I have seen some solutions for inrush current, but my case I think it's a little different. The point is that I need to limit the inrush current of 1500 volts peak rectified. In my project I need to keep charging and discharging a 200uF capacitor with 1500V every 50ms. So I could use a transformer and a bridge rectifier to do that like the following picture:

Simplified circuit

The switches would disconnect for a few microseconds the capacitor from the source and connect it to the load, then disconnect it from the load and finally connect it again to charge the capacitor. This loop keeps in 20Hz (I'll probably use power SCR's as switches).

The point is that the voltage is too high and the 50ms is too little time to charge the capacitor.

  1. If I use just a resistor of 30ohm to limit the current, the power peaks reaches 30kW, as you can see:

(The first time charging the capacitor followed by the switching period in the time 0.5s)

  1. If I use 1H inductor, the current keeps about 10A, that's also too much for it (The max current I've found is 1A for 1H inductor).

  2. I couldn't find NTCs thermistors that could handle such voltage, current and power. Also, the NTCs would probably keep always hot due to the high frequency of charge.

  3. Active components also can't handle the power here usually.

Do you guys have any suggestion?

  • 1
    \$\begingroup\$ How many pulses do you need to deliver in a day? You apparently need about 225 J per pulse (assuming you completely discharge the capacitor each cycle.) That's a run-rate of 4500 watts, using your 50 ms period as a guide. So it is a lot of power. But I'm wondering if you might charge up another capacitor (a bigger one) and then tap off from it during each pulse period, topping off your 200 uF capacitor from a much larger one that you pre-charge over a long period of time prior to your "work period." \$\endgroup\$
    – jonk
    Commented Apr 25, 2020 at 7:02
  • \$\begingroup\$ On top of what jonk said, at 4500 W there won’t be many cheap and easy solutions. Have you looked into a linear constant current pass transistor solution? It will need a massive heatsink though. \$\endgroup\$
    – winny
    Commented Apr 25, 2020 at 7:07
  • \$\begingroup\$ It's actually twice that power from the incoming supply due to capacitors only receiving 50% of the energy via a (or any) resistor. So, it's a 9 kW load you are developing (41 amps at 220 V). I think you should explain why you are trying to do this in case it's an XY problem. \$\endgroup\$
    – Andy aka
    Commented Apr 25, 2020 at 7:58
  • \$\begingroup\$ That's indeed a lot of power. I'm trying to generate huge magnetic pulses in a coil for an experiment. That's why I need to keep charging the capacitor. I'm gonna think about using another capacitor, but the switches are gonna be expensive too. I'm not sure about linear constant current, gonna take a look at it! Thanks for the answers! \$\endgroup\$ Commented Apr 26, 2020 at 5:47

1 Answer 1


The problem with a transformer/rectifier is that it's a low impedance voltage output power supply. Connected to a discharged capacitor, a very high current could flow. You can take the sting out of this by using a series resistor. However, this is not a 'nice' solution as it (a) wastes power and (b) provides a very low charge rate as the capacitor gets to a similar voltage to the supply.

Charging a capacitor is really a job for a current source. At low power, a conventional linear regulator configured for output current control would do. At high power, you need some sort of inductive energy storage somewhere.

One solution is a boost converter, configured for current output control. Related to this is a flyback. In both cases, you store energy in an inductor, and it feeds this out to a high voltage at a controlled current. They differ in isolation, and the degree of step-up that they can achieve efficiently.

An alternative is the semi-resonant voltage doubler. It's strongly related to the boost converter, but is automatically operating.


simulate this circuit – Schematic created using CircuitLab

Assume C1 starts off uncharged. Close SW1. The supply voltage appears across the inductor, and current starts to build. If you want to describe the waveforms for the current and C1 voltage, then L1 and C1 are most easily understood as a resonant circuit. After one quarter cycle of the L1C1 resonance, C1 voltage is equal to the supply, and the current has stopped increasing. The next quarter cycle of the resonance sees L1 continuing to push current into C1, whose voltage rises to twice that of the supply, while L1 current falls back to zero.

As L1 current tries to go negative, it's blocked by D1. The charging cycle is now finished. No energy has been dissipated. The peak current is controlled and predictable. The charging time is predictable and takes exactly one half cycle (hence the name semi-resonant) of the L1C1 natural frequency.

If C1 starts off with some initial voltage, then it doesn't rise to double the input voltage, only increases by twice the difference of the supply and its initial voltage.

  • \$\begingroup\$ Thanks for your answer. In this case I would still need a transformer + rectifier + filter to get 750V right? \$\endgroup\$ Commented Apr 26, 2020 at 5:42
  • \$\begingroup\$ @LucasRibeiroBarzotto Absolutely, but it can be a boring old constant voltage low impedance supply. \$\endgroup\$
    – Neil_UK
    Commented Apr 26, 2020 at 8:49

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