CREE XML-T6 LED emitter low current at tailcap

Hi all and just a quick question in regards to the figures i got when i measured the Current at the tailcap on my CREE XML-T6 LED emitter DX Drop in HERE and i got the below results:

Low : 165 mA
Mid : 725 mA
High : 1.65 A

It was running on fully charged Troch Light 3.8v 26650(Protected) 4000mAh Rechargeable Li-ion http://www.lck-led.com/p822/Troch-L...0mAh-Rechargeable-Li-ionTro/product_info.html

Input voltage needs to be higher.

It is in a 2D mag at the moment but i do have a 3D mag i might try it in on 32600 batteries .

What Lumens would i be getting at my highest current draw 1.65A

Would i notice a difference in brightness if i used the 3x 32600 batteries @: 11.1 Volts and would that be closer to the 3A DX rating.

I thought i would have got the max Brightness and Lumens at anywhere between the manufatures spec 5.8~12V

Unless you are looking at direct drive or say a linear regulator, tail current draw is not the same thing as LED drive current. I assume this DX drop-in has some sort of buck driver, based on the stated specs in the link you provided (Working voltage: 5.8~12V).

If you fed the drop-in 2xLi-ion at a nominal 7.4V and assume that the XM-L Vf is 3.35V at 3A drive and the driver efficiency is 85%, then

0.85*7.4V*Ibatt = 3.35V*3A = 10.05W

Thus, Ibatt = 1.60A, which looks suspiciously close to what you measured.

If you feed the drop-in 3xLi-ion at 11.1V nominal, your tail current measurement will decrease, not get closer to 3A.

Ok just looked at the data sheet http://www.cree.com/products/pdf/XLampXM-L.pdf and what you say above is correct and i have learnt something by your reply but can you explain how you got these figures " 0.85*7.4V*Ibatt = 3.35V*3A = 10.05W Thus Ibatt = 1.6A as i am trying to get my head around it so i know for future reference.

Also why do people measure at tailcap and wouldn't it be better to get a true reading at the Emitter end of what it is actually pulling in ( Amps ).

Am i also correct to say that DX website states 3000mA at max is that at the emitter side ? and is there a formular to work out what lumens it is throwing out at the above readings.

Appreciate your help and i am all johnny 5 .....more input all ears

Power from the driver = Power consumed by the LED

0.85 is 85%, which is a moderately high, but still reasonable, driver efficiency value.
7.4V is the nominal voltage for two Li-ions in series.
3A is the advertised drive current.
3.35V is the datasheet value for the XM-L's forward voltage (Vf) at 3A drive current (If).

Driver power sent to the LED is driver efficiency*Vbatt*Ibatt.
LED power consumed is Vf*If.

Thus, 0.85*7.4*Ibatt = 3.35V*3A. Solving for Ibatt gives you 1.6A.

People measure tail current probably because 1) it is easy to do, 2) it suggests what sort of run time they might get, and 3) it gives them an idea as to how hard the cells are being driven.

Measuring drive current at the LED requires access to the LED, which is not always possible (the head may be threadlocked and thus sealed). You have to measure the current in series with the LED, which generally requires that you de-solder the connection to hook your DMM into the circuit. That can be a hassle. If you screw up your connections with a boost driver and feed it an open load, you often will kill the driver.

I have no idea what DX's advertising copy is supposed to mean. All I can do is assume that the 3A refers to drive current.

To estimate lumens from drive current, refer to the LED's datasheet. There is typically a graph of relative luminous flux vs drive current. Find the value of relative luminous flux for the drive current of interest and apply that RLF factor appropriately. For the XM-L, at 3A, RLF ~325%. Multiply that by the lumens at the standard forward current. For the XM-L, you can see from the graph that this current is 700mA (that's where RLF is 100%). You also can infer this from the fact that the lumens rating for a T6 XM-L is given as 280 lumens at 700mA forward current.

280 lumens * 325% = 910 lumens. This is the estimated lumens at the emitter. You will have losses from the reflector or optics, the glass window, LED junction heating, blockage by the bezel, etc. Multiply the emitter lumens estimate by say 1/2 or 2/3 or 3/4 or whatever to get an estimate of the out the front lumens.

That explains alot thankyou for your time and knowledge Justin and even though it is not all sunk in yet it does make alot more sense.

