ece 663 so far, we saw how to calculate bands for solids kronig-penny was a simple example real...

ECE 663

• So far, we saw how to calculate bands for solids

• Kronig-Penny was a simple example

• Real bandstructures more complex

• Often look like free electrons with effective mass m*

• Given E-k, we can calculate ‘density of states’

• High density of conducting states would imply metallicity

States and state filling

Carrier populations depend on

• number of available energy states (density of states)• statistical distribution of energies (Fermi-Dirac function)

Assume electronsact ‘free’ with a parametrized effective mass m*

ECE 663

Carrier Statistics

ECE 663

Labeling states allows us to count them!

Si

ECE 663

dE

k

dE

dkx xxxxxxxx

E

k

For 1D parabolic bands, DOS peaks at edges

Where are the states?

ECE 663

dE

k

dE

dkx xxxxxxxx

E

kk = 2/L

(# states) = 2(dk/k) = Ldk/

DOS = g = # states/dE = (L/)(dk/dE)

Where are the states?

ECE 663

dE

k

dE

dkx xxxxxxxx

E

k

Analytical results for simple bands

k = 2/L

E = h2k2/2m* + Ec dk/dE = m/2ħ2(E-Ec)

DOS = Lm*/22ħ2(E-Ec) ~ 1/(E-Ec)

ECE 663

dE

k

dE

dk

E

k

In higher dimensions, DOS has complex shapes

............

............

dE

2kdk

# k points increasesdue to angular integralalong circumference,as (E-Ec)

dNs = 2 x 2kdk/(2/L)2

g~S(E-Ec), step fn

Increasing Dimensions

ECE 663

From E-k to Density of States

Use E = Ec + ħ2k2/2mc to convert

kddk into dE

dNs = g(E)dE = 2 2 (dk/[2/L]) for each dimensionΣ1 k

ECE 663

(mc/22ħ3)[2mc(E-Ec)]1/2 (Smc/2ħ2)(E-Ec) (mcL/ħ)/√2mc[E-Ec]

From E-k to DOS for free els

E

Ec

E

DOSDOS

E

DOS

3-D 2-D 1-D

ECE 663

Real DOS needs computation

VB CB

DOS

E (eV)

ECE 663

Keep in mind 3-D Density of States

32

)(2)(

cEEmm

Eg

In the interest of simplicity, we’ll try to reduce allg‘s to this form…

ECE 663

(Ev-E)

(Ev-E)lh

ECE 663

Such that number of states is preserved

What if ellipsoids?

E – EC = ħ2k12/2ml

* + ħ2k22/2mt

* + ħ2k32/2mt

*

1 = k12/a2 + k2

2/b2 + k32/b2

a = 2ml*(E-EC)/ħ2

b = 2mt*(E-EC)/ħ2

ab

ECE 663

What if ellipsoids?

a = 2ml*(E-EC)/ħ2

b = 2mt*(E-EC)/ħ2

Total k-space volume of Nel ellipsoids = (4ab2/3)Nel

where

k-space volume of equivalent sphere = (4k3/3)

k = 2mn*(E-EC)/ħ2

where

mn* = (ml

*mt*2)1/3(Nel)2/3

EquatingK-spaceVolumes

ECE 663

Valence bands more complex

where

ECE 663

Valence bands more complex

hh

lh

But can try to fit two paraboloids for heavy and light holes and sum

ECE 663

Valence bands more complex

hh

lh

4k3/3 = 4k13/3 + 4k2

3/3

k = 2mp*(EV-E)/ħ2

k1 = 2mhh(EV-E)/ħ2

k2 = 2mlh(EV-E)/ħ2

ECE 663

mn* = (ml

*mt*2)1/3(Nel)2/3

mp* = (mhh

3/2 + mlh3/2)2/3

gC(E) = mn*[2mn

*(E-EC)]/2ħ3

gV(E) = mp*[2mp

*(EV-E)]/2ħ3

So map onto 3D isotropic free-electron DOS

with the right masses

ECE 663

Density of states effective mass for various solids

mn* = (ml

*mt*2)1/3(Nel)2/3

mp* = (mhh

3/2 + mlh3/2)2/3

ml* mt

* Nel mhh mlh mn* mp

*

GaAs

Si

Ge

1

6

8

0.98 0.19 0.49 0.16

1.64 0.082

0.067 0.067 0.45 0.082

0.28 0.042

0.067 0.473

1.084 0.5492

0.89 0.29

But how do we fill these states?

