Wouldn't an analog voltmeter messure in true RMS? - Page 1
Forum section: electronics / beginners

author: Deactivated-1
author_rank: Frequent Contributor
timestamp: September 11, 2023, 11:14:16 pm
content: Wouldn't an analog galvenometer measure an AC signal in true RMS if it was rectified ofc.Since the peaks would move the needle up faster than the smaller peaks.

author: TimFox
author_rank: Super Contributor
timestamp: September 11, 2023, 11:25:21 pm
content: No.A galvanometer, or any other instrument that responds to DC and has a time constant much longer than 1/60 of a second, would measure the "average voltage", which is defined as themean absolute value, when driven from an ideal full-wave rectifier of the AC input (with zero DC value).A traditional Simpson 260 or an -hp- 400 AC VTVM measures AC waveforms in this manner.-hp- always labeled their moving-needle meters "average responding" for this circuit, meaning that the dial was calibrated for the RMS voltage of a sinusoidal waveform with this average voltage.They were cataloged as "average-responding, RMS indicating".By simple calculus, you can calculate the ratio between the RMS voltage and average voltage for different periodic waveforms with zero mean (no DC component).For a square wave, they are equal to each other and to the peak voltage.Note that the "iron-vane" meter is not a galvanometer (by the definition thereof), but the d'Arsonval meter is a galvanometer.https://electricalacademia.com/instrumentation-and-measurements/basic-analog-meter-movement-darsonval-iron-vane-meter-movement/The deflection of the iron-vane meter is always in the same direction, regardless of DC polarity or AC, but the galvanometer or d'Arsonval meter deflects positive or negative according to polarity.You can easily find discussions of this topic anywhere--this has been known since the 19th century.

author: bdunham7
author_rank: Super Contributor
timestamp: September 12, 2023, 12:09:20 am
content: You specified a galvanometer, so no. But if you devised an analog meter that had a heating element and responded to temperature, then perhaps.

author: MrAl
author_rank: Super Contributor
timestamp: September 12, 2023, 01:54:35 am
content: Quote from: ELS122 on September 11, 2023, 11:14:16 pmHi,As you probably know, you can use a "crest factor" for an analog movement to get RMS from the normal average reading. In many cases a multimeter will have this already built into the movement so it shows the RMS value for AC sine wave measurements.There are other crest factors you can use for other types of waves also, you could look them up.The math behind it for any waveform is as follows.For the average voltage it is:Vavg=integrate(v(t),t,0,Tp)/Tpand for RMS it is:Vrms=sqrt[integrate(v(t)^2,t,0,Tp)/Tp]What this means is that Vavg adds up many instantaneous values and divides by the period.For Vrms it adds up many values of the square of the instantaneous values and divides by the period, then takes the square root.There are certain waveforms where they will both give the same reading but not for a sine wave. A simple example is a DC voltage, which will give the same Vrms as Vavg.The other thing to think about is the frequency of the sine wave (or other type of wave). Most multimeters are made for 50Hz, 60Hz, up to maybe 400Hz. Anything above that it may not read correctly. Some meters are made for higher frequencies though like 300kHz.

author: TimFox
author_rank: Super Contributor
timestamp: September 12, 2023, 03:02:38 am
content: Quote from: bdunham7 on September 12, 2023, 12:09:20 amA once-common true-RMS analog meter uses two matched thermocouples with heaters in an insulated chamber, one heated by the input voltage and the other heated by a DC voltage from a servo amplifier using the difference between the two thermocouples as its error signal.Measuring the DC voltage gives you the RMS value of the arbitrary waveform.

author: Doctorandus_P
author_rank: Super Contributor
timestamp: September 13, 2023, 09:34:40 pm
content: This does not work because of the way the meter responds to input signals.Assume there is a short peak, which has an amplitude of 10x the "nominal" value.During that peak, the meter will have a 10x bigger current through it ,and therefore it's needle would experience a force 10x bigger, so it will start accelerating upwards.However, when the same peak is put into a resistor, then during that peak the current will also be 10x bigger, which is a 100x increase of power.As a result, those averaging meters show values that are too low for signals with high peaks.

author: golden_labels
author_rank: Super Contributor
timestamp: September 13, 2023, 10:17:36 pm
content: Maybe it is worth emphasising this part: the RMS value is about power, the average value is just voltage.In other words, RMS is related to the average of power on a hypothetical resistor seeing measured voltage. While average voltage is just average voltage.(Corrected, see TimFox’s note below)In other words, RMS is the average of power on a hypothetical resistor seeing measured voltage. While average voltage is just average voltage.

author: TimFox
author_rank: Super Contributor
timestamp: September 13, 2023, 10:25:52 pm
content: Quote from: golden_labels on September 13, 2023, 10:17:36 pm"MS" is the power (mean of voltagesquared), scaled to resistance."RMS" is the voltage (square root of mean power)."Average" is the mean absolute value of an AC voltage, which differs from RMS.

author: golden_labels
author_rank: Super Contributor
timestamp: September 14, 2023, 12:13:22 pm
content: In case this is an attempt to correct me: I did not say it is power. I said it isaboutpower.RMS voltage is a measure of voltage, but the averaging process is done for power.

