Coleman Max 115 lumens?

[alpg88]

The emitter is not receiving 560mA in your example. 560mA is the current draw from the batteries. After going through the resistor, the current to the emitter will be considerably less. The emitter can handle 1000mA.

lol, really??
current, on both sides of circuit, is equal with resistors, and linear regulators, the only time when it is different is when switching regulator is present, don't believe me??? do simple measurements.
well you might be somewhat correct, about actual current, but not in a way you described, the tailcapswitch isn't that well made, thin spring, cheap button... resistance of it might actually be higher than of the dmm i used, so actual current might be a bit less than measured. but not by much.


also. the calc. you linked doesn't take into account voltage sag of the cells, i can tell you if you use resistor and 2 rcr2 cells, you Will dissipate a bit more power on resistor than on led itself. even according to the link, power dissipation resistor will be almost 3 !!! watts. like buying foot long sandwich and throwing away more than half right away. not to mention there isn't enough room for 3w resistor.


another thing, heat transfer, since i took the light apart, i would not put more than 500ma on it, heat moves away from the star via thin edge of the body, half of the edge actually, may be enough for 500ma, but wont be enough for 1a, you'll fry the led, i have fried many leds\parts in decade or more i work with lights, by now i can see how such simple set up will work.

c,mon man you giving advice on thing you never seen, and bad one too.

 

[nikon]

also. the calc. you linked doesn't take into account voltage sag of the cells,.

No, the calculator doesn't take into account the voltage sag. I did that. Two Li-ion batteries = 8.4 volts (4.2 x 2). I entered 7.4 volts to account for sag.

I didn't say 8.2 Ohms of resistance is a good idea in this application. In fact, it's a bad idea. It would be much better to use a small resistor and a single li-ion battery. But 8.2 Ohms is what's in there, and 8.2 Ohms is what the question was about.

I'm not sure I understand whatever else you're going on about, as your attitude problem seems to have gotten in the way of your ability to speak clearly. Try reining it in.


UPDATE:

I just took the light apart and what I found was not one 8.2 Ohm resistor, but two. They're wired in parallel to give a total resistance of 4.1 Ohms. They're ceramic resisters and I don't know what power they're rated for, but in this application they only need to handle one watt.

I rated the two primary cr2 batteries at a total voltage of 5.6v. to compensate for voltage sag. I rated the voltage drop across the LED as 3.5v. and asked for a current of 600mA. The calculator recommended a resistor of 3.9 Ohms and 1 Watt power handling. The current draw is predicted to be 538.5 mA. The LED will dissipate 1885mW and the resistor will dissipate 1131 mW. Not the most efficient way of doing things but it works. Coincidentally, this is the same setup that streamlight used for the first version of the TL-2 flashlight using a Luxeon I LED. As soon as I get time I'll change the resistor to a much smaller one so that the light can run on a single Li-ion battery. It should be around 1 Ohm.

 

[nikon]

I finally got around to modding this light. I put a 1 Ohm resistor across the pair of 8.2 Ohm resistors, which gave a total of .8 Ohm. The light draws 550mA from a 14500 li-ion battery. This is a safe load and should give over an hour of runtime. The brightness appears to be greater than it was before the mod.