Ok put together another 2D Mag today with the below items and it is pulling virtually the same current as the drop in module from DX

Cree XLamp® XM-L LED - U2
http://www.lck-led.com/p865/Cree-XL...mm-Board-U2-Group,1000Lm@3A/product_info.html

Constant Current LED Driver Board - 3 Modes, 5.5-12V 3.0A
http://www.lck-led.com/p501/Constan...3-Modes,5.5-12v,3.0A,P7,MCE/product_info.html


Troch Light 3.8v 26650(Protected) 4000mAh Rechargeable Li-ion
http://www.lck-led.com/p822/Troch-L...0mAh-Rechargeable-Li-ionTro/product_info.html

Are you still only using 1 cell? Looks like you need to be using a second cell judging by the drivers input voltage

ljw2k said:

Ok put together another 2D Mag today with the below items and it is pulling virtually the same current as the drop in module from DX

Cree XLamp® XM-L LED - U2
http://www.lck-led.com/p865/Cree-XL...mm-Board-U2-Group,1000Lm@3A/product_info.html

Constant Current LED Driver Board - 3 Modes, 5.5-12V 3.0A
http://www.lck-led.com/p501/Constan...3-Modes,5.5-12v,3.0A,P7,MCE/product_info.html


Troch Light 3.8v 26650(Protected) 4000mAh Rechargeable Li-ion
http://www.lck-led.com/p822/Troch-L...0mAh-Rechargeable-Li-ionTro/product_info.html

Click to expand...

Ok, so what's the problem? As described above, when you run 2xLi-ion at a nominal 7.4V and a buck driver of approx efficiency of 85% to drive an XM-L at 3A and a Vf spec of about 3.35V, you get a calculated tail current around 1.6A, which is what you seem to be measuring.

Tail current is not the same as drive current for your buck driver case.

Using 2 cells and as Justin says it is Tailcap current which would be different at the led so both will be pulling around 2.8-3A at the LED i think correct me if i am wrong.

A few pics of what i have done so far :

Would i benefit from using a linear regulator over the buck driver i am using at the moment if so what are those benefits.

Ssorry for all the quetions ...> Johhny 5 more input all ears and all that.

7135 linear regulator recommended max input voltage is 6V. 2xLi-ion exceeds that by a lot. Stick with the buck driver. Much more flexible in terms of Vin, as long as you can feed it the minimum Vin to reach regulation.

Yeap 2 LiIon. I run mine with that and get 1.5 amps at 7.4 volts. Need higher voltage.

350xfire said:

Yeap 2 LiIon. I run mine with that and get 1.5 amps at 7.4 volts. Need higher voltage.

Click to expand...

What do you run with 2 Li-ion? A buck driven XM-L? 1.5A at the tail?

Tail current is not the same thing as drive current.

Justin Case said:

What do you run with 2 Li-ion? A buck driven XM-L? 1.5A at the tail?

Tail current is not the same thing as drive current.

Click to expand...

Yes, buck driver... And yes, I agree about the tail not being the drive, but should relatively be close minus losses and such!

So the only way to measure true current draw is after the driver in series with the LED.

Do you think i wil be getting the maxumin out of the LED 900-1040 Lumens with the Buck driver and batteries that i am using.

350xfire said:

Yes, buck driver... And yes, I agree about the tail not being the drive, but should relatively be close minus losses and such!

Click to expand...

Why should it be "relatively close"? Only if your driver is less than 50% efficient, which is a really bad driver. Basically, in the ideal case of 100% efficiency, with 2xLi-ion at 7.4V nominal and an XM-L with Vf of 3.3V at 2.8A drive, your Vbatt/Vload ratio is over 2:1 and your tail current vs drive current ratio will be the same. In reality, this is modified by the fact that drivers aren't 100% efficient. So let's assume an average efficiency of say 80%. Thus,

Power delivered by the driver = power consumed by the LED
80% * 7.4V * Ibatt = 3.3V * 2.8A = 9.24W

Thus, Ibatt = 1.56A, which is very close to your figure of 1.5A. If we increase the assumed driver efficiency to 85%, we get Ibatt = 1.5A.

ljw2k said:

So the only way to measure true current draw is after the driver in series with the LED.

Do you think i wil be getting the maxumin out of the LED 900-1040 Lumens with the Buck driver and batteries that i am using.

Click to expand...

I don't know anything about that driver, but the photo looks like it was soldered by a 1st grader.