jbeem wrote:
After about 30 seconds of having the newest 4th soundboard turned on, I once again smelled the burning rubber/plastic and discovered that it once again was melting stuff. Heres the pics for it:
Yeah, the driver shouldn't be related to melting up near the emitters. The LEDs run pretty much direct-drive anyway, so it's not much different than if you were to hook up the battery directly to the red and black wires coming back from the emitter. All the driver does is connect that electrical path when it thinks the blade should be on; it doesn't limit the amount of current flowing through. Current is controlled only by the resistor(s) in-line along the wire, combined with the way battery voltage sags under load and LED forward voltage rises, so where they meet is where the current settles.
Basically, say the battery is at 4.0V, and a typical LED might have a Vf (forward voltage) of 3.3V. The LED is designed to run at up to 1.5 amps. Connect them and the battery sags while the LED Vf rises. They meet at 3.7V, perhaps, and maybe 2 amps, and it settles. It'll likely be a little hot, running 0.4V and 0.5A hotter than desired, but not enough to cause problems. 2 amps is fine as long as the heat sinking is decent. I think this is what the "+" means in "12W+" -- it can exceed the spec'd power level a bit when the battery is full.
But red LEDs have a significantly lower Vf. So let's try the same thing with a red LED with a Vf of 2.0V. This is a much wider gap, so a lot more current flows. Maybe they meet at 3.2V this time, at 8 amps, and it's running 1.2V and 6.5A hotter than spec. The LED will most likely give up its ghost in short order unless some other component blows first. To avoid this, they put a resistor in-line to reduce the effective voltage and maximum current.
But the LED didn't fail, the wire and/or resistor did. So the electricity is likely flowing along some other path and bypassing the LED. But the battery's protection circuit didn't trip, so the short is probably still fairly small and doesn't allow enough amps for the cell to shut itself down. Or perhaps the wires are so long and thin (high resistance) that they can't pass more current than the battery allows.
With four different drivers doing it, that hints that the driver likely isn't the problem. I've seen things like this happen on a flashlight when the LED+ pad (which is electrically the same as BAT+) was accidentally touching a metal reflector which touched the aluminum body of the light, and the body was carrying BAT-. So, it shorted the battery to itself and the weakest highest-resistance part (usually a spring) would melt. It had nothing to do with the driver.
Have you opened up the emitter pill to make sure nothing up there is shorted? If both wires were touching the same piece of metal somewhere inside the pill, it would behave exactly like you showed.
Alternately, the resistor(s) could be the wrong value, allowing too much current to pass. But for that I'd expect the LED to fail from overload before anything else melted. Could still be possible though, since red LEDs have a much lower forward voltage so the resistor would have to soak up a lot more voltage than with other colors.
Usually I avoid that problem by using a bunch of AMC7135 current-control chips instead of a resistor. Each one allows 350mA to pass, and each one burns off any extra voltage as heat. It's only 350mA per chip though, so several chips are sufficient to spread the heat around safely, even for very high-amperage lights. Just keep adding chips until the amps are sufficient. Like, 3x7135 per die (12x7135 total) would allow 1.05A per die (or 4.2A total), a fairly reasonably ceiling for the 4xXQ-E setup SF uses. Plus, it keeps the brightness more steady throughout the life of the battery.
Anyway, I'd be looking for a new LED module, not a new driver.