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Flux Direction effect on Saturation, Quick Question

  1. Mar 26, 2015 #1
    If you've got an arbitary magnetic Flux1 and Flux2, travelling in the same or opposite directions, through a core, it doesn't matter with regards to the point on the saturation curve does it? Opposing or in the same direction it'll still be at the same point on the mag curve wont it?

    Thanks a lot, I just need a sanity check
  2. jcsd
  3. Mar 26, 2015 #2


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    The point at which the iron saturates will simply depend on the NET magnetic flux.
  4. Mar 26, 2015 #3
    Great, the magnetic moments don't fight eachother, once they're full, they're full. I thought so. Thanks for the confirmation.
  5. Mar 26, 2015 #4


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    It's no different to current in a wire in terms of how this superposition operates Tim.

    Say you had a wire in a circuit which you analysed using superposition, and you found that it had 100A flowing from left to right and 90A flowing from right to left. Then the net current (10A from left to right) is the only one that is physically a real current. No fuse or other circuit element will "care" about the 100A, nor the 90A, they don't really exist. It is only the net current that matters. Same deal with your flux analogy.
  6. Mar 26, 2015 #5
    Oh....oh....Well if that's true then say you had a 95A fuse on that wire, it wouldn't blow. So if I had a flux1 travelling in the opposite direction of flux2, and they're of equal magnitude, there'd be NO FLUX because the NET flux has cancelled out?? And it's only saturated when they're in the same direction?

    One reason I ask is because if you've got a Short Circuit secondary TX the flux from the primary will cause by lenz's law an current to flow which creats a flux to oppose the primary flux. So I imagine the net flux will be low or zero, does this mean this case is different to just going up the saturation B vs H curve? Because this would mean it has a LOW B, yet still has no inductance on the primary and so the primary is acting like a short circuit.

    Last edited: Mar 26, 2015
  7. Mar 26, 2015 #6

    jim hardy

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    uart speaks truth.

    C'mon tim, finish your thought, you're soooo very nearly there.....
    Indeed, more current will rush into the primary.

    What does that increased primary current do to flux?
  8. Mar 26, 2015 #7
    Ahaha, oh, yeah, the story doesn't end there, the primary will oppose the decrease in flux ....so where does B of the core end up in steady state? Somewhere in the middle? I'd have to guess the primary overwhelms the flux of the secondary, saturates the core AND THAT'S WHAT CAUSES heaps of current to be drawn by the primary, and flow in the secondary.
  9. Mar 26, 2015 #8

    jim hardy

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    Why would it overwhelm ?

    Why couldn't they balance?

    Maybe your trouble is in thinking about fluxes. While that's mathematically valid, as uart points out there's only one flux.

    Have you forgotten about mmf, magneto-motive-force, that pushes flux ? It's measured in amp-turns.

    Sum of the amp-turns is what's available to push flux around the core.
    So what balances is the MMF's. (Primary mmf - Secondary mmf )/ ℝeluctanceof core = flux

    How much flux must there be? Somewhere between zero and saturation, surely.
    Just enough to match voltage applied to primary (minus IR drop there)..
  10. Mar 26, 2015 #9
    Point 1: Ok, so under shorted secondary condition: the Primary MMF will be Number of primary turns * [ Voltage on primary / (resistance of primary wire + internal impedance of voltage source) ] ?
    as per the picture: t.png

    And the primary MMF is significantly greater than the secondary?

    Is it fair to say that any section of core with a winding on it can be modeled as a reluctance and an MMF, irrespective of which is infront of wich, so long as they're in series next to each other?

    Point 2: [below] Because I know if Nm is 10, then the flux in the middle section double the two outter sections, but what if Nm is higher or lower, what will the flux in the circuit be then? What if Nm was 5 or 20 turns?
    Will the current flowing through them make the flux through all the them be the same?
    figure out.png
    Last edited: Mar 27, 2015
  11. Mar 27, 2015 #10

    jim hardy

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    your red V = Vsource should read V = Vsource - I X (Rsource + Rwinding)

    Significantly? By just enough to magnetize the core, flux/reluctance. By an amount about the same as transformer's no-load magnetizing current X Nprimary.

    sounds fair

    I think so. mmf's in series add just as do emf's, and reluctances in series add just as do resistances.

