Inrush Current浪涌电流抑制.pdf

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1、14-1-14Inrush Current Elliott Sound ProductsBy Rod Elliott(ESP)Page Updated 25 August 2013Inrush Current MitigationMoShare|ShShShShArticlesMain IndexContentsIntroductionWarning1-What Is Inrush Current?2-Filament&Other Lamps3-Power Factor Correction4-Inductive&Transformer Inrush5-Capacitive Inrush6-P

2、assive Inrush Limiting6.1-Thermistor6.2-Resistor6.3-Bypass Systems7-Active Inrush Limiting7.1-Zero Voltage Switching7.2-MOSFET Limiting7.3-Ultimate Active Limiting(?)ConclusionReferencesIntroductionInrush current explained very simply is the current drawn by a piece of electrically operated equipmen

3、twhen power is first applied.It can occur with AC or DC powered equipment,and can happen evenwith low supply voltages.By definition,inrush current is greater than the normal operating current of the equipment,and theratio can vary from a few percent up to many times the operating current.A circuit t

4、hat normally draws1A from the mains may easily draw 50 to 100 times that when power is applied,depending on thesupply voltage,wiring and other factors.With AC powered equipment,the highest possible inrushcurrent also depends on the exact time the load is switched on.In some cases,its best to apply p

5、ower when the mains is at its maximum value(peak of RMS=nominal voltage*1.414),and with others its far better to apply power as the AC waveform passesthrough zero volts.Iron core transformers are at their best behaviour when the mains is switched on atthe peak of the waveform,while electronic loads(

6、rectifier followed by a filter capacitor for example)prefer to be switched on when the AC waveform is at zero CurrentThis is a surprisingly complex topic,and one that is becoming more important than ever before.Moreand more household and industrial products are using switchmode power supplies,and th

7、ey rangefrom just a few Watts to many hundreds of Watts.Almost all of these supplies draw a significant over-current when power is applied,and almost no-one gives any useful information in theirdocumentation.Please note that the descriptions and calculations presented here arefor 230V 50Hz mains.Thi

8、s is the nominal value for Australia andEurope,as well as many other countries.The US and Canada,alongwith a few other countries,use 120V 60Hz.This is not a problem-allformulae can be recalculated using whatever voltage is appropriate.Formost examples,the frequency is(more or less)immaterial.It is n

9、ot possible to provide a lot of detail for every example,so in many cases a considerable amountof testing or background knowledge may be needed before you will be able to make use of theinformation here.In addition,component suppliers do not always provide information in the sameway,and some info in

10、cluded by one supplier is omitted by others.This can make selection achallenge at times.WARNING:This article describes circuitry that is directly connected to the AC mains,and contact with any part of the circuit may result in death or serious injury.By readingpast this point,you explicitly accept a

11、ll responsibility for any such death or injury,and holdElliott Sound Products harmless against litigation or prosecution even if errors oromissions in this warning or the article itself contribute in any way to death or injury.Allmains wiring should be performed by suitably qualified persons,and it

12、may be an offencein your country to perform such wiring unless so qualified.Severe penalties may apply.1-What Is Inrush Current?Inrush current is also sometimes known as surge current,and as noted above is always higher thanthe normal operating current of the equipment.The ratio of inrush current to

13、 normal full-load currentcan range from 5 to 100 times greater.A piece of equipment that draws 1A at normal full load maybriefly draw between 5 and 100A when power is first applied.This current surge can cause component damage and/or failure within the equipment itself,blownfuses,tripped circuit bre

14、akers,and may severely limit the number of devices connected to a commonpower source.The following loads will all have a significant inrush current,albeit for very differentreasons.Incandescent lamps using a tungsten filament(AC or DC powered,any voltage)Fluorescent and other gas discharge lamps(inc

15、luding compact fluorescent types)Power transformers,especially toroidal types of 500VA or morePower supplies that obtain a low voltage AC from a transformerElectronic power supplies,as commonly used for personal computers,wall supplies,etc.Electronic power supplies with active power factor correctio

16、n(PFC)Electric motors of all types,with the greatest problems caused when starting under load fromrestCRT computer monitors and TV receivers.Inrush is deliberately created to operate thedegaussing coil(s)The list above covers a great many products,and with modern electronics infiltrating almost ever

