Let's look at the numbers if you use an 8.4V supply, 0.5 ohm coil, and a linear current regulator set to 3 amps. While this is turned on, the coil will be dissipating 4.5 watts. But the total dissipation will be 25.2 watts. So the current regulator will be dissipating 20.7 watts; clearly not good.CentiZen wrote:Paul:
But if I fixed the LM317's output current to be in line with the maximum safe current that the batteries could put out, wouldn't there be minimal heat lost in that process? Or would the low ohmage heater coil still cause too much current to be draw out of the batteries? To get the coil ohmage to match the ~25W of power that can be supplied by the battery I'd have to lengthen it by at least five times. I don't have access to any smaller guages of nicrome either. Is there any other simple way to control the amount of current that I want to be delivered to the coil?
A switching regulator gets around this, which is effectively what your PWMed FET is. With a low switching frequency it will work just fine for bringing the average current down to a safe level, but the peak current will be 16.8 amps. If the batteries can supply that kind of peak then this is okay, although you'll have to be careful to not inadvertently get the duty cycle too high or something will likely burn up. If the batteries cannot provide that high a peak then you'll have to either increase the heating element resistance or increase the PWM frequency to the point that it limits the peak current.
Another reason to increase the heating element resistance is to reduce the dissipation by the FET. That FET has an on resistance of 0.035 ohms typical (0.045 ohms max, at 25C - it goes up with temperature) with a 5V gate. That's a pretty significant fraction of the 0.5 ohm heater resistance. It's probably manageable but a higher resistance heater would help keep the FET cooler.
Anyway, there's nothing like actually experimenting to help give a more intuitive feel for all of this stuff. Just try not to burn yourself.