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Powering Studio Strobes Using An Inverter


dinsdale

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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

generator?

 

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?

 

Thanks,

 

TH

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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.

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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:

 

http://www.photo.net/bboard/q-and-a-fetch-msg?msg_id=000FCn

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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.

 

TH

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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 <em>huge</em>. 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.

<p>

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.

<p>

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.

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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.

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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.

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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.

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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.

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<I>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?</I><P>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.
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<cite>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.</cite>

<p>

Strobes aren't rated in Watts. Strobes are rated in Watt-seconds, (or Joules, which is just another name for Watt-seconds).

<p>

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.

<p>

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 <a href="http://www.photogenicpro.com/file.asp?F=CA51CFEAD91249EEAD3753631E201727%2Epdf&N=PowerLight1250%5F2500%5FOperators%5FManual%2Epdf&C=store_manual" target=_blank>Photogenic Powerlight 1250 and 2500 manual and specs.</a>. 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.

<p>

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 <em>slo-blo</em> fuse, which indicates that the designers expect them to occasionally draw short bursts of more than the nominal 15 amps.

<p>

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.

<p>

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.

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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.
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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.

 

,Grinder

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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

"

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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.

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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.

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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.

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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.

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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.

 

So,

 

[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.

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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.

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