24-11-2016, 05:58 PM
(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! However I felt it would be a better figure to compare with published data on power ratings.
I did include the peak current figure as well, though!
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.
Any speaker available can be used to check the amplifier although the power should be kept down if the impedance is lower than specified - particularly if it is a 4Ω one!
Higher impedance speakers will work without any restrictions but the output power will reduce as the impedance increases and would need to be taken into account.







