Chapter 3-1(II 2008-2009) [Compatibility Mode]

8/14/2019 Chapter 3-1(II 2008-2009) [Compatibility Mode]

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1. Intrinsic semiconductors

(1) Silicon crystal and energy band diagram

Electronic confi uration of Si: 3s23 2

hyb orbitalsValence

electron

Si ion core (+4e)

The four hybrid orbitals of silicon, each has one electron so that they are half-

occupied.

2

ere ore, a hyb or a o one a om can over ap a hyb o a ne g or ng

atom to form a covalent bond with two spin-paired electrons.


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Overlapping of two hyb orbitals

Antibonding orbitals Bonding orbitals

Two valence electronsNo electrons

occupy the bonding orbitals

Interaction, Valence bandConduction bandEnergy difference,Eg

3

Ev

: top of the VB, Ec

: bottom of the CB. The energy distance from Ec

to the

vacuum level, the width of the CB, is called the electron affinity .


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(2) Electrons and holes

VB is full and CB is empty

Onl the electrons laced in CB can freel

respond to an applied electric field.

Free electron-The effective mass of the

e ec rons n s ou e mo e .

If electrons in VB gains enough energy (E> g , ey can e exc e o .

A free electron in CB and a hole in VB after

an electron is excited to CB by photon.

The empty electronic state of the missing

4

.

Behave as a positive charged particle with

effective mass much different from electron.


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Thermal generation of electron-hole pair.

Thermal energy results in the vibration of atoms

In a certain region, the atoms, at some instant,

may be moving in such a way that a bond

the overstretched bond ru turin and hence

releasing an electron into the CB

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The free electron in the CB can wander around the crystal and contribute to the

electrical conductivit .

The hole, denoted as h+, can also wander around the crystal as if it were free. How

to wander around?

Jump-tunneling. The neighboring electron can fill the hole.

When a wanderin electron in the CB meets a hole in the VB the electron has

Two types of charge carriers in semiconductors: electrons and holes.

found an empty state of lower energy and therefore occupies the hole. This is called

recombination and results in the annihilation of an electron in the CB and a hole in

the VB.

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(3) Conduction in semiconductor

An electric field is applied to a semiconductor

=

(dV/dx) = -Ex

PE=(PE)i+ (PE)e

V= -Ax +B

(PE)e = - e V (x) = eAx-eB

All the energy level and thus energy

Under the action ofEx, the electron in the CB moves to the left and the holes in the

band must tilt up in thex direction.

7

.


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Since both electrons and holes contribute to electrical conduction, we may write the

current densityJ, from its definition, as,

dhde epvenvJ +=

,

VB, and vde and vdh are drift velocities of electrons and holes in response to anapplied electric fieldEx.

xede Ev=

xhdh Ev=

Drude model- free electron model in metal

e

ee

m

e =

The ideas on electron motion in metals can also be applied to the electron motion in

the CB of a semiconductor. We must, however, use an effective mass me* for the

electron in the cr stal rather than the mass m in free s ace. The effective mass

8

depends on the interaction of the electron with its environment within the

crystal.


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acceleration, that is inertia, there is no reason why the hole should not have a

mass.

*h

h

h m

e

=

The conductivity of a semiconductor is,

he epen +=

Where n and p are the electron and hole concentrations in the CB and VB,respectively. This is a general equation valid for all semiconductors.

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(4) Electron and hole concentrations

How to determine the e and h concentrations?he epen +=

(a) Energy band diagram.

(b) Density of states (number of states per unit energy per unit volume).

10

(c) Fermi-Dirac probability function (probability of occupancy of a state).

(d) The product ofg(E) andf(E) is the energy density of electrons in the CB (number of electrons per unit energyper unit volume). The area under nE(E) versus E is the electron concentration.


