《多路輸出隔離式開關(guān)電源設(shè)計》 Off_line_Converter_Design_and _magnetics

1、C H A P T E R3Off-line Converter Design and MagneticsC H A P T E R3Off-line Converter Design and Magnetics Off-line converters are derivatives of standard dc-dc converter topologies.For example,the yback topology,popular for low-power applications(typically<100W,is really abuck-boost,with its usu

2、al single-winding inductor replaced by an inductor with multiple windings.Similarly,the forward converter,popular for medium to high powers,is abuck-derived topology,with the usual inductor(“choke”supplemented by a transformer. Theyback inductor actually behaves both as an inductor and a transformer

3、.It stores magnetic energy as any inductor would,but it also provides“mains isolation”(mandated for safety reasons,just like any transformer would.In the forward converter,the energy storage function is fullled by the choke,whereas its transformer provides the necessary mains isolation.Because of th

4、e similarities between dc-dc converters and off-line converters,most of the spadework for this chapter is in fact contained in the preceding chapter(“DC-DC Converter Design and Magnetics”.The basic magnetic denitions have also been presented therein. Therefore,the reader should read that chapter bef

5、ore attempting this one.Note that in both theyback and the forward converters,the transformer,besides providing the necessary mains isolation,also provides another very important functionthat of axed-ratio down-conversion step,determined by the“turns ratio”of the transformer.The turns ratio is the n

6、umber of turns of the input(“primary”winding,divided by the number of turns of the output(“secondary”winding.The question ariseswhy do we even feel the need for a transformer-based step-down conversion stage,when in principle,a switching converter should by itself have been able to up-convert or dow

7、n-convert at will?The reason will become obvious if we carry out a sample calculationwe will thennd that without any additional“help,”the converter would require impractically low values of duty cycleto down-convert from such a high input voltage to such a low output.Note that theworst-case ac mains

8、 input(somewhere in the worldcan be as high as270V.So when this ac voltage is rectied by a conventional bridge-rectier stage,it becomes a dc rail of almost 2×270=382V,which is fed to the input of the switching converter stage that follows. But the corresponding output voltage can be very low(5V

9、,3.3V,or1.8V,and so on,so the required dc transfer ratio(conversion ratiois extremely hard to meet,given the minimum on-time limitations of any typical converter,especially when switching at high frequencies.Therefore,in both theyback and forward converters,we can intuitively think of the transforme

10、r as performing a rather coarsexed-ratio step-down of the input to a more129Chapter3amenable(lowervalue,from which point onwards the converter does the rest(including the regulation function.Flyback Converter MagneticsPolarity of Windings in a TransformerIn Figure3-1,the turns ratio is n=n P/n S,whe

11、re n P is the number of turns of the primary winding,and n S is the number of turns of the secondary winding. V INV IN +OFigure3-1:Voltage and Currents in a Flyback130Off-line Converter Design and Magnetics We have also placed a dot on one end of each of the windings.All dotted ends of a transformer

12、 are considered to be mutually“equivalent.”All non-dotted ends are also obviously mutually equivalent.That means that when the voltage on a given dotted end goes “high”(to whatever value,so does the voltage on the dotted ends of all other windings. That happens because all windings share the same ma

13、gnetic core,despite the fact that they are not physically(galvanicallyconnected to each other.Similarly,all the dotted ends also go“l(fā)ow”at the same time.Clearly,the dots are only an indication of relative polarity. Therefore,in any given schematic,we can always swap the dotted and non-dotted ends of

14、 the transformer,without changing the schematic in the slightest way.In theyback,the polarity of the windings is deliberately arranged such that when the primary winding conducts,the secondary winding is not allowed to do so.So when the switch conducts,the dotted end at the drain of the mosfet in Fi

15、gure3-1goes low.And therefore,the anode of the output diode also goes low,thereby reverse-biasing the diode. We should recall that the basic purpose of a buck-boost(which this in fact also isisto allow incoming energy from the source during the switch on-time to build up in the inductor(only,and the