ECE 663

Fermi-Dirac Function• Find number of carriers in CB/VB - need to know

– Number of available energy states (g(E))– Probability that a given state is occupied (f(E))

• Fermi-Dirac function derived from statistical mechanics of “free” particles with three assumptions:1. Pauli Exclusion Principle – each allowed state can

accommodate only one electron2. The total number of electrons is fixed N=Ni

3. The total energy is fixed ETOT = EiNi

ECE 663

Fermi-Dirac Function

kTEE FeEf

1

1)(

ECE 663

Carrier concentrations

• Can figure out # of electrons in conduction band

• And # of holes in valence band

top

c

E

Ec dEEfEgn )()(

v

bottom

E

Ev dEEfEgp )(1)(

El. Density StateDensity

Occupancyper state

gc(E)

f(E)

ECE 663

kTEE

ed

F

FNn

CFc

cC

/)(

1)(

)(2

0

21

21

21

Full F-D statistics

ECE 663

ECE 663

Carrier Concentrations• If EC-EF >> kT the integral simplifies – nondegenerate

semiconductors Fermi level more than ~3kT away from bottom/top of band

• We then have << -1, so drop +1 in denominator of F1/2 function

• For electrons in conduction band:

23

2

)(

23

2

*22

*22

hkTm

N

eNn

eh

kTmn

nC

kTEE

c

kTEE

n

FC

CF

Or equivalently

CB lumped density of states

ECE 663

Carrier Concentrations

• Can do same thing for holes (nondegenerate approximation)

23

2

)(

23

2

*22

*22

h

kTmN

eNp

eh

kTmp

pv

kTEE

v

kTEE

p

vF

FV

VB lumped density of states

ECE 663

Intrinsic Semiconductors• For every electron in the CB there is a hole in the VB

• Fermi level is in middle of bandgap if effective masses not too different for e and h

**

ln4

3

ln22

)(

)()(

p

nF

C

VVCF

kTEE

VkTEE

C

mmkT

midgapE

NNkTEE

E

eNeN

pnVFFC

ECE 663

Intrinsic Semiconductors

• Can plug in Fermi energy to find intrinsic carrier concentrations

• Electrical conductivity proportional to n so intrinsic semiconductors have resistance change with temperature (thermistor) but not useful for much else.

kT

E

VCkT

EE

VCi

GVC

eNNeNNnpn 22

)(

ECE 663

Doped Semiconductors

Boron has 1 less el than tetrahedral

It steals an el from a nb. Si to form a tetrahedron. Thedeficit ‘hole’ p-dopes Si

P has 1 more el than tetrahedral

Extra el loosely tied (why?)

It n-dopes Si (1016/cm3 means 1 in 5 million)

ECE 663

Carrier Concentrations - nondegenerate

kTEE

v

vF

eNp)(

kTEE

c

FC

eNn)(

2

//)(

i

kTEVC

kTEEVC

nnp

eNNeNNnp GCv

Independent of Doping(This is at Equilibrium)

ECE 663

A more useful form for n and p

For intrinsic semiconductors

kTEE

ci

iC

eNn)(

kTEE

vi

vi

eNp)(

n = nie(EF-Ei)/kT p = nie

(Ei-EF)/kT

EF

Ei

EF

Ei

ECE 663

Charge Neutrality

0/ k Poisson’s Equation

In equilibrium, E=0 and =0

0

AD

AD

NNnp

NNnpq

Charge NeutralityRelationship

Number of ionized donors:

kTEEDD

D

DFegNN

/)(1

1

(gD = 2 for e’s, 4 for light holes, 1 for deep traps)

-

-

ECE 663

Charge Neutrality

kTEEDD

D

DFegNN

/)(1

1

(gD = 2 for e’s, 4 for light holes, 1 for deep traps)

0

ED PN e-(EN-EFN)/kT

<N> = 0.P0 + 1.P1 = f(ED-EF) = 1/[1 + e-(EF-ED)/kT]“0”

“1”

ECE 663

Charge Neutrality

kTEEDD

D

DFegNN

/)(1

1

(gD = 2 for e’s, 4 for light holes, 1 for deep traps)

0

PN e-(EN-EFN)/kT

<N> = 0.P0 + 1.(P + P) + 2.P

ED

2ED + U0

ECE 663

Can equivalently alter ED to account for degeneracy

01

1

1

1/)(/)(

/)(/)(

kTEEA

kTEED

kTEEC

kTEEV

FADF

CFFV

egegeNeN

Charge Neutrality

n

ECE 663

ECE 663

Nondegenerate Fully Ionized Extrinsic Semiconductors

• n-type (donor) ND >> NA, ND >> ni

n ND

p = ni2/ND

• p-type (acceptor) NA >> ND, NA >> ni

p NA

n = ni2/NA

ECE 663

Altering Fermi Level with doping (… and later with fields)

• Recall:

kTEEi

kTEEi

Fi

iF

enp

enn/)(

/)(

Take ln

i

AFi

i

DiF

iFii

nN

kTEE

nN

kTEE

EEnp

kTnn

kT

ln

ln

lnln

n-type

p-type

ECE 663

In summary

• Labeling states with ‘k’ index allows us to count and get a DOS

• In simple limits, we can get this analytically

• The Fermi-Dirac distribution helps us fill these states

• For non-degenerate semiconductors, we get simple formulae for n and p at equilibrium in terms of Ei

and EF, with EF determined by doping

• Let’s now go away from equilibrium and see what happens