what happens to reflected energy ?
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.
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
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