16、n later,during the off-time,to“collect”all this energy(and no moreat the output.Note that this is the unique property that distinguishes the buck-boost (and theybackfrom the buck and the boost.For example,in a buck,energy from the input source gets delivered to the inductor and the output(during the

17、 on-time.Whereas,in a boost,stored energy from the inductor and the input source gets delivered to the output(during the off-time.Only in a buck-boost do we have complete separation between the energy storage and the collection process,during the on-time and theoff-time.So,now we understand why they

18、back is considered to be just a buck-boost derivative.131Chapter3Transformer Action in a Flyback,and Its Duty CycleClassic“transformer action”implies that the voltages across the windings of the transformer, and the currents through each of them,scale according to the turns ratio,as described in Fig

19、ure3-1.But it is perhaps not immediately apparent why theyback inductor exhibits transformer action.When the switch turns ON,a voltage V IN(the rectied ac inputgets impressed across the primary winding of the transformer.And at the same time,a voltage equal to V INR=V IN/n (“R”for reectedgets impres

20、sed across the secondary winding(in a direction that causes the output diode to get reverse-biased.Therefore,there is no current in the secondary winding when the primary winding is conducting.Let us calculate what V INR is.This voltage translation across the isolation boundary follows from the indu

21、ced voltage equation applied to each windingV P=n P ddt and V S=n S ddtNote that both windings enclose the same magnetic core,so theuxis the same for both, and so is the rate of change ofux d/dt for each winding.ThereforeV S=n S× V Pn PorV S=n S× V IN n P =V IN nV INRAlso,V P n P=V S n SV

22、PV S=nThis above equation represents classic“transformer action”with respect to the voltages involved.But we also learn from the preceding equation that the Volts/turn for any winding (at any given instantis the same for all the windings present on a given magnetic coreand this is what eventually le

23、ads to the observed voltage scaling.Note also that voltage scaling in any transformer occurs irrespective of whether a given winding is passing current or not.That is because,whether a given winding is contributing to132Off-line Converter Design and Magnetics the netuxpresent in the core or not,each

24、 winding encloses this entireux,and so thebasic equation V=N×d/dt applies to all windings,and so does voltage scaling.We know that energy is built up in the transformer during the on-time.When the switchturns OFF,this stored energy(and its associated currentneeds toyback/freewheel.We alsoknow t

25、hat the voltages will automatically try to adjust themselves in any possible way,so asto make that happen.So we can safely assume the diode will somehow conduct during theswitch off-time.Now,assuming we have reached a“steady state,”the voltage on the outputcapacitor has stabilized at somexed value V

26、 O.Therefore,the voltage at the secondary-sideswitching node gets clamped at V O(ignoring the diode drop.Further,since one end of thesecondary winding is tied to ground,the voltage across this winding is now equal to V O.Bytransformer action,this reects a voltage across the primary winding,equal to

27、V OR=V O×n. But the switch is OFF during this time.Therefore,under normal circumstances,the voltageat the primary-side switching node would have settled at V IN.However,now this reectedoutput voltage V OR,coming through the transformer,adds to that.Therefore,the voltage atthe primary-side switc

28、hing node eventually goes up to V IN+V OR(for now,we are ignoring the turn-off spike encircled in Figure3-1.Note:During the on-time,the primary side is the one determining the voltages across all the windings.And during the off-time,it is the secondary winding that gets to“call the shots”!We can cal

29、culate duty cycle from the most basic equation(from voltseconds lawD=V OFF V OFF+V INWe have the option of performing this calculation,either on the primary winding,or the secondary winding.Either way,we get the same result,as shown in Table3-1.Table3-1:Derivation of dc transfer function ofybackPrim

30、ary Winding Secondary WindingV ON V IN V INRV IN/nV OFF V ORV O×n V ODC Transfer FunctionD=V OFFV ON+V OFFD=V ORV IN+V OR D=V OV INR+V OD=nV OV IN+nV O133Chapter3Tond out the(absolutevoltage at the swinging end of any winding,we can use the following level-shifting rule:To get the absolute valu

