In message , Rob
writes
FranK Turner-Smith G3VKI wrote:
"gareth" wrote in message
...
"gareth" wrote in message
...
One thing that puzzled me for years was that the output impedance of
an amplifying stage could be much higher than the voltage drop
across the active element divided by the current through it.

Now, clearly as current sources, this would be true, and it could be
measured.,
by changing the loads.

But, how could you calculate such a value up front?

Interesting how the great blusterers display their ignorance, either
by not entering the discussion by the device of side-stepping with abusive
remarks*****, or trumpeting complete nonsense ....

Consider the case of your homebrewed 1.5kW linear being set up with QRP
for safety.

Let's say 6W output on a 12V supply. Some claim that this means an
impedance of 24 ohms that has to be matched to the 50 ohm antenna.

So, having done your matching, you now crank up the output to 1.44kW
by increasing the drive, and those some claimers will now say that the
ouput impedance is now 0.1 ohm.

Hang on there! By their argument, you will now be in danger of severely
damaging your pride and joy by driving into what you say is a complete
mismatch, set, not for 0.1 ohm but for 24 ohms.

You would need to increase the supply volts to at least 200V to give
headroom, unless you want a square wave output.

It is a widely distributed misunderstanding that a linear that has
been designed to delever its output in a 50 ohm load always has a
50 ohm output impedance. In fact it almost never has.

My understanding of RF design doesn't include how to calculate the
output impedance of PA stages. However, it's obvious that of the output
impedance was 50 ohms, the PA efficiency would be only 50% at best - and
clearly this is not so (certainly for class-C operation, which is
notionally 66%-ish). I've always understood that, in practice, it's
usually resistively low-ish (around 25 ohms?) and quite capacitive. An
expert opinion is needed!
--
Ian