Electrical engineer types: Lumen output of lamps with varying voltages.

It’s well known that if you have three different lamps of the same type and wattage, but different rated at voltages, they will have different light outputs. For instance, a 12 volt, 50 watt lamp will be very bright, a 120 volt, 50watt will be less intense, and a 277 volt, 50 watt lamp will be dimmer still.

Voltage, current and resistance being related as they are, this phenomenon makes sense in a somewhat mysterious way, but I’m not enough of an expert to explain the actual electrical machanics behind it. Assuming all other things are equal (varying manufacturing methods and so on), can anyone offer an explanation as to why this is? Thanks…

The total energy dissipated by a 50 W, 12 V bulb is almost exactly the same as that dissipated by a 50 W, 120 V bulb. The difference is what form that dissipated energy takes. Since perceived brightness depends on the amount of visible light being emitted, the hotter filament will appear brighter, due to the blackbody spectrum I’m sure you’re familiar with. The temperature of the filament, in turn, is primarily dependent upon the current flowing through it. The 120 V bulb has a lower current flow through it, so the filament dissipates most of its energy as infrared, which is invisible to the human eye and does not affect the apparent brightness of the lamp. The 12 V bulb, OTOH, has more current flowing and therefore gets hotter, releasing much more of its energy in the visible portion of the spectrum. This is a bit simplified, of course–filament geometry also plays a minor but important role here–but it should suffice to get the concept across.

QED – they’re all 50 Watt bulbs, though. In an ideal world, they’re all the same temperature, because the product of current and voltage have to be that same 50 watts in each case. If they’re the same temperature, they ought to have the very same output spectrum.

I’ve never heard of this phenomenon before. If the three bulbs are, in fact, putting out different spectra, then they’re probably different temperatures, which means that something isn’t operating as advertised – maybe that 50 Watts is approximate, not actual wattage, and you get slightly different wattages with different currents.

I have to admit I didn’t know this. Anyone have a cite?

And why would filament temperature depend on current? If you approximate it as a blackbody, the temperature should only be dependent on power output per unit area.

They’re not going to be the same temperature, though, because the thinner filaments needed for a 120 V bulb will cool off faster via radiation than the heavier filaments found in 12 V lamps of the same wattage. That’s the filament geometry I was referring to above.


Q.E.D.: Yes, that is exactly the type of explanation I was looking for. Makes perfect sense, and it will to anyone to whom I explain it. Thanks again, and thanks for giving me the benefit of the doubt on the blackbody spectrum. :o)

  • Ol Peculiar

What do you mean by “cool off”? I thought we were talking about a steady-state condtion.

Your other link makes sense, but it seems to be based on the assumption that all light bulbs with the same rated current use the same tungsten wire for the filament, with the only difference being the length. Is this accurate?

You’re correct, of course. I was trying to keep things simple and worded that rather poorly. By “cool off”, I mean the rate at which the filament loses thermal energy. At some point, the energy input will equal that thermal energy loss and the temperature will stabilize, assuming the current holds steady. The temperature at which that happens will depend on how fast the filament loses heat. That’s what I was trying to get at there.

Basically, this is correct. Of course, there’s a wide variation in filament geometry even among bulbs of the same voltage and wattage, but in general, low voltage bulbs must have a shorter and/or thicker filament in order to have a low resistance compared to a higher voltage bulb of the same wattage, which must have a higher resistance. All of that can be worked out from the power equation : P = I^2 x R or P = I x V. Obviously, in order to maintain the same wattage, a higher voltage bulb must have a higher resistance filament.

This still doesn’t look right – blackbody radiation is blackbody radiation, and the emissivity (which defines how fast something emits rasdiation, and in which direction) is an intrinsic property of the material, not its geometry. If it’s dissipating 50 watts, it oughtta be the same temperature, and if it’s tungsten, it ought to have the same emissivity if it’s a thin wire or a thick one. The current running throyugh it ought not to make a difference.

Of course, tungsten isn’t an ideal black body, but at high temperatures it’s surprisingly close to one. Perhaps the slight differences in geometry do make a difference. But I wouldn’t think it was a significant one.

I don’t think that’s quite right. If we have a 1" square of material being heated by 50 watts of power, and a 10" square of the same material also being heated by 50 watts, I’d expect the larger square to lose heat much faster. I’m probably wrong that the bulk of the loss is due to radiation; I imagine a considerable amount is due to conduction and convection, since bulbs are usually gas-filled. I’m not familiar enough with the physics to say for sure.

Seems the term we are looking for there is effective emissivity:

Bolding mine.

The difference between the bulbs, I think , has got to be the total area –

from this site: http://www.oshinolamps.co.jp/english/product/mikata.html

Because it’s a larger area, there’s more material to be emitting light , and it has more volume, which means more thermal mass, which means that different filaments won’t neccesarily have the same temperature.

With differing temperatures you get different spectra, even with perfect blackbodies. Therefore different numbers of lumens.

I think this is pretty close, but rather than area alone, I’d say surface-to-volume ratio plays a role here, as well. Thus, a long, thin filament with the same total area as a short, thick filament will still radiate off energy faster. No?


QED – You gots to read yourself a book on radiometry. A theoretical blackbody has the same radiance regardless of geometry. Real substances aren’t blackbodies, but at the same temperature the radiance of two patches is still the same, regardless of geometry. My problem was assuming something not obviously incorrect at first – that two filaments rated at the same power would have the same volume and mass. Given that they don’t, their temperatures won’t be the same. But a long thin filament and a short thick one at the same temperature will be radiating the same power per unit area.

Of course, if one filament has a greater surface area, the total amount of power radiated will be greater, and if that’s the whole of our disdagreement, then we really do agree. But the power per unit area ought to be the same.

Thanks, Cal. Any recommendations?

Oh, God, no, I can’t. The most widely available one is by my ex-advisor, and I hate his guts.

Hee, hee… I’m having a ball with this thread of mine. Or rather, yours. I have got to get you two together for a drink. I’m in Rhode Island - who’s in?

I don’t know anything about black body radiation. But I have a hunch Q.E.D. is correct.

Temperature is not the same as energy. (Temperature is also not the same as power.) An iceberg, for example, has a lot more energy than a lit match. But the temperature of a lit match is much greater than the temperature of an iceberg.

Let’s say I have the following two items:

  1. 120 VAC, 75 W light bulb.
  2. 25,566 ft. of 18-AWG copper wire. (The total resistance is 192 Ohms)

Let’s say I apply 120 VAC[sub]rms[/sub] to the light bulb, and 120 VAC[sub]rms[/sub] to the spool of copper wire.

The total power emitted by the light bulb will be 75 W.

The total power emitted by the 25,566 ft. spool of wire will be 75 W.

So CalMeacham… are you saying the temperature of the spool of wire will be the same temperature as the light bulb??

No, because the surface area of the copper wire will be vastly larger than the surface area of the tungsten filament. I think Cal is right in the case of a purely blackbody emission, but since most light bulbs are gas-filled, other factors come into play, too. I know that in air (or water), an object with a larger surface-to-volume ratio will lose heat faster all else being equal. What I don’t know (or know how to calculate) is the relative effect of radiation losses versus conductive and convective losses in a typical light bulb.