Review of Basic Semiconductor Physics - Cornell University · Review of Basic Semiconductor Physics...

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ECE 407 – Spring 2009 – Farhan Rana – Cornell University

Review of Basic Semiconductor Physics

In this lecture you will learn:

• Review of electronic states and energy band in solids• Structure of semiconductors• Semiconductor alloys • Electron and hole statistics in semiconductors• Doped semiconductors• Transport in semiconductors and band diagrams in real space


InP

850 nm

980 nm

1300 nm1550 nm

Semiconductors and their Alloys

2


Group IV Cubic Semiconductors

Periodic Table:

Silicon Lattice (Diamond lattice):

=


Group III-V Cubic Semiconductors

GaAs Lattice (Zinc Blend Lattice): FCC Lattice Structure

3


Group III-V Hexagonal Semiconductors

GaN Lattice (Wurtzite Lattice):


Electrons in Solids

rErrVm

2

22

Wavefunctions and energies of electrons in solids are obtained by solving the Schrodinger equation:

Periodic potential of the atoms

Solution can be written in terms of the Bloch functions:

ruerr knrki

kn ,,

Bloch functions have the following property:

reRr knRki

kn ,,

ruRru knkn ,,

Follow from the periodicity of the atomic potential

rkErrVm knnkn ,,

22

2

Bloch functions satisfy the Schrodinger equation:

Energy of the Bloch functions

4


Electrons in Solids

The First Brillouin Zone (FBZ):

The wavevectors are restricted to the first brillouin zone:

ruerr knrki

kn ,,

k

FBZ of the FCC lattice

Two labeling items:i) The wavevectorii) The band index “n”

k

Energy Bands:

The band index “n” stands for different energy bands


Energy Bands in Semiconductors

Germanium Silicon GaAs

Ev

Ec

Ev

Ec

cce

cc KkKkm

EkE 2

2

vvh

vv KkKkm

EkE 2

2

Parabolic expansion of energy band dispersion:

5


Energy Bands in SemiconductorsMore generally:

ceccc KkMKkEkE 12

2

Ev

Ec

zzzyzx

yzyyyx

xzxyxx

e

mmm

mmm

mmm

M

vhvvv KkMKkEkE 12

2

Parabolas in 3D can be anisotropic!


Counting Electronic States in Solids

kEr nkn

,

No two electrons can occupy the same quantum state (Pauli’s exclusion principle)

So how many states are there in the FBZ?

In a solid of volume V, the number of energy levels in one energy band in volume of the FBZ is:kd

3

33

22

kd

V

Two possible spin states of electrons

FBZ3

3

2

kd

Vk

Summations over electron state in FBZ can be written as integrals:

6



How many electron states per energy band in the entire FBZ?

FBZ of volume

2

2

222

33

3

FBZ

Vkd

VFBZk

But:

cell unit primitive the of volume

2FBZ of volume

3

crystal the in cells primitive of number 2

cell primitive the of volume2

V

How many electron states per energy band in the entire FBZ?

Each primitive cell contributes two states or energy levels to each band



Ev

Ec

Case of Silicon:

Each primitive unit cell contains 2 Si atoms Each Si atom contributes 4 electrons to the crystal

Therefore, each primitive cell contributes 8 electrons to the crystal

Each energy band has twice as many states as the number of primitive cells in the crystals

Therefore, 4 lowest energy bands will be full of electrons and the remaining ones will be empty!

7


Semiconductor AlloysElemental Semiconductors:

Example: Si, Ge, and C

Binary Alloys:

Example: Si(1-x)Gex

Alloy of two elemental semiconductors Consists of (x) fraction of Ge and (1-x) fraction of Si

Ternary Alloys:

Example: AlxGa(1-x)As

Alloy of two compund semiconductors Consists of (x) fraction of AlAs and (1-x) fraction of GaAs

Quarternary Alloys:

GaAsGaP1InP11InAs1PAsGaIn -1-1 xyyxyxyxyyxx

GaAs1AlAs xx


Semiconductor Alloys: Vegard’s Law

Vegard’s law says the lattice constant of an alloy is a weighted sum of the lattice constants of each of its constituents, and the weight assigned to each constituent is equal to its fraction in the alloy.

