier distribution is to draw the following diagram. This diagram represents n-type material since there are more electrons than holes.

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Another more useful way to convey the carrier distribution is to draw the following band diagrams. The position of EF with respect to Ei is used to indicate whether is n-type, p-type or intrinsic.

Visualization of carrier distribution (continued)