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gcb(E) as the density of states in the CB. The probability of finding an electron in

a state with energy E is given by the Fermi-Dirac f(E). Thus, The number of

c c ,

dEEEncE +

=

11

cE


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( ) ( )dEEfEgnc

c

E

Ecb

+

=

We assume that (Ec EF) >> kT,( i.e. EF is at least a few kTbelowEc) so,

( )

+=

kT

EEEf Fexp1

1

( )( )

=

kT

EEEf Fexp

The assum tion of >> kTmeans assumin that

E

For cthe number of electrons in CB is far less than the number

of states in this band.

electrons

f(E) decays rapidly with energy so thatgcb(E)f(E) 0near the top of the band. So we can take the upper limit

to be E = rather thanEc+.

12

f(E)( ) ( )dEEfEgn cE cb

=


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( ) ( )dEEfEgncE

cb

=

The density of statesg(E) is

2/3

228 E

h

Eg e

=

The energy of the bottom of

the band is zero in the

derivation.

2/3

( ) ( ) 2/12

2/1 )(28 ce EE

h

mEg

= For CB

2/3

( ) ( ) ( ) 2/122/128 EEhmEg ve

=For VB

13


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So,

( )( )

dEkT

EEEE

h

mn Fc

E

e

c

exp

28 2/13

2/3*

Which leads to, (detailed derivation see Appendix 1)

( )

=

kT

EENn Fcc exp

Where,

2/3*

22

= hmN ec

14

c s a - epen en cons an , ca e e e ec ve ens y o s a es a e e ge.


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Similarly, we can calculate the concentration of holes in VB.

The probability of the occupancy of the hole probability that

an electron is absent, isAssumption: (Ev

( ) ( )

=

+=

kT

EE

kT

EEEfF

Fexp

exp1

1

11

F) >> kT,( i.e. F is

at least a few kTabove Ev)

( )( )

=

vE Fv

e

dEkT

EE

EEh

m

p 02/1

3

2/3*

exp

28Why now use

Fermi-Dirac

statistics?

( )

=

kT

EENp vFv exp

2/3

2

*22

=h

kTmN hv

15


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( )

=

EENn Fcc exp

( )

=

EENn Fcc exp

( ) =

EEN vFex

( ) =

EEN vFex

It is interestin to consider the roduct n

TT

=

kTNNnp

g

vc exp

WhereEg =Ec-Ev is the bandgap energy. The right side of this eqn is a constant

that depends on T and the material properties, Eg, and not on the position of

the Fermi level.

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=

ENNnp

g

vc exp

This is a general equation that is valid as long as we have thermal equilibrium.

External excitation, such as photogeneration, is excluded. The product np is a

T-dependent constant.

If we somehow increase the electron concentration, then we inevitably reduce

the hole concentration.

An intrinsic semiconductor is a pure semiconductor crystal in which the electron

and hole concentrations are equal. By pure we mean virtually no impurities in. . , ,

which we can denote as ni, the intrinsic concentration.

Clearly in a pure semiconductor, electrons and holes are generated in pairs by

t erma exc tat on across t e an gap.

==

ENNnpn

g

vci exp2

17n p, extrinsic semiconductors.


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When an electron and hole meet in the crystal, they recombine. Recombination

.will be proportional to the number of electrons and also to the number of holes,

npR

The rate of generation G will depend on how many electrons are available for

excitation at v, that is, Nv; how many empty states are available at c, that is,

Nc: and the probability that the electron will make the transition, that is, exp(-

Eg/kT),

kT

ENNG

g

vc exp

In thermal equilibrium, we must have the rate of generation equal to the rate

of recombination, that is,

18

RG =


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( )

=

kT

EENn Fcc exp

( )

=

kT

EENn Fcc exp

( )

=

kT

EENp vFv exp

( )

=

kT

EENp vFv exp

==

kT

E

NNnnp

g

vci exp

2

The product ofnp is not dependent on position of Fermi level.

,

and hole concentrations. It serves as a mathematical crank to determine n

and p.