    Last edited by a moderator: Apr 28, 2017
  12. Mar 27, 2015 #11
    Ah! yes, thank you.

    These two questions are critical for some pieces of the puzzel falling into place:
    Riiiight, so the permeability mu is still high enough for the inductance of the core to be substantial? Even though the primary is drawing a lot of current? I perhaps made the incorrect assumption that the inductance would drop, but if the inductance is quite high than so is the VARs it's drawing?

    Ok Jim, so I can have an two equal MMF sources in a set-up like above, but one can have more flux passing through it than the other, there's no problem with that?

    For some reason I've been thinking each MMF source could only pass as much flux as it created.
  13. Mar 27, 2015 #12

    jim hardy

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    Maybe we're not using "inductance" in the same sense.

    Inductance is a property of a structure, determined by its permeability , length, area, and how many turns encircle it.
    L = μ0μrN2Area/Length

    Old books use another term "Induction" to describe how much magnetic flux is flowing in a magnetic circuit.

    Further complicate that by definition of inductance being NΦ/I , flux linkages per ampere, and we now have a mishmash of similar words that draw the mind to mix up basic concepts.
    That's the trouble with English versus Math..... Back to Lavoisier, below....

    When i feel and hear a transformer humming because of the flux in its core swapping direction 120 times a second, i think "Now there's induction. There's enough flux in there to support 120 volts on its primary".

    By contrast when i look at its windings i say " That'll have inductance because it's wires encircle some area, mu X n^2 X area/length"


    That's just fine.

    No such rule.
    You are still developing your basics of flux and MMF. Read that Hopkinson's page linked above.
  14. Mar 29, 2015 #13
    Thanks for the response, it was helpful.
    I was inafact using the definition of Inductance you gave first: L = μ0μrN2Area/Length
    However I should have perhaps used the term impedance. It is the specific concept of opposing (Net) Flux and how it affects the aformentioned Inductance that I'm addressing, I'll rephrase:

    So with a TX that has a shorted secondary, there are two opposing MMFs: (Primary mmf - Secondary mmf ) / ℝeluctanceof core = Net flux

    There is a large current in both primary and secondary. As you said this will result in the Secondary coil producing just enough flux to match the primary voltage. You also said "How much flux must there be? Somewhere between zero and saturation, surely." So there is a Net Flux, meaning that the Primary MMF is in some magnitude bigger than the Secondary. But surely it is more one end of the Mag curve than the other.
    You said that two opposing fluxes that Net near zero will have a mag density of B near zero.
    (Side observation: when a transformer saturates, the permiability drops and the impedance drops, so it draws heaps of current)
    What I previously was getting at was that, since the Primary current is quite high and there would be a number of turns, then the MMFs are quite high, But B in the core is quite low, then I was thinking the impedance of the transformer would still be quite high, and the inuctance would still be quite high.
    Thus it would draw a lot of VARs. I'll run through how I see it:

    Fluxfrom Pri = Vpri/ [Npri turns*wfreq of supply]
    this in turn makes the shorted secondary produce a counter flux of:
    Nsecondary turns*[(emf= Nsecondary turns*d(flux)/dt) / resistance of secondary]
    is Flux from Sec= Nsecondary turns*[(emf= Nsecondary turns*{Peak Fluxfrom pri/wfrequency of supply} / resistance of secondary]
    Net Flux = {Fluxfrom Pri - Flux from Sec} / Reluctance

    So since the Inductance of the TX is L = μ0μrN2Area/Length and μ depends on B which is low meaning μ is quite high, than L is quite high and X = 2wL
    then the Icurrent in primary*X is a lot of Vars?

