On 10/29/2014 3:06 PM, Lostgallifreyan wrote:
rickman wrote in :

Actually even single ended digital inputs don't have much hysteresis
unless they are designed for that.

Well, as a proportion if they only go high above soem fairly close approach
to V+, then low when close to 0V, then the dead band could be wide, the aim
was to eliminate false states so they ARE usually designed for it. I take
your point on very low volt systems, if the actual difference is small even
though proportionally it may not be.

Anyway, now I know that the supply is so small, your suggestion of discrete
transistors is almost certainly the way to go, unless there is enough similar
demand out there to have cause an off-shelf part to be made.

Normally I'd just look at how others are solving similar problems, so I guess
the question I can ask is: what is the signficant difference in this case
that prevents the nearest off-shelf answer from working?

What off the shelf answer? I have not seen any all digital receivers
for any frequency. I think it may only be practical for this case and
I"m not sure of that. lol

This signal is very unique in that it has a very low data rate. This
allows integration in the digital domain over a large number of samples.
Theoretically the signal would be detectable with a negative SNR.
There are actually a number of issues I need to solve to get a prototype
working. The big one is being able to get a large enough signal that
even statistically it is noticeable at the receiver input.

--

Rick

rickman wrote in :

What off the shelf answer?

I just meant in terms of interfacing. Never mind, one of my other replies
might be far more useful. While you can integrate digitally, why do so? It
seems to me (if I haven't missed something I shouldn't) that you might get
away with much less gain before analog integration, then you can boost the
resulting slow signals with much less struggle with gand bandwidth products
and slew rates for low power and such. If you can do it this way, the
resulting slow pulses can be boosted with CMOS which at those speeds will be
pretty much nanopower.

El 29-10-14 21:03, rickman escribió:
On 10/29/2014 7:45 AM, Wimpie wrote:
El 28-10-14 21:33, rickman escribió:
I have a project in mind that would need a very good antenna in the
frequency range of 60 kHz. Originally I looked at loop antennas and
liked the idea of a large shielded loop made of coax tuned with a
capacitor. My goal is to get as large a signal as possible from the
antenna and matching circuit to allow the use of a receiver with very
low sensitivity... in fact an all digital receiver.

I spent some time simulating antennas in spice and was able to get a
bit of a feel for the circuit, but I'm not convinced it would work the
way I want. Just before I set the project aside I was told I needed to
model the radiation resistance. That has the potential of wrecking the
Q of the circuit. I am counting on the high Q to boost the output
voltage. If the radiation resistance is at all appreciable I would
lose the high Q and need to start over.

Anyone have an idea of how to estimate the radiation resistance of a
tuned, shielded loop antenna?

The other factor I don't understand how to factor in is the
distributed capacitance of the coax. Is that a significant influence
on an antenna or is it in the noise compared to the tuning capacitor.
The coax is RG-6-Solid Coax Cable. The loop is made up from 50 feet of
this. The specs are 16.2 pf/foot and 6.5 mOhms/foot in the center
conductor, or would the resistance be a round trip measurement of both
inner conductor and shield? I assume the shield has a much lower
resistance than the inner conductor but I don't know that for sure.


To get some idea of the output voltage of a loop you need to know:

The fieldstrength of the desired signal at your area. This is probably
given in V/m (dBuV/m, etc). As a first guess use E/H = 377 Ohms to
convert this to H-field [A/m].

EMF = n*A*u0*w*H gives you the EMF for a loop with area A and n number
of turns, w = radian frequency, u0 = magn. permeability for air.

This is new to me. I guess I have been mistakenly using the E field
formula. The field strength at optimum times is estimated at 100 uV/m
at my location which is at the weak end of the CONUS map. I will plug
the numbers into your H field version of the equation.
Based on your 100 uV/m, H = 0.27 uA/m Using a coil with 2 ft
diameter, this would result in EMF = 35 nV for a single turn.


The EMF is boosted with the Q-factor of your tuned loop. Guessing the Q
is the difficult part. You can't just use resistive loss (even when
corrected for skin effect). As you have a multi-turn loop there is an
eddy current loss due to proximity of the turns (the so-called
proximity
loss). At these frequencies loss due to radiation is negligible, unless
you make very large coils.

