[Amps] This may be an oversimplification on my part, but it seems to work.

[Dennis12Amplify@aol.com]

In a message dated 2/22/05 9:23:47 PM Central Standard Time,  KD7QAE at ARRL.NET 
writes:

Dennis,

When I worked an example for 1.8MHz I got reasonable  numbers for C1 and 
L but C2 was 10x higher than what the PI-L program  yields.  If you 
divide C2 by Q it is about  right.

Tomm



Tomm,
 
 That sounds kinda backwords to me!
 
 I would have suspected that the Q would be higher as C2 was made  larger, 
but would have been determined primarily by the C1 Capacitance  value.
 
 Since my output capacitors don't unually have a 10 to 1 adjustment  range, 
and I was able to tune to full power output, I don't believe I ever had  that 
problem in actual operation. Maybe it was because the Q of my output L  network 
was lower than the Q of the input L network. 
 
 In any case, since C1 is the smallest value of capacitor (many times  
smaller than C2), I would guess that the overall tank circuit Q would be  determined 
primarily by that capacitor, and the C2 capacitor would only be there  as 
part of the capacitance voltage divider that was approximately equal to the  
impedance transformation ratio.
 
 Once again, I am not a math major, but it had worked for me in the  past. I 
was hoping someone more educated than I could explain why it  worked.
 
 I must admit though that I have never attempted to use it  below 14MHZ....
 
Regards,
 
Dennis O.