# Why even care about average voltage?

#### bloguetronica

Joined Apr 27, 2007
1,467
average emf is used for flux calculations (remeber using them for derivation in transformers)
but then again u can always use the form factor cant u?
That's what I use.

#### bloguetronica

Joined Apr 27, 2007
1,467
The average voltage of 230V RMS is 2*Vpeak/pi=2*325/3,14=207V
The 230V AC is an average value and has 325V peak (650V from positive peak to negative peak). That is the way in Europe. EOS.

The value of 0.707 was calculated and confirmed in a real life situation. In a transformer, measured 22V AC, but after rectification and capacitative filtering measured around 30V. Aprox the peak voltage with 1V subtracted, being the 1V the voltage across two diodes from the bridge. Not to offend you, but if you need more proff, go to Europe, buy a wall wart and open it (I don't know how it applies in the US or other countries). I cannot say more than that. Sorry if I was too harsh.

#### recca02

Joined Apr 2, 2007
1,214
in india 230V is rms (if we are talking distribution values)
so average rms/1.1.
while i know rms is used power does it not make sense to add in magnitude
the average power for AC rather than adding in sign?
will have a study on this and reply back.

#### bloguetronica

Joined Apr 27, 2007
1,467
Sources:

http://en.wikipedia.org/wiki/Alternating_current

I think the portuguese version is more correct. Have in account that A is the peak value. Also, using integrals, as I said before, had led me to the same conclusion. I prefer to call it Veff instead of Vrms, because is the effective (or average) voltage that we want.

Para 220 V CA, a tensão de pico VP ou A é, portanto, 220 V × √2 = 311 V (aprox.). O valor de pico-a-pico VP-P de 220V CA é ainda mais alta: 2 × 220 V × √2 = 622V (aprox.)

#### winmx

Joined May 24, 2007
14
The 230V AC is an average value and has 325V peak (650V from positive peak to negative peak). That is the way in Europe. EOS.

The value of 0.707 was calculated and confirmed in a real life situation. In a transformer, measured 22V AC, but after rectification and capacitative filtering measured around 30V. Aprox the peak voltage with 1V subtracted, being the 1V the voltage across two diodes from the bridge. Not to offend you, but if you need more proff, go to Europe, buy a wall wart and open it (I don't know how it applies in the US or other countries). I cannot say more than that. Sorry if I was too harsh.
No! I live in Europe and I have measured the output of a 230V wall outlet in a university lab with an oscilloscope and the peak voltage is 325V and the peak to peak is of course 650V. Using rms integration you get 230V rms and using average integration you get 207V average. I have seen it with my eyes talked about it with my professors so I do not think I can be wrong about that! Average voltage we get when we do a simple voltage integration over time
RMS voltage we get when we integrate the square of the voltage and calculate the square root of the integral
These are mathematical concepts that have been applied to electricity (since electricity is maths to an extent...)

#### bloguetronica

Joined Apr 27, 2007
1,467
So, how can be 207V an average? Do the integral of abs(sin(x)) from x=0 to x=2Pi, and tell me the value. As you know, integrals are used to calculate areas, and therefore, averages. Used them a lot.

I insist on this so my curiosity can be satisfied.

#### recca02

Joined Apr 2, 2007
1,214
over 2pi average will be zero(perhaps) so lets calculate for half a wave.
int(sin x )(dx) = cosx , cos 0 - cos (pi) = 2. dividing by period pi
we get V(av)= ~ V(peak)x.63
is that what u are looking for?

#### winmx

Joined May 24, 2007
14
Ups! recca02 made me think again and I have to correct myself. The average of a sine wave is ZERO!!! Sorry... But for a fully rectified 230V rms sine wave (or half a sine wave from 0 to pi) the average voltage is 207V and you get to that result the way recca02 describes.
2*325V(peak)/pi = 206.9V(average) or
325V(peak)*0.633=206.9V(average)

#### Ron H

Joined Apr 14, 2005
7,014
So, if you have a current with that waveform going across through a resistor, the power dissipated in the resistor is the same when it is crossed by a current with a constant potential of 3.53V? Sounds a little strange, so with a 5Ohm resistor:
Power when crossed by 5V: 5W
Average power (square wave regime): 2.5W
Power when crossed by 3.53V: 3.53^2 / 5 = 2.492W

Makes sense, but that is not the average voltage. So RMS is related to power, that is, RMS voltage is the constant voltage that causes the same amount of dissipation as the waveform. Right?
Yes. With pulse waveforms having insignificant risetimes, you can calculate the power at the two voltage levels and average them proportional to the duty cycle.

