Resistor value for 3 blue LEDs in PR base

Step 1, look at my logon name since it explains a lot


Now, my question: I've been violating some $1.99 photon wannabe's for the LEDs and batts (2 CR2016s per light). What resistor value would I use to run a PR-base with 3 blue LEDs in a 4 AA or 3 AA configuration? I can make dummy cells fairly easily, so if 3 is easier, that's not a problem. (This will most likely end up in a Dorcy Boss that I found for $3.00 at a discount store.)

Thanks!

Anyone?

Just use that neat little calculator at Click HERE

Klaus

Well, I'll stick my nose in.

3 AA alkalines will give you 4.5v, 4 would give you 6v. (note that rechargeable Nicads or Nimhs will give you a lower voltage as they are 1.2v per cell..)
So, if we go by 30mA per LED, in parallel, then the total current draw you want will be 90mA. LED voltage is 3.6v, so your resistor value will be:
battery voltage minus 3.6v, divided by .09A

So for 4.5v, it would be about 10ohms, you could go a little lower if you want to drive the LEDs a bit harder.
For 6v it would be about 27ohms, again lower for brighter..

Hope this helps..

Edit: Just saw that calculator Klaus mentioned. Well, that makes it easy..

Graham

I did a 2- nichia mod on a 3 cell torch using a PR bulb. I hnow the resistor value for 1 LED, for 2 I simply halved its value, so for 3 LED's divide it by 3.
To make construction of my LED bulb simple, I soldered the resistor to the central base then I attached the anodes of the LED's to the free end wire.Make sure your resistor is up to the job power wise, or your torch will be a SED (smoke-emitting) torch rather than an LED torch!
The cathodes were then soldered to the bulb base in gaps I filed out in addition to the original gap occupied by the bulb wire.
If using this bulb in a metal torch, the cap of the bulb provides some conduction thereby giving effective heatsinking for the entir assembly.
To prevent mishaps such as shorts occuring, fill the cavity of the base with an insulated potting or epoxy compound.