Let's suppose you have a vertical antenna with a base feedpoint
impedance of, say, 75 - j300 ohms (75 ohms resistance in series with 300
ohms of capacitive reactance), typical of a thin 5/8 wave vertical. A
network consisting of 1275 ohms in *parallel* with 319 ohms of
capacitive reactance has the same impedance. (Note how the Xc isn't much
different from the Xc of the series circuit in this case.) If we put an
inductor with 319 ohms of inductive reactance in parallel with the
antenna (that is, from the base feedpoint to ground), the reactance of
the inductor cancels out the capacitive reactance of the antenna, and
we're left with 1275 ohms of pure resistance from the antenna base to
ground (that is, across the inductor).

We can use the inductor as an autotransformer. If we tap up on the
inductor some fraction k of the whole way up, the impedance we'll see at
that tap will be very nearly 1275 * k^2, and it'll be purely resistive
(no reactance) because the impedance across the whole coil is purely
resistive. For example, half way up the coil we'll see 1275 * (.5)^2 =
1275 / 4 = 319 ohms. So to get 50 ohms, we tap up sqrt(50/1275) = 20% of
the way up the coil.

Roy Lewallen, W7EL

john doe wrote:
I'm trying to understand the shunt tapped inductor as a circuit.

How does it work??? As long as the inductor cancels out the reactance of
the radiator you just need to find the 50 ohm point on the coil ... or for
that matter any feedline impedance???

Pardon my lack of knowledge.

de ka2pbt

Thanks Roy!!!

Is it possible to cancel the capacitive reactance of the antenna by placing
an inductor in series with the radiator??

i.e center of coax connected to one end of a coil with the other end of the
coil connected to the radiator. Coax shield connected to ground plane
radials.

de ka2pbt


"Roy Lewallen" wrote in message
...
Let's suppose you have a vertical antenna with a base feedpoint impedance
of, say, 75 - j300 ohms (75 ohms resistance in series with 300 ohms of
capacitive reactance), typical of a thin 5/8 wave vertical. A network
consisting of 1275 ohms in *parallel* with 319 ohms of capacitive
reactance has the same impedance. (Note how the Xc isn't much different
from the Xc of the series circuit in this case.) If we put an inductor
with 319 ohms of inductive reactance in parallel with the antenna (that
is, from the base feedpoint to ground), the reactance of the inductor
cancels out the capacitive reactance of the antenna, and we're left with
1275 ohms of pure resistance from the antenna base to ground (that is,
across the inductor).

We can use the inductor as an autotransformer. If we tap up on the
inductor some fraction k of the whole way up, the impedance we'll see at
that tap will be very nearly 1275 * k^2, and it'll be purely resistive (no
reactance) because the impedance across the whole coil is purely
resistive. For example, half way up the coil we'll see 1275 * (.5)^2 =
1275 / 4 = 319 ohms. So to get 50 ohms, we tap up sqrt(50/1275) = 20% of
the way up the coil.

Roy Lewallen, W7EL

john doe wrote:
I'm trying to understand the shunt tapped inductor as a circuit.

How does it work??? As long as the inductor cancels out the reactance of
the radiator you just need to find the 50 ohm point on the coil ... or
for
that matter any feedline impedance???

Pardon my lack of knowledge.

de ka2pbt

"john doe" wrote in message
...
Thanks Roy!!!

Is it possible to cancel the capacitive reactance of the antenna by
placing an inductor in series with the radiator??

i.e center of coax connected to one end of a coil with the other end of
the coil connected to the radiator. Coax shield connected to ground plane
radials.

de ka2pbt

John DK2 PBT

I'm no Roy. But, I have thought about Smith Charts alot. I wonder if
you'd consider trying to identify where the antenna's input impedance must
be located so that it might be matched with a series inductor. I'd be way
out of line to assume that you want to look at a Smith Chart for knowledge
about the effectiveness of that series inductance. But, the Smith Chart
can be a real big aid in estimating 'what it takes' to match any antenna's
input impedance.

Jerry




"Roy Lewallen" wrote in message
...
Let's suppose you have a vertical antenna with a base feedpoint impedance
of, say, 75 - j300 ohms (75 ohms resistance in series with 300 ohms of
capacitive reactance), typical of a thin 5/8 wave vertical. A network
consisting of 1275 ohms in *parallel* with 319 ohms of capacitive
reactance has the same impedance. (Note how the Xc isn't much different
from the Xc of the series circuit in this case.) If we put an inductor
with 319 ohms of inductive reactance in parallel with the antenna (that
is, from the base feedpoint to ground), the reactance of the inductor
cancels out the capacitive reactance of the antenna, and we're left with
1275 ohms of pure resistance from the antenna base to ground (that is,
across the inductor).

We can use the inductor as an autotransformer. If we tap up on the
inductor some fraction k of the whole way up, the impedance we'll see at
that tap will be very nearly 1275 * k^2, and it'll be purely resistive
(no reactance) because the impedance across the whole coil is purely
resistive. For example, half way up the coil we'll see 1275 * (.5)^2 =
1275 / 4 = 319 ohms. So to get 50 ohms, we tap up sqrt(50/1275) = 20% of
the way up the coil.

