# Finding inductance and core loss for a powered iron torroid inductor in an "2TFC" SMPS

Z

#### zenerbjt

##### Guest
Hi
On page 35 of DER484 by Power Integrations (below), there is a torroid inductor (Made of "kool mu" powered iron) for the output of a Two Transistor Forward converter.

This torroid is 77930-A7 by Mag-Inc (as below). It is wound with 75 Turns.

Finding saturation flux density
Lookin at the B vs H curve here
...Kool Mu appears not to saturate at 600mT....but its not possible to find out from the website what is the saturation flux density of Kool Mu?

Find Inductance at 4.6A
I wish to first find the Inductance at its average current level of 4.6A.

I believe I have to do this by finding the new permeability at the current level of 4.6A, then find the Reluctance at this permeability level {using Reluctance = l/(uo.ur.A) }, then find inductance at 4.6A by L = (N^2)/Reluctance.

This is done by looking at the “Permeability vs DC Bias” curve of this…

https://www.mag-inc.com/Products/Powder-Cores/Kool-Mu-Cores/Kool-Mu-Material-Curves#kmpermvsdc

…in fact , better still, there is an equation here stating “Percentage of initial permeability” is 1/(a+b.H^c) [H in Oersteds]

The result is that Inductance at 4.6A is 283uH. Would you agree?

Finding Core Loss

We then have to find the core loss of this inductor. I take the peak and trough of inductor current as being 5.6A and 3.6A respectively. We then find the B at the peak and trough currents using B=uo.ur.H

The result is that the Bpkpk value is 0.013T. Is this correct? It seems kind of bizarre because the Flux density at 3.6A is greater than the flux density at 5.6A.
Then, taking this Bpkpk of 0.013T, we use it with the graph of “Core Loss Density curves” shown here…

https://www.mag-inc.com/Products/Powder-Cores/Kool-Mu-Cores/Kool-Mu-Material-Curves#kmcoreloss

..to find core loss..

The switching frequency is 66kHz, and from the above graph, the core loss is seen to be well under 30mW/cm^3 at 66kHz. (I.e 0.03W/1E-6 m^3) The torroid volume is 4150e-9 m^3. Therefore, the core loss is less than 120mW.

This seems ridiculously low for a powered iron core, which is known for having high core losses. Do you know what’s wrong here?

DER 484 document:

Torroid is 77930-A7 by Mag-Inc:

My workings are as attached

#### Attachments

• Sendust inductor for 2TFC from DER484.zip
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#### FvM

##### Super Moderator
Staff member
I see Bmin of 0.47T and Bmax of 0.59T in the B/H curve for the given current points, thus delta B is 10 times higher and loss 100 times.

Inductance drops to 30% at 4.6 A, the operation point is a bit outside the useful range, I think.

zenerbjt

Z
Points: 2
Z

#### zenerbjt

##### Guest
I see Bmin of 0.47T and Bmax of 0.59T in the B/H curve for the given current points, thus delta B is 10 times higher and loss 100 times.
Thanks, i must admit i'm still a little phased by the Power Integrations calculations for this inductor in their DER484 document...
https://ac-dc.power.com/sites/default/files/PDFFiles/der484.pdf

..on page 44, they get an inductance value of 229.4uH at 4.6A.
...Again on page 44, they get a DC flux of 0.2385T in this inductor.
...Also pg 44, they get an AC flux of 0.0575T in this inductor
...on page 55, they get a total core loss of 0.62W

Either Power integrations in DER484 are wrong, or i have , in my top post, badly violated the laws of physics, i dont know which it is yet.

#### FvM

##### Super Moderator
Staff member
Although some intermediate results in the calculation look wrong, the core loss result is in the right order of magnitude.

See below the calculation according to magnetics catalogue fits.

#### Attachments

• PowderCores.zip
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zenerbjt

Z
Points: 2
Z

#### zenerbjt

##### Guest
Thanks FvM for your excel doc.
I looked and your calculation accords exactly with the Mag-Inc website.
I noticed the Mag-Inc website did not specify whether "B" means Bpk or Bpkpk....so Mag-Inc is ambiguous here.

Your calculation for B at various current levels accords exactly with the equation on the Mag-Inc website...however, the value of 0.2385T given on page 44 of the DER484 document does not accord with the Mag-Inc equation.
--- Updated ---

See below the calculation according to magnetics catalogue fit
Thanks, i back-calculated the inductance at 4.6A from your value of H and B at 4.6A.......it came to 559uH.
This does not correspond to the inductance at 4.6A calculated using the Permeability vs DC Bias equation given by Mag-Inc...because that gives 283uH.

