You need first to realize that the "shield" IS the antenna. The whole
point of the "shielded loop" is that you can make it very symmetrical,
which is just what's needed to reject strong local electrical fields.
The symmetry does nothing to reject electromagnetic signals. BUT you
can make an "unshielded" loop which is as symmetrical as a "shielded",
if you are careful, and get the same advantages. If you really want to
build one like a classical "shielded loop" and maintain high Q, just
build the "shield" out of copper pipe and put the capacitor across the
gap. The wire inside the pipe is just the center conductor of a short
piece of coax connected to the feedpoint.

If you don't understand this, please see King, Mimno and Wing's
"Transmission Lines, Antennas and Waveguides." It's explained quite
nicely in the "antennas": chapter. It's also explained reasonably well
in Johnson and Jasik's "Antenna Engineering Handbook."

Cheers,
Tom

On 25 Oct 2005 17:31:02 -0700, "K7ITM" wrote:

You need first to realize that the "shield" IS the antenna. The whole
point of the "shielded loop" is that you can make it very symmetrical,
which is just what's needed to reject strong local electrical fields.
The symmetry does nothing to reject electromagnetic signals. BUT you
can make an "unshielded" loop which is as symmetrical as a "shielded",
if you are careful, and get the same advantages. If you really want to
build one like a classical "shielded loop" and maintain high Q, just
build the "shield" out of copper pipe and put the capacitor across the
gap. The wire inside the pipe is just the center conductor of a short
piece of coax connected to the feedpoint.

If you don't understand this, please see King, Mimno and Wing's
"Transmission Lines, Antennas and Waveguides." It's explained quite
nicely in the "antennas": chapter. It's also explained reasonably well
in Johnson and Jasik's "Antenna Engineering Handbook."


Hi Tom,

No, I don't understand this. I thought a shielded loop meant the loop
antenna wire was shielded by the copper (non-ferrous) surrounding the
wire. The shield tends to protect the wire from electrical field
inputs and allows it to only respond to magnetic field variations.

I thought the capacitance between the wire and the surrounding shield
material represented a loss in Q, therefore a loss in output voltage.
So, a loop that might have a Q of 100 in free space would have a much
lower Q if the loop wire was enclosed in a non-ferrous pipe.

There are countless horror stories about those attempting to use
surplus hardline as shielded loops on LF and VLF, all with
disappointing results. The predominate attitude was that the
capacitive coupling between the wire and the shielding material was
the cause. I don't say the predominate attitude is correct.but, if it
is a false assumption, then I am not the only one who needs
revision)

If the copper pipe IS the antenna, then why have the wire inside it at
all??

I must say I'm more confused now than I was before reading your
message.

I'm sorry, I have to leave now. The director of the asylum is
calling.......

T

On Tue, 25 Oct 2005 23:01:01 -0400, TRABEM wrote:

No, I don't understand this. I thought a shielded loop meant the loop
antenna wire was shielded by the copper (non-ferrous) surrounding the
wire.

It is not an effective antenna shield if it is wholly continuous - and
it is not, it has a gap opposite the mounting point which is generally
at ground/reference potential. Part of the point of being "shielded"
is to enforce a symmetry and that ground/reference is electrically
neutral as long as you guarantee it is equidistant both sides around
the loop to that gap.

The shield tends to protect the wire from electrical field
inputs and allows it to only respond to magnetic field variations.

There is no such thing as "only" magnetic fields variations.

I thought the capacitance between the wire and the surrounding shield
material represented a loss in Q,

Q is a simple relation between loss and storage. Lower Q for the same
storage (be it in a capacitor or an inductor) can only result from
resistive loss of Ohmic conduction or radiation. Any loss
attributable to a capacitor is conductive loss - hence the discussion
of ESR. You would have to go back to the stone age of electronics
with paper and wax dielectrics to find loss BETWEEN the plates.
Equivalent Series Resistance for garden variety capacitors, when
compared to radiation resistance, is not trivial. That is, unless,
you swamp that loss by putting your loop in the closet with your
mothballed summer wardrobe or burying it in the garden mud. Design
for failure is easily achieved if you need a rationale to ignore
simple considerations.

Consult:
http://www.w8ji.com/magnetic_receiving_loops.htm

There are countless horror stories about those attempting to use
surplus hardline as shielded loops on LF and VLF, all with
disappointing results.

