boost电感设计

1、Experiment 2Boost Switching DC-DC Power ConverterConverter power stage (Exp. 2, next week)V+g50 VPWMOpen-loop controller(Exp. 1)+V100 VCompVref+Feedback controller(Exp. 2, weeks 2 and 3)Power Electronics Laboratory1Experiment 2: Boost converterweek 1Todays lectureObjectivesUnderstanding the boost co

2、nverterWaveformsOperation of transistor and diodeAnalysis of boost circuitBasic approach and approximationsDesign equationsInductor designDesign constraintsDesign procedurePower Electronics Laboratory2Experiment 2: Boost converterweek 1Last weeks lecture:Buck converterSPDT switch changes dc componen

3、tSwitch output voltage waveformDuty cycle D:0 D 1complement D:D = 1 - DVg+vs(t) 0switchposition:1+2vs(t)Rv(t)VgDTs D Ts 0DTsTst121Power Electronics Laboratory3Experiment 2: Boost converterweek 1DC component of switch output voltagevs(t)Vg = DVgarea =DTsVg00DTsTstFourier analysis:Dc component = avera

4、ge value1Tsv =v (t) dtTsss0v= 1 (DT V ) = DVsTssggPower Electronics Laboratory4Experiment 2: Boost converterweek 1Insertion of low-pass filter to remove switchingharmonics and pass only dc component1L+V+2vs(t)CRv(t)gV Vgv vs = DVg001DPower Electronics Laboratory5Experiment 2: Boost converterweek 1Th

5、ree basic dc-dc convertersBuckBoostBuck-boosta)1LiL(t)+Vg+2CRvb)L2iL(t)+1+VgCRvc)12+Vg+iL(t)CRvLM(D)M(D)M(D)10.80.60.40.200543210000-1-2-3-4-5M(D) = D0.20.40.60.81DM(D) =11 D0.20.40.60.81DD0.20.40.60.81M(D) = D1 DPower Electronics Laboratory6Experiment 2: Boost converterweek 1Analysis:Inductor volt-

6、 second balance, capacitor charge balance, and the small ripple approximationActual output voltage waveform, buck converterBuck converter containing practical low-pass filterActual output voltage waveformv(t) = V + vripple(t)1iL(t)L+vL(t) iC(t)+2+CRv(t)Vgv(t)actual waveformv(t) = V + vripple(t)VDc c

7、omponent V0tPower Electronics Laboratory7Experiment 2: Boost converterweek 1The small ripple approximationv(t)actual waveformv(t) = V + vripple(t)v(t) = V + vripple(t)VDc component V0tIn a well-designed converter, the output voltage ripple is small. Hence, the waveforms can be easily determined by i

8、gnoring the ripple: vripple Vv(t) VPower Electronics Laboratory8Experiment 2: Boost converterweek 1Boost converter analysis:Inductor current waveformL2iL(t) + vL(t) original+1Vgconverterswitch in position 1LiL(t)+ vL(t) iC(t)+Vg+CRvVg+iC(t)CRvswitch in position 2LiL(t) + vL(t) iC(t)+CRvPower Electro

9、nics Laboratory9Experiment 2: Boost converterweek 1Subinterval 1: switch in position 1Lets find the inductor voltage and currentInductor voltage and capacitor currentL+v = ViL(t) + vL(t) iC(t)LgiC= v / R+VgCRvSmall ripple approximation:vL = VgiC = V / RKnowing the inductor voltage, we can now find t

10、he inductor current viavL(t) = LdiL(t)dtSolve for the slope:diL(t)=vL(t)=Vg The inductor current increases withdtLLconstant slope Vg/LPower Electronics Laboratory10Experiment 2: Boost converterweek 1Subinterval 2: switch in position 2Inductor voltage and capacitor currentLv = V v+iL(t)+ vL(t) iC(t)L

11、giC= iL v / RVg+CRvSmall ripple approximation:vL = Vg ViC = I V / RKnowing the inductor voltage, we can again find the inductor current viav (t) = L diL(t)LdtSolve for the slope:diL(t)=vL(t)Vg V The inductor current changes with andtLLessentially constant, but different,slopePower Electronics Labora

12、tory11Experiment 2: Boost converterweek 1Inductor voltage and current waveformsvL(t)VgDTsDTstVg Vdi (t)v (t)iL(t)L=LdtLIiLVgVg VLL0DTsTstPower Electronics Laboratory12Experiment 2: Boost converterweek 1Inductor current waveformduring turn-on transientiL(t)di (t)v (t)VgL=L=dtLLiL(nTs)iL(n + 1)Ts)Vg v

13、iL(Ts)LiL(0) = 02TsnTs(n + 1)Ts t0 DTs TsWhen the converter operates in equilibrium:iL(n + 1)Ts) = iL(nTs)Power Electronics Laboratory13Experiment 2: Boost converterweek 1The principle of inductor volt-second balance:DerivationInductor defining relation:v (t) = L diL(t)LdtIntegrate over one complete

14、 switching period:1Tsi (T ) i (0) =v (t) dtLL sLL0In periodic steady state, the net change in inductor current is zero:T s0 =vL(t) dt0Hence, the total area (or volt-seconds) under the inductor voltage waveform is zero whenever the converter operates in steady state. An equivalent form:1Ts0 =v (t) dt

15、 =vTsLL0The average inductor voltage is zero in steady state.Power Electronics Laboratory14Experiment 2: Boost converterweek 1Inductor volt-second balance:Boost converter examplevL(t)Total area VgInductor voltage waveform, previously derived:DTs t Vg VIntegral of voltage waveform is area of rectangl

