Target response tester

[Skippy]

Interesting, roughly what I was expecting: affects the fastest responses a bit too much, but quite acceptable for the longer TC targets. That's why I suggested the twin-output option, I felt that a little bit of LPF would clean up the short TC, and more filtering would be useful on some of the longer TC targets. Another option is to have a pair of sockets (eg. turned-pin IC sockets) on 0.2" spacing, and plug in an 'appropriate' leaded cap to suit the test target TC.
I was originally contemplating using the second opamp just like the first, see attached cct sketch. But you could simply add the two pin sockets across the existing 16K opamp feedback resistor. That would then leave the second opamp free, for different gain-scaled outputs etc.

Looking at those three charts, I would think a filter time-constant of 0.5 microsec would be more appropriate for the 10 usec Nickel coin. And filter TC of 0.1 usec for the foil (about 1.5 usec TC ? ). And hence, based on "Filter TC = 0.05 x Target TC", the 25c 'quarter'coin (TC about 100us) could suit a filter TC of 5 usec.
TC = 0.1 usec is 10k & 10pF, or 16k & 6p8
TC = 0.5 usec is 10k & 47pF, or 16k & 33pF
TC = 5 usec is 10k & 470pF, or 16k & 330pF

Just some ideas to play with. Maybe drop your 10k filter resistor to 4k7 or 3k3, then C values would be 2 / 3 times greater.

[green]

I've thought about the thick lead disc vs. the 1 ounce coin ....

the circulating currents in the flat coin will be predominantly in the one flat plane; they will be that way initially, and will stay that way as they decay, hence the uniform 'slope'. The bulky lead disc may start off with currents circulating mainly in the plane, because of the way the detectors' coil / magnetic field pass through the sample. But it seems likely that they may not stay circulating that way, and could easily end up flowing in loops in a perpendicular plane, ie. through the bulk thickness. The currents will still decay (as the lead has electrical resistance)... but the magnetic field they generate will not be picked up (or picked up much less) because their plane is perpendicular to the detectors coil. It's the "coin-on-edge effect", on a microscopic scale.
This is just a thought .... I'm not sure what particular experiments you could try, to understand the behaviour. Would a second coil, at right-angles to the flat disc/ 1 ounce coin give the answer? A drawing would help explain ... the second coil is on the same plane as the flat coin, but oriented so it sees the coin as 'on-edge'. Edit , attached pic.

------------

And regarding your response tester: I was thinking of other ways to clean up the noise on your charts.
It's not hard to implement a simple ( or not-so-simple ) digital filter in Excel.
A simple one to try is: Filtered sample value = (0.25 x Previous sample value + 0.50 x Sample value + 0.25 x next sample value)

Example: if you have samples s₃, s₄, s₅ etc chronologically
then your new samples would be: S₄ = 0.25 x s₃ + 0.5 x s₄ + 0.25 x s₅
You could create dummy values s₀ = s₁, and s_last+1 = s_last to make the first/last S data points less duff.


( did you try the low-pass filter on the 'reference current' input of the log-amp, by the way?)

First paragraph: I can simulate a target in spice with an inductor and resistor in parallel. Double inductance or halving resistance doubles time constant. A 1inch square of aluminum foil has a time constant near 1.75usec. Time constant doubles if width doubles(2inch square)or thickness doubles(2 layers). With the foil, is inductance doubling or resistance halving with the two examples? Or a combination of inductance increasing and resistance lowering? What is causing or why is the inductance and or resistance changing?

[green]

First paragraph: I can simulate a target in spice with an inductor and resistor in parallel. Double inductance or halving resistance doubles time constant. A 1inch square of aluminum foil has a time constant near 1.75usec. Time constant doubles if width doubles(2inch square)or thickness doubles(2 layers). With the foil, is inductance doubling or resistance halving with the two examples? Or a combination of inductance increasing and resistance lowering? What is causing or why is the inductance and or resistance changing?

Found some discussion about the subject. http://www.geotech1.com/forums/showt...229#post238229
With the foil I think I can see the resistance reducing to half by doubling the thickness or width if resistance is measured edge to edge not across the face. Don't know if that makes sense.

[Skippy]

The simplest way to model it, is to imagine it is a single loop of wire. Rather than round wire, it's flat, ie. thin and wide. But it still has resistance, and the inductance is similar to what you can calculate theoretically for a round wire. R of the loop is just the simple R = pl/A. So your theoretical TC is L/R.
For a first attempt, imagine your 25mm square has a 10mm square hole in the middle of it. So your 'wire' is 7.5mm wide, you've measured its thickness, so you can calculate an equivalent 'round wire' radius/diameter. Estimate the circular radius of the wire loop however you see fit. There are online loop inductance calculators, though the actual formula is pretty simple if you want to do it yourself.
I will give it a try on a few targets, later, just for curiosity.

