HIGH Q CAPS FOR VLF LOOP ANTENNA?
 

On Fri, 28 Oct 2005 00:15:03 -0400, TRABEM wrote:

OK, now we're making progress. I knew there had to be an explanation
for your insistence that the antenna presented a 2K impedance to the
feedline! I absolutely knew it could not be correct and you were
equally determined::

You are not off the hook yet.

Let me describe exactly what I hope to build, and you can enlighten me
regarding what the proper term is. Thanks for hanging in there, the
road was a little bumpy....

I'm thinking feedline attached to one end of the wire. The other end
of the wire is attached to the capacitor bank. The other side of the
capacitor bank is attached to the other feedline terminal. Or, stated
another way, the cap is in series with the wire and the 2 transmission
line terminals are connected to the left over cap and the unused wire
end.

Fine, but it doesn't really matter. The unloaded Q is one thing, the
loaded Q that is seriously depressed is another.

OK, but I hadn't thought this entered into the calculations of the
loop.....it is what it is and we all know it's too damn short and too
damn close to the ground to be efficient.

You have so little efficiency, that the proximity of ground hardly
matters (you already know this) - until we look at the lobe pattern,
balance, and that side of the coin.

or do I really need to look at this issue to proceed?

You do your best - you are not that far from it. The loop allows you
to end load it better than using a dipole with coils.

I've thrown together a quick model of your 5M on a side loop using
(and being generous) #1 wire. The bottom of the loop is 10M above
ground. In terms of performance relative to an isotropic antenna it
is down 60dB. It displays an impedance of:
Impedance = 0.05849 + J 10.37 ohms
An addition of a 0.2555µF capacitor draws this down to:
Impedance = 0.05851 + J 0.006073 ohms

Please note that the capacitor is perfect, no ESR whatever. In a
system with Zc = 0.0585, this antenna presents a 1.11 SWR. The half
power points of its resonance are only 600 Hz apart. Hence a Q of
100.


OK, are you telling me my receiver would need to have an (impossibly
low) input impedance of .06 ohms to work well with the antenna I've
planned?

Just compare that to the 2 Ohms you hoped for and see what happened.

OK, I'm not completely understanding the last paragraph. Have we
abandoned the .06 + j10 real loop impedance and/or are we talking only
about feeding a 2 ohm impedance loop into a 2 ohm impedance receiver

You have injected 2 Ohms into the series circuit - it is now part of
the antenna even if it inside the receiver.

(which seems to be a match made in heaven at first glance)?

Simply a slight of hand, mathematically. You have to pick a reference
you call Zc (characteristic impedance) of something, and your
resistance is wedded to the antenna. However, the reality of it is
that the native loop compared to the actual load is severely
mismatched. You don't lose 15dB gain on the stairway to heaven. Most
would parallel a loop's capacitor with a really big resistive input Z
match to preserve Q. You chose a path that is leading to grief.

Send me a sign, I sense an incoming lightening bolt::

ZAP

A series load in a circuit that is heavily laden with current requires
a very, very small value. Your switches are deadly elements even if
they are the best in the market.

That same circuit, sans series detection, also supports high voltages
courtesy of Q. At this frequency you have a world of components that
can tap them without seriously loading the circuit and killing the Q,
and the voltage, and the efficiency.

Hence a follower ACROSS the capacitors and AT the loop driving the
line back to the detector. A follower that presents 20K or even 2000K
Ohm load to the loop at 60KHz is a walk in the park. You aren't going
to recover that initial 60dB plunge from the size of the antenna, but
you don't have to grease the slide for another 15dB plummet.

Still, and all, there's always the practical reality that the design
will work out of the box. Wrist watches do it. It won't be the best,
but there's not much competition and this isn't DX we are talking
about, unless you want to work Europe's VLF (and even then you may
hear signals).

You still haven't bitten the bait on what makes the "I" and "Q"
channels useful beyond cramming them into black boxes to strip away
the only really useful angle (pun intended) they offer - phase
information. This is in fact the single most fascinating aspect of
the detectors.

73's
Richard Clark, KB7QHC

HIGH Q CAPS FOR VLF LOOP ANTENNA?
 

TRABEM wrote in message
...


I think you are heading down the wrong path with the series circuit
as your
fighting a loosing battle. Assuming a perfect coil and capacitor
you create
an infinite Q circuit. Now you hook it up in your circuit. First
there has
to be enough resistance to develop the voltage , and here is the
rub, as you
increase the resistance to develop a voltage you decrease the Q.
Yuk!

Go with a parallel circuit like the rest of the world uses and you
will be
going in the right direction.


I think I'm starting to get it. Am I cutting off my foot to spite my
face::

Comments made by you and a few others have nudged mein the right
direction.....

The higher I make the series resonant Q, the lower the impedance
goes,
hence it's almost impossible to get a lot of voltage out of it??

Not sure why it matters that much. But, I was under the impression
that a perfectly matched antenna and front end would only decrease
the
Q by a factor of 2.

Follow along with Richard's comments if you like and add your
comments
as I check here often and read everything, sometimes many mant y
times::

Regards,

T

PS:I had begun thinking that the higher imedance presented by a
parallel loop was harder to match with a balun, which is why I
started
thinking of a series loop. I'm gettin there, thansk again.

=======================================
Trabem,

This discussion is getting you nowhere very fast.
So let's summarise.

I don't have your exact dimensions but the following are good enough.

L = 27uH, Reactance = j10 ohms, Conductor loss = 0.05 ohms, ESR =
0.01 ohms, Radiation ohms = 0. Receiver input = 10 ohms, Ground
loss ohms = 0.01

The intrinsic Q of the loop is 10 / 0.05 = 200.

The working Q of the loop, when series connected, is Reactance divided
by the SUM of all resistances including the receiver.

Working Q = Reactance / ( 0.05 + 0.01 + 10 + .01 ) = 10 / 10.07 =
0.993

Take note of the ridiculous low value of working Q due to the loop
being in series with the receiver.
----
Reg.

HIGH Q CAPS FOR VLF LOOP ANTENNA?
 

OK Richard, it looks like the lights are beginning to come on and I'm
on a different road (a road that doesn't end in a fiery disaster).

At the risk of jumping to far ahead......

Let me know if this is correct or not......

Let's dump the series loop, it doesn't sound promising::

So I'm (now) using the same wire and the same resonating C, but it's
in a parallel configuration. My loop is 2K impedance, my receiver is
50 ohms (more conventional although maybe not necessarily tuned, TBD).

Is my mission to purposely isolate the loop so that anything happening
in the receiver doesn't impact the loops Q? To maintain the best
possible Q, would I mismatch the antenna to preamp impedance on
purpose, providing a 50K or even 100K input impedance buffer, so as
not to suck power out of the loop? And, should the output of the
buffer present an approximate 50 ohm output impedance, therefore
matching the input impedance of the RX (for best power transfer)? If
this is the case, an active buffer amp seems inevitable.

You gave me numbers for a 2 ohm loop fed into a 2 ohm RX.

If I do not buffer the loop from the RX, wouldn't a 2K loop fed into a
2K RX also cause similar loss of Q (just like the 2 ohm over 2 ohm
example you gave previously)?

I'm absolutely sure I'm not out of the woods yet, but does it look
like I'm on the right road yet?? It's a big change for me, can you
give me a brief indication please....

Thanks.

And, I am definitely not avoiding the I/Q issue. I know of successful
hardware handling examples of the I/Q and also of successful software
handling methods. I just haven't decided which one to use yet. The
stand alone hardware is a bit harder to implement, complex filters,
much more hardware to build, and a wideband phase shifting network are
needed. But the advantage is that it's a stand alone solution.

Software works well also, but leaves one dependent on the programmer
and requires a computer. Not sure which way I'll go, but it's not
important yet as I don't have a viable front end and antenna yet.

So, am I on the right road, or am I still very far off track with
regard to the antenna Q issue?

Regards,

T

HIGH Q CAPS FOR VLF LOOP ANTENNA?
 

Dear Trabem,

The input impedance seen looking into the series-connected loop is the
RF loss resistance of the loop, in your case about .05 ohms.

If 0.05 is impedance-matched to a 10-ohm receiver then the working Q
only falls to about 100. But it is not an easy matter to match 0.05
ohms to 10 ohms at 60 KHz. ( I do not know the precise input
resistance of your receiver but you get the idea.)

The working Q of any tuned circuit, either series or parallel
connected, when impedance-matched to a load, always results in the
working Q becoming equal to half of the tuned circuit's intrinsic Q.

This is rather obvious because the loss resistance of the tuned
circuit and the load (after being transformed to the tuned circuit
value) are equal to each other.

Of course, impedance-matching also results in maximum voltage and
maximum current being developed in a given load (or receiver). Which
is also a desirable condition.

