RMS voltage of a square wave

[panfilero]

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

 

[Tim Wescott]

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

 

[BobG]

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.

 

[krw]

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

 

[Bill Bowden]

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

 

[daestrom]

"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

 

[daestrom]

"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

 

[krw]

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

 

[krw]

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

 

[krw]

In article <69s303F342iddU1@mid.individual.net>,
philallison@tpg.com.au says...

"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 ?

No Phyllis, you're the only one who's perfect in this conversation.
Perfect asshole, but perfect none the less.

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

** ?????

Try reading it. ...only three words and an inequality,

It *is* used, though perhaps not in handheld DVMs.

** Bollocks .

You're wrong, as usual.

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.

The only point you have is between your shoulders, Phyllis.

BTW:

Fix your stupid settings so the name of the poster you arr replying to is
left visible.

You're as wrong as Dimbulb, Dimwit.


--
Keith

 

[krw]

In article <69s4h7F34hffqU1@mid.individual.net>,
philallison@tpg.com.au says...

"krw" = just another autistic cunthead


Fuck off and die

- you festering pile of sub human excrement.

I bet you say that to all the guys.

--
Keith

 

[Michael A. Terrell]

krw wrote:

In article <69s4h7F34hffqU1@mid.individual.net>,
philallison@tpg.com.au says...


"krw" = just another autistic cunthead


Fuck off and die

- you festering pile of sub human excrement.

I bet you say that to all the guys.

Just the ones he wants to date. I heard it was the last thing his
professor told him before he was tossed out on his knees.


--
http://improve-usenet.org/index.html


Use any search engine other than Google till they stop polluting USENET
with porn and junk commercial SPAM

If you have broadband, your ISP may have a NNTP news server included in
your account: http://www.usenettools.net/ISP.htm

 

[krw]

In article <69sdsuF34dl0sU1@mid.individual.net>,
philallison@tpg.com.au says...

"Salmon Egghead = Fuckwit "


** Go drop dead - septic cunt.

I think that means he loves you.

--
Keith

 

[krw]

In article <69uaidF34vba9U1@mid.individual.net>,
philallison@tpg.com.au says...

"Doug Miller"


** Go drop dead - septic cunt.

NO, Doug. He'd just start repeating himself.

--
Keith

 

[Michael A. Terrell]

Doug Miller wrote:

In article <MPG.22a3c36b27c95ab2989c7a@news.individual.net>,
krw@att.bizzzzzzzzzz says...
In article <69uaidF34vba9U1@mid.individual.net>,
philallison@tpg.com.au says...

"Doug Miller"


** Go drop dead - septic cunt.


NO, Doug. He'd just start repeating himself.

Yeah, he's not real imaginative, is he?

He was raised by dingoes, so what do you expect?


--
http://improve-usenet.org/index.html


Use any search engine other than Google till they stop polluting USENET
with porn and junk commercial SPAM

If you have broadband, your ISP may have a NNTP news server included in
your account: http://www.usenettools.net/ISP.htm

 

[D. Ismay]

Don Kelly wrote:
[...]

What I wanted to know was what was in these boxes in broad terms of their
terms of their operation as there are several ways to get a true rms
measurement (assuming component frequencies outside the meter's bandwidth
can be ignored).

The 'cover' PDF gives a fairly clear and thorough description of how the
RMS-DC convertor is designed, down to the level of a few basic
functional blocks.

Did you expect others to do your homework for you?

 

[Allen]

On Tue, 27 May 2008 07:49:37 +0200, Don Kelly <dhky@shaw.ca> wrote:

"Phil Allison" <philallison@tpg.com.au> wrote in message

<snip>

** So you missed the schematic in figure 9 entirely ???

Wanker.
-----------
Yes- I only got the first page -hence the question.
Now I have the full pdf and see the process.

Thank you for your "courteous" response.

He must like you. He didn't even use the word 'festering'!

 

[daestrom]

"Phil Allison" <philallison@tpg.com.au> wrote in message
news:6a15lkF355n6tU1@mid.individual.net...

"Ben Miller"
D. Ismay wrote:

The 'cover' PDF gives a fairly clear and thorough description of how
the RMS-DC convertor is designed, down to the level of a few basic
functional blocks.

Did you expect others to do your homework for you?

He obviously did not get the entire document, for whatever technical
reason.


** Try following the thread - pal.

All the info was in the FIRST link I supplied( to the AD636) and he could
see ALL of that.

Therefore, he didn't see the other information.


** BOLLOCKS !


Since you apparently don't frequent this newsgroup, you might want to do
some homework yourself before you direct comments like that at an
engineer who can run circles around you in terms of his electrical
knowledge.


** Shame how you cannot even follow a simple thread - let alone see
through that arrogant fake Kelly.

From reading this thread, and having conversed with Don on this group over
the past several years, it is easy to see who is the arrogant, foul-mouthed,
whiner, and who is a courteous, sincere individual.

<plonk>

daestrom