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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?).

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Sometimes you can find the type of the core printed on it. Does it have any markings? –  Szymon Bęczkowski Jun 21 at 17:53
    
Looks like a powdered iron core to me, maybe from Micrometals. Their cores are color coded for permeability, but unfortunately there's no way to be sure it's one of theirs. Like others said it's best to measure since the permeability could be anything. –  John D Jun 21 at 20:11
1  
No markings on this toroid. I am liking the cleverness of all of the answers, but none of those solutions work for me because I am very limited on equipment at the moment (all I have is a multimeter - I don't even have a soldering iron to add leads to the inductor). That is why I was hoping for an answer with a permeability. –  Ryan Jun 21 at 22:13

4 Answers 4

up vote 3 down vote accepted

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.

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I prefer the series-resonant method because of the very sharp peak that can be had at the LC junction, but you have to be careful not to let the Q get too high or you'll create a singularity there and destroy the world. –  EM Fields Jun 21 at 20:55

Ferrites vary in \$\mu_r \$ from maybe 10 to more than 10,000, so you're likely to do better measuring the inductance and calculating \$\mu_r \$ from that.

Here is an example inductance meter project (there are plenty more out there, no endorsement is implied).

In fact most of the ones on the net (especially those using an LM311) are unabashed copies (or copies of copies) of the original AADE design- he was kind enough to publish the schematic and algorithm used, making it easy to copy. If you can afford it (~$100), it's better to get the original.

You can use your formula to estimate the inductance of a new coil wound on that core, so all is not for naught.

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  1. Connect your inductor, a 100 ohm resistor, and about a 0.1µF cap in series, and across a function generator, as shown.

  2. Connect an oscilloscope across the inductor, as shown.

  3. Set the function generator for a volt or so of sine wave output at any convenient frequency.

  4. Adjust the function generator frequency for a peak in E1

enter image description here

When the peak occurs the circuit will be in resonance and the reactance of the cap will be equal to the reactance of the inductor.

Knowing the value of the cap, solve for its reactance with Xc = 1/2pi f C, then once you get that, solve for the inductance with L = Xc/2pi f, where F is the frequency of the peak in Hz, X is the reactance in ohms, C is the capacitance in farads, and L is the inductance in henrys.

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I think the best thing to do would probably be to put it in an LR circuit such as the following and then measure the current at short time intervals.

schematic

simulate this circuit – Schematic created using CircuitLab

We note that by Kirchoff's voltage law:

$$IR+L\frac{\mathrm{d}I}{\mathrm{d}t}=V$$

And therefore we have:

$$I(t)=\frac{V}{R}\left(1-e^{-\frac{R}{L}t}\right)$$

And so measuring the current \$I\$ using the ammeter at time \$t\$ and using a known resistance \$R\$ and voltage \$V\$, and using an exponential interpolating function to calculate \$\frac{R}{L}\$ you can calculate \$L\$ from the data.

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