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Old June 7th 10, 05:19 PM posted to rec.radio.amateur.antenna
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First recorded activity by RadioBanter: Jul 2006
Posts: 644
Default what happens to reflected energy ?

On Jun 6, 7:29*pm, walt wrote:
....
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


If you allow that the thing driving the pi network has an effective
source impedance different from the load that the pi network presents
to it, then clearly the output impedance seen at the other end (the
"50 ohm" end) of the pi network won't be 50 ohms. Try it with some
numbers; for example, assume a pi network that transforms your 50 ohm
load to a 4k ohm load to the amplifier output, and assume an amplifier
output stage that looks like a 20k ohm source. Design the pi network
for a loaded Q of 10. I believe you'll find that the source impedance
seen by the 50 ohm load is about 11+j18 ohms.

As Wim has pointed out, requiring an amplifier to be loaded and driven
in a very particular way unnecessarily dismisses some very important
classes of amplifier. What do you do, for example, with a linear
amplifier? What do you do with an amplifier that drives a voltage
very hard (and for which a simple pi network is inappropriate for
matching to a load)? Perhaps an even more basic question is: why
exactly do we tune a pi network to present a particular load to an RF
amplifier stage? Why should we operate a 6146 with, say, a 3000 ohm
plate load? Why not 1000, or 6000?

And what if I set up a tube and pi network for operation such that the
apparent output source impedance is 50 ohms (while driving a 50 ohm
load), and then I add feedback to the amplifier in such a way that the
operating conditions are not changed, but the impedance looking back
into the plate is changed?

How did we get to the source resistance of "the (tube) power amp"
being non-dissipative? I know there are some of us who don't buy into
that...

Cheers,
Tom

 
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