I have not seen the proximity effect taken into account in any
calculations for similar antenna, so I assumed it was also not
appreciable at this frequency. I'm not at all sure about the radiation
resistance. I will be plugging the numbers into the equation I have. I
assume this resistance would be in parallel with the inductor so a
high value is better. Or would it appear in series with the inductor
and a low value is better?
What are you going to make (a link to a drawing may be helpful)?
What equations do you have for the Q factor for your geometry?



Practically spoken you can't model the proximity loss in spice. In my
opinion you should measure the Q of your loop, or do some search on
Q-factor of VLF/MF coils for your coil geometry. That result you can
put
into spice together with the induced EMF.

I'm surprised you feel the Q can't be calculated. When originally
digging into this I found that the calculation of inductance is an
amazingly complex thing. There are lots of equations out there each of
which simplifies some aspect of the phenomenon and have different
applications. I would not expect the proximity effect to be any more
complex.
If calculation of L is very difficult, Q will be also, as they are
related. Many formulas for Q factor for certain geometry are (partly)
empirical. Formulas for Q for real coils take proximity into account.

You may know that Q-factor heavily depends on frequency.



At these frequencies, external (induced) noise is the dominant factor,
think of man made noise. Only the resistive loss part of the capacitor
generates thermal noise. Using a coaxial cable as tuning capacitance
will not give the highest Q as you have a long/thin conductor. A
parallel plate capacitor has less resistive loss.

Q is important, but not the only factor. The coax was chosen to be
inexpensive and easy to work with. RG-6 with an 18 ga solid center
conductor is just slightly bigger than the skin effect and so is about
as usefully large a conductor without it being hollow. So I'm not sure
what might be better. I suppose Litz wire could improve the Q, but I'm
already looking at a Q of ball park 100 or more. Once you get a very
high Q it become hard to use the device without ruining the Q.


Are you able to use good quality RG58? As far as I know RG6 for
consumer
CATV has low copper content and may have a CCS center conductor.

I picked an RG-6 with a solid center conductor. The specified
resistance is 6.5 mohm per foot. Funny, I'm sure most RG-6 is used for
cable TV where the center conductor is steel for strength with copper
plating for conductivity at high frequencies. One vendor argued with
me that solid copper cores were not available in RG-6. lol

BTW, I measured the resistance of my 50 foot of cable and it is in the
right ball park for 6.5 mohm/foot. The shield measured in the same
range as well. I thought the shield might have had a lower resistance
because it would amount to a larger cross section, but I guess not. I
don't think the shield resistance factors into the Q, but I'm not
certain of that.

If you use the cable dielectric as part of the tuning, it is good that
you have cable with solid copper instead of CCS, otherwise lots of the
current would be into steel instead of copper. Your DC resistance
value is correct for copper (assuming about 1 mm diameter).

Your probably found that turns should not touch (increases proximity
loss and loss due to the jacket) to get highest Q factor. A high Q
factor helps you rejecting out of band signals. What values of
inductance do you expect?

In parallel equivalent circuit, the loss resistance (Rp) equals:
Rp = XL*Q = w*L*Q.
When the output goes directly to the input circuitry, Zin Rp to
avoid reduction of Q.

--
Wim
PA3DJS
Please remove abc first in case of PM

On 10/29/2014 4:41 PM, Lostgallifreyan wrote:
rickman wrote in :

What off the shelf answer?

I just meant in terms of interfacing. Never mind, one of my other replies
might be far more useful. While you can integrate digitally, why do so? It
seems to me (if I haven't missed something I shouldn't) that you might get
away with much less gain before analog integration, then you can boost the
resulting slow signals with much less struggle with gand bandwidth products
and slew rates for low power and such. If you can do it this way, the
resulting slow pulses can be boosted with CMOS which at those speeds will be
pretty much nanopower.

Before integration comes demodulation. How would you demodulate and
integrate in the analog domain on a 100 uW power budget? The signal is
PSK. But that is not the real reason. My goal is to show it is
possible to do this entirely in the digital domain.

