CREE XM-L T6 in Surefire G2 -- ok with 2 x RCR123?

I purchased a couple of 1-mode CREE Q5 LED's over a year ago for my Surefire G2 lights (replacing the P60 bulbs) and they've been working great. I've also read that I shouldn't run a 18650 or 2 x RCR123 batteries with that setup since that's too much voltage for that Q5 LED.

I was going to pick up some "750 lm spec" CREE XM-L T6 bulbs for my Surefire, and they're rated 3.7v - 7.4v operation voltage, and from 450 - 1000mA working current. From what I can tell, that should work with a 18650, or at least a 16340 if that's the only thing that'll fit. Also, it should also work with 2 x RCR123 Li-Ion, but I'm not sure about the current if you put 2 x RCR123 800mA batteries in series with each other. Likewise, most 18650's are around 1600mA, so I'm wondering if that'll fry the Surefire.


I've searched and didn't find anything particular on this setup. Does anyone have a recommendation on the batteries I can use if I get the XM-L T6 for my Surefire G2? Can the Surefire wires handle 1600mA of current? Right now I'm using Surefire CR123 alkalines, but would love to use my RCR123 batteries.

An 18650 won't fit.

2 rcr123's = 8.4V so that wont work either...

You would need to run a 17650 battery!

I wouldn't run that setup in a G2 anyway, it would get too hot, and the G2 does not have sufficient heatsinking.

...BTW, an XM-L needs 3+ Amps to achieve 750 lumens. If it only runs 1000mAh (1.0Amps) it will be nowhere near 750 lumens, but that would be an appropriate drive current for a plastic host.

Also, the current draw is determined by the led's driver. The "1600 mAh" rating you mentioned is the capacity of the battery and 1600 is really low actually.

I have some XM-L dropins that I run at 4A (4000 mAh) and I use 18650's that are rated at 3100 mAh capacity!

I use all metal hosts, with massive heatsinking.

A couple of points...
1. First off, a 18650 won't fit in a G2 as the diameter is about 2mm wider. You can't get precisionworks to bore it out with a lathe either, as the metal tube is the thing that allows for the electrical connection. You can however get a bored 6P tube etc to fit in there if you wish. Even 17670s won't be guaranteed to fit either, as Dan from oveready measured the tolerances for Surefires and found it to be tighter than that of the AW17670. So the only 1 Li-On cell that's guaranteed to fit is the 14670 but AW won't have them for a while (i asked him).

2. You CAN use a 18650 or similar cell with your existing Q5 as the nominal voltage of 3.7v is lower than that of the 2 CR123s. It's the voltage that determines compatability.

3. 2 Li-on cells can be used with a 3.7v-7.4 nominal spec. And the tailcap draw current of 450mA with RCRs (current decreased as voltage increases) is within the 2C limit of LiCo chemistry. With a 450mA current, the wattage to the LED is around 4 watts, so the lumen output would be closer to around 300-400 OTF. This is why it would be advisable not to get those generic ebay chinese drop ins. Ask out Nailbender or Vinhnyugen to make you the real 750OTF deal.

4. Most 18650s are 2600maH and up, and some are 3100maH, and Panasonic promises that 3400 and 3900maH ones would be up in a couple of years. The ONLY 18650 with 1600maH capacity are that of LiMn chemistry, allowing fro higher draw currents. And 800maH RCR123s are fake, the highest is around 700maH measured.

There are also less known 16650 cells on the market, HKJ has done a review of them. They fit in a G2 and many oter 2*Cr123 lights and have more energy than the standard 17670 cells.

brandocommando said:

...BTW, an XM-L needs 3+ Amps to achieve 750 lumens. If it only runs 1000mAh (1.0Amps) it will be nowhere near 750 lumens, but that would be an appropriate drive current for a plastic host.

Click to expand...

Where are you coming up with that. No disrespect, but that mAh of the battery is not what you measure to see if it produces 750 Lumnes. The way to measure lumens is to first look at the specs for your bulb. In this case XM-L so here it is http://www.cree.com/~/media/Files/Cree/LED%20Components%20and%20Modules/XLamp/Data%20and%20Binning/XLampXML.pdf

What does this say? It says that it can produce "1000 lumens with 100 lumens per watt efficacy."
What is the watts in your flashlight? To figure it out, do the following and use the formula below.
Formulas to use:
V= Voltage I=Current (Amps) R=Resistance (Ohms) P=Power (Watts)
Ohm's Law says V = I R
P = I V

How to measure Watts.
1. Measure Voltage first. Put the two leads on the "COM" connection of the tester and "Volts" or "V" connection. Turn your dial to "DCV" and put the wires on the two sides of your battery. If your flashlight takes two in line, line them up and measure both batteries. So two 3 volts should measure 6 Volts.

