I’ve been spending a lot of time reading about S2+ triple builds and modding of flashlights.
For me, I enjoy learning and understanding all the little details about the things that I own, this one topic has me a bit confused.

Here is my question, if i were to build a direct drive flashlight with a one of the ME FET drivers and use a fully charged EFEST 18650 rated for 20A continuous, how many AMPS would 1 LED would pull from it. Lets assume ZERO resistance for the sake of this example.
Lets take a single 219B for example: 3.09V - 700mA ( I think this is the vf value)
What does this mean in relation to how much current the LED will draw.

Someone explained to me that vf is the voltage/amp required to turn on the LED.
I think there is more to it because i keep seeing vf mentioned when people are discussing direct drive and how many amps are pulled.
Your time and patience is appreciated. I’m really trying to learn here.

I know i could spend $50-$75 and build what ever triple i wanted…but i want to understand how all the factors play here.

Volts forward is the voltage of the led at a given amp measurement. So in your example: the led would measure a voltage drop of 3.09v at 700ma current draw. Maybe someone has a link or two of common led v/amps chart?

Amps increases with Vf.
It is rated at 700mA at 3.09v, but a fully charged 18650 will be 4.2v so the amps will be a lot higher.
You need to look at a Vf vs Amps measurement chart like that one to find out how much, because every LED has a different curve.

Vf is the voltage drop across the led at a given current. The example you give is more like the led being fed from a CVCC power supply than a flashlight but the easiest way to get an idea is to look up the information in Djozz’s tests since he’s looked at many different sizes and brands with links to the info with consistently laid out graphs. It might not be perfectly accurate but it’s very helpful in ball parking goals or expectations.

Even though the resistance in the light is not zero and 18650’s sag (drop voltage) under load you can expect roughly 4 volts to the led with direct drive depending on the led. If you look at one of Djozz’s many tests like this one you can see that an XP-L V6 at 4 volts draws about 5.5 amps and produces around 1900 led lumens.

The 219B you referred to has a much lower Vf and as you can see here will start to reduce lumens output long before 4 volts is reached.

That chart helps a ton. Where do we find these charts?
I want to confirm a few to make sure I understand
1.) The xm-l2 above would pull 8 amps. The voltage on the battery would sag of course, let’s say to 4 volts for the sake of this example, at which point, the LED would be pulling 7 amps. As the battery voltage dropped, so would the amps being drawn, and therefore, lumen output as well.

2.) if I built a triple xm-l2, the initial amp pull would be 24 amps. This is higher than the continuous amp discharge of the battery, and therefore, the battery at this point would be limiting (struggling) to keep up with the demand of the LEDs. Soon, voltage would drop below 4 volts at which point the battery may be able to keep up with amp demands from the LED? Once again, lumen output would be falling throughout the process.

Yes. I am starting to learn this all. Starting to realize the importance of understanding it all. Starting to realize things like my high drain efest battery isn’t the best choice for my stock s2+ for instance. Higher capacity, lower amp rating would be a better choice as it might be cheaper and provide longer run times. Every component matters. It’s a lot of fun.

Yes, all three drop. At 7 amps even with a good battery it can be quick. Check out HKJ’s battery tests and look at the discharge time in minutes chart in individual tests.

Right. I was talking in a situation with no other resistance. But that’s great to know. So the difference between 24 and 12-15 would be due to resistance. coppeer led board, driver, springs, wires and flashlight body?

So for the 219b, it seems like 6.0 amps is a sweet spot for output, and a single emitter would be well paired with a constant current driver for 6 amps. The excess voltage would be lost to heat. Sound right? Although…how would resistance factor into it. At 6 amps is it usually an issue?

1.) So for this particular LED (Nichia 219c) what would happen if you used a FET Driver with a fully charged 4.2V battery capable of 20amps to 1 LED? The chart doesn’t extend out to those voltages or amps. Does that LED just burn up? Does this chart represent an LED that cannot handle that sort of direct drive, while the XM-L2 can?