In a message dated 2/22/05 9:23:47 PM Central Standard Time, KD7QAE@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.
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