author: The Electrician
author_rank: Frequent Contributor
timestamp: September 14, 2023, 01:00:39 pm
content: Quote from: golden_labels on September 14, 2023, 12:13:22 pmYou said "In other words, RMSisthe average of power on a hypothetical resistor seeing measured voltage."

author: schmitt trigger
author_rank: Super Contributor
timestamp: September 14, 2023, 01:46:43 pm
content: I guess that if one really wants a galvanometer to read true RMS, one could somehow kludge an AD536 or similar to perform the RMS conversion first.But I sincerely doubt it would be a cost effective project. Those linear RMS converters are not inexpensive!

author: TimFox
author_rank: Super Contributor
timestamp: September 14, 2023, 02:03:17 pm
content: My comment about RMS was a clarification: I'm sorry you took offense without parsing my reply.Where I do a correction is when someone confuses mean power with RMS power: the latter can be calculated, but is a useless variable.When you measurethe RMS voltage across a resistor, that voltage squared divided by the resistance is the mean power (or average power), not the RMS power.

author: wasedadoc
author_rank: Super Contributor
timestamp: September 14, 2023, 02:42:51 pm
content: In plain English, the RMS value of an AC voltage across a resistor is equal to the value of a DC voltage that would generate the same heat in that resistor.

author: bdunham7
author_rank: Super Contributor
timestamp: September 14, 2023, 03:00:47 pm
content: Quote from: golden_labels on September 14, 2023, 12:13:22 pmI think we all know the reason that RMS is relevant to AC voltage and yes that is about power, but when you start talking about real systems it isn't hard to have a miscommunication, even with yourself.RMS, of course, is a simple mathematical operation that can be applied to any sample set. You can have RMS annual rainfall or RMS temperature and, of course RMS power, but those aren't useful AFAIK. Power is actually not any part of thedefinitionof RMS voltage, although it is certainly a part of understanding and explaining the utility of the concept as well as measuring it in a few cases.Other than gated digital sampling systems (DSOs, a few modern DMMs) the methods of measuring RMS voltage actually end up using a weighted moving average of some intermediate value. A typical TRMS chip does a continuous analog integration of the square and then takes the root of that intermediate averaged value. Thermal systems are unique in that they actually average the power.Quote from: wasedadoc on September 14, 2023, 02:42:51 pmThat's correct and is the common way of understanding and explaining it, as well as one way of measuring it, but it isn't thedefinition.

author: IanB
author_rank: Super Contributor
timestamp: September 14, 2023, 03:06:20 pm
content: Quote from: TimFox on September 14, 2023, 02:03:17 pmWe may note that "RMS power" is sometimes quoted for the performance of an amplifier, but this is a misuse of the English language. The term should really be "maximum sustained or continuous output power", but language is fuzzy, and sometimes things cannot be taken literally.

author: MrAl
author_rank: Super Contributor
timestamp: September 14, 2023, 03:12:26 pm
content: Quote from: schmitt trigger on September 14, 2023, 01:46:43 pmHi,For slow moving target levels you could probably use an Arduino. A lot of measurements are done at 50Hz or 60Hz and sometimes at 400Hz. 50 and 60Hz should be easy to do. These would be sine waves only or some other fairly slow changing waves.

author: MrAl
author_rank: Super Contributor
timestamp: September 14, 2023, 03:26:03 pm
content: Hi,A simple way of explaining RMS is by comparing the math behind RMS and the math behind calculating average power, at least for sine waves, and of the same phase which would be the case for a resistor load.The simplest way to understand it is in it's use. For a sine wave voltage with RMS value Vrms and sine wave current with the same phase with RMS value Irms the power is simply:P=Vrms*IrmsIf there is a phase shift in the current, we have to use a scaling factor which is often known as the power factor. Then we have:P=Vrms*Irms*cos(ph)where ph is the angle of lead or lag between the voltage and the current.This shows the basic use of RMS measurements. You can use it in other ways too but it gets a bit more complicated.In reality there is no such thing as RMS power, but we sometimes refer to it as that anyway (ha ha). We might be able to say that it is the result of multiplying the RMS voltage times the RMS current in circuits that have sine waves. It will get confusing though because what we really mean is average power, and that is also the power heating power, which relates to something physical.For example, if you want to heat up a cup of water you would use the average heating power to calculate the time it takes. This gets a little confusing too though because when we think of an average, we think of a lot of changing things that we add up and then divide by the number of things we have considered. The average power is similar, but we don't usually have to do that we just have to know the average power, and it is considered a constant thing when we talk about the ability to heat an object to a certain temperature in a certain amount of time. If the power changed during the heating process, then we would have to ALSO average the multiple power levels, and this might be confusing because the average power seems like the average power already, yet when it changes we could have several different average power levels to consider. For example, if we heat the cup of water at a power of 100 watts and then change to 200 watts and then down to 50 watts, we have to average those while also considering the time of each duration. So we end up averaging the individual averages.

author: TimFox
author_rank: Super Contributor
timestamp: September 14, 2023, 03:56:48 pm
content: Quote from: IanB on September 14, 2023, 03:06:20 pmAs an exercise, I once computed the "RMS power" for a sine wave.For a given voltage and resistance, and sinusoidal voltage, I calculated the ratio between RMS power and mean power (the correct term): /  = (3/2)1/2= 1.225:1This is almost 1 dB (power ratio), similar to the fraudulent "+/- 1 dB" power advertisements in the late 1960s that were banned by the FTC.