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?
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?
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.
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.
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.
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.
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.
He's right, we should use P also, because RMS wattage doesn't exist in reality... only RMS voltage or RMS current...
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)
For a square wave, power is peak-to-peak squared divided by four times resistance (that is, peak squared divided by 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.
Let´s stay with lgreens example for simplicity and less confusion:
60 volts square peak to peak into 8 ohms is P=112.5Watts.
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.
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
60Vp / 1.4 = 43V RMS
(43V^2)/ 8 ohms = 230W
I don´t like to repeat myself but why don´t we stay with lgreen´s example.
Now that would be if you still don´t get it : 60V PEAK-PEAK.
That is not 60V rails and neither 60V peak.
Now that would be if you still don´t get it : 60V PEAK-PEAK.
That is not 60V rails and neither 60V peak.
confused
Ok, this is getting confusing.
As for RMS, are not amplifiers rated to deliver so many watts per channel RMS???
When I say peak-to-peak, I mean it! 60 v peak-to-peak is the same as plus AND minus 30 volts, so on a scope at 10V/div, the wave would go up 3 div and down 3 div, for a total of 6 div. being occupied vertically by the wave on the screen.
Ok, this is getting confusing.
As for RMS, are not amplifiers rated to deliver so many watts per channel RMS???
When I say peak-to-peak, I mean it! 60 v peak-to-peak is the same as plus AND minus 30 volts, so on a scope at 10V/div, the wave would go up 3 div and down 3 div, for a total of 6 div. being occupied vertically by the wave on the screen.
Re: confused
"A serious and reliable measure of the power output of a loudspeaker or audio amplifier is average power, commonly called watts RMS. RMS stands for root mean square and is actually a misnomer here but has nonetheless become the common term. The "RMS" power is found by averaging the instantaneous power output over a long period of time, so it is actually the average power or mean power. The term RMS is used due to the fact that the mean power is calculated from the RMS voltage and current (or one of them and the impedance)."
lgreen said:
"A serious and reliable measure of the power output of a loudspeaker or audio amplifier is average power, commonly called watts RMS. RMS stands for root mean square and is actually a misnomer here but has nonetheless become the common term. The "RMS" power is found by averaging the instantaneous power output over a long period of time, so it is actually the average power or mean power. The term RMS is used due to the fact that the mean power is calculated from the RMS voltage and current (or one of them and the impedance)."
I think we (I ?) got a little careless as the thread went along regarding "P-to-P" vs "Peak". No doubt because when we build these things we often think in terms of the rail voltage which is usually in terms of +/- V since each rail is (hopefully) the negative of the other.
Hi,
the formulae you need are
P = V^2 / R = I^2 * R : eq1, 2
= Vpk^2 / (2 * R) = Ipk^2 * R / 2 : eq3, 4
= Vpp^2 / ( 8 * R) = Ipp^2 * R / 8 : eq5, 6
so the example from eq5 gives 60 * 60 / (8 * 8) = 56.25W
the formulae you need are
P = V^2 / R = I^2 * R : eq1, 2
= Vpk^2 / (2 * R) = Ipk^2 * R / 2 : eq3, 4
= Vpp^2 / ( 8 * R) = Ipp^2 * R / 8 : eq5, 6
so the example from eq5 gives 60 * 60 / (8 * 8) = 56.25W
Hi,
a squarewave puts out exactly twice as much power as a sinusoidal voltage if the squarewave bandwidth extends to infinity.
For reference the extra power of the square wave is all above the second harmonic which is why tweeters burn out in a clipping amp. So a 100w single frequency 1kHz signal into a two way passive speaker should put almost zero power into the tweeter but a similar drive voltage square wave will put 100w out at 1kHz and another 100w out at 3kHz and above (if perfect squarewave).
a squarewave puts out exactly twice as much power as a sinusoidal voltage if the squarewave bandwidth extends to infinity.
For reference the extra power of the square wave is all above the second harmonic which is why tweeters burn out in a clipping amp. So a 100w single frequency 1kHz signal into a two way passive speaker should put almost zero power into the tweeter but a similar drive voltage square wave will put 100w out at 1kHz and another 100w out at 3kHz and above (if perfect squarewave).
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