On 11/5/2015 12:35 AM, Jeff Liebermann wrote:
On Wed, 4 Nov 2015 23:50:38 -0500, rickman wrote:

2. L1 and L2 were over coupled. I reduced the coupling from 1 to
0.02. I intentionally did NOT overlap the resonant peaks so the
tuning is slightly off. It's fairly close to critically coupled.

Why is this over coupled?

When you couple together two tuned circuits, over coupling will result
in an overly broad peak (low Q) while under coupling will result in
low output. The degree of coupling also has some effect on whether
you see one or two peaks in case you really do want a broadband
design. For a 60 KHz loop, you want it as narrow as possible, even if
it means some additional loss.

For a power xformer, you always want as much coupling as possible with
as little stray fields leaving the transformer. However, for tuned
circuits, you want whatever coupling gives you the desired bandwidth.
Different goals, I guess.

How do you control the coupling in the real circuit? I was planning to
use a current transformer which I assume would be strongly coupled. Of
course, I was minimizing C2 which resulted in a high frequency second
peak far above the 60 kHz peak. I don't recall seeing a poor Q in the
circuit. Q is useful to minimize any nearby interference, but otherwise
my concern is max signal strength to get enough signal to be detected by
the crude FPGA comparator input.


3. Adjusted C1 and C2 for 60 KHz tuning.
4. Change frequency axis (.ac) parameters.

I like to have a major tick at the frequency I am interested in, 60 kHz
in this case.

So, add it. I spent about 15 minutes (mostly tuning L1 and L2) making
the changes and left out all kinds of goodies that would be nice.
Title block info, formatting L3/L4 to look like an xformer, etc. I
also didn't do a sanity check on any of the components. However, in
this case I can't help. I don't know how to add a frequency marker
and couldn't find any clues with Google.

You have to add a cursor which reads out in a small window, (and may not
show up in screen captures, can't recall) or you can do a measurement...
which you removed.


5. I got lazy and didn't add the usual title block stuff.
6. There are no values for Rs which needs to be considered.

What is Rs, the loss resistance?

Yes.

Hmmm, this must have been an older copy,

Yep, it appears to be missing some things.

I am sure I included that, possibly in one of the coils since that
is what it is from. I'm not sure I included radiation resistance as I
barely knew what that was. I recall someone said it should be in there
and gave me a rough value which was very small. I now understand it
better and the calculated number is 2.669E-010 ohms, so obviously it can
be totally ignored.

L2 has Rs=7 ohms. L3 has Rs=0.325 ohms. I think both are rather high
for a 60 KHz loop. The other coils have no value for Rs.

Uh, high or not, that is the circuit I was simulating. 50 feet of RG-6
coax, solid copper inner conductor and a current transformer I don't
have a part number for off the top of my head. I was looking for the
turns ratio to give the optimum output voltage from the current
transformer giving the load circuit. I'm not sure the simulation would
provide that given the strong dependance on Q which can be affected by
many unplanned effects. I have already built a frame for 8 turns of
coax, but am thinking more would be better to increase the voltage and Q.

BTW, it is hard to get much lower on the resistance (skin depth = 0.266
mm) so the Q is about as high as you can get unless you use *much*
bigger wire or tubing or add lots more turns. Since adding turns boosts
the signal strength I think that is better than the more exotic types of
conductors that are required for transmitting loops.

Remember, the absolute resistance isn't important, it's the ratio of
inductive impedance to resistance.


When I do an antenna, I usually have the design running in 4NEC2,
which provides me with various parameters including radiation
resistance, efficiency, etc. I don't know what a sane number would be
for a 60 KHz loop, but can probably find a WWVB antenna model that
would give a ballpark value. (However, not now).

Most WWVB antennas are ferrite loops. Good luck.


My real circuit had some other components at the output that complicate
the real circuit. The "receiver" is an FPGA with a very high input
impedance. To bias the input to the threshold of the input there is an
output of the quantized value which is filtered by an RC circuit and
used to bias the other side of the CT secondary rather than grounding
it. I haven't decided on the exact circuit for the digital side.

High impedance means high voltages. If you use a realistic value for
the input voltage instead of 1, it will show if you're going to
overload your FPGA A/D converter or whatever you're using for input.

Uh, I seriously doubt I can overload the input with just an antenna no
matter how well it is constructed. Look at the E field for the WWVB
transmitter and you will see it is very marginal receiving it at all on
the east coast.

I just want to fix a couple of typos in the formulas I posted earlier
for my own benefit. These help me see what is going on.

