Engineering Physics 2 Unit-2

7/29/2019 Engineering Physics 2 Unit-2

1/94

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it

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it

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it

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it2222

1

Materials

Materials

Materials

Materials


2/94

BasicS

emicondu

ctorCrys

talStruct

ure


3/94

INTROD

UCTION

Material

having

electrical

resistiv

ity

(or)

ele

ctrical

conductivityinbetween

aconductorandaninsulato

r.

(or)

Solidwhichhastheen

ergybandsimilartothatofan

insulator.

Itactsasaninsulatoratab

solutezeroandasa

conductorathightem

peraturesandintheprese

nceof

impurities

.

Examples:

Si,Ge,GaA

s,InP,GaPetc


4/94


5/94

Resistivityvariesfrom10-4to106m.(Conductor&Insulator).

At0K,it

behavesasinsulators.

At0Ktheyhaveempty

conductionbandandalmostfilled

valenceband.

PropertiesofSemicond

uctors

Bothelectronsandholes

arechargecarriers.

Haveneg

ativetemper

aturecoefficient

ofresistance.

=

(e+h)

1 T T

or

L NM M M

O QP P P


6/94

Band

Structure

s


7/94

Band

Structures

Plentyoffreee-s

Veryfew(or)no

freee-s

Fewfree

e-s

OverlappingVB&CB

La

rgeE

g(7eV)

NarrowEg(1eV)

Verysmallele

ctricfield

forconduction.

Verylargeelectric

field

forconduction.

Smallelectric

fieldfor

conduction.

+TCR(Temp

Co-effiof

Resistan

ce)

-TCR

-TCR


8/94


9/94


10/94

Semicon

ductor

Semicon

ductor

Semicon

ductor

Semicon

ductor

omosition

omosition

omosition

omosition

PuritPuritPuritPurit

Classif

ication

Classif

ication

Classif

ication

Classif

ication

10


11/94

ElementalS.C

Singleele

ment.(Ge,Si)

GroupIV

elements.

CompoundS

.C

Two

(or)moreelements.

(GaAS

,InP)

III-V

(or)II-VIelem

ents.

Composition

Composition

Composition

Composition

ovaent

on

ueto

bounding

ofneighboring

atomswith4valencee-s).

Indirectb

andgap

semiconductors.

(recombinationofe-sandho

les

takeplace

throughtraps)

(due

todifferencein

electro

negativity).

Directbandgap

semiconductors.

(reco

mbinationofe-sand

holestakeplacedire

ctly)


12/94

Elemen

talS.C

Heatisproducedduring

recombina

tion.

Lifetimeofchargecarriersis

C

ompoundS.C

Photonsareproducedduring

recom

bination.

Lifetimeofchargeca

rriersis

Composition

Composition

Composition

Composition

more.

Diode,tra

nsistor,etc.,

less.

LED,Laserdiodes,etc.,


13/94

TypesofSem

icon

ductors

TypesofSem

icon

ductors

TypesofSem

icon

ductors

TypesofSem

icon

ductors

Puresem

iconductorwithoutanyimpurities-Intrin

sicS.C.

Eg:Si,Ge,etc.In

trinsicS

emiconductor

Extrinsic

Semicondu

ctor

Impurefo

rmofsemiconductors-Extrin

sicS.C

*N-type:

(IntrinsicS.C

+

Pentavalentatoms(P,Ar,An,etc.,)

).

*P-type:(IntrinsicS.C

+Trivalentatom

s(B,A

l,Ga,etc.,

)).


14/94

IntrinsicSemiconducto

r

Semic

onductorsGe,S

ihavecrystallinestructurean

dfour

valenceelectronsino

utermostshell(tetravalent).

AtomicnumberofGe

is32.

(1s22s22

p63s23p

64s23

d104p2).28

electronstightlyboun

dand4electron

srevolveinoutermostorbit.

AtomicnumberofSiis14.

