On Thu, 13 Jan 2011 12:41:53 -0800, Richard Clark
wrote:

No doubt there is more to be extracted from this article.

Upon review of other correspondents to your plight, ditch the article
and pick up an EE sophomore book on circuit analysis. The coverage
should be encompassed in the section (from my own copy) called
"Transform Driving Point Impedance and Admittance (Immittance)."

73's
Richard Clark, KB7QHC

Richard Clark wrote:

If by being close to ground you mean the 58 Ohms finds itself invested
in the dirt, well, yes that might be the case.


:-)

"Greg Neill" wrote in message
m...
Richard Clark wrote:

For the 58 Ohm resistive value, it would have to be about 300 feet
tall - not the size of operation one usually comes to expect for a
Xtal radio aficionado.

If it is that tall, it would exhibit 200 Ohms Inductive reactance (one
fifth of what you report, and the opposite sign).

Something doesn't wash here.

Perhaps it's a longwire antenna, of crappy wire, parallel to
the ground?


The author said it was an antenna in his attic.
That's all I know.
Mikek

"Richard Clark" wrote in message
...
On Thu, 13 Jan 2011 12:42:01 -0600, "amdx" wrote:


"Richard Clark" wrote in message
. ..
On Thu, 13 Jan 2011 11:05:38 -0600, "amdx" wrote:

This is in regard to a crystal radio, so the match is for a low
impedance
antenna to a high impedance tank circuit.
The antenna: R=58 ohms C=1072 ohms at 1Mhz.

Hi Mike,

Is this a fantasy antenna?

It's an example from an article, I don't like the number either, seems
like
maybe
2 to 12 ohms would be more realistic. I think the capacitance is ok.

Hi Mike,

Then your intuition is firing on all cylinders, great.

Here's what is stated in the article:
" The concept is that at any given frequency, a parallel RC network has
an
equivalent
series RC network, and vise versa.

This is classically true. The frill of "at any given frequency" can
be discarded.

We can use this property to transform the
real component
of an impedance to a much higher or lower value.

This concept is not a "property," however; more a transformation (as
should be apparent in the original statement). In other words
concept
equivalent
property
transform
use four words to describe one thing - transform (plain and simple).

As long as Xc series R series.
Xc (parallel) = Xc (series)
R (parallel) = XC^2 (series) / R (series)
The Xc of a 17pf at 1Mhz is 9368 Ohms.
To rewrite Rp= 9368^2 / R = 1.513 Mohms

Well, what looks like hand-waving is probably close to the numbers one
could expect.

The article then goes on to say,
The utility of this equivalence can be seen by choosing a sufficiently
small
value of C series
(a large Xc series) A "small" resistance can then be transformed into a
"large" value.

The reason why I say hand-waving (and this is probably your gut
reaction as to "why") is that the five lines of operations you quote
starts with a presumed requirement and then proves it has been met.

What happens if Xc series R series?
What happens if Xc series R series?
What happens if Xc series = R series?
What happens if Xc series R series?

This somewhat clouds the mystery for you of understanding Parallel to
Series conversion. I wonder too. Are we dropping in a new component
and stepping back with a wave and Voila! to find the Parallel circuit
has suddenly been transformed? Yeah, that WOULD be a mystery.

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)
and the antenna has a low R and a C in the 100's of ohms.
So I don't know, is that () ? The question is retorical in nature.

And what is this Xc(parallel) and Xc(series) stuff?

I took a little liberty there, The article had a schematic diagram showing
a series RC, then the radio showing the Parallel equivalent antenna
connected.
The the actual labels were Xs, Xp, Rs, and Rp.
Sorry if I muddied it, the attempt was to make it clearer.

Xc is stricty a function of pi, capacitance, and frequency.

And what is this R(parallel) and R(series) stuff?

R is a function of its, well, resistance. No variables to be found.

No doubt there is more to be extracted from this article.