17、yhousehold and industrial item used it actually covers just about every product available.Few modernproducts are exempt from inrush current-at least to a degree.Some of the most basic items we usedo not have an issue with inrush current at all-most are products that use heating coils made Currentnic

18、hrome(nickel-chromium resistance wire)or similar.The current variation between cold and fulltemperature is generally quite small.This applies to fan assisted,column and most radiant heaters,toasters and electric water heating elements.Apart from these few products,almost everything elsewill have a s

19、ignificant inrush current.In some cases,we can ignore the inrush current because it is comparatively small,and/or extremelybrief.A few products may draw only double their normal running current for a few mains cycles,whileothers can draw 10,50 or 100 times the normal current,but for a very short tim

20、e(often only a fewmilliseconds).Some products can draw many times their normal current for an extended period-electric motors with a heavy starting load or power supplies with extremely large capacitor banksbeing a couple of examples.2-Filament&Other LampsAlthough they are being banned(either by dec

21、ree or stealth)all over the world,there are still manyincandescent lamps in use,and this will not stop any time soon.Most traditional filament(incandescent)lamps are known to fail at the instant of turn-on.This is for two reasons-the filament iscold so has much lower resistance than normal,and being

22、 cold it is also more brittle.When power is applied,there is a high current surge,along with thermal shock and rapid expansionof the tungsten.This doesnt affect the lamp initially,but as the filament ages it becomes thinner andmore brittle,until one day it just breaks when turned on.For very large l

23、amps used for theatricallighting(amongst other things),the solution is to preheat the filament-just enough power is applied tokeep the filament at a dull red.Full power is almost never applied instantly-it is ramped up so thelamp appears to come up to full brightness very fast,but this is a simple t

24、rick that works because theresponse of our eyes is quite slow.The cold resistance of a tungsten filament is typically between 1/12 to 1/16 of the resistance whenhot.Based on this,it might be expected that the initial inrush current for a cold filament would be 12 to16 times the current at rated powe

25、r.The actual initial inrush current is generally limited to somesmaller value by external circuit impedance,and is also affected by the position on the AC waveformat which the voltage is applied.I measured the cold resistance of a 100W reflector lamp at 41 ohms,and at 230V(assuming thepower figure i

26、s accurate)the resistance will be around 530 ohms-a ratio of 12.9:1 and comfortablywithin the rule of thumb above.The time for the initial inrush current to decay to the rated current is determined almost entirely by thethermal mass of the filament,and ranges from about 0.05 seconds in 15W lamps to

27、about 0.4seconds in 1500W lamps.1 This varies with the rated voltage too-a 12V 50W lamp has a muchthicker(and therefore more robust)filament than a 230V 50W lamp for example.If incandescentlamps arealways either faded up with a dimmer or use some kind of current limiter,they will typicallylast at le

28、ast twice as long as those that are just turned on normally.Traditional(iron-core ballast and starter)fluorescent lamps also draw a higher current during theswitch-on cycle.During the startup process,there are filaments at each end of the tube that areheated,and this draws more current than normal o

29、peration.Contrary to what you might hearsometimes,this startup current is typically only between 1.25 and 1.5 times the normal current,and itis not better to leave fluorescent lamps on than to switch them off when you leave the room.However,constant switching will reduce the life of the tube,so ther

30、e is a compromise that depends on theapplication.Power factor correction(PFC)capacitors are used in parallel with many fluorescent lamp ballasts,especially those designed for industrial use.These are necessary to minimise the excess currentdrawn by a relatively linear but reactive load.When power is

31、 turned on,the inrush current may be Currenthigh-typically up to 30 Amps or more depending on the exact point in the main cycle when power isapplied!This is many times the rated current of the PFC capacitor(as determined by thecapacitance,voltage and frequency).Many fluorescent tube lights are now u

32、sing the relatively new T5 tubes,and these are specificallydesigned to use electronic ballasts.Even the older T8 tubes will give more light output with a highfrequency electronic ballast,and we will eventually see the iron-core ballast disappear completely,The electronic versions can be made to be m

33、ore efficient,but they wont last anywhere near as long.Some of the efficiency gained will be lost again when the ballast(or the entire fitting)has to bereplaced because a$0.10 part has failed.Many other lamps also have(often very)high inrush currents,but these will not be covered here.3-Power Factor

