Magnetic Circuit Basis

8/6/2019 Magnetic Circuit Basis

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Massachusetts InstituteofTechnologyDepartmentofElectricalEngineeringandComputerScience

6.685ElectricMachinesClassNotes2MagneticCircuitBasics February11,2004c2003JamesL.KirtleyJr.1 IntroductionMagnetic Circuitsoffer, as do electric circuits, a way of simplifyingthe analysis ofmagnetic fieldsystems which can be represented as having a collection of discrete elements. In electric circuitsthe elements are sources, resistors andso forth which are represented as having discrete currentsandvoltages. These elements are connected together with wires andtheir behavior is describedbynetworkconstraints(Kirkhoffsvoltageandcurrentlaws)andbyconstitutiverelationshipssuchas Ohms Law. In magnetic circuits the lumpedparameters are called Reluctances (the inverseof Reluctance is called Permeance). The analog to a wire is referred to as a high permeancemagneticcircuitelement. Ofcoursehighpermeability istheanalogofhighconductivity.

Byorganizingmagnetic fieldsystems into lumpedparameterelementsandusingnetworkconstraintsandconstitutiverelationshipswecansimplifytheanalysisofsuchsystems.2 ElectricCircuitsFirst,letusreviewhowElectricCircuitsaredefined. Westartwithtwoconservationlaws: conservationofchargeandFaradaysLaw. Fromthesewecan,withappropriatesimplifyingassumptions,derivethetwo fundamentalcirciutconstraintsembodied inKirkhoffs laws.2.1 KCLConservationofchargecouldbewritten in integral formas:

df Jnda+ dv=0 (1)

volume dtThissimplystates that thesumofcurrentout ofsomevolume ofspace andrateof change of

freecharge inthatspacemustbezero.Now,ifwedefineadiscretecurrenttobetheintegralofcurrentdensitycrossingthroughapart

ofthesurface:ik = Jnda (2)

surfacek

andif

we

assume

that

there

is

no

accumulation

of

charge

within

the

volume

(in

ordinary

circuit

theorythenodesaresmallanddonotaccumulatecharge), wehave:

Jnda= ik =0 (3)k

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3 MagneticCircuitAnalogsIn the electric circuit, elements for which voltage and current are defined are connected togetherbyelements thoughtof as wires, orelements withzeroor negligible voltage drop. The interconnection points are nodes. In magnetic circuits the analogous thing occurs: elements for whichmagnetomotiveforceandfluxcanbedefinedareconnectedtogetherbyhighpermeabilitymagneticcircuitelements(usually iron)whicharetheanalogofwiresinelectriccircuits.3.1 AnalogytoKCLGaussLaw is:

Bnda=0 (8)whichmeansthatthetotalamountoffluxcomingoutofaregionofspace isalwayszero.

Now, we will define a quantity which is sometimes called simply flux or a flux tube. Thismightbethoughttobeacollectionoffluxlinesthatcansomehowbebundledtogether. Generallyit isthefluxthat is identifiedwithamagneticcircuitelement. Mathematically itis:

k = Bnda (9)Inmostcases,fluxasdefinedaboveiscarriedinmagneticcircuitelementswhicharemadeofhigh

permeability material, analogous to the wires of high conductivity material which carry currentin electric circuits. It is possible to show that flux is largely contained in such high permeabilitymaterials.

Ifallofthefluxtubesoutofsomeregionofspace(node)areconsideredinthesum,theymustaddtozero:

k =0 (10)k

3.2

Analogyto

KVL:

MMF

AmperesLaw is

Hd = Jnda (11)Where,asforFaradaysLaw,theclosedcontourontheleftistheperipheryofthe(open)surface

ontheright. NowwedefinewhatwecallMagnetomotiveForce, indirectanalogto ElectromotiveForce,(voltage).

bkFk = Hd (12)

akFurther,definethecurrentenclosedbya looptobe:

F0 = Jnda (13)ThentheanalogytoKVL is:

Fk =F0k

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NotethattheanalogisnotexactasthereisasourcetermontherighthandsidewhereasKVLhasnosourceterm. Notealsothatsigncountshere. Theclosedintegral istaken insuchdirectionso that the positive senseof the surfaceenclosed is positive (upwards)whenthe surface is to theleft of the contour. (This is another way of stating the celebrated right hand rule: if you wrapyourrighthandaroundthecontourwithyourfingerspointinginthedirectionoftheclosedcontourintegration,yourthumb ispointing inthepositivedirectionforthesurface).3.3 AnalogtoOhmsLaw: ReluctanceConsider a gap between two high permeability pieces as shown in Figure 3. If we assume thattheir permeability is high enough, we can assume that there is no magnetic field H in them andso the MMF or magnetic potential is essentially constant,just like in a wire. For the moment,assume that the gap dimension g is small and uniform over the gap area A. Now, assume thatsomefluxisflowingfromoneofthesetotheother. Thatfluxis

=BAwhere B is the flux density crossing the gap and A is the gap area. Note that we are ignoringfringing

fields

in

this

simplified

analysis.