2. Measure Amps. Put the batteries in your flashlight. Do NOT put the cap on. Now REMOVE the wire from your tester marked with "Volts" and place it on the connection marked with "A" or "Amps". Turn the dial to your Amps setting. Now place the other ends of the wires on the negative side of the exposed battery and the other wire on the place where your cap screws on. The flashlight should turn on. You should be able to read the Amps. Now take the Amps measure from the flashlight (note this has NOTHING to do with the mAh of the battery) and multiply that by the MEASURED voltage of your battery or batteries. Remember if two batteries in-line, then double the voltage. Now this number is Watts or POWER (P = I V). This is the watts in an ideal world, but your flashlight not being made out of a good conductor (Gold) will loose watts, so you must deduct 20% of the wattage. Now multiply that number by 100 since the Cree manufacturer (pdf) states that you get "100 lumens per watt".

lightphysics said:

Where are you coming up with that. No disrespect, but that mAh of the battery is not what you measure to see if it produces 750 Lumnes. The way to measure lumens is to first look at the specs for your bulb. In this case XM-L so here it is http://www.cree.com/~/media/Files/Cree/LED%20Components%20and%20Modules/XLamp/Data%20and%20Binning/XLampXML.pdf

What does this say? It says that it can produce "1000 lumens with 100 lumens per watt efficacy."
What is the watts in your flashlight? To figure it out, do the following and use the formula below.
Formulas to use:
V= Voltage I=Current (Amps) R=Resistance (Ohms) P=Power (Watts)
Ohm's Law says V = I R
P = I V

How to measure Watts.
1. Measure Voltage first. Put the two leads on the "COM" connection of the tester and "Volts" or "V" connection. Turn your dial to "DCV" and put the wires on the two sides of your battery. If your flashlight takes two in line, line them up and measure both batteries. So two 3 volts should measure 6 Volts.

2. Measure Amps. Put the batteries in your flashlight. Do NOT put the cap on. Now REMOVE the wire from your tester marked with "Volts" and place it on the connection marked with "A" or "Amps". Turn the dial to your Amps setting. Now place the other ends of the wires on the negative side of the exposed battery and the other wire on the place where your cap screws on. The flashlight should turn on. You should be able to read the Amps. Now take the Amps measure from the flashlight (note this has NOTHING to do with the mAh of the battery) and multiply that by the MEASURED voltage of your battery or batteries. Remember if two batteries in-line, then double the voltage. Now this number is Watts or POWER (P = I V). This is the watts in an ideal world, but your flashlight not being made out of a good conductor (Gold) will loose watts, so you must deduct 20% of the wattage. Now multiply that number by 100 since the Cree manufacturer (pdf) states that you get "100 lumens per watt".

Click to expand...

This approach is not a good way to estimate LED output.

The XM-L datasheet provides various graphs that are far more useful.

If you want 750 (emitter) lumens from a T6 flux bin XM-L, you will need a relative luminous flux of 750/280 = 2.7. The datasheet graph of relative luminous flux vs forward current tells us that you need about 2.3A drive. At that drive current, Vf looks to be about 3.25V, again based on the datasheet (see the graph of If vs Vf).

For a buck driver regulated light that uses 2xRCR123 to power the XM-L at 2.3A, let's generously assume 90% driver efficiency. That gives

Vbatt * Ibatt * 90% = Vf * If = 3.25V * 2.3A ~7.5W

Thus, Vbatt * Ibatt = 7.5W/90% = 8.3W

That means your power source (2xRCR123) has to be able to deliver about 8.3W of power to the driver, so that the driver can deliver 2.3A drive current to the XM-L to get 750 emitter lumens.

To figure out what 8.3W means in terms of Ibatt, a simple way is to assume at t=0 that Vbatt = 8.4V. Then at t=0+, the voltage rapidly sags to some lower voltage (let's say 8.0V), and finally holds at some quasi-steady-state voltage for most of the run time (let's say 7.4V).

Thus, Ibatt can start out at slightly less than 1A, increase to slightly more than 1A, and hold at about 1.1A for most of the run time. Let's just call it 1A.

1A is close to the limit for the current draw from a small RCR123 cell. If the driver efficiency is less than 90% (which is most likely the case), then the RCRs will have to deliver more than 1A. Stressing those little cells like that will sacrifice cycle life. I'd probably use IMR16340 cells.

If you want to power your G2 XM-L drop-in using 1x17670, then you'll probably use a drop-in based on a 7135-based linear regulator driver. The 7135 outputs a nominal 350mA drive per chip. If you get a driver that parallels 7 chips together, you get 2.45A drive, which is close to the estimated 2.3A needed to get 750 emitter lumens.

