How Do You Calculate Watts Per Channel?

Hk_fan, in case it's not clear my last point has nothing to do with impedance swings, and holds true even if the speakers are a perfect 8 ohm load at all frequencies.

To further clarify my last point, consider the situation in which the speakers are single-amped, using either a pair of 400W monoblocks or a 400W per channel stereo amplifier. Let's say that the speaker efficiency and volume setting are such that during peak passages of some recordings the full 400W capability of that amplifier is called upon.

Then let's change the setup to a pair of biamped 200W monoblocks per channel. Will that setup be able to play those same recordings at the same volume levels, without clipping or otherwise running out of steam?

The answer is that it will not, unless the frequency content of those musical peaks happens to be such that their power requirements at frequencies above and below the frequency of the crossover point are equal, which is very unlikely. So it would not be correct to consider the 2 x 200W biamped configuration as having the same power capability as a single-amped 400W configuration.

Regards,
-- Al

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So it would not be correct to consider the 2 x 200W biamped configuration as having the same power capability as a single-amped 400W configuration.

Al you precisely answered the question that I imprecisely asked. Thank you!
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In your first scenario the full 400W are being SHARED among all the drivers.

For sake of simplicity lets say there are 2 drivers, and assume that they have exactly the same impedance and efficiency. Therefore the two drivers will perfectly split the 400W, 200W per driver.

This is no different than individual drivers connected to a single 200W amp each.

Mitch, thanks!

08-06-13: Hk_fan
In your first scenario the full 400W are being SHARED among all the drivers.

For sake of simplicity lets say there are 2 drivers, and assume that they have exactly the same impedance and efficiency. Therefore the two drivers will perfectly split the 400W, 200W per driver.

Somehow my point is not getting across, and I'm not sure how to explain it any more clearly than I already have.

But in the case of your two-driver example with the 400W amplifier, the two drivers will NOT "perfectly split the 400W, 200W per driver."

The 400W will be divided between the two drivers based on the frequency content of the music, at any instant of time. That is what you seem to be overlooking.

If a bass drum beat is loud enough to require 400W, since most of its energy is at low frequencies the low frequency driver will get most of the 400W, assuming that no other loud notes occur at the same time. While if a cymbal crash is loud enough to require 400W the high frequency driver will get most of the 400W (just before it expires :-)), assuming that no other loud notes occur at the same time.

This is no different than individual drivers connected to a single 200W amp each.

It is very different. In the above situation the biamped 200W amplifiers will only be able to supply 200W to the low frequency driver when the bass drum beat occurs, and they will only be able to supply 200W to the high frequency driver when the cymbal crash occurs. Not 400W (or close to it, depending on the harmonic structure of the note), which could be supplied to that one driver by the single-amped 400W amplifier.

Regards,
-- Al

Somehow my point is not getting across, and I'm not sure how to explain it any more clearly than I already have.

Amen, Almarg!!
Your explanation couldn't have been more lucid....
thanks.