"Roy Lewallen" wrote in message
...
Rob Roschewsk wrote:

Roy, how did you come up with "1275 ohms in *parallel* with 319 ohms "
equatating to "75 ohms resistance in series with 300 ohms of capacitive
reactance" ??? Just wondering.

I got it from a routine I keep on my HP48GX calculator, which comes from
the following series-parallel transformations:

Let:

Rs = resistance of series equivalent circuit
Xs = reactance of series equivalent circuit
Rp = resistance of parallel equivalent circuit
Xp = reactance of parallel equivalent circuit

To convert a series circuit to a parallel circuit which has an identical
impedance:

Rp = (Rs^2 + Xs^2) / Rs
Xp = (Rs^2 + Xs^2) / Xs

To convert a parallel circuit to a series circuit which has an identical
impedance:

Rs = (Rp * Xp^2) / (Rp^2 + Xp^2)
Xs = (Rp^2 * Xp) / (Rp^2 + Xp^2)

These aren't very difficult to derive if you're comfortable with complex
arithmetic. They should be in the toolkit of everyone who works with
electrical circuits.

Important things to keep in mind when using these transformations:

1. Although frequency isn't explicitly involved in the conversions, when
you make an equivalent circuit from a resistor and inductor or capacitor,
Xp and Xs will change with frequency. Therefore a transformed circuit will
have the same impedance as the original only at one frequency. If the
frequency changes, new values of resistance and capacitance or inductance
have to be calculated for the equivalent circuit.

2. Because point 1, one circuit or the other will usually be better for
modeling a real circuit over a range of frequencies, because the
impedance of the real circuit will change with frequency more like one
or the other of the two equivalent circuits.

In the example, where the feedpoint Z = 75 - j300:

Rs = 75
Xs = 300

so

Rp = (75^2 + (-300)^2) / 75 = 1275
Xp = (75^2 + (-300)^2) / (-300) = -318.75

You can check this if you'd like. You'll find that the parallel
combination of 1275 and -j318.75 ohms is 75 - j300.

Roy Lewallen, W7EL

Roy

I like your method for computing the parallel equivalent of the series
circuit. I wonder if this is an appropriate place to suggest that the
Smith Chart can be used to quickly estimate the parallel equivalent of the
series circuits.
I sometimes overlay a "reversed" Smith Chart over "regular" Smith Chart to
identify the admittance parameters of the circuit.
I've been disassociated from any antenna discusions since 1969 so I may be
introducing information that everyone knows and I'm being an interferance to
this discussion.

My quick and dirty "overlay" of a Smith Chart indicated the parallel
circuit would be 1,250ohms resistive in parallel with 300 ohms of capacitive
reactance.

Jerry

Roy Lewallen wrote:
wrote:

Hi Roy, I'm OK with 319+j0 half way up the coil, but at 20% of the
way
up I keep getting 50+j50. I did it on the Smith Chart, and on a
Spice
based circuit analysis program. Too old to do this stuff by hand.
That's not a real good match for 50 ohm coax. The are other taps
that
will provide a better match, but no where did I find 50+j0.
In an earlier post I stated that it looked like the real part of
the
antenna impedance needed to be less than 50 ohms to get a perfect
match, using this method.
The impedance across the whole coil is not purely resistive, it
is at
the 50% point. Apparently I am modeling incorrectly, or missing
something.
Gary N4AST


In SPICE, what coefficient of coupling did you specify between the
portion of coil below the tap and the portion above the tap? The
autotransformer impedance relationship I gave is strictly true only
for
a coefficient of coupling = 1. A real inductor will be a little less,

but 1 is a decent approximation for a real solenoid of typical
proportions for this application.

I'm curious how you handled coupled inductors on the Smith chart -- I

don't believe I've ever seen it done.

Roy Lewallen, W7EL

Hi Roy, This is where I am probably missing the boat. In both models,
I used 2 separate inductors, the sum of both being the required
inductance, in this case .35uh. I assumed that the tapped coil could
be modeled this way. I see now that this is not an auto-transformer,
it is simply 2 inductors.
The Smith Chart program has a standard transformer feature, and after
you cancel out the reactance, a transformer with a ratio of 1:0.2 gives
50 ohms. The Spice program has coupled inductors, but I will have to
do a little research to see if I can apply them to this model.
Thanks for the explanation, and thanks to others who replied by email
with suggestions.
Gary N4AST

Thanks again!!

de ka2pbt

"Roy Lewallen" wrote in message
...
Rob Roschewsk wrote:

Roy, how did you come up with "1275 ohms in *parallel* with 319 ohms "
equatating to "75 ohms resistance in series with 300 ohms of capacitive
reactance" ??? Just wondering.

