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I'm trying to nail down a few concepts on inductance. Let me frame my confusion:

Take an inductor with an air core. As current is applied, energy is stored within the magnetic field that the inductor creates. Initially, the current is inhibited from rising quickly since energy is being used to build the magnetic field. Due to the finite series resistance, the current cannot infinity increase, and similarly, the stored energy in the inductor's field levels off. If the RL circuit is suddenly opened, the field collapses, and becomes a source of electrical energy for the system to maintain a continuous current. From what I've read, the details on how energy is precisely stored and how the field exactly collapses are very complex. I am okay with not getting into the weeds on that.

What I am confused about is the situation that arises when a ferromagnetic core is present. Now the small field created by the inductor can be amplified by the magnetic domains of the core. But there are a few edge cases that are tripping me up, so I'll walk through the same initial conditions and steady state conditions that I did with the air core. Initially, energy is being stored in the magnetic field of the core material, this occurs by setting the domains. If the core saturates, there are no more domains to be set, and therefore no more energy is stored in the field, aka the inductance drops drastically.

When the current is removed and the magnetic field collapses, it seems like this is analogous to saying that the magnetic domains 're-scatter' in the core, as a result of losing the magnetic field energy that bound them to be in line. However, there is residual magnetism due to the hysteresis bh curve. But if this remaining energy did not dissipate back into the circuit, why isn't there an assymetry to the inductance value when charging and discharging. Energy that was used to build the field did not get expelled back into the circuit. The charge and collapse of the field seems asymmetric with a core. What am I missing?

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  • $\begingroup$ Of course there is energy loss due to hysteresis. But I am not sure what you want answered as you seem to already have a handle on the issue. In most physics classes inductors with cores are treated as ideal to help simplify problems. $\endgroup$
    – Triatticus
    Jul 3, 2018 at 2:21
  • $\begingroup$ Well let's say that I'm trying to calculate a time constant for an LR circuit with a core. If the inductance changes due to the residual magnetism on the field collapses, then I have two different time constants for charging and discharging. This might matter in many engineering applications. $\endgroup$
    – ardarn
    Jul 3, 2018 at 2:39
  • $\begingroup$ Might need some empirical analysis, the residual field in the iron core is many many times smaller than the field that the inductor will have due to the core. I syppose it depends on how bad of a hysteresis iron has. Im upvoting to raise awareness. $\endgroup$
    – Triatticus
    Jul 3, 2018 at 2:50

2 Answers 2

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In most applications, the current through an inductor is cyclical and the magnetic core will have a reasonably symmetric hysteresis loop.

There will be energy losses in each cycle, but the losses will be similar in both direction of the loop.

This is applicable to continuous AC currents or pulses, with and without a DC offset.

So, as long as the current is not too high to get the core into saturation, the behavior of the inductor should be reasonably symmetric.

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  • $\begingroup$ "In most applications, the current through an inductor is cyclical and the magnetic core will have a reasonably symmetric hysteresis loop." I'm a bit confused about this claim if your saying that most applications don't enter saturation. If saturation is not entered then isn't there no hysteresis loop? I Agree that if saturation is not entered and there is no hysteresis loop, there is total symmetry of energy stored and released, but even if saturation is entered and there is an even hysteresis loop, it seems like it takes more energy to store the field than to release it. $\endgroup$
    – ardarn
    Jul 3, 2018 at 12:41
  • $\begingroup$ @AidanPhillips Hysteresis loop is present even if there is no saturation (HB curve going flat): just google it and look at different curves. The loop is a result of losses due to analog of internal friction of domains. Yes, more energy is spent to magnetize a core than can be extracted from it, but when magnetization and demagnetization (or magnetization in the opposite direction) is done in a cycle, similar losses happen in both directions and the shape of the loop reflecting the losses is reasonably symmetric. $\endgroup$
    – V.F.
    Jul 3, 2018 at 14:26
  • $\begingroup$ @AidanPhillips There is some good info and examples of hysteresis loops in this article: meettechniek.info/passive/magnetic-hysteresis.html $\endgroup$
    – V.F.
    Jul 3, 2018 at 14:40
  • $\begingroup$ Okay, so In the AC case I can buy that after the very first cycle (when b=0 & h=0), there is an equal time charging and discharging, some energy lost from hysteresis in there. This conclusion can be shown from the bh curve's symetry: the upper and lower curve represent time periods when the AC current is decreasing and increasing, respectively, and for that whole time energy is being stored, collapsed or dissipated as loss. Since they are symmetric the charging and collapsing inductances are eaual. However, in the DC case small deviations might exists due to the residual. $\endgroup$
    – ardarn
    Jul 3, 2018 at 16:56
  • $\begingroup$ @AidanPhillips Could you describe the DC case you have in mind? $\endgroup$
    – V.F.
    Jul 3, 2018 at 18:01
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I got stuck with the exact same question and spent long time reading and researching. I also wondered: Stored energy is 0.5 * L * I(t)^2, but where exactly is the energy and how is it transferred if there remains magnetism afterwards? I just want to understand the basic, one DC pulse. Finally I found a document explaining exactly this case, unidirectional excitement for just one pulse. Depending on the material, it's really the case that less energy is returned than you put in for this one pulse.

Page 10: http://www.idc-online.com/technical_references/pdfs/electrical_engineering/Eddy_Current_&_Hysteresis_Loss.pdf

There is also an application note of Würth (producing coils) showing it: https://www.we-online.com/catalog/media/o109035v410%20AppNotes_ANP029_AccurateInductorLossDeterminationUsingRedExpert_EN.pdf

After reading that, I noticed there are hard-magnetic cores where literally no energy is returned in each cycle. How can this work in, let's say, a DC buck converter? The answer is that coils with an air gap are used. Even though they sometimes look like toroidal coils with a closed core, they have a part of the core filled with non-magnetic material making them air-gap inductors with an open core in the end. This leads to the result that most of the energy is stored in the field across the airgap, hence the core losses will only be a fraction of the energy stored and returned in one cycle.

Hope you still get the reply as the thread is old, but I guess more people could be looking for it.

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  • $\begingroup$ Thanks for the response! I think maybe the question should be reoriented towards an application that would be impacted by having different inductor time constants for charging and discharging from a single pulse. I'm not sure there are many, but will do some thinking. $\endgroup$
    – ardarn
    Apr 3, 2022 at 15:07

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