To keep it simple you are comparing a DC power calculation to a AC (audio is a ac signal) power calculation. As the total supply voltage avalible is only 5vdc, the theoretical maximum audio power available where the whole peak to peak waveform can 'fit' within a 5vdc limit is: 5vpp = 2.5v peak = 1.767 volts RMS
I looked up RMS on Wikipedia and I found the following on it:
So what you're saying is if I input 0..5v into the amplifier it outputs -2.5..2.5v
And to calculate the power I need to calculate the RMS amplitude. So I take the amplitude which is 2.5v, and I divide that by sqrt(2) which is 1.414, and I get 1.768v for my RMS amplitude.
and then audio watts would be (ExE)/R or (1.767X1.767)/4 ohms = 3.12 watts rms. As the chip's amplifier is not 100% efficient the 2.75 watt figure says it's running at 88% efficiency, not bad at all.
You lost me.
1.76^2 / 4 = 0.78146
I can't figure out how you got from there to 3.12 watts rms. And if you made a mistake and the .78W figure is correct, why isn't that anywhere near the quoted 2.75W?
[edit]
Hm...
3.12 * 4 = 12.48 and the sqrt of that is 3.53, and half that is 1.76... So it looks like you did (1.76*2) ^ 2 / 4 = 3.12, but I don't know if that was a mistake or not, and if not, why you multiplied by 2, unless you're doubling the RMS amplitude because you need to account for both halves of the waveform?