# active bleeder circuit for capacitor discharge

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.

simulate this circuit – Schematic created using CircuitLab

• 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.
– jms
Jan 28 '16 at 2:31
• 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.
– jms
Jan 28 '16 at 2:36
• Could you please edit you schematic to show the position of all the capacitors on the power line?
– Dave
Jan 28 '16 at 4:18
• 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.
– JonB
Jan 28 '16 at 4:46
• On the schematic, the dashed box is an independent unit, of which there are 64 total -- each with a capacitor, etc.
– JonB
Jan 28 '16 at 4:47