[FullRangeMan]

Sorry, but this post is an incorrect application of the formula. In an audio system the power is determined by the current delivery capability of the amp into the load (speaker) being used. You cannot randomly have X amount of current flowing in any old load when applying voltage Y.

The formula you would use is V²/R where V is the voltage measured across the load and R is the nominal impedance of the load.

And yes, tube watts produces exactly the same heating effect (power) in a load as solid state watts. They are exactly equivalent. However tube amps conceal their distortion better because they don't clip hard. They just start sounding compressed.

Also, tube amps SOUND different into speaker loads as they have higher output impedance and therefore their frequency response will vary into the load more drastically than that of a solid state amp.

But it is also OK if you don't believe in physics,, each to his own 

Where is the error in these calcs below?
Example of different structures of a 30W figure:
Amp X 10V x 3Amp=30W
Amp Y 15V x 2Amp=30W
Amp Z 7.5V x 4Amp=30W
Do you would show us how is the calculations of these same 3 values above with your formula?

[Steve]

 "I've found that the gain specification of an amp is the least significant issue."

I have found this statement trouble some.

1. With lower amp gain, I have less of a ic "hum pickup" problem, especially with ultra low capacitance, non shielded ics.

2. I want a minimum of stages in my system, less sonic degradation. My amp has only 16db of gain, and two total stages.

My 11A has just 1 stage, 20db of gain. I also got rid of a gain stage in my cd player, and still have 1 vrms output. Why design a cheap, junk gain stage of 2 (6db) in the cd player? 

The total gain of my system still allows me to set my volume control to ~ noon, not 8am as in high gain systems.

Tube amps tend to clip with lower harmonics vs hard clipping of most SS amps. If one system has higher gain than another, the higher gain will be louder for the same volume setting. But that does not tell us which system has the highest power output rating.

cheers

steve

[raindance]

Where is the error in these calcs below?
Example of different structures of a 30W figure:
Amp X 10V x 3Amp=30W
Amp Y 15V x 2Amp=30W
Amp Z 7.5V x 4Amp=30W
Do you would show us how is the calculations of these same 3 values above with your formula?

Your formulas are correct for calculating power given voltage and current. But you cannot have current flow unless you have a load. You also can't measure current through a load at all possible frequencies easily. Voltage applied across a given load produces current flow. So measuring just the voltage and using a simplified representation of the complex impedance of the load (R), you can come up with a close approximation of power in watts.

Applying your examples:
Amp X - your load resistance would be V/I = 3.33 ohms. Using my formula (10²/3.33), we get 30 watts.
Amp Y - your load resistance would be 7.5 ohms. Using my formula we get 30 watts.
Amp Z - your load resistance would be 1.875 ohms. Same deal, 30 watts.

The first two are feasible for speaker loads, the last is not. Ohms law always applies. You cannot have a power calculation without taking the load into account.

[FullRangeMan]

Thanks Raindance. The Joule Law are 3 formulas that relate power, resistance, current and voltage, I used the first that are simpler also used to calc power transformer capacity, you used the second.

P = V x I or V x A

P = V2 / R

P = R x I2