Examples:

Ge1Si GeSi 1 axaxa xx

GaAs1AlAs AsGaAl 1 axaxa xx

GaAs GaP1

In11InAs1AsGaIn 11

axyayx

PayxaYxPa yyxx

8


Properties of Semiconductor Alloys

What about other material parameter like dielectric constants, effective masses, band gaps, etc?

If you don’t know any better, the linear rule law can be a good first approximation. But it does not always work very well for quantities other than the lattice constant.

For example, the band gap of AlxGa(1-x)As at the -point is given more accurately by:

2xxg xcbxaE AsGaAl 1

eV 42.1

300K at eV 0.438

eV 087.1

a

c

b

21 AsIn Ga cxbxaE xxg

The band gap of GaxIn(1-x)As at the -point is given by:

eV 324.0

300K at eV 0.4

eV 7.0

a

c

b


Properties of Semiconductor Alloys

Direct to indirect bandgap transitions in alloys:

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InP

850 nm

980 nm

1300 nm1550 nm

Semiconductors and their Alloys


Electron Statistics: GaAs Conduction Band

Consider the conduction band of GaAs near the band bottom at the -point:

e

e

e

m

m

m

M

100

010

0011

This implies the energy dispersion relation near the band bottom is:

e

ce

zyxcc m

kE

m

kkkEkE

22

222222

Suppose we want to find the total number of electrons in the conduction band:

We can write the following summation:

FBZin

2k

c kfN

10


The Fermi-Dirac Distribution

A fermion (such as an electron) at temperature T occupies a quantum state with energy E with a probability f(E-Ef) given by the Fermi-Dirac distribution function:

TKEEffe

EEf

1

1

Ef = chemical potential or the Fermi level K = Boltzmann constant = 1.38 X 10-23 Joules/Kelvin

fEEf

E0

1

fE

T = 0K

E0

1

fE

T > 0K

E0

1

fE

T >> 0K

~2KT

~2KT

0.5

0.5 The Fermi level Ef is determined by invoking some physical argument …(as we shall see)

fEEf

fEEf


fc

KTEkEc EkEfkf

fc

exp1

1

Where the Fermi-Dirac distribution function is:

We convert the summation into an integral:

KTEkE

kdVkfN

fckc

exp1

1

222

FBZ3

3

FBZ in

Then we convert the k-space integral into an integral over energy:

?

?FBZ3

3

exp1

1

22 fc

fc

EEfEgdEV

KT

EkE

kdVN

We need to find the density of states function gc(E) for the conduction band and need to find the limits of integration

FBZin

2k

c kfN

Another way of writing itEf


11


?

?FBZ3

3

exp1

1

22 fc

fc

EEfEgdEV

KT

EkE

kdVN

e

ce

zyxcc m

kE

m

kkkEkE

22

222222

Energy dispersion near the band bottom is:

Electrons will only be present near the band bottom

(parabolic and isotropic)

Since the electrons are likely present near the band bottom, we can limit the integral over the entire FBZ to an integral in a spherical region right close to the -point:

point3

2

point-3

3

FBZ3

3

8

42

22

22

fc

cc

EkEfdkk

V

kfkd

Vkfkd

VN

Ef



point

3

2

8

42 fc EkEfdk

kVN

Since the Fermi-Dirac distribution will be non-zero only for small values of k, one can safely extend the upper limit of the integration to infinity:

03

2

8

42 fc EkEfdk

kVN

emdkk

dkdE 2

ce EE

mk 2

2

and

We have finally:

e

ce

zyxcc m

kE

m

kkkEkE

22

222222

cEfcfc EEfEgdEVEkEfdk

kVN

03

2

8

42

We know that:

Ef


12


We have finally:

Where the conduction band density of states function is:

ce

c EEm

Eg

23

222

2

1

cEfcfc EEfEgdEVEkEfdk

kVN

03

2

8

42

E

Egc

cE

The density of states is the number of states available per unit energy per unit volume of the crystal

Ef



ce

c EEm

Eg

23

222

2

1

cEfc EEfEgdEn

fEEf

EfE

Egc Ef

KTfEE

eEEf f

If then one may approximate the Fermi-Dirac function as an exponential:

