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AN-1197Selecting Inductors for Buck Converters

ApplicationReportSNVA038B May2001 'r'.. r'.. May2001 RevisedApril2013AN-1197 SelectingInductorsforBuckConvertersSubmi tDocumentationFeedbackCopyright 2001 2013, , step-by-stepprocedureis ,allthepreviousstepsareconsolidatedin a singledesigntable,whichanswersthequestio n: Howwilltheselectedinductoractuallyperfor min a specificapplication? Theimportantpointtonotehereis thatthougheveryinductoris designedassumingcertainspecific designconditions', , ,thedesignermustknowhowtobeabletoaccurat elypredict,orextrapolate,theperformanceo ftheinductortoa newsetofconditions,whicharehisspecific applicationconditions'.

Now, for a buck regulator, we can show that the duty cycle is: (5) where VD is the forward drop across the catch diode (≅0.5V for a Schottky diode). ‘r' can be finally written as: (6) and L is therefore: (7) where f is in Hz. SNVA038B– May 2001– Revised April 2013 AN-1197Selecting Inductors for Buck Converters 3 Submit Documentation ...

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Transcription of AN-1197Selecting Inductors for Buck Converters

1 ApplicationReportSNVA038B May2001 'r'.. r'.. May2001 RevisedApril2013AN-1197 SelectingInductorsforBuckConvertersSubmi tDocumentationFeedbackCopyright 2001 2013, , step-by-stepprocedureis ,allthepreviousstepsareconsolidatedin a singledesigntable,whichanswersthequestio n: Howwilltheselectedinductoractuallyperfor min a specificapplication? Theimportantpointtonotehereis thatthougheveryinductoris designedassumingcertainspecific designconditions', , ,thedesignermustknowhowtobeabletoaccurat elypredict,orextrapolate,theperformanceo ftheinductortoa newsetofconditions,whicharehisspecific applicationconditions'.

2 It willbeshownthat intuition' detailedprocedureis requiredandis presentedin theformoftheDesignTable(Table2) andtheSelectionFlowChart(Figure2).2 Background:TheInductorCurrentWaveformRef ertoFigure1, whichshowsthecurrentthroughaninductorin continuousmodeoperation(boldline). is thegeometricalcenteroftheAC/rampcomponen t is theaveragevalueofthetotalinductorcurrent waveform is thecurrentintotheload,sincetheaveragecur rentthroughtheoutputcapacitor,asforanyca pacitorin steadystate,is IDC+ I/2,andit determinesthepeakenergyin thecore(e= L I2), whichin IDC- I/2anddeterminestheconstantresiduallevel ofcurrent/energyin dependsontheload,eventhoughit is.

3 IAC= I = IPEAK ITROUGH(1) theloadcurrentforthecaseshownin Figure1, IDC= r' is definedastheratiooftheACtoDCcomponents(c urrentrippleratio)evaluatedatmaximumload ,IO. Notethatbydefinition'r' is a constantfora givenconverter/application(asit is calculatedonlyatmaximumload),andit is alsodefinedonlyforcontinuousconductionmo de.(2)A highinductancereduces I andresultsin lower r' (andlowerRMScurrentin theoutputcapacitor),butmayresultin a ,formostbuckregulators, r' is chosentobein (atthemaximumratedload).SeeAppendixA. Oncetheinductanceisselected,aswedecrease theloadontheconverter(keepinginputvoltag econstant), I ,atthepointoftransitiontodiscontinuousmo deofoperation,theDClevelis I/2asshownin Figure1.

4 So: Thecurrentrippleratioatthepointoftransit iontodiscontinuousmodeis ,theupperlimitfor'r' is also2. Theloadatwhichthishappenscanbeshownbysim plegeometrytober/2timesIO. Soforexample,iftheinductanceis chosentobesuchthat r' is loadof2A, ,whichis :If theinductoris a swinging'inductor, May2001 RevisedApril2013 SubmitDocumentationFeedbackCopyright 2001 2013, (Amps)LDC( H)Et(V secs)DCR(nom)m Et100(V secs) = L dl/dt,wegetduringtheONtimeoftheconverter :(3)whereVINis theappliedDCinputvoltage,VSWis thevoltageacrosstheswitchwhenit is ON,D is thedutycycleandf is theswitchingfrequencyin I wecanwrite r' as:(4)Now,fora buckregulator,wecanshowthatthedutycyclei s.

