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RMS Voltage and Square Route over 2

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wuchy143

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I'm sure this is a re-post but have been a little confused about it and would like to get some resolution.

So basicly what is the calculation for RMS? Why do we do it? I've just accepted it for what it is...divide the amplide of the sinusoid by sqrt(2) and that's it.

Is a 20V RMS AC signal the same as a 20V DC signal? I would think not but have heard conflicting information even from professors. Also, I guess a better question would be if a hook up a load resistor to a 20V DC source will it heat up the resistor(dissipate as much energy) just as much as a 20V AC source.

Last question is what's the point of RMS voltage?

Thanks for any replies!

-mike
 

MikeMl

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...

Is a 20V RMS AC signal the same as a 20V DC signal? ...
If you put a resistor in a calorimeter, and measured the rate of temperature rise after heating it for equal intervals with 20Vdc and then repeated the experiment with 20Vrms, you couldn't tell any difference in the result...
 

Sceadwian

Banned
Just as an example, the RMS value of standard AC electric outlets in the US is 120 volts, but the peak voltage is 170 volts. RMS is 'effective' voltage. 20VRMS AC is the same total energy as 20VRMS DC but obviously not the same instantanious voltage.

And yes in situations where power needs to be measured measuring the heating of a resistor is actually one way of calculating RMS current especially at RF frequencies where a meter isn't practical. One side effect of the squaring process is the result is always positive.
 
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Nigel Goodwin

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The point is quite simply so you can compare it to any other waveform, including DC.

RMS simply gives the same heating effect as DC, so your 20V RMS across a resistor will heat it exactly the same as 20V DC.
 

wuchy143

Member
ok so I guess I had a better understanding than I thought. Although the math which we get there is a little arbitrary to me. Do you guys know of an easy way to explain why when you take A/sqrt(2) of a sinusoid it gives you basically the equivalent voltage of DC(ignoring calculus). I've looked online but everyone just goes thru the algebra without explanation. I may need to do a little more digging.

-mike
 

MikeMl

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You need a math method which accounts for the fact that when heating a resistor, the resistor doesn't care which direction the current is flowing through the resistor... With an AC waveform, the "average" value is zero, which if you thought that the average current through the resistor would predict how it was heated would obviously not work.
 
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crutschow

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ok so I guess I had a better understanding than I thought. Although the math which we get there is a little arbitrary to me. Do you guys know of an easy way to explain why when you take A/sqrt(2) of a sinusoid it gives you basically the equivalent voltage of DC(ignoring calculus). I've looked online but everyone just goes thru the algebra without explanation. I may need to do a little more digging.
It's not at all arbitrary. It involves using calculus to integrate the instantaneous power over time for the sinewave for one cycle period which gives the average power for that period. If you do the math the answer turns out to be equivalent to multiplying the peak voltage value by sqrt(2). Of course the answer is different for other waveforms such a triangle or sawtooth.
 

Sceadwian

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Stupid question.
When I simulate a sine wave in ltspice with a peak voltage of 170 volts why do I get an RMS voltage of 117 volts. But the 170\Sqrt(2) method tells me it's 120 volts?
 

MrAl

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Hi Sceadwian,


That could be because of several reasons, one being the increment time and
another one could be that it just has too much error in the calculation
somewhere. You could try a smaller increment time and see if it gets closer
to 120v.

For the OP:

RMS is a statement about the power producing ability of a waveform...
it is a measure of the effectiveness of a source in delivering power to
a resistive load.
It stands for "Root of the Mean of the Square", hence the "RMS".
To calculate this for arbitrary waveform, we first square the wave,
then compute the mean of that result, then take the square root.

Mathematically for a voltage wave it looks like this:

VRMS=sqrt((1/T)*integral[0 to T](v(t)^2 dt))

where
v(t) is the voltage wave function (and we squared it)
T is the period
and we used the mean value theorem to calculate the mean
and finally we took the square root of that result.

Now if we defined v(t) to be a sinusoid, we would end up with
a simpler formula that would be valid for sinusoids which happens
to be Vpk/sqrt(2).
 
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Sceadwian

Banned
Yep, that was exactly it thanks Al, the granularity on the simulation was too large. That's always bugged me, nice to know it's just a math error.
 

wuchy143

Member
wow thanks guys. I think I have a much better grasp of the concept. So, all we are doing is trying to account for the current going thru a load no matter which way it is going.

We need to square because of negative amplitudes on the voltage.(won't be negative if it's got a DC bias) The intergral comes in because you need to do an average(mean) of the period(from 0 to T) and then you take the square route because you squared the voltage to begin with. Does that describe

VRMS=sqrt((1/T)*integral[0 to T](v(t)^2 dt))?
 

MikeMl

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Most Helpful Member
Yep, that was exactly it thanks Al, the granularity on the simulation was too large. That's always bugged me, nice to know it's just a math error.
To get the RMS value of the attached LTSpice sim to give the exact answer, I had to turn off compression of the plot window, set the min step size during sim to 10usec, and make sure that the plot ended on a whole cycle of the waveform. Note I let LTSpice compute the peak amplitude, and the duration of the sim to be exactly 10 cycles...
 

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Sceadwian

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Yeah, that's what I did as well. I didn't have to set compression off though. I just had to set the maximum time step to around 1u. It may come up with better results if I set the maximum time step to some perfect fraction of the AC waveform but I don't care to try it, just something I'll have to keep in mind.
 
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Hero999

Banned
It's because: P = V²/R

Incidentally it's not √2 for all waveforms, for squarewave it's √duty.

For example, if you have a 12V light bulb and run it from a 24VDC supply but don't want to use a linear or switching regulator so opt for PWM.

duty = (Vout/Vin)² = (12/24)² = 0.5² = 0.25

So you'll need a duty cycle of 25%

Transposing the formula:
24√0.25 = 12V
 

Sceadwian

Banned
Wikipedia has the equations for most common waveforms.
 

Hero999

Banned
The example above shows how dangerous non-true sinewave DVMs can be.

Suppose the meter takes the average voltage and a current and that the bulb uses [email protected] and is purely resistive.

You measure the output voltage of the PWM unit with your Micky Mouse meter and it reads 6V, you measure the current and it reads just 0.5A when the peak voltage and current is 24V and 2A and the RMS voltage and current is 12V and 1A.

This might not seem to bad unless you're thinking or running a 6V device off it or if the peak voltage was 120VDC and you read just 30V so thought it was safe to touch.
 
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