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