I'm designing an over/under-voltage protection circuit that needs to have really fast shutoff (ideally in the low 10s of ns). The downstream circuit is connected/disconnected with a bidirectional mosfet switch whose gate is driven by an optocoupler (probably the VO1263). The optocoupler is needed to isolate the gate from the comparator circuitry. Otherwise the gate-source and gate-drain ratings would be greatly exceeded (input can go to +/-100V). I've attached an image showing this part of the circuit.
My question is: how can I make this shutoff as fast as possible? The turn-on time isn't critical (within reason, anything in the ms range, or even up to a second is fine). Here's what I've come up with so far. I'm using a P-channel JFET to switch in and discharge the charge stored on the MOSFET gates (the capacitance is significant, as these MOSFETs are designed for high VDS and low Rds(on)). This is discussed in the VO1263 and VOM1271 datasheets. Additionally, I'm planning to add a bleed resistor (unlabeled resistor in the schematic) to hasten the turn on time of Q4. And, I can use JFET with a low on-state channel resistance, or parallel several. There's a tradeoff here, since I don't want the JFET gate capacitance too high. I'll use the smallest possible bleed resistor such that the optocoupler can still bring the MOSFET gate voltage high enough (optocoupler short-circuit current is in the 10s of uA). To be able to make this bleed resistor smaller, I'm planning to parallel a number of these optocouplers (I think 10, but that's TBD). Another thing I'll do is to make the JFET gate threshold voltage as high as possible so it kicks in as quickly as possible when the gate starts to discharge.
Are there any other options? Have I overlooked anything in my various optimizations discussed above?