Right, Nick, let's go back several steps and DESIGN a basic output stage from first principles.
Audiophiles might want to look away, now.
Lessee... A nice table radio might do 4 watts, I suppose, and the HT might be around 250v running on load.
These two guesses do rather a lot for us. Let's assume that the resistance of copper wire forming the output transformer primary is fairly low, low enough to neglect the ohmic losses at the fairly low currents involved. This is a reasonable assumption. So the average voltage on the anode of the valve must be 250 volts! Running with a sinewave signal, the anode voltage will swing symmetrically to either side of 250v.
Time for another assumption. A real world valve isn't going to be turned on hard enough to pull its anode right the way down to zero volts, and if you got anywhere near zero, you'd have hellish distortion (other adjectives are available. A guitarist might even like it)
So let's make another assumption, let's say the anode only gets pulled as low as 50v above ground.
So the max voltage across the primary is going to be 250-50 = 200v
But as the waveform has to be symmetrical, when the valve is almost turned off, the primary voltage on the transformer must swing as far the other way... -200v across the primary, so the anode must go up to 450v.
But the power supply is only 250v, so where does 450v come from? Well the half of the wave where the valve is pulling its anode voltage down below the 250v HT passes current through the primary winding and charges up the field in the inductance that winding presents to the world. This field is stored energy, and when the valve is driven to lower than quiescent currents, the transformer gives up its energy and drives the anode voltage ABOVE the HT voltage. (This is why single ended power output stage transformers have a gap in their cores to stabilise the inductance value and to prevent saturation)
So our anode voltage is swinging sinusoidally between extremes of 50v and 450v, with an average value of 250v, so there is no dc component of voltage across the primary winding.
So we've made a 400v peak-to peak sinewave. That's 220v peak or 200 * 0.707 = 141.4 volts RMS across the primary.
Now say the speaker is 3 Ohms, and we want 4 Watts. Power = Vsquared over R. R is 3, power is 4 so voltage squared is 12, so the RMS voltage across the speaker is root(12) which is 3.5 as near as makes no odds. Now we have the turns ratio of our transformer because it's the same as the voltage ratio. 141.4/3.5 so we need a 40.4:1 ratio.
Easy?
Now what about that transformer? We want to have some nice bass. We don't want the transformer limiting whatever the speaker and cabinet can do. Let's decide to make the low frequency roll-off be 3dB at 30Hz which would make for a good quality table radio.
Now that 3 Ohm speaker seen through a 40.4 to one ratio transformer looks like the turns ratio squared times the speaker impedance if viewed from the primary.
so 40.4 squared, times 3 is 4896 Ohms. For that 3dB roll off we need the reactance value of the inductance of the transformer primary to equal this at 30Hz.
Reactance is 2 * Pi * F * L =4896 we know F is 30 so we can rearrange and calculate L
So L needs to be 26H which is quite an amount of inductance. If we relaxed that 30Hz choice, and allowed the bass to roll off higher, the inductance required would scale down pro-rata.
Hey, but here we are, we've just designed the requirements for the output transformer and no-one's even mentioned what type of valve we might be using
This is why output valve substitutions work so well without people changing output turns ratios. The turns ratio is set mostly by the choice of HT voltage, by the wanted amount of power and by the speaker impedance. Different valves might change that assumption of taking the anode voltage down to 50v, but that has a smaller effect on the transformer than the other parameters.
Now, what about anode current and biasing?
Power = Isquared * R
4W into a 3 Ohm speaker gives us 1.15 Amps RMS. now that is plus or minus 1.633 A on the peaks..... into the speaker.
At the transformer primary, this is reduced by the turns ratio so 1.633/40.4 gives us plus or minus 40mA on the peaks. Remember, this is the signal current component and adds with the quiescent current.
We don't want the valve current to cut right off on the peaks, or it flat-tops and we get nasty distortion again, but this calculation is with it going right to the 50v minimum anode voltage, so we can play just as close at the other end, but let's leave it 10% of the signal current, 4mA so to get 4ma left when the signal is -40mA, then the quiescent current needs to be 44mA!
And we still haven't involved any specific valve yet.
So we want one which can run at an average 250v and survive 450v peaks
We want one we can bias at 44mA and will handle an 84mA peak
The 250v and 44mA gives an average anode dissipation of 11 Watts
So how about that 6V6, then?
DC anode volage long term 315v that'll do!
Pek anode voltage 1200v (even if triode connected) that'll do!
Anode dissipation 12W yup, that's in.
Peak cathode current 105mA yup, that's plenty to allow some screen current.
Now we've got these, what about the transconductance? That essentially sets the gain you get, which tells you how much oomph you need in whatever drives the chosen valve.
The next step is to look at the curves and pick a cathode resistor to give us 44mA quiescent current.
The output-side work can be done by laying a load line across a set of characteristic curves, and it'll make the onset of distortion more visible, but these wet-finger calculations will do the job.
David