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22nd July 2005, 08:25 PM  #1 
diyAudio Member
Join Date: May 2003
Location: San Diego, USA

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? 
22nd July 2005, 08:44 PM  #2 
diyAudio Member
Join Date: Jun 2002
Location: Left Coast

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. 
22nd July 2005, 08:56 PM  #3  
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Join Date: Dec 2001
Location: Germany

Quote:
> 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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jens 

22nd July 2005, 09:34 PM  #4 
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Join Date: Apr 2004
Location: Hungary

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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Best regards, Danko 
22nd July 2005, 10:01 PM  #5  
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Join Date: Dec 2001
Location: Germany

Quote:
The solution to the integral BTW is not that dirty.
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jens 

22nd July 2005, 10:06 PM  #6 
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Join Date: Nov 2004
Location: Québec, Québec

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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DIYaudio for President ! 
22nd July 2005, 10:12 PM  #7 
diyAudio Member
Join Date: Jul 2005
Location: england

For a sine wave, power is peaktopeak squared divided by eight times resistance (that is, peak squared divided by twice resistance)
For a square wave, power is peaktopeak squared divided by four times resistance (that is, peak squared divided by resistance)
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Tim Martin 
22nd July 2005, 10:31 PM  #8  
diyAudio Member
Join Date: Jun 2002
Location: Left Coast

Quote:
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. 

22nd July 2005, 11:00 PM  #9  
diyAudio Member
Join Date: Dec 2001
Location: Germany

Let´s stay with lgreens example for simplicity and less confusion:
Quote:
60 volts square peak to peak into 8 ohms is P=112.5Watts.
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jens 

22nd July 2005, 11:45 PM  #10 
diyAudio Member
Join Date: Jun 2002
Location: Left Coast

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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