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  • #31
    Jase, the circuit in posting #28 calculates difference of two integrals.

    The first integral is calculated during the window of first sample. The value is a negative DC voltage in output because appears with inverted polarity of input voltage. Integrating capacitor is C6.

    The second integral is calculated during the window of second sample. Integrating capacitor is C5. During integration increases DC level in output. If the value of second integral overcomes negative value of the first integral, the output becomes positive, ie the target is accepted.

    Therefore to achieve acception of nonferrous metals, we can:
    1. Attenuate amplitude of first sample or increase amplification at second sample.
    2. Change sample delay of the second sample.
    3. Change integrating window (duration) of first or second sample.

    With these measures we can eliminate response of some targets - the diferrence of two integrals is zero.

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    • #32
      Thanks Mikebg
      Try to make some experiments.
      Jose

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      • #33
        Originally posted by mikebg View Post
        Maikl, for best results we need new value of TX coil. At PI metal detectors, we can not calculate inductances of TX and RX winding because we do not know the equivalent capacitances connected to it. They are shown below as equivalent circuit diagram.

        We need to determine suitable inductances L1 and L2 experimentally connecting coils with more turns than expected, then test operation of circuit, unwind one turn, again test etc.

        For the same reason it is necessary to determine experimentally and damping resistances, because they depend on capacitances, inductances and coil resistance (not shown).
        Hi mikebg. Probably would have still something to be changed except coil, would eventually get what it says in Tepco Post # 18. In any case, thanks for your reply.

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