Using a 7135 driver means that your 1x17670 will have a current draw (Ibatt) of 2.45A nominal, as long as it delivers enough voltage to run the driver in full regulation (basically, Vbatt has to be at least 3.25V + 0.12V = 3.37V, and realistically more than that). 2.45A draw from a 17670 should not be a problem. Run time in regulation is probably going to be fairly short, however. Total run time will be longer.

Why hasn't anyone mentioned heat? I'd be scared of not being able to cool down the XM-L in a G2. I used an XM-L in my 6P and it was far, far too hot, although it was at 3.1A.

That is an excellent analysis Justin!

We can complete the analysis by estimating what the junction temperature would be, although that Cree data sheet does not show the LED I/V curves for other temperatures. We can use a general rule of thumb for temperature dependence.

What do you recon is the thermal resistance of the whole module? For the 2.3A, it looks like the thermal resistance cannot exceed 14C/W and we will see a degradation in the lumen output.

-Desirider.

The XM-L datasheet says that the thermal resistance from junction to solder point is 2.5 C/W. If we reflow the XM-L onto an MCPCB, the thermal resistance of the MCPCB could be about 3-4 C/W. Then you might use Arctic Alumina to glue the MCPCB to a heat sink. The last I saw, AA was spec'ed at >4 W/m-K (they vendor seems to have stopped quoting a thermal conductivity for AA). If we assume that the AA glue layer is 0.001" thick, then we get a thermal conductance of 157500 W/m^2-K. The XM-L's thermal pad is 2.8mm x 4.8mm, or 1.344 x 10^-5 m^2. Multiplying the thermal pad area by the calculated thermal conductance, we get about 2 W/K, or a thermal resistance of about 0.5 K/W (K and C are equivalent for our purposes).

So approx total thermal resistance from the junction to the heat sink could be around 2.5 + 3.5 + 0.5 = 6.5 C/W.

This ignores the thermal resistance between the heat sink and the host and then the host to the ambient. I'm going to assume that this is small and ignore it (that probably means that you need a tight, large area fit between your heat sink and flashlight host body or metal bezel in the case of the G2 since the plastic G2 body is a poor thermal conductor).

If the total resistance from junction to ambient is 10 C/W, then the datasheet says that if you run the XM-L at 3A drive, then the max ambient temp allowed is about 35C. At 7 C/W, you can run the XM-L at 3A drive with a max ambient temp of around 70C.

I can understand the datasheet graph that shows 7C/W and 70C max ambient (although, I would have expected something closer to 80C max ambient), but I'm not sure I understand the 10C/W and 35C max ambient. At 3A drive, Vf is 3.35V according to the datasheet, or 10.05W of power draw by the LED. For 10C/W thermal resistance, that's a 100C temp rise for the junction above ambient. The max junction temp is 150C. To me, that means the max allowable ambient temp is thus about 50C, not 35C.

As for the junction temp for the case at hand, at 6.5 C/W and ~8W draw by the LED, you are looking at about 50C temp rise above ambient in the junction. If ambient is 25C, then the junction could reach 75C. The datasheet says that relative luminuous flux is about 90% at that point (i.e., you might lose about 10% of your output). Of course, if you use your light for short bursts, the light probably won't heat up to that extent. The max allowable ambient temp for operation looks to be about 100C before the junction temp reaches the 150C limit.

One of my G2s runs 1 X 17650 and a Chinese "XML-T6." I changed to an aluminum bezel. The light works fine, but the bezel does warm up. The dropin was not expensive so I plan on continue using it. I don't use that light for more than a few minutes at a time. Would I take that light into a cave or mine? Nope.

Great analysis once again.

If we assume the junction temperature is 150C, then I estimate the Vf to be 3.83V for 3A drive = 11.5W. For a 10C/W j-a thermal resistance, the maximum ambient temperature can be 150C - 11.5W * 10C/W = 35C, so it sounds about right.

I admit, I cheated. I back calculated what Vf should be at a junction temperature of 150C for a 3A drive. I couldn't fit the rudimentary diode equation I=Is * (exp(Vf/nVt)-1) to the I/V curve. LEDs are not exactly simple junction diodes, but the increased Vf from 3.35V@25C to 3.83V@150C for a 3A drive sounds reasonable.

Then there is the output degradation. We end up with only about 72% lumen output => 540 lumens, despite the increased input of 11.5W.

-Desirider.

Post deleted.

I missed the -ve temperature dependence. Thanks. In that case I speculate that Cree's ambient temperature plot assumes a junction temperature of 125C. Possible?

Post deleted.