(1s22s22

63s23p2).10electrons

tightlyboundand4electronsrevolve

inoutermostorbit.

When

twosemicondu

ctingatomsare

broughttogeth

er,each

positivecoreattracts

thevalenceelectronfromothe

ratomand

Soeachelectronissharedbytwoatoms.


15/94

Intrinsicsemi

conductor

Acovale

ntbondisform

edbetweenthe

2electronsofa

nytwo

atoms.Sosuchelectronpairsarenotav

ailableforconduction.

At0K,a

puresemiconductoractsasaninsulator.Energ

yneedto

breaksuchacovalentbo

ndis0.72eVfo

rGeand1.1eV

forSi.

Whenelectronescapesfromcovalentbond,anemptyspaceis

.

Onceah

oleiscreated,anelectronfromthecovalentb

ondofa

nearby

atomshiftsto

occupythishole.

Thisprocesscontinuesandtheholegoesonshiftingfromone

atomtoanother.


16/94

Intrinsicsem

iconductor..

Holea

ctslikeafreep

ositivecharge.

So,sem

iconductorhas

bothholesand

electronsascharge

carr

iers.

Dueto

thermalexcita

tion,whenelectronsaretransf

erredfrom

thev

alencebandto

theconduction

bandaneual

numberof

hole

sarecreatedin

thevalenceban

d.

Fermilevelatthemiddle

ofthegapindicatesthatthenu

mberof

electronsa

ndholesareequalinintrinsicsemiconductor.


17/94


18/94

Structure&

Latticeofa

'normal'purecrystalofSilicon.


19/94


20/94

E

xtrinsicSemiconductor

Processofaddingimpuri

tyatomstose

miconductoratoms(the

orderof1im

purityatomp

er10million(ormore)atom

softhe

semiconductor)iscalledDo

ping.(theratioofabout105:1).

Impurefo

rmofsemiconductors-Extrin

sicS.C

*N-type:

(IntrinsicS.C

+

Pentavalentatoms(P,Ar,An,etc.,)

).

*P-type:(IntrinsicS.C

+Trivalentatom

s(B,A

l,Ga,etc.,

)).


21/94

Arsenic,A

ntimonyorPh

osphorusatom

s-fivee-s-o

utermost

valanceband

share

with

otheratoms-"Pent

avalent"

impuritie

s.

Fourof

thefivee-s-bondwithits

neighbouringS

iatoms-

leavingone"freeelectron"tomo

ve-whenan

electrical

voltaeisa

lied.

N-typeSemicondu

ctor.

Eachimp

urityatom"do

nates"onee--

"Donors".

Antimony

-51e-s-5shellsaroundthenucleus-5valancee-s.

Excessof

currentcarryinge-s-withanegativecharge

"N-type"

material-e-sa

s"MajorityCa

rriers"&resultantholes

"Minority

Carriers".


22/94

Structure&

Latticeofthedonorimpu

rityatomAntimony.


23/94


24/94

MaterialswhichhavePentavalentimpu

rityatoms(donors)and

conductby"electron"m

ovement-N-ty

peSemiconduc

tors.

Inthesetypesofmaterialsare:

1

.Donorsarepentavalentatom

s.

2

.Therearealargenumberoffr

eeelectrons.

3

.Asmallnumberofholesinrelationtothenum

berof

Su

mmary-N-type(Antim

ony)

freeelec

trons.

4

.Dopinggives

:

negativelychargedfree

electrons.

5

.Supplyofene

rgygives:

negativelychargedfree

electrons.

positive

lychargedhole

s.


25/94

P-Type

Semiconductor

"Trivalen

t"(3e-)impurity-Aluminium,BoronorIndium.

Only3valencee-s-available-fourthco

valentbondcan

notbe

formed.

Complete

connectionis

notpossib

le

-

giving

rise

-

abundanceofpositively

chargedcarriers-holes.

