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Searching the previous questions I can find ones similar to this but nothing that answers this specifically.

I have a power converter which makes 115VAC/400Hz from 115VAC/60Hz and provides 2.5 amps. My problem is my load has an inrush of 5 amps for 100ms.

My question is: What size/type of capacitor and inductor would I need on the output of my power converter to compensate for the extra 2.5amps @ 100ms?

I will accept a different solution.

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    \$\begingroup\$ Have you considered an NTC on the input of the load? Otherwise E=deltaU^2*C/2. E=115*5*100m. Can you cut into the DC bus inside the inverter? \$\endgroup\$ – winny Jan 31 '17 at 22:37
  • \$\begingroup\$ No I can't get into the 60Hz/400Hz converter or into the load. I think I'll look at either an NTC or just making a soft start circuit. Thanks for the thoughts. \$\endgroup\$ – Sky Yarbrough Feb 1 '17 at 23:32
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insert in series with your load a 47 Ohm resistor that is shorted after the inrush by a delayed relay. A better alternative would be some fabricated soft start circuit, such as a NTC resistor just like in the PC power supplies that also have an inrush when connected to the mains voltage. A NTC from an old PC power supply at least should be tried. Just put it in series with your load.

If you need that 100 ms period + 100% extra power, then the extra must be stored to somewhere before your load is connected. In theory that can be a LC resonator, just in perfect tune for 400 Hz. But that requires so huge and low loss coil that it would be impractical. A better alternative would be a a motor with a substantial flywheel. The motor should be a sychronous one to avoid the phase difference. Its uprunning would be tricky.

What about an active electronic circuit that emulates the motor? Forget it! It woud be an extra power supply.

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  • \$\begingroup\$ I like your idea of the resistor and relay. That just might be my solution. If it works out I'll let you know. Thanks for the thoughts. \$\endgroup\$ – Sky Yarbrough Feb 1 '17 at 23:33

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