"Richard Clark" wrote in message
...
On Thu, 13 Jan 2011 19:51:21 -0600, "amdx" wrote:


"Richard Clark" wrote in message
. ..
On Thu, 13 Jan 2011 16:18:35 -0600, "amdx" wrote:


The article's focus is on matching a crystal radio tank to the antenna,
So as a general statement, the tank is a high impedance (mostly R)

Hi Mike,

There's your first mistake. Tank Z is never, ever "mostly R," or you
wouldn't be able to make the Q claim of 1000 (or even 10).

Ok, Richard that wasn't clear to me, I think at resonance the tank is
all R, but I put mostly R because I figured you would have an objection to
all R. So are you saying it is pure R at resonance?

Hi Mike,

Let's say there is absolutely no loss in the Tank (superconduction and
perfect dissipation values as it were); then we would have to ask
ourselves what happens to energy applied to this Tank at resonance? It
can never enter it, thus the Tank is, in effect, infinite in
resistance. But what about the circulating currents? The Tank is, in
effect, infinite in conductance.

Infinite Ohms & Zero Ohms simultaneously.

Is this the Z of the Tank? Is this the R of the Tank to which you are
matching? No, not even close and certainly it has nothing to do with
resonance - except the condition is a function of it being at
resonance. A low Z Tank or a high Z Tank each evidences the same
Infinite Ohms & Zero Ohms simultaneity given my initial condition of
absolute losslessness.

For the energy being applied to or drawn from the Tank, the Tank is in
parallel operation. For energy in the Tank, the Tank is in series
operation. Where is the Q in this duality? Q suffers by the nature
of what you call R. Q has two different values by this duality. One
is called "Loaded Q" and as you might guess, the second is called
"Unloaded Q." Consult Terman for the engineering design rationale for
optimal Qs as I suggested.

A lot there, but I didn't get anything out of it.
At resonance, does the tank look capacitive, inductive, resistive, or all
to the antenna you connect to it. (this makes the assumption attaching
the antenna didn't change anything, it did, let's say I adjusted back to
resonance)
And then my original question,
Are you saying it is pure R at resonance?
A yes or no will be fine, then I can try to reprocess the above that
didn't get anything out of.


When the discussion of "matching" seeks to employ R (pure resistance),
then the next step is toward a conjugate match and elaborations of
efficiency and maximum transfer of power. There is also an
alternative discussion called the Zo Match. This second match seems
to invite the same elaborations (many who post here try to force them
both into the same salad bowl and cover the illogic with dressing).

I'll bite,
I want maximum transfer of power, I'm still working with the tank as
a large pure R so I want the antenna to look like the same large R.
I realize there is still capacitance from the antenna to deal with.
Now are getting to conjugate :-)


When you offered the comment about "the tank is a high impedance
(mostly R)" it was steering the car off the cliff. Is this a Zo match
or a Conjugate match you are seeking? (I can already anticipate this
has gone over your head, as well as many readers. This and the
questions that follow are rhetorical.)

For instance, and returning to antennas (the purpose of this group's
discussion focus), you can have very high Z antennas with very low
resistance characteristics. Do you want a Zo Match, or a Conjugate
Match? Let me flip the antenna: you can have very high Z antennas
with very high resistance characteristics. Do you want a Zo Match, or
a Conjugate Match? Let's do this sideways: you can have very low Z
antennas with very low resistance characteristics. Do you want a Zo
Match, or a Conjugate Match? I could box the compass here, but the I
think I will let the reader off.


......which, then leads me to ask

"What do you really want?"

73's
Richard Clark, KB7QHC

I want to understand the use of an air variable to match an antenna
to the tank of a crystal radio, over the AMBCB frequency range.
With that, I found I need to understand the series to parallel
conversion,
which I now understand, just IS, it's not anything you do. A series RC has
a parallel RC equivalent.
I'm not sure how it can be both at the same time. But as long as that R is
transformed up, and minimally loads my tank, that's all good.
Then, I understand I still have C left that I can use as part of the C for
resonating my LC tank.
Mikek

John had some number issues with Tony's explanation, but the gist of
Tony's rational treatment should be your lesson as it provides for
your requested "why." It also implies (by my comments of the sudden
appearance of two new components) that our (Ham) tuners have been
designed to introduce the proper amounts of reactances in the proper
parallel/series relationships to enable the necessary transform
towards optimal Q and loading balance. The most elaborate of tuners
can change from Pi to T topologies, or series L parallel C (or series
C parallel L), or series LC, or parallel LC, or parallel C series L
(or parallel L series C)... and any of the other combinations I have
not enumerated (about 9 in all). Each shines for a particular
situation - you have named only one.


Because of the wavelength of the BCB antennas are usually short and
capacitive,
so simple tuning with a single series cap works. The problem begins when you
tune to the high end of the band and you have to much capacitance to get to
resonance with your tank. Then the inductor can be added parallel to the
series
antenna tuning cap. Alternately and you could increase the tank inductor
size.



It is not a trivial discussion by any means even when we are talking
about the addition of only two new components. So, your obtaining an
understanding is not going to be achieved at one sitting in front of
the "definitive" posting to a thread.

One problem of seeking the "definitive" posting is that it cannot be
born from a broken premise that article you were trying to figure out
is lame in the extreme.

Given everything you have revealed about it,
it didn't present a solution to its fantasy antenna.

I didn't rewrite the whole article here, there was a solution
with three equations, that, using the antenna finds the L with the
constrants
of highest operating frequency and lowest capacitance of your capacitor,
then a program is run that finds values for C (antenna) and C (tank) and
I'm not ready to go here yet C (load), he also tunes the diode/earphone
load for optimum with yet another capacitor.
The funs over for now go to get ready for work,
Thanks, Mikek
PS, he runs the program with another, what you call fantasy antenna,
and I agree...
73's
Richard Clark, KB7QHC