Loading ...
Sorry, an error occurred while loading the content.

Re: Naudin Retests MEG-3

Expand Messages
  • dtb1000 <dtb1000@yahoo.com>
    I don t agree that the current *magnitude* is correct. I have used that exact same model of ammeter clamp, and I m always having to double-check the polarity
    Message 1 of 13 , Jan 12, 2003
    • 0 Attachment
      I don't agree that the current *magnitude* is correct. I have used
      that exact same model of ammeter clamp, and I'm always having to
      double-check the polarity and range setting (it has two switchable
      gains, 10mA/mV and 100mA/mV). And don't forget that a conversion
      factor must be supplied to the scope for it to display the probe
      output voltage as current.

      I note that this time he has disclosed neither the setting of the
      current probe range switch nor his scope's voltage-to-current
      conversion setting.

      There's very good reason to suspect a phase shift within the amp
      probe, as one would not normally expect to see such a low power
      factor into a physically small resistive load at such a low
      frequency. But the amp probe is operating at near its upper frequency
      limit, and it is not rated or specified for phase shift. The scope
      manufacturer acknowledges this may happen.

      The best way to check for this is to compare the probe output
      waveform against that seen across a series shunt in the same
      circuit.

      --- In MEG_builders@yahoogroups.com, "Mark Coffman <mscoffman@y...>"
      <mscoffman@y...> wrote:
      > All;
      >
      > Despite implications, there is no way to confirm from these pictures
      > alone that MEG circuit measurements are being made correctly.
      >
      > ...In reference to URL;
      >
      > http://jnaudin.free.fr/html/meg31tst1.htm
      >
      > I just wanted to point out, that while the current probe shown in
      the
      > picture makes it extremely easy to know the exact *magnitude* of
      > current flowing in the MEG output it makes it extremely difficult
      to
      > deduce the correct polarity/phase of the current flow. A reversal
      of
      > current waveform polarity is just too easy to accomplish by:
      >
      > a) having the wire go through the jaws of the current probe in the
      > opposite direction
      > b) selecting the return wire rather than the source wire for
      > measurement
      > c) inverting the current probe instrument body.
      >
      > The cleanest way to rectify this problem would be to create another
      > picture in this set showing the voltage drop through some low value
      > shut resistor while having it's waveform superimposed with the MEG
      > output voltage waveform *while both scope minus leads are connected
      > to the exact same point in the circuit* and simultaneous dual
      channel
      > scope sweep.
      >
      > Despite assurances that the indicated current phase polarity is
      > correct, reversal of the polarity of the current waveform is
      > consistent with the lack of real power being disipated in the
      > conditioned resistor rather then that which is indicated
      numerically.
      > So I feel that this current waveform inversion error is indeed
      > occurring. Since this absolute phase validation would be so easy
      for
      > the original experimenter to accomplish, I think it is necessary for
      > him to demonstrate correctness of the current waveform polarity
      > before he should ask us to believe that there anything counter to
      > accepted electrical theory occurring in this particular MEG
      circuit.
      > Thanks.
      >
      > :MarkSCoffman
    • Mark Coffman <mscoffman@yahoo.com>
      ... Thanks for creating that excellent diagram Stan. ... (These things and their associated math are discussed technically in any copy of The ARRL Amature
      Message 2 of 13 , Jan 13, 2003
      • 0 Attachment
        ---Stan Mayer wrote:

        >
        > Anyway, for the heck of it I have used a graphics editing program
        > to create another version of Naudin's scopes traces that shows the
        > current inverted or flipped 180 degrees and it is attached.
        > Perhaps some of you readers who are technically qualified
        > would like to assess this new trace set and make comments.
        > Enjoy,
        > Stan
        >

        Thanks for creating that excellent diagram Stan.

        >
        > Interesting idea and yes very possible. However if the current
        > shown in Naudin's latest test is 180 degrees wrong, then the
        > difference in phase between current and voltage would be
        > something like 120 degrees. That's a lot and not something
        > that can be explained by normal RC or RL phase shifting IMO.
        >

        (These things and their associated math are discussed technically in
        any copy of The ARRL Amature Radio Handbook)

        Just to review; Every AC circuit has two kinds of "energy" flowing in
        it; the real part, which is the kind that heats resistors and the we
        pay for, and an imaginary part that represents energy trapped or
        stored in the circuit. The imaginary part could be called rotating
        energy or resonant energy. So an LC circuit, not compromised by
        resistors, automatically falls into a resonant condition after any
        driving signal is removed. The real part of the energy is the
        part of the current flow that is in phase with the voltage, rising
        when it does, while the imaginary part is anti-phase. So if you see
        voltage going up and current going negative one knows one has a
        resonant condition in the circuit.

        Stan as you have proposed in your previous post; it is easy to suspect
        that a conditioned resistor has become a capacitor and therefore will
        have it's own has capacitive reactance. The burning-in process is
        also called sintering and it is the same thing that is done to
        manufacture tantalum electrolytic capacitors. So one could suspect
        that the conditioned resistor is also a fairly good sized capacitor.
        Without being able to do sensitivity analysis and without an estimate
        component values on the MEG, I am unable to estimate whether the
        resonance between the capacitance of the conditioned resistor comes
        to dominate the MEG operating frequency. It is easy to assume though,
        that the capacitive reactance is the cause of any low impedance
        behavior observed in the load. So from this point of view since the
        capacitance of the output is unknown - so either chart could be
        correct.

        --Of course;
        The conditioned resistor is most likely a non-linear circuit element
        that adjusts itself based on it own internal temperature and therefore
        has variable reactance. Non-linear devices are often designated with
        a multiplication symbol. Most likely it is responsible for negative
        feedback AGC automatic gain control, it keeps the MEG core from
        becoming magnetically saturated and may be responsible for inter-
        gating of real energy and stored resonant energy.

        -> Therefore the reactive set points most likely *moves* between the
        a real dissipative and a "inactive/idle" resonant mode in the process
        of keeping energy levels stable inside the MEG. Only a times two (x2)
        change in reactance would be required to do this. Therefore the
        JNauddins picture is when the MEG needs to dissipate max real energy
        and the average case is what causes the heating in the load when the
        MEG is regulating. (Hopefully greater then the 6.5Watts input.)

        Other people have made the claim that coupling real energy out of the
        MEG is difficult and this shows why. To couple energy out of one
        circuit into another requires component changes but the conditioned
        resistor is monolithic. The heating that changes the conditioned
        resistor internal characteristics is also responsible for throwing
        away energy rather then redistrubuting it. The conditioned resistor is
        going to have to be replaced by a circuit model built of standard
        components and modified before it can both preform it's function and
        couple out energy.

        :MarkSCoffman
      Your message has been successfully submitted and would be delivered to recipients shortly.