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 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

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

Hi John,

I do not understand where your -2947j figure comes from. I see it appearing
nowhere in my calculations.

The antenna mentioned by Mikek has an impedance of 58R-1,072j which, according
to my spreadsheet, corresponds (at 1 MHz) to the parallel of 19862R and -1075j
(that is a 148,1 pF capacitor).

In any case, parallel -- series trasformations never result in a change of the
reactance sign; therefore it is not possible that a -2957j (negative) reactance
is transformed into a +2948j (positive) reactance.

73

Tony I0JX

On 1/14/2011 10:50 AM, Antonio Vernucci wrote:
Antonio - I think you slipped a decimal point. The parallel equivalent
of the series combo 58R-2947j is actually 149k+2948j.

Hi John,

I do not understand where your -2947j figure comes from. I see it
appearing nowhere in my calculations.


Well, it's not mine and it appears in you post as 54 pF. Isn't 58R in
series with 54 pF equal to 58-2947j? And isn't the parallel equivalent
of that equal to 149k ohms of resistance in parallel with -2948 ohms of
reactance (~54 pF)?

I'm pointing out that you slipped a decimal point or you would have seen
that 54 pF is too much it results in the parallel equivalent resistance
of 149k rather than 1.49M. Mike's figure of about 18 pF (17 pF series
combination) will do the job.

In any case, parallel -- series trasformations never result in a
change of the reactance sign; therefore it is not possible that a -2957j
(negative) reactance is transformed into a +2948j (positive) reactance.

You are correct. I allowed the sign of the suseptance to creep through.

73

Tony I0JX

Well, it's not mine and it appears in you post as 54 pF. Isn't 58R in series
with 54 pF equal to 58-2947j? And isn't the parallel equivalent of that equal
to 149k ohms of resistance in parallel with -2948 ohms of reactance (~54 pF)?

I'm pointing out that you slipped a decimal point or you would have seen that
54 pF is too much it results in the parallel equivalent resistance of 149k
rather than 1.49M. Mike's figure of about 18 pF (17 pF series combination)
will do the job.

Yes, at one o' clock in the morning, I slipped the decimal point. So, the total
antenna series capacitance should have been about 17 pF, not 54 pF. This
requires putting a 19-pF capacitance in series with the antenna, not 85 pF.

And the inductance resonating the residual parallel capacitance becomes 1,490 uH
instead of 470 uH.

Sorry for mistake!

73

Tony I0JX

On Fri, 14 Jan 2011 19:35:43 +0100, "Antonio Vernucci"
wrote:

Well, it's not mine and it appears in you post as 54 pF. Isn't 58R in series
with 54 pF equal to 58-2947j? And isn't the parallel equivalent of that equal
to 149k ohms of resistance in parallel with -2948 ohms of reactance (~54 pF)?

I'm pointing out that you slipped a decimal point or you would have seen that
54 pF is too much it results in the parallel equivalent resistance of 149k
rather than 1.49M. Mike's figure of about 18 pF (17 pF series combination)
will do the job.

Yes, at one o' clock in the morning, I slipped the decimal point. So, the total
antenna series capacitance should have been about 17 pF, not 54 pF. This
requires putting a 19-pF capacitance in series with the antenna, not 85 pF.

And the inductance resonating the residual parallel capacitance becomes 1,490 uH
instead of 470 uH.

Sorry for mistake!

73

Tony I0JX

Hasn't anyone pointed out that this a problem made for using a Smith
Chart?

(Since no one really seems capable of doing the math :-)

...Jim Thompson
--
| James E.Thompson, CTO | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| Phoenix, Arizona 85048 Skype: Contacts Only | |
| Voice480)460-2350 Fax: Available upon request | Brass Rat |
| E-mail Icon at http://www.analog-innovations.com | 1962 |

I love to cook with wine. Sometimes I even put it in the food.

On Jan 14, 6:41*pm, Jim Thompson To-Email-Use-The-Envelope-I...@On-My-
Web-Site.com wrote:
On Fri, 14 Jan 2011 19:35:43 +0100, "Antonio Vernucci"



wrote:
Well, it's not mine and it appears in you post as 54 pF. Isn't 58R in series
with 54 pF equal to 58-2947j? And isn't the parallel equivalent of that equal
to 149k ohms of resistance in parallel with -2948 ohms of reactance (~54 pF)?

I'm pointing out that you slipped a decimal point or you would have seen that
54 pF is too much it results in the parallel equivalent resistance of 149k
rather than 1.49M. Mike's figure of about 18 pF (17 pF series combination)
will do the job.

Yes, at one o' clock in the morning, I slipped the decimal point. So, the total
antenna series capacitance should have been about 17 pF, not 54 pF. This
requires putting a 19-pF capacitance in series with the antenna, not 85 pF.

And the inductance resonating the residual parallel capacitance becomes 1,490 uH
instead of 470 uH.

Sorry for mistake!

73

Tony I0JX

Hasn't anyone pointed out that this a problem made for using a Smith
Chart?

(Since no one really seems capable of doing the math :-)

* * * * * * * * * * * * * * * * * * * * ...Jim Thompson
--
| James E.Thompson, CTO * * * * * * * * * * * * * *| * *mens * * |
| Analog Innovations, Inc. * * * * * * * * * * * * | * * et * * *|
| Analog/Mixed-Signal ASIC's and Discrete Systems *| * *manus * *|
| Phoenix, Arizona *85048 * *Skype: Contacts Only *| * * * * * * |
| Voice480)460-2350 *Fax: Available upon request | *Brass Rat *|
| E-mail Icon athttp://www.analog-innovations.com| * *1962 * * |

I love to cook with wine. * * Sometimes I even put it in the food.

just plug it in and try it... if the volume isn't high enough get a
real radio!

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

"Antonio Vernucci" wrote in message
. ..
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

Thank you Tony, for the good description you made.
I wrote a short basic program to calculate the parallel conversion.
I can run the numbers for different frequencies, and different Cs and Rs for
the antenna.
Thanks for the tip about the parallel inductor.
Thanks again, Mikek