PWM开关电源分类buck, boost介绍及主要元件参数选择

ESR<= △Vo/IPK=0.05/0.91=55mΩ
Design Example
• Boost converter • Step3:selection of power switch(MOSFET) PMOSFET=Ipk^2*Rds(on)*Dmax+[0.5*Vin(max)*I pk*(tr+tf)*fs] Assume Ta=55 ° C RθJA=50 ° C/W Pd=0.91*0.91*0.0135*0.6+[0.5*7*0.91*0.3u*110k] =0.112W Tj=Ta+(RθJA*Pd)=60.6° C
Design Example
• Boost converter Step4:selection of power Rectifier
Pd=Ipk*Vd=0.91*0.5=0.455W Tj=Ta+(RθJA*Pd)=55+(15*0.455)=61.8° C
Advantage and Disadvantage
Advantage of Linear
Cost lower Low output noise and ripple Good outf Linear
Efficiency lower Need bigger transformer and heat sink Narrow input voltage range Low hold up time (~1ms)
Design Example
• Boost converter Step2:selection of the output capacitor Cout>=Io(max)*Dmax/fs*△Vo
=0.3*0.6/110k*0.05 =32.73uH Ipk=[Io(max)/1-Dmax]+Vin(max)*D/2*fs*L =0.3/1-0.6+7*0.6/2*110k*120u=0.91A
• DC-DC circuit Boost converter
iIN L iL iD D iC Q iS C RL Vo iO + Vi PWM control
Toff Ton
Switching circuit
• Boost converter 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 + VL D iIN iL iO
Design Buck converter
• How to select Cout? Esr=Vripple/2Io(min) Cout>=△ IL/8*fs* △Vo • How to select Diode? Vrrm>=1.25Vin(max) Ipk=Io(max)-Io(min)
Switching circuit
Step2: 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=0.6A fs=110kHz
So we can choose 22uH
Design Example
Step3:
Thermal Consideration
• Switching IC Thermal consideration: • Tj=Ta+ PLoss*RθJA Where Tjjunction temperature Taambient temperature PLosstotal dispation RθJA thermal resistance(die junction to ambient air)
Switching circuit
• DC-DC circuit Buck converter
iS Q iD D C L iL iC RL Vo iO +
Vi Ton Toff PWM control
VD +
-
Switching circuit
• • • • • Buck converter function description Q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Q off)
Thermal Consideration
• Select Heat sink RθSA=RθJA-RθJC-RθCS Where RθJCThe thermal resistance between IC chip(junction point) and package backside connecting to the heat sink RθSAThe thermal resistance of Heat sink RθCS The thermal resistance between package backside and the heat sink
Switching circuit
• Buck converter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
iS
Q
+ iL
VL L
iO + iC C RL Vo
Vi
D
-
Switching circuit
Design Example
• Buck converter Example:Vin=5~6V,Vout=3.3V Step1: Duty ratio D=Vo+Vd/Vin-Vds(sat)=Ton/Ts Assume Vd=0.5V,Vds(sat)=0.1V D=0.64(Vin=6V)
Design Example
How to separate Linear or Switching ?
Linear : 將一 DC 電壓經由電路元件,直接 線性轉換為另一DC 電壓. For example:
DIODE
TRANSFORMER
5V Vf ~= 0.7V
AP1117 5V
3 IN ADJ OUT 2
4.3V
12V 12 Turns GND
Design Example
Step5: Tj=Ta+(RθJA*Pd) =55+(50*0.5) =80° C Ta=Ambient degree Assume=55 ° C RθJA=50 ° C /WThermal resistance
Design Example
Step6: Rectifier Diode: Pd=Io*Vd*(1-Dmin) =3*0.5*(1-0.55) =0.675W(Vin=7V) Tj=Ta+(RθJA*Pd)=55+(15*0.675)=65 ° C
5V 5 Turns GND
3.3V
Cin 10uF
1
Cout 100uF
How to separate Linear or Switching ?
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.
Design Example
• Step7: Input capacitor: 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
Design Example
电源分类Classification of Power
Supply
线性电源Linear : Regulator or LDO ( Low drop out ) 开关电源Switching :
Buck conversion Boost conversion Buck-Boost conversion(SEPIC,Cuk…)
L Vi Q C
+ iC RL
Vo -
Boost converter Design
• How to select inductor?
L= Vin[(min)-Vds(sat)]*Dmax △IL*fs Vo/Vi=1/1-D D=ton/Ts
Thermal Consideration
• Switching IC Thermal consideration • Ploss=PQ+Psat+Ps where PQ:quiescent loss Psat:saturation loss Ps:switching loss
Cout>= △IL/8*fs* △Vo =0.6/8*110*10^3*0.033 =20.66uF ESR<= △Vo/ △Io=0.033/0.6=0.055Ω
Design Example
Step4: Power MOSFET select: Power dissipation(condition+switching losses) can be estimated as: 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
• How to select inductor L>=[Vin(min)-Vsat-Vout]*Ton(max) 2Io(min) Vo/Vi=D=ton/Ts
Design Buck converter
• How to select Cin? Cin電壓rating至少為 Vin 1.5倍 Iin的計算方式如下表所示:
• Boost converter Step2:
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) =111.36uH So we can choose 120uH
• Boost converter Example Vin=5~7V,Vo=12V Step1:
Duty ratio D=Vo+Vd-Vin(min)/Vo+Vd-Vds(sat) The Duty cycle for Vin=5,6,7 is 0.6,0.52,0.44
Design Example
L Vi Q iS C
+ iC RL Vo -
Switching circuit
• Boost converter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 - VL + iIN iL iD iO
Advantage and Disadvantage
Advantage of switching
High frequency operating to reduce size of inductor and capacitor Light weight and Small total size High efficiency and power density High input voltage range and hold up time(~25ms) Disadvantage of switching High ripple voltage Circuit design complicated Higher EMI interference
• Buck converter 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 - VL + Q iO iL
L Vi D C
+ iC RL Vo
-
Design Buck converter