How will the core loss (hysteresis and eddy current loss) in an LC circuit at high frequencies (about 800kHz) affect the inductance of the inductor. When I measure the frequency I can see that it will increase. Moreover the voltage will decrease (Why does this happen?).

I thought that the only cause for making this happen is that the inductance decreases since the frequency is defined as

$$f = \frac{1}{2 \pi \sqrt{LC}}$$

and the capacity stays constant which means that the only way for the frequency to increase is a decrease in inductance. What exactly will make the inductance decrease and how can the reduction of inductance be calculated?

  • \$\begingroup\$ It could be useful to specify the frequency range of the intended operation. Is it for audio, radio or mm-wave frequencies? Just to know which frequency dependent effect can be neglected. \$\endgroup\$ Jan 27, 2021 at 20:57
  • \$\begingroup\$ it is about 800 kHz \$\endgroup\$
    – ht332932
    Jan 27, 2021 at 21:05
  • \$\begingroup\$ "When I measure the frequency I can see that it will increase due to core losses." How are you measuring this? Are you actually inserting a core? What is the construction of your core & coil? \$\endgroup\$
    – TimWescott
    Jan 28, 2021 at 0:20
  • \$\begingroup\$ As mentioned previously, losses are a side effect. For instance, the heat lost due to eddy currents flowing in a conducting iron core are due to the eddy currents flowing through the electrical resistance of the iron core. Those losses are just losses and manifest as a resistor in the equivalent circuit. However, it is the eddy currents themselves that cause the inductance to change and, with lower resistive losses in the core, the more obvious that inductance change is. Eddy current losses are a side effect. Eddy currents themselves cause inductance to lower. \$\endgroup\$
    – Andy aka
    Jan 28, 2021 at 10:40
  • \$\begingroup\$ So, when you say this: When I measure the frequency I can see that it will increase due to core losses. - that is incorrect because it has nothing to do with "losses". A perfect conductor with no resistive losses will cause the perceived inductance to lower more than a conductor with high losses. \$\endgroup\$
    – Andy aka
    Jan 28, 2021 at 10:44

3 Answers 3


What exactly will make the inductance decrease

Consider the following tuned circuit that resonates at precisely 800.000 kHz: -

enter image description here

If I plot the resonance of Vout whilst changing R1 from 1k to 3k3 and then to 10k, the frequency doesn't move one little bit: -

enter image description here

It remains at exactly 800.000 kHz. It gets peakier as resistance increases but that's the Q factor changing; a well known phenomena. I'm showing this because it's important to realize that "resistive losses" of the type I've used above do not change the resonant frequency. Example 1.

Next, I introduce a 1 μH shorted inductor (L2) and lightly magnetically couple it to the main inductor (L1) via "K1". Initially k (coupling factor) is set to zero -

enter image description here

The coupling factor is varied from 0 to 0.8 in steps of 0.1: -

enter image description here

The left-hand plot is with zero coupling and, as expected, the resonant frequency is exactly 800.000 kHz. If I make the coupling 0.1, the resonant frequency gets a little bit higher but, as I increase coupling towards 0.8, the resonant frequency gets a lot higher.

The coupling of the 1 μH shorted inductor is the equivalent of introducing a solid conductor in the vicinity of the magnetic field produced by the main inductor L1. Eddy currents flow and, it is these eddy currents that shift the frequency. In effect, the shorted inductor (L2) is reducing the inductance of the main inductor due to transformer coupling.

Next, consider what happens when I short the inductor using a 3 Ω resistor: -

enter image description here

I've varied the coupling factor from 0 to 0.8 in 0.1 steps as previously.

The peak of resonance gets lower because losses increase as coupling increases but, importantly, the resonant peak shift to the right (as coupling increases) is less extensive compared to when I used a pure shorted inductor.

This tells us that it isn't the eddy current losses that shift the frequency, it's the actual eddy currents themselves and the action of transformer coupling. In fact, the losses seek to reduce the extent that the resonant frequency shifts.

