[Radio Wrangler]

There are a couple of questions you haven't asked, and as they set the foundations of later things, it helps to have them out in the open. Let's take a couple of steps back, and look at things from a different direction:

What is the function of the output transformer. Why have one at all?

1) It acts as a safety element, allowing the loudspeaker connections to be at or near ground potential, isolated from the high voltages needed by valves - even at flea power.

2) The loudspeaker impedance is inconveniently low compared to the optimum impedance that your valve(s) should be loaded with. The transformer lives up to its name and transforms the impedance your speaker presents into the impedance your design work says you should load the valves with. Your valves could drive the speaker impedance directly, but the power you could get would be very low and the efficiency would be lousy.

3) The transformer allows push-pull amplifier operation without needing (or faking) complementary devices. You can make push-pull amplifiers with one valve pulling up and another pulling down, but they are very complicated. Having a push-pull transformer is the easy way out.

Point (2) above raises the question; 'what is the optimum impedance to load your valves with?'

To answer this, you get a plot of the anode characteristics of the valve you've chosen and decide what is the maximum anode current you want to run the valve to, and what is the maximum anode voltage.

Now it's time to pin the tail on the donkey... well, two tails, actually.

For point 'A' find the horizontal level of the max current you've chosen, and pick a point along it where the anode current has come out of the curvy bit at the low voltage end, and is looking flatter.

For point 'B' the current is zero for a class B design, (or a bit above zero for class AB).
Pitch the voltage of point 'B' halfway between the voltage of point 'A' and your chosen max anode voltage. The voltage of 'B' is also the HT voltage from your power supply. B here is the name of a point on a graph (Americans use B+ as their name for the high voltage supply rail where Brits say HT. This is just coincidence, I could have named those points Fred and Charlie).

Using a ruler draw a line through 'A' and 'B'. Extend it rightwards up to your chosen max voltage. This will go into negative anode current regions on your graph, which looks silly because valves don't conduct in reverse. But this is a push-pull amplifier and the 'negative' current is current from the other valve, reversed by the transformer.

The slope of the line 'A' to 'B' shows a linear drop in voltage versus current... A linear resistance from the power supply.

It is this resistance that the output transformer transforms the resistance of your speaker to make.

Have a general read up on 'load lines' now and it should make things a lot clearer about what is going on. You are NOT trying to match the anode impedance of your valve. If it comes close, it is only coincidence, nothing more. You are setting the impedance presented to the valve. Given an HT voltage, and a number for the amount of power you want, the presented load impedance comes out of the maths with relatively little influence of your choice of valve. At first, this looks surprising, but it doesn't half make life easier.

Once you've found the load line impedance, you get your transformer turns ratio from it and your speaker impedance.

Next, you need to know how much inductance you need to go as low as you want in frequency. Primary inductancetimes 2 times Pi times Frequency = Load line resistance will give you your FL -3dB point. Pick a -3dB frequency and plug in the numbers to get your primary inductance..

You now need to design a transformer for this inductance, with a core big enough to be clear of saturation with a primary current od point A on your load line.

It all fits together like a jigsaw. I was a bit concerned that you were looking too intently at the pieces and not at the picture on the lid

David