Q=X/R in a series circuit or R/X in a parallel circuit. Since at resonance
there is 0 ohms X in the series circuit or infinite X in the parallel
circuit, The Q would be 0 as in zero.
No flywheel action here.
In a real world case, say 15,000 vdc and 5A plate current, the plate load
impedance would be about 1700 ohms. There would probably be 100 pf from
plate to cathode which is about -58 ohms at 27 MHz. So, looking into the tube,
there is a parallel circuit of 1700 ohms resistive and about 58 ohms
capacitive. The Q is 1700/58 = 29.
The series equal is 2 -j58. The antenna would need to look like 2 +j58 for
a match. That is not a very friendly number so more help is needed.
I will leave the rest of the design work to others.
73,
Gerald K5GW
In a message dated 4/8/2010 5:50:05 P.M. Central Daylight Time,
dezrat1242@yahoo.com writes:
ORIGINAL MESSAGE:
On Thu, 08 Apr 2010 17:09:08 -0400, Ron Youvan
<ka4inm@tampabay.rr.com> wrote:
> By adjusting the antenna length to be resonate at the exciter's
frequency the antenna
>is that "tank circuit," it's just lossy by the radiation resistance. A
variable coupling
>capacitor is not necessary.
REPLY:
Upon further consideration, if the antenna impedance matched the
tube's plate load impedance, wouldn't the Q be just one? I think you
still need the simple LC tank circuit for the flywheel effect.
73, Bill W6WRT
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