Roy Lewallen, W7EL

john doe wrote:
I'm trying to understand the shunt tapped inductor as a circuit.

How does it work??? As long as the inductor cancels out the reactance of
the radiator you just need to find the 50 ohm point on the coil ... or
for
that matter any feedline impedance???

Pardon my lack of knowledge.

de ka2pbt

john doe wrote:
Thanks Roy!!!

Is it possible to cancel the capacitive reactance of the antenna by placing
an inductor in series with the radiator??

i.e center of coax connected to one end of a coil with the other end of the
coil connected to the radiator. Coax shield connected to ground plane
radials.

de ka2pbt

Sure. In the example I gave, where the antenna has a 75 - j300 ohm
feedpoint impedance, you'd just need an inductor with 300 ohms of
reactance. Then you'd have a feedpoint impedance of 75 ohms resistive.
Of course, this would be useful in feeding an antenna only if the
resistive part of the feedpoint resistance is close to 50 ohms or some
other value your system can conveniently feed. Unlike the tapped coil
method, it doesn't give you any way of transforming the resistance.

If you need to match any arbitrary impedance, say 250 - j85 or
something, to 50 ohms or some other impedance, you need two things you
can adjust or choose, since you have two things (R and X or magnitude
and phase) which you need to transform. We could get 50 ohms resistive
using the tapped inductor scheme because we could choose the capacitor
and the tap position. Another very common solution is an L network,
which of course has two components. In theory, we can match (or
transform) anything to anything with an L network of some sort. In some
special cases, one of the elements of the L network is zero or infinite,
such as if we're transforming 50 - j200 to 50 + j0, where all we need is
a single inductor. But this isn't generally so.

A 5/8 wave radiator conveniently has a feedpoint resistance in the
neighborhood of 50 ohms, so it can often be matched well enough by
simply adding a series inductor as you suggest.

Roy Lewallen, W7EL

john doe wrote:
Is it possible to cancel the capacitive reactance of the antenna by placing
an inductor in series with the radiator??

That's exactly what happens with the 5/8WL vertical and
base coil.

coil 5/8 WL
GND---/////////--------------------------------------
^
|
50 ohm tap

The coil and 5/8 WL whip form a series-resonant circuit.
--
73, Cecil http://www.qsl.net/w5dxp


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Cecil Moore wrote:
john doe wrote:

Is it possible to cancel the capacitive reactance of the antenna by
placing an inductor in series with the radiator??


That's exactly what happens with the 5/8WL vertical and
base coil.

coil 5/8 WL
GND---/////////--------------------------------------
^
|
50 ohm tap

The coil and 5/8 WL whip form a series-resonant circuit.
--
73, Cecil http://www.qsl.net/w5dxp

Not quite. If you look at the analysis I posted regarding the tap feed,
you'll see that the inductive reactance required for feeding the antenna
this way isn't equal to the series capacitive reactance of the antenna.
They're often close (as in the example, where the feedpoint series X =
-300 and the required Xl = 319 ohms), but how close they are depends on
the Q (Xs/Rs) of the feedpoint Z. When you feed the antenna this way,
what you have is not a series resonant circuit but a parallel resonant
one, and the required inductance is different for the two. The
requirement for this type of feed is not series resonance, but that the
Z at the top of the coil be completely resistive, which series resonance
doesn't produce. As I pointed out in my most recent posting, you can
make a series resonant circuit consisting of the antenna and an
inductor, and put the source in series with that. But you need a
different size inductor.

Roy Lewallen, W7EL

Roy Lewallen wrote:
We can use the inductor as an autotransformer.

Someone should tell that to Kurt N. Sterba. This
month he says: "But for a half-wave or 5/8 wave
vertical you need a parallel tuned circuit at the
base if you want to feed it there with coaxial
cable."

He also says: "A reader is irate because Kurt has
stated that coaxial cable does not radiate. Actually,
Kurt is just quoting from textbooks."

He apparently forgot what he wrote previously: "The
Resonant Feedline Dipole takes advantage of the fact
that coaxial cable acts like a three conductor cable
for RF. There is the inner conductor, the inside of
the shield, and the outside of the shield. RF signals
travel down the inside of the cable with equal currents
on the inner conductor and the shield. The current on
the shield does not penetrate the shield."
--
73, Cecil http://www.qsl.net/w5dxp


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Roy Lewallen wrote:

We can use the inductor as an autotransformer. If we tap up on the
inductor some fraction k of the whole way up, the impedance we'll see
at
that tap will be very nearly 1275 * k^2, and it'll be purely
resistive
(no reactance) because the impedance across the whole coil is purely
resistive. For example, half way up the coil we'll see 1275 * (.5)^2
=
1275 / 4 = 319 ohms. So to get 50 ohms, we tap up sqrt(50/1275) = 20%
of
the way up the coil.

Roy Lewallen, W7EL

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

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

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