Luxeon Star 1W Resistor Question

[JediKnightBJ]

I am having trouble finding a chart that shows this, but what resistors would I use for 1W luxeons. I know there are several. If there is a chart, I would much appreciate a link.

1W-Royal Blue-4.5v
1W-Green-6v
1W-Cyan-6v
1W-Red-3v's, no resistor-hehe I know.

[Ambo]

I would personally just use a Micro Puck. I don't have any issues with saying this only due to the fact that i know that Tim (TCSS) doesn't carry 1W so it isn't taking away from any sales. The Micro puck gives you the best out of a 1 watt. Link is below.

http://www.ledsupply.com/micropuck.php

[JediKnightBJ]

I can't afford $10 a piece for 8 of them. $.66 is much more affordable. And I would buy the resistors from Tim.

[Jay-gon Jinn]

Some one posted a mathematical formula for figuring it out, but I can't remember who, or where it was posted.

[Novastar]

For the record, resistors CAN be used in PARALLEL... to achieve effectively "less" ohms, and more watt capability.

For simplicity, I will only mention "same type" resistors here...

Let's say you have two of these:
1ohm, 1watt

Put them in parallel (twist them together so it looks like a "double-barrel shotgun" style, lol), and you get:
0.5ohm, 2watt

Extremely useful (to me) when you want to get a little less ohms resisted, yet more watt capability.

[xwingband]

Some one posted a mathematical formula for figuring it out, but I can't remember who, or where it was posted.

Voltage drop/Amps desired. An Ohm is Volt per Amp.

So say your battery is 6V and the LED is 3.9V your "drop" is 2.1V. Then you want to give it 1A. You'd need a 2.1 Ohm resistor.

Novastar is correct on the parallel stuff. You add reciprocals to get the total. And in parallel the current is divided so the total capacity is upped. You may run into issue with unalike ones, but with like ones it's usually good.

For reference series resistors you add the ohms and the watt limit is the lowest one.

[Eandori]

when adding up parallel resistors, the resulting resistance is the formula (product/sum)

Say you have a 100 Ohm and a 50 Ohm. You wire them in parallel to each other. You get...
(100*50)/(100+50) = 5000/150 = 33.333 Ohms of resistance.

If your goal is a 2.2 Ohms of resistance, but you need more wattage. Here's the math.

(x*x)/(x+x) = 2.2
X^2/2x = 2.2
X^2/x = 4.4
X/1 = 4.4
x = 4.4 Ohms.

But that is not a standard value. I think 4.2 and 4.7 are if memory serves me right. So either use 2x 4.2 Ohms (slight overdrive), or 2x 4.7 Ohms (slight underdrive).

Make sense?

[JediKnightBJ]

One last question.

Which would be better/brighter. Running a Red Lux 3 with 4xAA batteries and a 1000ma buckpuck, or direct driving with just 2xAA?

[Novastar]

One last question.

Which would be better/brighter. Running a Red Lux 3 with 4xAA batteries and a 1000ma buckpuck, or direct driving with just 2xAA?

Short answer:
The buckpuck.

Long answer:
The buckpuck will provide an even 1A for the length of the cell(s) charge.

But direct-driving with 2AA will begin at a certain current, and gradually taper off to below nominal current. Depending on the cells, you MIGHT end up with a little "over-driving" from the beginning, or... you might not.

Ultimately, the only way to be certain with your resistor setup is to actually METER it while it is running. Meter it from a fresh set... then meter it later on after it's been on for a while (or pop in a quasi-used set of cells), and so forth.

Chances are--the 2AA will be UNDER-driving, because to get the full brightness from a Lux III... 2.95v is not quite enough. More like 3.2v. But again... depends on how your cells meter, and how much amperage they are able to "spit out".

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EDIT: BTW... 1A is not the "best" current for a Lux III red/red-o/amber. 1400mA is more like it. 1540mA is the maximum continuous current, but... lately many of us have been having issues with certain LIII LEDs that can't seem to handle that!!! Grrr....

[JediKnightBJ]

I'm still having trouble here. I can't afford the micropucks. So I'm back to resistors.

I want to use 6v supply, with 350ma current on LED's with 3.42Vf. According to my math and help from the equations above, I need an 8.2Ohm 1W resistor. But I can't even find one. Is my math correct?

Also, for a 4.5v supply, 350ma current for LED with 3.42Vf. Is a 3.3Ohm 1/2W resistor what I need. I just want to confirm my math before getting a bunch of useless resistors.