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Well, it's not much.
You don't have to go deep in physics of it, etc.
Think of it as a formula.
This is not even math, this is arithmetic.
And that table Hammond provided, is even easier. The principle is, know your input parameters (output impedance and the load impedance), find the closest in the table and match them by connecting the appropriate leads.
People who do not drink, do not smoke, do not eat red meat will one day feel really stupid lying there and dying from nothing.
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I'm pretty sure I understand now. I watched an Uncle Doug YouTube video that helped explain this in dummy terms.
https://www.youtube.com/watch?v=5jUitplchok I would hate to do all that measuring with my transformer. So if I use the plate load impedance of PP 42 tubes at 10K and a voice coil of 1.5 ohms I could just use the chart I posted earlier that would tell me to use the 4&5 connection. If I used the PM speaker with voice coil at 5.3 ohms I could use the 6 ohm column and find the nearest row to 10K. If that is so I don't need all that arithmetic?
Please if you would tell me what happens when the match is not perfect.
Just to help me understand a little more about the 42 tube, is my radio class A or AB?
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>>If that is so I don't need all that arithmetic?
Exactly. They did the work for you. In this particular case you don't.
Yours is AB.
Class A is strictly Single-ended. It is a bit less linear and the least efficient.
Class B is the pushpull, the most efficient but with Zero quiescent (through) current, due to which it introduces crossover distortion.
Then the small quiescent (through) current, that flows through even without signal, is introduced. This takes care of the crossover problem with a bit of less efficiency. This is the Pushpull Class AB.
People who do not drink, do not smoke, do not eat red meat will one day feel really stupid lying there and dying from nothing.