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Consider an intrinsic semiconductor, n = p = ni. We can solve for the Fermi

energy in the intrinsic semiconductor,E

Fi, that is,

( )( )

===

ENNnpn

EEN

g

vcivFi

v expexp2/1

= c

N11

vgvFi

N22

2/3*2 kTm e

+=

*

*

ln4

3

2

1

h

egvFi

m

mkTEEE

2

hc

2/3*

222

= hTmN hv

20


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+=*

*

ln31 e

gvFi

mkTEEE

h

IfNc = Nv or me* = mh*, then Fermi energy is in the middle of the bandgap.

gvFi 2/1+=

Normally, me* mh*, the Fermi energy is slightly shifted from midgap by an

* *

For Si and Ge, the hole effective mass is slightly greater than the electron

effective mass, so EFi is slightly above the midgap.

e h , g.

CB

c

Ev

EFiIntrinsic

21VB


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( )

=

EENn Fcc exp

( )

=

EENn Fcc exp

( ) =

EEN vFex

( )

=

EENp vFv exp

T

==

ENNnnp

gexp2

If we increase the electron concentration, by adding impurities into the Si.

n >

n-type semiconductor

2

innp =( ) =

EENn Fcc exp

( ) =

EENp vFv exp>

22vFFc EEEE


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vFFc EEEE


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If we increase the hole concentration, by adding impurities into the Si crystal

.

p-type semiconductor

t t erma equ r um

2

innp =( ) =

EENn Fcex

( ) =

EEN vFex

24

kT

vFFc

EEEE >

Ec

EFpEv


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What is the average energy of the electron in the conduction band of semiconductor?

,

dEEEECB =1

n CB

Where the integration must be over the CB. Substituting the proper expressions forcb c

of the band, the result is,

3c

2

Thus, an electron in the CB has an average energy of 3/2kT above Ec. Since we

25

know that an electron at c is free in the crystal, 3/2kTmust be its average kinetic

energy.


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FOav EE3

For metal

What is the reason for the difference between the metal and semiconductor?

The CB in a semiconductor is only scarcely populated by electrons, which meansthat there are many more electronic states than electrons and thus likelihood of

.

then neglect the Pauli exclusion principle. Boltzmann statistics.

,

comparable in magnitude. The description of electron statistics in a metal involves

the Fermi-Dirac function, which is based on Pauli exclusion principle. In a metal

the average energy of the conduction electron is 3/5 EF and, for all practical

purposes, Tindependent.

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2. Extrinsic semiconductors

(1) n-type doping

Small amounts of pentavalent (valence of 5) element from Group V, such as As, are

.

As+

e

When an As atom bonds with four Si atoms, it has one electron left unbonded.

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As+

e

,

ionizing the As impurity.

em e4

==

, ,

ho.

822

29


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eVem

E eb 6.1322

4

==o

If we wish to apply this to the electron around an As+ core in the Si crystal

env ronmen , we mus use ro ns ea o o, w ere r s e re a ve perm v y o

Si, and also the effective mass of the me* in Si crystal. Thus the binding energy is,

==

2

*

222

4* 16.13

8 re

e

or

eSi

bm

meV

h

emE

With r = 11.9 and me* 1/3me for Si, we find EbSi = 0.032 eV, which is

com arable with the avera e thermal ener of atomic vibrations at room

temperature, ~ 3kT(~0.07 eV).

So, the fifth valence electron can be readily freed by thermal vibration of the

30

Si lattice. Therefore it will be in CB.


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At 0KAt RT

The energy state introduced by doping As atom (Ed) is 0.032 eV belowEc.

As atom donates an electron into the CB, it is called donor atom.

Before doping, at thermal equilibrium, n =p = ni

After doping, ifNd is the donor atom concentration in the crystal, all the atoms

are ionized, so,

31

n = Nd + ni


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n = Nd + ni

Ifp does not change, then np > ni2, is it right?

No. At thermal equilibrium, we must obey np = ni2. How to obey?

