Author: panfileroDate: 10:37 23-05-08
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I was wondering if I take the RMS voltage of a messy looking square
wave, a noisy square wave, with some measurement device, and I want to
know the regular current of the thing can I take the RMS current and
divide it by the square root of 2 in order to get my regular current?
I know you can do this for sinusoidal ac signals but don't know if the
math still works out the same for other signals...
PS - I got my RMS current value by dividing the RMS voltage value by a
known resistance I have in my circuit
Much Thanks for any responses I get
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Author: Tim WescottDate: 11:06 23-05-08
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On Fri, 23 May 2008 07:37:50 -0700, panfilero wrote:
> I was wondering if I take the RMS voltage of a messy looking square
> wave, a noisy square wave, with some measurement device, and I want to
> know the regular current of the thing can I take the RMS current and
> divide it by the square root of 2 in order to get my regular current?
>
> I know you can do this for sinusoidal ac signals but don't know if the
> math still works out the same for other signals...
>
> PS - I got my RMS current value by dividing the RMS voltage value by a
> known resistance I have in my circuit
>
> Much Thanks for any responses I get
There is no formal definition of "regular current", so if you define it
right, sure.
What do _you_ mean by "regular current"?
With sinusoidal waveforms, the peak quantity is the RMS quantity times
the square root of two. With square waveforms, the peak and RMS
quantities are equal.
--
Tim Wescott
Control systems and communications consulting
http://www.wescottdesign.com
Need to learn how to apply control theory in your embedded system?
"Applied Control Theory for Embedded Systems" by Tim Wescott
Elsevier/Newnes, http://www.wescottdesign.com/actfes/actfes.html
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Author: Bob EldDate: 12:10 23-05-08
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"panfilero" <panfilero@gmail.com> wrote in message
news:f1496677-7937-43e6-9bbd-386d103b675d@p39g2000prm.googlegroups.com...
> I was wondering if I take the RMS voltage of a messy looking square
> wave, a noisy square wave, with some measurement device, and I want to
> know the regular current of the thing can I take the RMS current and
> divide it by the square root of 2 in order to get my regular current?
>
> I know you can do this for sinusoidal ac signals but don't know if the
> math still works out the same for other signals...
>
> PS - I got my RMS current value by dividing the RMS voltage value by a
> known resistance I have in my circuit
>
> Much Thanks for any responses I get
Yes the RMS current is the RMS voltage devided by R.
No, the "regualr curent" What ever that is, is not the RMS current divided
by the square root of two. With a sign wave, the peak current divided by the
the square root of two is the RMS current, but ONLY for a sign wave.
With any other waveform, the the square root of two relation does NOT hold.
It must be calculated from first principles or measured with a true RMS
meter.
The equation for the RMS current is: Irms = ((int i(t)^2 dt )/ T)^1/2
Where i(t) is the current as a function of time, T is the time over which
the integral is taken, ( 0 to T ), usually one period of the wave.
That is, in words, the RMS current is the square root of the average time
integral of the current squared.
For a square wave of current Ip going plus and minus over a period of T the
RMS value is:
Irms =( ( (Ip^2 *T/2 + (-Ip)^2 * T/2)) / T)^1/2 = ( (Ip ^2)/2 +
(Ip^2)/2)^1/2 = (Ip ^2)^1/2
Irms = Ip, The RMS equals the peak value for a square wave. It very much
more complicated if noise is present.
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Author: BobGDate: 14:04 23-05-08
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The RMS of an on and off wave is the average. Try it with 1V and 0V.
Then you have to believe that it also works with 5v and 0v.
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Author: Bob EldDate: 15:27 23-05-08
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Author: krwDate: 19:11 23-05-08
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In article <f1496677-7937-43e6-9bbd-
386d103b675d@p39g2000prm.googlegroups.com>, panfilero@gmail.com
says...
> I was wondering if I take the RMS voltage of a messy looking square
> wave, a noisy square wave, with some measurement device, and I want to
> know the regular current of the thing can I take the RMS current and
> divide it by the square root of 2 in order to get my regular current?
Sample the waveform at whatever frequency you like, square each
sample, average the squares, and take the square root (root of the
mean squares). It doesn't matter what the waveform looks like, this
algorithm works.
> I know you can do this for sinusoidal ac signals but don't know if the
> math still works out the same for other signals...
