lecture 3 density of states & intrinsic fermi 2012

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EE415/515 Fundamentals of Semiconductor Devices

Fall 2012

Lecture 3: Density of States, Fermi Level

(Chapter 3.4-3.5/4.1)

Density of States• Need to know the density of electrons, n, and holes, p, per unit volume

• To do this, we need to find the density of permitted energy states and then the probability (given by the Fermi function) that these states are occupied by an electron

• See Neamen Sections 3.4 and 3.5 for derivations, and compare these with alternative approaches given below

ECE415/515 Fall 2012 2J.E. Morris

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Density of States

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•See figure in “k-space.”

•Also: p=ħ k=(h/2π)k, i.e.•Direct linear relationship between (wave-vector) k-space and (momentum) p-space.

•The Neamen derivation (next slide) is based in k-space.•See also a slightly different approach based in p-space in following slide.

J.E. Morris

Density of States

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Consider electron confined to crystal (infinite potential well) of dimensions a(volume V= a3)

It has been shown that k=nπ/a, so ∆k=kn+1-kn=π/a

Each quantum state occupies volume (π/a)3 in k-space.

Number of quantum states in range k to k+dk is 4πk2.dk

and the number of electrons in this range k to k+dk is

gT(k)dk = 2(1/8).4πk2dk/ (π/a)3 = (k2dk/ π2)a3

where “2” comes from spin degeneracy, and “1/8” because E=(ħk)2/2m,

so negative k values do not add more energy states,

i.e. use kx, ky, kz >0 quadrant only

Converting k-space to energy ………..

J.E. Morris

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Density of States

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J.E. Morris

Density of States (alternate)

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Heisenberg says ∆p∆x=h,

(but see ħ in Neamen section 2.1.3 p30, and ħ/2 in Streetman)

and for crystal volume V=L3=LxLyLz

∆px∆x=h, ∆py∆y=h, ∆pz∆z=h,

and hence for a single electron

∆px ∆py ∆pz=h3/V for ∆x=Lx, etc

is the “volume” occupied in p-space by one electron.

Number of allowed conduction electron states g(p)dp from p to p+dp

[see Fig 3.26 (slide 3) in k-space with p = ħk]

is the volume of the p-space “shell” containing all states from p to p+dp

[4πp2dp]

divided by the p-space volume of each electron, i.e. ……

J.E. Morris

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Density of states

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J.E. Morris

(Compare 1D, 2D systems with 3D)

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And similarly for holes:-

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Now that we have the densities of states, we need the probabilities that these are filled.

J.E. Morris

Ex 3.3 Calculate the density of quantum states (/cm3) for a free electron over the energy range of (a) 0 ≤ E ≤ 2eV & (b) 1 ≤ E ≤ 2eV


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Ex 3.3 Calculate the density of quantum states (/cm3) in Si from (Ev-kT) to Ev at T=300 K


Fermi-Dirac statistics (see also text “derivation”)

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First, let’s find the Maxwell-Boltzmann distribution

Consider a system of identical particles with random collisions

Assume 2 particles of initial energies E1 & E2 collide with energy transfer δ to give final energies E3 & E4

Conservation of Energy requires that:

If p(E)=probability that particle has energy E, then:

Equilibrium: equal numbers of forward & reverse collisions for constant energy distribution, i.e.

)()( 214321 EEEEEE

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J.E. Morris

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MB continued & FD intro

Result suggests an exponential distribution, i.e.

Check: Aexp-α(Ei+δ)=λ. Aexp-αEi if λ=exp-αδ

For Fermi-Dirac statistics, repeat as for MB, but with the Pauli Exclusion Principle included, i.e.