Hence, when you say this: -

When I measure the frequency I can see that it will increase due to core losses.

You are mistaken. It isn't the losses, it's the eddy currents themselves and not any heat (loss) generated by those eddy currents).

Moreover the voltage will decrease (Why does this happen?).

That has absolutely got everything to do with eddy current losses as my answer has tried to explain.

how can the reduction of inductance be calculated?

Well, I've used a simulator above but, to give you better advise here I need to know exactly what your full circuit is. You should also study this answer and recognize that the change in frequency is due to transformer coupling and the changes brought about when coupling varies.

As an aside to the main story and focusing on how to derive the equivalent circuit of the coupled inductors, these three scenarios below (A, B and C) are all equivalent: -

enter image description here

And, if you do the math on scenario "C" putting the two inductors in parallel and then putting the combined value in series with L7 you find something quite revealing; no matter what the inductance is of the shorted turn in scenario "A", the net inductance becomes: -

$$L_P \cdot (1 - k^2)$$

Hence, for this simple lossless scenario, if the "new" resonant frequency implies an inductor decrease to (say) 0.9 of its original value then k = \$\sqrt{0.1}\$.

This simple formula works when the conducting core is non-magnetic. If the core is magnetic then, there are two opposing mechanisms; the eddy current that seeks to reduce the inductance and, the presence of ferromagnetic material that seeks to increase the inductance. It can get quite complex even before losses are considered.

Many years ago, I designed metal contaminant detectors for the food and pharmaceutical production markets and, it was known that some stainless steel materials of a particular size were very difficult to detect. The reason is because the effect of eddy currents and the effect of ferromagnetism totally cancelled each other at certain operating frequencies. What remained to be detected was a pure resistive signal but, many foodstuffs are highly resistive (think saltwater) and of course the metal detectors are desensitized to avoid signals that are purely resistive in nature hence, problems!

  • \$\begingroup\$ Thank you very much for the great answer. Now I really understand it. From your graphs I can see that both the frequency and the voltage do not have a linear relationship to the coupling factor. Is the relationship between them exponential and if so why is this the case? \$\endgroup\$
    – ht332932
    Jan 28, 2021 at 14:32
  • \$\begingroup\$ @ht332932 no, it's not as simple as that. No exponential stuff just transformer theory and damn hard theory. I suppose, I really ought to try and explain it better for the pure shorted inductor case (no losses) but it's two or three paper sides of theory if you are a beginner in these things. To try and explain it with losses is probably 6 sides of theory and this site isn't geared up for that detail. I'll think about it over the coming days to see if I can give the maths (presentable) for one worked example. This is why we use simulators! \$\endgroup\$
    – Andy aka
    Jan 28, 2021 at 14:39
  • \$\begingroup\$ @ht332932 I added a picture at the end that shows how to convert the coupled inductors to non-coupled inductors so this might help. \$\endgroup\$
    – Andy aka
    Jan 28, 2021 at 15:39
  • \$\begingroup\$ Interesting discussion, especially the last part. However, isn't it more akin to transforming your inductor into a shorted transformer with a diminishing reflected impedance? I don't see this happening with a ferrite core at 800kHz. A shift in resonant frequency can still arise purely from the losses as these appear as an ESR of the inductor in the LC tank. In this case, the ESR is the cause of the shift rather than diminishing it. Your first example places the resistance outside the LC tank and will not affect the frequency. \$\endgroup\$ Jan 29, 2021 at 9:37
  • \$\begingroup\$ @G.Bergeron I think you may be missing something here. Losses (i.e. things that cause heat) do not shift the inductance. \$\endgroup\$
    – Andy aka
    Jan 29, 2021 at 9:43

The core losses will show up as a series resistance in the inductor. Then, you don't have an LC circuit anymore, but rather an RLC circuit. In this case, depending on the topology, the resonant frequency and the amplitude can vary. See https://en.wikipedia.org/wiki/RLC_circuit. As an example, if the circuit was a parallel LC circuit, then the addition of a resistance in series with the inductor will cause a shift in the resonance frequency given by $$ \omega_0 = \sqrt{\frac{1}{LC}-\left(\frac{R}{L}\right)^2}. $$

As the oscillations will be damped by the additional losses, the amplitude will also decrease.

