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?

  • \$\begingroup\$ Flux must be shared and thus coaxial wound on the same core \$\endgroup\$ – Sunnyskyguy EE75 Oct 19 '16 at 23:39

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.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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