Actually a match is dertermined by the ratio of the antenna Z to the
tranzmission line Z. The generater Z has nothing to do with it. At least
that is what I remember. So all these amp specs are irrelvelent.
----- Original Message -----
From: <TexasRF@aol.com>
To: <dezrat1242@yahoo.com>; <amps@contesting.com>
Sent: Thursday, April 08, 2010 6:58 PM
Subject: Re: [Amps] CB amp directly coupled via cap to antenna
> 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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