So if you double the width of your foil / target, your loop length will double, but your cross-sectional area will double, too. I would thus expect R to not change (much). The loop inductance calculator will show L will double as r_loop doubles. The complicated natural log expression should be unchanged, as the loop radius and wire radius will go up in proportion.
So, R is unchanged, L is doubled, hence TC = L/R, is doubled.

And yes, there are more complex target models for you PI guys. Thankfully, for a VLF machine, L & R seem adequate for simulation. It's skin effect that makes VLF target modelling trickier.

I'm inclined to copy & paste this into the 'target modelling' thread.

[Skippy]

Some time later .... I'm quite pleased with the success of the simple model. Here's some maths:
I tried it out on a 25mm square aluminium piece cut from a drinks can, that I've plenty of measurements on.
Thickness = 0.100 mm. Conductivity = 38% IACS. Measured TC = 6.8 usecs (on VLF machine)
Assume a 10 mm hole in the centre, remaining metal 'wire' is 7.5 mm x 0.1 mm.
I estimated an equivalent loop radius of [(5+ 7.5/2) x fiddle-factor of 1.1 ] = 9.6 mm.
Effective circular wire diameter = 0.489 mm.

R = pl/A produces a figure = 3.58 x 10^-3 Ohms for loop resistance

Inductor calcs give L = 36.7 x 10^-9 Henry.

Hence Time-constant = L/R = 10.25 microsecs.

That aint too bad, it's +50% over my measured value, I suspect the fact the 'wire' isn't remotely close to being circular in cross-section, accounts for most of this difference.

To pursue this further, it would be sensible to choose/make circular, rather than square, targets. And if possible, (eg. drinks-can metal, of foil) try removing the centre to make a loop.

And regarding changing the metal thickness and it's effect on TC values:

Clearly, doubling thickness will double the loop cross-sectional-area, and halve the loop R value.

But the L value of the loop will also change (fall) a bit, because of the natural log bit of the L formula.

A quick calculation for the drinks-can sample: Doubling thickness (to 0.20mm) halves R, and lowers L to 89%. So TC will be 78% greater.
Unfortunately it would be hard to get double-thickness can metal. I have my selection of cooking-grade foil in varying thicknesses
so this will be used for comparing theory/practice.

[green]

Interesting, roughly what I was expecting: affects the fastest responses a bit too much, but quite acceptable for the longer TC targets. That's why I suggested the twin-output option, I felt that a little bit of LPF would clean up the short TC, and more filtering would be useful on some of the longer TC targets. Another option is to have a pair of sockets (eg. turned-pin IC sockets) on 0.2" spacing, and plug in an 'appropriate' leaded cap to suit the test target TC.
I was originally contemplating using the second opamp just like the first, see attached cct sketch. But you could simply add the two pin sockets across the existing 16K opamp feedback resistor. That would then leave the second opamp free, for different gain-scaled outputs etc.

Looking at those three charts, I would think a filter time-constant of 0.5 microsec would be more appropriate for the 10 usec Nickel coin. And filter TC of 0.1 usec for the foil (about 1.5 usec TC ? ). And hence, based on "Filter TC = 0.05 x Target TC", the 25c 'quarter'coin (TC about 100us) could suit a filter TC of 5 usec.
TC = 0.1 usec is 10k & 10pF, or 16k & 6p8
TC = 0.5 usec is 10k & 47pF, or 16k & 33pF
TC = 5 usec is 10k & 470pF, or 16k & 330pF

Just some ideas to play with. Maybe drop your 10k filter resistor to 4k7 or 3k3, then C values would be 2 / 3 times greater.

Some data with different filters. I coupled to the +input so I didn't have to use another pot to adjust log out to .4v/decade. Used US nickel for target. Series resistor(R11) 15k. Changed C2, 10p, 33p, 100p and 330p.


Thanks for the replies for the foil. I was thinking wrong.

[Skippy]

I presume the (15K + pot) was needed because your capacitor-discharge tests showed the scaling was off? I haven't scrutinised the LOG114 datasheet, but I guess there's a tolerance on the '375mV/decade' figure, like everything else.

Regarding low-pass filtering, two other possibilities I thought of: Use cascaded low-pass filters, one with the first opamp, and follow it with the second passive filter/buffer. Trouble is, these filters have a slow roll-off and lag the signal, so not really what you want for this application. The other approach would be to make a second-order LPF with the buffer amplifier. Attached circuit, and equation for 'maximally flat' Butterworth response.

For your 5c nickel trace, it looked like 15K/33p filtering was about right. This is F_c = 320kHz.

