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I've got a very large capacitor which is likely to be storing up to 5 kV. When it is turned off, I want to be able to discharge it using a bleeder resistor (or similar) so I don't risk injuring myself.

I want to reduce the losses due to the resistor while also discharging quickly. Obviously, these are conflicting requirements, as I'd need high R to reduce losses (due to less current flowing) but low R to discharge it quickly. To solve this problem, I was thinking of switching the circuit so the resistor can be connected and disconnected at will (by a microcontroller), however, it seems near impossible to get a solid state switching component that can operate at the voltage I need it to (for less than, say, $50). Is there any way to configure the circuit so that the switching element isn't affected by the high voltages. I could use a high voltage electromechanical relay, but I'm just not sure if that's the best option.

Another concern of mine is that while the bleeder resistor will be able to drop the voltage to 37% within τ=RC seconds, but it is going to take a while to actually drop down to a safe voltage (especially with a large resistor). Is there any way to reduce the resistance automatically as the voltage decreases? I was thinking of using a varistor, but the resistance of those increases as the voltage decreases. There will be a microcontroller managing the circuit, so it may be possible to control it that way if a controllable resistance (that can withstand 5 kV across it) exists. Is there any way to achieve this? In an ideal world, I'd also be able to automatically measure the voltage to determine if it is safe to touch, but I'm aware that it would be very difficult to do if the voltage is changing.

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  • \$\begingroup\$ What value capacitor? \$\endgroup\$
    – Andy aka
    Nov 5 '20 at 9:49
  • \$\begingroup\$ @Andyaka Around 30uF \$\endgroup\$
    – rutloj
    Nov 5 '20 at 9:59
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Using the "electromechanical relay", or a mechanical switch or lever controlled with a solenoid or motor is probably the best way to do it. You could experiment gas discharge tubes.

If you want the resistance to lower with a lower current you could use a PTC resistor. With high voltage the current raises, which heats the resistor, which causes the resistance to rise. When the voltage can't keep up the current the resistor will begin to cool, which causes resitance to drop and current to rise again.

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  • \$\begingroup\$ How quickly would a PTC resistor heat up? I probably can't have it being very low resistance initially (when the switch closes) and then increase to a higher resistance, otherwise there will be a very high amount of current passing through it, right? \$\endgroup\$
    – rutloj
    Nov 5 '20 at 10:22
  • \$\begingroup\$ Yes, the initial current is high. PTC itself won't damage as the damage would be caused by overheating, but heating also raises the resistance, which lowers the power. And you still have the problem of switching the power. \$\endgroup\$
    – Ralph
    Nov 5 '20 at 11:04

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