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Most inductance formulas seem to assume that the COIL cross sectional area is the same as the CORE cross sectional area. Many times, the coil is wound on a bobbin which slides over the core. In this case the core area is slightly less than the coil.

How is the difference in inductance related to the core to coil area ratio?

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How is the difference in inductance related to the core to coil area ratio?

It's a good question but there will be "nuances" that means this answer is not 100% correct for all situations. Start with magnetic reluctance \$\mathcal{R}\$ and apologies if the math goes round the hills a couple of times.

It is defined thus: -

$$\mathcal{R} = \dfrac{\ell}{\mu\cdot A}$$

Reluctance is the length of the core divided by the permeability x the cross sectional area. Reluctance is also (more traditionally) defined as: -

$$\mathcal{R} = \dfrac{N\cdot I}{\Phi}$$

Here, reluctance is the number of turns (N) multipled by the ratio of applied amps to the magnetic flux produced. This basically tells us that a higher reluctance produces less flux per amp. It's probably what most folk are used to when understanding reluctance.

If these two formulas are equated we get: -

$$\Phi = \dfrac{\mu\cdot A\cdot I\cdot N}{\ell}$$

If we differentiate flux w.r.t time we get: -

$$\dfrac{d\Phi}{dt}= \dfrac{\mu\cdot A\cdot N}{\ell}\cdot \dfrac{di}{dt}$$

  • We can use Faraday's induction law to equate V/L to \$\frac{di}{dt}\$
  • And We can equate V/N to \$\frac{d\Phi}{dt}\$
  • V is voltage, L is inductance

We now get the well-known formula for inductance: -

$$L = \dfrac{\mu\cdot A\cdot N^2}{\ell}$$

From the top we can substitute \$\ell\$, \$\mu\$ and \$A\$ for reluctance and we get: -

$$L = \dfrac{N^2}{\mathcal{R}}$$

Note that this formula is a slightly rearranged version of \$A_L\$, (core inductance factor) seen in ferrite data sheets with \$A_L\$ being the inverse of reluctance (permeance).

We can "estimate" the reluctance of the air between the ferrite core and the coils by calculating the area it occupies in the overall cross-section of the coil then applying it into the formula right at the top.

Then, noting that reluctances in parallel sum-together like resistors in parallel, we should be able get a composite value for reluctance comprising air and core material.

Use this composite value in the bottom formula and bingo.

Where this method needs work (and where my understanding lets me down) is in "estimating" the reluctance of the air within the coil's cross-section - it may not be as simple as calculating the overall area it occupies because there may be nuances about the air-shape that means it's not generally applicable.

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  • \$\begingroup\$ "...it may not be as simple as calculating the overall area it occupies..." It requires solving a partial differential equation in three dimensions, which can only be done for a limited number of problems. Generally this is done numerically, using finite element analysis. \$\endgroup\$ – TimWescott Nov 30 '18 at 18:12
  • \$\begingroup\$ @TimWescott yeah I thought there might be some nuances about resolving the reluctance of the air-space but that is what it boils down to in a nutshell; i.e. if you can do the diff equations then the OP has an answer. \$\endgroup\$ – Andy aka Nov 30 '18 at 18:15
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    \$\begingroup\$ Nice answer. I'll just add for OP's benefit that FEMM(Finite element magnetic modeller) is a free tool, so if (s)he desires they could model a mixed core inductor. I think it only does cut plane models though, so it still wouldn't figure out the full 3D. You can model things well above your skill level if you understand the basics well enough to get everything punched in. It's just a bit time consuming. \$\endgroup\$ – K H Nov 30 '18 at 18:48
  • \$\begingroup\$ @Andy aka Since R1 || R2 for R1 >> R2 is approximately R2, is the effect of the air gap around the coil minimal until the ratio of the gap/core get close μ of the core? If so, then for a core with a μ of 1000 you could have a significant gap with minimal effect. \$\endgroup\$ – crj11 Nov 30 '18 at 19:47
  • \$\begingroup\$ @crj11 totally right but many many hf cores have a perm of only ten or so. \$\endgroup\$ – Andy aka Nov 30 '18 at 21:40

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