Absolutely. ...sorta by definition. ;-)
> PS - I got my RMS current value by dividing the RMS voltage value by a
> known resistance I have in my circuit
Yes. Think about it this way... RMS is used to measure power
because power is proportional to the square of the voltage (or
current), so to get the "effective" voltage of an arbitrary waveform
you need to average the square of the voltages. This also has the
benefit of deriving real power with negative voltages. The average
of the voltages doesn't tell you much (the average of your wall
outlet is zero).
> Much Thanks for any responses I get
No problem.
--
Keith
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Author: BobGDate: 20:32 23-05-08
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On May 23, 7:11=EF=BF=BDpm, krw <k...@att.bizzzzzzzzzz> wrote:
> The average
> of the voltages doesn't tell you much (the average of your wall
> outlet is zero).
=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=
=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D
Not after sampling and squaring. The power is all positive, like the
squared samples.
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Author: Phil AllisonDate: 22:44 23-05-08
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"BobG"
>
> The RMS of an on and off wave is the average.
** Completely wrong !!
RMS current or voltage follow the square root of the duty cycle.
Eg: with a duty cycle of 0.5 ( square wave) the RMS value is 0.7071 of the
peak.
..... Phil
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Author: Bob EldDate: 23:49 23-05-08
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"Phil Allison" <philallison@tpg.com.au> wrote in message
news:69pdnvF344snbU1@mid.individual.net...
>
> "BobG"
> >
> > The RMS of an on and off wave is the average.
>
>
> ** Completely wrong !!
>
> RMS current or voltage follow the square root of the duty cycle.
>
> Eg: with a duty cycle of 0.5 ( square wave) the RMS value is 0.7071 of
the
> peak.
>
>
>
> ..... Phil
>
Vrms = ((vp^2) *T/2 + 0) *1/T) ^1/2 = ((Vp ^2) / 2 )^1/2 = Vp/ (2 ^1/2)...
Yep, for 50% duty cycle, Vrms = Vp/ (sqr 2) if the other half of the wave
is zero.
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Author: Phil AllisonDate: 00:04 24-05-08
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"Bob Eld"
"Phil Allison"
> "BobG"
>> >
>> > The RMS of an on and off wave is the average.
>>
>>
>> ** Completely wrong !!
>>
>> RMS current or voltage follow the square root of the duty cycle.
>>
>> Eg: with a duty cycle of 0.5 ( square wave) the RMS value is 0.7071 of
> the peak.
>>
>>
>
> Vrms = ((vp^2) *T/2 + 0) *1/T) ^1/2 = ((Vp ^2) / 2 )^1/2 = Vp/ (2
> ^1/2)...
>
> Yep, for 50% duty cycle, Vrms = Vp/ (sqr 2) if the other half of the wave
> is zero.
>
** Keep it simple:
For a rectangular wave,
V (or I) rms = sq.rt Duty Cycle ( expressed as a decimal fraction )
For a bi-polar, rectangular wave - first rectify the wave.
..... Phil
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Author: Bill BowdenDate: 00:27 24-05-08
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On May 23, 7:37 am, panfilero <panfil...@gmail.com> wrote:
> I was wondering if I take the RMS voltage of a messy looking square
> wave, a noisy square wave, with some measurement device, and I want to
> know the regular current of the thing can I take the RMS current and
> divide it by the square root of 2 in order to get my regular current?
>
> I know you can do this for sinusoidal ac signals but don't know if the
> math still works out the same for other signals...
>
> PS - I got my RMS current value by dividing the RMS voltage value by a
> known resistance I have in my circuit
>
> Much Thanks for any responses I get
Look at it this way: You have a 1KW space heater you run half the time
(1 hour on and 1 hour off). The average power is 500 watts. Say the
heater voltage is 100 VDC and the current is 10 amps, and the heater
resistance is 10 ohms. So, the question is what RMS voltage will give
you 500 watts at 10 ohms and 100 volts peak? Work that out and we find
the voltage to be 70.7 and the RMS current to be 7.07 amps.
So, it looks like for a square wave, the RMS voltage or current is the
peak divided by the square root of 2.
-Bill
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Author: Salmon EggDate: 00:32 24-05-08
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Author: Don KellyDate: 00:58 24-05-08
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----------------------------
"BobG" <bobgardner@aol.com> wrote in message
news:a0bfd275-76bf-4a30-8aee-fb158186b6bd@34g2000hsh.googlegroups.com...