Adjoining

freee--attrac

ted-willtryto

moveintothehole.

e-fillingtheholeleavesanotherholebehindit.

Thisintu

rnattractsanothere-wh

ichcreatesanother

hole.


26/94

Holesaremovingasapositivecharge.

Eachimp

urityatomgen

eratesahole-

"Acceptors"-asthey

arecontinually"accepting"extraelectro

ns.

Boron-trivalentadditive-5e-s-2sh

ellsaroundnucleus-3

-

Addition

ofBoroncaus

esconduction

-mainlyofpositive

chargeca

rriers-"P-type"material.

Positiveholes-"Majo

rityCarriers"&

freeelectrons-

"Minority

Carriers".


27/94

Structure&

Latticeoftheacceptorim

purityatom

Boron.


28/94


29/94

S

ummary-P

-type(Boro

n)

Materialsw

hichhaveTrivalentimpuritya

toms(acceptors)and

conductby

"hole"movem

entandarecalled,P-typeSemiconds.

Inthesetypesofmaterials

are:

1.Acceptorsaren

egativelycharg

ed.

(Readilyacc

epts).

2.Therearealargenumberofholes.

3.Asmallnumber

offreeelectron

sinrelationto

the

num

ero

oes.

4.Dopinggives:

negativelychargedacceptors.

positivelychargedholes.

5.Supplyofenerg

ygives:

positivelychargedholes.

negatively

chargedfreeelectrons.


30/94

Intrin

sic

Extrinsic

Extrinsic

Ntype

P-type

P

entavalent

Trivalent

Donoratom

Acceptoratom

Donatese-s

Acceptse-s


31/94

CarrierC

oncentration

inIntrinsicSemiconductor

CarrierConcentration


CarrierC

oncentration


CarrierConcentration


At0K,i

ntrinsicsemico

nductorbehavesasinsulator,andiftemp.

increase

d,e-sfromvale

ncebandexcitedintothecond

uctionband.

Number

ofchargecarriersperunitvo

lumeofthematerialis

calledCa

rrierConcentration.

esee

ectronseave

ereeparc

e.

Holescreatedbythese

e-sinthevalencebandalsobe

havelikea

freeparticle.

So,massoftheelectron

andholeisreplacedbyitseffectivemass

me*a

ndmh

*respectiv

ely.


32/94

CarrierCo

ncentrationinIntrins

icSemiconductor

CarrierCo


icSemiconductor

CarrierCo


icSemiconductor

CarrierCo


icSemiconductor


33/94

Densit

yofelectronsinconduc

tionband

Numberofc

hargecarriersperunitvolume

ofthematerial

(i.e.,

intrinsicsem

iconductor)isk

nownasCarrie

rConcentration.

(

)

(

)

dn

ZE

dEFE

=

1

NumberofelectronsinCB-integratingfrom

Ec

to

Pa

ge1.1

9eqn.

(12)

Densityofsta

tesZ(E)dEofelectronsintheconductionband

is

(

)

()

cE

n

dn

ZEFEd

E

=

=

2

*

3/2

1/2

3

4

(

)

(2

)e

ZEdE

m

E

dE

h=

3

me-me*

Page1.2

0

P

age1.1

9eqn.

(11)


34/94

Ec

Lowesten

ergy

levelofCB

.

Energy(E)forane-inCBequalto

(E-Ec).


35/94

Densityo

felectronsinC

B

Eqn.

3becom

es

*

3/2

1/2

3

4

(

)

(2

)

(

)

e

c

ZEdE

m

E

E

dE

h

=

4

WeknowF(E)

1

(

)

1

exp(

)F

F

E

E

EKT

=

+

5

E=(E-Ec).