L � r * ln(r) * N²
R � r * N
Q � N * ln(r)
V � r² * N * Q * ln(r)
V � r² * N² * ln(r)
V � (r * N)² * ln(r)
l � r * N
V � l² * ln(r)

V = voltage
l = wire total length
L = inductance
R = resistance
r = radius of loop
N = number of turns
Q = quality factor

--

Rick

On Thu, 5 Nov 2015 02:44:24 -0500, rickman wrote:

How do you control the coupling in the real circuit? I was planning to
use a current transformer which I assume would be strongly coupled. Of
course, I was minimizing C2 which resulted in a high frequency second
peak far above the 60 kHz peak. I don't recall seeing a poor Q in the
circuit. Q is useful to minimize any nearby interference, but otherwise
my concern is max signal strength to get enough signal to be detected by
the crude FPGA comparator input.

The Q is approximately set by the ratio of:
tuning_capacitor / coupling_capacitor
However, that doesn't work with inductive coupling where the Q is
controlled by the inductors individual Q. I guess Q is the wrong
term. When you critically couple a collection of LC circuits, as in a
multi-section bandpass filter, the curve goes directly through the 3dB
bandwidth points, no matter how many stages are coupled. In other
words, the Q is set by the Q of one section.
What does change is the filter shape factor, which is the ratio of:
30_dB_bandwidth / 3_dB_bandwidth
or ocassionally:
6_dB_bandwidth / 6_db_bandwidth
depending on which reference book you're following. The first is more
common. Adding additional criticially coupled filter stages doesn't
change the Q, but really changes the shape factor. I can fire up a
filter design program to illustrate how it works, but not now.

I'm also a bit worried about the way you're feeding your FPGA directly
from mag loop. The problem is that WWVB uses both an amplitude
modulated time code, as well as the new phase modulated time code.
Decoding the former is going to require some AGC (automatic gain
control) to insure that the FPGA A/D converter is not going to get
clipped, go non-linear, or offer too low a signal level to get a
decent SNR. The phase modulated signal doesn't have this problem, but
has patent issues if you're going to try an sell chips or devices.
https://en.wikipedia.org/wiki/WWVB#Phase_modulation

Gone for a hot chocolate break...
--
Jeff Liebermann
150 Felker St #D http://www.LearnByDestroying.com
Santa Cruz CA 95060 http://802.11junk.com
Skype: JeffLiebermann AE6KS 831-336-2558

On 11/5/2015 8:27 PM, Jeff Liebermann wrote:
On Thu, 5 Nov 2015 02:44:24 -0500, rickman wrote:

How do you control the coupling in the real circuit? I was planning to
use a current transformer which I assume would be strongly coupled. Of
course, I was minimizing C2 which resulted in a high frequency second
peak far above the 60 kHz peak. I don't recall seeing a poor Q in the
circuit. Q is useful to minimize any nearby interference, but otherwise
my concern is max signal strength to get enough signal to be detected by
the crude FPGA comparator input.

The Q is approximately set by the ratio of:
tuning_capacitor / coupling_capacitor

Not sure where you get this. Q is a measure of the energy stored
compared to the energy lost. If the coupling capacitor were the only
path of lost energy that might work, but I've yet seen a situation where
that is the case.


However, that doesn't work with inductive coupling where the Q is
controlled by the inductors individual Q. I guess Q is the wrong
term. When you critically couple a collection of LC circuits, as in a
multi-section bandpass filter, the curve goes directly through the 3dB
bandwidth points, no matter how many stages are coupled. In other
words, the Q is set by the Q of one section.
What does change is the filter shape factor, which is the ratio of:
30_dB_bandwidth / 3_dB_bandwidth
or ocassionally:
6_dB_bandwidth / 6_db_bandwidth
????
depending on which reference book you're following. The first is more
common. Adding additional criticially coupled filter stages doesn't
change the Q, but really changes the shape factor. I can fire up a
filter design program to illustrate how it works, but not now.

You are assuming the two filters are coupled in a useful way. If the
frequency of the parasitic filter is far above 60 kHz it can be ignored
other than the possibility that it picks up a radio station which
clobbers the WWVB signal.


I'm also a bit worried about the way you're feeding your FPGA directly
from mag loop. The problem is that WWVB uses both an amplitude
modulated time code, as well as the new phase modulated time code.
Decoding the former is going to require some AGC (automatic gain
control) to insure that the FPGA A/D converter is not going to get
clipped, go non-linear, or offer too low a signal level to get a
decent SNR. The phase modulated signal doesn't have this problem, but
has patent issues if you're going to try an sell chips or devices.
https://en.wikipedia.org/wiki/WWVB#Phase_modulation

I didn't see anything about patents.

You worry far too much about "overloading" the FPGA input (single
comparator). My concern is being able to detect a signal at all.


Gone for a hot chocolate break...



--

Rick