I got it from a routine I keep on my HP48GX calculator, which comes from
the following series-parallel transformations:

Let:

Rs = resistance of series equivalent circuit
Xs = reactance of series equivalent circuit
Rp = resistance of parallel equivalent circuit
Xp = reactance of parallel equivalent circuit

To convert a series circuit to a parallel circuit which has an identical
impedance:

Rp = (Rs^2 + Xs^2) / Rs
Xp = (Rs^2 + Xs^2) / Xs

To convert a parallel circuit to a series circuit which has an identical
impedance:

Rs = (Rp * Xp^2) / (Rp^2 + Xp^2)
Xs = (Rp^2 * Xp) / (Rp^2 + Xp^2)

These aren't very difficult to derive if you're comfortable with complex
arithmetic. They should be in the toolkit of everyone who works with
electrical circuits.

Important things to keep in mind when using these transformations:

1. Although frequency isn't explicitly involved in the conversions, when
you make an equivalent circuit from a resistor and inductor or capacitor,
Xp and Xs will change with frequency. Therefore a transformed circuit will
have the same impedance as the original only at one frequency. If the
frequency changes, new values of resistance and capacitance or inductance
have to be calculated for the equivalent circuit.

2. Because point 1, one circuit or the other will usually be better for
modeling a real circuit over a range of frequencies, because the
impedance of the real circuit will change with frequency more like one
or the other of the two equivalent circuits.

In the example, where the feedpoint Z = 75 - j300:

Rs = 75
Xs = 300

so

Rp = (75^2 + (-300)^2) / 75 = 1275
Xp = (75^2 + (-300)^2) / (-300) = -318.75

You can check this if you'd like. You'll find that the parallel
combination of 1275 and -j318.75 ohms is 75 - j300.

Roy Lewallen, W7EL

Roy,
Thank you for your concise description of the auto-transformer.
Now to inject a little spice into the equation;)
Perhaps you would like to tackle the auto-transformer
configured as; Ground|--\\\\\\\---50R J0(from Tx)
^------To Antenna
Thanks again,
73



"Roy Lewallen" wrote in message
...
Rob Roschewsk wrote:

Roy, how did you come up with "1275 ohms in *parallel* with 319 ohms "
equatating to "75 ohms resistance in series with 300 ohms of capacitive
reactance" ??? Just wondering.

I got it from a routine I keep on my HP48GX calculator, which comes from
the following series-parallel transformations:

Let:

Rs = resistance of series equivalent circuit
Xs = reactance of series equivalent circuit
Rp = resistance of parallel equivalent circuit
Xp = reactance of parallel equivalent circuit

To convert a series circuit to a parallel circuit which has an identical
impedance:

Rp = (Rs^2 + Xs^2) / Rs
Xp = (Rs^2 + Xs^2) / Xs

To convert a parallel circuit to a series circuit which has an identical
impedance:

Rs = (Rp * Xp^2) / (Rp^2 + Xp^2)
Xs = (Rp^2 * Xp) / (Rp^2 + Xp^2)

These aren't very difficult to derive if you're comfortable with complex
arithmetic. They should be in the toolkit of everyone who works with
electrical circuits.

Important things to keep in mind when using these transformations:

1. Although frequency isn't explicitly involved in the conversions, when
you make an equivalent circuit from a resistor and inductor or
capacitor, Xp and Xs will change with frequency. Therefore a transformed
circuit will have the same impedance as the original only at one
frequency. If the frequency changes, new values of resistance and
capacitance or inductance have to be calculated for the equivalent
circuit.

2. Because point 1, one circuit or the other will usually be better for
modeling a real circuit over a range of frequencies, because the
impedance of the real circuit will change with frequency more like one
or the other of the two equivalent circuits.

In the example, where the feedpoint Z = 75 - j300:

Rs = 75
Xs = 300

so

Rp = (75^2 + (-300)^2) / 75 = 1275
Xp = (75^2 + (-300)^2) / (-300) = -318.75

You can check this if you'd like. You'll find that the parallel
combination of 1275 and -j318.75 ohms is 75 - j300.

Roy Lewallen, W7EL

Sorry, I don't understand the diagram. Can you describe it in words?

