Winding inductors for audio....

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Well, i need a rather large inductor (between 500 or 1000 mH) for a passive notch filter (a "Varitone" control for my guitar, if you need to know). Now, these are rare to come by, specially on small sizes, so i'll have to wind my own one.

So, i found a small ferrite toroid (about 2cm diameter) from and old small PSU transformer, which is in great conditions. My question is, what's the magnetic permeability of such a material? I recall being something like 500 or 1000 times the one of vaccum (which means i need about 1000 turns, not fun but can be done), but i want to be sure. Thanks in advice.

And yes, i know this is not an electronics forum, but with all the knowledgeable people in here, i'd be fool not to ask 🙂 Ah, and i'm finishing the circuit/pcb design for my integrated amp... i'll keep you updated!
 
Lisandro_p

There are two common soft ferrite materials:

Manganese-Zinc types with initial perm. range 1600 to 15000 and
Nickel-Zinc types with initial perm. range 10 to 1000.

And there are other magnetics materials too like iron powder and some alloys.

I think the best way to make your inductor is to measure (with a precision resistor, an ac voltmeter and an audio generator you can do this) the inductance of your inductor with few turns of wire over the toroid and make a extrapolation (with some error of course) considering L proportional to the square of the turns. Leave some wire free after winding the total turns and measure again until find the correct value.

I think your toroid is a iron powder one (perm from 5 to 150), because you said it was from a power supply and it was small. This kind of toroid is usually used in the output filter. If it was in the input filter it might be an ferrite type.

Regards
 
I sorta figured that too. I did my calculations with a µ of 200 just to be on a safe side, but i presume it's a lot higher... the toroid is one of those ceramic ferrite one, very tough one it seems (painted yellow, if that qualifies for anything).

BTW, how would you measure inductance that way? Can't seem to figure it out.
 
Lisandro,
You're either going to love this or hate it, depending...
Your best bet is to cut and try. Inductance formulas...well, I could probably dredge you up six or eight of them in the next ten minutes, depending on which books I opened. The catch? None of them quite agree on how many turns yield how much inductance. It's not for lack of study over the years, it's for lack of control over the variables: Is it scramble wound, or wound neatly? The thickness of the insulation. The permeability of the core (usually air core for most audio applications--would it be feasible to use air core so as to have an 'audiophile' effects pedal?). The diameter (gauge) of the wire. The cross section of the wire (round usually, but I've seen a fair number of coils [and transformers, for that matter] wrapped with square cross section wire, and even one that had hexagonal cross section!). And on and on and on. Yuck.
It's usually just easier to sit down with a spool of wire and put a few dozen turns around the former, scrape a little insulation off, and test it. Polyurethane (most wire is coated with polyurethane) or shellac will re-insulate the wire should you need to wind more. Keep notes on how many turns it takes so that next time you can whip out an identical coil in minutes.

Grey
 
BTW the inductance for a toroidal inductor is L = µ . µo . S . n^2 / ( 2 . Pi . r ), being S = section, r = average radius (both in meters) and n = number of turns (checked it on a book, and it's correct), always assuming no losses, infintely thin wire, etc... dunno, this might help someone.
 
If you have an oscilloscope it will be easier and more accurate, but you can find the inductance making a series LR circuit. Apply an AC voltage (take care of the bandwidth of your voltmeter and use no more than 1 or 2 volts to the circuit and sweep the frequency until the voltage over the resistor be equals to Vgenerator/2, use the reactance module formulae with the value of the resistor and the frequency (X=2piFL) and you will find L. Increase N considering the inductance proportional to the square of L and so on until achieve the inductance that you want. You must use the frequencies that you will work in your design.

Regards
 
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