RMS and true RMS

This post is indecipherable!

Going just from the subject line, I can add the following. I hope that applies but I will never know.

Many multimeters are made using moving coil or d'Arsenval meters for measuring ac. These are dc meters and will not indicate ac although they can damaged by ac. To work on ac, a rectifier is used to convert the ac input into dc. The scales on such meters are marked to read rms values for pure sine wave inputs. They will read erroneously if harmonics (waveform distortion) are present in addition to any other error. A true rms meter will read true rms irrespective of waveform. A typical meter that will do that is a moving vane meter.

Bill

-- Ferme le Bush

you no longer have AC at this point, its unfiltered DC.

an application where you would use a true RMS meter would be the output of a TRIAC circuit when the average voltage to the load is critical i.e as in a tube filament.

Instrument with which I'm working is digital. Do you know how to mathematically describe algorithm of calculating RMS that instrument use.?

Thanks

Mine even has trouble with bus-loads of courgettes...;)

IIRC analog meters (with moving coil and pointer) and simple digital meters inherently read the average voltage. Since nobody wants to know the average voltage, the reading is scaled to RMS by multiplying the average reading by 0.707/0.636 . That works when the voltage is a simple sinusoid, but not for other waveforms, including halfwave rectified AC or a sinusoid distorted by harmonics.

(Scaling average to RMS can be done by converting to peak = average/0.636. Then convert peak to RMS - peak x 0.707 = RMS. Average-scaled-to-RMS = (averge/0.636) x 0.707 .)

True RMS digital meters require some sophisticated electronics, and I think always have a limit beyond which their accuracy breaks down - like trains of very narrow pulses.

bud--

RMS = average*(0.5 pi/sqrt 2) = average*1.11072

ns

Digital isn't always better! Just use a moving vane or dynamometer electromechanical instrument. For high voltage measurement, use the usual form of electrostatic voltmeter.

Bill

-- Ferme le Bush

Everybody tends to forget: Every measuring device requires a standard. For a moving coil meter, it is the magnetic field in which the coil moves and the pivot spring. In an electrostatic voltmeter, it's the physical size of the plates and the restoring force. In a digital meter, it must be a "built in" source of a standard voltage, against which the readings are compared. It doesn't matter how many digits are displayed, the accuracy can be no better than the standard.

Thus, when some program says your CPU is running at 3012.7 MHz, that is only what your CPU is telling the program! The "standard" is a cheap crystal somewhere on the motherboard. :-(

suppose you tell us:

RMS refers to the voltage you wish to measure. True RMS refers to the instrument you are using to measure the voltage. Many multimeters will not read the actual (true) RMS but usually the average. A True RMS meter reads the actual RMS value of the voltage you are measuring.

That's not true. Bud is right. It is true for the half cycle average for a sine wave (full cycle average is

Fill in some details. What meter are you using for your measurements. What are the actual readings you are getting? Are you using a power diode to rectify, and what is the load resistance value? Have you looked at waveforms on a scope to see what you actually have?

Theoretically, if a sine wave is half-wave rectified, the new (true) rms value will be, I think, 1/sqrt(2) times what it was before being rectified. If I'm thinking right, this would be because the area of the squared waveform is 1/2 of the original, and then the root of that area is taken.

What a meter actually does to give an "rms" value depends on the meter. Old time meters for ac rms measurements used to assume sine waves, rectify the wave with a diode, and measure the average of that, but with a scale that was painted to give rms instead of average.

Now days anything should be possible, and since I saw 50 Hz mentioned, we are not worrying about frequency. --Phil

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We have been here before. Of course the average of a sinusoidal signal (which we were referring to) over a cycle is zero, as half is negative and half is positive. However both negative and positive voltages can do work. For RMS, square the instantaneous voltages which makes everything positive, average this, and then take the square root.

My statement is true. Per Buds method with 100 volts average: Convert to peak, 100/0.636=157.23 Peak to RMS, 157.23*0.707 =111.16 or Average scaled to RMS, (100/0.636)*0.707=111.16

I stated: "RMS = average*(0.5 pi/sqrt 2) = average*1.11072"

Slight difference due to Bud rounding. So where is my statement not true?

ns

Addendum Please note I define average voltage as peak*0.63663, otherwise as the average of the instantaneous values of a half cycle (a quarter cycle if in a hurry :)

ns

I think what Don was saying is that this method of calculation is *not* true for a half-wave rectified signal. Your numbers are valid for a sine wave, but the current in half-wave rectified is not the same thing.

So the RMS current for one half the cycle is just as you say (0.707 of the peak, or 1.11072*average). But for the other half-cycle the RMS, peak and average is zero. So, for the full cycle....

peak is equal to peak of conducting half-cycle average is 0.636*peak / 2 RMS is..????

daestrom

The old ac meters, voltmeters and ammeters, gave true rms readings. They did not care if the waveforms were not sinusoidal. If you look at the scale quantitatively, you will see that the scale reading the needle points to is proportional to the square root of the deflection. To the extent that the meter impedance is resistive as a function of the frequencies present, the reading will include the effects of harmonics.

Bill

-- Ferme le Bush

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I think that you missed my point. For a sine wave, what you say is true, AS I INDICATED in the first line of my message. I have no problem with your statement for the value of rms/average for a sinusoid. You simply calculated it differently than Bud but it comes down to the same thing. Your saying"RMS = average*(0.5 pi/sqrt 2) = average*1.11072" is exactly the same as what Bud did except that he did it in two steps

For any other waveform, be careful as "tain't neccessarily so" and that is where your statement doesn't apply. Examples are sawtooth and square waves or any other waves which contain harmonics.

For a half wave rms/peak =1/2 and ave/peak =1/pi giving rms/ave=1.57 That is why I said Bud was right.

You said "That's not true". As pertaining to Bud's post (that I responded to, and quoted above) your statement simply does not hold regardless of following qualification. Even your next statement "Bud is right" implys I was wrong. Bud was clearly referring to sinusoidal waveforms, as was I. He even made it absolutely clear this was so by stating his math was, and I quote "not for other waveforms". Anyway and as you know, my work discussed here has never involved half-wave rectified signals, as they sound nasty and can in fact be downright aggravating to listen to. I have never allowed them on the premises. :)

ns

For 120 RMS volts half-wave rectified RMS = 84.852813 ?

ns