My first mod, and a fried LS...

Long story short:
Inretech module for AA Mag, wired to a PR base to try out in a PT Tec 40, hoping for a relatively bright flood, and "just to see".
Module needed shaping to fit through the hole (though the length was right, due to using a gasket between it and the PR base), but first, out of curiosity, I stuck the optic that came with it between the reflector and lens of the PT, and screwed it down. That worked, and it was about twice as bright as it had been in the 2AA Mag for a couple of minutes, at which point it went out.
I had originally grounded the Inretech to the side of the PR base, but had to file it off to be able to get it on and off the PT's spring, so it was only grounded with a small blob of solder on the edge of the base.
I thought twisting the head down might have broken the little spot of solder, and that losing the ground had made the light go out, but when I unscrewed the bezel, one sniff told me the LED was fried.
The "top" of the LED was blown off, and stuck to the back of the optic.
Should a 1w LS be able to take 6v direect, or do I need a resistor inline?
Guess it's a little late to ask that...
If that's the problem, is there a chart somewhere that will tell me the values recommended for different voltages/amperages?
Thanks.

I'm trying to get the "think first, act second" thing down, but inspiration struck, and wouldn't be denied /ubbthreads/images/graemlins/grin.gif

I would try a "dummy" AA in the 4AA battery holder, to keep things at 4.5V -- I did try a very high Vf 1W running on two CR123s in a Legend LX that I called the LGI Shorty and true to its name, its lifespan was indeed rather short, although it did last a relatively good while via intermittent use I must say (and the one I made for Fasty/chr00t did, too) before both of them finally expired...

Owen,
No, obviously a 1W LS won't take 6v direct. Here's a link to a calculator to figure the resistor needed.

www.quickar.com/starresistorcalc.htm

As I understand it, the Inretech AA adapters work because the vf (forward voltage) is low on their particular luxeons, able to work on 3v.

Supply voltage - Leds Rated Forward voltage give you the drop voltage... or how many volts you need to drop.

in a case like this... you should really use a DMM but... using the standard rated forward voltage of the LS... it would go something like this.

6volts - 3.42volts = 2.58volts

So you have to drop 2.58volts.

To figure out the value of resistor that will accomplish this you take that drop voltage value and divide it by your led rated forward current...in this case 350mA. You must convert the 350mA to Amps...which would be .35A /ubbthreads/images/graemlins/smile.gif Use that in the equation.. like this

2.58volts/0.35A = about a 7 Ohm resistor /ubbthreads/images/graemlins/smile.gif

so you would need about 7 Ohms of resistance.

jtice, one of the many brilliant minds here, built a program for this. The effort was supported by a team of other bright fellows namely, rothrandir, dat2zip and logicnerd411.

I've never used it until just now.

Using Negeltu's figures I plugged in:

350 Desired LED Current - mA
3.42 LED VF Voltage
6 Battery VF Voltage
1 # of LEDs, Series

The "LED Pro" program calculated 7.14 Resistance in Ohms and Resistor Wattage 0.87 .

Don't take my word for it 'cause I don't have a clue about electronics and if I did I would probably misplace it. As proof of this; I don't understand the answer. I would have thought Resistor Wattage would have been close to what Negeltu calculated. If I've used the program incorrectly, surely a pro will jump in and say so.

This program is offered to CPF modders in the second sticky in this forum...free.

On a side note: Using Mr. Bulk's suggestion and plugging in 4.5 as Battery Input Voltage the resistor answer becomes .38 . It would be unlikely, I think, that Mr. Bulk would put resistance in a well sinked 4.5 V/1W or 3W configuration.

Negeltu -

Did that sound right to you?

LED Pro team -

If I got it right, well, that was pretty darn cool!

*Edit*

OK, I rethunk it and I think it did work. 7.14 Ohms resistance is in line with Negeltu's calculation. .87 resistor wattage is a whole different critter.

IceBreak,
You used the program correctly. I arrived at the same answer but rounded down. ... /ubbthreads/images/graemlins/smile.gif 7 Ohms should be just fine...

Negeltu -

I was editing while you were replying. Thanks.

This LED Pro program is a helpful tool. Wish I had half your EE knowledge.

- Jeff

Thanks guys. That will remove any guess work, and undoubtedly save me some frustration in the near future /ubbthreads/images/graemlins/smile.gif

[ QUOTE ]
Icebreak said:
Negeltu -

I was editing while you were replying. Thanks.

This LED Pro program is a helpful tool. Wish I had half your EE knowledge.

- Jeff

[/ QUOTE ]

I had nothing to do with the LED Pro software.. /ubbthreads/images/graemlins/smile.gif

Right...jtice, rothrandir, dat2zip and logicnerd411.

Just thanking you for confirmation I operated it correctly and yes, I still wish I had half of your EE knowledge.