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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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