Powering Studio Strobes Using An Inverter

Discussion in 'Lighting Equipment' started by dinsdale, May 19, 2006.

  1. How much wattage does a Norman P2000 power pack draw? And why couldn't you run
    a strobe system like this on location using an appropriate inverter or portable

    Do you have to keep the engine running when using an inverter like this?

    The advantage of using a studio strobe like this is the addition of a modelling
    light, and the batteries that don't last forever.

    I can't see purchasing a low-wattage portable strobe kit for occasional use.

    Has anyone tried powering a studio kit power pack off of an inverter?


  2. a 200 Watt strobe at 100 volts would draw approx 18 amps.

    Watts / volts = Amps

    Amps X volts = Watts

    An inverter in the sense you reference is used usually to up a 12 volt auto system to 110 volts.

    A generator has to be running to work they generally only have a very small capacity reserve for surge protection or for instant on systems. You could carry a car battery and inverter but would have to charge your battery after every shoot A marine deep cycle would serve Nicly but they weigh alot 40-75 pounds.

    If we are talking about taking 110 down to 12 volts we need a TR Transformer Rectifier.

    This is very doable it may be hefty but it is workable from an electronic stand point.

    BTW some electronics do not like modified sine waves so cheap inverters may not work with all electronics. Also watch inverters listing max watts. Be aware of Peak vs Cont. Watt output.
  3. That should read 2000 watt strobe at 110 volts = 18 amps.
    I am sorry I did not proof that I knew what I meant but no one else did.
  4. I agree with your assesment of the advantages.

    Yes you should keep the engine running.

    If you mean an inverter that plugs into a cigarette lighter, you are asking for trouble in car's electrical system. there are heavier duty ones that attach directly to your engine and are seperate from your car's electrical system.

    Check this old photo.net thread for a deeper discussion. I don't know how much has changed since then:

  5. Thanks, Donald.

    I see that power inverters are rated in wattage, so if my power pack draws 18 amps, multiplied times 110 volts (I presume), I would need a 1980 watt inverter, or one rated at 2000 watts?

    Is this correct?

    A large battery would be a possibility, but it's not necessary. My vehicle would be parked right next to the shoot, so I figured I could run the strobes off of an inverter. I'm looking at one online at Home Depot.com, rated at 2500 watts, for $250.00.

    I just can't see investing in a battery powered strobe system, like a Lumedyne, or Norman 400b.

    Would using an inverter like I propose require me to keep the engine running while I shoot? That way there isn't too much of a demand on the battery?

    Thanks again.

  6. The average draw of a studio lighting system isn't much, only slightly more than the rating of the modelling lights in Watts. But the peak draw is huge. Right after the flash fires, the charging circuit draws enormous current to fill the capacitors as quickly as possible. As the flash recycles, the current draw drops rapidly.
    That makes it hard to just read the ratings and determine what will work. Most published generator/inverter ratings are for continuous current draw. It would be overkill and horrendously expensive to buy an inverter or generator with a continous rating sufficient to handle the peak draw of a typical studio lighting system. To predict in advance what'll work, you need to know the peak ratings of your inverter/generator, and you should also know how the studio lighting system and the inverter/generator will react if the studio lighting system temporarily attempts to draw more than that peak. Often, the voltage will drop slightly out of spec, but if this only happens for a brief time, it may work OK. Or not.
    It's possible to end up with a combination that will work for occasional firings of the flash, but which will overheat if you repeatedly fire the flash without giving enough time for things to cool down between shots. The peak power draw temporarily exceeds the equipment's ability to get rid of heat.
  7. Sounds like an inverter setup might be too much trouble for the expense. And it may not work!

    I don't need to destroy my vehicle's electrical system while on location. It's a long walk home.

    Thanks, everyone.
  8. Listen to Richard.

    This "2000 watt strobe at 110 volts = 18 amps" - is wrong.