  • \$\begingroup\$ Wouldn't this imply that the frequency will decrease since we subtract from its initial value? Does this mean that the inductance will stay constant? \$\endgroup\$
    – ht332932
    Jan 27, 2021 at 21:25
  • \$\begingroup\$ Well, the inductance will stay constant up until the point where the permeability is affected. But, yes, the frequency will decrease in this topology. \$\endgroup\$ Jan 27, 2021 at 21:27
  • \$\begingroup\$ Is it not the case that a ferromagnetic object will change the permeability? And won't the eddy currents change the permeability as well. Would it be correct to say that the inductance decreases because the magnetic flux of the coil is decreased due to the magnetic flux of the eddy currents opposing it? \$\endgroup\$
    – ht332932
    Jan 27, 2021 at 21:32
  • \$\begingroup\$ All losses, wether through eddy currents or core losses, will show up as an equivalent series or parallel resistance. The presence of eddy currents do not really impact the permeability. The magnetic properties of the core will certainly affect the permeability and thus the inductance, but I took your question to imply changing the core losses while keeping the inductance constant. Otherwise, changing the core in general will affect all properties of the inductor with an increase of the inductance corresponding to the increase in permeability. \$\endgroup\$ Jan 28, 2021 at 0:29
  • 1
    \$\begingroup\$ It might be interesting to mention that the ferromagnetic effect of iron (the ability to increase inductance) can totally cancel out the inductance-lowering effect of eddy currents and, the net effect is frequency not changing one bit! This is what I uncovered when designing metal detectors. Well I didn't uncover the phenomena; I just explained it to the company I worked for. \$\endgroup\$
    – Andy aka
    Jan 30, 2021 at 17:06

There are plenty of offline and online sources available to address this, if you need more depth.

Eddy current are due to induction, which is proportional to change of the magnetic flux. At higher frequency the magnetic field changes faster, so more eddy currents, and more loss. It's a linear relationship.

At even higher frequencies the electromagnetic field can not penetrate into the whole wires (metal wires are great, but not ideal conductors), and the current will flow only at the edge/skin of the wire ("skin effect"). This loss is proportional with the square root of the frequency.

If the inductor wire is in a dielectric environment, like isolation coating or it is a planar inductor, where the substrate has a given dielectric constant and loss factor, than the changing electromagnetic field tries to rotate the dipoles in the dielectric. It requires some energy, which manifests itself as a loss from your point of view. It is also linear with the frequency.

At very high frequencies even the surface roughness of the conductor could lead to losses.

Nevertheless the main paradigm shift you need is to think about the inductor as a complex component, with series and parallel elements (Rs and Cs), instead of an ideal inductor.

In the case of a capacitor it is a bit simpler. You have an ESR (=equivalent series resistance) and an ESL(=equivalent series inductance). These will make the effective capacitance frequency dependent.

Update: at 800kHz most probably only the series resistance of the wires and eddy currents in a ground plane could play any measurable role in the inductor. Plus ESR of the cap.

  • \$\begingroup\$ And how will the inductance change due to the eddy currents? How will these eddy currents affect the frequency of oscillations? \$\endgroup\$
    – ht332932
    Jan 27, 2021 at 21:34
  • \$\begingroup\$ Would it be correct to say that the inductance decreases because the magnetic flux of the coil is decreased due to the magnetic flux of the eddy currents opposing it? \$\endgroup\$
    – ht332932
    Jan 27, 2021 at 22:14

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