
Home  Forums  Rules  Articles  diyAudio Store  Gallery  Wiki  Blogs  Register  Donations  FAQ  Calendar  Search  Today's Posts  Mark Forums Read  Search 
Parts Where to get, and how to make the best bits. PCB's, caps, transformers, etc. 

Please consider donating to help us continue to serve you.
Ads on/off / Custom Title / More PMs / More album space / Advanced printing & mass image saving 

Thread Tools  Search this Thread 
31st March 2001, 01:55 AM  #1 
diyAudio Member

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! 
31st March 2001, 02:59 AM  #2 
diyAudio Member
Join Date: Feb 2001
Location: .

Lisandro_p
There are two common soft ferrite materials: ManganeseZinc types with initial perm. range 1600 to 15000 and NickelZinc 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 
31st March 2001, 03:06 AM  #3 
diyAudio Member

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. 
31st March 2001, 03:09 AM  #4 
diyAudio Member
Join Date: Feb 2001
Location: Columbia, SC

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 applicationswould 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 reinsulate 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 
31st March 2001, 03:16 AM  #5 
diyAudio Member

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.

31st March 2001, 03:17 AM  #6 
diyAudio Member

I know GRollins, i just need a LOT of inductance, and no accurate values. My problem is i might end up winding 4000 turns where 1000 could have done the trick easily... and winding ain't fun

31st March 2001, 03:31 AM  #7 
diyAudio Member
Join Date: Feb 2001
Location: .

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 
31st March 2001, 03:34 AM  #8 
diyAudio Member

I'll try that... thanks!

31st March 2001, 03:36 AM  #9 
diyAudio Member
Join Date: Feb 2001
Location: .

Please, considering L proportional to the square of N, not L. I type it wrong a mistake in my other post. Sorry.

Thread Tools  Search this Thread 


Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Winding your own tiny inductors  punchpeanut  Parts  10  1st January 2016 01:10 PM 
Winding your own inductors  Jason  MultiWay  17  26th October 2012 05:42 AM 
Winding Inductors  noodle_snacks  MultiWay  40  10th March 2010 12:58 AM 
Winding perfect lay inductors  tktran  MultiWay  3  2nd February 2004 05:35 AM 
Winding own foil inductors?  tpenguin  MultiWay  2  25th December 2003 11:05 AM 
New To Site?  Need Help? 