    You're thoughts?

  15. Mar 29, 2015 #14

    jim hardy

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    Good. There will be large opposing mmf's .

    You're still speaking of transformer with short circuited secondary? Yes flux will be low.

    slow down..... you take the stairs too many at a time..
    wiki has a pretty decent link... http://en.wikipedia.org/wiki/Transformer
    "" impedance of the transformer would still be quite high, and the inuctance would still be quite high ""
    Which impedance? Which inductance?

    Remember that applied voltage divides between primary losses and counter-emf ,
    so counter-emf is only that fraction of applied voltage not dropped as primary losses I X (Rp +jXp).
    So your Vpri = Vapplied - I X (Rp +jXp) which will be fairly small .
    i think the rest of that paragraph is okay though.

    That would be XM, magnetizing inductance,
    i dont think that's right.

    You've made a false leap . We all want the answer to be easy, and it is but it's several steps to get there not one.
    The mind will concatenate words and fool us.
    μ is quite high because the core is made from good ferromagnetic material.
    What is changing is not permeability and inductance. , but flux and counter-emf; induction..
    That leaves primary current at the mercy of winding resistances Rp & Rs, and leakage reactances Xp & Xs.
    Those are 'necessary evils' so kept low on purpose.

    L is not changed
    And a lot of watts.

    Primary current is high because there's hardly any counter-emf induced to oppose it
    so what limits Iprimary ?
    Winding resistances and leakage reactances.
    Both of those are 'necessary evils', kept low on purpose.
    That's why primary current is high.

    It's important to work your mental model until it leads you to the right formula.
    μ is not truly constant but for first approximations we treat it as constant, refining only when called for.

    You're getting there. Study that wiki equivalent circuit.
    Visualize that flux inside the core.
  16. Mar 29, 2015 #15
    Thanks Jim, you're practically my mentor :p
    Sorry I should have been more clear, the magnetising impedance Xm and the Inductance of the Primary.
    Hmm, ok, Hopefully you can help me sort out what's going on with the Counter-EMF, with regard to your diagam:
    ] oxy.php?image=http%3A%2F%2Fupload.wikimedia.org%2Fwikipedia%2Fcommons%2Fthumb%2F3%2F39%2FTREQCCT.jpg
    So the current IO through the core impedance won't change, but the counter EMF Ep is usually quite high, but with a secondary short, the counter EMF is low, so heaps of current flows through primary and secondary? (I didn't think that was right because I thought the part of Ep was the ideal transformer and no currnt flowed through an ideal transformer) but that's the only place where I see counter emf and where else could the current flow?
    [As a side note let me see if I remember what the counter EMF IS: As the Vapplied resulting Current goes into the primary, causes a flux, that flux goes round and comes back through the primary coils, the coils try to produce a counter flux to oppose the origianl flux's return, via Lenz's law, and the coil does this by inducing a counter EMF Ep. Is that right?]

    Sorry what I should have said was that μ is a property of the steel that we can observe from the B vs H relationship. (Good to know that inductance and Magnetising impedance is still as high as it would be if there was an open circuit secondary) But what is/was confusing me is that usually I'd think of the graph:
    And I'd think, 'not much flux, B is low, μ must be low', but I can't use that graph here. MMF H ~ NI, and since I is large, N is substantial and constant, then H is large, but B on the graph is NET B, which is low, thus μ is low. That's what I meant.

    I think I'm getting my mental model close, thanks to your help.
  17. Mar 29, 2015 #16

    jim hardy

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    observe IO is in proportion to applied voltage Vp minus I X(Rp + Xp)
    Xm is impedance due to primary inductance, and Rc is core loss, ohm's law applies

    Yes. Counter emf is low because secondary mmf has cancelled almost all the primary mmf . Not much flux but large opposing mmf's.

