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Boy, do I feel stupid!

Discussion in 'Electronic Basics' started by Rich Grise, Jan 17, 2006.

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  1. Rich Grise

    Rich Grise Guest

    I've been poking around with a salvated microwave oven transformer (MOT),
    and doing little diddly simple experiments:
    http://www.abiengr.com/~sysop/images/MOT-test.jpg

    And the 0.1 ohm resistor in series with the primary seems to be dropping
    about .145 volts, +/- .4; that's an amp and a half! At 115 volts, that's
    <brain refuses to do arithmetic> over 150 watts!

    But nothing got even warm. It was humming merrily at 60 Hz, but it never
    even got above room temperature. So, where are these 150 watts going?

    Then, finally, after all of these years of dabbling in things electronical,
    it hit me - that amp and a half is going through _an inductor_!!! With a
    DC resistance of 1.0 ohms +/- 0.1 ohm. That's one and a half watts of
    real power.

    DUH!!!!!

    Next, I'm gonna see what happens when I take the magnetic shunts out;
    http://www.abiengr.com/~sysop/images/MOT-primary2.jpg
    has anyone ever done this, and is it a good idea?
    [the penny is only for scale, but I did have this wild-ass idea that the
    induced eddy currents would make magnetic effects that would be physically
    observable, but no such luck. )-; ]

    And at only 1.4 amps, is it worth the bother to wind more primary
    turns?

    And are there other newsgroups that might want to participate in a MOT
    saga? ;-)

    Thanks,
    Rich
     
  2. You have rediscovered power factor current. The 1.5 amps indicates
    the storage of energy in the magnetic field of the MOT, and its return
    to the power company, twice a cycle. Talk about short term loans.

    The power company dislikes inductive loads for similar reasons. That
    current not only dumps a little power into the MOT winding resistance,
    but all the transformers and transmission lines all the way back to
    the generator. But the watt hour meter at the service entrance
    charges you for only 1.5 watts while this thing sits there and loads
    the grid with 150 VA.
    It is a good idea if you want to witness what happens.
    If you intend to operate the MOT with short bursts of full load
    (turning the primary voltage off between those bursts), probably not.
    If you want to operate it for hours on end with a wide range of
    loads, it probably is. What might you use it for?
     
  3. Rich Grise wrote:
    (snip)
    (snip)

    Before you make the measurements, I suggest you draw a picture of the
    core and try to imagine the magnetic flux paths, and the magnetic
    reluctance (magnetic flux resistance) for the various branches, and
    see if you can reason out what to expect from the experiment.
     
  4. Tim Williams

    Tim Williams Guest

    *Cough* 150 VA. Sober up, Rich ;-) (Noticed you were posting drunk
    earlier...)
    Well, you measured it on a resistor, so if it isn't heat, it's still going
    through the wires, and the power company doesn't like that. Oughta toss on
    a capacitor (ooh, power-on surge-ified!) or some more turns to reduce the
    B-field.
    Well, the shunts are bypassing a little magnetic field closer to the
    primary, but not much (depending on width), and various parts of the core
    are probably running near saturation anyway, so it wouldn't make much
    difference.

    I have an MOT in regular use, without shunts, but it also has sufficient
    turnage that it doesn't saturate much.

    You should check the current waveform on that resistor and see what it looks
    like... bet it's got some nice nipples (that oughta get you runnin' to the
    scope!).
    Nah, you need iron filings for that. Hope the primary is well insulated ;-)
    I would. 150VA is a lot of current you could be using for ___.
    No idea... is there a mad scientist or high voltage group?

    Tim
     
  5. kell

    kell Guest

    http://forum.4hv.org
    Don't tell them I sent you.
     
  6. Alan B

    Alan B Guest

    Used your AC meter, didja?
    Not Watts. VAR's - Volt Amps Reactive. The power transmitted that is
    not usable (billable) load.
     
  7. colin

    colin Guest

    AFAIK the shunts only realy come into play when there is a heavy load on the
    secondary and act as a current limiter, old type welding transformers have
    adjustable shunts to vary the maximum current, when the secondary is short
    circuited the current cuases a magnetic field wich oposes that by the
    primary wich then finds an easier path through the shunts.

    with no secondary at all i suspect they will have an undramatic effect,
    altering the effective magnetic path length/cross sectional area only
    moderatly.

