On 7 jun, 04:29, walt wrote:
On Jun 6, 10:06*pm, K7ITM wrote:



On Jun 6, 3:21*pm, Wimpie wrote:

Measuring method used: change in resistive load, from voltage change
you can calculate the current change, hence the output impedance. One
note, except two, all where solid state.

Of course, since the source impedance may be anywhere in the complex
plane, you need to change more than just the resistive part of the
load, I believe, to get an accurate picture...

All high efficiency designs (class E, D) that I did have output
impedance far from the expected load impedance. With "far" I mean
factor 2 or factor 0.5. I did not measure that (as it is not
important in virtually all cases), but know it from the overload
simulation/measurement and I did the design myself.

What I've seen in similar situations is that the source impedance is
likely to be strongly reactive, depending on the filter network you
use to get sinusoidal output. *In any event, the source impedance is
likely to have a reflection coefficient magnitude that is quite close
to unity. *That is exactly what you should expect: *there's nothing to
absorb reflections. *You could (theoretically at least) use feedback
to make the output look like 50 ohms, but just as you say, Wim ...
why?? *There's really almost never any point in doing so.

...

Cheers,
Tom

Again Wim, we're not on the same page, so let me quote from your
eariler post:

"Virtually all high efficient amplifiers work in voltage saturated
mode
and are therefore not operated at maximum available power, therefore
their output impedance doens't match the expected load."

I have not been talking about MAXIMUM available power, only the power
available with some reasonable value of grid drive. In the real world
of amateur radio operations, of which I'm talking, when we adjust the
pi-network for that given drive level, we adjust for delivering all
the AVAILABLE power at that drive level. When all the available power
is thus delivered, the output source resistance equals the load
resistance by definition. We're now not talking about changing the
load, phase or SWR--we're talking about the single condition arrived
at after the loading adjustments have been made.

Walt, this restriction was not given in the original posting, it is
added by you and you exclude many other practical amplifier solutions
(like push pull 3...30 MHz no-tune amplifiers, fixed tuned single band
amplifiers, emerging high efficiency designs for constant envelope
modulation schemes and/or AM with supply voltage modulation).

I fully agree with your statement on optimum matching given certain
drive and output impedance in case of (pi-filter) matching, no doubt
about this. However there are more flavors as I tried to explain.

Even during SSB modulation with an optimally tuned amplifier without
power supply modulation, the amplifier is most of the time operated
into current saturation mode (instead of optimally tuned output, given
a certain drive). For tetrode/pentode, assuming no voltage saturation,
the RF plate impedance is seldom equal to the conjugated load
impedance (load impedance will be lower). I think for a triode in
common grid operation, this will apply also because of the cathode is
not grounded for AC, hence plate impedance increases. I have to be
careful now to avoid that I have to do a lot of work to fulfill
Richard's requests. I dismantled my PL519 common grid amplifier (22
years ago), so a cannot measure it anymore.

All amateurs that do not have a tuner inside their PA, have to live
with non-optimum VSWR (hence amplifier not operating at optimal
tuning, even under CW) or have to insert a tuner. For the latter case,
the problem now reduces to a cable and tuner loss problem, as after
successful tuning, there will be now power reflected to the PA, hence
the PA's output impedance doesn't matter.

Maybe the JC could provide some background behind his question.


And Tom, why would the source be reactive when the pi-network is tuned
to resonance? And because the source resistance of the (tube) power
amp is non-dissipative, its reflection coefficient is 1.0 by
definition, and so it cannot absorb any reflected energy, and
therefore re-reflects it.

Walt, W2DU

Best regards,

Wim
PA3DJS
www.tetech.nl
without abc, PM will reach me