What is The Formula for RMS Wattage? - diyAudio
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Old 22nd July 2005, 07:25 PM   #1
lgreen is offline lgreen  United States
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Default What is The Formula for RMS Wattage?

What's the formula to convert peak to peak voltage into watts RMS for sine and for square waves?

So like 60 volts sine peak to peak into 8 ohms is ____ watts RMS? And 60 volts square peak to peak into 8 ohms is ___ watts?
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Old 22nd July 2005, 07:44 PM   #2
sam9 is offline sam9  United States
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Don't know about square wave. For sine first divide peak voltage by sqrt(2) the square that result and divve by resistance/impedance.


W= ( (Vp/sqrt(2)) * (Vp/sqrt(2)) )/ R

Example: Peak volage =42, assume the load is 8 ohms
RMS voltage = 42/ sqrt(2)= 42/1.4=30
W=(30*30)/3 = 112.5

This does not take into account losses the fact that if you are taking impedance rather than resistance, it often varies with frequency.
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Old 22nd July 2005, 07:56 PM   #3
joensd is offline joensd  Germany
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Quote:
W= ( (Vp/sqrt(2)) * (Vp/sqrt(2)) )/ R
Yep, but Vp would be Vpp/2 and so 30V.
-> P=56.6W

For square waves power would be the same as with DC voltages
and so P=((30V)^2)/8ohm=112.5W.
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Old 22nd July 2005, 08:34 PM   #4
Danko is offline Danko  Hungary
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excuse me guys, but Vpeak/sqrt(2) is the effective value of a sinusiodal wave, and if P=(Vpeak/sqrt(2))^2/R, then I (and my teachers at the high school) call "P" as the effective power. not RMS (root means square).
The "Vpeak/sqrt(2)" is the effective value of the sinusoidal wave. This sqrt(2) comes from a very "dirty" maths.... Some integrals are involved grrr :-)
At square wave the effective valua is simply the duty cycle of the square wave.
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Old 22nd July 2005, 09:01 PM   #5
joensd is offline joensd  Germany
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Quote:
excuse me guys, but Vpeak/sqrt(2) is the effective value
which is the RMS value also.
The solution to the integral BTW is not that dirty.
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Old 22nd July 2005, 09:06 PM   #6
simon5 is offline simon5  Canada
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He's right, we should use P also, because RMS wattage doesn't exist in reality... only RMS voltage or RMS current...
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Old 22nd July 2005, 09:12 PM   #7
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For a sine wave, power is peak-to-peak squared divided by eight times resistance (that is, peak squared divided by twice resistance)

For a square wave, power is peak-to-peak squared divided by four times resistance (that is, peak squared divided by resistance)
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Old 22nd July 2005, 09:31 PM   #8
sam9 is offline sam9  United States
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Quote:
For a sine wave, power is peak-to-peak squared divided by eight times resistance (that is, peak squared divided by twice resistance)
Which if you use my starting point of 42Vp comes out to 110.25. Virtually the same result.

I mention this not to be contentious but to keep the original poster from getting confused. We are really just manipulation Ohms Law and the definition of RMS to get the same result by a different sequence.

If you want more precision look at:

http://www.signaltransfer.freeuk.com/powerout.htm


The example I stated with will come out more like 100W in the real world.
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Old 22nd July 2005, 10:00 PM   #9
joensd is offline joensd  Germany
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Let´s stay with lgreens example for simplicity and less confusion:
Quote:
So like 60 volts sine peak to peak into 8 ohms is ____ watts RMS? And 60 volts square peak to peak into 8 ohms is ___ watts?
60 volts sine peak to peak into 8 ohms is P=56.25Watts.
60 volts square peak to peak into 8 ohms is P=112.5Watts.
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Old 22nd July 2005, 10:45 PM   #10
sam9 is offline sam9  United States
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If you build an amp with 60V rails and put a set of output devices suitable for a 56W amp (Class B or AB assumed) you are in for a real suprise.

60Vp / 1.4 = 43V RMS

(43V^2)/ 8 ohms = 230W
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