I’m working one some assignment like this circuit (AC-AC converter 3 phase, 89 degree firing period). I've done this on PSIM software with 110 Volts 3 phase voltage source.

AC-AC Converter 3 phase, 89 degree firing periode

I've been told to analyze the line to line voltage. The waveform and the value looks like this enter image description here

My question is, can somebody tell me what is the equation to solve the line to line voltage value? I've been googling around and found nothing related to this particular circuit.

  • \$\begingroup\$ Are you asking what the RMS voltage is of your last picture? \$\endgroup\$
    – Andy aka
    Apr 25 '20 at 7:49
  • \$\begingroup\$ yes sir @Andyaka \$\endgroup\$ Apr 25 '20 at 8:08
  • \$\begingroup\$ Does your PSIM not have the ability to analyse the waveform and give you the RMS? I bet it does. Can it do FFT on the waveform? If it can then the fundamental frequency (being the one relevant to a motor load) is likely the answer you might need. \$\endgroup\$
    – Andy aka
    Apr 25 '20 at 8:10
  • \$\begingroup\$ The PSIM already give me the RMS voltage. it shown on the bottom of your right hand side in the second picture. what i'm asking is how can i manually calculate this waveforms, sir? \$\endgroup\$ Apr 25 '20 at 8:12
  • \$\begingroup\$ Why would you want to if PSIM gives you the numbers? \$\endgroup\$
    – Andy aka
    Apr 25 '20 at 8:14

Start by minimizing the problem to firstly this: -

enter image description here

Then, because the negative half cycle is a mirror image of the positive half cycle, it will have the same RMS value, thus we end up with only needing to analyse this section to get RMS: -

enter image description here

Then it breaks down into 4 sections that you need to find the square of the RMS for each: -

enter image description here

So, find those individual squares then multiply each by their time duration and the final RMS value of your waveform is this: -

$$\sqrt{\dfrac{d_1V_1^2 + d_2V_2^2 + d_3V_3^2 + d_4V_4^2}{d_1+d_2+d_3+d_4}}$$

  • \$\begingroup\$ Thank you so much for the explanation sir, i really appreciate that \$\endgroup\$ Apr 25 '20 at 8:44
  • \$\begingroup\$ No problem. I got it wrong the first time I wrote it but corrected the final formula. Do you see that in converting a section to its RMS and squaring it I get power, then by multiplying power by its duration I get an energy contribution. Then adding all the energy contributions I get total energy then dividing that by overall time I get back to power then, finally, taking the square root gives me RMS voltage. \$\endgroup\$
    – Andy aka
    Apr 25 '20 at 8:47
  • \$\begingroup\$ that's awesome, once again thank you sir \$\endgroup\$ Apr 25 '20 at 8:51

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