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So I am connecting a full-wave rectifier to a half-bridge inverter as shown in the circuit below. I added a smoothing capacitor of 0.47 Farad to get a smooth DC output voltage (see the first graph), so why is the output voltage across the DC-link capacitors (C1 and C2) alternating (see the second and third graph)? As you can see in the fourth graph this also affects the waveform of the output voltage of the inverter.

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Edit 1: Changing the capacitance of the DC-link capacitors C1 and C2 to 40,000 micro-farad fixed the issue (See waveform below). Isn't there a formula I can use to determine this capacitance?

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  • \$\begingroup\$ Perhaps the simulator thought 0.47 was 0.47 microfarad? \$\endgroup\$ – user253751 Sep 25 '18 at 5:11
  • \$\begingroup\$ I don't think so. PSIM will only think of it as 0.47 microfarad if i add a u next to the number. For example 0.47u. Anyways, I tried this capacitance value and it made the dc output waveform of the rectifier full of ripple. I kept on increasing the capacitance until I reached 0.47 Farad \$\endgroup\$ – DigiNin Gravy Sep 25 '18 at 5:20
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You have a timeconstant of RLoad * (C1 + C2) = 20us, how does that compare with your switching frequency (Which I make to be only about 600Hz)?

C1,C2 need to be large enough to keep the timeconstant long compared to the period of a switching cycle otherwise you will the voltage across the caps changing significantly during a cycle, and as the load voltage is the +-(supply voltage - the voltage at the midpoint of the caps)....

This stuff is easy to calculate, but I would hint that in such a converter the ripple current is likely to be as big a design factor as raw capacitance.

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