Roy Lewallen, W7EL

John Doe wrote:
Roy,
Thank you for your concise description of the auto-transformer.
Now to inject a little spice into the equation;)
Perhaps you would like to tackle the auto-transformer
configured as; Ground|--\\\\\\\---50R J0(from Tx)
^------To Antenna
Thanks again,
73

ASCII art was never my forte;)

In this configuration the transmitter feeds the top of the
auto-transformer coil with the bottom of the coil at ground
and the antenna is tapped along the coil.
hope that makes more sense?

"John Doe" wrote in message
u...
Roy,
Thank you for your concise description of the auto-transformer.
Now to inject a little spice into the equation;)
Perhaps you would like to tackle the auto-transformer
configured as; Ground|--\\\\\\\---50R J0(from Tx)
^------To Antenna
Thanks again,
73



"Roy Lewallen" wrote in message
...
Rob Roschewsk wrote:

Roy, how did you come up with "1275 ohms in *parallel* with 319 ohms "
equatating to "75 ohms resistance in series with 300 ohms of
capacitive
reactance" ??? Just wondering.

I got it from a routine I keep on my HP48GX calculator, which comes from
the following series-parallel transformations:

Let:

Rs = resistance of series equivalent circuit
Xs = reactance of series equivalent circuit
Rp = resistance of parallel equivalent circuit
Xp = reactance of parallel equivalent circuit

To convert a series circuit to a parallel circuit which has an identical
impedance:

Rp = (Rs^2 + Xs^2) / Rs
Xp = (Rs^2 + Xs^2) / Xs

To convert a parallel circuit to a series circuit which has an identical
impedance:

Rs = (Rp * Xp^2) / (Rp^2 + Xp^2)
Xs = (Rp^2 * Xp) / (Rp^2 + Xp^2)

These aren't very difficult to derive if you're comfortable with complex
arithmetic. They should be in the toolkit of everyone who works with
electrical circuits.

Important things to keep in mind when using these transformations:

1. Although frequency isn't explicitly involved in the conversions, when
you make an equivalent circuit from a resistor and inductor or
capacitor, Xp and Xs will change with frequency. Therefore a transformed
circuit will have the same impedance as the original only at one
frequency. If the frequency changes, new values of resistance and
capacitance or inductance have to be calculated for the equivalent
circuit.

2. Because point 1, one circuit or the other will usually be better for
modeling a real circuit over a range of frequencies, because the
impedance of the real circuit will change with frequency more like one
or the other of the two equivalent circuits.

In the example, where the feedpoint Z = 75 - j300:

Rs = 75
Xs = 300

so

Rp = (75^2 + (-300)^2) / 75 = 1275
Xp = (75^2 + (-300)^2) / (-300) = -318.75

You can check this if you'd like. You'll find that the parallel
combination of 1275 and -j318.75 ohms is 75 - j300.

Roy Lewallen, W7EL

Now I understand what you're describing. That could match an antenna
having parallel transformed feedpoint R less than 50 ohms and moderate
capacitive reactance. But it's probably not the best way to do it. You
couldn't feed a 5/8 wave vertical this way, because the parallel
transformed feedpoint R is greater than 50 ohms. The portion of L below
the tap would be the value needed to resonate with the equivalent shunt
Xc of the antenna. Then the additional turns would simply be the rest of
the autotransformer to step up the equivalent parallel feedpoint R.

Roy Lewallen, W7EL

John Doe wrote:
ASCII art was never my forte;)

In this configuration the transmitter feeds the top of the
auto-transformer coil with the bottom of the coil at ground
and the antenna is tapped along the coil.
hope that makes more sense?

Yes Roy,
A bit off topic on my part for the 5/8 on 2M.
But as you described the step down character of
the auto-transformer so well, I though you may wish
to add a description on the on the step up characterises.



Now I understand what you're describing. That could match an antenna
having parallel transformed feedpoint R less than 50 ohms and moderate
capacitive reactance. But it's probably not the best way to do it. You
couldn't feed a 5/8 wave vertical this way, because the parallel
transformed feedpoint R is greater than 50 ohms. The portion of L below
the tap would be the value needed to resonate with the equivalent shunt
Xc of the antenna. Then the additional turns would simply be the rest of
the autotransformer to step up the equivalent parallel feedpoint R.

Roy Lewallen, W7EL

John Doe wrote:
ASCII art was never my forte;)

In this configuration the transmitter feeds the top of the
auto-transformer coil with the bottom of the coil at ground
and the antenna is tapped along the coil.
hope that makes more sense?

