In reply to:
It really makes perfect sense.
Except that: P = V^2 / R (or P = I^2 * R)
Sorry to pick nits, Semi...
At any rate, the max power output of a solidstate amp is basically limited by its swing voltage (rather than the current) that can never go beyond the power-supply rail voltage. So, for a given max swing voltage (which is higher in a more powerful amp), the lower the load ohms (impedance), the higher the max power. This is as long as the amp can tolerate supplying the larger and larger output current into the speaker.
As shown in the equation, if the output current is truly a non-issue for an amp, then the max power at 4-ohm and 2-ohm load should be twice and four times that at 8-ohm load, respectively. Some expensive power amps, such as the Krell KAV series, actually meet this relationship, indicating that they use a very well-regulated power supply.
