Hardy Landskov wrote:
> 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.
>
>
But there is no transmission line. So does the tube output, coupling
cap, or antenna become part of the transmission line?
But first to address the issue of Standing waves on the antenna: my old
engineering book, "if I can find it, and IIRC" outter end of the antenna
represents an *almost* infinite impedance due to the abrupt
discontinuity between the antenna and free space. This point on the
antenna or end of the antenna always represents a very high impedance
and reflects substantial power back toward the feed point. Every bump
in the impedance between the transmitter and the far end of the antenna
results in power being reflected between that point and every other
discontinuity on the system. Now when it comes to radiating a signal an
antenna doesn't care if it's resonant or not. It does affect the R and X
values though with R cycling between low and high values every quarter
wave.
X (J) cycles from 0 through + and - these excursions. If the antenna is
long a series cap will tune out the reactance, if short it takes series
inductance.
So if the antenna is made long a plate coupling cap of the proper value
could tune out the reactance. but that still leaves matching the R. So,
how about link coupling? What other methods could be used to either
bring the antenna R, up, the plate R down, or a combination to get a
reasonable match.
There are a couple of points that run counter intuitive and at first
sound like they violate the rule requiring impedances be matched for
maximum power transfer. Point one: An antenna will radiate all of the
power that gets to it. Point two, regardless of SWR ALL of the power
(save that lost by the dielectric, resistance, or radiation) of the
transmission line will get to the antenna. A plus B means that all of
the power generated should be radiated whether the antenna is matched or
not. BUT in the case of no feed line the antenna becomes part of, or
becomes the tank circuit which to me seems like a different story.
Another point discussed that is not relevant in this case is harmonics.
IOW I doubt they care about either signal cleanliness (IM) OR harmonics.
Power out is power out. It appears the power out and only the power out
is the goal so we can dispense with those little niceties we worry about
such as a clean signal and harmonic reduction.
Now I can understand the case of the multi tube amp where the plate
impedance = the radiation resistance of the antenna. I can understand
how to get there when they are close, but HOW do you get a tube or tubes
with a plate impedance in thousands of ohms to match a very low
resistance reactive load? (IE a mobile antenna) To me the reactance
would be the easy one to deal with. But remember also we are dealing
with powers far beyond with what we are familiar. Another thing we don't
have to deal with is the corona, at least not to that extent. That
corona is likely to have a strong effect on the antenna's impedance.
Also a low R is difficult for antenna tuners to handle so how do they do
it with direct coupling, or doesn't it matter. AS for "Q" remember we
may be dealing with a non resonant antenna and an antenna longer than
resonance can have or is likely to have a reasonably high "Q"...I think.
73
Roger (K8RI)
> ----- 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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