73's
Richard Clark, KB7QHC

"K7ITM" wrote in message
...
On Jan 13, 9:05 am, "amdx" wrote:
Hi All,
Please look at this in fixed font.
I'm looking for understanding of a series to parallel conversion for
antenna matching.
I'm pretty sure I'm missing something, so point it out to me.
This is in regard to a crystal radio, so the match is for a low impedance
antenna to a high impedance tank circuit.
The antenna: R=58 ohms C=1072 ohms at 1Mhz.
The tank: L=240uh C=106pf Q = 1000
Tank Z =~1.5 Mohms
Here's my understanding of what I think I'm reading.
I put a matching capacitor in series with the antenna.
Antenna-- -----R-----C--------Match cap-----tank------ground.
and this is supposed to transform the circuit to this.
( Maybe better said, equivalent to this)

l------------l
l l
Antenna--- R C LC---Tank
l l
l------------l
^
Ground

I calculate an 18.5pf cap for the match, making the antenna look like 58R
and 17pf.

So this; Antenna-- -----58R-----270pf--------Match
cap18.5pf-----1.5Mohms------ground.
This converts to;
l-----------------l
l l
Antenna---58R 17pf 1.5M---LC Tank at
l l Resonance
l-----------------l
^
Ground

And I now have a 1.5 Mohms source feeding a 1.5 Mohm load.
The purpose of which is to cause minimal loading of the tank by the
antenna.
I don't understand how adding a series capacitor makes a parallel
conversion.
What do I misunderstand or do just need to believe the numbers.
Thanks, Mikek

Seems like it doesn't much matter whether the antenna is real or
imagined; the point is rather what's going on when you couple a low
impedance source to a parallel-resonant tank through a small
capacitance...

Perhaps it will help you to think first about a low impedance source,
let's say 50 ohms at 1.591MHz (1e7 radians/sec), that you want to
couple to an (imaginary) inductor, 2.5mH with Qu=500 at that
frequency, and maximize the energy transfer to the inductor. The
inductance and Q implies that the effective series resistance of the
inductor is 50 ohms. That means all we need to do is cancel out the
inductive reactance, and we can do that with a series capacitance--
that happens to be 4pF. That assumes the ESR of the capacitance is
zero, or essentially zero.

But what if the inductor has 1/4 as much inductance, 0.625mH, same
Qu? Then its effective series resistance is 1/4 as much, or 12.5
ohms, and to get the same energy dissipated in it, you need the
current through it to be 2 times as large as before (constant R*I^2).
The total capacitance to resonate the tank is 4 times as much: 16pF.
If you divide that into two 8pF caps, one directly across the inductor
and the other in series with the 50 ohm source, (very close to) half
the tank's circulating current flows in each capacitor. For the same
energy delivered to the coil, the voltage at the top of the inductor
must be half as much as before. Since the coupling capacitor to the
source is twice as large as it was before, the load impedance the
source sees must be the same as before (50 ohms, as required for max
energy transfer).

Carry that another step, to an inductor with 1/16 the original
inductance and the same Qu=500, and you need 64pF to resonate it and
16pF to couple to it from the low impedance source. Now you divide up
the total capacitance with 3/4 of it (48pF) directly across the
inductor and 1/4 of it to the source. 3/4 of the tank's circulating
current flows in the 48pF capacitor, and 1/4 of it in the 16pF to the
source; the source once again sees a 50 ohm load (because the voltage
at the top of the inductor is 1/4 of the original, and the capacitor
coupling to the source is 4 times as large -- with the current in the
source unchanged).

The key to understanding this coupling, to me, is the division of the
tank's circulating current between the two capacitances, one directly
across the coil and one in series with the source. The smaller the
inductance (at constant Q), the more circulating current required for
the same energy dissipation, and the smaller
_percentage_of_total_resonating_capacitance_ for coupling to the
(constant impedance) source.

Cheers,
Tom

Tom I need to print this out and work with it.
Did you change the inductor on purpose or just miss a
factor of ten from the numbers I posted?
Thanks, Mikek

Mikek,

on of the main problem of this newsgroup is that people write answers without
carefully reading the question.

