Fuqua, Bill L wrote:
> 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.
>
>
Yup! <:-))
73
Roger (K8RI)
> 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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>>>
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>>
>>
>
>
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