# Air gapped transformer

#### FuseFuse

Joined Feb 8, 2017
37
Hello,
I'm studying an air gapped transformer with E core shape and I want to make a gap of 1 mm.
I can for sure search for two E ferrites with a central gap of 0.5 + 0.5 mm in order to reach the goal.
I'm asking myself what happens if I pick two E ferrites with no gap and I put a separator of 1 mm. Is it the same?
In this case my magnetic flux is interrupted not only in the central branch, but also in the two lateral ones. Should it still be of 1 mm, in order to have the same inductance?

Thank you

Last edited:

#### Lightium

Joined Jun 6, 2012
118
"I'm asking myself what happens if I pick two E ferrites with no gap and I put a separator of 1 mm. Is it the same?"

Yes.

#### FuseFuse

Joined Feb 8, 2017
37
"I'm asking myself what happens if I pick two E ferrites with no gap and I put a separator of 1 mm. Is it the same?"

Yes.
How can you explain that?
Is an interruption of the magnetic flux on two lateral branches equal to an interruption (of the same air gap) in the central branch of the transformer?

Thank you

#### ericgibbs

Joined Jan 29, 2010
18,234
hi FF,
You are correct.
E
Air-gap is one of the most crucial part of magnetic circuits, especially in high power inductors. It significantly modifies parameters of magnetic devices by increasing of saturation current, linearizes B-H curve of magnetic circuit and causes decreasing of the inductance.

#### Attachments

• 702.1 KB Views: 5

#### FuseFuse

Joined Feb 8, 2017
37
hi FF,
You are correct.
E
Air-gap is one of the most crucial part of magnetic circuits, especially in high power inductors. It significantly modifies parameters of magnetic devices by increasing of saturation current, linearizes B-H curve of magnetic circuit and causes decreasing of the inductance.
Hi ericgibbs,

yes, I agree on that.
There's still a pending question in my mind on how things change in case of interruption of the magnetic E-core in the central or lateral branch. Your paper is interesting but it depicts the case of a U-core ferrite.
In lateral branches of a E-core ferrite, B-field splits in two halves, as far as I know. I believe I can't dimension air gap in the same way as an interruption in the central branch, in order to get the same inductance.

In other words, let's just pretend you have a gap of 1 mm in the central branch. Now you want to make a gap in lateral ones, instead of the central one. How thick should it be, in order to get the same inductance?
A third option is to put a gap in all three branches. A new design would occur in such case.

Thank you for replies coming from you or from other users.

#### ericgibbs

Joined Jan 29, 2010
18,234
Hi FF,
If you cannot find any technical documentation for that application, you could try to get some empirical data.

Do you have a signal generator that can produce a Sine wave of say 10Vppk at the frequency of interest?

E

#### Ian0

Joined Aug 7, 2020
8,947
The ”gaps” are normally made of cardboard (Kraft paper to be precise) which has a permeability of about unity.
You can work out the reluctance of the gap:
ℝ=l/(μ0.Ae)
where Ae is the cross sectional area, and l is the thickness of the gap.
you can then work of the reluctance of rest of the core this time it is
ℝ=l/(μ0.μr.Ae)
where l is the magnetic path length (in the core datasheet) and
μr is the relative permeability of the ferrite.
Add up all the Reluctances in the magnetic path to get the total reluctance ( just like you would add up resistances in an electrical circuit)
Then
L=n^2/ℝ
where L is the inductance, and n is the number of turns.
It is not perfectly precise as there is “fringing flux” around the gaps.

Hey, AAC web-designers, can we have the proper script F and R for reluctance and magnetomotive force in the characters menu?

#### FuseFuse

Joined Feb 8, 2017
37
Hi FF,
If you cannot find any technical documentation for that application, you could try to get some empirical data.

Do you have a signal generator that can produce a Sine wave of say 10Vppk at the frequency of interest?

E
Yes, sure but... I'm sure there's already something on scientific literature.
I'm just wondering if someone can address me in the right way.

#### Janis59

Joined Aug 21, 2017
1,783
The basics for any air gap calculus is this: n*i/B=(Magnetic.path.length.core/mju.rel/mju.zero + h.airgap/mju.zero) where B is Teslas, i is Amperes, n is turn count, all lengthes is meters, mju zero is 12.6E-7 and mju rel is something between 600 and 3000 according the ferrite datasheet, of for pure steel 4400.

P.S. - Gaps of central branch and both side branches have summing effect.

#### Ian0

Joined Aug 7, 2020
8,947
A further point.
With an E core there are two magnetic circuits in parallel. The reluctances for each circuit are in parallel, and calculate exactly the same as resistances in parallel.

#### FuseFuse

Joined Feb 8, 2017
37
A further point.
With an E core there are two magnetic circuits in parallel. The reluctances for each circuit are in parallel, and calculate exactly the same as resistances in parallel.
Hello,

so, definitively, I can say that
case 1: single coil wound around central leg, with a central gap g => L is proportional to N2/g, where N is the number of turns of the coil
case 2: single coil wound around central leg, with a "left gap" (i.e. a gap on the left branch of the E-core) of g/2 and a "right gap" of g/2 => L(left) is proportional to N2/(g/2), L(right) is proportional to N2/(g/2). L(total)=L(left)+L(right) =N2/g =L(case 1)

Am I right?

Thank you

#### Ian0

Joined Aug 7, 2020
8,947
Hello,

so, definitively, I can say that
case 1: single coil wound around central leg, with a central gap g => L is proportional to N2/g, where N is the number of turns of the coil
case 2: single coil wound around central leg, with a "left gap" (i.e. a gap on the left branch of the E-core) of g/2 and a "right gap" of g/2 => L(left) is proportional to N2/(g/2), L(right) is proportional to N2/(g/2). L(total)=L(left)+L(right) =N2/g =L(case 1)

Am I right?

Thank you
Yes. Theoretically, the centre leg of an E-core always has twice the cross-sectional area of the two outer legs, so for gaps of the same thickness, it doesn't make any difference if the gap is in the centre or the outer legs.
Although it's not part of this slightly simplified theory, it does make some small difference if the gap is under the coil, or it isn't.

#### FuseFuse

Joined Feb 8, 2017
37
Yes. Theoretically, the centre leg of an E-core always has twice the cross-sectional area of the two outer legs, so for gaps of the same thickness, it doesn't make any difference if the gap is in the centre or the outer legs.
Although it's not part of this slightly simplified theory, it does make some small difference if the gap is under the coil, or it isn't.
OK, thank you.
One more thing. If I have an air gap on lateral legs, this means there is an air gap on central leg as well.
How things would change in such case? This means I have 3 gaps. How should they be dimensioned to have the same inductance?
one third of total gap each?

Thanks

#### Ian0

Joined Aug 7, 2020
8,947
OK, thank you.
One more thing. If I have an air gap on lateral legs, this means there is an air gap on central leg as well.
How things would change in such case? This means I have 3 gaps. How should they be dimensioned to have the same inductance?
one third of total gap each?

Thanks
if you think of it as two UU-core circuits in parallel, each one with two gaps, you can see how to work it out.
You then have a single circuit of the cross sectional area of the centre leg, which has two gaps.

#### MrAl

Joined Jun 17, 2014
10,909
Hello,
I'm studying an air gapped transformer with E core shape and I want to make a gap of 1 mm.
I can for sure search for two E ferrites with a central gap of 0.5 + 0.5 mm in order to reach the goal.
I'm asking myself what happens if I pick two E ferrites with no gap and I put a separator of 1 mm. Is it the same?
In this case my magnetic flux is interrupted not only in the central branch, but also in the two lateral ones. Should it still be of 1 mm, in order to have the same inductance?

Thank you
Hi,

If you are using a true E and I construction then the simple answer is that if you want a magnetic gap of 1mm when you need to use a separator of 0.5mm across the three flats and that would be the most stable way to do it.

The reason for this is because the flux passes up through the center leg then goes through the first layer on top of the center leg, then splits and goes left and right and the flux on each side then passes a second time through the same separator thickness, but because only half the flux goes left and half right, the half one the left and the half on the right add up to one more full pass through the separator. Thus, the total flux goes through the separator two times. That of course means if you want a theoretical 1mm you have to use a separator of 0.5mm across all three leg tops. If the "E" is standing up like that letter E, then the separator would be on the right side of that E across all three leg ends, and again would be 1/2 of what is calculated for the total gap length. You can use fish paper for the separator for example. You want it to be non -conductive electrically and magnetically, as well as stand up to higher temperatures.

As said elsewhere, when you add a gap you reduce inductance so you may have to add turns to make up for that. The benefit is it takes more DC current to saturate the core.

#### joeyd999

Joined Jun 6, 2011
5,035