As you can see in the image above, the magnetic field of the shielded inductor looks smaller, but does this mean the unshielded inductor can store more energy? (This all assuming the inductors are both the same Henry rating, coil size, etc.)

I thought maybe the shielded inductor has a smaller, but more intense magnetic field, and so it can store the same amount, but is this the case?

  • \$\begingroup\$ Can you supply the source from where you got the image? \$\endgroup\$
    – Voltage Spike
    Feb 9, 2022 at 19:55
  • \$\begingroup\$ @VoltageSpike we-online.com/web/en/electronic_components/news_pbs/blog_pbcm/… \$\endgroup\$ Feb 9, 2022 at 19:58
  • 1
    \$\begingroup\$ The energy in an inductor, for a given current, is \$\frac{1}{2}LI^2\$. If the inductance isn't changing, then the energy changes if the current changes. Are you applying the same current, or holding something else constant? \$\endgroup\$
    – nanofarad
    Feb 9, 2022 at 19:58
  • \$\begingroup\$ @nanofarad Okay I see, I guess a better way to phrase the question would have been "How does shielding an inductor affect the inductance rating, if at all?" \$\endgroup\$ Feb 9, 2022 at 20:06
  • \$\begingroup\$ If we assume the shield is non-conductive high permeability magnetic material which only has a caveat for the wire turns you'll get the same inductance with much fewer turns in the shielded structure (assuming the dimensions and materials are the same except a piece of the outer shell is missing in the non-shielded version). You can see it with magnetic circuit reluctance calculations. If you buy a core and wind N turns of wire by yourself for a wanted Inductance L, you can use a coarse formula L=k(N)^2 where k is given in the core specs. The formula is not exact, but useful. Shield=bigger k \$\endgroup\$
    – user136077
    Feb 9, 2022 at 21:08

3 Answers 3


Does a shielded inductor store less energy than an unshielded one?

A shielded inductor usually has a construction where the ferrite core that is surrounding the coils is much more complete (more encompassing) than that of a non-shielded inductor. This effectively means it does have an air gap but, that air-gap is quite small.

FYI - it doesn't mean that there is some type of faraday shield surrounding the inductor.

So, we could compare an inductor with a small air-gap to an inductor with a bigger air-gap or, we could compare an inductor with no apparent air-gap with an inductor having a small air-gap.

We know that an inductor with no apparent air-gap actually has a distributed air-gap in the ferrite itself. So, I choose, just for the ease of the explanation, to compare an un-gapped core-set with the same core set but with a small gap.

Let's do this numerically.

  • Let's say the shielded inductor is a full core with no air gap

  • Let's say the unshielded inductor is a full core (as the shielded core) but with an air-gap.

  • Let's assume they both have the same inductance (that seems fair doesn't it)

  • Due to the air-gap, the effective core permeability might be (say) 9 times lower for the "unshielded" inductor compared to that for the shielded inductor.

  • For both inductors to have the same inductance, the unshielded part needs 3 times as many turns as the shielded inductor. This fights against the gains made above.

  • This means that for the same current, the ampere-turns (magneto motive force) of the unshielded inductor is 3 times greater than the shielded inductor.

  • MMF converts to H field due to the mean length of the magnetic field (\$\ell\$)

  • And \$\ell\$ for the unshielded part may be be 3% longer due to the air gap which gives the unshielded part a slight unfair benefit in comparing saturation levels.

  • But, given that the shielded part has nine times the permeability, and only one-third of the MMF (or H field), it's the big loser when it comes to saturation losses.

It loses by 95% of 3.

In other words, if you place the onset of core saturation as a "limit" for comparing the energies stored in the unshielded and shielded parts then, the unshielded part can handle nearly three times the coil current and therefore handles about nine times the energy storage.

does this mean the unshielded inductor can store more energy?

Abso freaking lutely


If you're using an inductor, then only the value and saturation current affect how much energy can be stored. It is \$0.5LI_{sat}^2\$.

If you're designing an inductor, then it's important to realise that almost all the energy will be stored in the air-gaps. That's either discrete gaps introduced into an otherwise high permeability core, or the distributed gaps that you get in a 'powdered ferrite in epoxy' low permeability core.

Maximising the energy in the volume of air available means aiming for a magnetic field that's as consistent as possible. Large variations mean that the low field parts are not storing as much as their high field neighbours.

The magnetic field geometry is complicated, and without doing FEMM it's not possible to say which is better. Bear in mind that both types can vary their effective airgaps by changing their geometry and materials. I suspect that unshielded types will be cheaper!


Long time ago, when we made our (radio-frequency) inductors by hand, there was an option to put on a shield, or not (the inductor was otherwise the same). The cores were cylindrical, the shields were square-based or circle-based metallic prisms.

If you decided to put a shield on, the calculation procedure called for derating the inductance by some percent. It was 2..20% depending on the core and the shield geometry.

Sometimes it didn't matter as the inductor was expected to be fine-tuned afterwards. The shields had openings on the top where one could put a screwdriver and move the core in and out of the coil.

  • 1
    \$\begingroup\$ >"How does shielding an inductor affect the inductance rating, if at all?" A shield can be considered as a "secondary" shorted for the "transformer" (with a "light" coupling) made by "primary" (inductor) and shield ("secondary" side). So inductance is modified (reduced) as also Quality factor. \$\endgroup\$
    – Antonio51
    Feb 10, 2022 at 8:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.