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Ö(2Watts / Resistance) = Peak Current
in this case Ö(2 x 60 / 8) = 3.872 A
The second step: how much idle current is needed to ensure that none of the output devices ever moves into cutoff? The math is even simpler than before:
Peak Current / 2.
In our example, roughly, 2 amps is what is needed. For a transistor power amplifier, 2 amps is a huge amount of current considering that the average idle current is usually on the order of 100 mA. For a vacuum tube power amplifier, 2 amps is an astronomical amount of current considering that 50 mA per tube is what is usually used.
Furthermore, even the beefiest, low-rp vacuum tube requires at least 100 volts for its cathode-to-plate voltage in order to conduct enough current to be useful; 100 volts times 2 amps equals 200 watts, which must be doubled in the push-pull amplifier and results in 400 watts of heat at idle. If we divide 400 watts by the number of output tubes, we will know what each tube will dissipate at idle. Eight 6080's or 6AS7's are what is commonly used, which means that each 6AS7 must be dissipate 50 watts of heat at idle in order to make the 60 watts of Class A power output. This is madness: the tube is only rated for 20 watts. Notice the inefficiency, 400 watts of heat for 60 watts of power output, 85% of the energy is wasted and this does not include the 100 watts the heaters consume.
Is this the end?
I am temped to say that this will be the last look into the Circular / Bridge amplifier, but I know that circuits are like fractal images, they allow infinite zooming. Not doubt this circuit will reappear in our journal. (In fact, I find the whole topic of floating power supplies fascinating and I can see writing an article on just on the use of floating power supplies: in high voltage regulators, power amplifiers and even curve tracers.) //JRB
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