    Strobes are rated in WattSeconds or Joules, and not in Watts. (except any modelling continuos bulbs if there is any in them).

    There was a very lengthy dicsussion on the same subject last year and all came to right conclusion at the end, after quite a few wrong ansers.

    If you turn Off your modelling bulb, or remove it, your 2000 WattSeconds flash would operate safely on a inverter that is rated at about 500 Watts. (not WattSeconds). If you have huge modelling incandescent bulb in your strobe, than perhaps a bit larger inverter would be recommended. E.g. 500 Watts modelling bulb would eat all the power constantly, so use modelling smaller bulb, e.g. 150 Watts, or turn it Off.

    The instant surge of current loading empty capacitor is HUGE, and the supply current demant will not match that by any inverter, that is why you need to consider average values and not short peaks.

    Luckily that huge surcharge lasts only fraction of a second, and the current drops to low flow in another 1 or 2 seconds, and after that, the current is very minimal, and only for sustaining the operation of electronics, and occasional "tricle" charge to keep the capacitor at the top operating voltage.

    Average power used by strobes is low, unless you use fast repeating photography, that will change a bit. The Average power is the Key in this equation.

    I would suggest to make sure that fuses in the inverter are of SLOW-BLOW type, in an inverter of 500 Watts capacity, since FAST-Blow fuses could be blown away by the Strobe. But I believe standard fuses could be already Slow-Blow type, but cannot be sure about unknown inverter.
  9. Tom don't be discouraged. I was not aware your car would be so close. DC power is not efficient to transmit over long distances but you will not have that worry. Get an inverter that has alligator clips so you can hook it straight to the car battery (not through the cigertte lighter you could start a fire). As mentioned above the peak watts strobes use is only for a fraction of a second but the most dangerous thing for electronics is under voltage due to rapid heat build up just make sure the inverter exceeds the peak demand of the strobes.

    I think you can easily do this because strobes are only used for a fraction of a second each minute. I would not worry about the car electrical system so much, most car alternators are at least 85 amps some mini vans are over 130 amps. Think of all the stuff on your car powers radios, GPS, headlights, washers, heated seats. Etc ETc. Since your car can run them all at the same time and you would have none of this on during a shoot you have plenty of capacity. RV dealers can help you with inverter questions and should be knowledgeable. I would make sure I had enough capacity buy a little larger than needed. I own a 250 Watt inverter and my florescent shop light will not start with it not that 250 Watts is not enough it is that it requires a higher start or peak wattage to start.

    Good luck and I would love to see the setup in action some day keep us posted.
  10. Frank, 2000 watts was used as example. I do not own a norman p2000 and Tom did not know it's wattage. I certanly know the diffrence between peak and cont. The formulas are correct plug your own numbers in as needed.

