24-11-2016, 02:43 PM
(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"
In practice, the emitter resistors and the preceding electronics results in rather less output. E.g., a typical Hacker output stage running at 18V only gives 1.5W into 15 ohms, rather than the 2.4 or 2.7 watts predicted. These losses will generally increase when driving lower impedances; no amplifier doubles its power into half the impedance.
The AC128/176 pair is good for 1A. So on paper, anything up to 8V peak will be fine, suggesting that a 16V PSU rail would be a sensible max - however, in practice it's probably OK up to 20V or so, given the losses.
But let us not forget that the speaker is not an 8 ohm resistor. It will only be 8 ohms at some frequencies. For more, it'll be substantially higher, and either side of resonance, it'll be a little bit below. So the actual peak current delivered will be dependant on the spectral content of the music.
The lack of bass is most like because the drive unit has a much lower Q than the original. In the correct cabinet (right volume, perhaps with a port), it might well be a decent enough unit. But very different to the original. And I suppose the output coupling capacitor might be a bit on the low side for an 8 ohm loudspeaker, but that'll depend on what it is.
It'll be worth feeding in a signal from an MP3 player or similar to rule out a problem with the cartridge...
