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I've been measuring low DC-powered devices consumption, by measuring the current at the needed frequency (able to detect the fastest consumption changes). I can then have the mean, max, min. To get the mean, i just integrate these measure points over time and multiply the result with the supply voltage.

One collegue of mine told me that this was not correct and, because that the current was not constant, I should use RMS measurement tools like for measuring AC current...

Could you confirm my procedure is correct?

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Mean and RMS are two different measurements which both have a use. It's not that mean current is somehow "wrong" and RMS current is "right". The kind of measurement you need will depend on what you plan to do with the value.

RMS is useful for power measurements assuming a constant load resistance, because power is proportional to current squared (or voltage squared), so if you want to e.g. estimate the amount of heat generated by your device, you'll want to use RMS current. On the other hand, if you want to know how long your device can run on a battery of a given capacity in A*h, you'll likely want the average current.

If your supply voltage is constant, you should use the average current for power measurement.

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For a DC supply of near constant voltage, it is the average current taken by the load multiplied by that constant DC voltage that yields true power consumption. Any AC content in the current does not contribute to average power consumption on a stable DC supply. Why: because a sine wave multiplied by a DC value is still a sine wave and its average value is zero. Power consumption is an average quantity.

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  • \$\begingroup\$ But the OP didn't say that the time-varying component was sinusoidal. Are you trying to say that a correctly measured mean current is the best approach, where correctly measured depends on the shape of the current waveform? \$\endgroup\$ – Elliot Alderson Jun 14 '19 at 15:16
  • \$\begingroup\$ @ElliotAlderson any of the sinewaves that make up the composition of any waveform shaped supply current will not contribute towards the measurement of average power on a fixed voltage DC system and, if you have placed a downvote I’ll ask you to reconsider that action. \$\endgroup\$ – Andy aka Jun 14 '19 at 17:14
  • \$\begingroup\$ Sorry, but I think @Dmitry did a better job of explaining the difference between mean and rms, and why the mean value is important in this case. I still feel that you have muddied the water by introducing the notion of sinusoidal current and talking about multiplying a sinusoid by a dc value...the OP's current may not even be periodic. Perhaps it would have been better to talk about the mean value of the time varying current, and then say "If the time-varying component happens to be sinusoidal, then we know that the mean value will be zero... \$\endgroup\$ – Elliot Alderson Jun 14 '19 at 18:04
  • \$\begingroup\$ So, is your modus operandi to downvote all answers except the one you mostly agree with? \$\endgroup\$ – Andy aka Jun 14 '19 at 18:32
  • \$\begingroup\$ I tend to downvote answers that I think will confuse the OP, if there is another answer that seems to more directly and clearly address the question. Sometimes I upvote multiple answers. \$\endgroup\$ – Elliot Alderson Jun 14 '19 at 20:14

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