Electronic – Estimating Inductance

inductanceinductor

Here is an inductor I found lying around, and I have no idea what the inductance is.

Inductor

In order to estimate the inductance, I am using the equation \$L=\mu \dfrac{N^2 A}{l}\$. Here are the relevant measurements:

Outer diameter: 0.5" (0.0127 m)

Inner diameter: 0.3" (0.00762 m)

Width (outer radius minus inner radius): 0.1" (0.00254 m)

Thickness (depth): 0.2" (0.00508 m)

Cross-Sectional Area: 12.9 mm\$^2\$

Length: \$l=\pi \left( 0.01016 \text{ m} \right) = 0.03192\$ m.

Number of Turns: About 30?

Therefore \$L = \mu \dfrac{30^2 \left( 0.0000129 \text{ m}^2 \right) }{0.03192 \text{ m}} = 0.3638 \mu_r \mu_0\ = 4.572 \times 10^{-7} \mu_r\$

The only constant I am missing is the relative permeability \$ \mu_r \$ , which I am not completely sure what value to use because I do not know what material is being used. I would guess something ferrite-based, but the picture shows a very strange yellow-colored material (looks almost plastic?).

Best Answer

You could resonate it with a parallel or series capacitor and use a signal generator and o-scope for finding the resonant-frequency. You need to have a capacitor of at least 50 times it's likely self capacitance but, that can also be measured with a frequency generator and an o-scope. Then add a known capacitor (say 10nF) and you should see the resonant frequency drop at least ten if not 100 times. Use this formula: -

\$f_R = \dfrac{1}{2 \pi \sqrt{LC}}\$

I regularly build coils for transmitting power from fixed units to rotating electronics and the parallel resonance way is the most reliable for accuracy.

You should also note that depending on the material of the ferrite the inductance may change quite significantly with current passed through it - this is due to the onset of saturation but some ferrites are designed to be like this so, if possible try and run the test with an oscillator delivering enough voltage to impart the right amount of current into the coil.