Yes, but if there are two, three, or four channels driven then there is also 325 Watts continuous power available to each channel. When 5 channels are driven the number drops to 300 Watts available continuously to each channel, since there is only 1500 Watts that can be had out of the wall. My question is: If 1500 Watts can be split up between the channels so that there is approximately 300 Watts continuously given to each channel, then why can't there be 1500 Watts continuously given if there is only one channel driven? Or 750 Watts if only two channels are driven? Was this limited to 325 Watts continuous since there are some speakers that cannot handle more than than on a over a period of time, or am I missing something? I still don't really understand.
Ok, I can see where you are having trouble. It's a little difficult because there are two factors at play here. On the one hand we have this large power supply that can supply lots of "power" to how ever many channels need it. On the other we have to look at the amplifiers themselves. By definition, an amplifier cannot swing or deliver more than the available DC power supply voltage to a load (speaker). Once we take losses, including the amplifier's efficiency, into account, the available voltage swing is reduced even further. It's this voltage swing that determines the maximum, continuous output power that the amplifier can deliver into a given load, assuming that we have unlimited current available. So, even though we might be able to deliver 1500 watts from the power supply, the amplifier is limited by the rail voltage.
Time for some quick math:
Let's assume P is power, V is voltage, I is current, and R is resistance. I'm going to leave the complex reactive speaker impedance out of the discussion and assume the load is purely resistive!
P = V x I
From Ohm's Law we know that I = V/R
Substituting this equation into the first equation we get:
P = (V x V)/R
In the A1500 we are running +/-85Vdc rails, which means the peak to peak voltage is 170Vdc. However, we can only swing voltage in one direction or the other at a given time. This means we need to use our peak voltage, which is 85Vdc. Power is usually, somewhat erroneously, specified in watts RMS, so we need to convert peak voltage into RMS voltage. To do this we multiply the peak voltage by the square root of 2 (0.707).
Therefore, Vrms = Vpeak x 0.707.
In the A1500 this gives 85 x 0.707 = 60.095 Vrms.
Going back to our original power equation, our maximum output power into 8ohms is:
P = (60.095 x 60.095)/8 = 451.23 watts
But wait a second, this assumes that the entire amplifier is 100% efficient, which it's not. We need to apply the approximate efficiency to the RMS voltage, and this is 90% in the A1500. This gives an actual RMS voltage swing of
V = 60.095 x 0.9 = 54.0855
Redoing our power calculation now gives:
P = (54.0855 x 54.0855)/8 = 365.66 watts
So this is the maximum single channel output power, no matter how much current is available from the power supply. Why does Ian's table state 325 watts for one channel? Because we're still in the preproduction stages of these designs and we wanted to be conservative until the final numbers are in.
I hope that was not too hard to follow and helps to clarify the continuous power ratings.