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Transformer inductance a function of mu only?

  1. Mar 17, 2015 #1
    Hi,
    I know that the permiability (mu) changes as the core material saturates and I would have thought
    L1 = N12 / reluctance
    L2 = N22 / reluctance
    and they'd both share the same reluctance length, this would indicate to me that L1 and L2 are only a function of mu. Is this correct?
    I know also that L = N.B.A / i
    but if the aformentioned case was correct, does that mean that as i changes, B as a ratio with it, changes in lock-step, maintaining the ratio?

    Thanks

    Below is what I've been looking at:
    useful TX 2.PNG useful TX 1.PNG
     
  2. jcsd
  3. Mar 18, 2015 #2

    jim hardy

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    not sure i quite grasped the question....
    B.A is flux
    and flux is mmf/reluctance
    and mmf is amp turns

    so B/i is a measure of reluctance, actually reciprocal of reluctance i'd say.

    Where reluctance is constant indeed B and i are in lockstep proportion.
    As you approach saturation reluctance falls off. In transformers that requires more current to push flux up to cosine wave's peak, hence that infamous third harmonic distortion.

    It is useful to slow one's imagination down to near freeze frame pace and think of your sinewave or cosinewave as a sequence of DC stills. After all, at any instant current has only one direction(, and one slope.)
    Look at this waveform for sine wave voltage... how hard current must work to push flux up to cosine wave peak.
    02336.png
    I can almost hear it grunt.
    Were iron linear(constant reluctance) everything would be nice smooth sine&cosine waves.
    Transformers operate toward left side of this curve where it's fairly linear.
    silicon steel is reasonably straight and requires reasonable mmf up to a little over 1Tesla
    operating a transformer at less than rated voltage keeps one farther away from saturation.


    fermags-600x488.jpg

    It is interesting to put voltage and current traces on an oscilloscope and watch current sprout those peaks as excitation voltage increases.

    Try it with a your scope and a doorbell transformer.

    old jim
     
    Last edited: Mar 18, 2015
  4. Mar 19, 2015 #3
    Thanks for the reply Jim, fascinating. So for the approximately linear part of the curve B and i are in lock-step approximately.
    Great (new to me) way of phrasing the current flux relation.

    But my remaining question is simple, to rephrase it, is about the inductance of the TX while there is current in both coils; I imagine the equations:
    L1 = N12 / reluctance
    L2 = N22 / reluctance

    describe the inductance of each coil, so since they share the same core (magnetic path) the denominators (the reluctance) should be the same value?

    Cheers
     
    Last edited: Mar 19, 2015
  5. Mar 19, 2015 #4

    jim hardy

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    Well, i didn't see on those pages their derivation of L for those formulas ...
    but another hyperphysics page said this about mutual inductance, which seems to account for the number of turns in their M calc

    indmut2.gif
    So, i agree with your assessment.

    One can always start from the basics
    Definition of inductance is flux linkages per ampere, NΦ/I
    remember your friendly flux formula
    Φ = μNIA/L
    (and since ℝeluctance is L/μA)
    Φ=NI/ℝ

    multiply both sides by N/I
    and we get
    NΦ/I = N2/ℝ

    Inductance is just what you said, N2

    Each winding has its own inductance yet each creates flux linkages in the other. That interaction is the M term.

    if they're doing something different they owe us a heads-up.
     
  6. Mar 19, 2015 #5
    Actually it's me that owes the reader an apology, those equations, or rather, that reasoning, was a possit from me that was concurred with by a friend; not hyperphysics.

    H'mm, food for thought, you've indicated to me that I wasn't clear in my working model to myself. I've since realised that I should start from basics and clarify the conditions. For this I apologise again, I need to learn not to jump the gun. This is an area where I have found myself in a catch 22 before and may be again.

