field test unit no 001 "model T"

[chudster]

Hi Chudster,

sorry, but you are wrong. Please have a look at Ohms law and voltage dividers.

Aziz

Aziz,

Don't be sorry. Every time I am wrong I learn something new.

But I expect that even though I am a Junior, both of us actually know ohm's law really well. If there is a difference then it is in topology or other assumptions .

To that end find an LTSpice model and two simulations. One with a 500 ohm damping resistor and one with none (1 megaohm).

While my target model is perhaps simplistic (a 1nA pulse) I think it demonstrates that not having a damping resistor results in a higher target response, not 2000 times due to the a more realistic coil model but a peak output of 27uV with no damping vs 450nV with a 500 ohm damper so thats about 60 times improvement.

I know its simple, but I have attached plots and the model.

I understand you are the king of target simulation so I would appreciate how we relate this and (a proper target and coil) model to Moodz' latest post and traces which I have just seen and am about to examine right now. Notwithstanding that his circuit is not available to simulate I would have hoped that we could sim a target signal from a coil into no load/ nodamping resistor and end up with something like Moodz has just posted.

Chudster

[moodz]

...The Chudster is asking the right questions but not all the questions that need to be asked yet ...I cannot say if the answers he is getting are right or wrong at this point in time.

I have prepared a circuit that pretty much is the production version .... see below shots from the QED front end .... the coil is an 18 inch 0.4 ohm 300u Minelab type.

Below is the damped UN-AMPLIFIED coil signal ...( ie what the first amp sees ). The baseline is ground. The flyback is damped in around 4 us ... 6 us if you are a pessimist LOL. The target responses are excellent IMHO for a PI.

This is the simplest and lowest cost version of the circuit ... the DSP version is much better ..but not off the testbench yet.



Now we zoom up the CRO gain to 500mv / div so we can see a little more detail ....



Now we add a largish non ferrous target .... remember ... this not amplified yet
verrry niiice.



Now we superimpose the non target and target response ..... I was holding the target slightly closer to the coil so the response is a little larger ... LOL.




moodz

[Aziz]

Aziz,

I hear you but consider the following. If the eddy current response in the coil is taken to be a 1 nanoamp source then yes the 1 ohm does not attenuate relative to a 500 ohm damping resistor. But if there is no damping resistor, then the resistance may be high- say 1 megaohm.

1 nano amp into 500 ohms is 2000 times less of a voltage to measure than into 1 megaohm.

So IF the signal we are trying to detect is nanoamps and comes after flyback then having no damping resistance at that point as Moodz has done should help greatly. I did say IF which comes back to my original question.

Chudster

Hi Chudster,

sorry, but you are wrong. Please have a look at Ohms law and voltage dividers.

Aziz

[chudster]

Hi Chudster,

You say:By the way I found this little reef known as "Lasseter's" not too far from my house http://en.wikipedia.org/wiki/Lasseter's_Reef . I'll share its location if someone can show me the magic eddy current blip If we don't know the signal we are searching for, it is a solution that needs no detector.

I can show you with a simulation what the target response of a 1us target looks like. But, you and another zillion people are going to say: "Ah, simulations are just not like the real thing." And you are right, even if the simulation has been created to show what I have observed on real circuits.

This is why I have posted the full TX and RX circuit that can show you the peak signal of a 1us target. http://www.geotech1.com/forums/showp...&postcount=168

The HEX file that goes with the circuit is not optimized for a 1us target, but I will re-write the timing schedule for you if you build the circuit and want to test and observe specific features.

The circuit is very robust. It will work even on a bread board. The only parts that must be soldered is the TX due to the high voltage combined with the relatively high current.

Don't be fooled by the low power consumption of less than 200mA at 12V. The TX has a peak current of about 4A at several hundred volts.

All the best

Tinkerer

Tinkerer,

Thanks for your reply. I am to start shortly to build a design and I will look at the circuit as it sounds impressive. Before I build a circuit though there are basic engineering steps. These are:
1) What is the characteristic signal that I am to process?
2) What other signal/ noise is it embedded in and how do I extract it?
3) How may it be characterised so that I can build strategies and algorithms capture its salient features in hardware and then to accurately represent that digitally?
4) Model (your model is useful here but steps 1,2 & 3 are best completed first)
5) Design
6) Build and test.
7) Iterate 4-6 and revisit 1,2,3

So I am at stage 1 to 2. Present best understanding is that the signal is an impulse function embedded in the flyback signal seemingly at the end of flyback decay (if flyback was quick enough for the target). If it wasn't quick enough then its info was lost when the signal was clipping at kilovolts or if it was a slow target hopefully I can see it.

That is what I see is likely from Moodz' waveforms posted here which by reputation are pretty much state of the art and from my extensive reading of the forums.

