In article ,
(David Platt) wrote:

When you lengthened the hairpin, you added inductance... probably too
much, so you've not only cancelled out the capacitive reactance from
the DE, but have left some excess inductance shunted across the DE.

Backwards. Too much inductance to resonate with the effective (parallel)
capacitance would have resonated with a somewhat smaller capacitance
than you actually have. (The product of inductance and capacitance has
to be the same, for a given resonant frequency.) So you can think of the
actual capacitance present as consisting of however much would be needed
to resonate with the (too large) inductance, in parallel with more
capacitance which does NOT get cancelled out by the inductance. Result:
the actual hairpin, plus the effective (parallel) capacitance the too
short driven element presents, is capacitive, not inductive (in parallel
with the desired 50 ohms).

This is just the opposite from a series resonant circuit where too much
inductance gives an overall inductive result.

Otherwise, I agree with everything David Platt wrote.

David, VE7EZM and AF7BZ

--
David Ryeburn

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On 9/24/2013 9:04 PM, David Ryeburn wrote:
In article ,
(David Platt) wrote:

When you lengthened the hairpin, you added inductance... probably too
much, so you've not only cancelled out the capacitive reactance from
the DE, but have left some excess inductance shunted across the DE.

Backwards. Too much inductance to resonate with the effective (parallel)
capacitance would have resonated with a somewhat smaller capacitance
than you actually have. (The product of inductance and capacitance has
to be the same, for a given resonant frequency.) So you can think of the
actual capacitance present as consisting of however much would be needed
to resonate with the (too large) inductance, in parallel with more
capacitance which does NOT get cancelled out by the inductance. Result:
the actual hairpin, plus the effective (parallel) capacitance the too
short driven element presents, is capacitive, not inductive (in parallel
with the desired 50 ohms).

This is just the opposite from a series resonant circuit where too much
inductance gives an overall inductive result.

Otherwise, I agree with everything David Platt wrote.

David, VE7EZM and AF7BZ


No, David Platt is correct. You are assuming the effective capacitance
is in parallel. The antenna equivalent circuit is a *series* RC. Put in
too much *shunt* inductance and the combination looks inductive at the
feed point.

John KD5YI

On 9/25/2013 4:27 AM, John S wrote:
On 9/24/2013 9:04 PM, David Ryeburn wrote:
In article ,
(David Platt) wrote:

When you lengthened the hairpin, you added inductance... probably too
much, so you've not only cancelled out the capacitive reactance from
the DE, but have left some excess inductance shunted across the DE.

Backwards. Too much inductance to resonate with the effective (parallel)
capacitance would have resonated with a somewhat smaller capacitance
than you actually have. (The product of inductance and capacitance has
to be the same, for a given resonant frequency.) So you can think of the
actual capacitance present as consisting of however much would be needed
to resonate with the (too large) inductance, in parallel with more
capacitance which does NOT get cancelled out by the inductance. Result:
the actual hairpin, plus the effective (parallel) capacitance the too
short driven element presents, is capacitive, not inductive (in parallel
with the desired 50 ohms).

This is just the opposite from a series resonant circuit where too much
inductance gives an overall inductive result.

Otherwise, I agree with everything David Platt wrote.

David, VE7EZM and AF7BZ


No, David Platt is correct. You are assuming the effective capacitance
is in parallel. The antenna equivalent circuit is a *series* RC. Put in
too much *shunt* inductance and the combination looks inductive at the
feed point.

John KD5YI

I'm wrong. Please disregard.

Sorry.

John

On 9/24/2013 9:04 PM, David Ryeburn wrote:
In article ,
(David Platt) wrote:

When you lengthened the hairpin, you added inductance... probably too
much, so you've not only cancelled out the capacitive reactance from
the DE, but have left some excess inductance shunted across the DE.

Backwards. Too much inductance to resonate with the effective (parallel)
capacitance would have resonated with a somewhat smaller capacitance
than you actually have. (The product of inductance and capacitance has
to be the same, for a given resonant frequency.) So you can think of the
actual capacitance present as consisting of however much would be needed
to resonate with the (too large) inductance, in parallel with more
capacitance which does NOT get cancelled out by the inductance. Result:
the actual hairpin, plus the effective (parallel) capacitance the too
short driven element presents, is capacitive, not inductive (in parallel
with the desired 50 ohms).

This is just the opposite from a series resonant circuit where too much
inductance gives an overall inductive result.

Otherwise, I agree with everything David Platt wrote.

David, VE7EZM and AF7BZ


Actually, thinking more about this, I believe a hairpin is a shorted
transmission line. So, using a Smith Chart, I investigated a 25-j25 load
and played with the chart to see what happened.

It turns out that the feed impedance is indeed inductive if the stub is
too long. So, I have now reverted to agreeing with Mr. Platt.

I think that the stub (hairpin) will have no effect if its length is a
quarter wave. Shorter than that, it becomes inductive. Very short and it
is highly inductive. Longer is less inductive. See where this is going?

Cheers es 73,
John

On Thursday, September 26, 2013 12:08:20 AM UTC-5, John S wrote:
Very short and it is highly inductive.
Longer is less inductive.

John, the inductive reactance of an ideal shorted stub is a tangent function that increases from zero at zero length up to a maximum on the Smith Chart as the length approaches 1/4WL (undefined at 1/4WL). The equivalent inductance is proportional to the inductive reactance so - very short and it is slightly inductive (low reactance). Longer is more inductive (up to 1/4WL).

What you say above seems to be referring to capacitive reactance in open-circuit stubs where very short is highly reactive and longer is less reactive..

On 9/26/2013 7:11 AM, W5DXP wrote:
On Thursday, September 26, 2013 12:08:20 AM UTC-5, John S wrote:
Very short and it is highly inductive. Longer is less inductive.

John, the inductive reactance of an ideal shorted stub is a tangent
function that increases from zero at zero length up to a maximum on
the Smith Chart as the length approaches 1/4WL (undefined at 1/4WL).
The equivalent inductance is proportional to the inductive reactance
so - very short and it is slightly inductive (low reactance). Longer
is more inductive (up to 1/4WL).

What you say above seems to be referring to capacitive reactance in
open-circuit stubs where very short is highly reactive and longer is
less reactive.


Yep, you're right. I was only noticing that the length of the line was
longer, but that is a shorter stub. My mistake again.

In article ,
W5DXP wrote:

On Thursday, September 26, 2013 12:08:20 AM UTC-5, John S wrote:
Very short and it is highly inductive.
Longer is less inductive.

John, the inductive reactance of an ideal shorted stub is a tangent function
that increases from zero at zero length up to a maximum on the Smith Chart as
the length approaches 1/4WL (undefined at 1/4WL). The equivalent inductance
is proportional to the inductive reactance so - very short and it is slightly
inductive (low reactance). Longer is more inductive (up to 1/4WL).

Agreed. So the inductive susceptance of a short shorted stub will be
high, while the inductive susceptance of a longer (but less than a
quarter wavelength) shorted stub will be low. Put that longer stub in
parallel with a dipole a little bit shorter than a half wave long (and
thus with a small amount of capacitive reactance and a large amount of
capacitive susceptance) and the net result comes out capacitive. This is
what I meant when I said that a shorted stub a bit too long to resonate
the (too short) dipole makes the whole thing will come out capacitive,
whereas if the stub is a bit too short for resonance, the whole thing
will come out inductive.

David, VE7EZM and AF7BZ

--
David Ryeburn

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