Author Topic: RMS VOLTAGE  (Read 5491 times)

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Offline snipersquad100

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RMS VOLTAGE
« on: March 22, 2013, 08:59:01 pm »
Can someone please explain the difference between RMS and Peek to peek voltage.
Thanks   :-//

Offline w2aew

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Re: RMS VOLTAGE
« Reply #1 on: March 22, 2013, 09:14:08 pm »
The RMS voltage is the voltage of an AC waveform that would produce the same work/power as a DC voltage of the same value. 

For sinusoidal AC voltages, the RMS value is equal to the peak voltage divided by sqrt(2).  The peak-to-peak voltage is simply the difference in voltage between the most positive peak and the most negative peak voltage levels.

For non-sinusoidal AC voltages, there would be a different conversion (not the sqrt(2)).
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Offline c4757p

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Re: RMS VOLTAGE
« Reply #2 on: March 22, 2013, 09:17:20 pm »
Peak-to-peak voltage is simply the voltage between peaks. If a waveform extends up to 2V and down to -2V, then it is 4V peak to peak.

The RMS voltage is essentially the equal-power equivalent. 2V DC will dissipate 4W average in a 1 ohm resistor, and so will 2V RMS. If you do the math to figure out what AC voltage gives you a certain average power (remember that power dissipated in a resistor is V2/R):

  • average(PAC) = Pequiv
  • average(VAC2/R) = Vequiv2/R
  • average(VAC2)/R = Vequiv2/R
  • (Solving for Vequiv)
  • Vequiv2 = average(VAC2)
  • Vequiv = sqrt(average(VAC2))

So the equivalent DC voltage that would give the same amount of average power (Vequiv) is the Root of the Mean Squared AC voltage, or RMS.
« Last Edit: March 22, 2013, 09:19:49 pm by c4757p »
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Offline Rerouter

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Re: RMS VOLTAGE
« Reply #3 on: March 22, 2013, 09:20:35 pm »


RMS means root mean square extraction, essentially it is used to calculate the equivalent DC voltage that would pass the same wattage over 1 cycle,

now it doesn't even need to be a clean waveform, those are just examples, however the math behind calculating it on an arbitrary waveform is not something i know...
 

Offline snipersquad100

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Re: RMS VOLTAGE
« Reply #4 on: March 22, 2013, 09:31:40 pm »
thank you guys

Offline c4757p

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Re: RMS VOLTAGE
« Reply #5 on: March 22, 2013, 09:47:30 pm »
now it doesn't even need to be a clean waveform, those are just examples, however the math behind calculating it on an arbitrary waveform is not something i know...

The name is the formula - see my comment. Just take a bunch of measurement points, square them all, average that, then take the square root. If it's a continuous function and not a bunch of measurements, you can average it with calculus (integrate it over one period, then divide that by the period).

Here is Wolfram Alpha demonstrating the calculus version on a 2Vpp sine wave - the RMS voltage is indeed 0.707V.
« Last Edit: March 22, 2013, 09:58:00 pm by c4757p »
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Offline KJDS

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Re: RMS VOLTAGE
« Reply #6 on: March 22, 2013, 10:38:44 pm »
The average of a sine wave is not 0.63 but 0. The average of half a sine wave is 0.63 [/pedant]

Offline c4757p

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Re: RMS VOLTAGE
« Reply #7 on: March 22, 2013, 10:40:07 pm »
I said to square first, then average - unless I accidentally reversed it.

(Checks) Nope, didn't reverse it.
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Offline olsenn

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Re: RMS VOLTAGE
« Reply #8 on: March 22, 2013, 11:23:49 pm »
Quote
The average of a sine wave is not 0.63 but 0. The average of half a sine wave is 0.63

The purpose of squaring it and then taking the square root again at the end is to eliminate negatives. Even though the average value of a sinusoid (with no offset) is zero, power is an absolute value -- putting a negative voltage across a resistor doesn't cool it down, it is polarity independent.

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Online IanB

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Re: RMS VOLTAGE
« Reply #9 on: March 22, 2013, 11:38:22 pm »
I said to square first, then average - unless I accidentally reversed it.

(Checks) Nope, didn't reverse it.

KJDS was responding to Rerouter above who posted in Reply #3 that the average value of a sine wave was 0.637 of the peak value...  ;)
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Offline c4757p

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Re: RMS VOLTAGE
« Reply #10 on: March 22, 2013, 11:39:54 pm »
Oops. Sorry  :)
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Offline c4757p

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Re: RMS VOLTAGE
« Reply #11 on: March 22, 2013, 11:41:37 pm »
The purpose of squaring it and then taking the square root again at the end is to eliminate negatives.

It's not the "purpose", you could just use absolute value if that were the case. See my equations above - it's directly derived from the equations for power.
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Offline NiHaoMike

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Re: RMS VOLTAGE
« Reply #12 on: March 23, 2013, 01:43:53 am »
What I would like to see is a meter that can be switched between true RMS, average responding, and "rectifier" measurements. The latter is particularly important for power supplies, which actually run better on a (lightly filtered) square wave of the same peak voltage as opposed to a sine wave. Considering that it's all implemented in DSP nowadays, it really only takes software...
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Offline c4757p

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Re: RMS VOLTAGE
« Reply #13 on: March 23, 2013, 01:51:54 am »
Considering that it's all implemented in DSP nowadays, it really only takes software...

That annoys me to no end. If you're going to computerize something, computerize it! There's no reason to limit it to one function like the analog equivalent. Hell, on the higher-end bench meters, I ought to be able to upload a mathematical expression for whatever measurement mode I want.
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Offline cyr

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Re: RMS VOLTAGE
« Reply #14 on: March 23, 2013, 08:48:24 am »
What I would like to see is a meter that can be switched between true RMS, average responding, and "rectifier" measurements. The latter is particularly important for power supplies, which actually run better on a (lightly filtered) square wave of the same peak voltage as opposed to a sine wave. Considering that it's all implemented in DSP nowadays, it really only takes software...

Not quite sure what "rectifier" measurement is, but some meters can do various AC metrics. Keithley 2001/2002 for example do AC RMS, AC average, AC+DC RMS, crest factor, peak voltage etc...  That's still mainly analog processing though, I'm sure there are more modern meters that do even more.
 


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