So it appears its not like radar or sonar where we ping, measure delay and then look for the reflected signal using techniques to pull a small signal out of noise but this is basically how I see it being talked about in the forums. I am wondering if its just lore and everyone else can see the Emperor's clothes that I cannot see.

To my knowledge, while it may match the theory, no one has posted a scope trace of the discrete "eddy current collapse" signal that we seek and it does not seem to be leaping out even as flyback time is shortened. I have seen it in simulations but that is step 4. As such no disclosure of Lasseter's reef location yet . Lasseter himself was sure that it was there and lots of people went along with his thinking but its never been found because they all looked in the wrong place out in the desert. Its actually near my house .

To be devil's advocate, existing designs are detecting something and they are successful. Moodz will disclose his breakthrough in due course. It may well be the end of PI improvement as we know it or TINKERERS_SB_MO THE MOTION VERSION might be the last word.

Call me an "old fashioned engineer" but before I build a signal processing circuit I'd like to be sure of the signal I am looking for. Alternatively I will just need to build something and find out for myself (because I am speaking from practical ignorance) and clearly need to suffer multiple soldering burns like the gurus have to attain my enlightenment and right of passage from a junior. I can do that but was hoping to poll (and challenge) the gurus before that was done hence the posts.

Hope this is on topic but it is Moodz' work that has got me thinking.

Chudster

[Tinkerer]

Thanks for your reply and input. Maybe I am still not seeing it but 1 just pretty much looks like a longer flyback time due to more energy to get out than 2.

Hdphilip, I hear what you are saying but the bumps I took to be Moodz' explanation that he has let it ring a little. However if that is the feature needed to be measured then at 500KHz fundamental sampling need be better than 1MSPS if sampling before demod.

Not to take away from Moodz' amazing achievements to terminate and shape flyback leaving essentially an open circuit coil without a resistive damped signal (which I am sure will revolutionise detectors because they do work already and it should make them work better), but I've read a lot of theory that says its there somewhere. That the time constant for non ferrous materials is short so they decay and show up just after flyback (provided its gotten to quickly enough and current is terminated fast enough) but I am still looking. If we could get flyback down to 2uS or 1uS would it show up? Can anyone point out that magic target signal in terms other than as longer/ shorter flyback less/more energy?

By the way I found this little reef known as "Lasseter's" not too far from my house http://en.wikipedia.org/wiki/Lasseter's_Reef . I'll share its location if someone can show me the magic eddy current blip If we don't know the signal we are searching for, it is a solution that needs no detector.

With hdphilip replied, Aziz? Moodz? Carl? Other gurus? Anyone? Can you help? Can anyone post that trace showing a discrete post flyback signal for us to hunt?

Chudster

Hi Chudster,

You say:By the way I found this little reef known as "Lasseter's" not too far from my house http://en.wikipedia.org/wiki/Lasseter's_Reef . I'll share its location if someone can show me the magic eddy current blip If we don't know the signal we are searching for, it is a solution that needs no detector.

I can show you with a simulation what the target response of a 1us target looks like. But, you and another zillion people are going to say: "Ah, simulations are just not like the real thing." And you are right, even if the simulation has been created to show what I have observed on real circuits.

This is why I have posted the full TX and RX circuit that can show you the peak signal of a 1us target. http://www.geotech1.com/forums/showp...&postcount=168

The HEX file that goes with the circuit is not optimized for a 1us target, but I will re-write the timing schedule for you if you build the circuit and want to test and observe specific features.

The circuit is very robust. It will work even on a bread board. The only parts that must be soldered is the TX due to the high voltage combined with the relatively high current.

Don't be fooled by the low power consumption of less than 200mA at 12V. The TX has a peak current of about 4A at several hundred volts.

All the best

Tinkerer

[chudster]

BTW,

getting rid of the damping resistor across the coil does not help much to increase the target signal strength. The coil impedance is much much lower than the damping resistor and the signal won't be significantly attenuated.
Lets make a simple comparison:
Coils impedance: 1 Ohm
Damping resistor: 500 Ohm
See the relation of the impedances!!

Aziz

Aziz,

I hear you but consider the following. If the eddy current response in the coil is taken to be a 1 nanoamp source then yes the 1 ohm does not attenuate relative to a 500 ohm damping resistor. But if there is no damping resistor, then the resistance may be high- say 1 megaohm.

1 nano amp into 500 ohms is 2000 times less of a voltage to measure than into 1 megaohm.

So IF the signal we are trying to detect is nanoamps and comes after flyback then having no damping resistance at that point as Moodz has done should help greatly. I did say IF which comes back to my original question.

Chudster

[Davor]

You can't. It is coupling related, hence the effect changes with distance. OK, PI are not famous for accurate discrimination, but this could change with some clever designs such as the Tinkerer's. We'll see.

