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FERRITE CORE COIL CALCULATION

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Newton Brawn

Mon Dec 12 2011, 03:07AM

Registered Member #3343 Joined: Thu Oct 21 2010, 04:06PM
Location: Toronto
Posts: 311

Hi !

1323658889 3343 FT0 1 Core Ferrite Coil Jpg

I just need some help to calculate the inductance of a ferrite core coil.

It should be be noted that the ferrite rod is hollow, taking the shape of a ferrite tube.

Any information will be very appreciated.

Regards.

Newton

Sulaiman

Mon Dec 12 2011, 08:17PM

Registered Member #162 Joined: Mon Feb 13 2006, 10:25AM
Location: United Kingdom
Posts: 3141

There are graphs that help with such things - can't locate one at the moment,
but when I do I'll post a url here.
i doubt that an accurate result will be achieved with just two turns
my first guess is;
if you calculate the inductance of the coil alone
then multiply by about 5 you should be close.
There's no point in using a ferrite with relative permeability more than about 5x (length/diameter)

Newton Brawn

Tue Dec 13 2011, 12:46AM

Registered Member #3343 Joined: Thu Oct 21 2010, 04:06PM
Location: Toronto
Posts: 311

Hi Sulaiman !

Thanks for your help.
Please correct me if Im going in wrong direction.

The calculation of the coil alone without the ferrite core:
Using the Weller' s Formula: L[uH] = (D^2 x N^2) / (40xl +18xD)
D and l in inches

D = 22.7mm = 0.894"
l = 5.0mm = 0.197"
N = 2turns = 2

L = (0.894^2 x 2^2) / (40x0.197 + 18x0.894)

L= 0.133uH

Them multiplying by 5:

L=0.667uH ( a reasonable number)

I will wait for your graphics to confirm the value.

Regards

Newton

EDIT: Please read Wheeler, not Weller. Tks

Patrick

Tue Dec 13 2011, 01:38AM

Registered Member #2431 Joined: Tue Oct 13 2009, 09:47PM
Location: Chico, CA. USA
Posts: 5639

where did the "5" come from, is that a magic number?

If the 5 comes from the turn spacing? why would that matter?

Sulaiman

Tue Dec 13 2011, 01:15PM

Registered Member #162 Joined: Mon Feb 13 2006, 10:25AM
Location: United Kingdom
Posts: 3141

Not the one I'm used to but just as good

So for your core;
Ue = 125 (good choice for this frequency range (-61 ferrite?)
L/D = 1.677
from fig.3 Urod = 6 approx. ..............(this is the 'magic number')
from fig.4 multiply by about 2 ...........new magic number = 12
so the inductance of your coil should be about 12x more with the core than without.
Inductance will be quite dependant on turns spacing and position along coil, once done glue the assembly rigid.

Your Wheeler formula looks wrong, check but I think it's L(uH) = (D^2 x N^2) / (40xL + 36xD)
I remember it as L=(R^2.N^2)/(9.R = 10.L) ....Radius and Length in inches.

EDIT: I've assumed that the hollow core will be similar to a solid core, I'd expect not much difference in this application.

Newton Brawn

Tue Dec 13 2011, 11:20PM

Registered Member #3343 Joined: Thu Oct 21 2010, 04:06PM
Location: Toronto
Posts: 311

Thanks to correct my spelling...
Wheeler formula: (from your source):

L=(R^2.N^2)/(9.R + 10.L) ....Radius and Length in inches.

R = D/2

L=[(D/2)^2 . N^2] / [9 . (D/2) = 10.L]

L= [ D^2/4 . N^2] / [9 . D/2 +10.L]

L= (D^2 . N^2 ) / [4 . (9.D/2 +10.L]

L= (D^2 . N^2) / (18.D +40.L)

As I have posted. Both formulas guive same result.

Now following the Phlips Application Note: (ISO)

lr/dr = 1.68 so urod= 5

L = 0.2uH

but lc/lr = 1.79 so the coeficient = 2.5

L = 0.5 uH

This number is close to my first approach to the indutance calculation.
Lets assume that the inductance of the coil is 0.5uH, unless someone give us better figure.

Any other suggestion (s) ??

Regards

Sulaiman

Wed Dec 14 2011, 05:28PM

Registered Member #162 Joined: Mon Feb 13 2006, 10:25AM
Location: United Kingdom
Posts: 3141

Unless you can model in detail
you'll have to build/measure/adjust.

One question, why use an 'open' magnetic circuit
which is prone to receive and transmit rfi
rather than a 'closed' one such as a low permeability metal powder toroid?

Newton Brawn

Thu Dec 15 2011, 12:26AM

Registered Member #3343 Joined: Thu Oct 21 2010, 04:06PM
Location: Toronto
Posts: 311

The project is a coupling coil 2/10 or 2/20 turns.
The reason to use a ferrite rod is to allow the addition of 10 to 20 turns secondary that shall be insulated to 35kV

Also the secondary will be connected in series with a 160 amper welding buzzbox. The 160 A demands a secondary cross section wire of 16mm2, at least . The secondary requires a lot of space.

See the complete information at

I have done some coupling coils using C ferrite core however the C tight space is a limitation for the 35kV insulation.

The next step is calculate the maximum induction in the ferrite core when a 0.001 uF capacitor discharge 5000V on the 2 turns primary. I understand the core shall be not saturated during the discharge.
if you can help me it will be nice.

Regards

peter hansen

Thu Dec 15 2011, 01:48AM

Registered Member #4233 Joined: Sun Nov 27 2011, 09:25PM
Location:
Posts: 6

Don't know if this calculator will help As you are using ferrite rods. But for rings this one is really good.

Newton Brawn

Fri Dec 16 2011, 01:21AM

Registered Member #3343 Joined: Thu Oct 21 2010, 04:06PM
Location: Toronto
Posts: 311

Hi Peter,
The mini ring calculator is very helpfull for "closed magnetic path", but my coil is "open core mag path" and I understand it can not be applied in my case.
Thanks, Im taking this program for future calculations

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