Su

b.eqn.4and5ineqn2

()

(

)

cE

n

dn

ZEFEdE

=

=

=

1

/2

*

3/2

3

(

)

4

(2

)

1

exp(

)

c

c

e

F

E

B

E

E

dE

m

E

E

h

K

T

+

6

*

3/2

1

/2

3

4

(

2

)

(

)

1

exp(

)

c

e

c F

E

B

m

E

E

dE

h

E

EKT

+

n=


36/94

F

B

E

E

KT

1

F

B

E

EKT

exp

1

FB

E

EKT

1e

xp

e

xp

F

F

B

B

E

E

E

E

KT

KT

+

=

Foralltem

peratures,theen

ergyrequiredbyanelectrontomove

fromthev

alencebandto

theconduction

bandisalwaysgreater

than4KB

T.

1

1

exp

1

exp

exp

F

B

F

F

B

B

E

E

K

T

E

E

E

E

K

T

K

T

=

=

=

+

Eqn.6becomes

*

3/2

1/2

3

4

(2

)

(

)

exp

c

F

e

c

B

E

E

E

n

m

E

E

dE

h

KT

=

e

xp

FB

E

EKT


37/94

Eqn.6

becomes

*

3/2

1/2

3

4

(2

)

exp

(

)

ex

p

c

F

e

c

B

B

E

E

E

n

m

E

E

dE

h

KT

KT

=

7

Tosolveeqn7

theintegrallimitsalsogetchang

edaccordinglyasfollows

c

E

E

x

=

+

c

E

Ex

=

Letusassum

e

dEd

x

=

c

c

E

E

x

=

cEx

=

case:1

case:2

c

W

hen

E

E=

whenE=

0

x=

x=

(Lower

limi

t)

(Upper

limit)

Eqn.

7becomes

*

3/2

1/2

3

0

4

(2

)

exp

(

)

(

)

exp

(

)

C

F

e

B

B

E

x

E

n

m

x

dx

h

KT

KT

+

=


38/94

/

1/2

3/2

0

(

)

2

B

xK

T

B

xe

dx

KT

=

*

3/2

3/2

4

F

c

E

E

=

Usinggammafunction

Therefore,

9

/

*

3/2

1/

2

3

0

4

(2

)

exp

B

xKT

F

c

e

B

E

E

n

m

x

e

dx

h

KT

=

8

3

2

e

B

h

KT

Eqn.

(10)-

densityofelectronsintheconduction

band.

3/2

* 2

2

2

exp

e

B

F

C

B

mK

T

E

E

n

h

K

T

=

10


39/94

Asthep

resenceoftheholecan

be

regarde

dastheabsenc

eofanelectro

n.

Den

sityofHole

sinValanceband

Numberofholesperunit

volumeintheV

Bofasemicon

ductor

canbefou

ndinasimilarmanner.

F(E)repre

sentstheproba

bilityoffilled

state.

Asthemaximumprobab

ilitywillbe1,the

probabilityofunfilledstateswill

be1-F(E)


40/94

CarrierConcentrat

ioninIntrinsicSemiconductor

CarrierConcentrat


CarrierConcentrat


CarrierConcentrat



41/94

1

1

1

exp

F

B

E

EKT

=

+

F

B

E

E

KT

Here

1

()

FE

exp

1

exp

F

B

F

B

EEKT E

E

K

T

=

+

1

F

E

E

exp

1

F

E

E

B

1

exp

1

F

B

E

E

K

T

+

=

1

(

)

exp

F

B

E

E

F

E

KT

=


42/94

(

)[1

(

)]

dp

ZE

F

E

dE

=

*

3/2

1/2

3

4

(2

)(

)

exp

F

h

EE

dp

m

E

dE

h

KT

=

Takingthe

effectivemass

ofholesasmh*

Numbe

rofholesinth

eenergyintervalEandE+dE

intheVBis

11 12

*

3/2

1/2

3

4

(2

)

(

)

exp

F

h

v

B

EE

dp

m

E

E

dE

h

KT

=

*

3/2

1/2

3

4

(2

)

(

)

exp

vE

F

h

v

B

E

E

p

dp

m

E

E

dE

h

KT

=

=

Energy(E

)foranholesinVBequalto(Ev-E).

