Question: True RMS DVM versus Analog Volt Meter
James P. Meyer
On Mon, 27 May 1996, Two Legs wrote:
> I assume that a True RMS meter applies the formula for RMS to the
> peak to peak value of the AC (.707 x X, if I remeber correctly). My
> guess is that an Analog meter averages the peak to peak value. Am I
> correct?
>
Either meter will read RMS directly and without any math
conversion needed. But they will only agree when the waveform is a sine
wave. If what you are measuring departs only a little from a sine wave
or if you don't need true RMS readings (few folks do), then either meter
will be satisfactory.
The true RMS meter can not apply any formulas to the peak to peak
voltage of a sine wave, because it will continue to read RMS properly
even when the input is a square wave. Or any other waveform for that matter.
Jim
Two Legs
What are some of the differences between a True RMS Digital Volt
Meter (as in a Fluke 87) and an Analog voltmeter? How will
measurements of AC voltage and current differ between the two types of
meters? What's the formula for converting a True RMS meter reading
to an Analog Meter equivalent?
I assume that a True RMS meter applies the formula for RMS to the
peak to peak value of the AC (.707 x X, if I remeber correctly). My
guess is that an Analog meter averages the peak to peak value. Am I
correct?
Thanks for your help.
Frank
Henry Spencer
In article <31a8fa2d...@netnews.worldnet.att.net> Pica...@Worldnet.ATT.Net (Two Legs) writes:
> I assume that a True RMS meter applies the formula for RMS to the
Uh, no, that's exactly what it doesn't do. "True RMS" means RMS, not a
peak or average measurement times a fudge factor -- it means that the
meter really does do a Root-Mean-Square operation on the waveform, so
it gets the correct RMS value even when the waveform is non-sinusoidal
(in which case the fudge-factor approach fails).
>...My guess is that an Analog meter averages the peak to peak value. Am I
>correct?
If memory serves -- I haven't worked with the things in a long time --
ordinary analog meters normally *did* use the fudge-factor approach, since
they respond to the mean, not the RMS, value of the waveform.
--
Unix was a breakthrough. | Henry Spencer
Windows 95 is more like a smash-and-grab. | he...@zoo.toronto.edu
Kevin McMurtrie
In article <31a8fa2d...@netnews.worldnet.att.net>,
Pica...@Worldnet.ATT.Net (Two Legs) wrote:
> What are some of the differences between a True RMS Digital Volt
>Meter (as in a Fluke 87) and an Analog voltmeter? How will
>measurements of AC voltage and current differ between the two types of
>meters? What's the formula for converting a True RMS meter reading
>to an Analog Meter equivalent?
>
> I assume that a True RMS meter applies the formula for RMS to the
>peak to peak value of the AC (.707 x X, if I remeber correctly). My
>guess is that an Analog meter averages the peak to peak value. Am I
>correct?
>
> Thanks for your help.
>
> Frank
The analog meters are typicaly true RMS or close since they don't filter
the input. Cheap digital meters measure something between RMS and .707 of
peak. They run into all kinds of problems trying to measure the true
power of non-sine waves, such as square waves generated by power inverters
or dirtly household power. The non-RMS digital meters I've used went
bezerk on AC that was noisey or not a perfect 50%+ 50%- duty cycle.
Clyde Manning
Many of the 'true RMS' meters also ahve a crest factor limitation, above
which you mill not get true rms readings. Check your manual for the
number.
Ruben
On Mon, 27 May 1996 00:42:45 GMT, Pica...@Worldnet.ATT.Net (Two Legs)
wrote:
> What are some of the differences between a True RMS Digital Volt
>Meter (as in a Fluke 87) and an Analog voltmeter? How will
>measurements of AC voltage and current differ between the two types of
>meters? What's the formula for converting a True RMS meter reading
>to an Analog Meter equivalent?
For sinusiodial waveforms the readings will be identical. When the
crest-factor increases the analog will show an increasing error, while the
RMS-meter will give accurate measurements. (Crest-factor is the ratio
between peak and RMS.) Most analogs try to model peak * 0.707, which for
sinus is the right, but for other waveforms is wrong..
> I assume that a True RMS meter applies the formula for RMS to the
>peak to peak value of the AC (.707 x X, if I remeber correctly). My
>guess is that an Analog meter averages the peak to peak value. Am I
>correct?
