I'm working with a variable voltage inverter.

I have 3 phase AC voltage coming in at 480V RMS line to line. It hits a full wave SCR front end, has a DC bus link, and is inverted with IGBTs to created a six step output wave form.

The IGBTs are firing in 180 degree conduction mode and I'm measuring the output waveform with a true RMS meter.

Here's my confusion:

With a full wave SCR front end I'm aware that I should be able to approach Sqrt(2)*line to lone RMS incoming, and I'm seeing that on the DC bus.

But I can't seem to find the equation that dictates how to calculate the conversion from my DC bus voltage to line to line RMS voltage of my six step output wave.

I've found a million different equations about converting from DC to 3 phase AC but none of them seem to match what I'm seeing.

The output waveform line to line voltage in my tests is always right around .816*DC bus voltage. This conversion factor stays accurate from low voltage all the way up to max DC voltage at ~680V DC (480*1.414).

No where in my research have I found where this number comes from. The closest I've come is an equation on the wiki page for RMS voltage enter image description here

If I divide peak (DC bus) voltage by the sqrt(3/2) I in fact get the conversion factor I was seeing in my tests. But I have no idea where this comes from. And if I'm not mistaken in what I'm seeing this means that with 480VAC RMS in I can get around ~554 VAC RMS out (480*1.414= ~678. 678*.816=~554).

Since we obviously can't just create energy out of no where would this higher voltage come with a reduced current compared with the incoming? Or can someone explain what's going on here and where this Sqrt(3/2) conversion factor comes from please.


1 Answer 1


The conversion factor comes from the fourier analysis of the waveform. Here is an illustration from Bedford and Hoft, "Principles of Inverter Circuits," 1964. The explanation in the book is six and a half pages.

enter image description here

  • \$\begingroup\$ Do you have the pdf of those pages? \$\endgroup\$
    – Jacob
    Commented May 23, 2017 at 18:12
  • \$\begingroup\$ Does that mean that for a given input current the output current will be lower since the output voltage is higher than input? \$\endgroup\$
    – Jacob
    Commented May 23, 2017 at 18:16
  • \$\begingroup\$ The fundamental input current is calculated based on the conservation of energy: Power = V X I X sqrt 3 X power factor. Input power = output power plus losses. The total RMS input and output currents each include their own independent harmonic content. The harmonic currents have related losses, but the effective power factor of the harmonic content is quite low. The reactive component of the motor current is supplied by the DC bus capacitors and not reflected in the input current. The power factor of the input fundamental is quite high. \$\endgroup\$
    – user80875
    Commented May 23, 2017 at 19:21
  • \$\begingroup\$ So you're saying the output can be at a higher voltage with the same output current as the incoming current? Can you explain that, because to me that seems like power generation, not conversion \$\endgroup\$
    – Jacob
    Commented May 24, 2017 at 3:30
  • \$\begingroup\$ Yes. The output can have higher output voltage with the same or higher current than the input because some of the apparent output power is actually reactive power due to energy circulating back and forth between the motor and the DC bus capacitors. However the input harmonic current often offsets that effect. Conservation of energy always holds true, but power factor and harmonics make it difficult to calculate the true energy transfer. \$\endgroup\$
    – user80875
    Commented May 24, 2017 at 10:32

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