What resistor for Cree Direct Drive

I want to put a Cree in my Golston and I need to figure out which resistor to use. I need to drop the 6v to approx 3.2 and I'd like to keep the drive current between 700 and 750ma. According to one formula I saw I would need approx a 4 ohm resistor. Does this sound right? I think the stock Golston resistor is 1 ohm. I think that would overdrive the Cree too much. I'm trying to make it bright but not overdriven too much to keep the heat managable.
Hopefully I'll get a real driver for this soon, but I'd like to be able to play with my light while I wait for it to get here.

Walt175 said:

I want to put a Cree in my Golston and I need to figure out which resistor to use. I need to drop the 6v to approx 3.2 and I'd like to keep the drive current between 700 and 750ma. According to one formula I saw I would need approx a 4 ohm resistor. Does this sound right? I think the stock Golston resistor is 1 ohm. I think that would overdrive the Cree too much. I'm trying to make it bright but not overdriven too much to keep the heat managable.
Hopefully I'll get a real driver for this soon, but I'd like to be able to play with my light while I wait for it to get here.

Click to expand...

The value of the stock resistor is different between the 'original' Golston (about 1 ohm) and the "3.6V x 2" rechargable version (about 3.2 ohms). You should measure the resistor you have first, maybe you'll get lucky and it will already be what you need.

I was hoping to try the stock one, but I bought a used Golston, and the resistor is gone.

Walt, I have seen as much as a 0.5V difference in Vfs for a Cree XR-E driven at 700mA. That 0.5 Volts will make a HUGE difference in the drive current. Depending where you are running on the V/I curve, a 0.1V change in drive current can result in a 200mA current change.

You will have to do some trial and error. Here is some info to get you started:

The cells (assuming 2x CR123) will sag to roughly 2.8 Volts (each) under the desired current load (700mA).
->Battery voltage will then be about 5.6V

The XR-E's I tested were 3.03V and 3.53V (let's round off to 3 and 3.5V) at 700mA respectively.

Assuming your Cree falls into this range, you will have to drop between 2.1V (5.6-3.5) and 2.6V (5.6-3).

For a 2.6V drop at a current of 0.7A, you would need a 3.7 ohm resistor. (V/I=R, 2.6/0.7=R)

For a 2.1V drop at a current of 0.7A, you would need a 3 ohm resistor.

Now you have a starting point for tweaking. Good luck!

if theres space in the golston, y not try the luxdrive micropuck?

Thanks Chimo! Next problem is to find some. Radio Shack isn't what it used to be.

X2x3x2, I've been looking at some of the drivers at the shoppe, but at this time of year, I'd expect it to take a while to arrive. This way I still have a light to play with while I wait for a driver. Plus I can play around and try different drive levels and see what level works best for me.

It has been a while and I am going from memory, but I think the difficult part in getting a new resistor for the Golston is in finding one that both fits in the available space and can handle the current from 2x CR123's. I think the best thing you'll typically find at Radio Shack are 1/2 Watt resistors and you'll need more than that (the big wirewrap resistors are hard to fit in the available space from what I've read, but I haven't tried it). You'll probably end up with 3 or more resistors in parallel in order to be able to handle the current. According to my calculations below, even using 3x 10 Ohm, 1/2 Watt resistors in parallel would be very marginal. You may have to hunt around a bit to find some resistors that are rated for 1 Watt and use 2 of them. It may be easier to consider designing your mod to use a 17670 Li-Ion cell, since then you won't need to drop as much voltage in the resistor.

To expand on Chimo's numbers:

For a 2.6V drop at a current of 0.7A, you would need a 3.7 ohm resistor, rated for at least 1.813 Watts (P=I*I*R=0.7*0.7*3.7).

For a 2.1V drop at a current of 0.7A, you would need a 3 ohm resistor, rated for at least 1.47 Watts.

Besides the problem of real estate this proposition is also very wasteful of energy.
With the 3 ohm resistor you'll waste 37% of the energy in heat. With the 3.7 ohm it is over 46% lost as heat, almost as much energy wasted than the energy required to drive the LED.

A 17670 Li-Ion cell would be much more efficient.

Given a P4 and an R123 or 18650 cell at 3.6v.

When the led is run DD, how much current will it "pull" ?