16、es: =Tsv (t)dt =VDT+V VDTwith D = 1 Dgsgs0LAverage voltage isvL= D Vg + D Vg VsThe voltage conversion ratio isEquate to zero and solve for V:VgV11Vg (D + D) V D = 0 V =M(D) = Vg= D= 1 DDPower Electronics Laboratory15Experiment 2: Boost converterweek 1Conversion ratio M(D) of the boost converterM(D)5

17、432100M(D) = 1 =1D1 D0.20.40.60.81DPower Electronics Laboratory16Experiment 2: Boost converterweek 1The principle of capacitor charge balance:DerivationCapacitor defining relation:i (t) = C dvC(t)CdtIntegrate over one complete switching period:v (T ) v (0) =1Ts(t) dtiCC sCC0In periodic steady state,

18、 the net change in capacitor voltage is zero:1Ts0 =iC(t) dt = iCTs0Hence, the total area (or charge) under the capacitor current waveform is zero whenever the converter operates in steady state. The average capacitor current is then zero.Power Electronics Laboratory17Experiment 2: Boost converterwee

19、k 1Determination of inductor current dc componentusing capacitor charge balanceCapacitor charge balance:TsVViC(t) dt = ( ) DTs + (I ) DTs0RRCollect terms and equate to zero: VR (D + D) + I D = 0Solve for I:I =VD REliminate V to express in terms of Vg:VgI = D2 RiC(t) I V/RDTs DTs t V/RI(Vg / R)864200

20、0.20.40.60.81DPower Electronics Laboratory18Experiment 2: Boost converterweek 1Determination of inductor current rippleInductor current slope during subinterval 1:diL(t) = vL(t) = VgdtLLInductor current slope during subinterval 2:diL(t) = vL(t) = Vg VdtLLiL(t) IiLVgVg VLL0DTsTstChange in inductor cu

21、rrent during subinterval 1 is (slope) (length of subinterval):Vg2iL =L DTsSolve for peak ripple:VgiL = 2L DTs Choose L such that desired ripple magnitude is obtainedPower Electronics Laboratory19Experiment 2: Boost converterweek 1Determination of capacitor voltage rippleCapacitor voltage slope durin

22、g subinterval 1:dvC(t) = iC(t) = VdtCRCCapacitor voltage slope during subinterval 2:dvC(t) = iC(t) =I VdtCCRCv(t)V v VIVRCCRC0DTsTstChange in capacitor voltage during subinterval 1 is (slope) (length of subinterval): 2v = RCV DTsSolve for peak ripple:v = 2RCV DTs Choose C such that desired voltage r

23、ipple magnitude is obtained In practice, capacitor equivalent series resistance (esr) leads to increased voltage ripplePower Electronics Laboratory20Experiment 2: Boost converterweek 1Realization of SPDT switch using transistor and diodeBoost converter with ideal SPDT switchRealization using power M

24、OSFET and diodeInductor current forward-biases diode when MOSFET is offL2+iL(t)+ vL(t) 1iC(t)+VgCRvLD1+iL(t)+ vL(t) iC(t)Vg+Q1CRv+DTs Tsvds(t)V0Q1 conductsD1 conductstPower Electronics Laboratory21Experiment 2: Boost converterweek 1Designing the filter inductorBoost example: inductor current wavefor

25、mi(t) I iL0DTsTstIn an inductor, magnetic field H(t) is proportional to winding current i(t) via Amperes law. Flux density B(t) is proportional to integral of winding voltage v(t) through Faradays law.Must avoid saturation of core: B(t) Rg :ni RgPower Electronics Laboratory24Experiment 2: Boost conv

26、erterweek 1First constraint: maximum flux densityGiven a peak winding current Imax, it is desired to operate the core fluxdensity at a peak value Bmax. The value of Bmax is chosen to be less than the worst-case saturation flux density of the core material, Bsat.From solution of magnetic circuit:ni =

27、 BAc RgLet I = Imax and B = Bmax :lnImax = BmaxAc Rg = Bmax g0This is constraint #1. The turns ratio n and air gap length lg are unknown.Power Electronics Laboratory25Experiment 2: Boost converterweek 1Second constraint: obtain desired inductanceMust obtain specified inductance L. We know that the i

28、nductance isL = n2 = 0 Ac n2RglgThis is constraint #2. The turns ratio n, core area Ac, and air gap length lg are unknown.Power Electronics Laboratory26Experiment 2: Boost converterweek 1Third constraint: winding areaWire must fit through core window (i.e., hole in center of core)corewire bare areaT

29、otal area ofAWcopper in window:nAWArea available for windingconductors:KuWAThird design constraint:KuWA nAWcore window area WAPower Electronics Laboratory27Experiment 2: Boost converterweek 1The window utilization factor Kualso called the “fill factor”Ku is the fraction of the core window area that

30、is filled by copperMechanisms that cause Ku to be less than 1: Round wire does not pack perfectly, which reduces Ku by a factor of 0.7 to 0.55 depending on winding technique Insulation reduces Ku by a factor of 0.95 to 0.65, depending on wire size and type of insulation Bobbin uses some window area

31、Additional insulation may be required between windings Typical values of Ku :0.5 for simple low-voltage inductor0.25 to 0.3 for off-line transformer0.05 to 0.2 for high-voltage transformer (multiple kV)0.65 for low-voltage foil-winding inductorPower Electronics Laboratory28Experiment 2: Boost converterweek 1Fourth constraint: winding resistanceThe resistance of the winding isR = lbAWw