KTEE fc

KTEE

KTEE

EEf f

ff exp

exp1

1

KTEE

NEEfEgdEn fcc

Efc

c

exp

cE

Where:

23

222

KTm

N ec

Maxwell-Boltzman approximation

Effective density of states (units: #/cm3)


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Ef

Valence Band in GaAs: Heavy Hole Band and Light Hole Band

Light Hole Band

Heavy Hole BandSplit-off band


Valence Band and Holes

Ef

• At zero temperature, the valence band is completely filled and the conduction band is completely empty

• At any finite temperature, some electrons near the top of the valence band will get thermally excited from the valence band and occupy the conduction band and their density will be given by:

• The question we ask here is how many empty states are left in the valence band as a result of the electrons being thermally excited. The answer is (assuming the heavy-hole valence band):

• We call this the number of “holes” left behind in the valence band and the number of these holes is P:

KTEE

Nn fcc exp

FBZ in

12k

fhh EkEf

fhhk

fhh EkEfkd

VEkEfP

12

212FBZ

3

3

FBZ in

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Hole Statistics: GaAs Heavy Hole Band

parabolic approx.

fhh EkEf

kdVP

12

2FBZ

3

3

hh

vhh

zyxhh m

kE

m

kkkEvkE

22

222222

Energy dispersion near the top of the valence band is:

Holes will only be present near the top of the valence band

Since the holes are likely present near the band maximum, we can limit the integral over the entire FBZ to an integral in a spherical region right close to the -point:

point

3

2

18

42 fhh EkEfdk

kVP

Ef

fhh EkEf

kdVP

12

2point-

3

3


03

2

18

42 fhh EkEfdk

kVP

Since the Fermi-Dirac distribution will be non-zero only for small values of k, one can safely extend the upper limit of the integration to infinity:

point

3

2

18

42 fhh EkEfdk

kVP

hh

vhh

zyxhh m

kE

m

kkkEvkE

22

222222

We know that:

hhmdkk

dkdE 2

EEm

k vhh 2

2

and

We have finally:

vE

fhh

fhh

EEfEgdEV

EkEfdkk

VP

1

18

42

03

2

Ef


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Where the heavy hole band density of states function is:

EEm

Eg vhh

hh

23

222

2

1

E

Eghh

vE

We have finally:

vE

fhh

fhh

EEfEgdEV

EkEfdkk

VP

1

18

42

03

2

Ef



EEm

Eg vhh

hh

23

222

2

1

vE

fhh EEfEgdEp 1

fEEf

EfE

Egc

Ef

If then one may approximate the Fermi-Dirac function as an exponential:

KTEE vf

KTEE

KTEE

EEf f

ff exp

exp1

11

KTEE

NEEfEgdEp vfhh

E

fhh

vexp1

cE

Where:

23

222

KTm

N hhhh

Maxwell-Boltzman approximation for holes

vE

Eghh

Effective density of states (units: #/cm3)


16


In most semiconductors, the light-hole band is degenerate with the heavy hole band at the -point. So one always needs to include the holes in the light-hole valence band as well:

Ef

fEEf

EfE

Egc

cEvE

Eghh Eg h

v

v

vv

E

fv

E

fhhh

E

fh

E

fhh

EEfEgdE

EEfEgEgdE

EEfEgdEEEfEgdEp

1

1

11

KTEE

Np vfv exp

Where:

23

222

KTm

N hv

and 322323hhhh mmm

EEm

EgEgEg

vh

hhhv

23

222

2

1

Density of states effective mass

Hole Statistics: GaAs Heavy Hole and Light Hole Bands


Example: Electron Statistics in GaAs – Electrons and Holes

Ef

At any temperature, the total number of electrons and holes (including both heavy and light holes) must be equal:

c

vvcf

vcf

c

v

fcc

vfv

NNKTEE

E

KTEEE

NN

KTEE

NKT

EEN

np

log22

2exp

expexp

Because the effective density of states for electrons and holes are not the same (i.e. Nv ≠ Nc), the Fermi level at any finite temperature is not right in the middle of the bandgap.

