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I am struggling to understand what seems like a fundamental behavior of an inductor. I am looking at a chip mount LQM18PN2R2MFR inductor. One of the diagrams on the manufacturer's website shows a graph that describes the relationship between inductance and current. I assume this refers to DC current through the inductor.

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Since there is a ferrite core in the inductor, the DC current aligns some of the magnetic domains in the core material. Is this the reason the inductance drops? If so, if I had a signal with a DC current component of 400mA, then the AC component would see an inductance of 0.8uH?

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    \$\begingroup\$ A ferromagnetic material opposes an external magnetic field by aligning its own magnetic domains against it. The stronger the field gets, the more of the domains are already aligned and the less additional ones are available to increase resistance. A similar thing happens with dielectric materials in physically small capacitors. \$\endgroup\$
    – Stack Exchange Supports Israel
    Commented Mar 12 at 21:37
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    \$\begingroup\$ Fun fact: Based on this principle, one can use inductors as electronic switches. When a coil is unbiased, its impedance is high, when it's biased by a current, its impedance becomes low. This is called a saturable reactor and was widely used before modern power semiconductor. \$\endgroup\$
    – 比尔盖子
    Commented Mar 13 at 20:26
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    \$\begingroup\$ The first computer I worked on was analog. It flew an airplane. The electrical control of the hydraulic actuators was the output of a magnetic amplifier, which was based on the idea of a saturable reactor. So they are more than switches; they are also amplifiers. Then I had a boss who said, "How can you have a computer in an airplane? Where do you put the card reader and line printer?" \$\endgroup\$
    – richard1941
    Commented Mar 15 at 3:04

3 Answers 3

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Anytime you see a curve starting to plateau, it's a sign of saturation. So that curve is describing magnetic saturation in the core. This curve shows you the effective inductance for this inductor under various current loads.

Why it saturates is a more difficult topic but fundamentally, as current increases, the stronger magnetic flux you are creating in the core, therefore making permeability more difficult maintain and thus decreasing inductance.

It's a good question. Ideally, inductors shouldn't do this.

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    \$\begingroup\$ Chip inductors in particular have very large amounts flux leakage so different parts of the core will saturate at different currents. Compare with an inductor wound on a toroid where there is very little flux leakage. \$\endgroup\$
    – Kevin White
    Commented Mar 11 at 18:05
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    \$\begingroup\$ Current does not "plateau" in saturation. Rather it begins to rise, bounded only by the external circuit and the wire resistance. \$\endgroup\$
    – Math Keeps Me Busy
    Commented Mar 11 at 18:13
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    \$\begingroup\$ @mathkeepsmebusy That's a good point 🙂 Perhaps what I was thinking was sort of like an I-V curve on a transistor. \$\endgroup\$
    – Colin
    Commented Mar 11 at 23:17
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    \$\begingroup\$ I appreciate the feedback everyone. Answer has been edited. \$\endgroup\$
    – Colin
    Commented Mar 12 at 13:46
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    \$\begingroup\$ There are often a variety of ways of quantifying the "effective inductance" of even an ideal RLC network. If one judges the effective inductance based upon the size of ideal inductor required to allow the same amount of energy to be extracted within one second given a certain amount of current flowing through it, I would think even a simple network with an ideal resistor in series with an ideal inductor would yield a curve like what was shown, even without saturation. \$\endgroup\$
    – supercat
    Commented Mar 12 at 21:15
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The clue is in your statement - "current aligns some of the magnetic domains in the core material". When you run out of domains to align in the core the core is saturated and the inductor behaves more like an air-core inductor. There will be a B-H curve for the core material so you can pick a flux density that suits your purposes if you are designing an inductor or transformer. The curves will also show you other non-ideal behaviours- temperature dependence of permeability and hysteresis to mention a couple important ones.

An analogous electrostatic mechanism is responsible for the drop of capacitance in some MLCC capacitors as the voltage increases due to drop in dielectric constant of the dielectric. Class II dielectrics have enormous dielectric constants compared to air or plastic film, but that, like the magnet core material, comes with limitations.

If there is substantial DC current you may be able to bias the core (for example with a permanent magnet) so that the inductance is maximum at the expected DC current. Or you can use an electromagnet carrying current in the opposite direction (which is what a common-mode choke does, essentially). It does not seem to be very common in practice but this person got a Masters thesis out of it, for example.

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    \$\begingroup\$ For the second part of my question, is it fair to say that if my signal had a DC component to my signal, I need to look at the this graph to know the effective inductance the AC components will see? \$\endgroup\$
    – tronhawk
    Commented Mar 11 at 19:48
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    \$\begingroup\$ Yes, that’s correct, if the AC component was small in relation to the DC component. If it is large you can look at the peak current (DC + peak of AC in the less favorable direction). \$\endgroup\$
    – Spehro Pefhany
    Commented Mar 11 at 21:26
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As Colin says (good answer, give 'em a vote!), this is a material property.

Remember that magnetic materials are generally made of atoms that have an odd number of electrons on the "outermost shell", meaning that they have a net spin (electronic angular moment) that's not zero – depending on the direction of that "last" electron that doesn't have another electron with opposite spin and equal energy.

Ferrite cores are ferrimagnetic. What that means is that neighboring atoms tend to have opposite net spin, but that they have alternatingly high and low angular moments (due to different energies of the outer uncompensated electrons).

You can apply an external field (from an electromagnet, i.e. your coil!) and align the "stronger" atoms in opposite direction to the magnetic field, and start to suppress the angular moment of the "weaker" atoms.

You can only drive that so far – at some point, the weaker moment is completely suppressed, and you can't easily put any more energy into the atoms.

That's when the material is said to be saturated.

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