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1) How to calculate average power by having voltage and current waveforms of one oscillation period that looks like this?

iV waveform

Someone suggested me to use:

formula

Won't this integration result in "negative power" somewhere between 5 us and 10 us and then again after 15 us? Applying this formula to waveforms that overshoot results in negative power. Sine waves with purely resistive loads should result in zero power (hence, RMS?). Shouldn't an RMS variation of this integration be used for average power measurement?

2) What can one say about instantaneous power where voltage waveform is positive and current waveform is negative (and vice versa)?

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  • \$\begingroup\$ For real power convert negative values to their absolute values and sum them \$\endgroup\$
    – user105652
    Sep 27, 2018 at 2:01
  • \$\begingroup\$ The bigger problem is that it looks like most of the power is dissipated in a very short interval at t=6.5us. I think you need to zoom in on that area (so to speak) and get more data points so that you can make sure you calculate the power accurately at that point. \$\endgroup\$
    – mkeith
    Sep 27, 2018 at 3:47
  • \$\begingroup\$ For the most part, any place where the voltage is negative, the current is nearly zero. But you can use absolute value of voltage to get rid of the problem. Current is never negative in your plot, so not an issue. Note that after 15us, voltage doesn't matter because current is zero. \$\endgroup\$
    – mkeith
    Sep 27, 2018 at 3:51
  • \$\begingroup\$ @mkeith, this is a waveform of an electrical discharge. It would be great to be able to zoom in onto the part where most of the current begins to flow. There are some problems with it. The waveform is unstable and it changes its shape quite a lot. Sometimes there are two or three "power peaks" like the one at 6.5us. If I capture one peak like that, I might miss other peaks. Do you have any suggestions on how to deal with it? \$\endgroup\$
    – space bobcat
    Sep 27, 2018 at 3:56
  • \$\begingroup\$ @mkeith, that's true about using absolute values. I have other waveforms where current gets negative, too. In general case it would be necessary to take absolutes of both voltage and current waveforms for the integral. Thank you! \$\endgroup\$
    – space bobcat
    Sep 27, 2018 at 3:59

1 Answer 1

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1) How to calculate average power by having voltage and current waveforms of one oscillation period that looks like this?

Integrate current times voltage, as you have done.

Won't this integration result in "negative power" somewhere between 5 us and 10 us and then again after 15 us? Applying this formula to waveforms that overshoot results in negative power.

Yes.

Sine waves with purely resistive loads should result in zero power (hence, RMS?).

No, sine waves with purely resistive loads should not result in zero power, where did you get that idea?

Shouldn't an RMS variation of this integration be used for average power measurement?

No. RMS is a shortcut for doing the integration.

Negative power means power is coming out of the load, into the source. Whether you add this to the total power, or subtract it from the total power, depends on what you're calculating for.

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