Is it possible to run this? Also, anyone know the low voltage cutoff of this driver?
The 6v Zener modded drivers are only suitable for 6v emitters, such as MT-G2, XHP-70, etc.
An XP-E2 will pop if you use that driver.
That’s what I pretty much knew, but just figured I’d ask to confirm. I cannot find a 1.5a buck driver that will drop down to 3.7v anywhere.
Ax2002 driver. Single mode but does exactly what you’re looking for. Simon carries them as well as a few other places.
AFAIK, that’s not necessarily true (that the XP-E2 will pop @ 6 volts). IIRC, the emitter doesn’t care what the voltage is, as long as current is controlled at an acceptable level. What pops a LED is overcurrent, not overvoltage. (Feel free to correct me if I’m actually wrong on this) But, if you have a driver that controls current and keeps it in range of the XP-E2, the driver itself will most likely pop from the overvoltage, even with the zener mod!
With a diode, voltage = current, so yes, if you hook up an XP-E2 at 6 volts it will instantly pop. But the 7135s are current controlled linear regulators, so they regulate the voltage down to whatever is needed to maintain the set amount of output current.
If you try and run a linear regulator, like the 7135 chips, with that much voltage drop there will be massive amounts of heat to dissipate. The 7135s will overheat and fail quickly but the XP-E2 will probably live.
An LED is a little sliver of silicon semiconductor like any other diode. As long as heat is kept under control, too much voltage is what will kill the LED die, even if current is kept very very low (a few milliamps, even though the LED can handle several whole amps when at normal safe voltage). Over-current usually causes something else in the package to fail like the bond wires. There is a point where over-current will kill the die too, but something else usually fails first.
MT-G2 (also XT-E & XB-D?) with no bond wires is virtually immune from any kind of over-current failure but too much voltage will still pop the dies.
I really dont want to take the chance. I ideally need a 1.5a buck driver that can except running 2 18650’s stacked. They used to make such a driver, but it appears to be obsolete.
Actually most of LEDs are based on GaN,not silicone semiconductor. LEDs behave like a non-linear resistor,so voltage and current are strongly dependent on each other,how can you put too much voltage without putting too much current also?
My deepest apologies. I mean, I know so little about anything I’m still happy with stone-age direct drive PWM open-source drivers, for cryin’ out loud! I’ll try to remember not to butt in to highly advanced technical discussions like ‘can I run this 3v LED from 8volts???/’, I’ll just wait for you to show up and handle things. 
What’s wrong with my suggestion? http://www.mtnelectronics.com/index.php?route=product/product&path=67_115&product_id=521
Not a thing
More arguments(currently: 0),less whining please.
The lowest drop is 5v, and correct me if I’m wrong, thats still to much voltage for an XP-E2. Plus it would be triggering the low voltage protection way to premature.
Your reply makes no sense. Rest assured that finges/Rufusbduck are correct. I’m not certain what you mean by “The lowest drop is 5v”, but the only mention of 5v on that page is the input range in the title so I’m going to assume that you’ve misinterpreted that. Please look through the “Available Options” and “Description” sections. I think you’ll quickly see that this is a very good match for what you want to do.
The input range is 5-15v. I am running an emitter that can only handle a range of about 4v and some change. Usually i would use a driver set up to handle 1 18650, and wouldnt trigger the low voltage protection until it drops to about 1 volt, I believe. Using a driver between 5-15v, its going to trigger the low voltage protection much sooner than 1v, right? Therefore i would not get as long of a runtime as you should having the advantage of running 2 18650’s. Please correct me if im wrong. If i knew what i was talking about, i wouldnt be posting on here.
- Yes, you are wrong.
- LVP should be setup to kick in when each cell is in the 2.8v to 3.5v range, not 1v (someone else can get more specific about voltages if you need them to).
- Yes, normally a DD or Linear driver for a “3v” emitter like the XP-E2, XP-G2, XM-L2, XP-L, etc would be used with a single Li-ion cell. LVP would be setup to correspond to the voltages of the single cell.
- I recommend doing a little background reading, it seems that you are way out in left field currently. Lagman wrote a pretty big post a while back on the topic of “Understanding the difference between Linear, Buck, Boost and Direct Drive drivers”. I later revised that post significantly and posted it further down the thread: Understanding the difference between Linear, Buck, Boost and Direct Drive drivers - #56 by wight
Frankly I’m unable to understand your post #17 very well at all. It may be a communications failure, but it looks like you are under some serious misconceptions - maybe just about how LVP works? Unfortunately I’m not certain that the post I linked to will help out with that. Generally speaking in a flashlight with removable cells which are in series, LVP works by simply measuring total battery voltage and comparing that against a value which is known to be “low” for that number of cells. The primary shortcoming of this method is that it assumes that all cells are around the same voltage when they could actually be seriously imbalanced.
EDIT: Or maybe you are simply confusing the words “input” and “output”? (sorry, I simply do not know where the confusion stems from)
It’s a buck driver so input voltage is greater than led voltage. Some of the extra power is stored in the inductor and converted to current. The driver senses battery voltage and indicates when it is low for the number of cells used.
Appreciate the help, but no need to be kind of insulting about it. Sometimes, you got to put it out there in the simplest form to make sure every concern is addressed, even if you already have a general idea. I think I understand now. Thanks.