On May 23, 7:11?pm, krw <k...@att.bizzzzzzzzzz> wrote:
> The average
> of the voltages doesn't tell you much (the average of your wall
> outlet is zero).
============================================
Not after sampling and squaring. The power is all positive, like the
squared samples.
-----
But the average of the squared values is not the average value of the
voltage and taking the root of the average of V*2 will not, in general, give
you the average of V.
It will give the rms value which was devised originally as a way to get an
"equal Power" to DC with the same load- as krw indicates.
--
Don Kelly dhky@shawcross.ca
remove the X to answer
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Author: Don KellyDate: 00:59 24-05-08
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Bob Eld and Phil Allison have dealt with this topic except for the question
"how did you measure the rms voltage?".
If you are using a true rms meter which samples at a high enough rate- then
you are OK.
If you are using a cheaper multimeter on the normal "rms" scale, then it
simply assumes that the AC measurement is sinusoidal and scales from the DC
average accordingly and will not give the correct value except for a
sinusoid or for an On/Off cycle with equal on and off periods.
--
Don Kelly dhky@shawcross.ca
remove the X to answer
----------------------------
"panfilero" <panfilero@gmail.com> wrote in message
news:f1496677-7937-43e6-9bbd-386d103b675d@p39g2000prm.googlegroups.com...
>I was wondering if I take the RMS voltage of a messy looking square
> wave, a noisy square wave, with some measurement device, and I want to
> know the regular current of the thing can I take the RMS current and
> divide it by the square root of 2 in order to get my regular current?
>
> I know you can do this for sinusoidal ac signals but don't know if the
> math still works out the same for other signals...
>
> PS - I got my RMS current value by dividing the RMS voltage value by a
> known resistance I have in my circuit
>
> Much Thanks for any responses I get
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Author: Phil AllisonDate: 01:28 24-05-08
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"Don Kelly"
>
> Bob Eld and Phil Allison have dealt with this topic except for the
> question "how did you measure the rms voltage?".
> If you are using a true rms meter which samples at a high enough rate-
> then you are OK.
** ?????????
The " true rms" function on a DMM is *not* obtained by sampling. It is
normally obtained by the use of a " true rms to DC" converter IC -
generally one made by Analog Devices like the AD736.
> If you are using a cheaper multimeter on the normal "rms" scale, then it
> simply assumes that the AC measurement is sinusoidal and scales from the
> DC average accordingly and will not give the correct value except for a
> sinusoid or for an On/Off cycle with equal on and off periods.
** Most DMMs use a precision rectifier circuit followed by RC averaging of
the output and then DC scaling so that the final reading equals the rms
value of a sine wave input ** within** the particular meter's specified
frequency range.
Mostly, this range is very narrow - ie from about 30Hz to a mere 1kHz (with
+/- 1% accuracy).
So, unless you KNOW the energy spectrum of your wave falls inside this
narrow band, the reading can be way wrong -ie well below the actual value.
Unfortunately, most " true rms " DMMs have the same frequency range issue -
some of the more expensive ones work out to 20kHz or 100 kHz with good
accuracy.
The only * low cost * solution is to examine the wave on a scope and then
compute the rectified average and rms values - not possible with any
accuracy if the wave is noise like.
..... Phil
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Author: daestromDate: 09:29 24-05-08
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"panfilero" <panfilero@gmail.com> wrote in message
news:f1496677-7937-43e6-9bbd-386d103b675d@p39g2000prm.googlegroups.com...
>I was wondering if I take the RMS voltage of a messy looking square
> wave, a noisy square wave, with some measurement device, and I want to
> know the regular current of the thing can I take the RMS current and
> divide it by the square root of 2 in order to get my regular current?
>
> I know you can do this for sinusoidal ac signals but don't know if the
> math still works out the same for other signals...
>
> PS - I got my RMS current value by dividing the RMS voltage value by a
> known resistance I have in my circuit
>
*PROVIDED* you measured that voltage with a true RMS voltmeter, then yes,
you have the true RMS current. But typical voltmeters are not true RMS
devices. So beware, "Garbage In / Garbage Out"
daestrom
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Author: daestromDate: 09:33 24-05-08
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"Phil Allison" <philallison@tpg.com.au> wrote in message
news:69pnc3F33qck8U1@mid.individual.net...
