+1 
! My T10B MT-G2 build came out real nice using Richard's kit.
I bought two kits myself as the Zener mod is easier than my method.
Maybe, but cutting traces and all that is not necessary. Seems like it would be pretty easy to stand the regulator on its side with the "+ In" laying on the input pad for stock diode, then a short wire to the ground ring and a short wire for the MCU feed pad. Solder wires and trim off excess.
Yup, you could do it that way. My inner child wanted to see it on the pads.
I am not well versed on electronics. That said, I did not heat sink the chips on my Qlite board that is running the MT-G2 in my M8 with Bucks copper pill and it holds output very well at 6.77A. With a start otf reading of 3564 lumens it drops to 3260 in 30 seconds as the 20Rs take the load, then over the course of the next 2 1/2 minutes it drops to 2953 lumens. At this point the cells are down under 3.9V and there’s heat throughout the pill and into the top cell (114º on the bottom of the head). This is, of course, head down and in stagnant air while in the lightbox. A very respectful performance I think. At 4 minutes, the heat is up to 144º on the lower portion of the head where the copper screws in, still making 2856 lumens.
As a direct comparison, with a sliced-n-diced MT-G2 in the OEM pill with 1/2” of copper added inside the pill, the start value at 5.61A is 2424 lumens with a drop to 1870 at 30 seconds and showing 1308 lumens at the 3 minute mark. So without the 8 ounce copper heat sink there’s more than 1100 lumens lost at a lower current. With the aluminum pill and that chunk of copper showing 1276 lumens at 4 minutes, the head is 132º.
Is the loss due to the regulation chips or the emitter itself heating up? With Bucks copper pill I have 18 total chips on the Qlite.
I have no idea how to isolate the chips from the mass that’s absorbing heat from that huge emitter, but even with 8 ounces of copper and exposed fins the heat builds up pretty fast, obviously. So I guess it’s a moot point whether it’s the chips or the battery sag or the emitter suffering from it’s own irradiation…
It’s also been pointed out to me that the fewer chips on the board the more each chip has to dump in terms of the excess voltage so they’ll get much hotter trying to run “sane” levels than otherwise. In the case above, there’s 18 chips dissipating that ~2+ V. Reduce that to the 8 chips that are native on the Qlite and it’s quite a different story…
Somewhere, there’s a balance.
I think that the even bigger thing with adding more chips with the MT-G2 (same principle for 3V LEDs, but way less heat) is that even though they spread the load I think that the battery sags more which helps out the situation a lot. Oldlumens' idea of using copper to suck away the heat is probably the best I've seen, although it isn't needed in every situation.
Since the chips are in parallel the extra 2 volts go to each chip whether there is one or twenty-one. The difference is the higher Vf of the led at higher current and greater voltage sag in the battery both lead to a smaller excess voltage burned off by each chip. I think you’re right otherwise Dale, it seems very common for drivers to ramp down due to heat from the led(s) rather than from the chips.
Lol, I’ve had it explained various ways so of course none of it makes sense. There’s a value of 2V. There’s 20 chips, the 20 chips dissipate that single 2V value handling a tiny bit each. They don’t all handle 2V. That’s one way I’ve heard it said, from an Engineer. At any rate, they do all handle the heat that dumping 2V generates, so each one has less heat to deal with from the scrub, but not from the emitter, which toasts everything in the end anyway.
The driver doesn’t “step down” as you put it. The chips fall out of regulation due to less available current from the cells. The chips can only allow what current is available to flow, not a driver control issue as much as available supply.
One thing I do know. Take the light off the meters and boxes and measurement devices of all kinds and take it outside to use it and you don’t perceive those losses until the cells are waaaay down! The lightbox might say it’s dropping like a rock in a spring fed pond but in use you just don’t see that. So I guess for all but the taskmasters it becomes a moot point. We do, after all, (or at least I do) make these to scatter darkness so we can accomplish a task. Whether that be checking for a loose wire under the dash of the car or making daylight at 2AM to help you decide which varmint to shoot, they all have their purpose.
I’ve thought about getting rid of some, more than 30 is just too many lights. But which ones? lol I like most all of em each for a different reason, and have a couple more on the way even now. Oh well, time to build a cabinet…
Nice pics of the driver Richard! 
