[Amps] CB amp directly coupled via cap to antenna

[TexasRF at aol.com]

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 at yahoo.com writes:

ORIGINAL  MESSAGE:

On Thu, 08 Apr 2010 17:09:08 -0400, Ron  Youvan
<ka4inm at 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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