Somehow, Power Integrations calculate 229.4uH.....so all three of us calculate different inductances at 4.6A DC BIas.

#### Attachments

• PowderCores_ZenerbjtEdit.zip
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Staff member
zenerbjt

Z
Points: 2
Z

#### zenerbjt

##### Guest
Thanks very much.

We have now calculated the inductance of another Mag-Inc powdered iron core inductor, and again the result has come out wrong, even though we used the correct standard magnetic equations.

I am referring to the Powdered iron torroid inductor on page 5 of the Infineon 1200W PFC App Note…

Infineon 1200W PFC App Note:
https://www.infineon.com/dgdl/Infin...N.pdf?fileId=5546d4624a56eed8014a62c75a923b05

Its using two stacked “Kool mu” torroids (77083A7) and 64 turns. The inductance is calculated at 22.5Amps.

Infineon calculate 168.5uH . However, the equations on the Mag-Inc website calculate either 383.4uH or 152.6uH depending on which equation “route” you take. (as in the attached excel document)

I did the calculations twice and get the same answers. I also used well known magnetics equations such as L = N^2/Reluctance etc etc.

Kool mu torroid 77083A7:
https://www.mag-inc.com/Media/Magnetics/Datasheets/0077083A7.pdf

This is the Mag-Inc web page from which the calculations were done..

#### Attachments

• Inductor _2 x 77083A7 by MagInc.zip
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#### FvM

##### Super Moderator
Staff member
The problem is the difference between DC and AC permeability. To get the correct AC permeability from the B/H curve, you need to calculate the slope delta B/delta H.

zenerbjt

Z
Points: 2
Z

#### zenerbjt

##### Guest
The problem is the difference between DC and AC permeability. To get the correct AC permeability from the B/H curve, you need to calculate the slope delta B/delta H.

Thanks, as you know, the slope of the B/H curve is “uo.ur”.

I have calculated uo.ur at the relevant Magnetising force H, so , as you know, I surely have no need to calculate the slope, as the uo.ur value is already this slope?

Mag-Inc have a web page which offers an expression which allows calculation of ur at a given value of H. I faithfully used this expression to calculate the value of ur (at the particular value of H). I then put this ur value into an equation to find the reluctance….i then used reluctance and the number of turns to find the Inductance….but the answer always comes out wrong.

Its impossible to see what Law of Physics i am violating here?

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#### FvM

##### Super Moderator
Staff member
Thanks, as you know, the slope of the B/H curve is “uo.ur”.
Of course it's not, B/H curve is not a straight line.

Z

#### zenerbjt

##### Guest
Thanks, yes, what i meant was, at any given point on the B/H curve, the gradient at that point is uo.ur
(because ur varies as H varies)

Z

#### zenerbjt

##### Guest
********______****************______********
Hi, On the same note....

On this web page (below), Mag-Inc show a Kool Mu inductor being designed for a 650W PFC. Just after the heading for section 5 “Determine Number of Turns”, They divide the “wished for” inductance value of 598uH by 0.5, …Why do they do this? Where does the “0.5” come from?

https://www.mag-inc.com/Products/Powder-Cores/Kool-Mu-Cores/PFC-Boost-Design
********______****************______********

#### FvM

##### Super Moderator
Staff member
Where does the “0.5” come from?
As explained in the paper, accounting for 50% roll-off due to DC bias.
what i meant was, at any given point on the B/H curve, the gradient at that point is uo.ur
In the Excel sheet, you calculated u0*ur as B/H where you should calculate delta B/delta H.

zenerbjt

Z
Points: 2
Z

#### zenerbjt

##### Guest
Thanks..
********************************************_________________________
Infineon 1200W PFC App Note:
https://www.infineon.com/dgdl/Infin...N.pdf?fileId=5546d4624a56eed8014a62c75a923b05

In the above app note, on page 7 they try and calculate Core loss by equation 12. But surely equation 12 is wrong?...it should be

P = a.(B^b).(f^c)

Where:
B is flux density in Teslas
f is frequency in khz
a = 44.3 ; b=2 ; c = 1.541
(as per “fit formula” in “Core loss density curves” section of this….
********************************************_________________________

#### Easy peasy

It is an iterative process for powdered iron, you work out the turns for the desired L, then you calc the ampere-turns/metre for the core in question - then you see what the real L is at that H, then you either add turns ( usually ) to see if the drop in L is acceptable at full current or go to the next larger core size and repeat ...

