Bulb voltage vs. amp draw

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Kirk

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Hey All!
I am ignorant on quite a few things. Maybe someone out there can learn me something. I bought an old, but un-used, Siebe Gorman "Divers Light" that runs on 3 "D" batteries. I put new batteries in it and the light output was pretty pathetic and yellow. I removed the bulb to discover it was rated at 6 volts and .3 amps. My question is, was this done to make the bulb last longer, the batteries to last longer, or ??? I figure amps is amps and if you run a "6 volt" lamp on 4.5 volts you are still pulling .3 amps. Or am I confused? Any illumination would be greatly appreciated. Thanks in advance.
Kirk
 
Kirk, what you want is a bulb with a voltage rating of about 4v, and amp. rating of perhaps .8amps. If you go lower on the voltage the light will be brighter but the bulb life will go down. A 6v bulb sounds like the wrong bulb.
 
Maybe a 3.6 volt krypton bulb will be good? The bulb life will be short though....

And if you run 6 volt lamp on 4.5 volts, I don't think you are pulling 0.3 amps. Should be slightly less. Try and get at least a 0.5 amp bulb for resonable brightness?
 
Thanks for the info. I tried your suggestions; just happened to have a few #13 mini screw base bulbs rated at 3.7 volts and .3 amps. Much "gooder"! I still wonder why the manufacturer stuck in a 6 volt bulb; maybe so it will last like a thousand hours instead of 15 hours like my #13. After all, if you're stuck in a "hard-hat" diving suit 300 feet down you don't want to have to worry about bulb life! It seems all flashlight bulbs, the PR type anyway, are "over-driven"; a PR3 is made for 3 D cells (4.5 volts) but is rated 3.57 volts @ .50 amps. But you only get a 15 hour rating.
Kirk
 
Bulb's resistance varies and is higher as it heats up. In other words, they will ramp up to a certain current draw and after that, the increase in current draw is very little.
 
Originally posted by Nerd:
Bulb's resistance varies and is higher as it heats up. In other words, they will ramp up to a certain current draw and after that, the increase in current draw is very little.
Click to expand...
<font size="2" face="Verdana, Arial">If the filament resistance increases with temperature, then the current draw will decrease as the bulb heats up.
 
Filament resistance changes with temperature, by about a factor of 10 from room temperature to operating temperature.

The filament temperature is determined by the power going into it and the power escaping, either by conduction or through the photons escaping.

Because the total energy radiated in the form of photons scales as the temperature to the 4th power, one finds that over a pretty wide supply voltage range, the temperature change is pretty small, but _critical_. If you drop the supply voltage by 50%, the operating temperature might drop by 25%. But small changes in operating temperature mean _big_ changes in light output.

Some equations which one can find on a Welch-Allyn web which describe how a light acts under different voltages are:

(These equations are all normalized, meaning that you presume that the light is supposed to be operated at a voltage of '1' and a current of '1'; scale appropriately for a real light)
(V=voltage, I=Current)

I = V ^ (0.56)
power = V * I = V ^ (1.56)
light output = V ^ (3.56)
bulb life = V ^ (-13)

What this says is that as the voltage changes, the power changes less than Ohm's law would predict (since the resistance changes), than efficiency increases as voltage increases, but that bulb life falls into the toilet as voltage increases.

If you operate a bulb at 20% increased voltage, you will get 11% increased current flow, 33% increased power flow, 91% increased light output, but a bulb life that is about 10% that expected are 'proper' voltage.

Note that these equations are semi-empirical rules of thumb, but they help plan how to use a light.

-Jon
 
Jon,

These equations are excellent. Thanks for posting them. Where did they come from?

> I = V ^ (0.56)
> power = V * I = V ^ (1.56)
> light output = V ^ (3.56)
> bulb life = V ^ (-13)
 
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