[Carl-NC]

However, playing with Mr. Thévenin's theorem and the fact that coil/target system works as a (very non-ideal) transformer, that 500ohm would interfere with target taus.
It would not interfere greatly with target signal strength, but will surely screw up any chance of correct discrimination.

You just calibrate it out in the disc system.

[Carl-NC]

Carl,
Assuming we are chasing small gold, doing a thought experiment, if we could terminate flyback within 100nS or less would we see a discrete blip for the eddy current because it seems not to be so in practice as we reduce flyback time so far?

Yes, for an ideal step response.

Alternatively with a large ferrous target we terminate flyback quickly have some microseconds with no signal then bang we get the impulse. In other words the timing is not related to a physical time constant referenced to current termination. Its all bound up together as we remove energy.

Long-tau eddy responses have a delayed peak whether ferrous or non-ferrous.

In consideration it seems we are removing as much electromagnetic energy as fast as possible so that the target energy can be measured as it is additive and more apparent as we take away excitation but always returns on the coat tails of removing the excitation energy so there is no discrete eddy current event.

Basically.

If we remove energy too slowly we never see small gold which is why we aim to remove it fast.

Depends on the detector design. For tiny fast-tau gold you need a high slew rate TX signal just to induce a decently detectable eddy current. You can potentially "see" this response even if you don't "remove" the energy, which is what VLF gold detectors do.

- Carl

[Aziz]

However, playing with Mr. Thévenin's theorem and the fact that coil/target system works as a (very non-ideal) transformer, that 500ohm would interfere with target taus.
It would not interfere greatly with target signal strength, but will surely screw up any chance of correct discrimination.

No, the most important advantage is coming from early decay (being able to sample the high frequency target response).

But the issue with parasitic (coil) capacitance remains.

[Davor]

However, playing with Mr. Thévenin's theorem and the fact that coil/target system works as a (very non-ideal) transformer, that 500ohm would interfere with target taus.
It would not interfere greatly with target signal strength, but will surely screw up any chance of correct discrimination.

[Aziz]

BTW,

getting rid of the damping resistor across the coil does not help much to increase the target signal strength. The coil impedance is much much lower than the damping resistor and the signal won't be significantly attenuated.
Lets make a simple comparison:
Coils impedance: 1 Ohm
Damping resistor: 500 Ohm
See the relation of the impedances!!

Aziz

[Aziz]

Aziz you are right ... I can adjust the control loop to totally "neutralise" the "ringing" ... in fact there are twe parameters that can be adjusted to compensate .... the appropriate mathematical analysis is more Laplace / Heaviside step function than the critical damping analysis usually used.

Paul.

Hi Paul,

do you convert the coil energy into heat or do you recycle it back?

I think you are using more or less a similar concept of my "interrupted TEM transmitter" with switched capacitor using the zero-current-switch.

Anyway, I'm interested what you have found and whether it is novel. But the most important thing is, that's not from Candy!!

Aziz

[chudster]

The ideal target eddy response has the form t*e^(-t/tau). You can easily see it in a real detector if you use Labview to subtract the instantaneous response from a held response. Early sampling only helps for fast-tau targets like tiny nuggets. Sampling at 5us does nothing for a 50us target that 10us sampling can't do. In fact, it could hurt.

Carl,

Thanks for the reply. I believe you are saying that if we took Moodz two waveforms and subtracted then we would see an impulse like blip. I can visualise that this is so.

Assuming we are chasing small gold, doing a thought experiment, if we could terminate flyback within 100nS or less would we see a discrete blip for the eddy current because it seems not to be so in practice as we reduce flyback time so far?

Alternatively with a large ferrous target we terminate flyback quickly have some microseconds with no signal then bang we get the impulse. In other words the timing is not related to a physical time constant referenced to current termination. Its all bound up together as we remove energy.

The reason that I say this might be important since the metal detection problem could be turned into an energy timing problem. What energy went in when and where and what, where, when did it come out. Conservation of energy provides the basis for analysis. That is quite a different problem to a radar like approach where we excite/ reflect off targets and look for a signal later.

In consideration it seems we are removing as much electromagnetic energy as fast as possible so that the target energy can be measured as it is additive and more apparent as we take away excitation but always returns on the coat tails of removing the excitation energy so there is no discrete eddy current event. The reason it returns on the coat tails is that as we remove the current we also remove the excitation to the targets and they immediately start losing their energy according to their "time constant". If we remove energy too slowly we never see small gold which is why we aim to remove it fast.

Sorry if its old ground and understood by all but I have scoured the forums and not really been able to understand.

Am I making sense?

Chudster

[Carl-NC]

The ideal target eddy response has the form t*e^(-t/tau). You can easily see it in a real detector if you use Labview to subtract the instantaneous response from a held response. Early sampling only helps for fast-tau targets like tiny nuggets. Sampling at 5us does nothing for a 50us target that 10us sampling can't do. In fact, it could hurt.