I think I am beginning to understand all the many differences in regards to amps. So when one looks at amp reviews and they state how an amp sounds better than the other, it's not really that the amp sounds different, it's more how the amp interacts with the speakers? So for example if you take very high efficient speakers (>95dB), no matter what amp type you use it will most likely sound exactly the same.
But then let's take for example a pair of Mark and Daniel Ruby with the following specs.
Nominal Impedance: 3-6 Ohms
Average Efficiency: 82.5dB/2.83V/1m
Power Handling: ≧ 80 Watts per channel
As you can see they dip down very low and are not sensitive at all. So when you power them with a non class A amp, you are more likely to hear distortion from the amp as it does not have enough current to properly drive the speaker when it dips below 4ohm?
Unfortunately you still have not quite got it. I know this is difficult and may be I'm a poor teacher.
Class per se, does not have anything to do with an amps ability to deliver current into low impedance loads. Class just is a way of classifying the operating principles of amps, and to some degree their circuit topology.
I touched on some of the advantages and disadvantages of the classes of audio amplifiers.
An amps ability to deliver large amounts of current is determined largely by two factors.
The ability of the power supply to supply enough current to the output devices.
The characteristics, specification and number of output devices.
Please look at my math again in my first post in this thread last night. The big point is how, as the impedance drops, the amp has to maintain voltage at the output. The math will show you how much more current has to provided to maintain output voltage as the current drops. It also shows what happens if the power the amp can not maintain voltage in the face of falling impedance. You will see how this limits the amps ability to provide the power called for.
For Peng's benefit I used the term impedance rather than resistance. But most people are familiar wit Ohm's law and I hope I have not added another confounding factor. Just substitute I for R in the Ohm's law equation.
Please get out your calculator and try some examples for yourself.
I do hope this helps. I have noted so much confusion and misunderstanding on these issues in these forums. It is one of those topics we revisit again and again.
This forum needs some sort of search trigger that, before new threads are started on these difficult topics, what has gone before comes up. I just hope we don't have to keep rehashing this and other topics so many times. But may be I'm impatient and frequent revision is part of the solution.