# RMS power and phase angle control

We have a piece of industrial equipment using a phase angle controller (PAC), and are having some issues calculating the RMS power used by the load.

Personally I am of the opinion that you measure RMS voltage and RMS current on the line side of the PAC. This gives you line RMS * PACed current. Assuming a duty cycle of 50%, this gives you 50% of full power. My reasoning is that this follows Kirchoff's law, and that power in is approximately equal to power out on our PAC (efficiency is >90%).

My colleague suggests that we measure the RMS voltage and RMS current on the load side, after the PAC. Assuming a duty cycle of 50%, this gives 50% current and 50% voltage and we end up measuring 25% of full power. His reasoning is that the load isn't conducting for the off section of the cycle, so the applied RMS voltage is lower.

The former gives us a linear ramp from 0 to 100% duty cycle, the latter gives a non-linear response. Assuming our power factor is 1, our mains supply and PAC signals are clean, and the RMS calculation is correct at the crest factors we're working at (it better be, I designed the sensor), which calculation is correct?

Thanks!

• Have you looked at harmonics and made sure the waveform is clean? – Voltage Spike May 22 at 15:56
• @laptop2d that's not the question. We don't want our equipment troubleshot. We want to know the answer to the theoretical question. – Joe of Loath May 22 at 15:58
• I don't know anything about industrial hardware, but I do know that differences like this are usually in the way RMS is calculated and in the harmonics. mikeholt.com/technical-power-quality-harmonics.php – Voltage Spike May 22 at 16:01
• In fact there are not a lot of power engineers that frequent this site, just a handful, most of us are circuit designers. – Voltage Spike May 22 at 16:02

Let's analyse one cycle of an AC waveform. Here I've used a stepped waveform for ease of analysis. Figure 1. The voltage, current and power waveforms with RMS calculations on the V and I waveforms (using the root of the mean of the squares). The P waveform shows the instantaneous power value.

In figure 1 we can see that the average power is 12.5 W. As a matter of interest this is only 12.5 / 20 = 62.5% of the peak power. Those weak parts make a big difference. Figure 2. The same waveforms but with a turn-on delay of 90°.

In figure 2 we can see that the average power is 6.25 W. The average power has decreased to 50% of the full-wave power.

Personally I am of the opinion that you measure RMS voltage and RMS current on the line side of the PAC. This gives you line RMS * PACed current. Assuming a duty cycle of 50%, this gives you 50% of full power. My reasoning is that this follows Kirchoff's law, and that power in is approximately equal to power out on our PAC (efficiency is >90%).

Yes, power in = power out, but you can't calculate power in by measuring the voltage across a full cycle, current across half a cycle and multiply them together. If you did this with my Figure 1 VRMS and my Figure 2 IRMS you'd get P = 7.9 x 1.12 = 8.85 W. The correct answer is 6.25 W.)

My colleague suggests that we measure the RMS voltage and RMS current on the load side, after the PAC. Assuming a duty cycle of 50%, this gives 50% current and 50% voltage and we end up measuring 25% of full power. His reasoning is that the load isn't conducting for the off section of the cycle, so the applied RMS voltage is lower.

So using the V and I from my Figure 2 we get 5.6 V x 1.12 A = 6.25 W. Your colleague is looking good.

Your colleague, however, needs to beware. He is right in this instance because we've assumed a resistive load so current and voltage are proportional. In the case of a rectifier charging a capacitor or battery, for example, the current and voltage waveforms won't be the same. The only way then is to sample both voltage and current often enough, calculate the power and average that value. This is, effectively what my diagrams are showing.

• Thanks, that clarifies things a lot for us! – Joe of Loath May 23 at 10:56

"Assuming our power factor is 1, our mains supply and PAC signals are clean..."

If those assumptions were accurate, multiplying RMS current by RMS voltage on either side of the controller would provide the correct result for power and the two would differ only by the power dissipated in the controller.

In the real world, the harmonic current and voltage content will cause and error. In effect, the harmonic content reduces the power factor.

You also need to consider how your instruments interpret the waveforms. What you are measuring may not be the true RMS values. The only way to get an accurate power measurement is to multiply instantaneous voltage by instantaneous current for many points over one cycle of the waveforms. Then calculate the average power. There is no such thing as RMS power. RMS voltage X RMS current X power factor = average power.

There are electronic wattmeters on the market that will accurately measure power with the waveforms that you have. A turning disc type kilowatt-hour meter will also give accurate results if connected on the line side because it responds only to the fundamental component of the current waveform and the voltage waveform has little or no distortion.

For a resistive load, common sense tells us that for a symmetrical signal, 50% conduction will give 50% power. But since power also equals Vrms^2/R, the Vrms must be 70.7% at 50% conduction.

Assuming a duty cycle of 50%, this gives 50% current and 50% voltage...

So, this assumption is wrong, RMS voltage is 70.7% and the RMS current is also 70.7% at 50% conduction. 70.7% * 70.7% = 50% power.

The following graph was created using instantaneous voltage and current over every 1/2 degree. Charles described most everything else well, I will only disagree with half of one paragraph.

If those assumptions were accurate, multiplying RMS current by RMS voltage on either side of the controller would provide the correct result for power and the two would differ only by the power dissipated in the controller.

I don't think that this will apply for the input side of the controller. At any instant when the current is zero, the voltage can't contribute to the power.