I have had no trouble with Q=200 coils. I have a Boonton Q meter that
verifies this when the correct core material (especially in a toroid) is
used. At 30 MHz we do have to be somewhat persistent, though.

Best regards,

Bill W0IYH

"J M Noeding" wrote in message
...
On Sat, 05 Feb 2005 15:42:57 GMT, "William E. Sabin"
wrote:

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


interesting to see that you came to the same conclusion alltogether,
but I believe it is another problem to make a coil with Q=200, usually
the Q will be lower for readily available components, so the better
way is to make the coil first, measure it and then start the
calculation, hi

73
Jan-Martin, LA8AK
---
J. M. Noeding, LA8AK, N-4623 Kristiansand
http://home.online.no/~la8ak/c.htm