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I have a circuit with a large capacitor, a solenoid and an N-channel FET fed by a 60VDC supply. When I trigger the FET the capacitor discharge activates the solenoid. This all works as expected (there are additional components -- resistors and diodes and such).

Now I need to add a safety circuit such that if the power is turned off and the supply line quickly falls towards ground potential the capacitor will quickly (e.g. <2sec) discharge to prevent accidental firing of the solenoid. Due to various constraints a bleeder resistor will not work.

I have done a pretty thorough online search and haven't come across a simple circuit that does this. It seems that a P-channel "safety" FET connected across the capacitor in series with a bleeder resistor, with the gate connected to the power supply would do the trick -- when the supply is high the FET is off, and vice versa.

Is this a reasonable approach? Some of my concerns: What happens when the supply is turning on? Will the safety FET also be on causing a large load on the supply? What about limiting the source-gate voltage to something reasonable (or is there a FET that can do the job and handle 60V)? I'm trying to get away with as few components as possible due to severe space constraints.

Here is what I had in mind (solenoid and trigger FET not shown). A further constraint of the problem is that each of the 64 solenoid circuit "units" (the dashed box) is independent and, because of cabling/connectors and such, additional wires to it are not allowed. So, there can be no separate "discharge" signal to each unit -- the only notice a given unit gets is that power has dropped.

schematic

simulate this circuit – Schematic created using CircuitLab

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    \$\begingroup\$ The gate-source voltage ratings of most power MOSFETs are usually in the +/- 20 V range, so most likely the MOSFETs would immediately die after applying power since the source is initially held low by the charging cap while the gate is immediately brought to 60V. \$\endgroup\$ – jms Jan 28 '16 at 2:31
  • \$\begingroup\$ I would use N-channel MOSFETs in series with power resistors, with the gates connected in parallel. Then you would only need a single loss of power detector that would turn on all MOSFETs at once. Of course said circuit would require a capacitor of its own in order to ride trough the power loss. \$\endgroup\$ – jms Jan 28 '16 at 2:36
  • \$\begingroup\$ Could you please edit you schematic to show the position of all the capacitors on the power line? \$\endgroup\$ – Dave Jan 28 '16 at 4:18
  • \$\begingroup\$ Thanks for the feedback. A further constraint of the problem is that each solenoid circuit "unit" is independent and because of cabling/connectors and such, additional wires are not allowed. So, there can be no separate "discharge" signal to each of the units. The only notice a given unit gets is that power has dropped. \$\endgroup\$ – JonB Jan 28 '16 at 4:46
  • \$\begingroup\$ On the schematic, the dashed box is an independent unit, of which there are 64 total -- each with a capacitor, etc. \$\endgroup\$ – JonB Jan 28 '16 at 4:47
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Tried the circuit, with 300K resistor from 60V to gate, and 100K from gate to source, using an IRF9540 P-channel MOSFET. It worked but not as well as I'd liked. I changed the 60V to gate resistor to 100K, and replaced the gate-source resistor with back-to-back 5.1V zeners. This worked much better because it tried to maintain a high gate-source voltage even when the supply voltage bled down.

BTW: Disconnecting the supply output from the solenoid unit worked well, with fast discharge of the cap, but disconnecting the input side did not. I lowered the supply bleeder resistor on the solenoid unit from 10K to 1K, and that worked better for supply input disconnect, but also boosted the supply dummy load to a few watts for each solenoid unit. Something in between works fine for my needs.

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