I think this maybe just about right. :bigsmile:
I think I’ve confused more than just the red and blue lines but to guage how hard the 7135’s are working one would still need to know the battery voltage at the working current. I’d expect with a Vf that high the driver would be fine but the host still needs to be able to dissipate on the order of 20-25W of heat. No small task.
No, you're confused and looking for an answer to a question based on misunderstanding how this works. The 7135-based driver cannot ever supply more current than those cells can push through that emitter without the driver present. 7135 is just a current limiter, not a current generator.
It's not really a "6 volt LED". Look at the graph here: https://budgetlightforum.com/t/-/19600#comment-417559
Those numbers go way way beyond 6 volts. Even if you had magical cells infused with Unobtanium and they exhibited absolutely ZERO voltage sag under load and stayed exactly at 8.4 volts, the emitter would still survive. But since your cells are a product of the laws of physics in this universe and not some bizarre parallel universe, they will sag some even if very very good, and so the current will be limited to whatever the Vf graph shows at whatever that voltage ends up at. Just hook the damn things up direct, take a current measurement (and voltage too, get some good data), then you will know at what point adding more 7135s will yield diminishing returns. Do it. You won't zap the LED, I will buy you a new one if you find a way to kill it with only two cells.
I suppose even a 10A cell will sag somewhat so it’s a case of where the voltage sag crosses Vf at the drive current. A bit below that piont is the best you can hope for from fully charged cells. If you want any kind of run time then back off the current more.
Apologies for the hack job, maybe this will remove some of the irrelevant info...
At 6.0 volts, no more than 1.5A can ever possibly pass through the circuit, no matter what kind of current the source is capable of. At 7 volts, no more than 6.5 amps can ever possibly pass through the circuit. (that's current to the emitter, doesn't take into account driver losses, but for 7135 drivers those are minimal)
It is impossible to kill a MTG2 with only two lithium-ion cells. This is not even remotely risky with only two cells. If you had two banks of fifty cells each, each bank connected in series, for a working voltage of 8.4v but many many cells in parallel to minimize any voltage sag under load, it still would not kill the LED.
That does make it more clear on that front.
So looking at the cells performance, what does this graph tell us? (Thanks HKJ!)
Given 5A to 15A, my Samsung cell vs everything HKJ tested, the Samsung is the best at holding Voltage all the way up to and beyond 15A, even tested to 30A. The legendary AW IMR 1600mAh cell is actually bested by these. So if these cells can hold forward voltage to beyond this 15A that HKJ charted here, what would happen to the emitter then? Admittedly, I’m no expert at reading these charts….but it looks pretty obvious that these cells are capable of delivering in excess of 15A if even for only a few minutes considering more than 7Vf. Maybe I should have left 20 and 30A on the chart…
@comfychair
I am assuming your talking about if the led was heat sinked properly. So at what current would a MT-G2 go poof with good heat sinking. I thought they where rated at 3 amps max, was kind of guessing it might handle 9 amps with copper.
djozz repeatedly took the MT-G2 to 16A without failure.
I’m mounting one on a Noctigon, which will be re-flowed to a 7/8” x 1” copper heat sink in contact with the aluminum L2P host. This is why I’m asking. How efficient would this amount of copper be and at what point would I expect diminishing returns.
I’m successfully running the MT-G2 in this K3 head on an L2P at 4.40A with only a 1/4” thick x 3/4” diameter piece of copper under the bottom of the pill (glued under the aluminum shelf with Arctic Silver Thermal Adhesive). 5 minutes yields 133º at the first fin, 10 minutes only sees it go to 146.2º. Very hot yes, but with a larger piece of copper as stated above I’m trying to determine the limit to the net gains.
This may help explain a little.
Even if the led survives, at what point does the host become too hot to handle. Overheating the cells is likely more dangerous than overheating the led. We have reached a point where single emitter drive currents outreach the capacity of not just edc size lights but of larger hosts as well so the question of how many is too many bears as much on the host as on the led.
Do you know what is the Vf type you are using? Most likely this is the 9V Vf type where the typical is 8.55V and max of 10.5V.
Looks like you can use 3x 18650 in series?
They will melt the solder and slide off the MCPCB before they are electrically damaged like happens to other LEDs. Clean it up and stick it back on the board and it will still work.
How is his current setup running 4.40A through the LED with only 8.4v input if this were a 9V part? Obviously this is a 6V MTG2.
With my current set-up, I measured 6.42Vf with an amperage draw at the tail of 4.41A.
It’s a 6V emitter series.
I basically wanted to know if stacking more than 12 chips would have any detrimental effect in and of itself. I do not know what direct drive would supply with any of the cells that I have, that’s really not the question because I’m not trying to drive it to the max. I could set up 16 chips to push approx. 6A and most of the cells I have would support that. Certainly the Samsung’s will. But will the chips get too hot, will the host get too hot, and at what point (temp) do I worry about the cells themselves?
I pulled a set of Panasonic NCR18650PDs out when I ran a 10 minute thermal test and while the head was 146.2º the top cell at the positive end was 121º…this cannot be good for the cells!
So I am trying to figure out if I should allow this new version with more copper help the light to run cooler with no power gain, or will the cooling effect actually allow me to bump output slightly and still have cooler head temps.
From djozz testing looks like 12 amps would be best for the most lumens.
I’am thinking 2D Maglite with a copper heatsink and finned head. Could it get rid of the heat?
All said I think RBD was right from the get go, put 16 chips on it…take measurements and see how it’s handling it. Knock off a chip at a time until a happy medium is obtained if that’s necessary.
When you think about 3100 lumens in a small Solarforce L2P with K3 head, there’s a whole lot of heat being generated in comparison to the amount of metal present. Is my math right that 28 watts of energy is being consumed? Pretty small light for that, isn’t it? Chunk of copper inside or no.
Is this a big enough piece of copper to get the job done?
2” long shown mounted, it’s 1” long cut to fit into the light. The 7/8” diameter is what fit’s inside the Solarforce L2P host, top to bottom wall to wall copper. Can’t be bigger unless the entire pill is machined from copper, and since I don’t have a lathe, this is about the most I could do. Turned in a Makita cordless drill, shaped with a Sears Craftsman rotary tool that’s 18 years old. See the white at the end of the copper? JB Weld Water Weld holding the copper bar stock to a .45 Long Colt pistol cartridge, which is in turn chucked into the drill. 
Admittedly an inexpert opinion but it seems to me a larger sink might act as a better buffer for the LED but the heat already produced is not dissipated by the surface area you have. I would think that to safely increase output you would need to increase the surface area of the ligtht as with your Photon or the BBC.
is in the leaves?
No, wait, I mean in a bigger light! 
Might be I should leave this one at 4.4A with the benefit of the added copper to help it run cooler. Focus my power needs instead on the much larger S2200! 3 cells, more mass, a much better target for bumping amperage.
Thanks everyone for the help, I think I’m seeing the light on this one, pun intended.
That was my thoughts also, you have a huge piece of copper to absorb the heat but you still have the same area to get rid of the heat. It might be a little more efficient but I wouldn’t think by much. That’s why I was considering 2D Maglite as a host.
Then it could be ramped up to 9A pretty easily. Perhaps a 3D Mag, with the top portion filled with a plug of copper to further expedite the sink. A “D” cell sized plug, with the 2 26650’s behind that and a tail clicky for power. Carbon fiber tube to keep heat from transferring to the cells.
It’s only money, right?