To get this response with the second-order filter:

R = 7k5, C = 100pF, C/2 = 47pF
or
R = 10K, C = 68pF, C/2 = 33pF
R = 15K, C = 47pF, C/2 = 22pF

It would be interesting to see what difference it made. But it's not so practical, as two caps ( or two R's) need swapping to tinker with the response, eg. if you were doing a 25c coin test, the C values should be 1n0 and 470pF.

[green]

I presume the (15K + pot) was needed because your capacitor-discharge tests showed the scaling was off? I haven't scrutinised the LOG114 datasheet, but I guess there's a tolerance on the '375mV/decade' figure, like everything else.

Regarding low-pass filtering, two other possibilities I thought of: Use cascaded low-pass filters, one with the first opamp, and follow it with the second passive filter/buffer. Trouble is, these filters have a slow roll-off and lag the signal, so not really what you want for this application. The other approach would be to make a second-order LPF with the buffer amplifier. Attached circuit, and equation for 'maximally flat' Butterworth response.

For your 5c nickel trace, it looked like 15K/33p filtering was about right. This is F_c = 320kHz.

To get this response with the second-order filter:

R = 7k5, C = 100pF, C/2 = 47pF
or
R = 10K, C = 68pF, C/2 = 33pF
R = 15K, C = 47pF, C/2 = 22pF

It would be interesting to see what difference it made. But it's not so practical, as two caps ( or two R's) need swapping to tinker with the response, eg. if you were doing a 25c coin test, the C values should be 1n0 and 470pF.

The junction of R1 R2 should be about 1/100 of 2.5v or 25mv. With 25mv at input R3 the output should be zero. With 2.5v at input the output should be .8v for .4v/decade. With 250uv at input the output should be -.8v(offset adjust R6). When I had 10k and 1meg for the reference input and 10k input resistor the 16k feedback worked as calculated. When I changed the reference and input resistors I had to change R9 to a 1k pot to adjust calibration. Don't know why.

Could lower amplifier gain to reduce noise. With the 1.5inch Rx I get more than enough signal for a lot of the targets but not the ground sample I have. To get 1 decade below zero(2.5mv)the input(amplifier out)should to be adjusted to less than 1mv offset, I've been trying for less than .5mv.

[green]

Some time later .... I'm quite pleased with the success of the simple model. Here's some maths:
I tried it out on a 25mm square aluminium piece cut from a drinks can, that I've plenty of measurements on.
Thickness = 0.100 mm. Conductivity = 38% IACS. Measured TC = 6.8 usecs (on VLF machine)
Assume a 10 mm hole in the centre, remaining metal 'wire' is 7.5 mm x 0.1 mm.
I estimated an equivalent loop radius of [(5+ 7.5/2) x fiddle-factor of 1.1 ] = 9.6 mm.
Effective circular wire diameter = 0.489 mm.

R = pl/A produces a figure = 3.58 x 10^-3 Ohms for loop resistance

Inductor calcs give L = 36.7 x 10^-9 Henry.

Hence Time-constant = L/R = 10.25 microsecs.

That aint too bad, it's +50% over my measured value, I suspect the fact the 'wire' isn't remotely close to being circular in cross-section, accounts for most of this difference.

To pursue this further, it would be sensible to choose/make circular, rather than square, targets. And if possible, (eg. drinks-can metal, of foil) try removing the centre to make a loop.

And regarding changing the metal thickness and it's effect on TC values:

Clearly, doubling thickness will double the loop cross-sectional-area, and halve the loop R value.

But the L value of the loop will also change (fall) a bit, because of the natural log bit of the L formula.

A quick calculation for the drinks-can sample: Doubling thickness (to 0.20mm) halves R, and lowers L to 89%. So TC will be 78% greater.
Unfortunately it would be hard to get double-thickness can metal. I have my selection of cooking-grade foil in varying thicknesses
so this will be used for comparing theory/practice.

Looks interesting. I could try something with my PI tester. Would you like to move to the 'target modelling' thread.

[green]

Chart with some Heavy duty aluminum foil(kitchen). Square appears to have higher signal strength than round. At start of decay the hole doesn't seem to effect amplitude but decays faster.

[Skippy]

That's pretty much spot on what I was expecting. The hole has only a modest influence, as the currents that circulate in the periphery dominate. Hence why I was happy to 'remove the centre' in order to work out the simple L-R model. And a square target looks like a circular one but 5% -> 10% larger. I imagine the currents don't flow so much in the corners of a square .... though if you make a square hole in the middle, you'll force them to do so, which probably explains why the square with/without hole are similar TC, but the circular with/without hole are a bit more distinguishable.

Just for fun: I assume your 'regular' foil sample, with TC = 1.75 us was the 0.0178mm thick stuff. So this heavy-duty foil, with a TC of 2.54 us must have a thickness about (2.54/1.75) x 0.0178 = 0.026mm ?

