Sorry, I clicked on the wrong email.
I was responding to Gerald's (K5GW) email where he was in error about resonance
and Q. Also, most of us are familiar with R,X and Q and also Bandwidth, Center
Freq. and Q. But there is another relationship which has to do with the energy
stored in a resonant system relative to the energy flowing through it during
each cycle.
Q is proportional to the stored energy in the tuned circuit divided by the
energy going into it on each cycle. That is why high Q circuits get hot or have
large wire sizes etc. Also, that is why the in the old days a tuned circuit was
often called a "tank" circuit. It actually stored oscillating energy like a
tank. I think some of the SKs from many years back understood resonance better
that today's engineers.
73
Bill wa4lav
________________________________________
From: Roger [sub1@rogerhalstead.com]
Sent: Saturday, April 10, 2010 3:57 AM
To: Fuqua, Bill L
Cc: TexasRF@aol.com; dezrat1242@yahoo.com; amps@contesting.com
Subject: Re: [Amps] CB amp directly coupled via cap to antenna
Bill Fuqua wrote:
> If what you say is true then no resonant circuit has Q.
>
Not quite. It's true the formula is either X/R or R/X and at resonance
Xc and Xl are equal. But the formula says to use EITHER Xc or Xl, not
both which would be zero at resonance.
73
Roger (K8RI)
> 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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>
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