"arie" wrote
- Frequency = 1.8 Mhz
- RF power = 1400 W
- RF peak amplitude = 5400 Volts
- External load res = 50 ohms
- Phase shift = 90 degrees (to get a Q of around 12)
- intrinsic Coil Q = 100

I get:

C1 = 122.5 pF
L = 63.80
C2 = 122.6 pF

Is sounds very strange to me that both C1 and C2 have the same
value whereas the Load resistance at anode is around 10414 ohms
or at least much higher than 50 ohms and an external resistance
of 50 Ohms

What about the Load resistance at anode, I would think this
should be somewhere aound 2800 ohms, considering the Power
output and anode voltage swing.

====================================
Hi Arie,

Circuit-Q is a dependent variable. It depends on nearly all program input
parameters. It's value cannot be entered in the program because you don't
know it. You know only what you would like it to be. It can be calculated
accurately only AFTER parameters such as efficiency have been calculated.
You may not be directly interested in phase shift but by entering it in the
program it does allow you INDIRECTLY to control circuit-Q.

To design an amplifier starting with values of Circuit-Q, DC Plate volts,
Plate current operating angle, etc., there is no alternative to using the
program to produce a multi-dimensional table of results. And even then they
would be approximations.

A phase shift of 90 degrees through a symetrical Pi-Network should set bells
ringing. With all 3 reactance magnitudes being identical the network
behaves as a 1/4-wave transmission line transformer. Terminating impedances
can be anything you like provided Line Zo = Sqrt(Z1*Z2).

You will notice you are not allowed to enter a phase shift as high as 180
degrees. That would correspond to 1/2-wave length of line which matches any
impedance Z only to itself. It then makes itself redundent. The program
would object to being asked to do something silly and would abort into the
operating system. The program user would then blame the long-suffering
author and not himself. ;o)
----
Reg, G4FGQ