    If you use to small of an inverter you will either

    1. Blow a fuse (Good)

    2. Damage a really expensive peice of equipment (Bad)

    3. Severly shorten the life of the Equipment (Bad)

    4. Or equipment will not work.

    Other than that yes you make a valid point. We need to know the peak draw of a norman p2000 with modeling light on.
  11. Tom here is a pretty good http://www.prophotocommunity.com/ubbthreads/showflat.php?Number=38764 to people mucking with generators. I can not find the specs other than the 1600 Norman power pack is rated to 15 Amps. So it is deffintly peaking under 15. Good luck. Do you have any friends that have an inverter you could try. Or a friend with an RV.
  12. Would using an inverter like I propose require me to keep the engine running while I shoot? That way there isn't too much of a demand on the battery?
    As I said earlier, yes it does require keeping your engine on unless you like having a dead battery. And has others have said it is the transitory peak voltage which will fry your car's electrical system. If you choose not to believe me, then find out the hard way. There is nothing like empirical knowledge gained from personal experience.
  13. Frank, 2000 watts was used as example. I do not own a norman p2000 and Tom did not know it's wattage. I certanly know the diffrence between peak and cont. The formulas are correct plug your own numbers in as needed.
    Strobes aren't rated in Watts. Strobes are rated in Watt-seconds, (or Joules, which is just another name for Watt-seconds).
    Watt-seconds are a unit of energy consumed in a single pop of the flash. Watts are a unit of power, or energy per unit time.
    I can't find the specs for that Norman unit on-line, but for an example of how you can't use the Watt-second rating of a strobe to figure current draw, see the Photogenic Powerlight 1250 and 2500 manual and specs.. Note that both the 500 W-s Powerlight 1250 and the 1000 W-s Powerlight 2500 are speced to draw 15 amps at 120V when they're charging (but they only draw a minuscule 0.2 amps standby). That's 1800 Watts for either model, even though one model is twice as powerful as the other, and neither model is rated at 1800 Watt-seconds. The 1800 Watts doesn't appear anywhere in the specs, but has to be arrived at by multiplying the spec'ed 15 amps by the spec'ed 120 Volts. The difference in power consumption between the two lights? The more powerful model has a slower recycling time, so it draws that charging current for a longer period of time for each pop of the flash.
    I actually own a couple of PL1250s, and I've put them on a meter to check current consumption. I've measured the peak at around 20 amps, but for such a very short time that I can't vouch for the accuracy of my meter reading. Note that those lights need a 15 amp slo-blo fuse, which indicates that the designers expect them to occasionally draw short bursts of more than the nominal 15 amps.
    Also, despite the fact that each light is spec'ed to require 15 amps, I routinely put two of them together on the same 15 amp household lighting circuit with no problems. That's because the peak draw is for such a short period of time that it doesn't trip the breaker, overheat house wiring, or cause any other difficulty. Strictly speaking, I'm overloading the circuit, and if you asked me, I wouldn't recommend anyone else doing this. But I know it works at my house.
    Some systems (e.g. Speedotron Blackline) have a switch to let you slow down the recycling speed. Why would anyone want to slow down the recycle rate? It cuts the peak power consumption. That might help when running off a generator or inverter, or even may prevent tripping a house breaker if you've got a bunch of packs plugged into an underrated circuit.
  14. nz


    If the car is so close why not stick a couple of batteries in the trunk? Wire them up so they charge while the car is running. Put a switch in so they don't drain the starting battery when the car is stopped. With big enough batteries you'd have no worries about needing to keep the car running. Plus batteries are the cheap part of this whole setup. A good sine wave inverter will set you back a chunk. But you can get some fairly large ones that can handle anything that your normal wall outlet can handle. Some have pretty good peak outputs to. Making it work isn't that hard IMHO. The question is it worth the cost? Speedotron has a 1500 watt second battery pack on thier website now. It's not cheap but I don't think you'll save much money buying the parts yourself and building something similar.
  15. Richard no crap who said they where rated in watts, I dont care how many WattSeconds the stobes are I dont care if they are a billion I was looking for there current draw in watts or amps. His question is about power consumption and inverter or generator use. I know the diffrence thanks for the electrical lesson though.

    I think most of this discussion would have went well in person but we are losing something in a message board format. We are all kinda on the same page just reading to much into what is written or not enough.

  16. jmf


    While everybody speculates, I just wrote Norman and one of their techs replied about this (about a month ago). 20a*120v=2400w

    HI Jim

    Most inverter do not have enough current for the packs. The packs can work on a modified wave but will work better on a true sine wave. The inverter will have to be able to put out more than 20 amps.

    Scott Osada
  17. Jim, Wow 20 amps thats alot of power. Thanks for the info.
  18. The battery packs that buff (White Lightning) sells are 150 Watt true sinewave inverters. You can buy these from invertersrus.com for $100.

    Buff claims they will power multiple units. I borrowed one once and found it worked fine for a single unit.

    Pure sine wave converters wont drop voltage when they reach current max. The modified wave converters will. Your recycle time will be less using the inverter but probably not by much.