34、 CorrectionThis is such an important topic that some explanatory notes are essential.Why is it essential to know,you may well ask.Simply because so few modern loads are resistive,and power factor correction(PFC)is(or will be)used in a vast array of products.Many loads that currently have little or n

35、o PFCwill be required to perform very much better in the future,and this has already happened with somecategories of equipment.Many PFC circuits draw very high inrush current when switched on.If youwant a more in-depth explanation of power factor,see Power Factor-The Reality.Power factor is not well

36、 understood by many people,and even some engineers have great difficultyseparating the causes of poor power factor.Simply stated,power factor is the ratio of real power(inWatts)to apparent(or imaginary)power(in Volt-Amps or VA).It is commonly believed(but onlypartially correct under some specific ci

37、rcumstances)that power factor is measured by determiningthe phase angle between the voltage and current(commonly known as Cos(Cosine Phi-thecosine of the phase angle).This is an engineering shorthand method,anddoes not apply with anyload that distorts the current waveform(more on this shortly).An in

38、ductive load such as an unloaded transformer will draw current from the mains,but will consumealmost no power.Fluorescent lamps use a ballast-an inductor that is in series with the tube.Similararrangements are used with other types of discharge lighting as well.For the sake of simplicity,wewill use

39、a resistive load of 100 ohms in series with a 1H(1 Henry)inductor.Voltage is 230V at 50Hz,so the reactance of the inductor is 314 ohms.Total steady-state circuit current is shown in Figure 1,for both the inductive and capacitive sections.The inductive current lags the applied voltage by about72,and

40、the capacitive currentleads the voltage by 90(voltage is not shown as it would make thegraph too difficult to read).Figure 1-Test Circuit With Voltage And Current W CurrentWithout the capacitor(C1),the mains current is 698mA(700mA near enough)in this circuit-anapparent power of 161VA.However,the pow

41、er consumed by the load(R1)is only 48.7W-698mAthrough 100 ohms.Therefore,the power factor(PF)is.PF=P R/P AWhere PR is real power and PA is apparent powerPF=48.7/161=0.3This is considered very poor,because the power company and your wiring must supply the full700mA,but only a small fraction is being

42、put to good use(about 213mA in fact),and only about 70Vof the input voltage is available for the 100 ohm load.The majority of the current is out of phase withthe voltage,and performs no work at all.This type of load is very common(all inductive loads in fact),and is easily fixed by cancelling the in

43、ductance with a parallel capacitor.Scams that claim that a sillypower factor correction capacitor will make motors run cooler are obviously false-the inductivecurrent is not changed!For the above circuit,the capacitance needed is about 9uF and it will draw around 650mA(again at230V,50Hz).Because the

44、 capacitive and inductive currents are almost exactly 180 out of phasewith each other,the reactive parts cancel as shown by the graph.As a result,the generator onlyneeds to supply the 48.7W used by the load,and the supply current falls to 213mA-exactly the valueneeded to produce 48.7W in a 100 ohm l

45、oad(ignoring losses).The current we measured in theinductor(698 mA)does not change when the capacitor is added.The difference is that the majorityis supplied by thecapacitor and not the mains.One of the greatest problems with the idea of power factor is that many of the claims do not appear tomake s

46、ense.The above example being a case in point-it seems unlikely that adding a capacitor todraw more current will actually cause it to fall.To understand what is going on requires a goodunderstanding of reactive loads,phase shift and phase cancellation-even though some of it mightseem nonsensical,its

47、all established science and it does work.For example,a leading phase angleimplies that the current occurs before the voltage that causes it to flow,and while this might seemimpossible,it is what happens in practice.It usually only takes a few cycles to set up the steady stateconditions where this oc

48、curs.Where some of the old-timers(and the not-so-old as well)get their knickers in a twist is when the loadcurrent is distorted.It has been argued(wrongly)on many a forum that the voltage and current are inphase,so power factor is not an issue.This is completely wrong-those who argue thus haveforgot

49、ten that the Cos method is shorthand,and only applies when both voltage and current aresinewaves.It has also been argued(and again wrongly)that the capacitance following the bridgerectifier creates a leading power factor.It doesnt!By definition,a reactive load returns the unusedcurrent back to the m

50、ains supply,but this cannot happen because of the diodes.Non-linear circuitshave a poor power factor because the current waveform is distorted,not because of phase shift.Note in the graph below that there actually is a displacement,with the maximum current peakoccurring slightly before the voltage p