Reply
 
LinkBack Thread Tools Search this Thread Display Modes
  #31   Report Post  
Old March 17th 05, 11:59 PM
Roy Lewallen
 
Posts: n/a
Default

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



  #32   Report Post  
Old March 18th 05, 02:34 AM
john doe
 
Posts: n/a
Default

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



  #33   Report Post  
Old March 18th 05, 03:02 AM
Jerry Martes
 
Posts: n/a
Default


"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





  #34   Report Post  
Old March 18th 05, 06:30 AM
Roy Lewallen
 
Posts: n/a
Default

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
  #35   Report Post  
Old March 18th 05, 04:55 PM
Cecil Moore
 
Posts: n/a
Default

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


----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==----
http://www.newsfeeds.com The #1 Newsgroup Service in the World! 100,000 Newsgroups
---= East/West-Coast Server Farms - Total Privacy via Encryption =---


  #36   Report Post  
Old March 18th 05, 05:05 PM
Cecil Moore
 
Posts: n/a
Default

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


----== Posted via Newsfeeds.Com - Unlimited-Uncensored-Secure Usenet News==----
http://www.newsfeeds.com The #1 Newsgroup Service in the World! 100,000 Newsgroups
---= East/West-Coast Server Farms - Total Privacy via Encryption =---
  #37   Report Post  
Old March 18th 05, 06:31 PM
Roy Lewallen
 
Posts: n/a
Default

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
  #38   Report Post  
Old March 18th 05, 11:43 PM
 
Posts: n/a
Default


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

  #39   Report Post  
Old March 19th 05, 12:34 AM
Rob Roschewsk
 
Posts: n/a
Default

"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.


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.

de ka2pbt


  #40   Report Post  
Old March 19th 05, 02:11 AM
Roy Lewallen
 
Posts: n/a
Default

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
Reply
Thread Tools Search this Thread
Search this Thread:

Advanced Search
Display Modes

Posting Rules

Smilies are On
[IMG] code is On
HTML code is Off
Trackbacks are On
Pingbacks are On
Refbacks are On


Similar Threads
Thread Thread Starter Forum Replies Last Post
Yaesu FT-857D questions Joe S. Equipment 6 October 25th 04 09:40 AM
LongWire Antenna Jim B Shortwave 5 March 2nd 04 10:36 AM
EH Antenna Revisited Walter Maxwell Antenna 47 January 16th 04 05:34 AM
Poor quality low + High TV channels? How much dB in Preamp? lbbs Antenna 16 December 13th 03 04:01 PM
Poor quality low + High TV channels? How much dB in Preamp? lbbs Shortwave 16 December 13th 03 04:01 PM


All times are GMT +1. The time now is 10:55 AM.

Powered by vBulletin® Copyright ©2000 - 2024, Jelsoft Enterprises Ltd.
Copyright ©2004-2024 RadioBanter.
The comments are property of their posters.
 

About Us

"It's about Radio"

 

Copyright © 2017