Homogeneous Semiconductors

• Dopants

• Use Density of States and Distribution Function to:

• Find the Number of Holes and Electrons.

Energy Levels in Hydrogen Atom

Energy Levels for Electrons in a Doped Semiconductor

Assumptions for Calculation

Density of States (Appendix D)

Energy Distribution Functions(Section 2.9)

Carrier Concentrations(Sections 2.10-12)

GOAL:

• The density of electrons (no) can be found precisely if we know1. the number of allowed energy states in a small energy

range, dE: S(E)dE “the density of states”

2. the probability that a given energy state will be occupied by an electron: f(E)

“the distribution function”

no = bandS(E)f(E)dE

For quasi-free electrons in the conduction band:

1. We must use the effective mass (averaged over all directions)2. the potential energy Ep is the edge of the conduction band (EC)

For holes in the valence band:

1. We still use the effective mass (averaged over all directions)2. the potential energy Ep is the edge of the valence band (EV)

2

1

2

*

2

2

2

1)( C

dse EEm

ES

2

1

2

*

2

2

2

1)( EE

mES V

dsh

Energy Band Diagram

conduction band

valence band

EC

EV

x

E(x)

S(E)

Eelectron

Ehole

note: increasing electron energy is ‘up’, but increasing hole energy is ‘down’.

2

1

2

*

2

2

2

1)( C

dse EEm

ES

2

1

2

*

2

2

2

1)( EE

mES V

dsh

Reminder of our GOAL:

• The density of electrons (no) can be found precisely if we know1. the number of allowed energy states in a small energy

range, dE: S(E)dE “the density of states”

2. the probability that a given energy state will be occupied by an electron: f(E)

“the distribution function”

no = bandS(E)f(E)dE

Fermi-Dirac Distribution

The probability that an electron occupies an energy level, E, is

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

– where T is the temperature (Kelvin)– k is the Boltzmann constant (k=8.62x10-5 eV/K)– EF is the Fermi Energy (in eV)

– (Can derive this – statistical mechanics.)

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

All energy levels are filled with e-’s below the Fermi Energy at 0 oK

f(E)

1

0EF

E

T=0 oKT1>0T2>T1

0.5

Fermi-Dirac Distribution for holes

Remember, a hole is an energy state that is NOT occupied by an electron.

Therefore, the probability that a state is occupied by a hole is the probability that a state is NOT occupied by an electron:

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

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

The Boltzmann ApproximationIf (E-EF)>kT such that exp[(E-EF)/kT] >> 1 then,

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

exp[-(E-EF)/kT] …the Boltzmann approx.

similarly, fp(E) is small when exp[(EF - E)/kT]>>1:

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

exp[-(EF - E)/kT]

If the Boltz. approx. is valid, we say the semiconductor is non-degenerate.

Putting the pieces together:for electrons, n(E)

f(E)

1

0EF

E

T=0 oKT1>0T2>T1

0.5

EV EC

S(E)

E

n(E)=S(E)f(E)

Putting the pieces together: for holes, p(E)

fp(E)

1

0EF

E

T=0 oK

T1>0T2>T1

0.5

EV EC

S(E)

p(E)=S(E)f(E)

hole energy

Finding no and po

2/3

2

*

/

2/3

2

*

2(min)

0

22...]/)(exp[

2

2

1)()(

kTmNwherekTEEN

dEeEEm

dEEfESp

dshVVFV

kTEEEv

Vdsh

Ev

Ev

pF

2/3

2

*

/

2/3

2

*

2

(max)

0

22...]/)(exp[

2

2

1)()(

kTmNwherekTEEN

dEeEEm

dEEfESn

dseCFCC

kTEE

Ec

Cdse

Ec

Ec

F

the effective density of statesat EC

the effective density of statesat EV

Energy Band Diagramintrinisic semiconductor: no=po=ni

conduction band

valence band

EC

EV

x

E(x)

n(E)

p(E)EF=Ei

where Ei is the intrinsic Fermi level

nopo=ni2

Energy Band Diagramn-type semiconductor: no>po

conduction band

valence band

EC

EV

x

E(x)

n(E)

p(E)