This confuses me, since some transformers say 230V RMS instead of 230V AC, and they are refering to the average voltage, because I know the peaks are about 325.22 (230 x sqrt (2)). Unless RMS and average voltage are the same for a 230V AC waveform. Also, this definition of average voltage for AC current exists on Malvino (Peak = Average x sqrt(0.5)). I calculated myself the average voltage of the module (I'm refering here to abs, because in Portugal we call it módulo, and I´m not sure about around here) of a sin function. For example the integer from x=0 to x=Pi of sin (x) is 0.707/2. So the integer from x=0 to x=2Pi of abs(sin(x)) would be 0.707, the value of sqrt(0.5).
I think your average calculation is wrong. The integral of sin(x) for x=0 to x=pi is 2. To get the average, divide by pi. The average is 0.6366. For 0 to 2pi, the average of abs(sin)x) will be the same.

#### winmx

Joined May 24, 2007
14
For 0 to 2pi, the average of abs(sin)x) will be the same.
Are you sure about that last part. I just calculated it and it is zero... I think (unless of course the wave is rectified) #### bloguetronica

Joined Apr 27, 2007
1,467
Are you sure about that last part. I just calculated it and it is zero... I think (unless of course the wave is rectified) It is abs(sin(x)) and not sin(x). The areas of sin(x) will subtract and the integral will equal 0. abs(sin(x)) corresponds more or less to a rectified wave.

It seems I must review my calculations. Now it is giving 4/2pi = 0.6366. Strangely, it is almost close to sqrt (0.5)/1.1 . I am almost convinced. I must see what kind of math I did before to get the sqrt (0.5) value. See if 1.1107 as RMS to average conversion factor is correct. If it is, then I'm convinced.

#### recca02

Joined Apr 2, 2007
1,214
rms= 1.11*average
1.11 is form factor.

#### winmx

Joined May 24, 2007
14
@cumesoftware Hope you are convinced Now, back on topic Any other views on average voltage? #### bloguetronica

Joined Apr 27, 2007
1,467
As you point out peak voltage is useful - no doubt about that. But why not use the rms voltage to calculate nominal dissipation of a voltage regulator instead of the average? Would it not yield correct results?
I know I answered this wrong the other time. But I can give you a hint:
Preg = (Vin - Vout) x Iacross.

For Vin I like to consider the average voltage. RMS voltage is not usable in there since someone said correctly it aplies to resistive loads. Regulators don't behave quite like that, as you can see in the formula above. Also, I need to know the voltage going in in absolute terms. Hence I use the aprox: Vin = Vpeak - 0.5Vripple. I it will be simply calculated considering the predicted load R at the constant voltage Vout.

#### bloguetronica

Joined Apr 27, 2007
1,467
As you point out peak voltage is useful - no doubt about that. But why not use the rms voltage to calculate nominal dissipation of a voltage regulator instead of the average? Would it not yield correct results?
I know I answered this wrong the other time. But I can give you a hint:
Wreg = Vin - Vout / Iacross.

Vor Vin I like to consider the average voltage. RMS voltage is not usable in there since someone said correctly it aplies to resistive loads. Regulators don't behave quite like that, as you can see in the formula above. Also, I need to know the voltage going in in absolute terms. Hence I use the aprox: Vin = Vpeak - 0.5Vripple.

Also, thanks for the enligthment. I'm convinced.

#### gootee

Joined Apr 24, 2007
447
Getting back to the original question, somewhat, isn't it true that a low-pass filter will produce approximately the average value of a rectified signal? (and not only for sinusoids)

I'm thinking in terms of detectors; for example, a classic opamp-based "ideal" half-wave rectifier, followed by a low-pass filter, to produce a quasi-DC "amplitude tracking" signal. If the filter's cutoff frequency is low-enough, the filter's output is DC at, I believe, the average value of the rectified signal (assuming unchanging input signal amplitude and shape for a long-enough time). And with a "real" filter, both the AVG and RMS values of the output's "DC + ripple" approach the AVG value of the rectified signal, as the low-pass cutoff frequency is lowered. (And the output's AVG value approaches it more quickly than the output's RMS value.)

I haven't recently worked through the math, for this. But I seem to have verified it with an LTspice simulation, just now, for both triangle and sine waveforms.

I have such a setup as part of a feedback loop, to try to precisely control the amplitude of a low-voltage triangle waveform: An amplified version of the triangle output waveform is AC-coupled into a half-wave rectifier, the output of which is sent through an RC low-pass filter. The difference between the resulting "DC" voltage (which has a slight ripple) and a precise DC reference voltage is integrated. The voltage output of the integrator determines a current that controls a "current-controlled resistor" (LED encapsulated with photocell) that is part of a voltage divider for the original triangle input.

In terms of the original question, I guess we could say that in a case like the one above, we care about the average value, rather than the RMS value, because the average value is what the reference voltage is compared with, to set the eventual output's amplitude.

- Tom Gootee

http://www.fullnet.com/~tomg/index.html

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#### winmx

Joined May 24, 2007
14
Thank you so much for your responses! I will have a closer look at it tommorow (it's late at night here in Europe) but things are starting to make sense! 