> The results of modeling with a serial inductance from plate to PiNet
> are in...sorta. I still couldn't find a combination that lowers the Q.
> Here's a table of results. The following table is for a single 813 at
> 2250V and 145 ma, class AB, plate load of 10K. Cout for the tube is 18
> pf, L1 is the series L (uH), C1 is the PiNet tune cap (pf), L2 the pinet
> inductor (uH), and C2 the pinet load capacitor (pf). In each case below,
> the network matched from 50 Ohms to 10K.
I think I see now what you are doing. You are counting the output C
of the tube as the exclusive component input for the pi input, then
you have solved for an output of 50 ohms.
When I'm speaking of a minimum Q system, I'm talking about an
adjustable network with two variable capacitors allowing operation
on multiple frequencies and multiple bands.
Solving for L in the case I was describing gives and L of 1.8 uH,
followed by a C of 254 pF, followed by an L of .127uH, followed by
another C of 254 pF.
This, if the capacitors are adjustable, gives you a minimum Q
network that is adjustable and switchable in the "normal" way
amplifiers are. The L section has a Q of 31, the pi-section a Q of
about 2 (I forgot the exact value).
Obviously if you have no requirement to move the PA to different
load impedances or frequencies, a simple system with only an L
and C will be the lowest Q solution.
The solution I'm thinking of has a Q almost the same as the two
component system (the "virtual" pi you are thinking of), except
greater flexibility. If you have a switch and tuning cap hanging on
the tank, then the solution I gave provides the lowest Q.
I haven't looked past these two solutions.
73, Tom W8JI
w8ji@contesting.com
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