Some of electrons in the CB will combine with holes to reduce the number of the

o e. n a so, e num er o e e ec rons n s a so re uce . owever, e

reduction of the number of the electrons should be less than ni.

i

N

np

2

=

d i, d,

The conductivity will then be,

32

edh

d

ied eN

N

neeN

+=


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At low temperatures, however, not all the donors will be ionized and we need to

.

fd(Ed), of finding an electron in a state with energy Ed at a donor. This probability

unc on s s m ar o e erm - rac unc on d excep a as a ac or o

multiplying the exponential term,

( )( )

+

=

kT

EEEf

Fd

dd

exp

2

11

1

Thus, the number of ionized donors at a temperature T is given by, Nd+ = Nd x

(probability of not finding an electron atEd) =Nd[1-fd(Ed)]. Thus,

( ) =+

EE

NN

dF

dd

33

kT


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(2) p-type doping

- ype sem con uc or s cause y op ng w sma amoun s o r va en a omsuch as B, Al etc..

When B shares its three electrons with four Si atoms, one of the bonds has a

missing electron, which of course is a hole.

A nearby electron can tunnel into this hole and displace the hole further away from

the boron atom.

The binding energy of this hole to the B- ion can be calculated as ~0.05 eV, so at

room temperature the thermal vibrations of the lattice can free the hole away from

the B- site.

Bh+

B

34

Free

(a)(b)


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Batomsites every 106Si atomsElectron energy

intocrystal

Ev

Ea~0.05eVh

+

VB

At 0 K At RT

The escape of the hole from the B- site involves the B atom accepting an

~ . v.

.

The B atom introduced into the Si crystal therefore acts as an electron

35

, , .


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If the concentration of acceptor impuritiesNa in the crystal is much greater than the

ni, then at room temperature all the acceptors would have been ionized and thus p

a.

The electron concentration is then determined by the mass action law, n = ni2/Na,

w c s muc sma er an p, an consequen y e con uc v y s s mp y g ven

by = eNa

h.

Examples of donor and acceptor ionization energies (eV) in Si

Donors Acceptors

P As Sb B Al Ga

0.045 0.032 0.039 0.045 0.057 0.072

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(3) Compensation doping

Compensation doping -the doping of a semiconductor with both donors and

What happens when a semiconductor contains both donors and acceptors?

acceptors to control the properties.

For example, ap-type semiconductor doped withNa acceptors can be converted to

- dexceedsNa.

When both acceptors and donors are present, the mass action law is obeyed.

np = ni2

(1) More donors:NdNa >> ni, n = (NdNa) andp = ni2

/(NdNa).

(2) More acceptors:Na Nd>> ni,p = (Na Nd) and n = ni2/(Na Nd).

37

These arguments assume that the T is sufficiently high for donors and acceptors to

have been ionized.


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3. Temperature dependence of conductivity

(1) Carrier concentration temperature dependence

Consider an n-type semiconductor doped withNddonors whereNd>> ni.

The dependence of the electron concentration on temperature thus has three regions:

CB

T


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CB

T


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Ts ni, This condition remains unchanged

until T= Ti when ni, which is Tdependent, becomes equal to Nd.

This temperature range is the range utilizing the doping properties of semiconductorin pn junction device application and is often referred to as the extrinsic range.

40

T > T


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T> Ti

As+EF

As+ As+ As+

(c) T= T3

3. High-temperature range (T > Ti). The concentration of electrons ni generated by

thermal excitation across the bandgap is now much larger than Nd, so the

.

intrinsic range

E2/1

=kT

n vc2

exp

41

31


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2

3

2

*22

=kTm

N ec

( ) ( )31

4

3

2

1

TNc

42v

Exponential part dominates

( )

kT

Tng

2

exp2over or

( )

E

Tn exp4

3

n = Nd

42The temperature dependence of the electron concentration in an n-type semiconductor.


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The slope of the straight line is

The graph can also be used to find

whether the dopant concentration at

given temperature is more than the

intrinsic concentration.

43

The temperature dependence

of the intrinsic concentration


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(2) Drift mobility: Temperature and impurity dependence

The temperature dependence of the drift mobility follows two distinctly different

temperature variations.

