Determining HMI draw.

Phil Gerke · March 14, 2008

Hey all,

I have been getting mixed info on how many amps HMI pull. The standard deviding volts by watts to get amps does not exactly work with HMI does it? How does the strike factor in?

K5600 states their 800 joker (which I love) pulls 12.5 amps. Is this factoring the strike, which is higher than the running draw right?

Is there an equation for quickly determining current? I don't believe Mr. Box gets into in his book either.

Thanks a lot!

Phil

Christopher Santucci · March 14, 2008

The formula is P=I x E

Which means wattage equals current times voltage, or current equals wattage divided by voltage.

As for draw during strike, check on the ballast to see if it's marked or call the manufacturer.

Edited March 14, 2008 by Christopher Santucci

Michael Collier · March 14, 2008

12.5 amps? That does seem a little high, they might be quoting striking amperes (plus overhead to cover themselves). When determining HMI current draw you have to factor in the efficiency of the ballast. Sure the head uses 800 watts, but the ballast is not 100% efficient. Typical efficiency of any transformer device is around 85%. Electronic balasts would be less efficient. Striking current I don't know for sure, but I would plan 50% overhead during strike for saftey. This doesn't mean you need 50% over all units, just the biggest device you have on any given circuit. You can strike them in staggered fassion.

In practice I have personally used 3 575's (1725 rated, 2029 at 85% e) on a 20 amp breaker, and a 1200 and a 575 on a single 20 amp breaker. all that is with typical stinger run (100-300ft) Any other combo of lights you should plan to use a xstal sync genny to run. I suppose you could run two 800's, but I have never personaly run that off household current.

When running towards the edge of power limits, or with long runs, you may experience strange shut downs. the other weekend I struck a 1000w tungsten only to see my 575 on the same circuit go cold. The breaker didn't trip, but the voltage drop of the tungsten unit switching on dropped the voltage enough to turn it off. If you have a dimmer, slowly dim any tungsten units up to avoid that voltage drop on initial striking. (don't put an HMI on a dimmer of course)

Phil Gerke · March 15, 2008

The formula is P=I x E

Which means wattage equals current times voltage, or current equals wattage divided by voltage.

As for draw during strike, check on the ballast to see if it's marked or call the manufacturer.

Right:) Got mixed up a bit. Should have read before I posted.

Michael thanks for the info. Basically what it boils down to is that you really have to go by experience and knowledge on a per fixture basis. Not every 1.2K is going to behave exactly the same, from manufacture to manufacture at least.

85%, thanks for that figure. So assume about an increase of 50% at the strike and operation at about 85% efficient when running. Though I will be most likely using electronic ballasts, what do you think?

Can you use P=IxE for the heads? At least in their relationship with the ballast. In the instance of the 800, is the head pulling 6.6 amps from the ballast, and the ballast being less efficient pulls more from the power source? Or is it not that simple? HMI (and flouro for that matter) don't quite work that way do they?

Great tip on the voltage loss! I could see that certainly happening.

Oh, one more question. In your example of the 3 575's. You added the 3 and got your 85% by multiplying by .85 right? Now, that is just for the sake of determining a draw, the fixtures total is still only 1725 right? The inefficiency of the ballast is not causing a greater output (lumens) right?

Thanks a lot for the help!

p.s. I realize there are about 5 questions in there, tackle what you wish.

Edited March 15, 2008 by Phil Gerke

Christopher Santucci · March 15, 2008

Can you use P=IxE for the heads? At least in their relationship with the ballast. In the instance of the 800, is the head pulling 6.6 amps from the ballast, and the ballast being less efficient pulls more from the power source? Or is it not that simple? HMI (and flouro for that matter) don't quite work that way do they?

Well, the draw is ultimately created by the heads and most electrical devices have tags on them with information printed that shows how much the device requires. And, I'd guess since there is a step up in voltage at the ballast that you are actually drawing less than you normally would for a 120 volt fixture.