500W PAR64 Lamp or 1000W Lamp with 50% Dimmer

[djwiltsh]

Hi All

 

Simple question:

 

Is there a visible difference between a 1000W PAR64 lamp at 50% power, and a 500W PAR64 lamp at 100% power?

 

Thanks

 

Dan

[niclights]

Maybe the 1000W at 50% will have lower colour temp than 500W at 100%? Maybe nonsense!

[Modge]

I'd say definitely not nonsense: The colour temp will be different. This can look good if you do it on purpose but if e.g. you have a row of 5 500W bubbles and a 1Kw bubble at 50% in the middle it may look somewhat silly. Brightness wise also be aware you'll find you need to muck around getting the level right: 50% on the desk is highly unlikely to be exactly the same as a 500W at 100%.

 

Or no it's not the same.

[David A]

50% power is at about 30% on the fader and not even close in intensity, the efficiency increases dramatically near working voltage.

[Trunker]

50% power is at about 30% on the fader and not even close in intensity, the efficiency increases dramatically near working voltage.

 

 

All depends on your dimmer curve though doesn't it!?

[Ekij]

Another alternative would be to wire two 1000W in series splitting the voltage (and hence the power) evently across the two 1kW units.

However as others have said these won't look the same as a 500W unit, dimmer and redder.

[bruce]

Another alternative would be to wire two 1000W in series splitting the voltage (and hence the power) evently across the two 1kW units.

 

Are you sure?

 

If you stick half the voltage across the unit, you'll deliver 25% of the power

[mbthegreat]

Now I'm probably being an idiot here but doesn't P=IV?

[bruce]

Now I'm probably being an idiot here but doesn't P=IV?

 

Yes.

 

But if you take a lamp which is rated at 1000W at 240V, and as suggested put 2 of them in series, you half the voltage across each. But you also double the total resistance, which means the current is halved.

 

So each lamp will have half the original voltage and half the current. Since P=IV, the power dissipated by each lamp decreases by a factor of 4.

 

Or to put it another way, for a constant resistance, P=V^2/R

 

 

Of course, since there are now 2 lamps, the total power output by the 2 "1000W" lamps will be 500W. Which might be what ekij was suggesting....

[mbthegreat]

Which reminds me how close my physics exam is.

[David A]

Trunker-I said ABOUT 30%.

What everyone is not allowing for is the huge change in resistance over the fade of the lamps, the resistance increases by ROUGHLY 10x from cold to full temperature so simple Ohms law equations do not apply as you would need to feed in the resistance at the various levels and it would need a very complex equation to give an accurate result.2 lamps in series would have a lower temp therefore lower resistance and draw more than half power.

[adam2]

Trunker-I said ABOUT 30%.

What everyone is not allowing for is the huge change in resistance over the fade of the lamps, the resistance increases by ROUGHLY 10x from cold to full temperature so simple Ohms law equations do not apply as you would need to feed in the resistance at the various levels and it would need a very complex equation to give an accurate result.2 lamps in series would have a lower temp therefore lower resistance and draw more than half power.

 

True

 

If two similar 1000 watt lamps are wired in series on a 240 volt supply, each lamp will have only 120 volts accross it, and the result will be a very dim yellow orange light of rather limited use.

 

In the case of perfect resistors, two wired in series would draw half the current, and therefore in total half the power, therefore each resistor would only disipate one qaurter of the power.

 

Lamps however are not perfect resistors, if a lamp is used on a reduced voltage, the filament will run cooler and therefore have a lower resistance than at full voltage.

Therefore a lamp run at reduced voltage will allways draw less current than at full voltage, BUT the reduction in current will be less than the reduction in voltage.

 

As an example, a lamp run 50 % voltage will draw about 65% of the current that it would consume at full voltage.

Practical test

A 12 volt 50 watt MR16 on 12 volts used 3.9 amps, on 6 volts it used 2.6 amps, a lot more than 50%

 

No great accuracy is claimed for this test since neither the 6 volt nor the 12 volt supplies were regulated, just batteries I had to hand.

It does clearly demonstrate that a lamp on reduced voltage DOES use less current, but that the the current DOES NOT decline in proportion the voltage.

 

As for dimming a 1000 watt lamp to give the same light as a 500 watt lamp, the power consumed would be far in excess of 500 watts since at the reduced power the efficiency is much lower.

[paulears]

Moderation: Perhaps we should remember that djwiltsh asked if there was a difference between two lamps - which every response says yes. If members think the topic worthy of a split to allow the maths to e debated, please let the mods know by using the report function.

[church]

Moderation: Perhaps we should remember that djwiltsh asked if there was a difference between two lamps - which every response says yes. If members think the topic worthy of a split to allow the maths to e debated, please let the mods know by using the report function.

 

and we haven't even begun to address the fact that the power a lamp takes does not necessarily relate to the light output or the colour temperature or even the manufacturer or even the lamp type. Not to mention that there is a difference between the light output from a lamp measured in a lab and the light from the front of the fixture due to fixture optics and the lamp filament tpe.

 

But true to say a 500W PAR will not look the same as a 1000W PAR run at 50% and there is unlikely to be any position on the dimmer for an open white fixture where it does look the same - suspect with a saturated filter that you can find a dimmer setting where most people wouldn't notice but it won't be 50%