[green]

That's pretty much spot on what I was expecting. The hole has only a modest influence, as the currents that circulate in the periphery dominate. Hence why I was happy to 'remove the centre' in order to work out the simple L-R model. And a square target looks like a circular one but 5% -> 10% larger. I imagine the currents don't flow so much in the corners of a square .... though if you make a square hole in the middle, you'll force them to do so, which probably explains why the square with/without hole are similar TC, but the circular with/without hole are a bit more distinguishable.

Just for fun: I assume your 'regular' foil sample, with TC = 1.75 us was the 0.0178mm thick stuff. So this heavy-duty foil, with a TC of 2.54 us must have a thickness about (2.54/1.75) x 0.0178 = 0.026mm ?

Posted some measurements awhile back when I got my new micrometer. Couldn't find the reply so I measured and tested some regular and HD foil. Folded foil for 16 layers, measured and divided by 16. 241/178*1.8=2.44usec calculated instead of 2.55usec measured TC. Could be different foil brand, measurement errors or something else?

Some thing I forgot to mention, the offset pot(R5) doesn't effect the capacitor discharge.

[green]

I've thought about the thick lead disc vs. the 1 ounce coin ....

the circulating currents in the flat coin will be predominantly in the one flat plane; they will be that way initially, and will stay that way as they decay, hence the uniform 'slope'. The bulky lead disc may start off with currents circulating mainly in the plane, because of the way the detectors' coil / magnetic field pass through the sample. But it seems likely that they may not stay circulating that way, and could easily end up flowing in loops in a perpendicular plane, ie. through the bulk thickness. The currents will still decay (as the lead has electrical resistance)... but the magnetic field they generate will not be picked up (or picked up much less) because their plane is perpendicular to the detectors coil. It's the "coin-on-edge effect", on a microscopic scale.
This is just a thought .... I'm not sure what particular experiments you could try, to understand the behaviour. Would a second coil, at right-angles to the flat disc/ 1 ounce coin give the answer? A drawing would help explain ... the second coil is on the same plane as the flat coin, but oriented so it sees the coin as 'on-edge'. Edit , attached pic.

------------

And regarding your response tester: I was thinking of other ways to clean up the noise on your charts.
It's not hard to implement a simple ( or not-so-simple ) digital filter in Excel.
A simple one to try is: Filtered sample value = (0.25 x Previous sample value + 0.50 x Sample value + 0.25 x next sample value)

Example: if you have samples s₃, s₄, s₅ etc chronologically
then your new samples would be: S₄ = 0.25 x s₃ + 0.5 x s₄ + 0.25 x s₅
You could create dummy values s₀ = s₁, and s_last+1 = s_last to make the first/last S data points less duff.


( did you try the low-pass filter on the 'reference current' input of the log-amp, by the way?)

reply #93


Was reading ITMD again today. Test with zinc penny and one with 5/8 hole drilled in it. They could detect the one with hole 1 to 1.5inches farther. Didn't see much difference with the round HD foil with and without hole when charting awhile back. Drilled a zinc penny with a 1/2 inch hole and recorded some data. Don't think I could drill one with a 5/8 inch hole with the tools I have, could try if anyone thinks that would cause a fair increase in signal. Don't see that much difference in signal strength between the penny with hole and one without. Maybe the 5/8 inch hole or the target being farther away rom the coil would cause a bigger difference? Need to get my detector working since I zapped the bench circuit so I could check for detection distance. Any other reasons why the one with hole might cause a higher signal?

While playing with the penny I noticed a couple other things. Recorded and charted the coin in different locations on the coil. 45deg1, most field passes thru coin edge. 45deg2, most field passes thru coin center. For 45deg1 coin position is critical, 45deg2 coin position isn't critical. Wonder if the currents are excited in both plains. Like your idea for the Rx coil 90deg to Tx. Need to make one and try it to see if it shows something different.

zinc penny_US penny, 3/4 inch diameter

[KingJL]

... Drilled a zinc penny with a 1/2 inch hole and recorded some data. Don't think I could drill one with a 5/8 inch hole with the tools I have, could try if anyone thinks that would cause a fair increase in signal. Don't see that much difference in signal strength between the penny with hole and one without. Maybe the 5/8 inch hole or the target being farther away rom the coil would cause a bigger difference?

You better watch your actions with all those zinc pennies... the Fed's will be coming for you!!!

[Skippy]

Got to say, any kind of plated/clad/non-homogenous target would be a poor choice, are proper bronze/brass cents getting scarce these days?

Re: the second pickup coil, any old coil would do, is the 133mm one still available? I'm not entirely sure what to expect from that experiment, you would likely need to obtain log amplitude plots to see any 'evidence'.