That last statement must be wrong. The average of peak-to-peak sinus is
zero, assuming a pure sinus, without DC etc.
A true RMS squares the input, averages that, and takes the root to get to
display. Display=(Root(Mean(Square(Input)))) No other processing takes
place. No peak, no average, no nothing.
---
Ruben
Kevin AstirCS 1U KO0B
On Mon, 27 May 1996 00:42:45 GMT, Pica...@Worldnet.ATT.Net (Two
Legs)
wrote:
> What are some of the differences between a True RMS Digital Volt
>Meter (as in a Fluke 87) and an Analog voltmeter?
The true-rms reads the square root of the average of the square of the
voltage (I just checked, and I can say that in one breath!) It does
this with either analog computing circuits, or by buffering the
voltage, applying that to a heater, and seeing how much DC it takes to
bring a similar heater to the wame temperature (thermal bridge
approach). See LTC application note by Jim Williams for excellent
discussion of latter approach.
The analog meter indicates .707 x average |V|. This requires a
rectifier (abs value), a low-pass filter (average), and an
appropriatly calibrated meter scale (.707x)
Which, as other respondants have indicated is the same for a siusoidal
waveform, but different for most others.
> What's the formula for converting a True RMS meter reading
>to an Analog Meter equivalent?
The formula depends on the shape of the waveform.
You apply the above discriptions to the waveform in question.
So you must integrate the absolute value of voltage over 1 period
(average) and multiply by .707 This tells you the reading on a
averaging (analog) meter.
Then you must integrate the square of the voltage over one period and
take the square root.
You do both sybolicly, keeping one thing which might vary reading
(duty cycle, or Vpeak) as a symbol.
You solve the equation for one symbol in terms of one meter reading,
then plug that expression into the equation (solved integral) for the
other meter type.
This is a nasty job unless the waveform is really simple. Here is a
real world example from my experience:
I once had to work this out for a waveform which was a 325V (peak) 400
Hz, sine-wave which was duty cycle modulated symetrically about the
90 and 270 degree peaks. This was used to control a spoiler actuator
motor in the wing of a B1-B.
The device was designed in 1970's and a USAF acceptance test procedure
(ATP) based on averaging type voltmeter readings, had to be re-written
to account for 1980's automated test equiptment which had a true-rms
voltmeter. I had to prove that the two sets of readings were
equivilent, and show that the much brader range of readings obtained
on true-rms meter was not indicative of a sloppier control circuit.
It took about a day and a half (well I did have to check my work!)
with integral, and trig identity tables People kept interupting me,
so I put on hearing protectors, and hid in enviromental test area
(lots of noisy blowers running) for half of three days.
This was only possible because the peak of the sine wave was a known
(closely regulated, and measured as part of ATP) value. The duty
cycle changed, to vary the motor voltage.
The closed formula took up a full page of paper. It was published as
an appendix to ATP. It took many iterations with publications
department before they got all the parenthesis, squares, radicals, and
trig functions properly arrainged.
After it was published, USAF questioned the derivation. Fortunatly, I
had kept my handwritten "proof" (about 12-15 pages) which I sent
them. Never heard anything more on that!
Lemme guess.....You have my old job, and the Air Force is asking
again!
-KF-
Kevin AstirCS 1U KO0B
kfer...@aquilagroup.com (Kevin AstirCS "1U" KO0B) wrote:
>The analog meter indicates .707 x average |V|. This requires a
>rectifier (abs value), a low-pass filter (average), and an
>appropriatly calibrated meter scale (.707x)
Error, constant above should be 1.11 not .707. Because average of
1V(peak) sine wave is 2/pi, not 1. Also, the "filter" mentioned above
is often just the mechanical (and magnetic) dampening of the movement
itself.
Can't believe I managed to fix this before someone else pointed it
out.
-Kevin
Roy McCammon
kfer...@aquilagroup.com (Kevin AstirCS "1U" KO0B) wrote:
>constant above should be 1.11 not .707. Because average of
>1V(peak) sine wave is 2/pi, not 1. Also, the "filter" mentioned above
>is often just the mechanical (and magnetic) dampening of the movement
>itself.
And to be really explicit,
1.11 = sqrt(.5) / [ 2/pi ]
Opinions expressed herein are my own and may not represent those of my employer.