Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Problem with core saturation

  1. Jun 14, 2015 #1
    Over the years I've learnt a lot about TXs etc. But I'm a bit rusty. Any thoughts on general reminders about core saturation.

    I know there are huge issues, but the specifics I can't remember that well. I vaguely remember hearing something like when there is a DC component on a transformer the hysteresis curve would get shifted up. But what's the difference between say if you were using an air core or a saturated core in practice? (as far as V and I are concerened)
    So when the core saturates and the permeability drops away to near air permeability and the flux isn't increasing that much or hardly at all, what's the big deal? Still better than using an air core isn't it? Or are there spikes in current or something I'm forgetting that happen at saturation that are dangerous, that wouldn't happen if it was just a linear air core?


    Edit: actually one thought occurs to me, if the primary voltage is going up, but the dΦ/dt isn't going up much because it's saturated, then V2 = N2*dΦ/dt
    the voltage on the secondary will be kind of a square wave won't it?
    And I suppose there is little back EMF on the primary so it will draw heaps more current?
    But how does this contrast to if the TX was DESIGNED to be air core, instead?
    Last edited: Jun 14, 2015
  2. jcsd
  3. Jun 14, 2015 #2
    I'm not sure, but isn't there some non-linearity near saturation? While this might be useful in a few applications, in most applications it will lead to unwanted signal (or even power:nb)) mixing.

    Plus there's the loss of the typically carefully designed efficiency with the magnetic flux spreading all cattywhumpas.

    I like your idea of using a transformer in saturation to square up a wave. Clever.
  4. Jun 15, 2015 #3
    Yeah I looked into it and there are transformers that do actually do that, although it's a bit complicated.

    But I'm still confused about what is happening on the secondary, so the current drawn from the primary spikes and the excitation voltage on the primary drops because the resistive voltage on the coil gets bigger, but rather than flattening the top off the secondary excition more, wouldn't since it's proportional to the derivative of flux the secondary excitation actually drop right away to just be really low....
  5. Jun 16, 2015 #4

    jim hardy

    User Avatar
    Science Advisor
    Gold Member

    I'm not sure the phasing is correct on this one,
    saturation doesn't have to occur at sinewave peak
    but how nicely his picture conveys the derivative relation between flux and Vs !
  6. Jun 16, 2015 #5
    Hi Jim!
    I did read a bit about peaking TXs but I didn't see that page, that's great. Peaking TX has an extra secondary winding that would Resonate with a capacitor, to capture the harmonics, from memory.
    So that picture is pretty indicative of how any es will react in any transformer? Not just a peaking transformer?
    I have some issues to nit-pick about that picture, see where the flux goes up linearly and the derivative of that is a steady es, wouldn't in reality that section of flux increase still look like a sineusoid going from -90o to 90o, and the flux wouldn't be flat, it'd just be a very shallow increase. See my interpretion of a normal saturating transformer:
    my interpretation.png
    I still need to read that page you linked to better, but what do you think?
  7. Jun 16, 2015 #6

    jim hardy

    User Avatar
    Science Advisor
    Gold Member

    i agree with you

    search on saturated transformer and you'll get photos of actual oscilloscope traces.

    glad to see you're working derivatives in your head now !

    ""Progress through patience , perfection is an asymptote......"
  8. Jun 16, 2015 #7
    I'm happy you agree, thanks.

    I have another question, which is, since (neglecting leakage flux) the Epri = N dΦ/dt
    and Esec = N dΦ/dt

    it is the change in Φ (which is net Φ from the Pri and Sec MMFs), which causes the back EMF on the Pri to limit the current drawn from the primary coil.
    What does inductance actually have to do with a transformer? it seems like it's just faraday's law (as stated above) that explain transformer operation, I can't imagine how stored energy in inductance is relevant? I know that some magnetising current is needed to create the flux, but is this relevant to inductance?

    Is inductance or stored inductive energy relevant to a TX, or is it just a side affect? Could you have a TX with no inductance and still set up the flux?
    Last edited: Jun 16, 2015
  9. Jun 16, 2015 #8

    jim hardy

    User Avatar
    Science Advisor
    Gold Member

    What is inductance ? The physics definition ?

    Inductance is flux linkages per ampere,
    which is NΦ/I

    Doesn't that determine what is magnetizing current?
    Try and work that out with algebra.

    What's inductance of an ideal transformer primary ? Number, please...
  10. Jun 16, 2015 #9

    Edited: Yeah so that's what I'm now thinking, it's not possible to have a TX with no Inductance because the higher the inductance, the higher the flux (Li/N). So if there is no leakage flux to disipate energy in eddy currents in objects near by. What happens because of, or to, the energy stored in the inductance of the core?
    Does it go back to the primary, get transfered to the secondary, or get burnt in magnetising losses?

    EDIT: (Relevant extract from a message)
    As far as I've been thinking about it, you need a little bit of magnetising current to set up the flux (which means that the PF will be less than 1), But I'm not sure how much? I always thought the amount of flux is dependant on the reluctance of the core (but that's only if you know the MMF), then the amount of flux in the core would determine how much the back EMF was, thus determining how much real in-phase current the primary side would draw? But maybe I'm thinking about it wrong if you only know the applied voltage: is the amount of magnetising current that the TX needs determined by the inductance reactance: Imag = Vpri/jω(L)
    thus Φ = I*L/N,
    thus Back EMF = N*dΦ/dt

    What do you think?
    Also, since there is in reality a leakage flux, does this mean that the bigger the leakage reactance the lower the power factor?