It is a serious mistake to think in terms only of volts-input to the
receiver. Or only current-input to the receiver. Receiver S-meters
are POWER meters. That's why they can be calibrated in decibels or in
terms of 6dB per S-unit. Or S9 plus so many decibels.

For example, with a 50-ohm receiver, the reference level S9 = 50
pico-watts receiver input power.

Please accept my apologies for digressiing from 5-metre square loops
at 60 KHz.
----
Reg.

HIGH Q CAPS FOR VLF LOOP ANTENNA?
 

On Fri, 28 Oct 2005 11:02:18 -0400, TRABEM wrote:
Is my mission to purposely isolate the loop so that anything happening
in the receiver doesn't impact the loops Q?

The advantage of Q is that it multiplies I and V giving you
sensitivity. As I have pointed out before, your current design could
work without any changes. I cannot answer this for myself much less
you and the advice I would have to offer is that you build your
receiver with flexibility in mind. We are not talking big changes in
components. That is the long answer. The short answer is yes.

If this is the case, an active buffer amp seems inevitable.

Easy enough to include, or remove depending on need.

If I do not buffer the loop from the RX, wouldn't a 2K loop fed into a
2K RX also cause similar loss of Q (just like the 2 ohm over 2 ohm
example you gave previously)?

Certainly, but not similarly. The Q is not going to plunge to 2 or 3.

And, I am definitely not avoiding the I/Q issue. I know of successful
hardware handling examples of the I/Q and also of successful software
handling methods. I just haven't decided which one to use yet.

That is the point of my questions. They are veiled implications, not
tests of knowledge. No one in your list of links, much less those
I've read over the years knows the PRACTICAL implication of the "I"
and "Q" channels. So, I may as well drop the other shoe.

One does the demodulation of AM signals, the other provides
demodulation of FM and SSB signals. I'm not sure which and what
particular arrangement of supporting circuitry is required beyond
simple AM amplifiers because my construction for that application was
back in 68-69. Building tube models and guaranteeing design
considerations was not as simple as the Tayloe circuit offers now.

However, one of the fascinating characteristics of this style of
detector is that you can feed each channel to the earpieces of a
stereo headset. "I" for one, "Q" for the other earpiece. This gives
you the chance to use your wet-ware instead of someone's software and
hardware.

The brain does all the necessary fourier analysis automatically and in
real time. The upshot of it is that when listening to a CW signal,
and hearing the field of signals around it, you perceive those signals
in a mind-space. The signal that is center tuned sounds like it is
between your ears, in the middle of your, as I described it,
mind-space. Those signals that are above it in frequency sound as
though they are coming from the right, and those signals that are
below it in frequency sound as though they are coming from the left.

The advantage of this detector, in this configuration, with this kind
of perception, is that your mind is separating the signals
psychologically. Even though the signals you hear on the left and
right are in equal amplitude to the center, you can exclude them
mentally. Imagine taping a conversation in room full of people and
the microphone is not at your, or your partners lips, but between you,
and you are both standing off a couple of feet talking over the crowd.
You full know that you could understand your partner at the time of
the recording, and you probably know that the tape would be a bitch to
make sense of, even though it makes a faithful record of the
conversation in that free-for-all.

The difference is that your binaural perception with its phase
separation capability could be brought to bear to ignore the field of
noise to concentrate on your partner. When you hear the mono
recording, the phase information is lost and your partner's
conversation merges with the background noise.

I cannot personally vouch for this effect because the payoff in my
construction back then didn't come down to finally evidencing this
effect for myself. This wet-ware characteristic was reported to me to
be one of the attractions of building for my professor. I have also
played with bucket-brigade delay lines to create this effect. At one
time Paul McCartney was using it with his music. Aural phase
relationships have a strong psychological information content that is
taken for granted.

73's
Richard Clark, KB7QHC

HIGH Q CAPS FOR VLF LOOP ANTENNA?
 

The mixed-up confusion along this extended thread is due to the
inability of contributors to describe in plain English exactly what
they mean about a relatively simple matter. It's a breeding ground for
baffle-gab, confusing nonsense and old wives.

To avoid wasting more time I respectfully suggest Trabem obtains a big
bunch of capacitors of various values and gets on with the job. We
will all be very interested in the outcome.

Now perhaps we can return to which part of a 1/2-wave dipole does the
most radiating - the middle bit or the ends?
----
Reg, G4FGQ.

