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  • Carl-NC
    replied
    Originally posted by h9361 View Post
    Regarding to your knowledge and experience, do you have any suggestions for increasing power of TX?
    1. Increase the TX power supply voltage
    2. Use a transformer to increase the voltage drive to the coil:
      Click image for larger version

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    3. Use an impedance transformer as used on the MXT:
      Click image for larger version

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      These are tricky to get right and only provide a 2x voltage boost at best.
    4. Reduce the TX coil inductance. For a given coil size, transmitted field energy is proportional to the ampere-turns NI. I is proportional to V/L and L is proportional to N2 so, overall, TX field strength is proportional to N*V/L --> N*V/N2 --> V/N. So reducing N should boost field strength.

    Leave a comment:


  • Smirnov-Arta
    replied
    Originally posted by pito View Post
    ....
    Naturally.
    If a hypothetical composite signal consists of a target value of 10 and a ground component value of 90, it makes absolutely no difference whether the air-test range for that target is 30 cm or 50 cm. This is because the ground component value (the "90") must be filtered out and subtracted—and this must happen dynamically (!). In other words, you could get 50 cm in the air and 25 cm in the ground for the same target, or 30 cm in the air and still get that same 25 cm in the ground. Put simply, increasing air range is completely pointless without proper filtering of the ground component. However, if it’s a marketing ploy, then of course... you have to boost those air-test figures.​)

    Leave a comment:


  • pito
    replied
    Originally posted by Smirnov-Arta View Post
    So, I’ll say it again: I see no sense in achieving a 50 cm air-test range on a 25 mm target with a 25 cm coil if those "specs" are completely negated once the coil is used on actual ground.
    The 50cm range in air tells us what range can be achieved in soil; this is more dependent on the quality of the ground balance than on the quality of the soil = mineralization; there is something called ground tracking which is a major factor.

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  • Smirnov-Arta
    replied
    Originally posted by h9361 View Post
    ...
    Hello. That is exactly what I wrote to you: I see no point in air testing. Since increasing aerial detection range—and doing so significantly—is not difficult (the methods, such as increasing the L/Q ratio or gain, have long been documented and proven), that "air performance" becomes virtually meaningless in real-world soil conditions. The real challenge lies in dynamically subtracting the ground component from the signal. So, I’ll say it again: I see no sense in achieving a 50 cm air-test range on a 25 mm target with a 25 cm coil if those "specs" are completely negated once the coil is used on actual ground. But, of course, that is up to you.

    Leave a comment:


  • h9361
    replied
    Originally posted by Smirnov-Arta View Post
    I respect your work. However, I do not understand the purpose of doing this. The search environment is completely different; consequently, all readings obtained in the air will be radically different—and significantly lower.
    Dear Smirnov,
    Ground subtracting will work when your coil is over soil/ground, not Air.
    Then, GB channel will detect ground and the subtraction will act.
    But, you have not GB if your Coil is in Air mode.
    Thus, i choiced Air mode testing for making an identical situation and understanding its depth (target channel).
    Yes you right, we have a low depth on soil because of GB channel and subtracting.
    Off course, all can be related to GB position. Some detector may put its GB position to lower or higher point.
    But, i think that this lower/higher point may affect on both small and big targets, not only big target.

    I tested several VLF detectors,
    All of them detect coins from an identical depth but some can detect big target from a 50cm more depth!

    Leave a comment:


  • h9361
    replied
    Originally posted by Carl-NC View Post
    Even with a low-impedance driver and a parallel cap you are still limited by the voltage compliance of the driver. If the driver supply is 8V then the peak-to-peak coil current is limited to

    i(t) = 8V/(2*PI*6600*580uH) = 333mA

    If this were not true, then it would imply that you could simply reduce the 47Ω resistor to get more current. But what you will see instead is an increase in waveform distortion as the current becomes more like a triangle wave, while the peak current does not increase.
    Thanks dear Carl for your nice explanation.
    Regarding to your knowledge and experience, do you have any suggestions for increasing power of TX?

    Leave a comment:


  • Smirnov-Arta
    replied
    Originally posted by h9361 View Post
    ....
    I respect your work. However, I do not understand the purpose of doing this. The search environment is completely different; consequently, all readings obtained in the air will be radically different—and significantly lower.

    Leave a comment:


  • h9361
    replied
    Originally posted by Smirnov-Arta View Post
    What is surprising or new about that? It is simply a combined target/ground signal; you subtract the magnetic/ground component, all while keeping in mind the masking effect caused by the transmitting coil's field (specifically the field itself) in the near-surface zone.
    Thanks for your replying.
    All of my tests was Air test.
    Subtracting magnetic/ground in Air test?!
    All testing was based on identical situations, coil size, coin and big targets.

    Leave a comment:


  • Smirnov-Arta
    replied
    Originally posted by h9361 View Post
    Thanks...
    What is surprising or new about that? It is simply a combined target/ground signal; you subtract the magnetic/ground component, all while keeping in mind the masking effect caused by the transmitting coil's field (specifically the field itself) in the near-surface zone.

    Leave a comment:


  • Carl-NC
    replied
    Even with a low-impedance driver and a parallel cap you are still limited by the voltage compliance of the driver. If the driver supply is 8V then the peak-to-peak coil current is limited to

    i(t) = 8V/(2*PI*6600*580uH) = 333mA

    If this were not true, then it would imply that you could simply reduce the 47Ω resistor to get more current. But what you will see instead is an increase in waveform distortion as the current becomes more like a triangle wave, while the peak current does not increase.

    Leave a comment:

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