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I was searching for few hours on this topic but I was unable to find any formula/research so I'm asking here. Please if this question is a duplicate let me know.

Right now I do not have the equipment to prove or deny the following formula I came up with.

I have two (same) tht axial inductors. Those in a 1/4w resistor-like shape (with PVC coating). I do not know the core material or the number of windings. But I'm trying to estimate the Mutual inductance M.

Knowing that inductance L of each inductor is given by:

L = u0 * ur * N^2 * A / l

(Better here: http://www.electronics-tutorials.ws/inductor/inductance.html)

And having the M impedance formula (found here: http://www.electronics-tutorials.ws/inductor/mutual-inductance.html)

M = u0 * ur * N1 * N2 * A / d

(Distance referred as 'l' in the article replaced with 'd' for clearance)

Assuming the following:

ur - all cores are of same material
N1 = N2
A - all cores areas are same

I came up that M of two same axial inductors with inductance L, length 'l' at distance 'd' is:

M = L * l / d

...sounds too simple, to be real... But is this correct? It mean that my 47uH inductors with l=2.9mm at a distance d=6mm has a mutual inductance M=22uH sounds huge?

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  • \$\begingroup\$ Flux must be shared and thus coaxial wound on the same core \$\endgroup\$ – Sunnyskyguy EE75 Oct 19 '16 at 23:39
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That would only be true if you had no air gap between the two. In practice, µr of your setup is near 1 because your air gap is enormous. d isn't "clearance" but coil length.

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  • \$\begingroup\$ Do you have any formula/literature to refer me to? \$\endgroup\$ – Newbie Oct 19 '16 at 23:45

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