PWM开关电源分类buck boost介绍及主要元件参数选择

电源分类ClassificationofPowerSupply,线性电源Linear:RegulatororLDO(Lowdropout)开关电源Switching:BuckconversionBoostconversionBuck-Boostconversion(SEPIC,Cuk),HowtoseparateLinearorSwitching?,Linear:將一DC電壓經由電路元件,直接線性轉換為另一DC電壓.Forexample:,3.3V,5V,Cin,10uF,AP1117,3,1,2,IN,ADJ,OUT,Cout,100uF,HowtoseparateLinearorSwitching?,Switching:電源電路中使用一個或多個開關元件,來切換能量之儲存或轉移,同時可利用各種電路架構來達到升壓.降壓.極性反轉或直流電壓轉交流電壓,交流電壓轉直流電壓等要求.,AdvantageandDisadvantage,AdvantageofLinearCostlowerLowoutputnoiseandrippleGoodoutputregulationDisadvantageofLinearEfficiencylowerNeedbiggertransformerandheatsinkNarrowinputvoltagerangeLowholduptime(1ms),AdvantageandDisadvantage,AdvantageofswitchingHighfrequencyoperatingtoreducesizeofinductorandcapacitorLightweightandSmalltotalsizeHighefficiencyandpowerdensityHighinputvoltagerangeandholduptime(25ms)DisadvantageofswitchingHighripplevoltageCircuitdesigncomplicatedHigherEMIinterference,Switchingcircuit,DC-DCcircuitBuckconverter,Switchingcircuit,BuckconverterfunctiondescriptionQ控制能量儲存與傳送方向L傳送與儲存能量及濾除電流雜訊C傳送與儲存能量及濾除電壓雜訊D提供放電迴路(當Qoff),Switchingcircuit,Buckconverter工作原理說明:Q導通,D截止,電感器之電壓為VL=Vi-Vo故電感器iL會開始持續線性上升,其交流成分iL(ac)流經電容器C產生漣波,直流成分會供應至負載RL,Switchingcircuit,Buckconverter工作原理說明:Q截止,電感器瞬間電流不變,故為感一反電動勢VL=-Vo而促使D導通,此時C原來所儲存的能量可經由D,L釋放至負載,L上的電流呈線性衰退,DesignBuckconverter,HowtoselectinductorL=Vin(min)-Vsat-Vout*Ton(max)2Io(min)Vo/Vi=D=ton/Ts,DesignBuckconverter,HowtoselectCin?Cin電壓rating至少為Vin1.5倍Iin的計算方式如下表所示:,DesignBuckconverter,HowtoselectCout?Esr=Vripple/2Io(min)Cout=IL/8*fs*VoHowtoselectDiode?Vrrm=1.25Vin(max)Ipk=Io(max)-Io(min),Switchingcircuit,DC-DCcircuitBoostconverter,Switchingcircuit,Boostconverter工作原理:Q導通,D截止,電感器之電壓為一固定値VL=Vi,故電感器電流iL會開始持續線性上升補充能量,此時電容器瞬間電壓不變開始釋放能量致負載,Switchingcircuit,Boostconverter工作原理:Q截止,D導通,電感器瞬間電流不變,故會感應一反電動勢-VL=Vi-Vo,此時輸出電壓將會備昇壓,電感與電源同時釋放能量至負載及電容器C上,電感之電流iL開始持續線性下降釋放能量至一定値,BoostconverterDesign,Howtoselectinductor?L=Vin(min)-Vds(sat)*DmaxIL*fsVo/Vi=1/1-DD=ton/Ts,ThermalConsideration,SwitchingICThermalconsiderationPloss=PQ+Psat+PswherePQ:quiescentlossPsat:saturationlossPs:switchingloss,ThermalConsideration,SwitchingICThermalconsideration:Tj=Ta+PLoss*RJAWhereTjjunctiontemperatureTaambienttemperaturePLosstotaldispationRJAthermalresistance(diejunctiontoambientair),ThermalConsideration,SelectHeatsinkRSA=RJA-RJC-RCSWhereRJCThethermalresistancebetweenICchip(junctionpoint)andpackagebacksideconnectingtotheheatsinkRSAThethermalresistanceofHeatsinkRCSThethermalresistancebetweenpackagebacksideandtheheatsink,DesignExample,BuckconverterExample:Vin=56V,Vout=3.3VStep1:DutyratioD=Vo+Vd/Vin-Vds(sat)=Ton/TsAssumeVd=0.5V,Vds(sat)=0.1VD=0.64(Vin=6V),DesignExample,Step2:L=Vin-Vds(sat)-Vo*D/IL*fs=6-0.1-3.3*0.64/0.6*(110*103)=21.6uHIL=2*10%*Io=2*0.1*3=0.6Afs=110kHzSowecanchoose22uH,DesignExample,Step3:Cout=IL/8*fs*Vo=0.6/8*110*103*0.033=20.66uFESR=Vin(min)-Vds(sat)*DmaxIL*fsIL=2*Io(min)*Vo/Vin(min)=2*0.05*12/5=0.24L=(5-0.1)*0.6/0.24*(110*103)=111.36uHSowecanchoose120uH,DesignExample,BoostconverterStep2:selectionoftheoutputcapacitorCout=Io(max)*Dmax/fs*Vo=0.3*0.6/110k*0.05=32.73uHIpk=Io(max)/1-Dmax+Vin(max)*D/2*fs*L=0.3/1-0.6+7*0.6/2*110k*120u=0.91AESR=Vo/IPK=0.05/0.91=55m,DesignExample,BoostconverterStep3:selectionofpowerswitch(MOSFET)PMOSFET=Ipk2*Rds(on)*Dmax+0.5*Vin(max)*Ipk*(tr+tf)*fsAssumeTa=55CRJA=50C/WPd=0.91*0.91*0.0135*0.6+0.5*7*0.91*0.3u*110k=0.112WTj=Ta+(RJA*Pd)=60.6C,DesignExample,BoostconverterStep4:selectionofpowerRecti