But at zero temperature, the Fermi-level is exactly in the middle of the bandgap

fEEf E

fE

Egc

cEvE

Egv

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Example: Electron Statistics in GaAs – Electrons and Holes

Ef

At any temperature, the total number of electrons and holes (including both heavy and light holes) must be equal:

where ni is called the intrinsic electron (or hole) densityinnp

Note that the smaller the bandgap the larger than intrinsic electron (or hole) density

KT

ENNn

nKT

EENN

nKT

EEN

KTEE

N

nnp

nnp

gcvi

ivc

cv

ifc

cvf

v

i

i

2exp

exp

expexp

2

2

2


Electron and Hole Pockets in GaAs

Ef

• At any non-zero temperature, electrons occupy states in k-space that are located in a spherically symmetric distribution around the -point

• This distrbution is referred to as the “electron pocket” at the -point

• At any non-zero temperature, the holes (heavy and light) also occupy states in k-space that are located in a spherically symmetric distribution around the -point

• This distribution is referred to as the “hole pocket” at the -point

Hole pocket

Electron pocket

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Electron and Hole Statistics in Degenerate Semiconductors

cEfc EEfEgdEn

23

221

23

2

22

2

1

1

π

kTmN

kT

EEF

π N

eEE

m2dEn

ec

cfc

KTEEce

E fc

fEEf

E

fE Egc

cEvE

Eghh

fEEf

E

fE

Egc

cEvE

Eghh

vE

fhh EEfEgdEp 1

2

3

221

2

3

2

22

2

12

π

kTmN

kTEE

Fπ

N

Ef EEm

π dEp

hv

fvv

vh

Ev


Donor Impurities in Semiconductors

Consider a semiconductor (say GaAs) in which one Ga atom site is occupied by a Si atom, as shown:

• Silicon has one more electron in the outermost shell compared to Ga (4 in Si compared to 3 in Ga)

• Since only 3 electrons are needed to form co-valent bonds with the nearby As atoms, the extra electron does not participate in bonding and can drift away leaving behind a positively charged Si atom

Ga As

As

Ga As

As

Si As

As

Ga As

As

Ga As

As

Ga As

As

Ga As

As

Ga As

As

Ga As

As

+

-

Ga

Ga

Ga

Ga

Ga

Ga

Ga

Ga

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Donor Impurities in Semiconductors: N-Type DopingEnergy

k

5.9 meV

• At very low temperatures the electron resides in the donor energy level and the donor atom is neutral

• At room temperature, the electron in the donor energy level can acquire enough energy to jump to the conduction band

When this happens the donor is said to have ionized

• Once in the conduction band the electron can move around and is no longer localized at the donor atom

• Donor impurities can therefore be used to dope semiconductors n-type

Donor ionization


Acceptor Impurities in Semiconductors: P-Type Doping

Consider a semiconductor (say GaAs) in which one As atom site is occupied by a carbon atom, as shown:

Ga As

As

Ga As

As

Ga As

C

Ga As

As

Ga

As

Ga

As

AsAs As

_Ga

Ga

Ga

Ga

Ga

Ga

Ga As

As

Ga As

As

Ga As

C

Ga As

As

Ga

As

Ga

As

AsAs As

_Ga

Ga

Ga

Ga

Ga

Ga• C has one less electron in the outermost shell compared to As (4 in C compared to 5 in As)

• Since 4 electrons are needed to form covalent bonds with the nearby Ga atoms, the required electron is taken from the valence band resulting in a negatively charged C atom and a hole in the valence band

+

20


Acceptor Impurities in Semiconductors: P-Type Doping

Energy

k

Acceptor ionization

• Acceptor atom gives rise to hydrogenic energy levels near the valence band maximum

• At very low temperatures the hole resides in the acceptor energy level and the acceptor atom location is overall neutral

• At room temperature, the hole in the acceptor energy level can acquire enough energy to jump to the valence bandWhen this happens the acceptor is said to have ionized

• Once in the valence band the hole can move around and is no longer localized at the acceptor atom

• Acceptor impurities can therefore be used to dope semiconductors p-type


Donor Ionization StatisticsEnergy

k

EdDonor ionization

Ef

In the grand canonical ensemble the probability of a system to have total particles N and total energy E is:

KTNEE feAENP ,

The donor level can have the following possible states:

