I need some insight on the best way to set up the inductor-capacitor portion of our circuit. My goal is to smooth the output from our diode bridge which rectifies the AC current input into DC current.

Here are some specifics for the output of the full-wave rectifier:

  • The current going will be about 180Amps. The current coming out should preferably be maintained around 180Amps.
  • The voltage going into the capacitor and inductor section will be around 16.5v to 20v at its peak voltage.
  • The frequency from both the positive and negative outputs of the diode bridge will be 120Hz (since it is doubled with the rectifier).

What specific setup would you recommend to smooth this voltage and current? I know it’d be best to split the capacitors into parallel to divide the current, but I am not sure what specific capacitance they should have or the best set up to arrange them in. It is at a different scale with the high amperage and low voltage compared to other typical circuits that use capacitors to smooth the rectified DC current. If you could propose a way to do this effectively without adding too much load on the circuit, I would appreciate it.

Additional information:

This power supply is for my fusion reactor which uses strong external magnetic fields to trap a well of a electrons under a vacuum, known as a Polywell. It needs a smoothed output because if it oscillates to 0v on it's 120Hz positive DC output the inductors creating the magnetic fields in my reactor will be weaker and not trap the electrons as well.


Since it seems very difficult to smooth the 180A output at a steady voltage, would a possibility be to split the 180A current into several parallel circuits and then use capacitors to smooth each output with less current? What is the maximum current at 20v that a single, cost-effective capacitor could smooth? Maybe I could split the circuit into 90 parallel wire sections each with one capacitor (preferably small sized and low cost) that could smooth 2A and 20v, and then bring the smoothed outputs back to 180A at a relatively steady 20v. Is this a realistic idea?


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    \$\begingroup\$ (1) What is this power supply for? (2) Why does it need to be smoothed? (3) Is three-phase power available? Put the information in the question and not in the comments. Welcome to EE.SE. \$\endgroup\$
    – Transistor
    Jun 28, 2016 at 0:05
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    \$\begingroup\$ At 180A, you will likely need impractically large amount of bypass capacitance if you need a stable voltage at the output of the device. An offline switching supply would be recommended, but designing a 180A, 15-24V supply is not a small task. \$\endgroup\$ Jun 28, 2016 at 0:32
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    \$\begingroup\$ Do you have three-phase available? The ripple before filtering is a lot less with full-wave rectified three-phase. \$\endgroup\$
    – Mark
    Jun 28, 2016 at 1:23
  • \$\begingroup\$ Quick math: For 180A, 120Hz, 1V ripple, you'd need about 180A/(120Hz*1V) = 1.5F. At about 20V, the ESR must be an order of magnitude or two below 20V/180A~0.1Ohm. \$\endgroup\$
    – JimmyB
    Jun 28, 2016 at 11:47

2 Answers 2


Before you design your smoothing capability, you need to specify what you want it to do.

'The current coming out should preferably be maintained around 180Amps.'

This is not a specification, and around is not a tolerance. It does sound like you want a constant current output, rather than a constant voltage. Or are you just using 'current' loosely? Would a constant voltage be OK?

At the risk of being obvious (this is to illustrate the need for a specification), you can approach zero voltage ripple on the output as you approach infinite capacitance. The two are precisely in inverse proportion. Less capacitance, more ripple. So, as capacitance costs money and takes up space, you can work out exactly how much capacitance you need, from how much ripple your application can tolerate.

The easiest way to produce a high ripple, high value capacitor, is to parallel several lower value caps. So yes, your idea of 90 off 2A capable caps to handle 180A is OK. You may not need as many as 90.

There are better ways to reduce the output ripple than simply racking up the capacitance, once the ripple becomes a critical specification.

The first is use of three phase power. The problem with single phase is the long period with no effective power coming from the mains, that has to be held over by the capacitors. Three phase reduces the ripple from raw rectified mains to a few percent, rather than the 100% of single phase.

The alternative is to use a SMPS, switched mode power supply. At this power, getting for 4kW, it should be power factor corrected. The design of a SMPS at this power level is non-trivial, and you would do best to buy something.


A large SMPS would work as would a three phase supply as has been stated by Niel and Mark .There are some workarounds for the extremely large cap requirement.One which I have tried on a smaller scale is to use two transformers and two bridges .This should not be a penalty at your power levels .Now if you retard the phase of one by using a lamp ballast choke is series with the primary and advance the other by using a mains capacitor in series with its primary .Get the advance and retard close to 45degrees and the result is that you have a 90 degree phase difference between the two transformers.When you run your two bridges into your common cap bank you will see on the scope twice the ripple frequency with current pulses of half the amplitude .This can get the size and cost of the cap down but at the penalty of a mains film cap and a mains inductor .At your low voltages you should win cost wise because electrolytic cap size gets much worse at low DC voltages.Added bonuses of this scheme are better power factor and distortion factor.Another workaround that I have not tied is a series bank of ultra capacitors .They could work too.


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