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



Well, that is nominally considered as "matching" in that the coupling
from the antenna is minimal so as to not load the parallel resonant
circuit; a tradeoff actually between energy transfer and loading.
A better way might be to have a low Z tap on the inductor; a method
used in radios since the 60's i think.

On Fri, 14 Jan 2011 14:45:25 -0600, "amdx" wrote:

Thanks, Mikek

Hi Mike,

You are welcome.

73's
Richard Clark, KB7QHC

"Robert Baer" wrote in message
net...
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



Well, that is nominally considered as "matching" in that the coupling
from the antenna is minimal so as to not load the parallel resonant
circuit; a tradeoff actually between energy transfer and loading.
A better way might be to have a low Z tap on the inductor; a method used
in radios since the 60's i think.

Yes that can be done, but my inductor will be 660/46 litz wire.
I'm sure it's possible to make a tap, but with the cost of litz I don't
want to.
Mikek

"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

amdx wrote:
"Robert Baer" wrote in message
net...
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



Well, that is nominally considered as "matching" in that the coupling
from the antenna is minimal so as to not load the parallel resonant
circuit; a tradeoff actually between energy transfer and loading.
A better way might be to have a low Z tap on the inductor; a method used
in radios since the 60's i think.

Yes that can be done, but my inductor will be 660/46 litz wire.
I'm sure it's possible to make a tap, but with the cost of litz I don't
want to.
Mikek


So what is wrong with winding a "second" layer with only a few truns
(say 1/20 of larger coil) and connecting it series aiding, use whole
"tapped" (now) coil for the LC resonance and the use that (new) tap?
Besides, if it is not would, the amount of the wire would be no
different wether you put a tap in or not.

amdx wrote:
"Robert Baer" wrote in message


Well, that is nominally considered as "matching" in that the coupling
from the antenna is minimal so as to not load the parallel resonant
circuit; a tradeoff actually between energy transfer and loading.
A better way might be to have a low Z tap on the inductor; a method used
in radios since the 60's i think.

Yes that can be done, but my inductor will be 660/46 litz wire.
I'm sure it's possible to make a tap, but with the cost of litz I don't
want to.
Mikek



Saving a few cents on Lits wire will cost you almost half of the power
you could deliver to the tank.

This won't give you maximum power transfer.

"Robert Baer" wrote in message
...
amdx wrote:
"Robert Baer" wrote in message
net...
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



Well, that is nominally considered as "matching" in that the coupling
from the antenna is minimal so as to not load the parallel resonant
circuit; a tradeoff actually between energy transfer and loading.
A better way might be to have a low Z tap on the inductor; a method
used in radios since the 60's i think.

Yes that can be done, but my inductor will be 660/46 litz wire.
I'm sure it's possible to make a tap, but with the cost of litz I don't
want to.
Mikek
So what is wrong with winding a "second" layer with only a few truns
(say 1/20 of larger coil) and connecting it series aiding, use whole
"tapped" (now) coil for the LC resonance and the use that (new) tap?
Besides, if it is not would, the amount of the wire would be no
different wether you put a tap in or not.

Nothing wrong with that at all, there are many ways to build a crystal
radio.
Here's just 77 of them.
Have fun, Mikek

joe wrote:


Saving a few cents on Lits wire will cost you almost half of the power
you could deliver to the tank.

This won't give you maximum power transfer.

Correction: You'll lose much more than half the power.

On Sat, 15 Jan 2011 07:50:13 -0600, joe wrote:

joe wrote:


Saving a few cents on Lits wire will cost you almost half of the power
you could deliver to the tank.

This won't give you maximum power transfer.

Correction: You'll lose much more than half the power.

Hi Joe,

Care to go deeper? Would that "more than half the power" be 3.1dB?
6dB? 20dB?

Why "half the power" as a round number instead of a tenth?

73's
Richard Clark, KB7QHC

Richard Clark wrote:
On Sat, 15 Jan 2011 07:50:13 -0600, joe wrote:

joe wrote:

Saving a few cents on Lits wire will cost you almost half of the power
you could deliver to the tank.

This won't give you maximum power transfer.

Correction: You'll lose much more than half the power.

Hi Joe,

Care to go deeper? Would that "more than half the power" be 3.1dB?
6dB? 20dB?

Why "half the power" as a round number instead of a tenth?

73's
Richard Clark, KB7QHC

Why, because I chose not to go into detail.

If you look only at a resistive source impedance, the most power we can
get is

((Vsource/2)**2)/Rsource

Now the example starts with Rsource = 58 ohms and is trying to transform
that to match a 1.5 Meg load. The power available from a source with 1.5
M source impedance is much less than that from a 58 ohm source.

Hint : 58/1.5M

So, doing something to match the tank to the load could be much more
effective. A tapped coil or additional winding has been suggested.