How to identify voltage of an unmarked bulb?

Do you mean a bulb/flashlight combination, or just a bare bulb lying around that you are thinking of using in a flashlight?
 
Put it on a variable DC bench voltage supply and turn it up until it blows /ubbthreads/images/graemlins/evilgrin07.gif At that point the voltage was too high /ubbthreads/images/graemlins/grin.gif OK. j/k j/k. Turn it up until it goes from yucky yellow to nicer yellow to bright yellow-white to intense white. And also record the voltages on the meter, of course.

Or if you don't have access to a bench supply, start by driving it with 1 NiMH or NiCd cell, then 2, then 3, then 4, etc.

Or send it to me and I'll figure it out.

Or do some careful looking and calculating:

The voltage of a tungsten filament lamp is directly proportional to the length of the filament. Filaments are almost always wound in a circular spiral coil (OK, not a great description, but you know what I mean). Find the circumference of the end of the coil and multiply it times the number of turns, and you have the approximate length of the filament. I'm sorry but I don't have the conversion factor in my poor brain, so you will need to do this little math exercise for both the unknown lamp AND for one of which you know the voltage for the drive-level you want. Then take the length of the unknown lamp filament and divide by the length of the known lamp filament and multiply the result by the voltage of the known lamp and VOILA you have the voltage of the unknown lamp.

PM me for my address if you want me to figure it out for you. It shouldn't take more than 15 minutes, so I'm sure I could turn it around in a day or two tops.
 
Oops. Just noticed that I said "find the diameter of the end of the coil . . ." Not right. I edited my post above to say "find the circumference of . . ." That's what I meant to say.
 
js, thanks for the info and offer; very good of you. Using the NiMH cells, I've determined it to be a 3 cell bulb.
 
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js said:
Put it on a variable DC bench voltage supply and turn it up until it blows


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This is actually a very good method. The bulb blows at the filament's melting point. And staying a little bit below that will give the highest efficiency.
Running a bulb at the edge needs a doft start and a very constant filament temp (=power delivered to the bulb).

The best method will be to measure the filament's temperature with an appropriate radiation thermometer.

But it also could be done by a side by side compare with a known bulb.

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Filaments are almost always wound in a circular spiral coil (OK, not a great description, but you know what I mean). Find the circumference of the end of the coil and multiply it times the number of turns, and you have the approximate length of the filament.

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All the filaments I know are not wound in a spiral way, they use a helix instead .-)
Further most bulbs use a double helix, that means what you think is the filament, is actually another helix.
Phillips one time even had a triple helix.

And even further on, there is no real 'nominal voltage' for an incandescent bulb. It always depends on the lifetime you want to have. What a 12V bulb is for one, it is a 15V for the other .-)
And sometimes the nominal voltage is given for the general public and has little to do with the voltage the bulb is usually running at (see the relabeling of the Petzl bulbs and the usual car bulbs labeled for 12V, but rated for 14V)

And it is not necessarly true that, for example, a 12V bulb has twice the filament length as a 6V bulb (both for the same current, glass envelope size and lifetime).
 
PeLu,

Not true. I have never seen a double wound filament for a 12V or lower bulb. A filament is double wound like that for very high voltage lamps, such as 120V projector lamps. I can see very clearly that the WA and Osram lamps I have are single wound.

As for "nominal voltage" you are absolutely correct. Which is why I stated that you needed the length of the filament for a lamp for which you know BOTH the voltage and lifetime, i.e. "drive level".

And yes, it is necessarily true that a 12V bulb has twice the filament length as a 6V bulb, all other things being equal. This is one of the laws of filament physics: voltage is directly proportional to the LENGTH of the filament, and current is related to the THICKNESS of the filament, all other things being equal.
 
Jim,
Right. Double wound are quite uncommon in lower voltage bulbs. I have an 82V MR-16 lamp that is double wound.

Perhaps it's more accurate to say that for a given filament composition, the voltage is proportional to the length. I'm thinking that alloys of different resistance values could generate different voltage/length curves. Of course, maybe all filaments are made with the same tungsten alloy and in that case, my statement is an uneccessary generalization.

Wilkey
 
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js said:
I have never seen a double wound filament for a 12V or lower bulb.


[/ QUOTE ] OK, you got me .-) Did work too much with 230V bulbs....


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And yes, it is necessarily true that a 12V bulb has twice the filament length as a 6V bulb, all other things being equal.

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What I meant was: assuming the same current, same glass envelope, same lifetime:
When the 12V bulb would have twice the filament length as the 6V, it will run on a higher temp, as some 'losses' are constant (conduction). That means that the 12V bulb will have a shorter lifetime, as it runs on a higher temp (and will give more than twice the light).
That means that filament lentgh is only roughly proportional the 'nominal voltage' (just for example lets define the nominal voltage as the voltage where you get a median life of 1000 hours).
I used to know how much the difference is, but I completely forgot the numbers.
Of course all this is mainly academic, for practical use the light colour compare will be the best method (assuming the glass envelope is not coloured).
I still have several of these sun eclipse glasses, which do a good word for comparing burning filaments and similar things.
 
PeLu,

All other things equal, doubling the filament length doubles the voltage and keeps the current the same, because it's just like two of the very same filament in series! It will not run any hotter. The resistance of the tungsten filament is a function of the cross- section and the temperature and the LENGTH.

Of course, if you double the length, it's hard to keep everything else the same, e.g. the glass bulb usually has to get bigger. But still. The double length filament will have an almost exactly doubled voltage for the same filament temperature and cross section and fill gas.
 
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js said:All other things equal, doubling the filament length doubles the voltage and keeps the current the same, because it's just like two of the very same filament in series!

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Jim, I think we just talk in a different direction.
What I meant is, that the conductive losses are not the same and therefore the 12V filament will run hotter.

I'm far from beeing shure there but I remember some 10-15% of conductive losses for these small envelope bulbs.

If I'm wrong in here (with my numbers), please correct me.
Maybe my numbers are only valid for relatively low temp bulbs. The higher the filament temperature gets, the lower the relative conductive losses are.
 
Lookup "Pilot Lamp Data". The chart shown lists the volts, amps, bulb type, and the color of the insulating bead at the bottom of the filament support wires. The bead color is the important identifier.
 

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