If what you say is true then no resonant circuit has Q.
In series resonance Q= Xc/R or you can use inductive reactance. they
are equal at resonance.
In parallel Q=R/Xc
73
Bill wa4lav
At 09:58 PM 4/8/2010 -0400, TexasRF@aol.com wrote:
>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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