Everyone tends to "convert" the question toward issues he is able to write
something about, Unfortunately those issues often have little to do with the
original question.

I am offering you my answer, which may be clear, so and.so, or hardly
understandable ... I do not know. But be sure that at least I paid maximum
efforts to appreciate your question (though my numbers do not always match
yours). I'll try to explain the issue in the easiest way I can, assumimng that
all components behave in an ideal manner.

Let us first resume your hypotheses:

- the antenna has an impedance equal to the SERIES of a 58-ohm resistance and a
capacitive reactance of -1,072 ohms which, at 1 MHz, corresponds to a 148.5-pF
capacitance
- your aim is to transform that complex impedance into a 1.5-Mohm
purely-resistive impedance, that would match that of your tank circuit.

That said, you must first appreciate that your antenna can be visualized in two
ways:

- as the SERIES of R=58ohm, C=148.5pF (as said above)
- or, by applying the series-to-parallel transformation formula, as the PARALLEL
of R=19,862ohm, C=148.1pF

These are just two fully equivalent ways of describing the same physical
antenna. You can freely use the one which suits you best. For our purposes let
us here visualize your antenna as the parallel of R=19,862ohm, C=148.1pF.

That said, assume for a moment that you are able to eliminate in some way (i'll
tell you after how) the 148.1-pF parallel capacitance. What would then remain is
a 19,862-ohm resistance, a value which unfortunately does not match the 1.5-Mohm
figure you wish to get.

So, how to get just 1.5Mohm instead?

Playing with the transformation formulas you would realize that, if the SERIES
representation of your antenna would hypothetically be R=58 ohm, C=54 pF
(instead of R=58 ohm, C=148.5 pF as it is in the reality), the corresponding
PARALLEL representation would then become R=1.5Mohm, C=53.9 pF. Just the
resistance value you wish to get!

But modifying the SERIES representation of your antenna according to your needs
is very easy: if you put an 85-pF capacitance in series with the antenna, its
total capacitance would change from C=148.1 pF to C=54pf. And the antenna SERIES
representation would then become R=58 ohm, C=54 pF, as you were aiming at.

Once you have put such 85-pF capacitance in series with your antenna, its
PARALLEL representation becomes R=1.5 MHohm, C=53.9 pF, as said earlier.

For removing the 53.9-pF residual parallel capacitance, just resonate it with a
470-uH parallel inductance. The trick is then done: what remains is just the
1.5-Mohm resistance you wanted to get!

In summary:
- put a 85-pF in series (i.e. in between your antenna and the tank circuit)
- put a 470uH inductance in parallel to the tank (in practice this just means to
increase the tank inductance by 470uH with respect to its nominal value).

73

Tony I0IX
Rome, Italy

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).

Please note the distinction between Z (which includes R) and R in
isolation. Further, read Terman's material on Tank Circuits in his
classic "Electronic and Radio Engineering" to clear up the clouds that
obscure the view of their design rationale.

and the antenna has a low R and a C in the 100's of ohms.

This is the sad (and useful) fate of short antennas, yes.

So I don't know, is that () ? The question is retorical in nature.

It is usually rhetorical, yes, insofar as not being enumerated. Often
in technology it is a shorthand for a 10:1 ratio. Perhaps 100:1.

And what is this Xc(parallel) and Xc(series) stuff?

I took a little liberty there, The article had a schematic diagram showing
a series RC, then the radio showing the Parallel equivalent antenna
connected.
The the actual labels were Xs, Xp, Rs, and Rp.
Sorry if I muddied it, the attempt was to make it clearer.

I could follow Xc(parallel/series) easily enough, but what you
neglected to mention was there are two schematics embodied in the
single one you brought to the discussion.

Making questions simpler often leads to Byzantine answers.

Another problem in this simplification is that there is more than one
agenda being served, and they sometimes conflict with any single
solution. As often happens, when I ask "What do you really want?" I
get a response, I offer a solution, and I am immediately rebuffed that
it does not serve the other agenda - which, then leads me to ask "What
do you really want?"