    I now have an optima blue top battery with an isolator installed in my car. It is how secondary batteries are installed in RV's. I can run the battery dead and still have the primary to start the car with. I have a 300 watt pure sine inverter and run old novatron 440 plus pack with it. Works great.
  19. I have five of the White-Lightning CU-150 inverters that they use in their Vagabond kits. I power them with a 12volt, 12 amp/hour battery.
    They are pure sine-wave, so they won't damage the flashes like a modified sine-wave inverter will. If I understand correctly, they can power a 2000 watt flash because the flash only draws this for a fraction of a second, the inverter does not have a "shut-down" feature if it sees this amount of current. I have powered three flashes with one of these without any problem, the only drawback is recycle time increases. The whole power pack weighs only a little more than ten pounds. Before these were available, I powered the flashes with a 3000-watt sine wave inverter that cost me $1200. The inverter is not that large in size, but the battery needed to make it work was really too heavy to be really practical. The CU-150 units mentioned above are really the way to go.
  20. You ask Norman, and will get the answer that you expect. That depends how you ask, that is what you get. If you asked for 2000 Watts, you got the right answer, if you ask for 2000 WattSeconds you will get a different answer, or most probably NONE.

    The largest Norman unit seems to be the P2000D unit. Certainly the Norman P2000 is a huge and heavy monster. It provides 2 sections of 1200 WattSeconds capacitors. It can use up to 7 lampheads, with modelling lamps of 150 Watts each, and for all of that it has 15 Amps fuse built-in, at 110 Volts AC.

    All the modelling lamps alone require 10 Amps fuse from 110 VAC supply, and the entire unit has 15 Amps fuse bult-in for circuit protection, when powering all modelling lamps and the strobe. So, you can see that the evarage power requirements for the strobe alone (less moodelling lamps - e.g. lamps turned OFF), is not that gigantic, as some of you suggest or get an answer for, and it just could be around the 500 Watts.

    (15Amps fuse built-in - 10 Amps needed for modelling bulbs = 5 Amps left for the strobe). 5 Amps x 110 volts = 550 Watts max power needed to power the strobe alone, that is when all 7 modelling lamps are Off).

    I used an Adorama monolight that was 1800 Wattseconds, with 500 Watts power inverter, using a SLOW-BLOW fuse, and the inverted did not even warm up, and worked OK. The inverter had a self resetting max current limitting circuit, in addition to the fuse.

    But what worked for me, may not work for you, so get much bigger inverter, if there is any.

    I think 1000 Watts inverter should work OK with 2000 WattSeconds flash system, with sufficient safety margin, if only few (e.g. less than 5) modelling lamps are turned ON, (at 150 Watts each). I would start testing with turning all the modelling lamp Off, or use just 2.
  21. The question asked to Norman could have been about the power ?, and the answer received was on a different subject - that is about the power, and not about energy, so your received answer is for the Power values.

    "While everybody speculates, I just wrote Norman and one of their techs replied about this (about a month ago). 20a*120v=2400w"

    Asking about WATTS OF POWER, has little or no direct relation to the WATTSECONDS OF ENERGY that we need to consider for strobes.
  22. "I just wrote Norman and one of their techs replied about this (about a month ago). "

    Norman has great techs!, that were able to provide the answer "a month ago", to the question that was "just wrote".
  23. Oh gee. I can't help but mess with this a little. It is Freshman Physics 101 again. Sweat! Sweat!

    Energy = 1/2 x Capacitance x Square (Voltage);
    Power = Current X Voltage.


    [2 x Energy x Square (Current)]/Capacitance = Square Power.

    Thus, if you know the capacitance of the capacitors in the Norman pack, its maximum watt-seconds or Joules, the amperage, the voltage, you get its maximum power draw. The amperage referred to here is NOT its recharging current draw. It is the RMS current provided by the power source.

    As you can see, Power is indeed related to Energy. You cannot charge capacitors i.e. store energy without using Power. Frank, what possessed you to make an erroneous statement like that? It just ain't physics, my man.

