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[The_Driver1 (The_Driver)]

@led4power

@The_Driver

@led4power

VTC5A looks good for 30Amps according to HKJ tests, much better than LG HBx. Most of 1x18650 lights would get hot and OTP would kick in <30sec anyway, so this is not continuous 30Amp draw.

One needs to consider all the contact resistances in the flashlight though. I think even the Samsung 20S won’t give you 30A for all that long (even if it is capable).

Max current depends on LEDs and battery most, max. current I got with 6x LH351D was about 24A due to relatively high Vf of LEDs, 30Amps will be “useful” only for 6x LuxeonV and probably 4x MZ.

I have thought about this some more. Initially I wanted to use just three Luxeon MZs to get maximum throw at 30A. Getting a constant 30A does seem a bit unlikely though.

At 10A the Luxeon MZ has a Vf of around 3.24V (see here ). Since I will be using the C8TT head, I will need to use an 18650. The best 18650 at 30A is the Samsung 20S. At this current the average voltage under load is around 3.35V (see here ).

So the resistance of all the electrical connections between these two points needs to be smaller than 0.0037Ohm = 3.7mOhm for a constant 30A in the turbo bursts. Your switch has 1-2mOhm, so that leaves 1.7mOhm for the rest. This seems very unrealistic.

Using 4 LEDs instead (which should be possible because the Cute4 optic has the same dimensions as the Cute3) will help a bit. Each LED will get around 7.5A and thus the Vf will be around 3.135V. The resistance of all connections between the battery and the LEDs now needs to be lower than 12.2mOhm. Thats better, but it still seems unrealistic.

Edit:
Here you state that a bypassed sping has a resistance of 1-2mOhm. So for two springs we can assume 4mOhm. This leaves us with 6mOhm for the wires and the flashlight (and the solder connections).

I estimate the wires will be about 4cm long in the C8TT head.
Using 0.5mm^2 would mean the resistance of each wire piece is 0.00156Ohm = 1.56mOhm. So we need to add 3.12mOhm to the total resistance.

This leaves us with 2.88mOhm for the flashlight, the pcb traces, the FET (another 0.7mOhm?) and and the solder connections.

Edit:

Of course I can use larger wires. The hole in the pcb has a diameter of 5.5mm and the one in my C8TT head was enlarged to 6mm. I can fit two 1.5mm^2 wire pieces (outer diameter: 2,5±0,15mm) through there or even eight 0.5mm^2 wires. The wires I’m talking about here are special, they have a very thin silicone insulation.

Using 1.5mm^2 would mean the resistance of each wire piece is 0.000532Ohm = 0.53mOhm. So we need to add 1.1mOhm to the total resistance. Using 8 0.5mm^2 wires would be even better. This would leave us with ~4mOhm for the flashlight and the pcb traces.