capacitormosfet
Can 3 ganged 120 A rated N-Channel MOSFETs discharge a charged cap dumping 300 A?
Will the FETs just explode or can they repeatedly discharge the cap?
What specific ratings and characteristics are important to determine if a FET is up to the task?
Granted the FET's Vds voltage would slightly exceed the cap voltage rating.
No, also for P-channel MOSFETs it's charging them. The confusion probably stems from the voltage you see at the lower side, which goes to ground (or near it). But that voltage isn't important, a capacitor's charge is determined by the voltage across it:
\$ Q = \Delta V \cdot C \$
So decreasing the gate voltage increases the gate-source voltage difference, which increases the charge of the capacitor.
When you switch off T1, there's current flowing from +12 V through R2 to the gate to discharge it's capacitance.
edit re the update of your question dd. 2012-07-09(*)
Turning off means that you discharge the gate to +5 V, and this happens by current through R2 and D3. So you bypass R? but R2 is still the limiting factor. A solution would be to swap R2 and T1, so that there's more current/less resistance to discharge the gate than to charge it.
\$ \$
(*) I'm using the ISO 8601 standard date format here. We have user from all over the world and for some 9/07 means 9 July, for others it's 7 September. ISO 8601 is unambiguous.
The best method is using a power resistor (3...5 W, 30...100 k\$\Omega\$), insulated with heat shrink tubes over the body and most of the wires.
If you are absolutely sure you will not connect the resistor across the capacitors of a power supply that is still turned on, lower resistor values may be useful for faster discharging. With 330 \$\Omega\$, your discharge current will start at something like 320 V / 330 \$\Omega\$ = 0.97 A, a safe value for most electrolytic capacitors.
Just remember that this resistor will dissipate (320 V)2 / 330 \$\Omega\$ = 310 W for a brief duration when used for discharging only (safe!) - but the power will be dissipated continuously when your resistor is used over a live power supply's capacitors (shock hazard from the voltage, burn and fire hazard from the large power dissipated by a relatively smallish resistor!).
Here's a discharge resistor I built for developing and fixing switching power supplies. The wire ends are (hollow) ferrules as used for connecting stranded wires into screw-on terminal blocks. They connect really well to the ends of the PCB-mounted capacitors' wires pointing through the solder connections - and they won't slip off and produce shorts to neighboring components. Y'all, feel free to copy the design and work safely ;-)
Also a neon lamp with a series resistor of ca. 100 k\$\Omega\$ is a fairly good indicator of lethal voltages. It will glow above ca. 70 V. Ok, 70 V is more than SELV, but the lamp is still better than nothing.
Shorting the electrolytic capacitors with a screwdriver may cause a very high discharge current, possibly damaging them. See here for some details. A plain short is really quite the redneck way to go...
Best Answer
Weather they can handle the discharge depends on a few things.
1) How much energy is in the capacitors. E = 0.5 * C * V^2
2) Where do you expect the energy will be dissipated during the short. And what fraction of the total energy is going into the FET. Energy may be dissipated in the following places.
a) In the internal ESR of the capacitors.
b) In the drain source resistance of the FET
c) In an external series resistor put in the circuit to absorb energy during the short
d) In the wiring resistance
3) What pulsed energy can the FET handle. You can usually find this in the Safe Operating Area graph for the FET.
For example...
C = 1mF
V = 30V
ESR = 15mOhms (from capacitor datasheet)
Therefore...
E = 0.5 * 1mF * 30V * 30V = 450mJ
Lets say that the datasheet for the FET lists Rds_on = 10mOhms.
Lets also say that we have 5mOhms of wiring resistance.
The peak curent would then be 30V / (10mOhms + 5mOhms + 15mOhms) = 1000A.
Lets say that the FET datasheet safe operating area graph says that the FET can handle 1000A with Rds = 30V for 10us. The pulsed energy rating would then be 0.3J. If the graph didn't say it could handle a 1000A pulse we would need to add a series resistor.
The energy dissipated in the FET is 450mJ * 10mOhms / (10mOhms + 5mOhms + 15mOhms) = 150mJ.
So in this case the FET would survive because 150mJ is less than 0.3J.
Lets try another example...
C = 1mF
V = 50V
ESR = 15mOhms (from capacitor datasheet)
Therefore...
E = 0.5 * 1mF * 30V * 30V = 1.25J
Lets say that the datasheet for the FET lists Rds_on = 10mOhms.
Lets also say that we have 5mOhms of wiring resistance.
The peak curent would then be 50V / (10mOhms + 5mOhms + 15mOhms) = 1666A.
Lets suppose that the FET datasheet says the FET can't handle more than 100A peak. Therefore we add a 0.5 ohm resistor in series to limit the current.
The peak current is now 50V / (10mOhms + 5mOhms + 15mOhms + 0.5 ohms) = 94.34A.
Lets also say that the SOA graph says the FET can handle 100A at 50V for 100us. 100A * 50V * 100us = 0.5J.
The energy dissipated by the FET during the discharge is...
1.25J * 10mOhms / (10mOhms + 5mOhms + 15mOhms + 0.5 ohms) = 23mJ.
In this case adding the series resistor greatly reduced the energy dissipated by the FET. Since 23mJ is much less than the 0.5J rating the FET in this example would survive.
One thing to point out in your specific case is that you can't depend on the Rds_on of the FETs being matched. That is you can't assume that each FET will share current evenly. Instead of 100A, 100A, 100A, you could get 70A, 130A, 100A or similar. You can only guarantee that the current sharing is even if you add some external resistance that is significantly higher than the FETs Rds_on.