Such disasters that arise are one of two possible scenarios:
1. They don't have a gap (short circuit city);
2. They don't guarantee symmetry (poor balance, poor tuning, poor
response).

The predominate attitude was that the
capacitive coupling between the wire and the shielding material was
the cause. I don't say the predominate attitude is correct.but, if it
is a false assumption, then I am not the only one who needs
revision)

We get that traffic - yes. They suffer the same learning slope.

If the copper pipe IS the antenna, then why have the wire inside it at
all??

Because you have to have a conductor pair back to the receiver. The
grounded "shield" serves as one half of the pair, the other spans the
gap connecting to the other half's "shield" (it looks like you are
shorting the inner conductor to ground) to thus pick up the opposite
potential. The voltage across the gap is thus sensed and it only
takes one wire. Look closely at any such standard "shielded" loop.
The sense of what is being shielded is THAT conductor which you
contrive to keep in a controlled environment (a coaxial shield) and
away from the imbalance of nearby capacitive couplings. The
"shielded" inner conductor spans a very small distance whose opposite
poles' capacity is balanced to all neighboring paths to ground. That
is, unless you push one side up against the wall.

Stretch out the gap of the shield loop and you have a conventional
dipole. A conventional dipole exhibits high Z and high V at its tips.
The middle of such a dipole has a low Z and a high I. With respect to
both ends, the middle is neutral and strapping it to a conductor does
nothing to change that topology (and is a common tower mounting
benefit). Being curved into a loop does not change this and allows
you to connect your transmission line to both sides without greatly
exposing a significant length of the transmission line (and thus
forcing an unbalance and upsetting the applecart).

This dipole is obviously very small with respect to its wavelength and
thus some form of end loading is required. Thus the capacitor arrives
on the scene. The circulating currents and potentials become
astronomic for progressively smaller antennas. Those currents flow
through and to the plates of the capacitor. If you don't choose the
right components for that capacitor (and manufacturers of HF loops
like to crow how they achieve this) then your design efficiency goes
TU. Hence to speak of capacitor Q is not appropriate as the correct
term is D (dissipation factor). It is certainly related (an inverse
relation) and despite comments to the contrary, D is resolvable with
standard bridges (although those bridges are of considerable design
sophistication in maintaining balance and their own shielding - not a
trivial matter).

There are simpler ways of achieving the same thing by building a
completely exposed loop with capacitor (still paying attention to the
ESR and keeping the whole shebang out of the mud), and simply building
a shielded coupling loop. Reg has adequately described this before
many times.

73's
Richard Clark, KB7QHC

Thanks Richard,

I have a bunch of reading to do. Appreciate your time and the
explanation that you gave.

My receiver has a 2 to 3 ohm input impedance, my hope was to use a
series tuned loop that was entirely floating (with no ground anywhere)
and feed it to the house with a balanced line. Since it's a short run
to the house (in terms of wavelength, I had hoped the impedance
mismatch between the 90 ohm impedance twisted pair transmission line
would not produce a big loss.

The receiver input is also untuned, so the only selectivity available
to it will be the antenna selectivity, which I think I can get away
with since the antenna is relatively high Q.

Again, thanks for the info and for the web resource. I have some more
studying to do!

Regards,

T




On Tue, 25 Oct 2005 23:01:01 -0400, TRABEM wrote:

No, I don't understand this. I thought a shielded loop meant the loop
antenna wire was shielded by the copper (non-ferrous) surrounding the
wire.

It is not an effective antenna shield if it is wholly continuous - and
it is not, it has a gap opposite the mounting point which is generally
at ground/reference potential. Part of the point of being "shielded"
is to enforce a symmetry and that ground/reference is electrically
neutral as long as you guarantee it is equidistant both sides around
the loop to that gap.

The shield tends to protect the wire from electrical field
inputs and allows it to only respond to magnetic field variations.

There is no such thing as "only" magnetic fields variations.

I thought the capacitance between the wire and the surrounding shield
material represented a loss in Q,

Q is a simple relation between loss and storage. Lower Q for the same
storage (be it in a capacitor or an inductor) can only result from
resistive loss of Ohmic conduction or radiation. Any loss
attributable to a capacitor is conductive loss - hence the discussion
of ESR. You would have to go back to the stone age of electronics
with paper and wax dielectrics to find loss BETWEEN the plates.
Equivalent Series Resistance for garden variety capacitors, when
compared to radiation resistance, is not trivial. That is, unless,
you swamp that loss by putting your loop in the closet with your
mothballed summer wardrobe or burying it in the garden mud. Design
for failure is easily achieved if you need a rationale to ignore
simple considerations.