    Ok, so, say that there is no self inductance because the core is so good all the flux from one coil passes through the other coil. Therefore L11 = (self flux linkage = 0) / i1 = 0 and likewise for L22 = 0 (no leakage flux)

    But now I'm wondering that, since L12 = L21 = K*(L11*L22)0.5 for a "homoeneous medium of constant permeability" then they = 0 which confuses me. (P.S lacking measurements I don't know what K would be so I'll assume it's 1)

    Cheers
     
  7. Mar 19, 2015 #6
    I guess what I mean here is that, looking in from either the primary or secondary side, what would the inductance look like? Rather than the zero Henrys, the maths above indicates to me (though I might be misinterpretting it) could you just calculate it as L = 2pi*frequency* 'calculated magnetising impedance'
    as it might appear to eather side?
     
  8. Mar 19, 2015 #7

    jim hardy

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  9. Mar 19, 2015 #8
    Thats right, but what I was saying is that I'm unsure as to how to calculate L1 and L2,
     
  10. Mar 19, 2015 #9

    jim hardy

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    what's wrong with your

    ?
     
  11. Mar 19, 2015 #10
    Ok cool, yeah that's what I wanted to varify; because the eqation in my book for Mutual Inductance is different to yours in that rather than having L1 and L2 in under the square root, it had L11 and L22 self inductances under the square root, which would be zero because it states that self inductance is different to coupled inductance. This is why it's so confusing for me. But you're saying rather than Self inductance being zero (because it's all coupled), I could calculate: rated reluctance of the core
    and then just calculate from the number of turns on that coil (N1) that the self inductance was: L11 = N12 / core reluctance, and same for L22, thus giving a non zero 'M' value?
     
  12. Mar 20, 2015 #11

    jim hardy

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    Doesn't M have to be determined by measuring effect of one coil on the other?

    That's quite a separate thing from inductance of individual coils.

    You'd either have to calculate how much flux from one coil goes through the other
    or make measurements

    and i suspect in a magnetic path involving part air the calculations could be , well, "downright burly" .
     
  13. Mar 20, 2015 #12
    You're not doubt right; I'm just talking theoretically, using your defined value of M and L1 and L2 from the turns of each coil2 over the rated reluctance of the core.
     
  14. Mar 20, 2015 #13

    jim hardy

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    did i define M ? I dont even like M....
     
  15. Mar 20, 2015 #14
    Lol, in your post #7.
     
  16. Mar 20, 2015 #15

    jim hardy

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    Well sure enough i did post the Hyperphysics page in my second post, #4.

    Actually i do like inductance even with its pesky M.

    Just i couldn't see how you got an M value out of this:
    you'd have to do measurements or intimidating(for me) calculations to come up with M.
     
  17. Mar 20, 2015 #16
    Thanks for the reply;
    Oh, so you couldn't do: M = [(N12/ Reluctance)*(N22/ Reluctance)]0.5
    Because I thought you said that you could calculate L1 and 2 that way.
     
  18. Mar 22, 2015 #17

    jim hardy

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    L1 and L2 ? surely each is N^2/ ℝeluctance

    but M ? No, M depends on geometry.

    Might some flux take a shortcut through air here?
    If so M is less than 1.

    http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/indmut.html

    You'd have to measure volts on one coil vs di/dt on other.
    That's a pretty easy measurement to do with sinewaves and ordinary meters thanks to the derivative relationship for sine/cosine.
    For triangle waves i think you'd want a scope.

    Thought experiment:

    A one amp RMS 60hz sinewave has di/dt of (120pi) X 1.414 cos wt
    Current through a 120 watt 120 volt incandescent lamp ought to be close.

    I made myself a SAFE test rig, an extension cord with portable outlet mounted in a box. White Neutral wire is brought outside the box and is long enough to wrap around my clamp-on ammeter. Plug 120 watts of incandescent lamp into the outlet and i have returning through the (insulated) neutral wire one amp at 60 hz.
    So by wrapping neutral around a core i can apply 60hz mmf, one amp-turn per turn.

    Or ΔI/Δt of 377amps/sec RMS per turn.
    To get more precise one could add an ammeter to his test rig.

    (Actually i made that test rig to measure current draw of my refrigerator as its compressor was dying. When i called Sears they said i had 6 hours left on the warranty. Whew !.)
     
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