>
> "Don Kelly"
>>
>> Bob Eld and Phil Allison have dealt with this topic except for the
>> question "how did you measure the rms voltage?".
>> If you are using a true rms meter which samples at a high enough rate-
>> then you are OK.
>
> ** ?????????
>
> The " true rms" function on a DMM is *not* obtained by sampling. It is
> normally obtained by the use of a " true rms to DC" converter IC -
> generally one made by Analog Devices like the AD736.
>
>
>> If you are using a cheaper multimeter on the normal "rms" scale, then
it
>> simply assumes that the AC measurement is sinusoidal and scales from the
>> DC average accordingly and will not give the correct value except for a
>> sinusoid or for an On/Off cycle with equal on and off periods.
>
>
> ** Most DMMs use a precision rectifier circuit followed by RC averaging
> of the output and then DC scaling so that the final reading equals the rms
> value of a sine wave input ** within** the particular meter's specified
> frequency range.
>
> Mostly, this range is very narrow - ie from about 30Hz to a mere 1kHz
> (with +/- 1% accuracy).
>
> So, unless you KNOW the energy spectrum of your wave falls inside this
> narrow band, the reading can be way wrong -ie well below the actual value.
>
> Unfortunately, most " true rms " DMMs have the same frequency range
> ssue - some of the more expensive ones work out to 20kHz or 100 kHz with
> good accuracy.
>
> The only * low cost * solution is to examine the wave on a scope and then
> compute the rectified average and rms values - not possible with any
> accuracy if the wave is noise like.
>
>
How about a precision resistor placed inside a calorimeter :-) The heat
dissipated is a direct measure of RMS current squared.
Wasn't this done with some high-freq RF stuff at one time? Measured the
temperature or wattage of a resistor or something like that?
daestrom
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Author: krwDate: 20:03 24-05-08
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In article <483818ea$0$31724$4c368faf@roadrunner.com>,
daestrom@NO_SPAM_HEREtwcny.rr.com says...
>
> "Phil Allison" <philallison@tpg.com.au> wrote in message
> news:69pnc3F33qck8U1@mid.individual.net...
> >
> > "Don Kelly"
> >>
> >> Bob Eld and Phil Allison have dealt with this topic except for the
> >> question "how did you measure the rms voltage?".
> >> If you are using a true rms meter which samples at a high enough rate-
> >> then you are OK.
> >
> > ** ?????????
> >
> > The " true rms" function on a DMM is *not* obtained by sampling. It
is
> > normally obtained by the use of a " true rms to DC" converter IC -
> > generally one made by Analog Devices like the AD736.
> >
> >
> >> If you are using a cheaper multimeter on the normal "rms" scale,
then it
> >> simply assumes that the AC measurement is sinusoidal and scales from the
> >> DC average accordingly and will not give the correct value except for a
> >> sinusoid or for an On/Off cycle with equal on and off periods.
> >
> >
> > ** Most DMMs use a precision rectifier circuit followed by RC averaging
> > of the output and then DC scaling so that the final reading equals the rms
> > value of a sine wave input ** within** the particular meter's specified
> > frequency range.
> >
> > Mostly, this range is very narrow - ie from about 30Hz to a mere 1kHz
> > (with +/- 1% accuracy).
> >
> > So, unless you KNOW the energy spectrum of your wave falls inside this
> > narrow band, the reading can be way wrong -ie well below the actual value.
> >
> > Unfortunately, most " true rms " DMMs have the same frequency range
> > ssue - some of the more expensive ones work out to 20kHz or 100 kHz with
> > good accuracy.
> >
> > The only * low cost * solution is to examine the wave on a scope and then
> > compute the rectified average and rms values - not possible with any
> > accuracy if the wave is noise like.
> >
> >
>
> How about a precision resistor placed inside a calorimeter :-) The heat
> dissipated is a direct measure of RMS current squared.
Calorimetry is notoriously difficult. Ask Pons and Fleishman. ;-)
> Wasn't this done with some high-freq RF stuff at one time? Measured the
> temperature or wattage of a resistor or something like that?
