LD-1 driver (5A pwm-less linear) info thread

If I want to drive 2 XM-L2 LEDs wired in parallel, driven by a single 26650 cell, and I don’t need to go over 5 amps total (2.5 amps to each LED) will there be a problem? Would the thermal load on the driver be any different driving 2 parallel LEDs at 5A total vs a single LED at 5 Amps?

The thermal load would be significantly increased since the "effective" vF is now much lower with the two parallel emitters. To maintain 5A constant current with the same input voltage requires the driver to burn off more heat. Linear drivers like this one (this one is linear, just without PWM) work best where the Vin and Vout are pretty close together. The farther apart you get, the more heat has to be generated to maintain constant current output.

So, why does it run more efficiently at lower levels but this heat issue rears it's head between 1A and 4A?

I guess that wasn’t directly intuitive but it is a good point. At 2.5 amps per emitter (assuming XM-L2), the Vf is only ~3.25V and at 5A to a single emitter, the Vf is well over 3.5V (I lost the interpolation graph) so this is roughly another 1-1/2 watts across the FET.

I don't think the driver itself is much more efficient (if any), but I could be very wrong about that since I haven't measured it at a given current vs. anything else. I wonder how it pans out compared to a single 7135 at moonlight levels.

The LED is running much more efficiently due to the lower drive current. On a normal 7135 5A driver for moonlight you are still pulsing 5A into the emitter, which is still pretty inefficient. This is delivering a constant low current, which at the emitter level is definitely more efficient than a high current PWM'd.

The other big benefit to this driver is the lack of PWM. No noises, no flickering, even on the lower levels.

I had to dig up the post from djozz where he tested emitters at high power on different MCPCBs so I could figure this out.
At 2.5 Amps, he measured 3.25V.
At 5 Amps, he measured 3.60V

Under a 5 Amp load most cells are going to drop below 4V real quick, so:

  • A single LED running 5 Amps, the voltage drop across the driver will be 4.0V-3.60V=0.4V. Power dissipated by the driver will be 0.4V*5 Amps = 2 Watts.
  • Dual LEDs running 5 Amps total, the voltage drop across the Driver will be 4.0V-3.25V=.75V. Power dissipated by the driver will be .75V*5A=3.75 Watts.

Wow, that’s almost double the power across the driver. I didn’t expect that.

Its in the math…

XM-L2 @ 1A = 2.95Vf >> 4.2Vs - 2.95Vf = 1.25Vdrop >> 1.25V*1A=1.25Wdrop @ FET

XM-L2 @ 2A = 3.15Vf >> 4.2Vs - 3.15Vf = 1.05Vdrop >> 1.05V*2A=2.10Wdrop @ FET

XM-L2 @ 3A = 3.35Vf >> 4.2Vs - 3.35Vf = 0.85Vdrop >> 0.85V*3A=2.55Wdrop @ FET

XM-L2 @ 4A = 3.55Vf >> 4.2Vs - 3.55Vf = 0.65Vdrop >> 0.65V*4A=2.60Wdrop @ FET

XM-L2 @ 5A = 3.75Vf >> 4.2Vs - 3.75Vf = 0.45Vdrop >> 0.45V*5A=2.25Wdrop @ FET

Although this is highly mathematic, you see how the curve changes in the watts calculations while linear in the input values. Not only that, the cell’s internal resistance will barely hold 4V with the best of cells for only a short period of time. At lower current draws, this will be closer to the 4.2V than at higher draws for longer periods of time. At 5 amps, I suspect this driver will be at direct drive into a single emitter within minutes unless multiple IMR parallel cells are used. Only the FET’s internal Vf losses will be added to the heat source.

So the primary yield will be in a light that might need 5A for Turbo level but will be used the majority of the time in a lower capacity, which will gain longer net run times at the lower end by almost 2X, right? A general purpose light or a walking light, seems ideal for that.

Trying to adapt my way of thinking from Absolute Max to...well, whatever else someone would want to use a light for. ;)

My wife and I were walking last night and she had her 501B with an XP-G and I had the SK68. We were having fun pointing at building to find that the Sk68 was great for lighting up distant signs where the 501B scattered before it got there. By the time we got home, the sk68 was still really cool and lit up the path just fine… but my wife, a non-flash-a-holic (or closet flashy?) was complaining of the 501B getting warm. I think a conservative “normal” mode is a good thing for unsuspecting users.