Z

#### zenerbjt

##### Guest
It is an iterative process for powdered iron, you work out the turns for the desired L, then you calc the ampere-turns/metre for the core in question - then you see what the real L is at that H, then you either add turns ( usually ) to see if the drop in L is acceptable at full current or go to the next larger core size and repeat ...
Thanks, Mag-Inc offer a formula as follows…

% of initial permeability = 1/(a+b.H^c)
Where a , b and c are given and H is in Oersteds

…this equation is found here….
https://www.mag-inc.com/Products/Powder-Cores/Kool-Mu-Cores/Kool-Mu-Material-Curves#kmpermvsdc
…under the “Permeability vs DC Bias curves” heading

It seems like this is a way for us to get to the required_inductance_at_particular_current in a one-er?

Ie, you get your “new ur” value from the above equation, then from that you calculate reluctance, then from that and the number of turns , calculate the inductance_at_the_particular_current.

I am postulating that this is what the Mag-Inc website is inviting people to do?

However, the values that it gives do not correspond to what various Application Notes calculate.

In fact, Mag-Inc do a worked example of a powdered core inductor calculation for a 650W Boost PFC, and they calculate the wrong peak inductor current (they get 8.5A peak at 65VAC input rather than 12A, considering a 598uH inductance). So in other words, you cant check things on the Mag-Inc website…and when you then use their contact form to ask them, it doesn’t work....

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#### FvM

##### Super Moderator
Staff member
I presume, you did wrong calculations, as before along this thread.

The curve fits given by Magnetics aren't perfect and seem to change between catalogue issues. But for me, they gave useable estimations up to now.

After fixing the Excel calculation you posted in #7 to use correct delta B slope, it gives 165 µH from B/H curve and 153 µH from permeability curve, a fair congruence, I think.

Z

#### zenerbjt

##### Guest
Thanks, i stupidly took B/H instead of delta B/delta H as you say.....but that in itself was me being even more stupid, because Mag-Inc give an equation for calculating B from H...and also provide the permeability adjustor so L can be calculated from Reluctance. Or are we saying that it is worth while to actually calculate "delta B/delta H" by painstakingly drawing out a B by H curve and finding the gradient at a point?

Z

#### zenerbjt

##### Guest
***********************________________________***************************
Hi,
In the DER-484 document, Power Integrations detail the design of a 280W Two Transistor Forward converter. This concerns the design of its output inductor using a powdered core torroid.

DER-484:
https://ac-dc.power.com/sites/default/files/PDFFiles/der484.pdf

…On page 44, they state that they choose a 77930 powdered core torroid by Mag-Inc. They expect it to have an inductance of 229.4uH at the DC bias current of 4.6A. They use 75 Turns.

Why have they chosen the 77930 torroid?

For a start, the 77930 core datasheet shows in a chart that its only for consideration up to 250 Ampere.Turns. Clearly, the actual ampere turns is 75*4.6A = 345 Turns.

Also, Mag-Inc have a “Kool mu” core selection chart on page 26 of the Mag-Inc Powder cores catalog. This allows you to put in the “L(mH) * i^2” value and read off what torroid is suitable. The “L(mH) * i^2” value is 5.99. Reading off from the chart, this means a 77586 torroid is suitable. The 77930 torroid doesn’t even appear on the chart.

Therefore, do you know why Power Integrations chose the 77930 Torroid?

Mag-Inc powder cores catalog:

77930 torroid datasheet:
https://www.mag-inc.com/Media/Magnetics/Datasheets/0077930A7.pdf

77586 torroid datasheet:

***********************________________________***************************

#### Easy peasy

Welcome to the vagaries of the Mag Inc design material -

I usually start by assuming the A-t will pull the AL ( aka Ueffective ) down by 50%, pick that point on the graph for the L ( and NI ) I want - then check the Bripple losses ...
--- Updated ---

As to the 880uH DC choke - they likely wanted a design that had high inductance at low currents to extend the CCM range - it doesn't matter if the L falls off at high current as they don't need the energy storage there - hence the design that runs the core to high H and keeps the size relatively low.

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zenerbjt

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