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  #42   Report Post  
Old March 19th 05, 03:03 AM
Jerry Martes
 
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"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


  #43   Report Post  
Old March 19th 05, 03:45 PM
 
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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

  #44   Report Post  
Old March 19th 05, 05:45 PM
Rob Roschewsk
 
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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



  #45   Report Post  
Old March 20th 05, 12:24 AM
John Doe
 
Posts: n/a
Default

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





  #46   Report Post  
Old March 20th 05, 01:04 AM
Roy Lewallen
 
Posts: n/a
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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

  #47   Report Post  
Old March 20th 05, 02:52 AM
John Doe
 
Posts: n/a
Default

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





  #48   Report Post  
Old March 20th 05, 03:49 AM
Roy Lewallen
 
Posts: n/a
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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?

  #49   Report Post  
Old March 21st 05, 03:19 AM
John Doe
 
Posts: n/a
Default

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







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