A Pi network is back to back L's.
My assumption has always been that the dip happens because you pull less current at resonance than you do on either side. When your running out of resonance, you're still driving the tube to get X amount out of it (usually grid current measured here). Transistors are the same way. Don't tune X out of the broadband xformer, and you end up with blown xsisters... Either from much too much I being pulled, or too much V (reactance can cause wild voltage excursions).
Pretty much sounds like your answer, once it's down on paper, too.
Input impedance is usually found either on the internet, by doing the math, or data sheets. If you have an MFJ antenna analyzer, you can tune the amp up for max output, then connect the mfj to the ANT jack, and put a pot on the tubes (all voltages off here, unplugged, relay energized (manually, if you ask me). Tune the pot for 50 ohms indicated on the MFJ. Remove pot, measure value. That's your loaded value of the input impedance for X amount of tubes. Simple shyt, really.
--Shane
Well thats true, in the past I found that to work as long as the max tube currents were adhered to.
But I want to understand the exact principles behind "dipping the plate". My understanding is that the PI network
is really an impedance transfer device to match the tubes 5K output impedance to a 50 ohm antenna system.
So the explanation my boss gave me about tuning out reactive current hence the dip seems the most plausible so far.
Also I would like to know how to measure the input impedance so I can try different approaches with the 160 meter L circuit and not just double values. Does a tube spec sheet give this info?
I suppose I could measure the cathode RF voltage and current and calculate.
Regards
Q, W1QWT
