25-11-2016, 01:11 AM
(24-11-2016, 05:58 PM)Terry Wrote:(24-11-2016, 02:43 PM)Mark Hennessy Wrote:(24-11-2016, 01:14 PM)Terry Wrote: To calculate the peak output power, take the supply voltage and divide by 3 (strictly speaking it should be 2.828 but the supply will almost certainly sag at maximum output) and square the answer. Then divide by the speaker impedance:
P(W) = ((Vs/3)^2)/ZΩ
That's not the peak power; rather it's the average sine wave power. What some folk erroneously call "RMS Power"
I was well aware of that Mark!
So why not say "take the supply voltage and divide by 2" rather than what you actually said (which amounts to "2-root-2 plus a bit")?
Clearly, you had "Watts RMS" in mind when you wrote that. Not "peak output power".
(24-11-2016, 05:58 PM)Terry Wrote: However I felt it would be a better figure to compare with published data on power ratings.
At that stage, we hadn't looked up the published ratings. That didn't happen until I did that (post #11).
(24-11-2016, 05:58 PM)Terry Wrote: I did include the peak current figure as well, though!
Yes, which is what matters here for the sake of the output transistors. Which I didn't question, BTW. Indeed, I provided the analysis that you could've done...
(24-11-2016, 05:58 PM)Terry Wrote: I ignored emitter resistors but I'd also rounded up the RMS conversion figure so the answer won't be 100% accurate anyway but as it will, in fact, be a higher figure than actually achievable it allows a degree of safety.
Yes, but even with the "engineering rounding" - which I have no problem with - you were talking about "Watts RMS" rather than "Peak Power".
Either way, the question has been answered.