Seeing it from the side like that, the layers in the G10 board, gave me a vision…I have a piece of G10 that is 12” in diameter and 2” thick, what a pcb huh? lol This raw, green laminate is made in the bigger style to mount surgical equipment on for stability. Same stuff, just much more of it!
They all drop the 2V with each one passing up to 350-380 mA. They divy up the total excess wattage but each one sees all the voltage difference between Vbattery and Vled.
Well see, there’s the second definitive answer I’ve gotten and each are opposing. I was told 10 chips would divvy up the 2V between them and each handle 1/10th, probably just didn’t understand it as it’s all elusive to me.
So why is it that a base Qlite with 8 chips fares so poorly doing this mod? Of the 7 that I’ve done the ones 4A and more have been the best. 6.77A is the highest I’ve done with a stock 3.04A being the lowest.
Less voltage sag and lower led Vf at lower drive current means a greater voltage drop across each chip. With pwm, on is on and off is off so for any single chip power consumed is 350-380mA x the voltage drop. Stacking more chips lowers the voltage drop for each and every chip. Not by sharing but by increasing the Vf of the led and decreasing Vbattery.
When I said the same thing a while back I was corrected by HKJ - well, not exactly corrected, I just got a simple "No." and no explanation (yes, it was actually both bold AND red in the original). So I wish I could now give the correct explanation, but I still don't know where I got it wrong.
The power consumed is divided because the current is divided but the voltage drop is the same for each chip on a given driver.
Well I was just guilty of imprecise wording. When I/we say 'burn off the excess voltage' it's not actually burning off voltage, it's just a way to make it understandable by us non-propellerheads. Of course the voltage is the same everywhere because they're all in parallel but the amount of current each chip has to handle in order to get Vin down to the required Vout is less per chip when there are more chips in the circuit. Which is just about the clumsiest thing I think I have ever written, so I think I'll stick with 'burn off the excess voltage' to preserve my own sanity.
Maybe this diagram will clear things up...
It’s one of those “once you understand it” things you go “DOH!” and slap your forehead. PilotPtk and Texaspyro schooled me and though I’m no “A” student it’s pretty clear in a general sense what goes on. At lower drive currents the forward voltage of the MT-G2 is much lower and there is also much less voltage sag in the battery. These two combine to greatly increase the voltage drop forced on each chip so increasing the current changes the amount of excess voltage each chip sees.
Higher currents also lead to higher voltage drops at every other point in the circuit as well.
If you want to run the MT-G2 with a 3A driver just add a high wattage/low ohm resistor to lessen the load on the chips. To pick the right resistance value find the difference between the Vf at your drive current and Vb with depleted cells. Just for kicks I’ll pick some random numbers:
Vf= 4.5V @ 3A
Vb= 5.5V depleted
So 5.5-4.5 = 1V
3A x (?) = 1V, answer = 1/3 ohm.
In this example, no matter what the battery voltage is, at 3A 1V will drop across the resistor and the rest across the chips. A higher value resistor will lessen the load on the chips but raise the “depleted” voltage at which point the chips fall out of regulation.
So, would it be more efficient in the Zener modded lights to use Li-Po cells with their lower initial V? This would more closely match the Vf of the emitter and not have the waste scrubbed off by the regulation chips, right? And, don’t the Li-Po cells hold current better, with higher capacities?
So, a LiFePo4 18650 cell like this one, with 3.2V nominal and 3.8V Max would have a potential 7.6V full charge and deliver 18A continuous current discharge, so would this be a better bet for a Zener modded light? Capacity is low, but not all that much lower than a 20R and if it holds it’s own it’d come out ahead, would it not?
If a 2 cell tube were made for the HD2010, couldn’t these A123 120A LiFePo4 cells be utilized? They have a charge rate of 10A, so they’d quick charge easily and deliver up to 70A continuous discharge at 2500mAh capacity with an MT-G2 in a 2 cell HD2010, would that be like a totally optimized light?
It would probably be more efficient at lower drive currents. As soon as you load most of the LiCo cells with 4.5+ amps though they tend to sag enough within a short period of time to bring the voltage difference between the input/output of the 7135s pretty close. For this reason Pana. 3400s, etc. might actually be better from an efficiency standpoint with the zener modded 105C. The closer Vin is to Vout the less heat the 7135s will generate.