# Disagreement between theoretical and experimental inductance of a coil

I have a coil of 20 AWG enameled copper magnet wire with approximately 270 turns: 9 layers of 30 turns. The coil is tightly packed with an overall height of about 30 mm, an overall outer diameter of 48mm, and an "air core" of diameter 24.5mm. I used this online calculator to find an inductance of 1.46 mH as shown below. I also modeled this as a homogeneous multiturn coil using the magnetic fields module of COMSOL Multiphysics and found an inductance of 1.52 mH, which agrees nicely!

However, the coil itself comes from the manufacturer (Jantzen Audio of Denmark) with a specified inductance of 5.00 mH. I measured the inductance with a handheld meter and found it to be 5.05 mH. I also built an RLC circuit and used the resonance condition $$\omega = \sqrt{1/LC}$$ to determine that inductance is 4.96 mH. These are all very self-consistent.

Update: I have now completed another experiment detailed here. At frequencies ~1kHz and above, the measured inductance was a consistent 5.1 mH. (At lower frequencies the inductance was higher, but that is an expected artifact due to the $$8\Omega$$ impedance of the AC source I used, and the $$2\Omega$$ resistance of the coil itself.)

So, theoretically and via finite element analysis, the coil has an inductance of 1.5 mH, but experimentally it's 5 mH. What could be causing the discrepancy? I am aware of the self-capacitance effect which can lead to erroneous LOW inductance readings, but I do not think that this effect can cause HIGH readings. Am I doing something wrong in my experiment? Am I interpreting the theory incorrectly? Any guidance or ideas to explore are greatly appreciated!

• The manufactured device probably has a ferromagnetic core and the calculations probably assume an empty core. Those are both just guesses
– Dale
Commented May 30 at 19:19
• The device comes with no core at all, and there was no core present (besides air!) for either of my measurements. I assumed an air core in my FEA simulation as well. Commented May 30 at 19:22
• Respect to the manufacturer! Goodluck back engineering this. Commented May 30 at 21:30
• This won't account for the huge discrepancy, but I'm puzzled by your figure for radial pitch. I'd have thought it would be less than the horizontal pitch assuming that for each layer (except the innermost) the turns fit into the grooves in the layer underneath. This would give the radial pitch as $\tfrac{\sqrt 3}2 d$ in which $d$ is the wire diameter. I expect I've misunderstood. Commented May 30 at 21:44
• 20AWG = 0.81mm diameter and the resistivity of copper is $\rm 1.724\times 10^{-8}\,\Omega m$. From the manufacturers data (coil 000-1982) the resistance is $1.950\Omega$ which gives a length of wire of ~58 metres as compared with the 30 metres when using the calculator?? Commented May 30 at 23:26

From the picture in your link we can see that, as you write, these coils do indeed have tightly packed windings:

The radial pitch therefore is about equal to the axial pitch. The latter you chose as $$1\,\text{mm}$$, giving 30 turns per layer (probably visible from the outside). Consequently, we need the radial pitch to be about the same. With the given inner and outer diameters of $$25\,\text{mm}$$ and $$48\,\text{mm}$$, a radial pitch of 1mm would give 23 layers, and therefore 690 windings, not the 270 you mention.

Of course, there can be some reason why the radial pitch is larger, but your claim that "the coil is tightly packed" is then only "loosely true", so to say.

Nevertheless, let's take a larger radial pitch of $$1.4\,\text{mm}$$, the value you also use in your screenshot. The inner and outer diameters of $$25\,\text{mm}$$ and $$48\,\text{mm}$$ will then give you 16 layers and the number of turns will be 480, again definitely not the 270 you use! If we replace $$N=270$$ by $$N=480$$ and use the $$N^2$$ behavior of inductance, the value will become $$3.16$$ times higher, so the discrepancy is solved.

In conclusion:

1. The value $$N=270$$ must be wrong. Note that the manufacturer's image is a generic picture for all coil types, it does not show the coil type you describe.
2. Your calculator input in the screenshot with a radial pitch of $$1.4\,\text{mm}$$ and 9 layers is inconsistent with the given inner and outer diameter. With pitch $$1.4\,\text{mm}$$ there should be about 16 layers.
3. If I would have to guess, the error could be that you used the manufacturer's image as input, but that isn't your coil type! Or else it is most likely that you erroneously counted only half of the layers of the coil you have, because the even and odd layers have some offset, making half of them less visible.
• Thank you very much Jos. You are absolutely correct. (The problem was: "you erroneously counted only half of the layers of the coil you have, because the even and odd layers have some offset, making half of them less visible"). Indeed, there were 16 layers--not 9. I'm a doofus. Using a value of N = 480, I was able to recover L = 4.8 mH in COMSOL. Commented May 31 at 16:30
• Perhaps a bit higher L will result if we take the conductor diameter smaller (after all it's enameled wire). 5% error is still not completely satisfactory of course. But then again, 20 AWG as was used in the calculator already should include this... Commented May 31 at 17:07