"Ian White, G3SEK" wrote
Reg Edwards wrote:

The only problem is that caused by entirely
unnecessary complications
introduced by people being too clever.

First design a single-ended, class-A, audio
amplifier like a 6V6
beam-tetrode tube plus a transformer to drive an
8-ohm speaker.

Then calculate the source resistance seen from the
speaker looking back
into the amplifier.

To save time searching for manufacturers data,
assume:

DC plate supply is 250 volts.
Peak signal at plate 200 volts.
Internal plate resistance, from mnfr's curves, is
100 K-ohms. Audio
power output = 5watts into 8-ohms load. Peak volts
across load = 8.944
volts peak. Transformer turns ratio = 22.36

Then ask yourself "Is the source resistance of the
amplifier in the
same ball-park as the 8-ohm speaker ?"

Is there anything like a conjugate match ?

Which is what it's all about.

For Class-B and Class-C conditions, just multiply
internal plate
resistance by a constant which depends on plate
current operating
angle. Curvature of tube characteristics can be
taken into account for
a higher order of accuracy.


Kind of as expected: Reg chops the problem down to
the easy bit,
explains that in detail, and airily dismisses the
hard part in one final
paragraph.

And all this was sorted out in the early 1920's.

And that problem-solving method - make the problem
fit the solution, and
chop the rest off - was invented even longer ago, by
an ancient Greek
bandit.
========================================

Life would have been made less complicated for you if I
had stopped at "Which is what it (the foregoing) is all
about."

After all the years of haggling on this newsgroup are
you not yet convinced the internal resistance of radio
transmitters is not 50 ohms?