Hi John,
> Absolutely correct Tom. We choose to use 50 ohms as a
convenience because
> transmission lines are usually 50 ohms today.
>
> However consider a high power audio amplifier driving a load of 4 ohms.
> The output Z of the amp isn't 4 ohms - more like approaching zero ohms to
> be able to deliver the required current!
That's right John. The design criteria is different.
> I think where some folks are going wrong is that they have lost sight of
> what Z actually is. You can (within reason) make a 50 ohm Z from a myriad
> of combinations of L, C and R but in only case - the series resonant
> condition - the resistive portion is actually 50 ohms and the reactance
> of the L & C cancel out. For lots of others the value of the apparent R
> is very low - like the audio amplifier example above. Power cannot be
> dissipated in a reactance, only in a resistance. So the "dissipative Z"
> theory immediately falls flat on its face because the R portion isn't 50
> ohms! Clearly in a practical world ALL components have some value of R -
> due to loss mechanisms. So defining the efficiency of a rf PA isn't easy.
> I am going through that scenario at work at the moment!
It is easy. Just use a high power 50 ohm attenuator and tweek the
PA with a little reverse RF near the operating frequency (so the
tank or harmonic filter is not in cut-off and shows the designed
impedance) and measure standing wave voltage of the reverse
generator on the line between the attenuator and the PA at several
points.
You'll be able to calculate SWR and impedance by measuring that
voltage while the PA is operating.
Or you could do a load pull.
> I had this exact discussion with Prof Abrey of Pretoria University some
> years ago and he explained that depending on the configuration of the
> amplifier the output Z could either very low or very high. Transistor
> amplifiers are predominately low Z and have very low collector impedances.
> Tube amplifiers on the other hand usually have highish impedances - as
> witnessed by the load-line resistance.
I have no idea what he is talking about, unless it is the plate
resistance which is something you don't match into anyway.
If indeed the source impedance is very low, power would increase
steadily when load impedance is reduced. It does not do that in
conventional PA's.
If the source impedance were very high, power in the load would
steadily increase in power if load impedance was raised. It does
not do that either.
A load pull is a valid way to determine source impedance, and it
certainly fails to confirm the fact high efficiency PA's are greatly
mismatched. As do reverse power voltage measurements along a
line.
One shouldn't lose sight of all the stray reactances in a tube, the grid
> to kathode and plate to grid capacitances can be fairly high and at high
> frequencies these come into the equations. Similarly the inductive
> reactances, although not often considered at hf, are still there.
One shouldn't loose sight of the fact the tube is a time-varying
resistance that is out of circuit for much of the cycle in a non-class
"A" PA, and that it is power limited by the maximum available
power in the energy conversion process. That power limitation does
NOT need to be from a dissipative resistance in series with a
limitless voltage source, or a resistor in parallel with a limitless
current source.
The system is non-linear at the tube end of the tank, and that rules
out using any theorems on the tube side of the tank. Class "A"
PA's would be an exception, except the model still won't give
dissipation because it makes no allowance for quiescent current.
All this stuff about something being 50% efficient when conjugately
matched, or adjusted for maximum power transfer, is nonsense.
73, Tom W8JI
w8ji@contesting.com
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