The loading on the resonator is 1050 ohms in parallel with 5000 ohms, which
is 868 ohms. A coil or capacitor reactance value that is 868 / 5 ohms
(173.6 ohms) produces a Q of 5.

The initial value of L is

L = 173.6 / (2 * pi * 30MHz) = 0.92 uH

An additional loading is due to the resonator itself, which is the product
of the coil Q times coil reactance.Let's assume a coil Q of 200. The
resistance value of the coil itself is

Q * XL = 200*173.6 = 38720 ohms

The resistance loading by the capacitor can usually be neglected if it is
very small.

The loading on the output side by the combination of the 5000 ohm load and
the 38720 is 4428 ohms.

Go back to the beginning and get a slightly revised value of the resistive
loading = 849 ohms that includes the loading by the coil itself.
Recalculate L and C for this slightly different value of loading.

This iterative process can be stopped here.

Connecting the 50 ohm source directly to the coil through a 1000 ohm
resistor produces an additional power loss which is equal to

20 * log[(1000+4428) / 4428)] = 1.8 dB.

This additional loss may not be acceptable in some applications. A low-loss
coupling transformer (e.g. tapped coil) or low-loss tuned network is a
better approach.

Bill W0IYH

"gudmundur" wrote in message
...
The source is 50 ohms, I am feeding through a 1000 ohm resistor to
my tuned circuit at 30mhz, and the load is over 5000 ohms (FET input).

I want to use a parallel resonant circuit to ground to form a pass
filter. If I want a Q of around 5, would I need 5000 ohm XL/XC ??

This would seem nearly correct. Have I calculated correctly?