1) No electrons present

2) One spin-up electron present

3) One spin-down electron present

4) Two or more electrons present

AENP 0,0

KTEEd

fdeAEENP ,1

KTEEd

fdeAEENP ,1

0,1 ENP

KTEEKTEE

fdfd

eAeA

21

1121

Sum of all probabilities should equal unity:

Coulomb repulsion does not allow it

21


Donor Ionization StatisticsEnergy

k

EdDonor ionization

Ef

Probability that the donor level is ionized

Probability that the donor level has no electrons

A

ENP

0,0

KTEE fde

21

1

If the total donor impurity concentration is Nd then the concentration of ionized donors N+

d is equal to:

KTEEd

dfde

NN

21

For acceptors we have a similar relation:

KTEEa

afae

NN

21

EaAcceptor ionization


Carrier Statistics in Doped Semiconductors

Energy

k

Acceptor ionization

Donor ionization

Consider a semiconductor that is doped with both donor and acceptor impurity atoms

• The total charge must be zero:

The above equation can be used to find the position of the equilibrium Fermi level since every term depends on the Fermi level position (one equation in one unknown)

0 npNN ad

KTEEd

dfde

NN

21

KTEEa

afae

NN

21

22


Carrier Transport in Semiconductors

Electron current density (units: Amps/cm2) can be written as:

rnDqrErqnrJ eee

Diffusion componentDrift component

KTq

De

e

Einstein relation:

Hole current density (units: Amps/cm2) can be written as:

Diffusion componentDrift component

KTq

Dh

h

Einstein relation:

rpDqrErpqrJ hhh


Carrier Number Conservation Equations

rRrGrJqt

rneee

1

rRrGrJqt

rphhn

1

Electron-hole generationrate

Electron-hole recombination rate

Electron-hole generationrate

Electron-hole recombination rate

rGrG eh

rRrR eh

23


Poisson Equation

rNrNrnrpqrEr ad

rNrNrnrpqrEr

rRrGrJqt

rp

rRrGrJqt

rn

rpDqrErpqrJ

rnDqrErnqrJ

ad

hhh

eee

hhh

eee

1

1

The Five Shockley Equations


Minority Carriers and Recombination Rates

In a p-type semiconductor, electrons are the minority carriers and holes are the majority carriers: pn

In a n-type semiconductor, holes are the minority carriers and electrons are the majority carriers: np

We consider a non-equilibrium situation in a p-type semiconductor in which the electron and hole concentrations are slightly perturbed from their equilibrium values

Equilibrium values: onop

e

oee

nnGR

Net recombination rate:Minority carrier lifetime

We consider a non-equilibrium situation in a n-type semiconductor in which the electron and hole concentrations are slightly perturbed from their equilibrium values

Equilibrium values: onop

h

ohh

ppGR

Net recombination rate: Minority carrier lifetime

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Band Diagrams in Real Space - I

For devices, it is useful to draw the conduction and valence band edges in real space:

cE

vE

fE

x

cE

vEfE

x

N-type semiconductor

Energy

k

fE

cE

vE

Energy

k

fEcE

vE

P-type semiconductor

KTEEc

fceNn

KTEEv

vfeNp


Band Diagrams in Real Space - II

Electrostatic potential and electric field:

cE

vE

fE

xN-type semiconductor

An electrostatic potential (and an electric field) can be present in a crystal:

The total energy of an electron in a crystal is then given not just by the energy band dispersion but also includes the potential energy coming from the potential:

Therefore, the conduction and valence band edges also become position dependent:

rrEr

and

kEn

rekEkE nn

reEEreEE vvcc

Example: Uniform x-directed electric field

xeExExE

xExr

xErE

xcc

x

x

0

0

ˆ

xErE x ˆ

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Non-Equilibrium Situations and Current Flow

rnDqrErqnrJ eee

rEq

rE c

1

rErEKT

rnrn

eNrn

fc

KTrErEc

fc

rErn

rErErnrErn

rErEKT

rnDqrErnrJ

fe

fcece

fcecee

Non-uniform Fermi level implies current flow

Similarly:

rErp

rErErprErn

rErEKT

rpDqrErprJ

fh

fvhvh

fvevhh

Non-uniform Fermi level implies current flow

Review of Basic Semiconductor Physics - Cornell University · Review of Basic Semiconductor Physics...

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