Can only get transfer of energy E1, E2 → E3, E4

if E1 & E2 are occupied AND if E3 & E4 are unoccupied

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)(.)( ii EpEp

kTEA

EAEpMB

/exp.

exp.)(

J.E. Morris

Fermi functionSo the collision probability is

and applying Equilibrium and Energy Conservation again:

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MB/FD/(BE) comparisons

Maxwell-Boltzmann Bose-Einstein Fermi-Dirac

Particles distinguishable

Particles indistinguishable (wave functions overlap – cannot find individual particle)

Any number of particles can be in the same energy state

Only one particle in any energy state

(Pauli) but note spin degeneracy

Particles are statistically independent

Atoms, molecules, etc BosonsPhotons, phonons,

4He nuclei (integral spins)

FermionsElectrons, protons,

neutrons, 3He nuclei (½-odd-integral spins)

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1

J.E. Morris

Fermi distributions

Electrons in solids obey Fermi-Dirac statistics due to the Pauli exclusion principle (each state can have only one electron – but remember spin!)

Probability density function gives the ratio of filled to total allowed states at a given energy.

Using statistical mechanics to count states we find the Fermi-Dirac distribution function:• f(E) = {1 + exp[(E-Ef)/kT]}-1

• k is Boltzmann’s constant = 8.62x10-5 eV/K

= 1.38x10-23J/K

• EF is the Fermi level, [where f(EF)=½]


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The Fermi-Dirac distribution function.

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2

1)(f T allFor

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EE

EEKT

J.E. Morris

At T = 0 K


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At T > 0 K


Fermi levels

The Fermi distribution gives the probability that an available state will be occupied.

• Within the band gap there are no available states.

• Even with non-zero f(E) there is no occupancy in the gap. (An energy state at the Fermi level would have a probability of ½ of being occupied, if there was one.)

At 0K every available state up to the Fermi level is occupied and every state above the Fermi level is empty.

Fermi function is “symmetric” about the Fermi level

For holes:

ECE415/515 Fall 2012 20

kT

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exp1

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J.E. Morris

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ECE415/515 Fall 2012 21

Fermi function symmetry:electrons and holes

J.E. Morris

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Approximate FD → MB for (E-EF)>>kT

See Example 3.6:f(EF+3kT) ≈ 5% for 300K

So f(E) varies quickly

Example 3.8:(fMB-fFD)/FFD=5%

at E-EF=kT ln(1/0.05) ≈ 3kT

Note f(E) = {1 + exp[(E-Ef)/kT]}-1

≈ exp -[(E-Ef)/kT] for

“(E-EF)>>kT” condition where

exp[(E-Ef)/kT] >> 1

J.E. Morris

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Ex 3.6 Assume EF=EC-0.3eV and T=300 K. Determine probability of state occupied at (a) E=EC+kT/4 & (b) E=EC+kT


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Ex 3.7 Assume EF=EC-0.3eV. Determine T at which the probability of state at E=EC+0.025eV is occupied is 8x10-6.

J.E. Morris

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Ex.3.8 Calculate the energy, in terms of kT and EF, at which the difference between the FD and MB distributions is 2% of the FD value.


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Electron & hole densities

n(E) = gC(E).f(E)

p(E) = gV(E).[1-f(E)]

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Effective density of states & Fermi integral

ECE415/515 Fall 2012 27

band valencein the holesfor )(gfor result similar a giveseatment similar trA

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J.E. Morris

Effective density of states

ECE415/515 Fall 2012 28

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NC is the effective density of states as if all electrons at conduction band edge EC. Only limiting assumption is that EC-EF>>kT; if so, result holds for intrinsic and extrinsic. (Note: EF→EC for N-type, but approx usually still OK.)

2

3

2

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h

kTmN n

C

J.E. Morris

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Electron Concentrations

Simplify: represent states throughout the conduction band by an effective density of states Nc located at Ec.

n0 = Ncf(Ec)

If Ec – EF > 3kT then

f(Ec) ~ e-(Ec-EF)/kT (Maxwell-Boltzmann)

kT = 0.026 eV at 300K, so usually OK

n0 = Nc e-(Ec-EF)/kT ; Nc = 2[2πmn*kT/h2]3/2

Termed non-degenerate.

If EF within 3kT of either band edge or inside band degenerate and we need to use the full Fermi function.