This

neglect

often

requires

correction

in

practice.

Since

thepermeabilityof freespace is0,(assumingthegap isindeedfilledwithfreespace),magneticfield intensity is

BH=

0andgapMMF isjustmagneticfield intensity timesgapdimension. This,of course, assumesthatthegap isuniformandthatsoisthemagneticfieldintensity:

BF = g

0Whichmeansthatthereluctanceofthegap istheratioofMMFtoflux:

F gR= =

0A

yArea A

x

g

Figure3: AirGap

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3.4 Simple CaseConsiderthemagneticcircuitsituationshowninFigure4. Herethereisapieceofhighlypermeablematerial shapedto carryfluxacross a single air-gap. A coil iswoundthroughthe window inthemagneticmaterial(thisshapeisusuallyreferredtoasa core). Theequivalentcircuit isshowninFigure5.

Region 1

Region 2

I

Figure4: Singleair-cappedCoreNotethatinFigure4, ifwetakeasthepositivesenseoftheclosed loopadirectionwhichgoes

vertically upwardsthrough the leg of the core through the coil and then downwards through thegap,thecurrentcrossesthesurfacesurroundedbythecontour inthepositivesensedirection.

+

F = N I

Figure5:

Equivalent

Circuit

3.5 FluxConfinementThe gap in this case has the same reluctance as computed earlier, so that the flux in the gap issimply=NI. Now,byfocusingonthetworegions indicatedwemightmakea fewobservations

Raboutmagneticcircuits. First,consider region1asshowninFigure6.

Figure6: FluxConfinementBoundary: ThisisRegion1

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Inthispicture,notethatmagneticfield H paralleltothesurfacemustbethesame insidethematerial as it isoutside. ConsiderAmperesLaw carriedout aboutaverythin loopconsistingofthetwoarrowsdrawnatthetopboundaryofthematerialinFigure6withveryshortverticalpathsjoiningthem. Ifthereisnocurrentsingularityinsidethatloop,theintegralarounditmustbezerowhichmeans themagnetic fieldjust insidemustbethesameasthemagnetic fieldoutside. Since

B=H,and highlypermeablemeans isvery large,thematerial isveryhighlypermeableand unless

B

is

really

large,

H

must

be

quite

small.

Thus

the

magnetic

circuit

has

small

magnetic

field

H andtherforefluxdensitiesparalleltoandjustoutside itsboundariesaeralsosmall.

B is perpendicula

Figure7: GapBoundaryAtthesurfaceofthemagneticmaterial,sincethemagneticfieldparalleltothesurfacemustbe

verysmall,anyfluxlinesthatemerge fromthecoreelementmustbeperpendiculartothesurfaceasshownforthegapregion inFigure7. This istrueforregion1aswellas forregion2,butnotethat the total MMF available to drive fields across the gap is the same as would produce fieldlines from the area of region 1. Since any lines emerging from the magnetic material in region 1wouldhaveverylongmagneticpaths,theymustbeveryweak. Thusthemagneticcircuitmateriallargely confines flux, with only the relatively high permeance (low reluctance) gaps carrying anysubstantiveamountofflux.3.6 Example: C-CoreConsidera gappedc-core asshown inFigure8. This is two pieces of highlypermeablematerialshapedgenerally like Cs. Theyhave uniformdepth inthedirectionyoucannotsee. WewillcallthatdimensionD. OfcoursetheareaA=wD, wherew isthewidthatthegap. Weassumethetwogapshavethesamearea. Eachofthegapswillhaveareluctance

gR=

0ASupposewewindacoilwithN turnsonthiscoreasshowninFigure9. Thenweputacurrent

I inthat coil. Themagnetic circuit equivalent is shown inFigure 10. Thetwo gaps are in seriesand,ofcourse, inserieswiththeMMF source. Sincethetwo fluxesarethesameandtheMMFsadd:

F0 =N I=F1+F2 =2Randthen

N I 0AN I= =

2R 2g7


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g

Area A

Figure8: GappedCore

NTurns

I

Figure9: Wound,GappedCoreandcorrespondingfluxdensity inthegapswouldbe:

0N IBy =

2g3.7 Example: CorewithDifferentGapsAs a second example, consider the perhaps oddly shaped core shown in Figure 11. Suppose thegapontherighthastwicetheareaasthegaponthe left. Wewouldhavetwogapreluctances:

g gR1 = R2 =

0A 20ASincethetwogapsare inseriesthefluxisthesameandthetotalreluctance is

3 gR=

20AFluxinthemagneticcircuit loopis

F 20AN I= =

R

3

g

andthefluxdensityacross,say,the lefthandgapwouldbe: 20N I

By = =A 3 g

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F +

Figure10: EquivalentMagneticCircuit

NTurns

I

Figure11: Wound,GappedCore: DifferentGaps

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Magnetic Circuit Basis