Electrical lightbulb question
 

I have one MR-11 6v 15 W Halogen bulb running off a 2900 mAh battery overvolted at 7v. Can I run a 12v 30W bulb at 7v?

That would seem to work although it's going to be a lot dimmer and run half as long since it can only produce 7v for a 12v rated bulb.

Forget the overvolting thing, how does it work? I'm trying to wrap my head around this v=ir thing. You're saying it will draw twice the current (edit) while being half as bright?

I'm trying to use cheaper bulbs for my bike light. The 15W 6v ones are like $20, the 12V ones are much cheaper no matter the wattage.

P1 = V1^2/R
P2 = V2^2/R
P1/P2 = V1^2/V2^2

(^ is exponent, in this case squared)

Say P1 = 7.0V, P2 = 6.0V

P1/P2 = 7x7/6x6 = 49/36 = 1.36

On 7.0V, the 6.0V lamp will consume 1.36 times its rated power, so the 30W bulb will consume 41W.

It will be brighter, by about a linear ratio of 1:1.36.

The life expectancy is an inverse square ratio. (1/1.36)^2, or 75% x 75% or about 50% of rated life.

Brighter, for a lot less time.

Old flashbulbs worked on this principle, a lot of power for a really short time!

Best Regards,
Jim

I can't personally vouch for the quality of the information, but have you looked at this article?

http://www.myra-simon.com/bike/lights.html

So Jim you're saying that if I run a 12v 20 watt bulb with 6v I'll get 5W output? That's terrible.

Man I just bought this lamp. It's one of these but I got it with the NiMh battery.

http://www.cygolite.com/light/Photos...ts/8ExpNcd.jpg

http://www.cygolite.com/light/products/8ExplorerNiCad.htm

I'm just gonna bite it when the time comes to replace the bulb. It's not gonna happen for years to come anyway. By that time, maybe HID technology will be more affordable. ;)

OK I get it.

I think.

This is the bulb I have but with a glass cover.

http://www.topbulb.com/find/Product_Description.asp_Q_intProductID_E_46050

I can't believe they want $20 for that thing. A 20W 12v is half the price.

Is "MB" embossed somewhere on the housing? :rolleyes:

Least it's not Lucas. :)

Interesting calcs . .
 

but not quite correct! The formula is :

Pr = E1^2/E2^2 x R2/R1

where Pr = power ratio between two dissimilar bulbs,
R2 = resistance of lower voltage bulb,
R1 = resistance of higher voltage bulb.

You can't ignore the resistance of either bulb since, as the formula shows, the resistance ratio will further worsen the power ratio because the lower rated bulb will have higher resistance. This will increase the power ratio between two dissimilar bulbs.

So for your example of "7V" and "6V" bulbs, the resistance ratio would make the ratio equal to 1.36 x R2/R1. Thus the 'final' ratio will be larger than "1.36" by the ratio of the two bulb's resistance. The only time you can ignore the resistance ratio is when the two bulbs are equal but this a 'trivial' case. :)

Bottom line: the stress on the two dissimilar bulbs is greater than just the ratio of operating voltages. You need to include the resistance. But in any case, the consequence is a very short, but very bright, life.

JimF, as I understood Kuan, he wanted to know what would happen if the voltage on the same bulb was changed from its base rating. That is what I based my formulas on, applying 7.0V to a 6.0V bulb. In such a case, I believe the formulas are correct as I wrote them.

You are correct if we introduce two voltages and two different bulbs.

Best Regards,
Jim

Another way . .
 

Add a rectifier diode in series with the 7v power; it will drop the voltage by 1v so that you don't overstress the 6v bulb.

Radio Shack: 276-1141 3amp 'barrel' rectifier diode.

JimH: missed the same bulb; thought that there were two different bulbs. But the general formula can be used in any case: if the bulb is the same then R2/R1 = 1. You know that but this is just to clarify for other readers.