Hole Concentrations

Concentration of holes in valence band

p0 = Nv [1 – f(Ev)]

If EF – Ev > 3kT then Fermi function reduces to Maxwell Boltzmann and

p0 = Nv exp[ -(EF – Ev)/kT ]

Nv = [2π mp*kT/h2]3/2

p0 = ∫gV(E)[1-f(E)]dE

1-f(E)=1-[1+exp((E-EF)/kT)]-1

=[exp (E-EF)/kT)]/[1+exp((E-EF)/kT)]

=1/[1+exp((EF-E)/kT)] ≈ exp-(EF-E)/kT


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Ex 4.1 Determine the probability that a quantum state at E=EC+kT is occupied by an electron, and calculate the electron concentration in GaAs at T=300 K if EF=EC-0.25eV.


(Nc from Table 4.1)

Ex 4.2 (a) Calculate the equilibrium hole concentration in Si at T=250K if EF=EV+0.27eV. [NV(300K)=1.04x1019/cm3] (b) What is p0(250K)/p0(400K)? (See Example 4.2)


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Intrinsic Concentrations EF = EFi for intrinsic material

n0=ni = Ncexp[-(Ec-EFi)/kT] pi = Nvexp[-(EFi-Ev)/kT]

n0 p0 = constant.

Plug in above => n0 p0 = ni2 = [NcNv]exp[-(EC-EV)/kT]

ni = [NcNv]1/2 exp[-Eg/2kT]

= 2[2πkT/h2]3/2(mnmh)3/2 exp[-Eg/2kT]

Remember n0p0 = ni2 and ni = 1.5x1010 cm-3 for Si

You’ll use both often! Also for Si: 6 different conduction band valleys (2 each x,y,z)

Note: varies in different sources

See Appendix F for more detail (complications) on effective mass

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2

12

3* 6)( tn mmm

*nm

J.E. Morris

Ex 4.3 (a) Calculate the intrinsic carrier concentration in GaAs at T=400 K and T=250 K, assuming Eg=1.42eV. (b) What is ni(400K)/ni(250K)?


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Intrinsic concentrations

In general, if EF ≠ EFi, then re-write these as

n0 = ni exp[(EF – EFi)/kT]

p0 = pi exp[(EFi – EF)/kT]

Also used frequently and helpful for visualization

With each problem consider what information you have and what you are looking for.

Law of Mass Action:

• n0p0=ni2

• Independent of EF, (i.e. of doping)

• So applies to both intrinsic and extrinsic at equilibrium


Temperature Dependence

ni T3/2 exp[-Eg/2kT]

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J.E. Morris

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Physical constants (@300 K unless noted)

Boltzmann’s const (k)

1.38 * 10-23 J/K 8.62 * 10-5 ev/K

Planck’s constant (h) 6.63*10-34 J-s 4.14*10-15 eV-s

Room temp value of kT

0.0259 eV Electron rest mass (m0)

9.11 * 10-31 kg

Electronic charge (q)

1.60 * 10-19 C Permittivity of free space (ε0)

8.85*10-14 F/cm

Density of states Nc Nv

2.8*1019 cm-3

1.04*1019 cm-3

Relative permittivity of silicon (εr)

11.9

Speed of light (c) 3*1010 cm/s Electron volt (eV) 1.6*10-19 J Band Gap Si Eg 1.11 ev Band gap Ge Eg 0.67 eV


Effective density of states & masses

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See Appendix B for m* values

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Intrinsic EF position

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.** .*

*

.*

*

VFi

00

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3 ..

.4

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2

1

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2

1

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m

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pnpn

J.E. Morris

Ex 4.4 Find the position of EFi at T=300 K wrt the band-gap center for (a) GaAs, and (b) Ge.


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Summary: Fermi function

Fermi function describes probability of a given state being occupied at temperature T.

f(E) = {1 + exp[(E-EF)/kT]}-1

For intrinsic semiconductors EF is mid-gap.

If E – EF > 3kT (non-degenerate, with EF in the forbidden gap) then

f(Ec)~e-(E-EF)/kT

Use effective density of states Nc located at Ec or Nv located at Ev. (still non-degenerate)

n0 = Nc e-(Ec-EF)/kT ; Nc = 2[2πmn*kT/h2]3/2

p0 = Nv e-(Ev-EF)/kT; Nv = [2π mp*kT/h2]3/2


lecture 3 density of states & intrinsic fermi 2012

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Transcript of lecture 3 density of states & intrinsic fermi 2012