Yes, I used an HP true RMS voltmeter when I was in college. It was
a marvelously expensive widget and they didn't like mere students
using it. ;-) It's likely the best way to measure true RMS voltage
at high frequency. I suppose you could read current with the same
meter. ;-)
--
Keith
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Author: krwDate: 20:08 24-05-08
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In article <69pnc3F33qck8U1@mid.individual.net>,
philallison@tpg.com.au says...
>
> "Don Kelly"
> >
> > Bob Eld and Phil Allison have dealt with this topic except for the
> > question "how did you measure the rms voltage?".
> > If you are using a true rms meter which samples at a high enough rate-
> > then you are OK.
>
> ** ?????????
>
> The " true rms" function on a DMM is *not* obtained by sampling. It is
> normally obtained by the use of a " true rms to DC" converter IC -
> generally one made by Analog Devices like the AD736.
Slick, though not perfect.
> > If you are using a cheaper multimeter on the normal "rms" scale,
then it
> > simply assumes that the AC measurement is sinusoidal and scales from the
> > DC average accordingly and will not give the correct value except for a
> > sinusoid or for an On/Off cycle with equal on and off periods.
>
>
> ** Most DMMs use a precision rectifier circuit followed by RC averaging of
> the output and then DC scaling so that the final reading equals the rms
> value of a sine wave input ** within** the particular meter's specified
> frequency range.
Normally <> is not
It *is* used, though perhaps not in handheld DVMs. IT is used in
Kill-a-Watts.
> Mostly, this range is very narrow - ie from about 30Hz to a mere 1kHz (with
> +/- 1% accuracy).
>
> So, unless you KNOW the energy spectrum of your wave falls inside this
> narrow band, the reading can be way wrong -ie well below the actual value.
>
> Unfortunately, most " true rms " DMMs have the same frequency range issue
-
> some of the more expensive ones work out to 20kHz or 100 kHz with good
> accuracy.
>
> The only * low cost * solution is to examine the wave on a scope and then
> compute the rectified average and rms values - not possible with any
> accuracy if the wave is noise like.
>
How do you get better than 1% accuracy on a scope? Sampling works
to any accuracy one desires and can go fairly high in frequency.
Even higher if the waveform is repetitive.
--
Keith
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Author: Phil AllisonDate: 22:59 24-05-08
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"krw"
Phil Allison
"Don Kelly"
>> >
>> > Bob Eld and Phil Allison have dealt with this topic except for the
>> > question "how did you measure the rms voltage?".
>> > If you are using a true rms meter which samples at a high enough rate-
>> > then you are OK.
>>
>> ** ?????????
>>
>> The " true rms" function on a DMM is *not* obtained by sampling. It
is
>> normally obtained by the use of a " true rms to DC" converter IC -
>> generally one made by Analog Devices like the AD736.
>
> Slick, though not perfect.
** Err - so just like you ?
>> If you are using a cheaper multimeter on the normal "rms" scale, then
it
>> > simply assumes that the AC measurement is sinusoidal and scales from
>> > the
>> > DC average accordingly and will not give the correct value except for a
>> > sinusoid or for an On/Off cycle with equal on and off periods.
>>
>>
>> ** Most DMMs use a precision rectifier circuit followed by RC averaging
>> of
>> the output and then DC scaling so that the final reading equals the rms
>> value of a sine wave input ** within** the particular meter's specified
>> frequency range.
>
> Normally <> is not
** ?????
> It *is* used, though perhaps not in handheld DVMs.
** Bollocks .
>> Mostly, this range is very narrow - ie from about 30Hz to a mere 1kHz
>> (with
>> +/- 1% accuracy).
>>
>> So, unless you KNOW the energy spectrum of your wave falls inside this
>> narrow band, the reading can be way wrong -ie well below the actual
>> value.
>>
>> Unfortunately, most " true rms " DMMs have the same frequency range
>> ssue -
>> some of the more expensive ones work out to 20kHz or 100 kHz with good
>> accuracy.
>>
>> The only * low cost * solution is to examine the wave on a scope and then
>> compute the rectified average and rms values - not possible with any
>> accuracy if the wave is noise like.
>>
> How do you get better than 1% accuracy on a scope? Sampling works
> to any accuracy one desires and can go fairly high in frequency.
> Even higher if the waveform is repetitive.
** You are not paying attention to the point at issue - fuckwit.
BTW:
Fix your stupid settings so the name of the poster you arr replying to is
left visible.
...... Phil
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