Hello, i have a toroid core. Inner diameter: 10mm, outer diameter: 20mm, it's made of the material "N30" (so it says in the catalog). Now the question itself: how is the number of turns calculated to get x amount of inductance. I would need an inductor of 12µH, how many turns does that equal? With a current of about 5A

Thank you.

 

Find the AL value in the specs (the inductance with one turn) and then I can help you

 

If you look in the catalogue and find your particular core, there should be a value for "AL", this is the inductance in uH for 100 turns (sometimes given in mH per 100 turns).

If AL is in uH per 100 turns, then the number of turns for a given inductance is:

N = 100 x SQRT(L/AL)

Where N = number of turns
L = inductance in uH

JimB

 

Check this one link out, it's a pdf of someones thesis on inductors, look on page 22 for equations about a toroidal inductor althgouh what the gy above me said might be easier.

http://www4.ncsu.edu/~mbs/vitae_thes...an_ms_2003.pdf

__________________
never lost, merely waiting for the place I\'m supposed to be to find me

 

Thank you all for your help. I found the AL value for my toroid (google roks!) AL = 4160 -+ 25%. I did some more searching and found a page http://www.ee.surrey.ac.uk/Workshop/advice/coils/#turn there is a formula

N = (10^9 × L / Al)^½

According to this I would require (10^9 * 12*10^-6 / 4160)^½ = 1.55 turns, that sounds somewhat too little.

According to JimB: N = 100 x SQRT(L/AL)

so: 100*(12/4160)^½ = 5.3 turns, now that souds better, but it is not the same answer I got before, which one is correct?
(Sqrt(x) = (x)^½, just incase you didnt know, (it looks prettier, that's why I used it))

I will take a look at that pdf psu_EE_guitar_nut.

Thank you all for your help! MUCH appreciated!

Whoa! That pdf didn't look so friendly

 

Your core have a high AL value. You can either make a cut in the toroid (gap it) to lower the AL value (but then you won't know the AL value unless you have a inductance meter and you can measure the L/turns value) Another way is to use a smaller core of the same material that will have a lower AL. Yet another way is to use a powdered iron core if your application is for power application like a switcher. These have lower AL in general for the same size ferrite core.

 

Hantto

I think the problem comes from the units used for AL.

If AL is in uH per 100 turns, use the formula N=100*SQRT(L/AL).

If AL is in mH per 1000 turns, use N=1000*SQRT(L/AL).

Manufacturers tend to use uH per 100 turns for iron powder torroids, and mH per 1000 turns for ferrites.

Check that catalogue and recalculate.

JimB

 

Actually it is in nH, Al = 4160nH, then what?

N = 10*(L/Al)^½ ? and L given in nH?

12µH = 12000nH

so 10*(12000/4160)^½ = 17 ? or?

 

N=SQRT(12uH/4160nH), gives 1.7 turns

 

If you follow through on the reference you gave,

http://www.ee.surrey.ac.uk/Workshop/advice/coils/#turn

(and if I had followed it and looked at THEIR definitions for AL)

we see that AL is in nH, and we can just use the formula N = sqrt(10^9 x L /AL), with L in H.

In your case 12uH so

N = sqrt(10^9 x 12 x 10^-6 / 4160) = sqrt(2.88) = 1.69 turns.

Which leads to another problem, on a toroid you can only have whole turns, if the wire goes through the toroid, it is one turn.
So if you were to put two turns on your toroid you would have:

L = AL x N^2 = 4160 x 2^2 = 16600 nH = 16.6uH

JimB

 

One must always try and get a good fill on a toroid otherwise coupling will be bad and in case of power applications using switchers leakage inductance. Go to a smaller core or something with much lower AL

 

Ok, thank you all, I will use a core with Al 1290, that will give me 3turns to get 12µH.

Your help has been much appreciated. Thank you once more.