    Last edited: Jun 16, 2015
  11. Jun 16, 2015 #10

    jim hardy

    User Avatar
    Science Advisor
    Gold Member

    keep going
    Φ = I*L/N
    so EMF = N*dΦ/dt as you said
    EMF = N * d(I*L/N)/dt
    L and N are both functions of geometry and iron properties not of time so they're constants

    EMF = (N*L/N)* dI/dt = the familiar E=LdI/dt, so we have agreement with a well known formula. That sorta confirms our logic.

    Seems so to me, it's in series with the winding so adds voltage drop IjXleakage, shifting phase a little.
  12. Jun 17, 2015 #11
    Ok, let's just map this out to refine the process>
    There are two ways of modeling the TX:
    Through MMF if you know the current through the primary coil.
    Through knowing the applied voltage to the TX primary coil, by which is determined in this order:
    Know the inductance so work out the magnetising current, know the magnetising current to find the flux, find the flux to find the primary Back EMF and the secondary excitation voltage.

    How's that? But that still doesn't address if the 'Energy' stored in the inductance does anything, or where it goes?

  13. Jun 17, 2015 #12

    jim hardy

    User Avatar
    Science Advisor
    Gold Member

    ?? You're not thinking again.....

    That's basic to ohm's law for AC .
    When Z is reactive, voltage and current are not in phase .

    Draw yourself two sine waves one above each other 90 degrees out of phase.
    Look at instantaneous power flow
    Observe that
    whenever volts and amps have same sign their product is positive, ie power flows into the reactance
    when their signs differ power flows back out of the reactance

    that's what reactance does, cycles power back and forth between itself and source,

    Have you never made this sketch?

    hence the old adage "Inductors and capacitors draw only imaginary power:.. "
    Hogwash !
    They draw very real power but it's only borrowed for a quarter cycle then returned.
    So it averages zero.
  14. Jun 17, 2015 #13
    Ha Ha! Brilliant.
    NO! I've never actually made that sketch, but I like it a lot! Ok, humour me though, say there are two inductors, one inductor is the inductance of the leakage flux and one inductor is the inductance of the core.
    So there are two stored inductive energies. Both these energies come from the primary excitation, so are both these stored energies returned to the primary excitation supply?
  15. Jun 17, 2015 #14

    jim hardy

    User Avatar
    Science Advisor
    Gold Member

  16. Jun 17, 2015 #15
    That's actually going to be handy for me to revise, cheers.
    Although in saying that, I guess one thing that also bugs me is, there are formulas for magnetic (hysteresis and eddy current) loss are caused from the flux that is in the inductor, so not all the energy can be returned to the primary (if that's where to goes to), unless the inductor energy and core losses are different...?
  17. Jun 17, 2015 #16

    jim hardy

    User Avatar
    Science Advisor
    Gold Member

    Of course, inductance is inductance...

    Of course, that's why we use the concept of ideal circuit elements so we can think one step at a time.
    A real inductor is not quite lossless.
    So we add to our ideal inductor an ideal resistor in series to represent resistance of the wire,
    and another in parallel to represent losses from yanking the iron molecules around,
    and we arrive at a model that's closer to reality.

    You need to practice step-wise thinking. :smile:
    Last edited: Jun 17, 2015
  18. Jun 17, 2015 #17
    Inductor energy always comes back (at least ideally). Look at Jim's graph. Inductors borrow energy for a time then return it. (Some tiny amount of the flux links to other circuits as a parasitic transformer or RF noise, but that is rarely modeled.)

    Core losses are resistive. Core losses include eddy currents and the like which create I2R losses as heat. They include I2R losses in the windings. (Technically those aren't core losses, but they are real resistive losses.) So an inductor continually borrows and returns the same energy, but each time this happens parasitic core losses take their toll. Generally it's not good to have purely reactive power flowing around causing I2R losses.
  19. Jun 17, 2015 #18
    Yeah, the problem I was conceptually having was with where the energy goes. Jim confirmed that it goes back to the primary source.
    as for your next paragraph I'll address that with my next reply to Jim.
  20. Jun 17, 2015 #19
    Ah of course *>hits head<*, I totally forgot about the parallel resistance to the magnetising Xm, oh dear, my mind is like a sieve when I don't practice this in a few months. (yeah so the Rc takes care of the magnetising losses. I suppose if you wanted to improve the Ideal TX model to make it more realistic, you could put a resistor in parallel to the Leakage Reactances to model any small losses caused by the leakage flux inducing magnetic losses in the real world outside of the TX.

    Thanks for confirming that the inductance is only relevant ot the primary side of the ideal TX.
  21. Jun 17, 2015 #20

    jim hardy

    User Avatar
    Science Advisor
    Gold Member

    Seems thinking has converged ..... rewarding to see "the light come on", isn't it ?
  22. Jun 17, 2015 #21

    jim hardy

    User Avatar
    Science Advisor
    Gold Member

    It's worthwhile spending the time to become conversant in the model of a real transformer

    this one at Wiki is just one of many similar

    when you can explain what physical effect each component represents ,
    and how you might measure it
    you are well equipped in the transformer department.
  23. Jun 17, 2015 #22
    Rewarding coupled with a feeling of foolishness :p
  24. Jun 17, 2015 #23

    jim hardy

    User Avatar
    Science Advisor
    Gold Member

    dont be so hard on yourself. Enjoy your success !

    help somebody else. " Pay it Forward" as the kids say (ahh, Helen Hunt - sigh)

    old jim
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook