DC watts and AC watts

[rzenc]

Hi,
I am trying to understand how does DC watts relates to AC watts.... I have heard things like:
"... You must, at least, supply 2.8 times DC watts for X AC watts to obtain proper frequency reproduction without audible distortion on low end..."
Others have suggested at least 4 times DC watts for X AC watts...
So, for a practical example, a 100W tube amp would need:

1) 100W AC = 280W DC
2) 100W AC = 400W DC

I must say it's all non-technical talk, i.e., no references... I goggled DC watts to AC watts with no luck either.
As far as tubes goes, DC watts could be considered heat generated on plate/screen??
I grabbed a 6550 typical operating point from G.E. data sheet; values are for tube two tubes

Plate supply    = 450VDC @ 0.150A = 67.5W @ zero signal
Screen supply =310VDC @ 0.009A = 2.8W @ zero signal
Plate + Screen = 70.3W @ quiescence.
When pushed, data sheet says it can be:
Plate supply      = 450 @ 0.295A = 132.75W
Screen supply   = 310 @ 0.038A = 11.78W
Plate + screen = 144.53W

According to data sheet, it should be capable of providing 77W. It turns out that we seem to have 144W of heat to produce 77W of audio power?
So, when designing a power supply I must predict abuse and prepare it to withstand max signal conditions? How much headroom (extra current demanding) should be giving for real world use and abuse? Is necessarily bigger better??
I know it’s a helluva questions but I have being scratching my head with such crazy statements...
Also, if you have a reference, I will be glad if you can share it.

Thanks in advance.
Best Regards
Rzenc

[PRR]

> we seem to have 144W of heat to produce 77W of audio power?

More.

Add 6.3V*3A= 20W for cathode heaters.

Add 2W-4W for the driver tube to boost and invert "normal" signals to 30V peak in each of two 50K 6550 grid resistors.

Add 3%-30% for filtering losses from raw AC to smooth DC.

> we seem to have 144W of heat to produce 77W of audio power?

If we had perfect devices and measured as Square Wave, we could hit 100% efficiency (77W DC in for 77W square-wave out).

We don't have perfect devices. The 6550 is an excellent tube but electrons really do not like to flow through vacuum in large numbers.

A screen grid helps the electrons flow but uses some power which does not get to the output.

The audio world standardized on Sine Wave Tests. A sine is mid-way most of the time. Simple linear amplifiers do this by wasting the excess voltage.

Efficency of an ideal amp on Sine Wave is 78%. Rounding to 77%, we see that the DC demand for your 77W out is 100W.

The 6550 won't pull-down lower than ~~50V, which is ~~10% of the supply, so there is additional tube loss over 10%.

The screens take 12% of plate current, that's another 12% on top.

The THD number in that sheet is very low; it is likely there is 80W-89W output available at very little more DC power, if you don't mind higher THD.

> How much headroom (extra current demanding) should be giving for real world use and abuse?

Cheat. Find a happy commercial amp with output like what you want. Weigh (or eye-ball) the heavy iron. Get iron about that size.

Note: same-application commercial amp! Hi-Fi amps not regulated by US FTC and not exposed to "full power sine" testing can be built MUCH lighter. Industrial motor drivers were often built quite heavy, because they could carelessly be over-loaded for hours, and the cost of more iron is less than the cost of breakdown and lost production. A guitar amp may be asked to output more than Sine Wave power (gross square clipping) but rarely for more than a few bars a minute. i.e. if building stage amps, steal from Fender, Marshal, Sunn.