Internals melting, please help

[mrkhar83]

Need some help and I am not sure what I am doing wrong here. I am basing my work off this video of with a GGW and NBv4:
https://www.youtube.com/watch?v=v7BdiKfMkro&t=511s

For the GGW, just like the video, I have the two positive green LEDs wired in parallel and attached to one .5ohm 3W Resistor, as the video suggested I also used a .5ohm 3W resistor on my W FOC positive pad. The result is this, the resistor to the FOC positive pad overheats like crazy, to the point that the solder is coming off the joint. I've tried switching it to a 1.2 ohm 3W resistor but with the same result ... is this the result of my beginner soldering skills or am I missing something here?

Also, I am using a 3.7 Trustfire Li-Ion2400 mah battery.

[Forgetful Jedi Knight]

Each LED should have its own resistor.

[mrkhar83]

Okay I can get that, is that a difference between the NB3 that the video shows how to wire and the NB4? Because according to the math (that I am still learning with Omh's Law) the Forward Voltage of the green's are around 3.7 and the white is 3.1, so with a 3.7V battery I didn't think EACH LED would need a resistor ...

[Forgetful Jedi Knight]

I’d have to research the various wiring diagrams, but I don’t think so. You add the resistor to help protect the LEDs themselves, and also, a 50 cent resistor is way cheaper than replacing a GGW Tri-Cree and/or melting internals.

[mrkhar83]

Agreed. But I guess where I am having trouble wrapping my mind around it is that when I to the math I come up with a negative value for R:

Vfbattery = 3.7
Green LED#1 FV=3.7
Green LED #2 FV=3.7 so
R= 3.7- (3.7+3.7)
R= 3.7 - (7.4)
R= -3.7/I where I=1
R=-3.7omhs

And then for the wattage:
P=R*I^2
P=-3.7 * 1^2
P= -3.7 watts??

I know that this is all new to me, but I can't figure out what to do with these negative values because the LED calculator gives me something different ....

[jbkuma]

You should use 4v for the battery. Full is 4.2 and although conventional wisdom says it drops quickly, that isn't true for modern batteries.

There it no such thing as negative current or resistance. I'm not sure what your math is supposed to be, but it's not right. Can you explain what formulas you think you are using?