Question about FET Direct Drive Lights

[nightwalker]

Been doing some reading on LED Vf and FET direct drive current. There are lots of threads here and on Reddit, but something isn’t clicking for me. Apologizes if this has been discussed ad nauseum. Here are two scenarios (two popular flashlights with FET that I own) that I’m trying to wrap my head around.

Scenario #1 (Emisar D4V2 w/ SST20 4000K): Four parallel LEDs capable of drawing 5A+ each for a total of 20A from a 30Q or VTC6 rated for 15A. Is LED thermal regulation all that is keeping the cell from being in the overcurrent state for too long? Repeated turbo activations over time would result in a lot of strain on the cell, wouldn’t it? And I know folks here stop charging their cells at 4.0V or lower to try to eek out every last charge cycle that they can, but I’ve never heard anyone mentioning overcurrent discharge straining the cell in a D4V2.

Scenario #2 (Sofirn SP36 BLF w/ LH351D 5000K): Four parallel LEDs capable of drawing 8A+ each for a total of 32A, but this time pulling from three cells. Current draw per cell is 32A / 3 = 10.67A, well within the current limit of the 30Q or VTC6. Here, the limiting factor will certainly be LED thermal regulation, so this seems slightly more acceptable. If you use the stock Sofirn cells, we’re (probably) back to scenario #1 with cell overcurrent.

Now that I write this all down, I’m not sure where I wanted to go with this, except to say, I’d like to swap the SP36 LH351Ds for 519As. I’ve found two folks on Reddit that have done it. But I’ve also read that the SP36’s turbo would burn out the 519As, because of their lower Vf, allowing for more current to be drawn, than with the LH351Ds. This is where I get stuck. I don’t understand how one would interpret that bit of information from the graph in the Reddit post I linked above. I understand that if voltage goes down, current must go up, all else being equal. I understand Ohm’s Law (or do I?). lol but how would one “simply understand” (from that graph) that the Vf is too low for the 519As to support FET operation? How low is too low?

If you made it this far, thanks for reading!

[Haukkeli]

When led is given more current it’s voltage goes up. When amperage is drawn from battery it’s voltage goes down. At some point voltages meet. Then led can’t draw any more current. If VF is lower that meeting point needs more amperage. → low VF emitter draws more current.

[Jeffgoldblum]

Not an expert but here’s my understanding.

So you see the point on the graph when lumens starts to slope down despite voltage increasing? Thats where we know the LED is definitely being damaged because we’re putting more power into it but it’s making less light. So not only is it at the limit of performance its past the limit, now it’s just breaking down and any extra power is going right into making heat. Even if it doesn’t straight burn out the emitter its being damaged and probably wont work well again.

For the lh351d that point is at 3.6Vf but for the 519a it’s at 3.38Vf. So any current that puts the 519a past 3.38Vf is just burning it up. To get the 519a to 3.6Vf would take 12A, the lumens are ~580 here and the emitter is already damaged beyond repair. At 13A it’s toast. The 519a has a low enough Vf that it’s really easy to get to these points.

[laser]

@nightwalker

I understand that if voltage goes down, current must go up, all else being equal. I understand Ohm’s Law (or do I?).

This is true for systems with constant power, but the LED isn’t one. Take this voltage vs current graph for a Cree XHP50.2:

As you increase the voltage, the current through the LED goes up.

Scenario 1: The thermal regulation of the LED driver keeps the system from overheating, and the internal resistance of the battery (and springs, etc) keep the current from adding up linearly. For example, suppose the SST20 draws 5A with a fully charged cell. Four SST20s in parallel would not draw 20A from the cell because the cell’s internal resistance would lower its voltage as more current is drawn. The total current is maybe around 10-15A, but certainly below 20A.

Scenario 2: One way to mitigate the drop in current due to internal resistance is to use cells in parallel, as in the SP36. In the SST20 example in scenario 1, using three cells in parallel would boost the current from 10-15A to closer to 20A. For the SP36 with four 519A LEDs and three cells, it certainly is possible to draw too much current for the LEDs.

[laser]

Here’s a voltage vs current plot for two LEDs. In this case, the LH351D is analogous to the blue curve and the 519A to the pink. At a given voltage, say 3.0V, the 519A is drawing more current than the LH351D, so the 519A may be damaged whereas the LH351D would be fine. This plot shows the effect of different forward voltages.

So it would not be recommended to swap the LEDs to the 519As, which have a lower forward voltage.

Using worse cells can help, where they do not provide enough current to damage the LEDs.

[DB_Custom (Dale)]

Even if you don’t actually poof the emitters the whole diminishing returns scenario pretty much makes it an act of futility.

Too much heat, too little run time, reduced output, all for what end?

I have a bad habit of doing that more often than not. Worked a good part of the day today making a smaller light do around 3000 lumens on the power hungry SBT-90.2 just because…. And even though I left it with 22ga wires it still gets too hot to hold in mere seconds. In part because of all the copper I stuffed in it but still…

I’ll switch to a lesser cell and it’ll be ok… 25S 18650’s are tough on a smallish FET light.

[ToyKeeper]

As shown in the pic you linked…

When you push more amps through a LED, its voltage increases. And for most of its range, it also makes more light. But push it too far, and the light output levels off or even starts going down.

As Haukkeli explained, the battery does kind of the opposite. More amps = less voltage. The voltage “sags” under load. And if it has too much current pulled from it for too long, it can damage the battery.

Direct drive hits the intersection point between those two curves. Like, if a LED is 3V plus whatever it gains from more amps, and a battery is 4V minus voltage sag, there’s a point somewhere in the middle where they meet. Like, maybe at 4.5A, both of the voltages are at 3.5V, where the LH351D makes 1250 lumens. So that’s where the system stabilizes.

Then as the battery drains, its voltage drops more and more… so the intersection point gets lower too. Fewer amps, less light, lower voltage, until finally the control circuit’s low-voltage protection kicks in and turns off the light.

Use the wrong combination of LED and battery though, and it’ll push way too many amps through the LED, causing it to be damaged or even turn into smoke. Like, maybe with 519A in the same light, the intersection point would be at 7A and 3.4V, which is past the 519A’s peak or safe output levels. So if you’re going to use direct-drive, it’s important to match the battery and LEDs to make sure that won’t happen. Lower-Vf LEDs or less-saggy batteries (or higher-voltage batteries) can both increase the total amps drawn before it reaches equilibrium.

Using similar graphs of batteries, like from HKJ’s database, and LED graphs like you already found, it’s possible to get an approximate idea of where the curves would meet for any combination of LED and battery. But of course, it’s hard to say for sure where the curves will meet without actually trying it.

Anyway, with the lower Vf on that graph, it definitely runs a higher risk of drawing too much power and damaging the LEDs. So if it runs near the limit with higher-Vf LEDs, I’d be wary of doing that modification. It might need weaker cells or something. But then, that might just age the cells faster instead.

Or you could just not use turbo. That would be an easy way to prevent damage. It’s also how some manufacturers ship lights – limited to 60% power or 80% power or whatever, to reduce the chance of damage.