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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\$ – Sparky256 Sep 27 '18 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 '18 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 '18 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 '18 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 '18 at 3:59
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Negative power is still power. A negative 1,000 volts would still shock a person badly. The problem lies in simple Dv/Dt equations where you sum all values. For a sine, square or triangle waveform the long term result of say 1,000 cycles that are symmetrical around zero volts, the sum is zero volts or a fractional value < one.

For real power, not just numerical polarity, you must use the {ABS} function to include negative values as having a positive potential. Real power. In your case 6.5 uS of it.

Now your average power is the sum of the RMS values during period Dt. In cases where voltage and current are not in phase you still use ABS value and sum them together, correcting for phase angle or offset.

If this were put through a ideal bridge rectifier it would give you a single instantaneous ABS peak potential for a given instant in time. Add a capacitor to integrate over time and you have a crude average of the power, both voltage and current combined.

Notice I did not elaborate over instantaneous power. Many engineers consider it a marketing term, like for car stereos. It is an arbitrary term where you can imply high power levels over a very short period of time, but this power level cannot be sustained without damage and is far below allowed average sustained power, and much less joules.

NOTE: Whether you need them or not the other members felt my answer was incomplete, based on the below comments. So I am adding a few simple equations here that were in comments. Q & A's are archived over time, but not typical comments.

A) ( by mkeith) Instantaneous power is V(t) * I(t). RMS power is root of the mean of the square of instantaneous power. It is mathematically reasonably well defined.

B) ( by immibis) IIRC rms(abs(V)abs(I)) gives "apparent power" - "real power" means abs(rms(VI)).

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    \$\begingroup\$ Negative 1000 volts will result in negative milliamps through the person which is positive power. \$\endgroup\$ – immibis Sep 27 '18 at 4:51
  • \$\begingroup\$ @immibis. I never disputed that, and ABS value covers that issue. \$\endgroup\$ – Sparky256 Sep 27 '18 at 4:55
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    \$\begingroup\$ Instantaneous power is V(t) * I(t). RMS power is root of the mean of the square of instantaneous power. No marketing. It is mathematically reasonably well defined. \$\endgroup\$ – mkeith Sep 27 '18 at 5:09
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    \$\begingroup\$ I am not being picky. Negative power is power that came back out of the circuit after it went in. You can't just say it's the same as positive power that went into the circuit, without explanation. IIRC rms(abs(V)*abs(I)) gives "apparent power" - "real power" means abs(rms(V*I)). Instantaneous power is also relevant to plenty of things. \$\endgroup\$ – immibis Sep 27 '18 at 5:43
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    \$\begingroup\$ For example, what's the average power through a 1uF capacitor if connected to a 240Vrms 50Hz sinusoidal power supply? It's zero, even though the average absolute power is about 18 W. All the power that enters the capacitor goes out again a moment later - if you plot the instantaneous power over time you'll see that it's positive half the time and negative half the time. \$\endgroup\$ – immibis Sep 27 '18 at 5:51
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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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