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Detection distance for a US nickel and quarter

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  • Ferric Toes
    replied
    I can see that I will have to fire up one of my later test units where I can vary the flat topped pulse width from 100 - 1000uS and plot the decay for nickel and quarter. Ideally I will look for a low conductivity coin the same size and thickness as the nickel to remove the size variable.

    Eric.

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  • green
    replied
    Originally posted by Carl-NC View Post
    Thanks, I was going to add that somewhere there is a thread on this topic with lots of sims, but then my Internet went out.

    BTW, it's not enough to flat-top the current; it has to remain flat-topped at least 5x longer than the longest target tau you are dealing with. That could be a very long time, by PI standards.

    https://www.geotech1.com/forums/atta...4&d=1588518934 chart from reply #7. Tx is 20us ramp to .5A with constant .5A for 4980us. Should be long enough? Maybe I'm doing something wrong?

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  • Carl-NC
    replied
    Originally posted by Qiaozhi View Post
    To demonstrate what Carl is saying, please see the attached LTspice simulation results.
    Thanks, I was going to add that somewhere there is a thread on this topic with lots of sims, but then my Internet went out.

    BTW, it's not enough to flat-top the current; it has to remain flat-topped at least 5x longer than the longest target tau you are dealing with. That could be a very long time, by PI standards.

    Leave a comment:


  • Ferric Toes
    replied
    Originally posted by Skippy View Post
    I'm struggling with this statement of Eric's:
    "with the result that the voltage has to rise to a higher value for the high resistance target than it does for the low resistance one to try and maintain the same current and field"
    It is the current that creates the magnetic field, which is measured by the detector coil. So the different voltage on the target is an interesting side-effect, but not one that is being measured. ?
    You are right in that we are not measuring the voltage induced in the target, but the the rate of change of the current with time and the voltage this induces in the now RX coil (assuming a mono). This is explained in a paper I have somewhere, but still packed in a box after last August's move. From memory, two identical solid spheres are described with the only difference being the conductivity. The low conductivity sphere exhibits a high starting voltage as seen as a voltage across a Rx coil, with a fast decay; while the high conductivity sphere starts with a low voltage and a long decay. The area under the curves (energy dissipated) is the same for the two spheres.

    This effect is also exhibited by the coil itself, which becomes its own target at switch off. Suddenly open circuiting the coil causes the voltage across it to rise to several hundred volts as it tries to maintain the magnetic field. This energy is partly dissipated in the damping resistor.

    Eric.

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  • Qiaozhi
    replied
    To demonstrate what Carl is saying, please see the attached LTspice simulation results.

    The simulation does 4 runs:
    1. Series resistor 3R3, TX pulse width 50us, target tau 50us
    2. Series resistor 6R86, TX pulse width 250us, target tau 50us
    3. Series resistor 3R3, TX pulse width 50us, target tau 10us
    4 .Series resistor 6R86, TX pulse width 250us. target tau 10us

    Detection Depth Sim: shows all 4 simulation plots together.
    Detection Depth Sim - zoomed in: shows a closeup view of all 4 plots.
    Detection Depth Sim - flat top: shows the plots for the 50us and 10us TC targets with flat-topping. In this case you can see that the strongest signal comes from the target with a tau of 50us.
    Detection Depth Sim - no flat top: shows the plots for the 50us and 10us TC targets without flat-topping. In this case you can see that the strongest signal comes from the target with a tau of 10us.
    Attached Files

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  • Carl-NC
    replied
    I'm gonna offer a different explanation. Suppose that the TX pulse width is wide enough that the coil current has flat-topped and target reverse eddies have decayed to zero. The TX current then steps to zero, and this step induces a step EMF in the target. The induced EMF is the same regardless of the target (EMF = -dPHI/dt). The EMF produces an eddy current i = EMF/Z (where Z is the "impedance" of the target, which is a combination of metal conductivity, skin depth, and size/shape) and the eddy current produces the reverse magnetic field sensed by the coil as another EMF (the voltage you see with the scope).

    Nickels have a higher "impedance" than quarters so you would expect the initial eddy current to be lower, and thus the initial reverse field to be lower. But all this assumes a starting eddy current of zero at the TX step. If the TX pulse is not wide enough, then targets will have a reverse eddy current flowing at the TX turn-off step and this directly subtracts from the t=0+ forward eddy. High tau targets require a much wider TX pulse to kill the reverse eddies so for a given TX pulse width you should see a progressively lower starting RX voltage for progressively higher taus.

    A way to test for this is to widen the TX pulse and see if the higher conductors start increasing. Eventually they should surpass the nickel.

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  • Skippy
    replied
    I'm struggling with this statement of Eric's:
    "with the result that the voltage has to rise to a higher value for the high resistance target than it does for the low resistance one to try and maintain the same current and field"
    It is the current that creates the magnetic field, which is measured by the detector coil. So the different voltage on the target is an interesting side-effect, but not one that is being measured. ?

