Need resistor value for cree mod.

connortn

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I am modding a streamlight twin task with a cree LED and need a little help. The switch in the TT is a single pole double throw used to turn on the 3 led's or the incadesent bulb. I was going to short the two independant poles togather so the led would have only "on-off" and have also concidered using a resistor across one of the poles to have a "off-high-off-low". I have Lighthounds 3.7v to 9v drop-in cree LED and mostly use 123 cells but am looking at using the rechargable 3.7 18650 (I think that's the correct number).

What I need is the value of the resistor to drop the output to somewhere around 10-20 lumens. If you are familiar with this drop-in let me know what would work.

Thanks...
Connor
 

LightForce

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50 mA forward current = 15 lumens. For 18650 cell you'll need ~ 60 ohm resistor.
100 mA = 30 lumens, and ~ 30 ohm.
Vf will be around 3.0V.

D.
 
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TMorita

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LightForce said:
50 mA forward current = 15 lumens. For 18650 cell you'll need ~ 60 ohm resistor.
100 mA = 30 lumens, and ~ 30 ohm.
Vf will be around 3.0V.

D.

This calculation is incorrect for two reasons I can think of, offhand:

1) LED brightness is not linear with currrent

2) In order to halve the current, you need to double the total resistance of the LED + resistor, not just the resistor itself.

Toshi
 

Calina

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Not as simple as it looks! You will probably have to take into account the variability between LED themselves. The VF of cree LEDs is quite variable.

The "easy" way out would be to mount a circuit with a suitable variable resistor. Play with the pot and find the light level you want, measure the resistance and replace with one of fix value.
 

Tronic

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The 18650 cell don't physically fit in the Streamlight TT-2L. (Is this the light you want to mod?)
You have to use the 17650.

The 5 digit number is deciphered like : 17650 = 17mm ( diameter ) X 65mm ( length ) where the last ' 0 ' stands for cylindrical cell
 
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