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Need help converting flashlight bulb to LED

  1. Jun 23, 2010 #1
    Ok so I have a flashlight that is 19.2 volts. It used to have a bulb that was 19.2 volts and .6 amps, but I broke it, and couldn't find a replacement, so I decided to try to use an LED and a resistor. The LED is 20 mA (.02 Amps) and 3.5 Volts. If I am correct, to find the resistor I need I would use:
    (Vsupply - Vneeded) / Ineeded = Resistance
    (19.2V - 3.5V) / .02A = 785 Ohms (or 800 to round up for safety)

    Is this right? Thanks
  2. jcsd
  3. Jun 24, 2010 #2
    That looks to be about right. but my concern is this 19.2 volts is kind of a weird voltage and a LED that is 20mA is not going to be too bright for a flash light. most of the high power LED flash lights are using 1W 3W or higher LED powers and they can be set around 350mA or more. Also what ever current you are using in the LED is dissipated in the resistor. Pdis = V*I or I^2 *R or 15.7V * 0.02A = .314 W not too much for a small LED but if you do this with a power LED you get 15.7 * .35 = 5.5 W in the resistor only! That would be a large resistor that got really hot. L
  4. Jun 24, 2010 #3


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    19.2V sounds like a nominal value of 18 or so. If this happens to be from one of the Ryobi / Ridgid / DeWalt line of rechargeable battery-powered work lights, the DeWalt DW9083 should work (they all use the same type of xenon bulb) you should be able to find it, or something similar at your local hardware store. Or here:

    The Craftsman / Porter Cable / Ingersoll-Rand lines of battery-powered tools all use 19.2V battery packs, but I wouldn't be too put off by the extra volt or so through a xenon bulb (just bring the burnt out bulb with you to the hardware store to make sure you're buying the right bulb--that's what I did!)

    Most of these are all made by the same company. However, the battery packs are usually keyed slightly differently (and the build quality / warranty varies a fair bit).

    EDIT: And welcome to PhysicsForums!
  5. Jun 24, 2010 #4
    Yeah you're right. I tried it with 1k Ohm and it looks like even with 800 Ohms it will probably be much too dim.
    So do flashlights designed to use LEDs have lower voltages then? It seems like there is no way to have a high voltage powering an LED without losing a lot of efficiency. Is there a way to step down the voltage the flashlight produces without increasing resistance? (Which would presumably boost efficiency)?

  6. Jun 24, 2010 #5
    Thank you!
    It is actually a Kawasaki battery powered flashlight, but I don't think it has a xenon bulb (it has a fillament). Either way, I'm sure it wouldn't be too hard to find one, I just thought this would be a good opportunity to try to do a little science project and see if I could retrofit a flashlight to be LED, but it doesn't look like it.
    Also, what do you mean by the battery packs are "keyed" slightly differently, like they aren't interchangeable between brands?

    Thanks for your response!
  7. Jun 24, 2010 #6
    There are a couple of things you can do. 1. you can string a few 20mA LED in series until you get the brightness you want, but with a Vf of 3V or so you will not be able to get very many in there. 2. go to digikey and find something with just a bit more kick. 3. go for the gusto and build a switcher / LED driver for a powerful LED like a Luxen 3W star. There are plenty of LED flash lights out there you may be able to get one and modify it for the 19V. In general for a down and dirty switcher you need an inductor , capacitor and probably a resistor a transistor like a mosfet and a timing circuit like a 555 timer. Linear tech I'm sure has LED driver circuits.
  8. Jun 24, 2010 #7
    Ok I'll probably just go with a bigger LED then. What would make for a bright light (it'll be used mostly for working under cars from about 1-3 feet away, so a big spread would be better, probably like 30 degrees I think) So I was thinking maybe a .5 A light?

    Thanks for your input!
  9. Jun 24, 2010 #8


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    There's a filament, and they're still incandescents, but instead of being mostly vacuum, there's a small amount of inert xenon gas inside which helps to reduce the evaporation of the filament:

    I wish Ryobi (and it looks like Kawasaki) would release some 18/19.2V fluorescent lights (brighter and longer-lasting) in North America, but that doesn't look like it's going to happen anytime soon.

    Yes, the packs often look similar, but they're not interchangeable (at least, not without some 'modification', which usually isn't worth it). I think the contacts are sometimes arranged a little differently as well.
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