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I modeled a DCDC boost converter, and I would like to measure the power at the output of the converter, I do not understand much about simulation times, sampling, and this sort of thing, so I do not know how to properly evaluate the actual output power of that converter, but I think it's not just V * I. Could someone give me some light?

The first method I used was to calculate Irms and Vrms and do the multiplication.

The second method was to get the Vrms to squared and divide by the Output Resistor

The method I used to calculate the average value of the power was to take the instantaneous value (just the multiplication of V and I) and go through the block "Mean" using as fundamental frequency 25kHz which is the switching frequency.

Are any of these methods correct, all wrong, or is there any better way to do this in simulink?

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  • \$\begingroup\$ You probably meant V*I, which is just that, the (instantaneous) power. If it's the output of a boost converter, the values will be DC, with some ripple. If the ripple is to great to let you analyse the DC power, then you should average that quantity. A cheap lowpass will do it; other, more complicated methods exist. \$\endgroup\$
    – a concerned citizen
    Commented May 11, 2018 at 13:03
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    \$\begingroup\$ Is it correct to measure the average value of the voltage and current with the block "Mean" using the fundamental frequency (Hz) the value of the inverter switching? For example 25kHz. \$\endgroup\$
    – SrnLord
    Commented May 11, 2018 at 13:18
  • \$\begingroup\$ @SrnLord yes, it is correct. \$\endgroup\$
    – MathieuL
    Commented May 11, 2018 at 15:01

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The method I used to calculate the average value of the power was to take the instantaneous value (just the multiplication of V and I)

That's how digital power meters do it and then average the result using a low pass filter for instance. This works for both AC and DC circuits providing harmonic distortion does not cause aliasing artefacts.

However, it is likely that the DC voltage will have very little ripple current so you can make a fair-to-reasonable estimate of average power by multiplying voltage by low-pass filtered current.

RMS of voltage or RMS of current are irrelevant to measuring power unless pure sine waves are involved and you know the power factor BUT this technique applies to AC circuits and is pointless for DC circuits.

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  • \$\begingroup\$ I disagree RMS value are useful to compute the apparent power of a converter. Your first sentence seem to imply that the RMS value are completly useless for switching circuit. \$\endgroup\$
    – MathieuL
    Commented May 11, 2018 at 15:07
  • \$\begingroup\$ @MathieuL this question is about DC power and not apparent power (AC). My first sentence doesn't mention RMS but explains how digital power meters calculate power. Remember this question is about DC circuits. \$\endgroup\$
    – Andy aka
    Commented May 11, 2018 at 15:12
  • \$\begingroup\$ But You said "this works for both AC and DC". So what you mean ? \$\endgroup\$
    – SrnLord
    Commented May 11, 2018 at 15:20
  • \$\begingroup\$ Measuring power using instantaneous voltages and currents works for DC circuits and AC circuits @SrnLord. It is the de facto method for power measurement. \$\endgroup\$
    – Andy aka
    Commented May 11, 2018 at 15:23

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