    This doesn't tell you the charging characteristics of the capacitors. For that you will have to look at the capacitor charging curves which would its charging behaviour. That has to do with dielectric type and construction.

    Since that information is not readily available (unless you break open the pack, find the manufacturer of the capacitor and write to it for the curves and that precludes what Norman may have done to control the inrush of current in the pack i.e. its recharging characteristics), we fall back on RULES OF THUMB or conventional wisdom which is to use a VA rating three times that of the pack's power rating. That varies from design to design but three times gives a usually safe headroom. For example, when the Norman tech said 20A for its pack, it is safe to provide a headroom of 60A for when it recharges.

    We're photographers after all and not many would like to muck around with the physics so conventional wisdom, not necessarily based on scientific rigour but on what works, will have to suffice.
  24. As an addendum, some packs allow slow recycle. This would allow for a more gentle recharging curve and consequently less headroom in the inverter or generator would be required. This is something that only a manufacturer of the particular pack knows best.
  25. You guys are getting way to carried away.

    The WS rating of a flash has nothing to do with load. If you don't believe me just charge you flash and unplug it from power. Most will still flash! You only need to ask your vendor or use an amp meter to find out what resting load and max load are for your units.

    An example. A cheap portable setup would be something like a Novatron 240 pack and lights. The resting load is based on if you have the modeling lights on or off. Max load is 4A @ 120v. That is just under 500 watts. A good pure sine inverter rated at 150 Watts will have a peak load of 500 watts. This will handle a Novatron 240 with three heads and model lights off.
  26. nz


    To take this the other way. Why wouldn't a light designed to be plugged into a wall outlet draw all the power it could expect to find there? 15 amps shouldn't be a problem on normal North American outlets if I understand right. The more power it draws the faster it's ready for the next firing. No?

    I know Speedotron claims my monolight has an average draw of 8 amps with a peak of 13amp. That's a full recharge.

    Why would a bigger WS power pack draw less?
  27. Without getting too technical, capacitors have what is called ramp time. It is the rate they charge and discharge.

    If you attempt to charge one too fast it will explode.

    Best advice is still to have someone who knows electricity to put an amp meter on the power cord and find out the draw at standby and recharge.

    And use a pure sinewave inverter. Things will work better.
  28. I am glad that some scientific explanation was given.

    SHIVER uses the equations, yet comes up with "the RULES OF THUMB or conventional wisdom"? - that conventional wisdom seems taking over the scientific equations? or are the equations not applicable?

    So the scientist comes with suggestion of: "it is safe to provide a headroom of 60A for when it recharges". Yes, that will certainly work OK, safely.
  29. jmf


    The question I asked (specificly) was what it would take to power a P2000D or P2000XT (I have both packs). So the answer from Norman is a specific answer to the original question. If you think you're smarter than the manufacturer so be it. Perhaps you'll grace us with your own line of pack lighting in the near future.
  30. jmf


    My apoligies for the formating, the forum won't let me use the <pre> tag

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    HI Jim

    Most inverter do not have enough current for the packs. The packs can work on a modified wave but will work better on a true sine wave. The inverter will have to be able to put out more than 20 amps.


    -----Original Message-----

    From: jim feldman
    Sent: Wednesday, April 05, 2006 3:37 PM
    To: PCC-Info
    Subject: dc power inverter for Norman Packs

    I've got some Norman packs (P800D, P2000D, P2020) that I may occasionally want to run off of DC power by way of a power inverter. Do the packs (and heads) need pure sine wave output or can they handle modified sine wave?

    jim feldman
  31. Jim, You had asked question to Norman, and received your answer,
    that certainly did not satisfy you?, since seems you keep asking again.

    If I am smarter than Norman's tech? - nobody really can tell, but chances are that perhaps I am, since I can answer your questions, so you will stop searching for anwsers.

    I only know the Norman P2000D, and do not know the other models, but perhaps it is the most powerfull of them ?

