What size cap can an LM393 for example discharge, without a series resistor. The cap would be charged to no more than 15 volts. Or is that a bad idea, in which case I might ask, how big a cap charged to 15 volts can you use a 2N7000 to discharge, without any additional resistance in series.
The LM393 data sheet does not specify the maximum current the output can carry, when not saturated, only a minimum it will pass while the voltage is below some maximum. But the unsaturated maximum as something around 10 or 20 mA. At 15 volts drop, 20 mA would produce about 15V *.02A = .3 watts of heat in the chip, which is pretty high, to handle continuously. But the chip so rated for something like 570 mW power in an ambient temperature of 25C, so one half might survive this. If you add a series resistor that limits the current to 20 mA (750 ohms), the chip power would cut way down. I wouldn't worry much about dumping a capacitor that was half discharged by 20 mA in a second, or about 2700uF.
It is a poor idea, because of the large variation in unsaturated output current from device to device. If you intend t0 make only one, it may work fine, if your chip is not one with an exceptionally high output current capability. Adding the resistor makes it consistent.
Assume all the energy in the cap (.5*V^2*C) gets dumped into the fet die. Look at figure 16 and 17 on page 6 of:
I think a .001 second pulse is fast enough to assume that no heat escapes the die. For the TO-92 case (figure 16) the thermal resistance for a single pulse (assumes the die starts out at ambient) is about .035 of the 312.5 degrees C per watt for the steady state case. In other words, The temperature rise during the millisecond pulse will be only .032 of 312.5 degrees per watt 0r about 10 degrees per watt. But you need to convert from power to total energy, since that is what you know from the capacitance and voltage.
The peak die temperature cannot exceed 150C. Lets say the ambient is 50C, just to be conservative. so the die can experience 100C rise during the discharge without failing. At 10 degrees rise per watt (averaged over a millisecond), and 100 degrees of rise available that means the average power during one millisecond could be 10 watts. But 10 watts dumped over a millisecond is 10 milliwatt seconds (joules) of energy. So starting at 15V, .01 joule= .5*15*15*C, so C max is about 88 uF. The reason this is lower than for the comparator alone is because all the energy will end up in the die, because the discharge will be much faster (something like 3 or 4 amperes peak with 10 volts of gate drive), instead of 20 mA. This leaves little time for the heat pulse to spread out from the die. (About
660 us, so well within the 1 millisecond the estimation was based upon.)
But again, if you added a resistor with more resistance than the fet, most of that heat would be transferred to the resistor. !0 ohms would be high enough to do that, while slowing the discharge to only about 2 milliseconds, or allowing a larger capacitor to be safely dumped.