    Colin =^.^=
     
  8. Derek Potter

    Derek Potter Guest

    The effect of a shunt magnetic path is simply to introduce a series
    inductance in the transformer, which will have the effect of a crude
    current limit. It sometimes seems odd that a *shunt* magnetic path
    should introduce a *series* impedance, but it's quite logical - think
    in terms of the secondary current producing a flux that does not
    contribute to transformer action but still creates a back EMF in the
    secondary.

    Same physical system, different way of describing it :)
     
  9. You're not the first to wonder about this. There's a whole buncha Mad
    Scientist types who actually need to know about it:

    http://www.pupman.com/

    Specifically:

    http://www.google.com/custom?q=mot+...ains=www.pupman.com&sitesearch=www.pupman.com

    Don't tell them I sent you.


    Mark L. Fergerson
     
  10. Derek Potter

    Derek Potter Guest

    And there are others who like to connect up several MOTs and try to
    get a few KW out of a magnetron in order to create plasma balls. Or
    to kill burglars and stray cats, I forget which.
     
  11. Derek Potter

    Derek Potter Guest

    Not unless it's running far too hot. By using more turns, you just
    waste the magnetic capacity of the core by reducing the flux. You're
    also adding series copper which will increase the losses.

    You want minimum turns for the highest acceptable flux density and
    then use as much copper as the core will take. That gives you amximum
    power throughput.
    Not yet..... newgroup alt.mot.flash.flash.flaaaaaaaaa
     
  12. Rich Grise

    Rich Grise Guest

    [John's excellent comments on power factor snipped]
    ....
    I haven't really decided yet. I'm entertaining thoughts of a spot
    welder, or just a general purpose bench supply; in any case I'm
    basically going to just play with it for awhile. :)

    Thanks!
    Rich
     
  13. Rich Grise

    Rich Grise Guest

    Uh, yeah, right after I finish that course in transformer physics. ;-)

    I don't even know how the length of the flux path relates to inductance
    yet! )-;

    I would think that there'd be a better coupling factor, since more of
    the flux goes through the secondary, but when I start to think about
    what this does to the primary current, my brain starts to hurt. 8-|

    I once worked for a guy who had many years' experience with transformers,
    and he designed ferroresonant units for battery chargers, and basically
    did it by the seat of his pants with liberal doses of Black Magick. ;-)

    Thanks!
    Rich
     
  14. Rich Grise

    Rich Grise Guest

    Thanks Everybody! This thread has been very enlightening! :) :)

    Thanks!
    Rich
     
  15. Inductance is proportional to the total flux created by a given
    current. The easy flux path through the primary coil is the solid
    metal path through the center of the core and back around through both
    outside legs. The shunts (with their series air gaps) add just a tiny
    bit of additional flux path in parallel to the solid metal loop out
    and back past them. This is like putting a 10k resistor in parallel
    with a 10 ohm resistor. Very little additional (flux) conductivity
    results.

    Now, if there were a secondary on the other side of that shunt, and it
    was circulation big ampere turns that produced a field that bucked the
    flux that passed through it, then the alternate flux path through the
    shunts and their air gap would provide the primary with a reduced but
    not zero amount of flux per ampere turn, regardless, so the primary
    inductance could not be forced effectively toward zero, but would
    retain a minimum inductance, and thus, a minimum impedance across the
    line. That is the current limiting effect of the shunts.
     
  16. Rich Grise

    Rich Grise Guest

    Thank you! I must admit, John (may I call you John?) when you describe
    something, you really do it good! When are you going to write your
    book? You could probably get a lot of material right off google! ;-)

    Thanks again!
    Rich
     
  17. ehsjr

    ehsjr Guest

    Keep on posting as you play ... er ... experiment.
    What do you expect to get out of the secondary, once
    you wind it?

    Ed
     
  18. Derek Potter

    Derek Potter Guest

    2 oz of boiling resin, 8 oz of molten copper, 1/2 litre dioxin vapour,
    3 joules of green light and a dead cockroach travelling at 180 mph.
     
  19. Ron Hubbard

    Ron Hubbard Guest

    Try some of the Tesla coil groups; they just love to blow things up or
    do tricks with high voltages (and sometimes currents).

    Ron
     
  20. Rich Grise

    Rich Grise Guest

    Well, I _expect_ it to act like an ordinary transformer, but that's
    what the play^H^H^H^Hexperimenting is for. ;-)

    Cheers!
    Rich
     
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