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:3.3V5VCin10uFAP1117312INADJOUTCout100uFHowtoseparateLinearorSwitching?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.AdvantageandDisadvantageAdvantageofLinearCostlowerLowoutputnoiseandrippleGoodoutputregulationDisadvantageofLinearEfficiencylowerNeedbiggertransformerandheatsinkNarrowinputvoltagerangeLowholduptime(~1ms)AdvantageandDisadvantageAdvantageofswitchingHighfrequencyoperatingtoreducesizeofinductorandcapacitorLightweightandSmalltotalsizeHighefficiencyandpowerdensityHighinputvoltagerangeandholduptime(~25ms)DisadvantageofswitchingHighripplevoltageCircuitdesigncomplicatedHigherEMIinterferenceSwitchingcircuit

DC-DCcircuitBuckconverter

SwitchingcircuitBuckconverterfunctiondescriptionQ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RLSwitchingcircuitBuckconverter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

Howtoselectinductor

L>=[Vin(min)-Vsat-Vout]*Ton(max)2Io(min)Vo/Vi=D=ton/TsDesignBuckconverterHowtoselectCin?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)SwitchingcircuitDC-DCcircuitBoostconverterSwitchingcircuitBoostconverter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SwitchingcircuitBoostconverter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)]*Dmax

△IL*fsVo/Vi=1/1-DD=ton/Ts

ThermalConsideration

SwitchingICThermalconsiderationPloss=PQ+Psat+PswherePQ:quiescentlossPsat:saturationlossPs:switchingloss

ThermalConsideration

SwitchingICThermalconsideration:Tj=Ta+PLoss*RθJA

WhereTjjunctiontemperatureTaambienttemperaturePLosstotaldispationRθJA

thermalresistance(diejunctiontoambientair)ThermalConsideration

SelectHeatsinkRθSA=RθJA-RθJC-RθCSWhereRθJCThethermalresistancebetweenICchip(junctionpoint)andpackagebacksideconnectingtotheheatsink

RθSAThethermalresistanceofHeatsinkRθCS

ThethermalresistancebetweenpackagebacksideandtheheatsinkDesignExample

BuckconverterExample:Vin=5~6V,Vout=3.3V

Step1:DutyratioD=Vo+Vd/Vin-Vds(sat)=Ton/TsAssumeVd=0.5V,Vds(sat)=0.1VD=0.64(Vin=6V)

DesignExampleStep2:L>=[Vin-Vds(sat)-Vo]*D/△IL*fs

=[6-0.1-3.3]*0.64/0.6*(110*10^3)=21.6uH△IL=2*10%*Io=2*0.1*3=0.6Afs=110kHz

Sowecanchoose22uHDesignExampleStep3:Cout>=△IL/8*fs*△Vo

=0.6/8*110*10^3*0.033=20.66uF

ESR<=△Vo/△Io=0.033/0.6=0.055ΩDesignExampleStep4:PowerMOSFETselect:Powerdissipation(condition+switchinglosses)canbeestimatedas:Pd=Io^*Rds(on)*Dmax+[0.5Vin*Io*(tr+tf)*fs]=(3*3*0.035*0.78)+[0.5*5*3*0.3*10^(-6)*(110*10^3)=0.5W(Vin=5V)Rds(on)=35mΩCEM4435

tr+tf=300ns

DesignExampleStep5:Tj=Ta+(RθJA*Pd)=55+(50*0.5)=80°CTa=AmbientdegreeAssume=55°CRθJA=50°C/WThermalresistance

DesignExampleStep6:RectifierDiode:Pd=Io*Vd*(1-Dmin)=3*0.5*(1-0.55)=0.675W(Vin=7V)Tj=Ta+(RθJA*Pd)=55+(15*0.675)=65°CDesignExampleStep7:Inputcapacitor:Iin(rms)=√[D*(Io(max)+Io(min)*(Io(max)-Io(min))+(△IL^2)/3]=√[0.78*(3+0.3)*(3-0.3)+0.36/3]=2.67A

DesignExampleBoostconverterExampleVin=5~7V,Vo=12VStep1:

DutyratioD=Vo+Vd-Vin(min)/Vo+Vd-Vds(sat)

TheDutycycleforVin=5,6,7is0.6,0.52,0.44

DesignExampleBoostconverterStep2:

L>=[Vin(min)-Vds(sat)]*Dmax

△IL*fs

△IL=2*Io(min)*Vo/Vin(min)=2*0.05*12/5=0.24

L=

(5-0.1)*0.6/0.24*(110*10^3)=111.36uHSowecanchoose120uHDesignExampleBoostconverterStep2: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=0.91A

ESR<=△Vo/IPK=0.05/0.91=55mΩ

DesignExampleBoostconverterStep3:selectionofpowerswitch(MOSFET)PMOSFET=Ipk^2*Rds(on)*Dmax+[0.5*Vin(max)*Ipk*(tr+tf)*fs]AssumeTa=55°CRθJA=50°C/WPd=0.91*0.91*0.0135*0.6+[0.5*7*0.91*0.3u*110k]=0.112WTj=Ta+(RθJ