I have a Solarstorm SP03 with 3 dedomed XM-L2 U2’s driven by a FET. Here are some pictures of the SP03 on the right compared to a normal SRK to show size.
I am considering modding my SP03 to be 4S 18650’s and 3P XHP35 HI’s. Would that be a step forward or a step backward? I’m thinking I will have a bit bigger hotspot, but with more lumens and possibly higher lux. Am I on the right page here? What I do not want to happen is for it to become much more floody like my Thorfire S70.
I normally answer this question for myself, but it seems the jury is still out on the XHP35’s? Has someone done a direct comparison/same conditions type test showing throw and output vs 3v emitters?
This is my understanding. Properly driven (2.5A), the XHP35 HI has similar luminance to a direct drive XPL HI, so it would throw similarly. This is based on DB customs’ mods with XHP35s in the L2 and L6. The advantage of course is that there is ~1.5x more lumens. I’ve measured the die size to be 2.50mm across, so the beam will be just a bit bigger than dedomed XML.
You might not be able to beat the lux of your current LEDs, but it would prob be close.
Not positive that this will help but here is a couple of pictures taken by Dale with the Maxtoch 2X with the factory dedomed XM-L2 U4 1A in it.
(also had the Olight M3XSUT Javelot with an HI emitter in second link)
Then the Maxtoch 2X after he put the XHP35 in it with the proper driver and added an extension tube.
Helpful links robo, thanks. In the second link, I can’t see much difference in throw between the XHP35 Maxtoch and the Olight. If anything the Olight might light up the barn a little more? The Maxtoch has a slight advantage in reflector size though,I believe.
I wonder if dale ever measured the throw number on them?
So if the throw is about the same or even a little less than a dedomed XM class emitter, why are the big-name dedicated throwers using them? Just for the lumens bump? I don’t remember stock lights with throw ratings like the K70 until the 35 arrived
My opinion is that the XHP35 is a good emitter to use for a thrower because it is a good compromise between output and throw. Though it is a lot of trouble to drive it correctly because it’s 12V, and you can’t just use a FET driver even if you have 4s cells because it will die from the high current.
Big manufacturers are using it partly just because it’s new and will sell because it’s the latest thing.
good question, been wanting to know this as well. it would seem a dedomed xml2 and xpl should throw further than the xhp35 HI.
i still think dedomed xpg2 throws the best and is almost as bright as the dedomed xpl’s and xml2’s.
you will have a bigger hotspot for sure with the xhp 35’s i believe, and will be noticably brighter.
the throw maybe slightly less but all in all, i think it would be a good upgrade.
I did an estimation of this recently. It looks like even with 3 or 4 XHP35s in parallel, there is still probably too much current to each emitter. This is based on a rough extrapolation of the forward voltage from cree’s graph on their datapage and considering the voltage drop from the cell IRs and other circuit resistances. I estimate that an additional ~0.2 Ohms of resistance (like adding a coil of wire somewhere) would bring the current to around 2.5A to each emitter.
I agree adding some resistance, although at first glance not the most elegant solution, is a good and easy solution. A linear driver is another possible solution. In either case the driver or extra resistance would have to burn off a lot of power, so it is not super efficient. With a triple XHP35, (2.5A(3))^2(0.2Ohms)=11.25W dissipated in the extra resistance or linear driver. But, this is not huge compared to the ~100W going to the LEDs.
0.2 Ohms is actually a lot of resistance and one would need to think about how best to add it to the circuit. For example, 24awg wire has 84 mOhm per meter so it would take over 2m of this wire to add 0.2 Ohms.
If you are going to do this project I can do a more precise estimate of the amount of resistance you have to add. I have an XHP35 that I can measure the forward voltage of.
4 (li-ion) batteries in series would be ~16.8V, and theoretically, that would be regardless of whether you put “x” emitters that are in parallel. That’s in theory though and doesn’t account for battery (voltage) sagging and any resistance from wires (v=ri, so if big resistance, the voltage drop across the wire would be higher. Then again, also, with ‘x’ emitters in parallel, the resistance on each path might also be slightly different.
There’re a lot of variables, so answering exactly what “would” happen vs. what “could” happen is very difficult.
I’m with you. A linear driver would probably burn up anyway unless it was very well heatsinked.
This sounds like a fun project; I will think a bit about the best way to add that resistance. Something like a homemade wire wound resistor. Like a bunch of thin wire encapsulated in some thermally conductive epoxy or something.
I suppose I could always just PWM a FET below max output, that’d be the cheapest. Don’t like FET-only drivers though, too much variance based on voltage. How much heat do you think a 7135 can handle with a big copper heatsink?
Yeah, good idea about just PWMing down a FET driver.
I imagine a 7135 could drop a lot of voltage if it was heatsinked well. But heatsinking 20 7135 chips (for 7.5A) seems difficult. You could also combine the two approaches; add some resistance somewhere else to lessen the load on the 7135 chips.