    Leave a comment:


  • waltr
    replied
    Originally posted by Ferric Toes View Post
    Another way of looking at it, is that the Tx sets up a given value of magnetic field which we can take as a step from a plus value down to zero. The current path in the nickel is a high resistance relative to the low resistance in the quarter. There is a basic law (Maxwell's?) that states that current will circulate in a conductive target to try and maintain the magnetic field at the value it was just prior to switch off. For either target the field is the same, with the result that the voltage has to rise to a higher value for the high resistance target than it does for the low resistance one to try and maintain the same current and field. The end result is that the induced voltage in the Rx mode also starts at a higher value for the low conductive (nickel) than for the quarter.

    Eric.
    That is what I was trying to say. Well stated Eric.

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  • Skippy
    replied
    I meant to add: When I stated the skin depth in Cu-Ni was greater than copper / silver , this would be for a large sample of the metals. Coins have a finite thickness, which obviously puts a limit on the depth currents can flow. They can't flow 3mm deep in a 5c coin, because it's not that thick.

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  • Ferric Toes
    replied
    Another way of looking at it, is that the Tx sets up a given value of magnetic field which we can take as a step from a plus value down to zero. The current path in the nickel is a high resistance relative to the low resistance in the quarter. There is a basic law (Maxwell's?) that states that current will circulate in a conductive target to try and maintain the magnetic field at the value it was just prior to switch off. For either target the field is the same, with the result that the voltage has to rise to a higher value for the high resistance target than it does for the low resistance one to try and maintain the same current and field. The end result is that the induced voltage in the Rx mode also starts at a higher value for the low conductive (nickel) than for the quarter.

    Eric.

    Leave a comment:


  • Skippy
    replied
    Skin depth for the 5c nickel will be roughly 4 times deeper than for high-conductor coins like a 25c. Remember skin depth was inversely proportional to the square-root of conductivity of the metal. And Cu-Ni as used in the 5c has IACS conductivity about 5% ( 4.8% from memory?), so its skin depth is 4.5 times that of pure 100% IACS copper.
    I assume if more metal has currents flowing in it, it will give a stronger response ? Negated somewhat by the smaller diameter/thickness of the 5c.

    Obviously you would need a selection of coins the same size but different conductivities to really do any scientific analysis. Using test blanks would be one way, but using real coins, there is some choice with British coinage. A number of our pre-decimal coins were struck in 925 silver, 500 silver and Cu-Ni ( same alloy as the US 5c ). If you include the New Zealand CuNi threepence, you can obtain a weight-proportional family of coins: 3d , 6d, 1shilling (=12d), 2shilling(=24d) and 2s6d and 5s if needed.

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  • Koala
    replied
    There's two composions of the Eisenhower dollar and two for the nickel.

    Are we sure that it just a size difference

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  • green
    replied
    Originally posted by waltr View Post
    The Nickle signal (Ta) is larger ONLY at a much shorter time since it has a short TC (low conductivity). After 10-30 usec the Nickle signal is LESS than the higher TC targets.

    Possibly in a low conductive target the the magnetic field is stronger, for a shorter time, due to Eddy currents collapsing much faster.
    Another thing to play with is the TX ON time verse various targets.

    The nickle and other low TC target only require a 50-80usec TX pulse (3-5 TC Tau) to obtain largest signal whereas, high TC targets, like a Quarter, requires a 180-200 us or more TX pulse.
    Thanks for the reply. Trying to understand why the nickel has a higher signal at start of decay. TRT_58 used a constant current Tx 5000us on time so all targets should have decayed before turn off.

    Tried another test to see what happens during Tx on with constant rate Tx at 16700A/sec. With the same Tx signal the quarter has a higher signal during Tx on, the nickel has a higher signal at beginning of Tx off. Decay for the nickel is noisy during Tx on but probably good enough to see the decay slope.

    IDMD discusses a coin and a coin with a hole using a VLF to compare signal strength. My PI tester shows the coin with a hole has a higher signal at start of decay TRT_58 reply #7, not much higher. Don't know if coin with hole relates to the nickel quarter comparison.
    Attached Files

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  • waltr
    replied
    Originally posted by green View Post
    Was looking at some charts I've posted in the past. Why is the signal for the nickel greater than the quarter or the Ike dollar? The nickel is smaller, I could guess but then it would be a guess.
    The Nickle signal (Ta) is larger ONLY at a much shorter time since it has a short TC (low conductivity). After 10-30 usec the Nickle signal is LESS than the higher TC targets.

    Possibly in a low conductive target the the magnetic field is stronger, for a shorter time, due to Eddy currents collapsing much faster.
    Another thing to play with is the TX ON time verse various targets.

    The nickle and other low TC target only require a 50-80usec TX pulse (3-5 TC Tau) to obtain largest signal whereas, high TC targets, like a Quarter, requires a 180-200 us or more TX pulse.

    Leave a comment:


  • green
    replied
    Was looking at some charts I've posted in the past. Why is the signal for the nickel greater than the quarter or the Ike dollar? The nickel is smaller, I could guess but then it would be a guess.
    Attached Files

    Leave a comment:

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