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Cree Led Conversion

illfixit

Jan 7, 2014
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Hi All. I have a question and want to get the right element. I'm attempting to convert a spot light to use a (RED) Cree Led instead of a White Led element. My question is what RED element would be the BRIGHTEST possible to solder in place of the original element. This unit uses a 20mm element with FULL and MOON stamped on it as it is a high and low beam. Also it runs on 3 AA 1.5v batteries that would be hard to change looking at the bat. pack configuration. I emailed a LED company and this is what they recommended...
creexpe-red-1-1.jpg

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BobK

Jan 5, 2010
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Simply replacing a white LED with a red one may or may not work depending on the dirving circuitry. The forward voltage of the white LED is likey 3.3-3.6V whereas the red one is going to be about 2.2 to 2.5V. So you might simply blow it out if you replace it with no other modifications. If you could show us what is connected to the the red and black wires in your pic, we might be able to help.

Bob
 

illfixit

Jan 7, 2014
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Ok, this is a picture of the wires, they just go directly to the switch. I don't see a resister or anything, what else should I look for.
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Arouse1973

Adam
Dec 18, 2013
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I agree with Bob. You can't just swap packages or colours. Looking at the module they have suggested being the INDUS STAR which is specked at 295 lumens for 700mA. To put this in context in the UK 100W light bulb is about 1300 lumens. The XL and XM white lights are up to 1040 lumens and the MKR a massive 1769 lumens, but then it does have 4 LEDs inside. So changing to red might not do what you want it to do. So if you do want to go red then basically what you need to look for is the LED with the highest mcd out put with the largest viewing angle, this will give you the greatest amount of luminous flux (lumens)
Adam
 

BobK

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The picture does not help much, I was hoping to see a circuit board.

Bob
 

Arouse1973

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I think the only option if you want red is the INDUS STAR module but heed what Bob said you risk damaging the module if it is not driven correctly. And this will be no where near as bright as the white one you have, I think you will be disappointed.
Adam
 

illfixit

Jan 7, 2014
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I found the switch board, does this help any.
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(*steve*)

¡sǝpodᴉʇuɐ ǝɥʇ ɹɐǝɥd
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Jan 21, 2010
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It looks like this is one of those special circuits that uses the internal resistance of the cells to limit the current.

I guess the nice thing is that it would warm your hand on a cold day...

edit: maybe not. A series 4.7 ohm resistor, and two other resistances(?) that I can't quite read. Does it have 2 brightness settings?
 
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BobK

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So it looks like it is simply using a resistor to limit the current. In that case, you should just replace the resistor board with a series resistor calculated based on the new red LED. The equation is:

R = (Vb - Vf) / If

Where Vb is the battery voltage, Vf is the foward voltage of the LED and If is the forware current at that voltage of the LED.

For example, if the battery is 4.5V and the LED forward voltage is 2.5V at 350mA you would get:

R = (4.5 - 2.5) / 0.35 = 5.7 Ohms. A 5.6 would be the closest standard value.

Bob
 

illfixit

Jan 7, 2014
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It looks like this is one of those special circuits that uses the internal resistance of the cells to limit the current.

I guess the nice thing is that it would warm your hand on a cold day...

edit: maybe not. A series 4.7 ohm resistor, and two other resistances(?) that I can't quite read. Does it have 2 brightness settings?

Yes it does have two brightness settings, the switch operates first click bright and second dim.

So it looks like it is simply using a resistor to limit the current. In that case, you should just replace the resistor board with a series resistor calculated based on the new red LED. The equation is:

R = (Vb - Vf) / If

Where Vb is the battery voltage, Vf is the foward voltage of the LED and If is the forware current at that voltage of the LED.

For example, if the battery is 4.5V and the LED forward voltage is 2.5V at 350mA you would get:

R = (4.5 - 2.5) / 0.35 = 5.7 Ohms. A 5.6 would be the closest standard value.

Bob

Bob, Your so far over my head I'd need the shuttle to catch up. This looks like a bit more involved than I planned. I think I'll purchase a light that works for me.
 

BobK

Jan 5, 2010
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It is really not all that involved. All you have to do is replace the board that connects to the two red wires with a single resistor. The resistor is calculated via the formula I gave once you know the specifications of the red LED that you are using.

Bob
 

illfixit

Jan 7, 2014
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It is really not all that involved. All you have to do is replace the board that connects to the two red wires with a single resistor. The resistor is calculated via the formula I gave once you know the specifications of the red LED that you are using.

Bob

That seems simple enough, I'll try to find the match up and see what I can create. This is a really handy and packable light for those long cold hikes in the woods at night, I hope to get it working.
Thanks
 
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