"John Doe" wrote in message
...
ASCII art was never my forte;)

In this configuration the transmitter feeds the top of the
auto-transformer coil with the bottom of the coil at ground
and the antenna is tapped along the coil.
hope that makes more sense?

"John Doe" wrote in message
u...
Roy,
Thank you for your concise description of the auto-transformer.
Now to inject a little spice into the equation;)
Perhaps you would like to tackle the auto-transformer
configured as; Ground|--\\\\\\\---50R J0(from Tx)
^------To Antenna
Thanks again,
73



"Roy Lewallen" wrote in message
...
Rob Roschewsk wrote:

Roy, how did you come up with "1275 ohms in *parallel* with 319 ohms
"
equatating to "75 ohms resistance in series with 300 ohms of
capacitive
reactance" ??? Just wondering.

I got it from a routine I keep on my HP48GX calculator, which comes
from
the following series-parallel transformations:

Let:

Rs = resistance of series equivalent circuit
Xs = reactance of series equivalent circuit
Rp = resistance of parallel equivalent circuit
Xp = reactance of parallel equivalent circuit

To convert a series circuit to a parallel circuit which has an
identical
impedance:

Rp = (Rs^2 + Xs^2) / Rs
Xp = (Rs^2 + Xs^2) / Xs

To convert a parallel circuit to a series circuit which has an
identical
impedance:

Rs = (Rp * Xp^2) / (Rp^2 + Xp^2)
Xs = (Rp^2 * Xp) / (Rp^2 + Xp^2)

These aren't very difficult to derive if you're comfortable with
complex
arithmetic. They should be in the toolkit of everyone who works with
electrical circuits.

Important things to keep in mind when using these transformations:

1. Although frequency isn't explicitly involved in the conversions,
when
you make an equivalent circuit from a resistor and inductor or
capacitor, Xp and Xs will change with frequency. Therefore a
transformed
circuit will have the same impedance as the original only at one
frequency. If the frequency changes, new values of resistance and
capacitance or inductance have to be calculated for the equivalent
circuit.

2. Because point 1, one circuit or the other will usually be better
for
modeling a real circuit over a range of frequencies, because the
impedance of the real circuit will change with frequency more like one
or the other of the two equivalent circuits.

In the example, where the feedpoint Z = 75 - j300:

Rs = 75
Xs = 300

so

Rp = (75^2 + (-300)^2) / 75 = 1275
Xp = (75^2 + (-300)^2) / (-300) = -318.75

You can check this if you'd like. You'll find that the parallel
combination of 1275 and -j318.75 ohms is 75 - j300.

Roy Lewallen, W7EL

wrote:
In an earlier post I stated that it looked like the real part of the
antenna impedance needed to be less than 50 ohms to get a perfect
match, using this method.

Maybe it's the same as the parallel tuning coil installed
at the base of a 75m mobile antenna which has a real part
equal to 12-25 ohms (depending on the quality of the antenna).
--
73, Cecil http://www.qsl.net/w5dxp

----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==----
http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups
----= East and West-Coast Server Farms - Total Privacy via Encryption =----

wrote:
In an earlier post I stated that it looked like the real part of the
antenna impedance needed to be less than 50 ohms to get a perfect
match, using this method.

EZNEC comes with VERT1.EZ, a 40m vertical. I just ran it
at 19 MHz and EZNEC reports a 50-j160 ohm feedpoint impedance
with a maximum gain of 2.23 dBi and a TOA of 16 deg. Seems all
that is needed for matching to 50 ohms is a series base coil.
--
73, Cecil http://www.qsl.net/w5dxp

----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==----
http://www.newsfeeds.com The #1 Newsgroup Service in the World! 120,000+ Newsgroups
----= East and West-Coast Server Farms - Total Privacy via Encryption =----

Cecil Moore wrote:
wrote:
In an earlier post I stated that it looked like the real part of
the
antenna impedance needed to be less than 50 ohms to get a perfect
match, using this method.

Maybe it's the same as the parallel tuning coil installed
at the base of a 75m mobile antenna which has a real part
equal to 12-25 ohms (depending on the quality of the antenna).
--
73, Cecil http://www.qsl.net/w5dxp

----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet
News==----
http://www.newsfeeds.com The #1 Newsgroup Service in the World!
120,000+ Newsgroups
----= East and West-Coast Server Farms - Total Privacy via Encryption
=----

If you use an "L" network consisting of two inductors as a matching
device, the R must be less than 50 ohms. I was incorrectly modeling
the tapped coil as a two inductor "L" network.
Gary N4AST