5 (5)whereVDis theforwarddropacrossthecatchdiode( Schottkydiode). r' canbefinallywrittenas:(6)andL is therefore:(7)wheref is in May2001 RevisedApril2013AN-1197 SelectingInductorsforBuckConvertersSubmi tDocumentationFeedbackCopyright 2001 2013, 12 Vata ( 15mV).WeassumeVSW= ,VD= = 150, ,forloopstabilityreasons,weshouldnotusea nyoutputcapacitoroflessthan100m , andsincewedonotwishtouseanLCpostfilter,o ur I mustbe:(8)So r' is:(9)Therequiredinductanceis:(10)L = 127 HTherequiredenergyhandlingcapabilityis ,thepeakcurrentis:(11)IPEAK= (12)Therequiredenergyhandlingcapability e' is:(13)whereL is in :(14)NOTE.

6 Duringa hardpower-up(nosoftstart)orabnormalcondi tionslikea shortcircuitontheoutput,thefeedbackloopi s noteffectivein actuallygoingtohittheinternalcurrentlimi tofthedevice,ICLIMin Figure1, theinductorhassaturated,andif theinputDCvoltageishigherthan40 Vthecurrentcouldslewupata ratesohighthatthecontrollermaynotbeablet olimitthecurrentatall, , ,theinductorshouldbesizedtohandlethewors tcaseenergyeCLIM:(15)whereL is in H andICLIMis theinternallimitoftheregulatorin May2001 RevisedApril2013 SubmitDocumentationFeedbackCopyright 2001 2013, secsor Et'whichis simplythevoltageacrossthewindingoftheind uctortimesthedurationin secsforwhichit is :Currentrampsuptothesamepeakvaluewhether V (theappliedvoltageacrossinductor)islarge butt (thetimeforwhichV is applied)is small,orwhetherV is smallbutt is whatreallycounts.

7 (Theonlyexceptiontothisis theCoreLosstermsincethisdependsdirectlyo ntheabsolutevalueofthefrequencytoo,notju sttheEt).Also,Etcanbecalculatedduringthe ON-time,(V secsgained),orduringtheOFF-time(V secslost).Bothwillgivethesameresultsince thereis nonetchangein V secspercyclein ,rememberthatthoughV secsis relatedtotheenergyin thecore,it secsgivesinformationonlyabouttheACcompon entof r', thatis, , it givenapplicationis completelydefinedbya certainIOandEt(andfrequencyforthecorelos sterm), L (or r')andwefixit accordingtotheguidelinesin = L dl/dt,wecanwritethatV dt= L dtis ,bydefinition:Et= V t = L I V secs(16)whereL is in H.

8 R' canthereforebewrittenas:(17)SolvingforL: (18) :(19)tON= secsSoEtis:Et= (VIN- VSW- VO) tON= (24- 12) secsEt= secsL is therefore:(20)(21)L = 127 Hwhichgivesusthesameresultasin Example1 May2001 RevisedApril2013AN-1197 SelectingInductorsforBuckConvertersSubmi tDocumentationFeedbackCopyright 2001 2013, Aninductanceof127 H (orgreater,basedonmaximum r' ) DCloadof1A(toensureacceptabletemperature rise,specify T)ORsteadystateEnergyhandlingcapabilityo f84 J (toruleoutcoresaturationif DCinputvoltage 40V)ORpeakenergyhandlingcapabilityof1016 J.

9 (MaxCurrentLimitofLM2593 HVis ) Etof38V secs Frequency150kHzThesecanbecommunicateddir ectlytoa vendorfora Figure2, thefirstpassselectionis partfromPulseEngineeringbecauseitsL andIDCareclosetoourrequirements,eventhou ghtherestdoesnotseemtofit ourapplication(seeTable1 andthefollowingbullets).Inparticularthef requencyforwhichtheinductorwasdesignedis 250kHz, , ourcase,andthatis whyit isimportanttofollowthefullprocedureprese ntedbelow. Intuition' : Theinductoris suchthat380mWdissipationcorrespondsto50 C risein temperature. Thecorelossequationforthecoreis 10 18 is in HzandB is in Gauss.

10 Theinductorwasdesignedfora frequencyof250kHz. Et100is theV secsatwhich B'is :Forcorelossequationsit is ,likemostvendors,the B'aboveactuallyrefersto mindin :a)ACComponentofCurrent:Thiscanbeeasilyc alculatedfrom:Et= L I V secswhere:L is in :(22)b) r':Sothisinductorhasbeendesignedforthefo llowing r':(23)r = )PeakCurrent:(24)IPEAK= May2001 RevisedApril2013 SubmitDocumentationFeedbackCopyright 2001 2013, )RMSC urrent:(25)(26)IRMS= )CopperLoss:Thisis:PCU= IRMS2 DCRmWwhere:DCRis in m .Inmostcases,toa closeapproximation,wecansimplyuseIDCinst eadofIRMSin , 387mWPCU= 385mWf) TheACComponentoftheB-Field:Thisis secsproduces100 Gauss(B).


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