LED Resistor

[dark_hawk]

Didn't know that asking a simple question will result in such an informative discussion, will make sure to ask so often.

I had the voltage divider and current limiting resistor concept fuzzy in my head, Thank you Ian for the clarification.
This Joule Thief circuit sounds like fun.

The power source are 4 1.5V Type C batteries so there is no fear of an over-voltage running through circuit.

The LEDs I got are generic with no datasheet, burned some to see their limit and they seem to pop at around 50-60ma with decent brightness at 3.5V at 20ma. Applying the full LED loads (18 5mm white + 5 x 10.5mm white) results in approximately 0.5A and voltage drop across a new set of batteries from 6.4V to 5.7V and an old set of batteries from 5.5V to around 4.4V. So correct me If I'm wrong the 130ohm resistor seems adequate.

[cthree]

An lm317 is going to require a heat sink sourcing 360mA and I think the op said space was an issue.

[MacAttak]

Most typical LEDs are rated for 20 mA max, that doesn't mean they will pop at 21mA. It just means that once you exceed that rating the lifespan of the component is reduced. It won't "pop" until you greatly exceed that rating, but it will certainly be damaged.

I'm not sure I understand your method for measuring voltage drop. A typical white LED should be something in the range of 3.5v. When lit, you should be able to read roughly a 3.5v difference between the anode and cathode. If your fully-charged battery source is at 6.4v then that means you need to drop the remaining 2.9v with the resistor. R = V / I, so the desired resistance is 2.9v/0.02A = 145 ohms. Anything less than that and you risk overdriving the LED. And even though the current rating is usually 20mA, most are efficient enough to where you almost can't tell the difference between ~15mA and ~20mA.