1、電源分炎亠觸sificatiorL of PowerSupply線性電源Linear : Regulator or LDO ( Low drop out )開關電源Switching :>Buck conversion> Boost conversion> Buck-Boost conversion(SEPIC,Cvik.)How to separate Linear orSwitching ?Linear :將一 DC電壓經由電路元件，直接 線性轉換爲另一DC電壓.For exam pie:How to separate Linear orSwitching ?Switch

2、ing :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Advantage and DisadvantageAdvantage of LinearCost lowerLow output noise and rippleGood output regulationDisadvantage of LinearEfficiency lowerNeed bigger transformer and heat sinkNarrow input voltage rangeLow hold up

3、time (-1ms)Advantage and DisadvantageAdvantage of switchingHigh frequency op crating to reduce size of inductor and capacitorLight weight and Small total sizeHigh efficiency and power densityHigh input voltage range and hold up Ume(25ms)Disadvantage of switchingHigh ripple voltageCircuit design com

4、plicatedHigher EMI interferenceSwitching circuit DC-DC circuitBuck converterVoSwitching circuit Buck converter function description-QT控制能量儲存與傅送方向-LT傳送與儲存能量及濾徐電流雜訊-CT傳送與儲存能量及濾除電壓雜訊-DT提供放電迴路(當Q off)Switching circuit+ Buck converter工作原理說明：Q導通,D截止，電感器之電壓爲VL=ViVo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

5、會供應至負載RLSwitching circuit+Vo Buck converter X作原理說明：-Q截止,電感器瞬間電流不變，故爲感一反電動勢VL=VO 而促使D導通，此時C原來所儲存的能量可經由DL釋放至 負載,L上的電流呈線性衰退Design Buck converter How to select inductor">L>=Vin(min)-Vsat-Vdut*Ton(max)2Io(min)Vo/V i=D=to n/TsDesign Buck converter How to select Cin?TCin電壓rating至少爲Vin 1.5倍lin的計算

6、方式如下表所示：Calculationstep-down (buck) requlator6TonTon 于 Toff)IPK1 MMZXmro I /Al/Jim IB)K/ mwinuii + / /.MWdtiu)? X Z1腫Design Buck converter How to select Cout?Esr=Vri pp le/2Io(min)Cout>=A Ii78*fs* Vo How to select Diode?Vrrm>= 1.25Vin(max)Ip k=Io(max)-Io(min)Switching circuit DC-DC circu讓Boost

7、 converterLZINzoVi Toff TonPWM controlQisicVoSw讓ching circuit Boost converter 工作原理：Q導通,D截止，電感器之電壓爲一固定値VL二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+VoSwitching circuit Boost converter工作原理：+VQ截止Q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Boost conv

8、erter Design How to select inductor?L= Vin(min)-Vds(sat) * DmaxlL*fSVo/Vi=l/1-DD=ton/TsThermal Consideration Switching IC Thermal consideration P1oss=PQ+Ps£U+PswherePQ：quiescent lossPsat:saturation lossPs：switching lossThermal Consideration Switching TC Thermal consideration: Tj=Ta+ PlossRojaWh

9、ereTj->junction ternpenUureTa->ambient ternperaturePLoss total dispationRoja T thermal resistance(die junction to ambient air)Thermal Consideration Select Heat sinkResA=RejARwcRecsWhereR0JC->The thermal resistance between IC chipQunction point) and package backside connecting to the heat si

10、nkRosAThe thermal resistance of Heat sinkRocs ->The thermal resistance between package backside and the heat sinkDesign Exam pie Buck converterExample: Vin=5-6V,Voul=3.3 VStepl:Duty ratio D=Vo+Vd/Vin-Vds(sat)=Ton/TsAssume Vd=0.5V,Vds(sat)=0.1VD=0.64(Vin=6V)Design Exam pieStep2:L>=Vin-Vds(sat)-

11、Vol*D/ AlL*fs=6-0.1-3.3*0.64/0.6*(110*l)3)=21.6uHAIi2*10%*Io=2*0.1*3=0.6Afs=110kHzSo we can choose 22uHDesign Exam pieStep3:Cout>= AlL/8*fs* AVo=0.6/8*110*103*0.033=20.66uFESR<= aVo/ aIo=0.033/0.6=0.055QStep4:Power MOSFET select:Power dissipaiion(condition+switching losses) can be estimated as

12、:Pd=lo*Rds(on)*Dmax+t0.5Vin*lo*(tr+tf)*fsJ =(3*3*0.035*0.78)+0.5*5水39.3*10人( 6)*(110*103)=0.5W(Vin=5V)Rds(on)=35m Q T CEM4435tr+tf=300nsDesign Exam pieStep5:Tj=Ta+(RojA* Pd) =55+(50*0.5) =8(rcTa=Ambient degree Assume=55 °CRejA=50 °C /WThermal resistanceStep6:Rectifier Diode:P d=Io* Vd*( 1

13、-Dmin) =3*05*(l-0.55) =0675W(Vin=7V)Tj=Ta+(RojA*Pd)=55+( 15*0.675)=65 °CDesign Exam pie Boost converterStep2:L>=Vin(min)-Vds(sat)*DmaxlL*fSAli2*Io(niin)*Vo/Vin(niin)=2*0.05*12/5=0.24L= (5-0.1)*0.6/0.24*(l 10*103) =111.36uHSo we can choose 120uHDesign Exam pie Boost converterStep2:selection o

14、f the output capacitorCout>=lo(max)*Dmax/fs*AVo=0.3*0.6/110 時 0.05=32.73uHIpk=Io(max)/l-DmaxJ+Vin(max)*D/2*fs*L=0.3/1 0.6+7*0.6/2* 110k*l 20u=0.91AESR<= AVoZIPK=0.05/0.91=55m QDesign Exam pie Boost converter Step3:selection of power switch(MOSFET)PM OSFET=Ipk 八 2*Rds(on)*Dmax+0.5*Vin(max)*I p k*(tr+tf)*fsAssume Ta=55 °CRejA=50 °C/WPd=0.91 *0.91 *0.0135*0.6+0.5*7*0.91 *0.3u* 11 Ok =0.112WIj=Ta+(R0jA* Pd)=60.6°C Boost converterSte