[AMPS] Series Tuned Tank?

[Tom Rauch]

> 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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