HolesinVB

calculated-inte

grating-

toEv

13


43/94

En

ergy(E)foran

holesinVBequalto(Ev-E).

Ec

Lowestenergy

levelofCB.

Ev

Higheste

nergy

levelofVB.


44/94

v

E

E

x

=

E=

Letusconside

r

Tosolveeqn.(14)

vE

Ex

=

dE

d

x

=

*

3/2

1/2

3

4

(2

)

exp

(

)

ex

p

vE

F

h

v

B

B

E

E

p

m

E

E

dE

h

KT

KT

=

14

v

v

v

E

E

x

=

(

)

v vE

x

E

x

=

+

=

ase:

x

=

(Lo

wer

limit)

0

x=

(Upper

limit)

Eqn.

(14)becomes

0

*

3/2

1/2

3

4

(2

)

exp

e

xp

(

)

v

F

h

B

B

E

x

E

p

m

x

dx

h

KT

KT

=


45/94

0

*

3/2

1/2

3

4

(2

)

exp

ex

p

(

)

v

F

h

B

B

E

x

E

p

m

x

dx

h

KT

KT

=

0

*

3/2

1/2

3

4

(2

)

exp

exp

(

)

v

F

h

B

B

E

E

x

m

x

dx

h

KT

KT

=

Toremov

evesignintegrallimitsis

changed

/

*

3/2

1/2

3

0

4

(2

)

ex

p

B

xKT

v

F

h

B

E

E

m

xe

dx

h

KT

=

1/2

/

3/2

0

(

)

2

xkT

B

x

e

dx

KT

=

Using

gammafu

nction


46/94

* 2

2

2

exp

h

B

V

F

B

mK

T

E

E

P

h

K

T

=

aboveequationcanbesolve

dwiththehelpofgammafunctionand

then

15

qn.

-nu

m

ero

oes

orvacance

sn

evaence

an.


47/94

e-sinCB

HolesinVB

F(E)dE

1-F(E)dE

(E

-EF

)>>K

BT

(

E-EF

)

1

exp

exp

a

F

a

F

B

B

E

E

EE

KT

KT

+

=

exp

1

a

F

B

E

E

K

T

>>


76/94

(

)

exp

F

a

a

B

E

E

F

E

K

T

=

exp

F

a

a

B

E

E

nN

KT

=

Thereforeeqn(2)

becomes

Atequilibrium

(

)

a

a

nN

FE

=

3

Densityofh

olesin

VB(eqn.1)

Densityofelectronsinacceptor

energylevel.

(eqn.

3)

=

3/2

* 2

2

2

exp

exp

V

f

h

B

F

a

a

B

B

E

E

m

KT

E

E

N

h

KT

KT

=


77/94

3/2

* 2

2

2

exp

exp

h

B

V

F

F

a

a

B

B

m

KT

E

E

E

E

N

h

KT

KT

=

(

)/

3/2

(

)/

* 2

2

2

V

F

B

F

a

B

E

E

KT

a

E

E

KT

h

B

N

e e

mKT

h

=

((

)

2

)/

3/2

* 2

2

2

V

a

F

B

E

E

E

KT

ah

B

N

e

mK

T

h

+

=

(

)/

3/2

* 2

22

V

F

a

F

B

E

E

E

E

KT

ah

B

N

e

m

KT

h

+

=


78/94

Takinglogonbothsides

3/2

* 2

(

)

2

log

2

2

V

a

F

a

B

h

B

E

E

E

N

KT

m

KTh

+

=

3/2

* 2

2

(

)

log

2

2

a

F

V

a

B

h

B

N

E

E

E

K

T

m

KT

h

=

+

(

)

2V

a

F

E

E

E

+

=

3/2

* 2

(

)

log

2

2

2

2

V

a

a

B

F

h

B

E

E

N

KT

E

m

KT

h

+

=

0

,

0,

AtK

whenT

=


79/94

Expressionfor

Expressionfor

Expressionfor

Expressionforp

p

p

p

ininininVBVBVBVBintermsof

intermsof

intermsof

intermsofNNNNaaaa

Whent

empincreases,

more&moreacceptoratomsionize.

