What is The Formula for RMS Wattage? - diyAudio
 What is The Formula for RMS Wattage?
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 22nd July 2005, 07:25 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, 07:44 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.
diyAudio Member

Join Date: Dec 2001
Location: Germany
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.
__________________
jens

 22nd July 2005, 08:34 PM #4 diyAudio Member   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. __________________ Best regards, Danko
diyAudio Member

Join Date: Dec 2001
Location: Germany
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.
__________________
jens

 22nd July 2005, 09:06 PM #6 diyAudio Member   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... __________________ DIYaudio for President !
 22nd July 2005, 09:12 PM #7 diyAudio Member   Join Date: Jul 2005 Location: england 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) __________________ Tim Martin
diyAudio Member

Join Date: Jun 2002
Location: Left Coast
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.

diyAudio Member

Join Date: Dec 2001
Location: Germany
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.
__________________
jens

 22nd July 2005, 10:45 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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