In the high-temperature region, it is observed that the drift mobility is limitedby scattering from lattice vibrations.

At low temperatures the lattice vibrations are not sufficiently strong to be the

major limitation to the mobility of the electrons. The scattering of electrons by

ionized impurities is the major mobility limiting mechanism.

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At the high temperature region (lattice scattering dominates):

*

eme =

sthNSv

1=

2

al=u

Where Sis the cross-sectional area of the scatter;

vth is the mean speed of the electrons, called thethermal velocity; andNs is the number of scatters

u

For metal,

.

ElectronTa 2 TS

T

1

FOav EE53 vth is insensitive to T.

45

T

1


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*

e = 1=

e sthv

For semiconductor:

Ta 2 TS

kTvm the22

= 2/1Tv th

L ,

11

( ) ( )( )2/12

=TTNva sth

L

TL

46

L s a a ce v ra on sca er ng m e mo y.

At l t t tt i f l t b i i d d i iti d i t


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At low temperatures, scattering of electrons by ionized donor impurities dominates.

KE> |PE|e

KE= 1/2mev2

KE - |PE|

e o an e ectron at a stance rfrom an As+ ion is due to the Coulombic

attraction, is

As+

c

r

e

PE ro4

2

=

KE< |PE| Ifr> rc, it has not been scattered.

Ifr= rc,KE |PE|, so,

( )cro

cr

erPEkTKE

42

3 2===

From which rc = e2/(6orkT). The scattering cross section S = rc2 is thus given by,

4

47( )

2

26

= TkTr

Scro

22/1

T


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2 TS2/1

Tv th

T 2/311 T2/3

( )( ) IIIthI

NNTTNSv 2/12=

I

IN

I is the ionized impurity scattering limited mobility, which decreases withincreasing ionized dopant concentration NI, which itself may be temperature

dependent. Indeed, at the lowest temperatures, below the Ts, NI will be stronglyTdependent because not all the donors would have been fully ionized.

The overall temperature dependence of the drift mobility is:

LIe

111+=

48

(3) Conductivity temperature dependence


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( ) y p p

log()Semiconductorit

y

een =

IntrinsicMetal

ale R

es

istiv ,

in the ionization range,

2/1

Extrinsic

ithmicSc

=kT

NNn dc2

exp2

T2/3

Ionization

T -3/2 T3/2

Lattice

scatteringLoga

I

IN

scattering

1/TLow TemperatureHighTemperature

.

49Schematic illustration of the temperature dependence of

electrical conductivity for a n-type semiconductor.


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een =

log()Semiconductor

istivity

At the extrinsic range,

log(n) T

Metal

icSca

le Re

dNn =

Extrinsic

IonizationLattice

scatterino

garithm The conductivity follows the

temperature dependence of

drift mobility.

log()T -3/2 T3/2

Impurity

scattering


50


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( )en +=

log()Semiconductor

stivity

At the high temperature, in

the intrinsic ran e

log(n)

Intrinsic

T

Metal

Scale R

esi

( )

=

kT

ENNn

g

vci2

exp2/1

Extrinsic

IonizationLattice

g

arithmi

2/3 TL

log()T -3/2 T3/2

Impurity

scattering

L

Concentration dominates.


51

(4) Degenerate and nondegenerate semiconductors


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(4) Degenerate and nondegenerate semiconductors

( )

=

kT

EENn Fcc exp

Assumption: Ec >> EF (Ec is at least several kTaboveEF).

Fermi-Dirac statistics Boltzmann statistics.e num er o s a es n e ar excee s e num er o e ec rons ere, so e

likelihood of two electrons trying to occupy the same state is almost nil and Pauli

exclusion principle is neglected.

Those semiconductors for which n n, the above discussion is valid.

sem con uctors. ey essent a y o ow a t e scuss ons a ove an ex t a

the normal semiconductor properties outlined above.

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When the semiconductor has been excessively doped with donors, then n may be so19 20 -3, ,

c

.

