In a series resonant circuit, at resonance it is equivalent to a dead short
(disregarding the R of the circuit). Series resonant circuits are usually
used as traps. To develop a voltage one needs a parallel resonant circuit at
the resonant frequency, The Q will simply determine how quickly the voltage
falls off each side of resonance.

Next there are two types of Q, first the calculated unloaded Q and second
the in circuit or loaded Q.

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.

TRABEM wrote in message ...
On Thu, 27 Oct 2005 14:55:32 -0400, "Fred W4JLE"
wrote:

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

Sure, it's a fair question.

I have a simple receiver with a low impedance input that is few with a
toroid transformer and a tuned circuit to match the impedances and to
keep out of band signals out.

I want to convert the receiver from HF to VLF (60 KHz) and to use a
series tuned loop of high Q as an antenna. In order to simplify the
receiver input, I have mentioned as an option to eliminate the 50 ohm
matching transformer and the tuned circuit in the front end of the
receiver....and to feed it directly with my low impedance loop. In
this way, the loops high Q would serve as the only means of preventing
out of band signals from getting into the receiver.

In order to make sure that actually happens, I suggested making the
loop Q as high as possible.

Hence my 'almost fanatical adherence to Q'

Not sure if it will work, but wanted to run it past the group.

Regards,

T

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 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.

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.

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.

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

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.

Trabem,

Without wishing to detract you in any way from your objective of a
matched series tuned loop I would like to describe how I would do a
similar job with the usual parallel tuned, multiturn loop.

I do not understand the type of receiver you propose and I am not
seriously interested. But I should say the theoretical working
bandwidth of my proposal is about 1/2 of yours.

Actual bandwidth of both your and my proposals is indeterminate
because of the uncertainty of ground proximity and nearby
environmental loss. The working bandwidths could be very similar.

Using similar size loop dimensions to yours, ie., 5.3 metres square -
Frequency = 60 Khz.
5 turns of close wound 2mm diameter enamelled wire.
Inductance = 710 micro-henrys.
Tuning capacitor = 0.01 uF approx.
Reactance of L and C = 268 ohms.
Conductor resistance loss = 2.5 ohms.
Intrinsic coil Q = 107
Matched working Q =53
3dB working bandwidth = 1.12 KHz.

Impedance match to 50-ohm receiver obtained via small coupling loop,
in the same plane, about 1 metre square. Working Q = 53 or less
depending on height above ground.

The working Q may not be high enough for your particular application.
I describe the antenna for you to see what is possible in comparison
with your series-tuned proposals.
----
Reg, G4FGQ.

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


I think that's a high compliment, considering how totally messed up I
was at the start of this.

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

If a little is good, is more better?

And a very BIG + THANKS from me. Thanks for hangin' in there.

Regards,

T

PS:Read your previous example, which closely parallels some existing
real life loops I found on the Internet last evening. Thanks for the
example as well and I think it's time to start soldering.

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