There was ambiguity in my initial question, and my overall circuit schematic was not provided which led to misunderstanding. I have rephrased my question below, hoping to make it more straightforward and providing more detail.

Basically, I need to discharge my filter supercapacitor bank for smoothing the high current output from a rectifier. After getting some comments it seems like the best way to do this would be to use a bleed resistor in parallel with the supercapacitors (please correct me if I’m wrong).

What is the best way to safely discharge the capacitors after the circuit has been turned off? If I use 2kΩ bleed resistors will the capacitors be discharged quickly enough or will they take a long time to discharge?

To be clear of the context the filter supercapacitor bank takes place in (also see diagram below):

  • The output from the full-wave rectifier can be adjusted (by changing the Variac setting) in the range of 0-200Amps and 0-10v (high current, low voltage)
  • The full-wave rectifier output divides into 10 parallel circuits each with one supercapacitor rated at 2.7v, 100F
  • The current given into one of these supercapacitors will therefore be between 0-20Amps (I will start at 0 and gradually increase, they may not be able to handle full 20Amps)
  • The voltage given into one of these supercaps will be below their rated voltage, probably around 2.5v
  • The idea behind this is that the supercaps will each smooth the DC waveform individually (at lower currents), and when the parallel circuits are recombined again there will be the high current output like before, except smoothed, not hitting the 0v at 120Hz
  • Note: I understand there are better ways to smooth a high current output (these options are too expensive and complicated, ex. using a three phase power supply)

This bank of supercapacitors needs to be discharged after the circuit is turned off. The current and equipment being used is of course VERY DANGEROUS, and I am well aware of that (which is why I am asking questions here and talking to other electricians).

enter image description here

The polywell, in short, is a device that uses external magnetic fields to confine electrons and deuterium ions for fusion reactions. To create the magnetic fields a high current power supply is needed. It is a novel fusion concept that is underfunded and has great potential for growth (and I’m definitely in a useful position to do useful experiments with it once my circuit is worked out- I have 3d printed a polywell frame, have vacuum chamber set up, have a thermionic emission set up, and have a high voltage pulsing power supply… @Russell McMahon).

The outputs of the entire circuit described above are ideally a somewhat smoothed DC waveform, high current up to 200A and low voltage ~ 2.5v. This output is going to be used alone for some experiments with the polywell. However, in other experiments, I am planning on lowering the output current for this circuit for the use of charging eight capacitors in parallel rated at 450v and 1500uF. This bank of capacitors will be used to provide a single pulse of very high current (~2.5kA) with a SCR and trigger pulse circuit (circuit provided below). Protection diodes will be used (they are placed backwards in parallel) for protecting these capacitors. In theory, this bank of caps does not need to be discharged like the filtering supercaps because it will be drained with the SCR pulse already. But to be extra safe, it is best to have a discharge circuit set up for these caps in case the SCR does not give the pulse we need or they are not fully discharged for some reason.

enter image description here

Relating back to my question above, what is the best way to safely discharge these capacitors (after the SCR discharges them initially)? And what is the best way to discharge the smoothing supercapacitors?

  • 3
    \$\begingroup\$ Connecting one end of your capacitor bank to ground (through a resistor or not) isn't going to discharge the caps. You need to connect the resistor across the caps for that to happen. \$\endgroup\$
    – brhans
    Commented Jul 28, 2016 at 21:57
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    \$\begingroup\$ To elaborate on @brhans comment, with the earth switch the circuit is fully isolated and floating with respect to ground. Adding the earth references one point of the circuit to ground but no current can flow because there is no return path for it into the circuit. \$\endgroup\$
    – Transistor
    Commented Jul 28, 2016 at 22:04
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    \$\begingroup\$ @JeremyAdams If there is 2V across the caps, then perhaps one side is at 18V (relative to ground) and the other side is at 20V (relative to ground). If you connect the first side to ground then now it's 0V and 2V respectively. That doesn't help you discharge the caps. \$\endgroup\$ Commented Jul 28, 2016 at 23:32
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    \$\begingroup\$ @JeremyAdams Following your apparent reasoning, by connecting one side to ground you've only discharged that side. (Note this is not correct wording but seems to match the way you are thinking about the situation) \$\endgroup\$ Commented Jul 28, 2016 at 23:33
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    \$\begingroup\$ 2300 Joules at 400V. Yee HA !!!!!!!!!!!!!!!!!!!!!! . Death would be easy. ||| IMPORTANT IMPORTANT IMPORTANT It is not obvious how you intend to use the supercaps to reduce ripple, but it sounds like you have a major misunderstanding of how they should and can be used and you can VERY easily destroy them with the voltages in use. | Please provide a circuit that shows where the supercaps are connected and how they are intended to function. [You do not say what the actual voltage for the Polywell is but as you are using 450V caps it sounds like HV is in use]. ... \$\endgroup\$
    – Russell McMahon
    Commented Jul 29, 2016 at 1:06

