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I am planning on buying a toroid core from Digi-Key. I want to be sure that this is the right type and that I can achieve 170 uh inductance with 22-18 AWG wire. How can I wind this and what is the formula so I can calculate this myself in the future. If this doesn't work which one can I buy from Digi-Key to get the right inductance. (my budget is 4 dollars or under for the toroid core) Last, I want this to handle up to 10 amps of current so tell me if I can't use 18 AWG wire.

edit-fixed broken link, i guess it was just because it was linked directly to my shopping cart

From a comment It said to buy my own I can't find any already pre-wound 10 amp 170 uH toroid and the only thing close was like 10 dollars so I would like to wind it myself!!!

Gustavo Litovsky
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skyler
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  • link to toroid is broken! – placeholder Mar 18 '13 at 22:04
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    And yet again! How is knowing how to wind your own toroidal inductor NOT part of EE? What is with the down votes and close vote? – placeholder Mar 18 '13 at 22:06
  • It's admirable how much interest and desire to understand the fundamentals you have at your age. That said, I would shy away from trying to wind your own inductors. I say this because the math required to understand reactive components is at least Algebra II level, and to really understand what is happening, and why the Laplace transform is so nice, at least one semester of Calculus is needed. Save yourself some frustration for now, and just buy the parts you need. Digikey's search tool makes it easy. – Matt Young Mar 18 '13 at 22:08
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    @MattYoung - I'd strongly disagree, no advanced math is needed to wind torroids to approximate values. There are even tables in Ham radio books - they are actually quite user-friendly parts. – Chris Stratton Mar 18 '13 at 22:13
  • @ChrisStratton I don't disagree with you. That was a general comment to skyler based off a couple of his comments I've seen elsewhere recently, and it sort of fit here. – Matt Young Mar 18 '13 at 22:26
  • sorry the link i guess was part of my shopping cart so it didnt work on other computers. I fixed the link now!! – skyler Mar 18 '13 at 23:32
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    And plus i'm in algebra 2 next year and my math this year is easy so HA! – skyler Mar 18 '13 at 23:37
  • Sorry, but your project described in comments to the removed answer does not sound safe. Learning about toroids by winding them for low power circuits sounds like a great idea, but high power and (on the other side of the transformer, high voltage) is a bit too much to tackle without a more thorough understanding. – Chris Stratton Mar 19 '13 at 02:52
  • I know you said your budget is four dollars, you want a torroid, and you want to wind it yourself. But just for reference, this looks like it would meet the current and inductance specs: http://www.digikey.com/product-detail/en/1140-221K-RC/M8379-ND/774919 – Stephen Collings Mar 19 '13 at 12:51

3 Answers3

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Magnetics design is hard. Bear with me as I walk through some of the considerations.

On the surface, it's actually fairly easy to figure out how many turns you need to get a given inductance on a given toroid.

First, check the datasheet for the core and see what the \$A_L\$ (inductance factor) is:

enter image description here

Your core's inductance factor is 1760nH +/- 25%.

The actual ratio that inductance factor stands for is:

\$ A_L = \dfrac{nH}{(turns)^2} \$

So, to get the number of turns you need, it's simple algebraic manipulation:

\$ turns = \sqrt{\dfrac{nH}{A_L}} = \sqrt{\dfrac{170000}{1760}} = 10\$

To figure out if your wire is going to fit, you need to consider the inside dimension of the toroid and figure out how many diameters of your wire can fit inside it.

Now the hard part.

To figure out whether or not the inductor is suitable for the application (i.e. will it saturate out) you need to know some things so you can calculate \$B_{max}\$:

  1. The operating frequency
  2. The expected peak-to-peak AC current
  3. Is there a DC component to the current
  4. The material characteristics

Number 4 is a biggie here. Why? The material for your choice of toroid is N30, which is suitable for frequencies from 10 to 400kHz according to the datasheet. Is this important?

That's the thing with inductors. The core material has a huge impact on what you can do with the part - it's not just the inductance. The core material dictates how much loss will be generated, at what flux density the core will saturate ... essentially everything.

You're not in a position to figure out the best number 4 unless you know number 1, 2 and 3. That means lots of calculations / predictions / simulations, and potentially lots of trial and error, before the magnetics are done 'right'.

So, will your 18AWG wire be OK for 10A? Most likely. The core? It depends on many things which you haven't specified in your question (like operating frequency, peak-to-peak ripple, etc.) so I cannot say for certain.

Adam Lawrence
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I believe 18 awg is ok up to 16 amps.

Toroidal inductor calculator here -> http://hyperphysics.phy-astr.gsu.edu/hbase/magnetic/indtor.html

You will have to find the permeability of the core material (ferrite? iron?) to enter into the calculator.

Cheers.

Mark B
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  • Is it dc current rating that is being requested because core-losses at not-stupidly-high frequencies can be the killer performance wise. – Andy aka Mar 19 '13 at 08:42
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The next step after MMGM's excellent answer is to put a few numbers from his datasheet into the calculator from Mark B's answer at

Averaging the inside and outside diameter (6mm and 10mm) we get radius 0.4cm and MMGM's 10 turns. Datasheet has "Ae=7.83mm^2" so enter 0.0783 (cm^2) in the "Area" box and it will calculate a coil radius. Enter 4300 for relative permeability (datasheet calls it ui, calc calls it k, these things happen!) and the calculator confirms inductance 0.168mh, pretty close... So far so good.

Now the crucial question : will the coil take 10 amps?

There is another calculator to answer that on the same site... Enter the radius (0.004m this time!) 10 turns, k=4300 again. And new, the "Flux density near saturation" from the N30 data sheet - B = 380mT = 0.38T, and click the link to "current" above.

For this core size and material, with these turns, and this saturation flux density, the calculator says "0.177 amps".

So, no...

As an experiment, try a 4cm radius, 1cm^2 area, 9 turns, same material. The first calculator says 0.174mh, again pretty close. The second now says 1.96 amperes which is heading in the right direction, but a MUCH bigger coil...

So, as MMGM says, magnetics design is hard.

But that was a first step. Now try some different core materials (lower ui=k, larger cores, lower inductances, and see where you get.

(Also bear in mind that 10A DC may translate to 20A or more at AC. Try designing for 1A,5V until you have something working)