EF

]/)(exp[0 kTEENn FCC

nopo=ni2

Energy Band Diagramp-type semiconductor: po>no

conduction band

valence band

EC

EV

x

E(x)

n(E)

p(E) EF

]/)(exp[0 kTEENp VFV

nopo=ni2

A very useful relationship

kTEVC

kTEvEcVC

VFVFCC

geNNeNN

kTEENkTEENpn//)(

00 ]/)(exp[]/)(exp[

…which is independent of the Fermi Energy

Recall that ni = no= po for an intrinsic semiconductor, so

nopo = ni2

for all non-degenerate semiconductors.(that is as long as EF is not within a few kT of the band edge)

kTEVCi

ikTE

VC

g

g

eNNn

neNNpn2/

2/00

The intrinsic carrier density

kTEVCi

ikTE

VC

g

g

eNNn

neNNpn2/

2/00

is sensitive to the energy bandgap, temperature, and (somewhat less) to m*

2/3

2

*

22

kTm

N dseC

The intrinsic Fermi Energy (Ei)

]/)(exp[]/)(exp[ kTEENkTEEN ViViCC

For an intrinsic semiconductor, no=po and EF=Ei

which gives

Ei = (EC + EV)/2 + (kT/2)ln(NV/NC)

so the intrinsic Fermi level is approximatelyin the middle of the bandgap.

Higher TemperaturesConsider a semiconductor doped with NA ionized

acceptors (-q) and ND ionized donors (+q), do not assume that ni is small – high temperature expression.

positive charges = negative charges

po + ND = no + NA

using ni2 = nopo

ni2/no + ND

= no+ NA

ni2 + no(ND-NA) - no

2 = 0

no = 0.5(ND-NA) 0.5[(ND-NA)2 + 4ni2]1/2

we use the ‘+’ solution since no should be increased by ni

no = ND - NA in the limit that ni<<ND-NA

Simpler

Expression

Temperature variation of some important “constants.”

Degenerate Semiconductors1. The doping concentration is so high that EF moves within a few kT of the band edge (EC or EV). Boltzman approximation not valid.

2. High donor concentrations cause the allowed donor wavefunctions to overlap, creating a band at Edn.

First only the high states overlap, but eventually even the lowest state overlaps.

This effectively decreases the bandgap by

Eg = Eg0 – Eg(ND).

EC

EV

+ + + +

ED1

Eg0Eg(ND)

for ND > 1018 cm-3 in Si

impurity band

Degenerate SemiconductorsAs the doping conc. increases more, EF rises above EC

EV

EC (intrinsic)available impurity band states EF

Eg

EC (degenerate) ~ ED

filled impurity band states

apparent band gap narrowing:Eg

* (is optically measured)

-

Eg* is the apparent band gap:

an electron must gain energy Eg* = EF-EV

Electron Concentrationin degenerately doped n-type semiconductors

The donors are fully ionized: no = ND

The holes still follow the Boltz. approx. since EF-EV>>>kT

po = NV exp[-(EF-EV)/kT] = NV exp[-(Eg*)/kT]

= NV exp[-(Ego- Eg*)/kT]

= NV exp[-Ego/kT]exp[Eg*)/kT]

nopo = NDNVexp[-Ego/kT] exp[Eg*)/kT]

= (ND/NC) NCNVexp[-Ego/kT] exp[Eg*)/kT]

= (ND/NC)ni2 exp[Eg

*)/kT]

Summary

non-degenerate:nopo= ni

2

degenerate n-type:

nopo= ni2 (ND/NC) exp[Eg

*)/kT]

degenerate p-type:

nopo= ni2 (NA/NV) exp[Eg

*)/kT]