    An ideal transformer needs no magnetizing current, that's why Xm in model is is to left of it.
    But an ideal transformer has primary current equal to secondary current(load current) X turns ratio.

    Under shorted secondary, what limits current? Look at the model.

    remember we operate transformers below saturation.
    Where's saturation on your graph?

  18. Mar 29, 2015 #17

    jim hardy

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    Here's a good link for magnetizing iron.
    http://phys.thu.edu.tw/~hlhsiao/mse-web_ch20.pdf [Broken]

    see fig 20.13
    Last edited by a moderator: May 7, 2017
  19. Mar 30, 2015 #18
    Aah, ok Let me see If i've got this straight: so on that diagram, even though it is an ideal transformer on the right, the bulk of the large current DOES flow through it and Ep which is through (Np), and is limited only by: Rp, Rs and the small counter-EMF ofcourse. And THATs where the lion's share of the Power is burnt. As the magnetising branch will only use:
    (Vp - I*(Rp + Xp))2/ Xm Watts
    Which is less power in the mag branch than for a non-shorted secondary because I is so much higher through the coils and thus the voltage on them detracts from the voltage on Xm.

    Right so "counter-emf ndΦ/dt in the primary" is the counter emf in the secondary depending on the turns of the secondary. That's faraday and Lenz's laws I gather.

    Saturation on my graph is at the top right, and I imagine opperating a shorted secondary TX is below the maximum tangent of your hysteresis quadrant. What I was saying before was that when I first started this thread I was thinking that there were huge magnetising forces (H was really high) however they're opposing eachother, so NET H is far to the left of the graph, likewise for B.

    If what I wrote above regarding the power is correct then I think you may have cracked it for me, this is a fundementally different way I'm viewing where the current is going than when I started the thread. THANKS
  20. Mar 30, 2015 #19

    jim hardy

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    Secondary turns see same dΦ/dt, but we don't call its voltage counter emf.

    Glad it helped.
    Feels good, doesn't it? Work transformers in your head next few weeks until it's intuitive how they balance mmf's.,

    Then take a break and read that Lavoisier introduction to treatise on chemistry i linked way back.... When i was a young man he affected my thinking process for the better. Pay particular attention to his Abbe de Condillac quote, it's outrageously funny, and happens all the time

    old jim
  21. Mar 30, 2015 #20
    Sorry I should have said the mag branch (Xm, Rc) will use less VARs, not watts.

    Over the next week I will read Lavoisier's treatise, thanks.
    If this is wrong that correct it, if not, then no need to reply: Just to be doubly clear, you did say that the current that flows into the 'ideal' transformer part is IP , where IS = IP/a
    when the transformer is not Open circuit.

    Great, again another misconception I was opperating under was that the ideal transformer had no current in it, but what you said was that the ideal transformer had not 'magnetising current' flowing through it (the mag branch was taken out so that could happen elsewhere), it will still have Ip and Is flowing through it. I'm not sure why I get muddled up like this but hopefully that'll be the last time. Thanks again Jim, have a good one!
  22. Mar 31, 2015 #21

    jim hardy

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    You're quite right.

    And i missed something in that model... look at the little trick they used, it's not obvious at first (wasn't to me anyway) :

    They made their ideal transformer 1::1 and moved secondary side leakage inductance Xs and secondary resistance Rs over to primary side.
    Then they re-named the current flowing into ideal transformer's primary winding IS !
    I'd have named it Ip - (IC + jIM) , and shown IS on secondary side with Xs and Rs there too.

    BUT since the transformer is ideal, their relocation of secondary impedances to primary after correcting for turns ratio, works just fine.
    Kirchoff tells us that my Ip - (IC +jIM) equals their IS .

    So i stand corrected.

    I think you have it now
    good job and thanks for sticking with it.
    To help young folks over the same hurdles i struggled with makes an old guy feel a little less useless.
    Today's curricula seem more math intensive than in my day.
    I hope they still have labs.

    Best of luck in your studies, and in your career.

    old jim
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