31、e of the voltage at the swinging end of any winding,we must add to the voltage across the winding,the dc voltage present at its“non-swinging”end.So,for example,to get the voltage at the drain of the mosfet(swinging end of primary winding,we need to add V IN(voltage at other end of winding,to the vol

32、tage waveform that represents the voltage across the primary winding.That is how we got the voltage waveforms shown in Figure3-1.Coming to the question of how currents actually reect from one side of the transformer to the other,it must be pointed out that even though thenal current scaling equation

33、s of ayback transformer are exactly the same as in the case of an actual transformer,this is not strictly“classic transformer action.”The difference from a conventional transformer is,that in theyback,the primary and secondary windings do not conduct at the same time.So in fact,it is a mystery why t

34、heir currents are related to each other at all!The current scaling that occurs in ayback actually follows from energy considerations.The energy in a core is in general written asE=12LI2We know the windings of ouryback conduct at different times,but the energy associated with each of the currentows m

35、ust be equal to the energy in the core,and must therefore be equal to each other(we are ignoring the ramp portion of the current here for simplicity. Therefore,E=12L P I P2=12L S I S2where L P is the inductance measured across the primary windingwith the secondary windingoating(no current,and L S th

36、e inductance measured across the secondary windingwith the primary windingoating.But we also know thatL=N2×A L×109Henry134Off-line Converter Design and Magnetics where A L is the inductance index,dened previously.Therefore in our case we getL P=n P2×A L×109L S=n S2×A L×

37、109Substituting in the energy equation,we get the well-known current scaling equationsn P I P=n S I SorI PI S=1 nWe see that analogous to the V olts/turns rule,the ampere-turns also need to be preserved at all times.In fact,the core itself doesnt really“care”which particular winding is passing curre

38、nt at any given moment,so long as there is no sudden change in the net ampere-turns of the transformer.This becomes the“transformer-version”of the basic rule we learned in Chapter1that the current through an inductor cannot change discontinuously.Now we see that the ampere-turns of a transformer can

39、not change discontinuously. Summarizing,transformer action works as followswhen reecting a voltage from primary to secondary side,we need to divide by the turns ratio.When going from the secondary to the primary side,we need to multiply by the turns ratio.The rule reverses for currentsso we multiply

40、 by the turns ratio when going from primary to secondary,and divide in the opposite direction.The Equivalent Buck-boost ModelsBecause of the many similarities,and also because of the way voltages scale in the transformer,it becomes very convenient(most of the timeto study theyback as an equivalent d

41、c-dc(inductor-basedbuck-boost.In other words,we separate out the coarsexed-ratio step-down ratio and incorporate it into equivalent(reectedvoltages and currents.We thereby manage to reduce theyback transformer into a simple energy-storage medium,just like any conventional dc-dc buck-boost inductor.I

42、n other words,for most practical purposes,the transformer goes“out of the picture.”The advantage is that almost all the equations and design procedure we can write for a conventional buck-boost now apply to this equivalent buck-boost model.One exception to this is the leakage inductance issue(and ev

43、erything related to itthe clamp,the loss in efciency due to it,the turn-off voltage spike on the switch,and so on.We will discuss this exception later.But other than that,135Chapter 3all other parameters like the capacitor,diode,and switch currents for example,can be more readily visualized and calc

44、ulated if we use this dc-dc model approach.The equivalent dc-dc model is created essentially by reecting the voltages and currents across the isolation boundary of the transformer to one side.But again,as in the case of the duty cycle calculation (see Table 3-1,we have two options here we can either

45、 reect everything to the primary side,or everything to the secondary side.We thus get the two equivalent buck-boost models as shown in Figure 3-2.We can use the primary-sideequivalent model to calculate all the voltages and currents on the primary side of the original yback and the secondary-side eq

