Continue to Site

Welcome to our site!

Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

  • Welcome to our site! Electro Tech is an online community (with over 170,000 members) who enjoy talking about and building electronic circuits, projects and gadgets. To participate you need to register. Registration is free. Click here to register now.

Toroid core, how many n = x µH?

Status
Not open for further replies.

hantto

Member
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
 
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 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,



(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.
 
I am new in transformer design can anybody give me guidelines of designing a high frequency transformer.
 
I am new in transformer design can anybody give me guidelines of designing a high frequency transformer.

Please, start a new thread. This thread "Toroid core, how many n = x µH?" is six years old and is not about transformers. You will get a better response if you start a new thread.
 
The two most common spec methods is nH/N^2 and mH/(1000 turns)^2. They are equal since 1000^2 = 1,000,000 so 1 mH/1,000,000 = 1 nH.

nH/N^2 is more intuitive.

Current handling depends on saturation characteristics of particular core and number of turns on it.

N30 is not a particular good choice for a switcher coil. It is used for EMI chokes.

Look at this link for all calculations you will need and source of torroid cores.

Ferrite Data : CWS ByteMark, largest supplier of toroids, ferrite cores, iron powder cores, MPP cores and RF cores
 
Last edited:
Status
Not open for further replies.

Latest threads

New Articles From Microcontroller Tips

Back
Top