    And the answer is: there are many many power inverters that will provide sufficient power, just do not go to Radio Shack for them.

    Some inverters from Power Master Technology will work OK with Norman P2000D.

    If you turn Off 7 all modelling lamps, inverter PM-650L will work for you. It only provides 650 Watts continuos, and nearly 2000 Watts short duration peak (surge that will sustain HUGE capacitor loading time of 1 to 2 seconds). E.g. I think this inverter could handle 2 modelling lamps (at 150 Watts each), in addition and any number (up to 7) of the strobe heads ( 5 of them without modelling lamops).

    If you want to power continuosly all your 7 x 150 Watts modelling lamps, then you may need a bigger inverter, and that would be model PM-1200L with continuous power of 1200 Watts (1000 for 7 modelling lamps, plus spare power for the 7 strobe heads). This unit has 3600 watts surge power, and will be more than you need for the P2000D Norman Strobe.

    I really did use successfully a 500 Watts inverter with single 1800 WattSeconds monolight and single modelling light of 150 Watts. Recommendations to provide an inverter supplying 60 Amps of current (at the output level 110 Volts) are simply expression of ignorance.
  32. Show me how I have been ignorant, Frank, with physics, not bullshit opinion?

    If you can't, you are the ignorant one.
  33. I think the guy from Norman is smokin' something. His statement of The inverter will have to be able to put out more than 20 amps. means that it cannot be used in any normal household outlet without blowing the breaker.
    Do yourself a favor and just spend the 5 minutes to get it tested.
  34. Shiver,
    Take it easy, you are not ignorant, we all do fantasize, sometimes...

    You presented 2 equasions:
    1. - Energy = 1/2 x Capacitance x Square (Voltage); Power = Current X Voltage.
    2. - [2 x Energy x Square (Current)]/Capacitance = Square Power.

    yet you could not figure what to do with them ?

    and found them totally useless in providing usable advice,

    then you resorted to a "RULE of THUMB" and "common wisdom", and came up with a 60 Amps, why not 100 Amps ?
  35. Thank you, Frank. I know that given enough rope you will soon hang yourself.

    So, you cannot prove that I am ignorant and think that by muddying the waters, you can get away with it. Not only cannot you not prove that I am ignorant, you also prove that even with equations biting you in the ass, you still cannot figure it all out. In short, it is you who don't know what you are looking at and what to do with the equations.

    The equations relate Power to Energy which in your assertion are qualities which are not related. I proved your ignorance beyond belief. I dragged out the equations to prove that you do not know what you are talking about.

    And if your reading comprehension is up to scratch, I have said in my posting UNLESS YOU KNOW THE CHARACTERISTICS OF THE CAPACITORS AND DESIGN OF THE PACK, these equations do NOT describe the RECHARGING behaviour of a flash pack which is another kettle of fish.

    Rules of Thumb are guidelines which have been tested and known to work SAFELY out in the field by those who practise their trade everyday. If you do not have the requisite knowledge of the equipment at hand, you fall back on Rules of Thumb. Engineers do it all the time as do scientists and professionals. It is conventional wisdom and common sense guidelines which are tried, tested and work SAFELY.

    The 3X times Rule of Thumb came about because manufacturers of these packs know that capacitors from different capacitor makers come in a wide range of tolerance with different charging characteristics.

    The ignorant are the ones who would tell you to do it because it worked for the ignorant and thus it should work for you.

    So, tell me, Frank, where is your scientific proof that I am ignorant. You talk a good game but where is your proof?

    As expected, you have NO PROOF. You are not only not as smart as you believe, you are actually a pretty dim bulb. So stop pretending that you know better than those manufacturers who design the equipment and dispensing advice. You, more intelligent than that Norman tech? Don't make your children blush.

