Calculating true RMS current

Malabaristo

Well-Known Member
A rough guild
convert
RMS to Peak is 1.414
peak to RMS is .707

This is where meters that are not true RMS go wrong: they essentially measure the peak voltage and display "RMS" based off a similar calculation. That's only accurate when measuring a pure sine wave. With dirty power or a dimmer at anything other than full (without any voltage regulation), the number it gives you is meaningless and not directly related to the actual RMS value.
 

DrewE

Well-Known Member
This is where meters that are not true RMS go wrong: they essentially measure the peak voltage and display "RMS" based off a similar calculation. That's only accurate when measuring a pure sine wave. With dirty power or a dimmer at anything other than full (without any voltage regulation), the number it gives you is meaningless and not directly related to the actual RMS value.
Most non-true-RMS meters measure the average voltage (rather than the peak voltage), and scale that to give the RMS equivalent assuming a sine wave. The general concept is the same, in so far as the displayed value is only correct for a very few select waveforms, and incorrect to various degrees for any other ones. (Of course, for a given specific waveform, the relationship between displayed values and the actual RMS value is consistent, but to use that fact you have to know in advance what the waveform is and what conversion factor it requires.)
 

DrewE

Well-Known Member
I think most (decent) switch mode power supplies these days have some form of power factor correction built-in, as I believe is required by many certifications/codes. For ones that do, the current waveforms for the supply will be approximately sinusoidal, and the neutral currents well-behaved.

The Source 4 Series 2 Lustr, for one example I chose more or less random, has a claimed power consumption of 167W, and a current consumption of 1.44A at 120V, giving an apparent power of 172.8 VA, and a power factor of 96%. That's really close to being a purely resistive load so far as the AC supply is concerned.
 

David Ashton

Well-Known Member
Power factor is not the issue, the issue is that switch mode supplies tend to draw more current at the peak of the waveform and this will lead to higher neutral currents, so has anyone with a large LED rig actually measured it.
 

DrewE

Well-Known Member
Power factor is not the issue, the issue is that switch mode supplies tend to draw more current at the peak of the waveform and this will lead to higher neutral currents, so has anyone with a large LED rig actually measured it.
Consuming excess current at the peak of the waveform will necessarily lead to a poorer power factor. Switch mode power supplies without power factor correction have pretty poor power factors due to the peaky nature of their current consumption, and power factor correction works by smoothing out the current peaks--either through the use of passive components such as a filter inductor, or (far more commonly these days) by having an active device modulate the incoming line current throughout the AC cycle, forcing it to be a nice sine wave that is in phase with the line voltage.
 

David Ashton

Well-Known Member
Consuming excess current at the peak of the waveform will necessarily lead to a poorer power factor. Switch mode power supplies without power factor correction have pretty poor power factors due to the peaky nature of their current consumption, and power factor correction works by smoothing out the current peaks--either through the use of passive components such as a filter inductor, or (far more commonly these days) by having an active device modulate the incoming line current throughout the AC cycle, forcing it to be a nice sine wave that is in phase with the line voltage.
OK could you link to this "active device' for my information? I totally fail to see a mechanism for an inductor to fix this problem but I would like to be enlightened, I recently had a case of an harmonic filter, a very clever device which completely stuffed up the dimmers it was supplying. Why does consuming excess current at the peak of the waveform necessarily lead to a poorer power factor, which relates to apparent power vs real power ? I'm always keen to learn something new.
 

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