Whentempfurtherincreases,allacceptoratomsionized.

Furtherincreasesintemp,

-

Asare

sult,fermilevelmoves

toward

sintrinsicferm

ilevel.

a

.

Thusat

highertemp,intrinsic

behaviourismaintaine

d.


80/94

Substituting

valueofE

Fineqn1

1

3

/2

* 2

2

2

exp

h

B

V

F

B

mKT

E

E

n

h

KT

=

3/2

* 2

(

)

log

2

2

2

2

V

a

a

B

F

h

B

E

E

N

KT

E

mK

T

h

+

=

3/2

*

3/2

2

* 2

log

2

2

2

2

2

2

xp

V

a

a

B

V

e

h

B

h

B

B

E

E

N

KT

E

mK

T

h

mKT

p

e

h

K

T

+

=


81/94

3/2

*

3/2

2

* 2

log

2

2

2

2

2

2

xp

V

a

a

B

V

e

h

B

h

B

B

E

E

N

KT

E

m

KT

h

mK

T

p

e

h

K

T

+

+

=

3/2

*

3/2

2

1

2

xp

log

h

B

V

a

a

e

mK

T

E

E

N

p

e

=

+

3/2

*

3/4

2

* 2

2

2

xp

exp

log

2

2

2

a

h

B

V

a

e

B

h

B

N

mK

T

E

E

p

e

h

K

T

m

KT

h

=

2

2

2

B

h

B

m

KT

h


82/94

3/2

*

3/4

2

* 2

2

2

x

p

2

2

2

a

h

B

V

a

B

h

B

N

m

KT

E

E

p

e

h

KT

m

KT

h

=

3/4

* 2

2

2

xp

2

h

B

V

a

a

B

mK

T

E

E

n

N

e

h

KT

=

E-E

=In

iz

in

nr

fA

r

E=Amou

ntofenergyrequiredtotransfe

rane-fromEvtoEa.

3/4

*

2

2

2

2

BEKT

h

B

a

m

KT

n

N

e

h

=

Eqnisvalid

onlyatlowtemperatures.Butathightemp,wemust

taketheintrinsiccarrierc

oncentrationalongwiththis.


83/94

Variationof

Variationof

Variationof

VariationofEEEEFFFFwithwithwithwithtemp

andan

dan

dan

dNNNN

aaaa


84/94

Variat

ionof

Variat

ionof

Variat

ionof

Variat

ionofnnnneeee&&&&

nnnnhhhhwithwithwithwith

temptemptemptemp


85/94

Variationof

Variationof

Variationof

Variationofelectrical

electrical

electrical

electricalcond

cond

cond

condwithwithwithwithte

mp

te

mp

te

mp

te

mp


86/94

HallEffect

Measurem

entofconductivitywillnotdeterminewhetherthe

conduct

ionisduetoelectronorholesan

dthereforewilln

ot

distingu

ishbetweenp-ty

peandn-typese

miconductor.

typesofcarriersandtheirc

arrierdensitiesan

disusedtodetermine

themobilityofchargecarriers.

Halleffe

ctwasexplained

by

E.Hallin

1879.


87/94

H

allEffect

Whencon

ductor(metalo

rsemiconductor)carryingacu

rrent(I)

isplace

dinatransversemagneticfield(B),an

electric

field(EH)isproducedinsidetheconductorina

direction

normaltoboththecur

rentandthemagneticfield.

ThisphenomenonisknownasHallEffect.

Thegeneratedvoltageisc

alledHallVoltage.

I

B

EH

+++++

+++++++++++++++++

+++++

+++++++++++++++++

EH

B

I


88/94

N

-Type

P-Type

-

-

Chargemov

ement

Righttoleft.