Consult:
http://www.w8ji.com/magnetic_receiving_loops.htm

There are countless horror stories about those attempting to use
surplus hardline as shielded loops on LF and VLF, all with
disappointing results.

Such disasters that arise are one of two possible scenarios:
1. They don't have a gap (short circuit city);
2. They don't guarantee symmetry (poor balance, poor tuning, poor
response).

The predominate attitude was that the
capacitive coupling between the wire and the shielding material was
the cause. I don't say the predominate attitude is correct.but, if it
is a false assumption, then I am not the only one who needs
revision)

We get that traffic - yes. They suffer the same learning slope.

If the copper pipe IS the antenna, then why have the wire inside it at
all??

Because you have to have a conductor pair back to the receiver. The
grounded "shield" serves as one half of the pair, the other spans the
gap connecting to the other half's "shield" (it looks like you are
shorting the inner conductor to ground) to thus pick up the opposite
potential. The voltage across the gap is thus sensed and it only
takes one wire. Look closely at any such standard "shielded" loop.
The sense of what is being shielded is THAT conductor which you
contrive to keep in a controlled environment (a coaxial shield) and
away from the imbalance of nearby capacitive couplings. The
"shielded" inner conductor spans a very small distance whose opposite
poles' capacity is balanced to all neighboring paths to ground. That
is, unless you push one side up against the wall.

Stretch out the gap of the shield loop and you have a conventional
dipole. A conventional dipole exhibits high Z and high V at its tips.
The middle of such a dipole has a low Z and a high I. With respect to
both ends, the middle is neutral and strapping it to a conductor does
nothing to change that topology (and is a common tower mounting
benefit). Being curved into a loop does not change this and allows
you to connect your transmission line to both sides without greatly
exposing a significant length of the transmission line (and thus
forcing an unbalance and upsetting the applecart).

This dipole is obviously very small with respect to its wavelength and
thus some form of end loading is required. Thus the capacitor arrives
on the scene. The circulating currents and potentials become
astronomic for progressively smaller antennas. Those currents flow
through and to the plates of the capacitor. If you don't choose the
right components for that capacitor (and manufacturers of HF loops
like to crow how they achieve this) then your design efficiency goes
TU. Hence to speak of capacitor Q is not appropriate as the correct
term is D (dissipation factor). It is certainly related (an inverse
relation) and despite comments to the contrary, D is resolvable with
standard bridges (although those bridges are of considerable design
sophistication in maintaining balance and their own shielding - not a
trivial matter).

There are simpler ways of achieving the same thing by building a
completely exposed loop with capacitor (still paying attention to the
ESR and keeping the whole shebang out of the mud), and simply building
a shielded coupling loop. Reg has adequately described this before
many times.

73's
Richard Clark, KB7QHC

I could perhaps scan the relevant pages of the references I
mentioned...

How are you planning to couple your 2-ohm load to your loop without
doing really bad things to its Q? (And just what sort of detector do
you have that represents a 2 ohm load?)

Cheers,
Tom

"K7ITM" wrote in message
ups.com...
I could perhaps scan the relevant pages of the references I
mentioned...

How are you planning to couple your 2-ohm load to your loop without
doing really bad things to its Q? (And just what sort of detector do
you have that represents a 2 ohm load?)

Cheers,
Tom


He probably measured the DC resistance at the antenna input connector. If
there's an inductance path to ground, then that's probably what he measured.
The DC resistance is NOT the RF impedance of the input.

Cheers!!!!


--
Dave M
MasonDG44 at comcast dot net (Just substitute the appropriate characters in
the address)

Never take a laxative and a sleeping pill at the same time!!

On Wed, 26 Oct 2005 20:58:26 -0400, "DaveM"
wrote:

"K7ITM" wrote in message
oups.com...
I could perhaps scan the relevant pages of the references I
mentioned...

How are you planning to couple your 2-ohm load to your loop without
doing really bad things to its Q? (And just what sort of detector do
you have that represents a 2 ohm load?)

Cheers,
Tom


He probably measured the DC resistance at the antenna input connector. If
there's an inductance path to ground, then that's probably what he measured.
The DC resistance is NOT the RF impedance of the input.


NO.

Cheers!!!!