The devices I have available are not 100% optimized for low power at low
clock rates, but they are pretty good. If I can find devices that have
lower quiescent current the digital design has potential of being lower
power than the analog approach.

--

Rick

rickman wrote in :

Before integration comes demodulation. How would you demodulate and
integrate in the analog domain on a 100 uW power budget? The signal is
PSK. But that is not the real reason. My goal is to show it is
possible to do this entirely in the digital domain.


Low Vf diode in feedback loop of op-amp? I'm curious though, it's an
interesting thought, doing it all in digital equipment, but why? The main
drive behind me 'off-shelf' remark is that I suspect the best answer already
exists in many forms. I'm curious about what makes a need to keep searching.
I'm not denying it, far from it, there's usually more than one good way to
do something, I'm just not sure what the differentiating factor is in this
case.

rickman wrote in :

The signal is PSK.

I missed that bit. I thought it would be simple AM.. If the integrated
signal (after feedback diode demod) differ enough in amplitude (or AC
content) with frequency, threshold detection might be enough. I'm just
pondering it though, I have no idea if it can be done with less power than
you can give it.

rickman wrote in :

The signal is PSK.

My sight isn't very good. That's Psk, not Fsk... Phase? What did I miss.
I've been hung up on the notion that this is an MSF time signal thing, and I
just looked at the spec for the UK one which is a simple switch on/off of a
carrier, so easy to detect efficiently. Yours is something else entirely, but
what? You may need to lay a lot more cards down before you find an answer you
can use, unless you hunt in the dark. (No reason not to, I usually do, on
most things I do, as the net usually makes some light at greatest need).

On 10/30/2014 1:02 PM, Lostgallifreyan wrote:
rickman wrote in :

Before integration comes demodulation. How would you demodulate and
integrate in the analog domain on a 100 uW power budget? The signal is
PSK. But that is not the real reason. My goal is to show it is
possible to do this entirely in the digital domain.


Low Vf diode in feedback loop of op-amp? I'm curious though, it's an
interesting thought, doing it all in digital equipment, but why? The main
drive behind me 'off-shelf' remark is that I suspect the best answer already
exists in many forms. I'm curious about what makes a need to keep searching.
I'm not denying it, far from it, there's usually more than one good way to
do something, I'm just not sure what the differentiating factor is in this
case.

I don't know about "best" but you can buy a time code receiver chip that
spits out a demodulated signal to be decoded by an MCU. At that point
the data rate is pretty low so an MCU can run at very low power levels,
likely dominated by the quiescent current.

When you suggest an op amp, we already covered that ground and they
aren't low power enough. I'm curious how they amplify the signal in the
receiver chip with the whole circuit drawing a very low power level.

--

Rick

On 10/30/2014 1:23 PM, Lostgallifreyan wrote:
rickman wrote in :

The signal is PSK.

I missed that bit. I thought it would be simple AM.. If the integrated
signal (after feedback diode demod) differ enough in amplitude (or AC
content) with frequency, threshold detection might be enough. I'm just
pondering it though, I have no idea if it can be done with less power than
you can give it.

The signal is also AM, but the PSK is supposed to be detectable at lower
signal levels.

--

Rick

On 10/30/2014 2:01 PM, Lostgallifreyan wrote:
rickman wrote in :

The signal is PSK.

My sight isn't very good. That's Psk, not Fsk... Phase? What did I miss.
I've been hung up on the notion that this is an MSF time signal thing, and I
just looked at the spec for the UK one which is a simple switch on/off of a
carrier, so easy to detect efficiently. Yours is something else entirely, but
what? You may need to lay a lot more cards down before you find an answer you
can use, unless you hunt in the dark. (No reason not to, I usually do, on
most things I do, as the net usually makes some light at greatest need).

I have not studied the international time signals extensively, but I
believe they all use AM. The US located beacon added PSK a few years
back to make the signal easier to receive. The US is large enough that
reception is poor in some of the east coast areas. I am east coast and
would like to see just how much I can do to optimize the antenna to make
this work well.

--

Rick