RMS VOLTAGE - Page 1
Forum section: electronics / beginners

author: snipersquad100
author_rank: Frequent Contributor
timestamp: March 22, 2013, 08:59:01 pm
content: Can someone please explain the difference between RMS and Peek to peek voltage.Thanks

author: w2aew
author_rank: Super Contributor
timestamp: March 22, 2013, 09:14:08 pm
content: 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)).

author: c4757p
author_rank: Super Contributor
timestamp: March 22, 2013, 09:17:20 pm
content: 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) = Pequivaverage(VAC2/R) = Vequiv2/Raverage(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 theRoot of theMeanSquared AC voltage, or RMS.

author: Rerouter
author_rank: Super Contributor
timestamp: March 22, 2013, 09:20:35 pm
content: 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...

author: snipersquad100
author_rank: Frequent Contributor
timestamp: March 22, 2013, 09:31:40 pm
content: thank you guys

author: c4757p
author_rank: Super Contributor
timestamp: March 22, 2013, 09:47:30 pm
content: Quote from: Rerouter on March 22, 2013, 09:20:35 pmThe 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).Hereis Wolfram Alpha demonstrating the calculus version on a 2Vpp sine wave - the RMS voltage is indeed 0.707V.

author: KJDS
author_rank: Super Contributor
timestamp: March 22, 2013, 10:38:44 pm
content: The average of a sine wave is not 0.63 but 0. The average of half a sine wave is 0.63

author: c4757p
author_rank: Super Contributor
timestamp: March 22, 2013, 10:40:07 pm
content: I said to square first, then average - unless I accidentally reversed it.(Checks) Nope, didn't reverse it.

author: olsenn
author_rank: Frequent Contributor
timestamp: March 22, 2013, 11:23:49 pm
content: QuoteThe 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.What is the speed of dark?

author: IanB
author_rank: Super Contributor
timestamp: March 22, 2013, 11:38:22 pm
content: Quote from: c4757p on March 22, 2013, 10:40:07 pmKJDS 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...

author: c4757p
author_rank: Super Contributor
timestamp: March 22, 2013, 11:39:54 pm
content: Oops. Sorry

author: c4757p
author_rank: Super Contributor
timestamp: March 22, 2013, 11:41:37 pm
content: Quote from: olsenn on March 22, 2013, 11:23:49 pmIt'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.

author: NiHaoMike
author_rank: Super Contributor
timestamp: March 23, 2013, 01:43:53 am
content: 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...

author: c4757p
author_rank: Super Contributor
timestamp: March 23, 2013, 01:51:54 am
content: Quote from: NiHaoMike on March 23, 2013, 01:43:53 amThat 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.

author: cyr
author_rank: Frequent Contributor
timestamp: March 23, 2013, 08:48:24 am
content: Quote from: NiHaoMike on March 23, 2013, 01:43:53 amNot 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.