    In lieu of a scientific proof that I am ignorant, you are proof positive that you are just a dim wit blowhard.
  36. I had to type that hastily as I had to meet a client. Upon re-reading it, I found that I had stooped to ad hominem attacks against Frank Skomial which were unnecessary and regrettable.

    Frank asked why 60amps and not 100amps? That is because in studying the characteristics of most capacitors used in the construction of flash packs or generators (as they're commonly called in Europe), if the flash is rated for a 20amp continuous or RMS draw, 60amps i.e. would provide sufficient headroom for the recharging surge or ramp characteristic of the caps. You can provide 100amps and all the power to you for providing headroom in excess of what is required. Nobody is saying you cannot do that but it is more than necessary.

    There is another factor which I had neglected to mention and that is Power Factor. In a purely reistive load, 1W = 1VA. However, in a reactive load like a flash pack where there are capacitors and resistors and possibly inductors, phase shift occurs. And where there is phase shift, there is a change in power factor. For example, 1W = 0.7 VA or 1W = 0.259VA, depending on the reactive load of the flash pack. That is where the Rule of Thumb comes in (and I was not smart enough to invent this rule of thumb): use a 3X VA rating of your pack if you do not know its design characteristics. Thus, if it draws 3 amps RMS, use a power supply that can swing 9amps.

    Vince, an inverter is of really quite a different creature from the mains power supply. I don't think you should imply that the Norman tech was on some hallucinogenic substance until you find out more about the difference between inverters and the mains. An inverter converts DC to AC. For example, for a 600W continuous rating inverter (with a 1200W surge rating) running off a 12V battery supply, the iverter makes 60amps draw at 12V DC, and 120amps draw at peak surge power of 1200w at 12V DC. And we have not even gone into the gauge of wire required to handle this draw.

    I apologise to Tom Hanser who asked this question; I think he is even more confused now. I have not helped him any but I had hoped to steer him away from erroneous though well-meaning advice put out by a few. Internet discussions have a tendency to degenerate and I have contributed to it, regrettably.

    All this is more involved than I would want to get into. As I said from the beginning, we are photographers, not flash engineers, and I defer to the knowledge of the people who designed these things. For the here and now, The Rule of Thumb is conventional wisdom which works SAFELY.
  37. Listen Guys, i'm glad to see everyone discussing such a valuable topic. I appreciate that you all have an opinion- and that all the information is based on a hint of knowledge and experience.

    The only thing you all lack is a common agreement on a single effective solution. Until now.

    Monday morning I read all the posts in the hope I could become enlightened.

    Today I will share my solution with you all - as I spent yesterday building a mighty black suitcase with the power of the gods and the bite of a hungry lion.

    It's a power supply from hell - with firepower to eclipse those field battery operated strobe kits you pay through the nose for.

    Enter Jaycar electronics.

    With the help of JayCar in Sydney - I took along my most power-hungry BronColor strobes and proceeded to churn through box after box of "high" powered inverter, testing each one on the store bench.

    I found the following in my quest:

    1: A PURE SINE wave inverter is much more equipment friendly than anything else and wont blow up your strobe
    2: A 1500Watt inverter with 3000Watt surge abilty - thats what you'll need.

    3: The JayCar model is a BEAST. 24 Volts is produced by a 'in-series' joining of two 12Volt GEL batteries of about 60 Amp hour rating.

    These provide 24 Volts to the Inverter, whilst being able to disconnect and re-connect in parallel for recharging by car or normal 12v power supply.

    When fully charged, this unit (which fits into a large black and silver hard-case, has passed my load test > Hairdryer+ 3 bron color studio strobes + bar fridge all at once on full load startup cycle.

    It didn't miss a tick and the recycle time for full-burst is under one second even with all appliances on.

    Now thats a real life fact without the bickering over watts and amps etc.

    My complete kit including the giant batteries, cables and inverter, oh and the case to hold everything in, came to AUD $700.00

    And it will last forever and power a small village if ever required.

    Hope it helps ;)

    If anyone wants specs or pics of the beast please email me.


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