73's
Richard Clark, KB7QHC

On 1/13/2011 6:05 PM, Antonio Vernucci wrote:
Mikek,

on of the main problem of this newsgroup is that people write answers
without carefully reading the question.

Everyone tends to "convert" the question toward issues he is able to
write something about, Unfortunately those issues often have little to
do with the original question.

I am offering you my answer, which may be clear, so and.so, or hardly
understandable ... I do not know. But be sure that at least I paid
maximum efforts to appreciate your question (though my numbers do not
always match yours). I'll try to explain the issue in the easiest way I
can, assumimng that all components behave in an ideal manner.

Let us first resume your hypotheses:

- the antenna has an impedance equal to the SERIES of a 58-ohm
resistance and a capacitive reactance of -1,072 ohms which, at 1 MHz,
corresponds to a 148.5-pF capacitance
- your aim is to transform that complex impedance into a 1.5-Mohm
purely-resistive impedance, that would match that of your tank circuit.

That said, you must first appreciate that your antenna can be visualized
in two ways:

- as the SERIES of R=58ohm, C=148.5pF (as said above)
- or, by applying the series-to-parallel transformation formula, as the
PARALLEL of R=19,862ohm, C=148.1pF

These are just two fully equivalent ways of describing the same physical
antenna. You can freely use the one which suits you best. For our
purposes let us here visualize your antenna as the parallel of
R=19,862ohm, C=148.1pF.

That said, assume for a moment that you are able to eliminate in some
way (i'll tell you after how) the 148.1-pF parallel capacitance. What
would then remain is a 19,862-ohm resistance, a value which
unfortunately does not match the 1.5-Mohm figure you wish to get.

So, how to get just 1.5Mohm instead?

Playing with the transformation formulas you would realize that, if the
SERIES representation of your antenna would hypothetically be R=58 ohm,
C=54 pF (instead of R=58 ohm, C=148.5 pF as it is in the reality), the
corresponding PARALLEL representation would then become R=1.5Mohm,
C=53.9 pF. Just the resistance value you wish to get!

But modifying the SERIES representation of your antenna according to
your needs is very easy: if you put an 85-pF capacitance in series with
the antenna, its total capacitance would change from C=148.1 pF to
C=54pf. And the antenna SERIES representation would then become R=58
ohm, C=54 pF, as you were aiming at.

Antonio - I think you slipped a decimal point. The parallel equivalent
of the series combo 58R-2947j is actually 149k+2948j.

Once you have put such 85-pF capacitance in series with your antenna,
its PARALLEL representation becomes R=1.5 MHohm, C=53.9 pF, as said
earlier.

For removing the 53.9-pF residual parallel capacitance, just resonate it
with a 470-uH parallel inductance. The trick is then done: what remains
is just the 1.5-Mohm resistance you wanted to get!

In summary:
- put a 85-pF in series (i.e. in between your antenna and the tank circuit)
- put a 470uH inductance in parallel to the tank (in practice this just
means to increase the tank inductance by 470uH with respect to its
nominal value).

73

Tony I0IX
Rome, Italy

On Fri, 14 Jan 2011 01:05:17 +0100, "Antonio Vernucci"
wrote:

In summary:
- put a 85-pF in series (i.e. in between your antenna and the tank circuit)
- put a 470uH inductance in parallel to the tank (in practice this just means to
increase the tank inductance by 470uH with respect to its nominal value).

Hi Tony,

Great walk-through, excellent solution. And the appearance of Two
additional, unstated components. Are they found in the original text
of the article? Possibly not and thus the source of mystery (and a
cautionary tale about what might be found as knowledge on the
Internet).

