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https://www.electronics-tutorials.ws/accircuits/average-voltage.html

https://www.electronics-tutorials.ws/accircuits/rms-voltage.html

I understand that the RMS of an AC voltage is the magnitude of the equivalent DC voltage in terms of power produced,so I expected that when the AC voltage is rectified, the output DC voltage would be equal to the RMS of the original AC (assuming no other voltage drops for diodes.)

When used to compare the equivalent RMS voltage value of an alternating sinusoidal waveform that supplies the same electrical power to a given load as an equivalent DC circuit, the RMS value is called the “effective value” and is generally presented as: V eff or I eff.

In other words, the effective value is an equivalent DC value which tells you how many volts or amps of DC that a time-varying sinusoidal waveform is equal to in terms of its ability to produce the same power.

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V DC = 0.9 * V RMS

V DC = 0.9 * V RMS

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The \$\boxed{\text{average of }|x|}\$ is \$\color{red}{\text{not mathematically the same as the}}\$ \$\sqrt{\text{average of } |x|^2}\$

One computes an average value of a signal but the other computes the power associated with that signal.

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The average of a full-wave rectified sine wave of peak voltage A is 2A/\$\pi\$

The RMS value of a full-wave rectified sine wave of peak voltage A is A/\$\sqrt{2}\$

You can easily verify these numbers by computing the definite integral of the relevant function over a half cycle.

So the ratio of average to RMS (for a sine wave!) is 2\$\sqrt{2}/\pi\$ or about 0.9003163

Consider a 10VDC supply going into a 10 ohm resistor. Average = RMS = 10VDC and power is 10W.

Now consider a 1000V pulse with 1% duty cycle. The average is 10V, however the RMS is 100V and the power is 1000W (100kW for 1% of the time).

The "spikier" the waveform the higher RMS is to average.

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  • \$\begingroup\$ I think the problem is with the example - and I don't quite get it either. If your goal is to supply a 1 kOhm resistive load at 220 VDC with the output of the rectifier you usually would look for the same power or you have to limit the peak voltage to the DC value. I don't see the use case for the AC average voltage as given in the example. \$\endgroup\$ Commented May 20, 2020 at 11:07
  • \$\begingroup\$ @Arsenal didactics? \$\endgroup\$ Commented May 20, 2020 at 11:08
  • \$\begingroup\$ Teaching things that most people understand differently? Or is my understanding of that example completely off? We had some of those questions during the bachelor course and I don't remember ever using the average of the AC voltage for anything. \$\endgroup\$ Commented May 20, 2020 at 11:10
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    \$\begingroup\$ @Arsenal It's useful in only one context that I can think of, off the top of my head, in measuring sinusoidal AC voltage using a precision rectifier (but not a true-RMS measurement). To read equivalent to RMS for a sine wave input the gain has to be adjusted to make it read higher than the actual filtered voltage by about 11% .. 1/0.9. \$\endgroup\$ Commented May 20, 2020 at 11:13
  • \$\begingroup\$ Okay, thanks. I almost expected that a big hole of missing education was opening up, but it doesn't seem too bad. \$\endgroup\$ Commented May 20, 2020 at 11:15

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