46、uivalent model for calculating all the currents and voltages on the secondary side of the originalyback. Primary-side equivalent model Secondary-side equivalent modelV IN vin i_in cin I vsw vo i_out center co vd Duty Cycle Current RippleRatio I IN V INR =V IN /nI INR =I IN xnC IN n 2xC INL p L S =L

47、P /n 2V SW V SW /nV OR =V O xn I OR =I O /n I OR /(1D=I O /nx(1DC O /n 2V D xn D r V OI OI O /(1DC OV DDrFigure 3-2:The Equivalent Buck-Boost Models of the Flyback136Off-line Converter Design and Magnetics We know that voltages and currents reect across the boundary by getting either multipliedor di

48、vided by the turns ratio.In fact,thereected output voltage,V OR,is one of the mostimportant parameters of ayback,as we will see.As the name indicates,V OR is effectivelythe output voltage as seen by the primary side.In fact,if we compare the switch waveformof theyback in Figure3-1with that of a buck

49、-boost,we will realize that to the switch,itseems as if the output voltage is really V OR.As an example,suppose we have a50W converter with an output of5V at10A,and aturns ratio of20.The V OR is therefore5×20=100V.Now,if we change the set output tosay10V and reduce the turns ratio to10,the V OR

50、 is still100V.We willnd that none ofthe primary-side voltage waveforms change in the process(assuming efciency doesntchange.Further,if we have also kept the output power constant in the process,that is,by changing the load to5A for an output at10V,all the currents on the primary side willalso be una

51、ffected.Therefore,the switch will never know the difference.In other words,theswitch virtually“thinks”that it is a simple dc-dc buck-boostdelivering an output voltageof V OR at a load current of I OR.As mentioned,the only difference between a transformer-basedyback that“thinks”it isproviding an outp

52、ut of V OR at the rate of I OR,and an inductor-based version that really isproviding an output of V OR at the rate of I OR,is theleakage inductanceof theybacktransformer.This is that part of the primary side inductance that is not coupled to thesecondary side and therefore cannot partake in the tran

53、sfer of useful energy from the input tothe output.We can conrm from Figure3-1,that the only portion of the primary-side(switchvoltage waveform that“doesnt make it”to the secondary side is the spike occurringjust after the turn-off transition.This spike comes from the uncoupled leakage inductance,as

54、we will soon see.Note that in the equivalent buck-boost models,the reactive component values also getreectedthough as the square of the turns ratio.We can understand this fact easily fromenergy considerations.For example,the output capacitor C O in the originalyback wascharged up to a value of V O.S

55、o its stored energy was1/2C O V O2.In the primary-sidebuck-boost model,the output of the converter is V OR,that is,V O×n.Therefore,to keep the energy stored in this capacitor invariant(in the dc-dc model,as in theyback,the outputcapacitance must get reected to the primary side according to C O/

56、n2.Note also fromFigure3-2how the inductance reects.This is consistent with the fact that LN2.The Current Ripple Ratio for the FlybackLooking at the equivalent buck-boost models in Figure3-2,the center of the ramp on thesecondary side(average inductor current,“I L”must be equal to I O/(1D,as for abu

57、ck-boost(because the average diode current must equal the load current.Thissecondary-side“inductor”current gets reected to the primary side,and so the center of theprimary-side inductor current ramp is“I LR,”where I LR=I L/n.Equivalently,it is137Chapter3equal to I OR/(1D,where I OR is the reected lo

58、ad current,that is,I OR=I O/n.Similarly, the current swings on the primary and secondary sides are also related,via scaling(turns ratio n.Therefore,we see that the ratio of the swing to the center of the ramp is identical on both sides(primary-and secondary-side dc-dc models.We are thus in a position to dene a current ripple ratio r for theyback topology toojust as we did for a dc-dc converter.We just need to visualize r in a slightly different manner this timein terms of the center of the ramp(switch or diode,rather than the dc i