73's
Richard Clark, KB7QHC

On Jan 13, 2:27*pm, "amdx" wrote:
"K7ITM" wrote in message

...
On Jan 13, 9:05 am, "amdx" wrote:



Hi All,
Please look at this in fixed font.
I'm looking for understanding of a series to parallel conversion for
antenna matching.
I'm pretty sure I'm missing something, so point it out to me.
This is in regard to a crystal radio, so the match is for a low impedance
antenna to a high impedance tank circuit.
The antenna: R=58 ohms C=1072 ohms at 1Mhz.
The tank: L=240uh C=106pf Q = 1000
Tank Z =~1.5 Mohms
Here's my understanding of what I think I'm reading.
I put a matching capacitor in series with the antenna.
Antenna-- -----R-----C--------Match cap-----tank------ground.
and this is supposed to transform the circuit to this.
( Maybe better said, equivalent to this)

l------------l
l l
Antenna--- R C LC---Tank
l l
l------------l
^
Ground

I calculate an 18.5pf cap for the match, making the antenna look like 58R
and 17pf.

So this; Antenna-- -----58R-----270pf--------Match
cap18.5pf-----1.5Mohms------ground.
This converts to;
l-----------------l
l l
Antenna---58R 17pf 1.5M---LC Tank at
l l Resonance
l-----------------l
^
Ground

And I now have a 1.5 Mohms source feeding a 1.5 Mohm load.
The purpose of which is to cause minimal loading of the tank by the
antenna.
I don't understand how adding a series capacitor makes a parallel
conversion.
What do I misunderstand or do just need to believe the numbers.
Thanks, Mikek

Seems like it doesn't much matter whether the antenna is real or
imagined; the point is rather what's going on when you couple a low
impedance source to a parallel-resonant tank through a small
capacitance...

Perhaps it will help you to think first about a low impedance source,
let's say 50 ohms at 1.591MHz (1e7 radians/sec), that you want to
couple to an (imaginary) inductor, 2.5mH with Qu=500 at that
frequency, and maximize the energy transfer to the inductor. *The
inductance and Q implies that the effective series resistance of the
inductor is 50 ohms. *That means all we need to do is cancel out the
inductive reactance, and we can do that with a series capacitance--
that happens to be 4pF. *That assumes the ESR of the capacitance is
zero, or essentially zero.

But what if the inductor has 1/4 as much inductance, 0.625mH, same
Qu? *Then its effective series resistance is 1/4 as much, or 12.5
ohms, and to get the same energy dissipated in it, you need the
current through it to be 2 times as large as before (constant R*I^2).
The total capacitance to resonate the tank is 4 times as much: *16pF.
If you divide that into two 8pF caps, one directly across the inductor
and the other in series with the 50 ohm source, (very close to) half
the tank's circulating current flows in each capacitor. *For the same
energy delivered to the coil, the voltage at the top of the inductor
must be half as much as before. *Since the coupling capacitor to the
source is twice as large as it was before, the load impedance the
source sees must be the same as before (50 ohms, as required for max
energy transfer).

Carry that another step, to an inductor with 1/16 the original
inductance and the same Qu=500, and you need 64pF to resonate it and
16pF to couple to it from the low impedance source. *Now you divide up
the total capacitance with 3/4 of it (48pF) directly across the
inductor and 1/4 of it to the source. *3/4 of the tank's circulating
current flows in the 48pF capacitor, and 1/4 of it in the 16pF to the
source; the source once again sees a 50 ohm load (because the voltage
at the top of the inductor is 1/4 of the original, and the capacitor
coupling to the source is 4 times as large -- with the current in the
source unchanged).

The key to understanding this coupling, to me, is the division of the
tank's circulating current between the two capacitances, one directly
across the coil and one in series with the source. *The smaller the
inductance (at constant Q), the more circulating current required for
the same energy dissipation, and the smaller
_percentage_of_total_resonating_capacitance_ for coupling to the
(constant impedance) source.

Cheers,
Tom

*Tom I need to print this out and work with it.
Did you change the inductor on purpose or just miss a
factor of ten from the numbers I posted?
* * *Thanks, Mikek

Mike, I don't care what your values are. I'm interested in showing
you in the generalized _concept_: how it works. Once you understand
that, you can work with whatever component values you want. If you
want to actually build something that works, you better pick values
you can realize.

Cheers,
Tom