is that real ? this can drive 1 XML by 2800mA ?

Hi all

i was wondering if this driver http://www.fasttech.com/product/1122302 can really drive 1 XML by 2800mA ?

if not so what driver you suggest to run 1 xml by about 2500 to 3000 mA ? from single 18650 or even 2x 18650 connected in parallel

If you use decent cells or a power supply, it should be able to.

At it's maximum mode it's simply 8x 350mA constant current sinks in parallel, so yeah it will sink 2.8A, and yeah it will work on a single 18650, with provisos as the boxing monkey said; the battery needs at least 3.3V available for an XM-L to be driven at that current (see "electrical characteristics" graph in datasheet), and the battery needs to be capable of delivering that much current (most 18650s are, though some could trigger the protection circuit). When the battery gets below 3.3V(ish) the max current will drop, and the LED will start to dim.

As there are no inductors on there, this is a linear regulator.

So let's say you're hooked up to a 4.2 volt cell, driving 3 A into a XM-L at the typical forward voltage for that current: 3.35.

(4.2 - 3.35) * 3 = 2.55 watts.

I'd heatsink that PCB, but I don't doubt that it'll do what you're asking of it here (It's got 8 regulator IC's on it.)

BTW, this regulator would only be ~75% efficient at these numbers. You're putting about 10 watts into the LED. It gets better as the cell voltage drops, obviously.

It looks like there is an input diode where the positive lead comes in. You could pop that off for more efficiency, just don't cross the wires. =)

papershredder said:

It looks like there is an input diode where the positive lead comes in. You could pop that off for more efficiency, just don't cross the wires. =)

Click to expand...

There is a diode, but it's just there for reverse polarity protection for the microcontroller.

Mike S said:

There is a diode, but it's just there for reverse polarity protection for the microcontroller.

Click to expand...

Just don't hook it up warsdback. =)

papershredder said:

Just don't hook it up warsdback. =)

Click to expand...

Right, but what I meant was that the diode isn't in series with the LED. Removing it won't make the driver more effecient.

thank u all for your inputs i did order it, when it comes isa i will feedback

gamezawy - any results? I bought a few and did a quick setup with an XM-L and a variable PS. 4.25vdc gave me 3.0 mA to the LED. Seemed like the available current to the LED was only limited by the supply current. I only had a few minutes to mess with it so I'll try to get back to it again. If anyone wants to know anything specific I'll do what I can (given available time and skill set!).

papershredder said:

BTW, this regulator would only be ~75% efficient at these numbers. You're putting about 10 watts into the LED. It gets better as the cell voltage drops, obviously.

It looks like there is an input diode where the positive lead comes in. You could pop that off for more efficiency, just don't cross the wires. =)

Click to expand...

Yes, 75% at full battery charge, which only lasts a short time. It will rapidly drop below 4.0V, where the efficiency would be 84%. At 3.7V (so-called nominal voltage), 91%, and it will drop out (current comes out of regulation and begins to drop), which should be around 3.5V (regulator needs 0.12-0.15V) it will hit peak efficiency of 96%. As the voltage continues to drop, the light will gradually dim, but efficiency will remain above 94% until you get tired and shut it off or the protection circuit kicks in.

It's essentially impossible to calculate, but the average efficiency over discharge can be estimated at around 90%. It will depend on temperature, cell condition, depth of discharge, etc. That's pretty good for a cheap-a__ regulator!