In that case the Pauli exclusion principle becomes important in the electron statistics.

Fermi-Dirac statistic is needed. Such a semiconductor exhibits properties that are

more metal-like than semiconductor-like. Semiconductors that have n >Nc orp >Nv

are called degenerate semiconductors.

Degenerate n-type semiconductor. Large number ofdonors form a band that overlaps the CB.

53

egenera e sem con uc ors as many mpor an app ca ons suc as aser

diode, zener diode, ohmic contacts in IC and as metal gates in many

microelectronic MOS device.

4 Recombination and Minority Carrier Injection


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4. Recombination and Minority Carrier Injection

Recombination: a free electron in CB is returned into a hole in VB.

(1) Direct and indirect recombination

Energy conservation: The excess energy of the electron is lost as a photon ofenergy hv =Eg.

Momentum conservation: , are the wavenumbers of

electrons in CB and VB,vbcbvbcb kkkk ,,hh =

Direction recombination occurs when both

energy and momentum conservation are satisfied.

(LEDs) is the direct recombination.

Indirection recombination: the momentum

54

.

process is through a third body (recombination

center) such as impurity, defects..

F h l l i d Si d G k k i i h i th iddl f th


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For the elemental semiconductor, Si and Ge, kvb = kcb is right in the middle of the

. ,

can satisfy kvb = kcb, and so direct recombination in Si and Ge is impossible.

For some compound semiconductors, such as GaAs and InSb, the states with kvb

= kcb are right at the top of the VB. Consequently, an electron in the CB of GaAscan drop down to an empty electronic state at the top of the VB and maintain

vb cb . .

55

Indirect recombination


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The recombination center provides a localized electronic state below Ec in the

bandgap. When an electron approaches the center, it is captured. The electron is

then localized and bound to this center and waits there for a hole with which it can

recombine. In this recombination process, the energy of the electron is usually lost to

lattice vibrations (as sound) . Emitted lattice vibrations are called phonons.

CBc

Recombinationcenter

Phonons

Ev

Er Er Er

(a) Recombination

56

Charge carrier trapping


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Trappingcenter

Et

CB

EcEt Et

VBv

(b) Trapping

The electron falls into the trapping center at Et and becomes temporarily removed

from the CB. At a later time, due to an incident energetic lattice vibration, it becomes

exc e ac n o e an s ava a e or con uc on aga n. us rapp ng

involves the temporary removal of the electron from the CB, whereas in the

case of recombination, the electron is permanently removed from the CB since

the capture is followed by recombination with a hole.

In general, defects that give localized states near the middle of the bandgap tend to

act as recombination center.

57

(2) Minority carrier lifetime


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( ) y

In an n-type semiconductor, electrons are the majority carriers and hole are the

minority carriers:

nno is defined as the majority carrier concentration (electron concentration in

an n-type semiconductor) in thermal equilibrium in the dark. These electrons,constituting the majority carriers, are thermally ionized from the donors.

pno is termed the minority carrier concentration (hole concentration in an n-

type semiconductor) in thermal equilibrium in the dark. These holes that

In both cases the subscript no refers to an n-type semiconductor and thermal

.

equ r um con ons, respec ve y. erma equ r um means a e mass

action law is obeyed and nnopno = ni2.

58

no no i


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uppose a e e ec ron an o e concen ra ons a any ns an are eno e y nnandpn, which are defined as the instantaneous majority (electron) and minority (hole)

concentrations, respectively.

nn is the excess electron (majority carrier) concentration: nn = nn-nno is the excess hole minorit carrier concentration: = - .

59

Photoexcitation creates electron-hole pairs (EHPs) or an equal number of


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p ( ) q

electrons and holes, So pn = nn, and obviously the mass action law is notobeyed: nnpn ni2.

nddn nn =pddp nn =

tt tt

Since nno andpno depend only on temperature.

