电流变压器设计步骤

1、Thesolutionofcurrentdetectionusecurrenttransform電流的檢出利用電流變壓器（CT）可以檢出與負載電流成比率關係的的切換器初級圈電流（切換電流）。將切換電流轉成電壓，然後使用峰值檢波電路，获取直流電壓。半波式檢波電路(Fig.1)適用於前向變換器（有半個週期無法檢出的問題）。檢波器電路為單純的電容器輸入型整流電路。由RC值決定檢出的反應時間，此方式的反應時間比並連電阻型慢，但是，此一方法可以檢出功率切換晶體或整流二極體故障所產生的過大電流。Fig.1CTandhalfbridgerectifiercircuit1/52020/4/13Thesolut

2、ionofcurrentdetectionusecurrenttransformCoreSize&MagnetizingcurrentforalltypeofcurrenttransformerIngeneral,forcurrenttransformers,Thelargertheinductance.Thesmallerthemagnetizingcurrent.Moreaccuratethemeasurement.Themagnetizingcurrentcomponentincreaseduringthepulsedurationandwillbesubtractedfromthequ

3、antitytobemeasured.Consequently,attheendoftheconductionpulse,themagnetizingcurrentshouldbesmallcomparedwiththemeasuredquantity.Forcurrentlimitapplication,magnetizingcurrentof10%isatypicaldesignlimit.2/52020/4/13ThesolutionofcurrentdetectionusecurrenttransformCurrentTransformerDesignProcedureStep1Cal

4、culate(orobserve)Peakprimarycurrent.Theslopedi/dt（i）onthetopthecurrentwaveformThiswillbeusedtocalculatetheminimumcurrenttransformerinductance.Step2CurrenttransformersecondaryvoltageSelectcurrenttransformersecondaryvoltageatthelimitingcurrentvalueTypically1Vincludingthediodedrop.Step3Ahigh-permeabili

5、ty（uofmaterial）sothatalargeinductancewillbeobtainedfortheminimumnumberofturns.2becauseL=uAN/lAlowremanenceBrThiswillensurethatthecorewillnotsaturateafterafewcyclesofoperation.Note：Thisrapidcorerestorationallowscloselyspacedforwardcurrentpulsestobemonitoredaccurately,withoutsaturationofcore.Clearly,t

6、hevalueofr1willbechosentogettherequiredminimumrecoverytime,andthevoltageratingofD1mustbeselectedtoblockthereverseflybackvoltageacrossr1showninFig2.Fig2.Recoverypathacrossr1Theresetvoltageandtimearedeterminedbyr1andtheresettimemustbelessthantheshortest“off“time.3/52020/4/13Thesolutionofcurrentdetecti

7、onusecurrenttransformUnidirectionalCurrentTransformerDesignExampleStep1CalculatePrimaryAmpere-Turns（NI）Intheexample,iftheprimarycurrentisXAinasingleturn.ThisgivesaprimarymagnetizingforceofXampere-turnsduetoforwardforcingcurrentthatflowsintheprimaryoftheforwardconvertertransformer.Step2DefineSecondar

8、yTurnsandCalculateSecondaryCurrentThisdesignwillusethepreferredwindingnotexceeding100turnsof#34AWGinsinglelayer（togiveasmallcoresizeandgoodhigh-frequencyperformance）.Therefore,toprovideanequaldepolarizingmagneticforce0f10ampere-turns,thesecondarycurrentwillbe100mA.IfIp=10A、N1=1&N2=100Formula：Is=(N1/

9、N2)Ip（10A）=100mAN2IsStep3DefinetheRequiredSecondaryVoltageThesecondaryvoltagewillbe0.8V,madeupofthe0.6Vdiodedropandthe0.2VcurrentanaloguesignaldevelopedacrossR2.Consequently,with100turnsonthesecondaryandasignalcurrent-linkedturnontheprimary,thevoltagedropontheprimarywindingwillbe8mV.0.81/100=8mVThep

10、rimaryinsertionlossisthusverysmall.Forforwardconverterapplication,thepeakvalueofprimarycurrentatendofanperiodisrequiredcurrent-limitingvalue（thisindicatesthepeakcurrentflowingintheoutputinductor）.Hencethesecondarycurrentmustreflectarisingcurrentduringtheandthis“on”periosetsalimitonthemagnetizingcurr

11、ent.apulse-by-pulsevoltageanalogueofthecurrentisgivenbythevoltageacrossR2.Henceafast-actingcurrentlimitmaybeintroducedatthispointinthecontrolcircuit,Usingthepeakanaloguevoltageinformation,orthecurrentsignalmaybeusedinacurrentmodecontrolcircuit.4/52020/4/13Thesolutionofcurrentdetectionusecurrenttrans

12、formStep4CheckMagnetizingCurrentCheckthatthemagnetizationcurrentisacceptablysmall.Tocalculatethemagnetizationcurrent,thesecondaryinductanceisrequired.IftheAlvalueofthecoreisknown,thentheinductancewillbecalculatedfromthebasicformula：L=N2AlIftheAlvalueofthecoreisnotknown,thentheinductancewillbecalcula

13、tedfromthe2basicformula：L=(roAeN)/leN=numberofsecondaryturns=ro7o=410r=permeabilityofcoreifr=7500L=31mHTheslopeofthemagnetizationcurrent,di/dt,canbecalculatedfrome=-Ldi/dtthereforedi/dt=e/Le=0.8Vdi/dt=25.8A/secIfprimarycurrent（Ip）attheendofthe10spulse，Imag=0.258mAThemagnetizationcurrenttransformstotheprimaryas25.8mA,anegligibleerrorin10A.Thesecondarycurrentwillbeatruetransformoftheprimarycurrentwithalossofonly1/4mAin100mAThisindicatesthatasmallercorecouldbeused,itcanbedifficulttowindthesecondaryonacoremuchsmallerthanthis.ThevalueoftheburdenresistorR2requiretodevelopthechosensig