(electrons)

Lefttoright.(H

oles)

Magneticfield

applied

Z-dir

Z-dir

Voltageproduced

Y-dir

Y-dir

Duetofield

e-smovetowardsface1

w

ithvelocityv

holesmovetowards

face1withvelo

cityv

Thereforeap

.doccursb/wf

ace(1)andface

(2)-givesEH

-Ydir.


89/94

N-Type

P-Type

Forceduetop.d

=

(-)eE

H

=eE

H

Forceduetomagfield

=(-)Bev

=Be

v

AtE

quilibrium

(-)eE

=(-)

Bev

eE=Bev

EH=Bv

EH=Bv

Currentdensity(Jx)inX-dir

Jx

=(-)n

ee

v

Jx

=nhev

v=(-)Jx/

nee

v=Jx/nhe

(1)

(2)

Su

b.

(2)in(1)

EH=(-)BJx

/nee

EH=BJ

x/nhe

RHHallCo-efficient

Negative(-veY

dir)

Positive(+veYdir)

EH=(-)BJ

xRH

EH=B

JxRH

RH=1

/nee


90/94

H

allco-effintermsofHallVoltage

EH=

BJxRH

RH

=EH/BJx

HalfCoefficient(RH)isdefinedastheHallfield(EH)

developedperunitcurrentdensity(Jx

)perunit

applied

magneticfield(B).

Ifthethickn

essofthesample

istandthevoltagedevelopedisVH,

thenHallvo

ltage

EH=

BJxR

H

(1)

VH=

EH.t

(2)

u.

n

,wege

VH

=BJxRH

t

(3)

Ifbisthewidthofthesample,

thenareaof

thesampleis(b.t)

Therefore,c

urrentdensity

=Jx

=Ix/(b.t)

(4)

H

V

H

x

RIBt

bt

=

Sub.

(4)in(3

),weget

H

HallVoltageV

H

x

RIB

b

=

H

HallCoeffi

H xVb

R

IB

=


91/94

ExperimentalDeterminationofH

allEffect

H

allEffect

H

allEffect

H

allEffect

Asemiconductorslabofthicknesstandbrea

dthbistakenandcurrent

ispassedusingthebattery.

Theslabisp

lacedbetweenthepolepiecesofanelectromagnets

othat

currentdirectioncoincideswithx-axisandma

gneticfieldcoincideswith

z-axis.

magneticfieldalong

z-axis.

twoprobesatthecenterofthe

topand

bottomfaces

oftheslab(y-axis).

IfBismagn

eticfieldapplied

andtheVH

istheHallv

oltageproduced,

thenthe

Hallcoeff

icientcanbecalculatedfrom

H

H

allCoeffi

H xVb

R

IB

=


92/94

Mobilityof

chargeCarriers

Mobilityof

chargeCarriers

Mobilityof

chargeCarriers

Mobilityof

chargeCarriers

H

1

HallCoeffi

R

ne

=

Weknow

O

nlyfordriftvelocity

H

3

1

HallCoeffi

8

R

ne

=

Semi.Cond-Avgvelocity

H

1.1

8

R

=

(3)

(2)

Weknowtheconductivityforntypeis

e

e

ne

=

1

e e

ne

=

Eqn.(3)canberewritten

as

H

1

1.1

8R

n

e

=

(

4)

(5)


93/94

Substitutingeqn.

(5)in(4

)weget,

1.1

8e

H

e

R

=

H

HallCoeffi

H xVb

R

IB

=

Sin

ce

(6)

1.1

8

e

H

e

R

= 1

.18e

H

e

xVb

IB

=

therefore

Thusbyfind

ingHallvoltag

e,Hallcoefficientcanbecalcu

latedand

thusthemob

ilityofthecharg

ecarriers(e

&

h)canalsobedetermined.

(7)


94/94

94

Engineering Physics 2 Unit-2

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