HIGH Q CAPS FOR VLF LOOP ANTENNA?
 

OK, maybe you're beginning to understand. Q can be calculated as
reactance (at resonance) divided by the effective series resistance, or
as effective parallel resistance divided by reactance at resonance.
For a loop where you know the series resistance, it's easiest to use
that first relationship. If you put your 10 ohm receiver input in
series with your 10 ohm reactance loop, you've ruined all that effort
to get to a very low loop conductor resistance and obviated the need
for high-Q capacitors. And we're all having a very hard time seeing
how you will couple your 10-ohm receiver input to EITHER the
parallel-tuned loop OR the series tuned loop, without having nasty
consequences for your holy-grail Q.

You might think it's best to impedance match ("conjugate match") to
your load, so you transfer the most power to the load. However, that
may not be optimum from a system design standpoint. If you already
have enough signal (along with atmospheric noise) that the receiver
doesn't contribute significantly to the overall SNR, then you may be
better off by intentionally mismatching so that the Q remains high, if
that's important to you. (I personally think you've overrated it, but
that's up to you to decide.) But even if you're wanting to get the
lowest noise contribution from your electronics, the appropriate match
is generally not the conjugate impedance match that results in highest
power transfer. For example, an MMBT2222 NPN transistor running at
about 100uA collector current in a common-emitter configuration with no
feedback will have a low-frequency (e.g. 60kHz) input resistance around
50kohms, but the optimal source resistance from a noise standpoint--the
source resistance which will yield the lowest noise figure for the
amplifier--will be about 2kohms. At optimal source resistance, you can
get a noise figure well below 1dB from an MMBT2222--and from many other
bipolars.

One reason that people like to use FET amplifiers across their
"parallel-tuned" loops is that the amplifier input resistance is quite
high, but (using appropriate FETs) the noise contribution of the
amplifier is negligible. And with proper design, the distortion
contribution can be considerably lower than the distortion of your
detector. For high source impedances, JFETs can give noise figures
that are a small fraction of a dB.

Cheers,
Tom

HIGH Q CAPS FOR VLF LOOP ANTENNA?
 

OK Reg,

Hi Reg,

Read both of you responses, and it's very clear now that I was
seriously out in a fantasy world with respect to the topic. I feel a
lot closer to reality now.

If you can, check the latest comments between Richard and myself. I
think I'm getting it, or at least the first approximation::

You provided a key piece of info when you gave me the verbiage about
the loaded Q formula in a series tuned loop. When I started out, I had
no idea that the loaded Q could possibly drop so much when connected
to a receiver!

The working Q of any tuned circuit, either series or parallel
connected, when impedance-matched to a load, always results in the
working Q becoming equal to half of the tuned circuit's intrinsic Q.

I knew this already!

My big problem was in realizing that the loop impedance was so very
very low. Once Richard got me a closer approximation of the actual
number, it became VERY clear to me that there was no impedance match
in my original configuration! Richard suggested the impedance of the
loop was 2K, I guessed it was 2 ohms, but the actual figure was in the
milliohm range.

I feel SO MUCH better now and I think I'm much better off thanks to
your (and Richard's) patience. Thank you so much for helping me to get
to this point!

T

HIGH Q CAPS FOR VLF LOOP ANTENNA?
 

By the way, I consider the most sensible and understandable
contributions to this thread have been the questions asked by the
originator, Trabem.

I am now half-way down a bottle of South African red plonk. It's
supposed to be good for the arteries.
----
Reg, Hic.

HIGH Q CAPS FOR VLF LOOP ANTENNA?
 

"TRABEM" bravely wrote to "All" (28 Oct 05 00:42:15)
--- on the heady topic of " HIGH Q CAPS FOR VLF LOOP ANTENNA?"

TR From: TRABEM
TR Xref: core-easynews rec.radio.amateur.antenna:219360

This should require a matching transformer otherwise not enough
current can flow to take advantage of the potentially huge Q. Without
current flow there will be no energy storage in a series circuit. In a
parallel circuit the current flow is internal between the parallel
reactances but in a series circuit it must be external. A series
circuit is naturally current driven so the transformer would basically
be converting current into a voltage that the receiver input can use.
If you measure your loop with a test signal you should find it
requires a lot of current and little voltage. From your
investigations, calculate the effective resistance and use this as
your Rs to find the actual Q and the matching turns ratio required for
an autotransformer, for example.

Good luck,

A*s*i*m*o*v

.... Why is Brassiere singular and Panties plural???