First,I apologize if I didn't answer to some questions(thread and PM),I'll do that ASAP.

@Texas , if you want to drive 2 xm-l2s at 5Amps,you can use expander board. Order PCBs from OHSpark,you'll need MOSFET and sense resistor.You can use sense resistor from LD-1 PCB since it must be removed,actually if you have hot air station and skills(or you have a friend who has that),you can "transplant" even MOSFET from LD-1 board,because if you use expander pcb,mosfet on LD-1 doesn't have any function,but it's impossible to remove it without hot air,so you can leave it on board and buy other MOSFET for expander board.

It's interesting that this driver is actually less efficient on all currents compared to direct drive pwm DD type.

But "here's the whole picture": imagine that you have two identical lights,one with LD-1,other with PWM DD driver,and both run at x.xx Amps and with identical battery.

Total (average) power consumed from battery is Vbatt*I,so total dissipated power is the same both lights.But, in case of pwm DD driver VledDD~Vbatt,so ALL power is dissipated in LED: VledDD*I (well,let's say that resistance of DD driver is negligible so dissipated power is ~0),while LD-1 takes (Vbatt-VledLD)*I part of power and dissipates that in heat,and LED only has to dissipate VledLD*I.

Note that total flashlight heat is the same because:

DD: Ptot=Pdr+Pled=0+VledDD*I=Vbatt*I

LD: Ptot=Pdr+Pled= (Vbatt-VledLD)*I+VledLD*I=Vbatt*I-VledLD*I+VledLD*I=Vbatt*I

So,one of good things that LD-1 does is that takes that extra heat on itself("sacrifice" itself),and this way helps LED by reducing its heat dissipation.And it's much better idea to dissipate that extra heat in mosfet,than in those tiny gold wires,sensitive GaN die and phosphor(LEDs efficiency drop at high current proves that).

So when you combine facts that LEDs are much more efficient on lower currents,and that linear drive reduces heat generated in LED ,overall lm/W efficiency is better with linear drive.

The DD is a “convergence” between battery resistance and LED current. The cell itself becomes the passive R-device and I am sure it generates heat accordingly. But it has a better heat dissipating mass and surface area than the little FET.

I guess there really is no better driver than one with a true voltage converter with a current sense to determine output. Of course, by its very nature, this is a pulsed output set by the frequency of the switching control circuit.

We are fortunate enough to have reached the point of designing to “convergence” in single emitter, single cell (and parallel source) lights. but the advent of the 6V MT-G2 has asked for more. A 1.2-2.4V overhead is something to contend with when using 2 cells. Or better yet, a robust boost driver.

I like the LD-1 as an elegant solution including the innovative coding on the firmware. With all the new e-switch host offerings out there, it is a great alternative. I would want one just to have one even though I have no e-switch hosts (yet).

Just curious… how hard is to isolate the low voltage components from the power circuit in the LD-1? I would think you need a low power step-down converter to drive the logic circuit and a low power step-up converter to drive the output up to drive the FET gate pin. The step-up may even be do-able with a divider network.

Got the proto still in a Y3, ordered 3. Look'n forward to get 1 or 2 in a e-switch EDC. Like'n the ZY-T11 clone (UltraFire) as of late, so one is destined there, probably a SupFire L5, and not sure yet on the third.

Led4power, please take an unbiased look at your driver.
As LED- and LED+ are not marked, and neither the sales thread nor this thread give any hint about it: Where would you - quite intuitively - solder the LED wires to…
Solution is only to be found in the preliminary thread… you might update the Connection Diagram section. :wink:

EDIT: typo ( - and + )

Sorry,I'll update OP with connection diagram.

just imagine my immediate reaction when I connected the wires, the LED, the shunt, the battery, and then…
took me some minutes to figure out what went wrong
but it seems the driver has survived this, all 3 show similar behaviour
it might just be helpful to prevent others from doing the same

I would have done it wrong straight up! :bigsmile:

copied and shared…

e-switch has a plus and minus?

No,but one side of switch is connected to ground,so just convenient labeling,switch of course doesn't have polarity.

Okay, so the switch could be connected to Common (host body) on the minus end.

just a theoretical question: what if someone did a horrible job soldering the eswitch negative connection? Would the resulting solder bridges fry anything permanently?

And what if you just bridged the sense resistor? would you loose all modes?