2 Answers 2


Your circuit designs don't make sense, and won't work.

Your first circuit won't work because there is nothing in place to discharge the capacitors. There will be a current through the coil while the capacitors charge, then nothing once they've charged.

Your second circuit won't work for several reasons:

  • You cannot charge a capacitor to a higher voltage than it is being supplied with. If the voltage present across your bridge rectifier were 2.5V, the capacitor bank will charge to 2.5V, and no higher.

  • The "smoothing supercapacitors" present the same problem as the capacitor bank in the first circuit. They will act as an open circuit once charged.

  • If the primary capacitor bank actually charged, discharging it would lead to a flyback current that would probably blow out the bridge rectifier and protection diodes.

As others have noted, you are in way over your head here. The voltages and currents you're describing are extremely hazardous. You are at serious risk of injuring or killing yourself or someone else. Step back and get some more experience in basic electronics before you attempt to proceed.


This answer, which will need to 'mature' as the question does, is as of early Sunday 31 July NZ time so sometime on Saturday 30th in most of the real world. (That's to allow match between question and answer to be checked.)

A very comprehensive word picture of what you think you are doing with voltage and current and the Polywell is in order. At present it is not obvious what you wish the poser supply to achieve, but almost certain that what you have shown will not achieve it.

With the updated question it is still not obvious that you understand how capacitors affect rectified DC or how capacitors act if separate or combined. It is not obvious how you intend capacitors to be separated or combined or whether you intend them to be all connected in parallel at all times or in series on some occasions.

It would be "unusual" to charge a capacitor bank 'through' the load as shown in your first diagram. It is not the way they would usually be used to provide smoothing of voltage ripple and it is not obvious either what you would achieve or seek to achieve with this arrangement.

Drawing energy from a capacitor (or a number in parallel or series) causes their voltage to fall. If the Polywell or other load takes a significant portion of the energy the voltage will be much reduced. Energy content is proportional to voltage squared or voltage is proportional to the square root of energy content. So eg if a capacitor has 100% of target energy at 2V it will contain 25% as much energy at 1V ( (1/2)^2 = 1/4 = 25%) and 0.25% as much energy at 0.1V ( ((0.1/2)^2 = 1/400 = 0.25%).

Capacitors at 2.5V do not constitute a shoick hazard. They do constitute an energy discharge hazard. Energy stored in a capacitor = 0.5 x C x V^2.
A 100F capacitor at 2.7V contains 0.5 x 100 x 2.7^2 Joule ~= 360 Joule. 10 of these = 3600 Joule. (About 310J and 3100J at 2.5V) That's around the energy range of a top personal firearm (high powered rifle & similar) and a short circuit into a metal surface could throw dangerous amounts of melted metal.

Discharge time constant is given by t = R.C
So eg 1 2000 Ohm resistor and a 100F capacitor have a time constant of 200 x 100 = 20,000 seconds ~= 5+ hours. The time constant is the time for Vend to fall by ~= 63% of Vstart so you need a period of several time constants to approach zero Volts reasonably well.
A 1 Ohm resistor will have a time constant of 100 seconds.

To provide 310 Joule at 200A and 2.5V takes t = E/(V.I) = 310/(2.5x 200)= 0.6s or 6s to charge 10. So the drain from a 1 Ohm resistor is small compared to the charge current. (Charge = 1200 amp.second.)

The following will be aannoying. I apologise but its better said now so you have a chance to fix things. No intention to be rude (again): You cannot start to achieve what you are trying to do without learning basic electrical concepts. At present you really genuinely come very close to 'not having a clue' - ie your fundamental understandings are absent or wrong - and that has to change. Putting this aside and looking at some basic training material is essential.


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