Let us assume aweak illumination, which causes only a 10 percent change in nno,

(nno = 5x1016 andpno = 4.5x10

3), that is,

.. == cmnn non

316105.0 == cmnp nn

Under illumination, the minority carrier concentration is

nnnon pcmppp =+=+= 163163 105.0105.0105.4

60That is, pn pn, which shows that although nn changes by only 10 percent, pnchange drastically, that is, by a factor of ~ 1012.


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when the li ht is switched off, the condition = must eventuall revert back tothe dark case wherepn = pno.

In other words, the excess minority carriers pn and excess majority carriersnn must e remove . s remova occurs y recom nat on. xcess o esrecombine with the electrons available and disappear.

61

How long will the recombination process take?


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n or er o escr e e ra e o recom na on, we n ro uce a empora quan y,denoted by h and called the minority carrier lifetime (mean recombination time).

h is the average time a hole exists in the VB from its generation to itsrecombination, that is, the mean time the hole is free before recombining with anelectron.

If the minority carrier recombination time is 10 s, and if there are some 1000 excess

holes, then it is clear that these excess holes will be disappearing at a rate of1000/10s = 100 per second.

The rate of recombination of excess minority carriers is simply pn/h. IfGph is therate of hoto eneration then clearl the net rate of chan e of is

nph

n pGpd

=

62

h


63/68

h

nph

n Gdt =

This is a general expression that describes the time evolution of the excess minoritycarrier concentration given the photogeneration rate Gph, the minority carrier lifetime

=h, . nnno).

, ,crystal defects, temperature and so forth.

junction, others require longer recombination time e.g. persistent luminescence.

63

5. Diffusion and conduction equations


64/68

Diffusion : the concentration gradient of electrons results in a net diffusion motion of

the electrons in the direction of decreasing the concentration. This is described by Ficks

law:

2

and;l

Ddx

dnD eee ==

e = electronsflux, De = diffusion coefficient ofelectrons, dn/dx = electrons n

concentration gradient. l=mean free path, =mean free time between scattering events

dx

dneDeJ eeeD ==,

D, e = electric current density due to electron diffusion,

Similarly, the electric current caused by the hole diffusion is :

64

dx

dpeDeJ hhhD ==,

Electric field electron drift and


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Electric field- electron drift and

hole drift

Concentration gradient - electron

.

Charge motion involves bothdrift and diffusion

The total current density due to the electrons drifting, driven by Ex and also

diffusion driven by dn/dx, is given by

dx

dneDenJ exee += E

dp

Similarly, the total current density due to the hole drifting driven by x and

diffusion driven by dp/dx, is given by

65

dxeep hxhh =

Diffusion coefficient is a measure of the ease with which the diffusion charge

carriers move in the medium.


66/68

move in the medium.

Althou h one is driven b concentration radient and the other is driven b

electric field, the diffusion coefficient and drift mobility are related by

Einstein relation:

kTDh =kTDe =

he

De = diffusion coefficient of electrons, e = electron drift, Dh = diffusion coefficient

of the holes, h = hole drift mobility

66

Appendix 1: derivation of the electron


67/68

:FunctionGamma

( ) ( ) ( )

=

=+=

2/3*

0

1

2

1;1;dtte t

=

3

,'

exp

SoEEt

dEkT

Eh

n

c

E

Fc

e

c

( )

+

2/3*

0

2/1

3

2/3*

'

'exp'28

dtkT

EEtt

h

mn Fce

0

2/1

3

'exp'exp dtkT

tkTh

Fce

67

2/3*

/'= kTtt


68/68

2/3*

) ( ) 2/12/32/3*03

28

'expexp

EEm

dtkTttkTkThn

tFce

Fce

2/1

03

113=

=

=

dtet

kTh

t

( )( )

( )2/33

2/3*

0

exp28

2222

EEkT

mn Fce

( )2/3

2

*

exp2

2

kT

EE

h

kTm Fce

( ) 2/32

*22where;exp

=

=

h

kTmN

kT

EENn ec

Fcc

68

Documents

Chapter 3-1(II 2008-2009) [Compatibility Mode]