Comparing accuracy of true RMS meter with Averaging type and waveform quality - Page 1
Forum section: electronics / projects

author: Lightages
author_rank: Unknown
timestamp: October 01, 2011, 04:56:10 pm
content: I am working on a number of things right now and I have been trying to compare the accuracy of reading between two meters, one with true RMS and one without. My question is; What is the effect of sine wave distortion on the accuracy of the averaging type meter? In other words, if I have a sine wave with let's say 0.7% total distortion, what would be the effect in % accuracy on the reading of this meter?I suspect that at this level of distortion that the effect would be almost nil.

author: jahonen
author_rank: Super Contributor
timestamp: October 01, 2011, 05:38:38 pm
content: I think that i depends how much the crest factor will change, there is not single answer for given distortion. But for small distortion figures, I also guess that the effect is quite "small". IIRC, even for square wave, the reading will be off "only" by 11%.Regards,Janne

author: Zero999
author_rank: Super Contributor
timestamp: October 01, 2011, 05:59:17 pm
content: The best method to test a true RMS meter is to measure a squarewave with 50% duty cycle with a fast rise/fall time: the RMS voltage should be equal to V/sqrt(2). A 5V squarewave should read 3.536V. You could use a CD4060 oscillator/counter run off a 5V regulator to get a squarewave with a duty cycle of exactly 50%.

author: Bored@Work
author_rank: Super Contributor
timestamp: October 01, 2011, 08:07:56 pm
content: Quote from: Hero999 on October 01, 2011, 05:59:17 pmOf course not. V/sqrt(2) is the formula for sinusoidal voltages only.QuoteA 5 V peak-to-peak square wave should read 2.5V.

author: vk6zgo
author_rank: Super Contributor
timestamp: October 02, 2011, 02:19:03 am
content: Yes,an averaging meter scaled to read as RMS will read 3.5356 volts--or at least,an analog one will try to!(probably readable to 3.5!)I'm not altogether sure if the relationship stays correct with an averaging DMM for a square wave--never tried it.Usually use an Oscilloscope,+ a HP22s--or guess!VK6ZGO

author: IanB
author_rank: Super Contributor
timestamp: October 02, 2011, 02:59:53 am
content: Quote from: BoredAtWork on October 01, 2011, 08:07:56 pm...and this is because if you rectified the square wave you would get a level 2.5 V DC with no AC components.

author: vk6zgo
author_rank: Super Contributor
timestamp: October 02, 2011, 04:06:30 am
content: Quote from: IanB on October 02, 2011, 02:59:53 amA 5 volt peak to peak"Square wave"centred around zero voltswillgive you,assuming an ideal full wave rectifier,a dc signal of 2.5volts.If you rectified a 5v "TRUE" square wave(not centred around zero volts),you would have a 5v square wave,again assuming an ideal full wave rectifier.One half of the rectifier will not turn on during the time the waveform is at zero volts,so you could replace the rectifier with a direct connection & get the same result. .You need to use a filter network,or average it using a mechanical (analog) meter movement, or sample & average over a longer time period as in a DMM.Both of these signal look very similar,but they may yield different readings.RMS is really just a "workaround" to allow us to use the classic P=E^2/R or I^2R formulas with non-dc power sources.VK6ZGO

author: IanB
author_rank: Super Contributor
timestamp: October 02, 2011, 05:14:14 am
content: Quote from: vk6zgo on October 02, 2011, 04:06:30 amIt is fair to assume that allACwave forms are centered around zero volts unless stated otherwise. A square wave with peaks at 0 V and 5 V is a 2.5 V AC square wave with a 2.5 V DC offset, and is an AC+DC wave form.Thiswaveform would read 3.536 V on a true RMS basis.In this thread ambiguity has arisen because some important parameters of the wave forms under discussion have not been specified.The 3.536 V result above can be obtained by the square law of addition when given superimposed AC+DC signals:2.5 V DC = 2.5 V RMS2.5 V AC (square wave) = 2.5 V RMSVtot^2 = Vdc^2 + Vac^2 = 2.5^2 + 2.5^2 = 12.5Vtot = sqrt(12.5) =3.536 V RMS

author: Lightages
author_rank: Unknown
timestamp: October 02, 2011, 06:02:58 am
content: This seems rather counterintuitive. If 2.5VDC = 2.5VRMS where a DC voltage can be considered a 100% duty cycle square wave from 0V to 2.5VDC, then 0VDC to 5VDC 50% duty square wave should provide the same average AND RMS value.2.5VDC at 1 amp for one hour will produce 2.5 watt/hours. A 0VDC to 5VDC square wave at 50% duty cycle at 1 amp for one hour will produce 2.5 watt/hours. The suggested 3.536 VRMS at 1 amp will produce 3.536 watt/hours with the same energy input as 2.5VDC at 1 amp. This cannot be right.And back to my question, how much will some 2nd order and 3rd order distortion adding up to around 0.7% distortion affect the reading on a non-true RMS type AC meter?

author: amspire
author_rank: Super Contributor
timestamp: October 02, 2011, 06:44:20 am
content: Quote from: Lightages on October 02, 2011, 06:02:58 amIt is correct. I think you are forgetting the for half the cycle the DC and AC are cancelling each other out.It is intuitive if you think of it this way:5V Dc is 5V 100% of the time. 2.5vDC + 2.5VAC squarewave is 5V half the time, 0v the other half.So the DC + AC power is half the 5Vdc power.If it is half the power of the 5VDC, then volts have to be lower by 5V* (1/2)-2= 3.536V RMS. (Power is proportional to the square of the voltage).QuoteThere is no answer for this, unless you define the nature of the distortion.For example 3rd harmonic distortion could increase the peak voltage, reduce it, or leave it unchanged.The error would be less then +/-(2 x 0.7%) = 1.4%, but with specially chosen distortion it could be zero.Also non-RMS voltmeters can use rectified average or rectified peak detectors and they will both give different errors, so you have to define the method used to convert AC to DC.Richard

author: Lightages
author_rank: Unknown
timestamp: October 02, 2011, 07:05:50 am
content: DOH!Of course I was thinking backwards. I need to remember the load is assumed to be constant! I cannot decree the same current for the two different waveforms. Again.... DOH!Thanks for correcting my blunder, I know better than this....