On 26 Oct 2005 09:27:44 -0700, "K7ITM" wrote:

I could perhaps scan the relevant pages of the references I
mentioned...


No, I can get them att he schools library I think. thanks for the
offer.

How are you planning to couple your 2-ohm load to your loop without
doing really bad things to its Q?

Is it better to convert the loop to a higher impedance just to feed it
into the house? It appears that anything I do is going to knock the
heck out of the antennas Q though.

I have not decided whether to mount the receiver at the antenna yet,
or whether to run the twisted line directly into the house from the
antenna (since it's a short run). Most likely it will have a short run
of cat 5 cable going from the antenna to a 1 to 1 toroid transformer
located in the receiver. The only selectivity for the receiver will be
the antenna itself. The receiver is very small, and uses very little
power, so it's pretty feasible to mount the entire receiver at the
antenna and run a balanced line feed of the audio into the house.

(And just what sort of detector do
you have that represents a 2 ohm load?)

Is it better to convert the loop to a higher impedance just to feed it
into the house?

It's an analog switch input, modified by my neighbor that gave me one
of them. The switch vendor says the switch series resistance should be
around 3 ohms, but it measures around 2.5 ohms. Probably is a little
lower than expected due to the integrating capacitors (.1 uF) which
are hung on the output of each switch. The .1's go to ground.

I measured it twice, once with a 1:1:1 isolation transformer and once
with a 6:1:1 isolation transformer....The tester looses accuracy at
low impedances, so we repeated the measurement with the generator
feeding the high impedance side of the a transformer also. I got
nearly the same reading after correcting for the transformers
impedance step down value, since both readings agree pretty well with
the switch vendors ratings, it's very likely that the receiver input
impedance is around 2 ohms.

(And just what sort of detector do
you have that represents a 2 ohm load?)

Cheers,
Tom

The point is that the loop's inductive reactance is on the order of 10
ohms in the frequency range you're talking about. When resonated with
a capacitor, if the Q is, say, 300, then the impedance at resonance
will be about 3000 ohms, resistive, as seen across the capacitor.
Reg's program gives you an estimate of what it will be. If you put a
low-resistance load across that, the Q will drop drastically. And if
you put your 2 ohms (which it won't be at the received frequency, if I
understand what you have) in series with the loop and capacitor, it
will also drastically lower the Q. So my question remains: how will
you couple to the loop and maintain the Q?

When you measure the input impedance of your detector, you should do it
while the detector is operating, and do it versus frequency. I expect
you'll see a large increase in impedance at the operating frequency.

Cheers,
Tom

May I ask, what is with the almost fanitical adherence to Q?

TRABEM wrote in message ...
On 26 Oct 2005 09:27:44 -0700, "K7ITM" wrote:

I could perhaps scan the relevant pages of the references I
mentioned...


No, I can get them att he schools library I think. thanks for the
offer.

How are you planning to couple your 2-ohm load to your loop without
doing really bad things to its Q?

Is it better to convert the loop to a higher impedance just to feed it
into the house? It appears that anything I do is going to knock the
heck out of the antennas Q though.

I have not decided whether to mount the receiver at the antenna yet,
or whether to run the twisted line directly into the house from the
antenna (since it's a short run). Most likely it will have a short run
of cat 5 cable going from the antenna to a 1 to 1 toroid transformer
located in the receiver. The only selectivity for the receiver will be
the antenna itself. The receiver is very small, and uses very little
power, so it's pretty feasible to mount the entire receiver at the
antenna and run a balanced line feed of the audio into the house.

(And just what sort of detector do
you have that represents a 2 ohm load?)

Is it better to convert the loop to a higher impedance just to feed it
into the house?

It's an analog switch input, modified by my neighbor that gave me one
of them. The switch vendor says the switch series resistance should be
around 3 ohms, but it measures around 2.5 ohms. Probably is a little
lower than expected due to the integrating capacitors (.1 uF) which
are hung on the output of each switch. The .1's go to ground.

I measured it twice, once with a 1:1:1 isolation transformer and once
with a 6:1:1 isolation transformer....The tester looses accuracy at
low impedances, so we repeated the measurement with the generator
feeding the high impedance side of the a transformer also. I got
nearly the same reading after correcting for the transformers
impedance step down value, since both readings agree pretty well with
the switch vendors ratings, it's very likely that the receiver input
impedance is around 2 ohms.

(And just what sort of detector do
you have that represents a 2 ohm load?)

Cheers,
Tom