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I should design an inductor which has these features:

  • 3F3 magnetic ferrite material
  • Temperature is between 35-70 celcius
  • Copper filling factor, kc = 0.5
  • Max 10A DC or 10A AC 50kHz
  • U type inductor
  • 12 microhenry

I should find these:

  • flux density
  • current density
  • turn number
  • gap space (if there is)
  • turn area

I considered using this resource, page 300.

Am I on right way? Do you have any trick point for me? Like advising, resources, etc.

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  • \$\begingroup\$ There are no written page numbers in that pdf document. Copy and paste the page you refer to. Why do you ask for advertising? \$\endgroup\$
    – Andy aka
    Commented Jan 19, 2021 at 9:36
  • \$\begingroup\$ You can use pdf editor page settings to go to page 300. if you use google chorme to open it, just write 300 to page counter. I am looking for advertising because I don't have any experience with this situation. Maybe there are pratictal ways that I don't know. \$\endgroup\$
    – Piko
    Commented Jan 19, 2021 at 14:24
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    \$\begingroup\$ "Advertising" is the useless fluff written by the marketing department. I think we have a translation problem. What language do you normally speak, and what word are you translating as "advertising?" \$\endgroup\$
    – JRE
    Commented Jan 19, 2021 at 15:37
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    \$\begingroup\$ @JRE I think it is highly likely that he meant "advising" \$\endgroup\$
    – Chachmu
    Commented Jan 19, 2021 at 21:08
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    \$\begingroup\$ Sorry for my mistake. I am not native, my main language is Turkish and I am always confused with advertising and advising, they are so similar for me. @Chachmu you are right, thank you for fix my mistake. \$\endgroup\$
    – Piko
    Commented Jan 20, 2021 at 7:49

1 Answer 1

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Max 10A DC or 10A AC 50kHz

You should consider this first and recognize that 10 amps AC is an RMS specification. This means that the peak current is 14.14 amps (for a sinewave). This exceeds the DC specification and so, it should be used to test a prospective design to see if there is significant core saturation.

In that requirement you said or but, if you meant and you would need to consider the worst case as being 24.14 amps.

However, there are many inductor applications that work with square waves and, a 10 amp RMS square wave has a peak current of 10 amps. So you need to define what you really want because not defining what you want will usually end in disappointment.

3F3 magnetic ferrite material

3F3 means you will use a Ferroxcube part for the transformer so, go through the parts available (there are literally hundreds of specific part types that use 3F3 material) and find the core shape that you want.

Then pick one that you think might be appropriate - Ferroxcube has some great resources so use them to narrow down the search for an appropriate core size of the right type in the 3F3 material you want.

I will add this - 3F3 is good for over 500 kHz so you might be being steered down a path that isn't necessary. For instance, 3C85 or 3C90 is probably better material at 50 kHz in some respects.

U type inductor

If you actually looked at Ferroxcube U cores, you will find that none are available in 3F3 material. They are generally available in 3C90 material. So, do some thinking and decide what you do want.

12 microhenry

So, once you have chosen a core type, you can make an guesstimate of the core size and "test" to see if it's likely to saturate. You do this by calculating the number of turns to produce 12 μH. Then with number of turns and peak current (maybe 14.14 amps) you can calculate the ampere-turns (magneto motive force). That's simply amps x turns.

You turn that into the H-field level (magnetic field strength) by dividing it by the mean length of the core (it will be specified in the core-set data sheet).

Once you have H-field you can convert that to peak flux density (Bmax) by using the BH curve for the material you have chosen (you say 3F3) - there will be a data sheet for 3F3 material so use that.

If the Bmax value is greater than around 300 mT then you are likely to be on the knife-edge of good/bad performance. Then, you can make a decision to add a gap to the core. The gap will reduce the effective core magnetic permeability and, in turn, this reduces inductance and saturation levels. So, if you need a gap, you must add more turns to get to the required inductance of 12 μH. What you will find is that if the inductance due to the gap has fallen by 4, you only need to double the turns to restore it to the right value.

Recalculate H-field - this time you use the new value for turns. H-field will obviously be bigger (with the gap) because there will be more turns but, due to the permeability reduction, there will be a clear benefit when gapping and the Bmax level will reduce.

Then, if the Bmax value is below (say) 200 mT (a bit of a rule of thumb), try to estimate how much of the available space you have for winding your turns. Remember to choose a copper wire diameter that easily handles the RMS current without significant overheating.

Iterate down the path above a couple of times and hey presto, you've designed the inductor.

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  • \$\begingroup\$ That is my homework so, not important whether it is realistic or not. I will try to handle this situation with your steps. \$\endgroup\$
    – Piko
    Commented Jan 19, 2021 at 14:27
  • \$\begingroup\$ what about tempreature ? \$\endgroup\$
    – Piko
    Commented Jan 19, 2021 at 14:33
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    \$\begingroup\$ @Piko if you look at the characteristic of ferrites, they are usually good to go up to (generally) 150 degC or more. The data sheet for 3F3 material will show how temperature affects permeability in a graph. But, when you have a gap, the relationship is much more stable plus, a gap reduces peak flux density hence... a gap is a pretty big winner in most cases. However, watch out for copper losses once you have curtailed Bmax to below 200 mT. \$\endgroup\$
    – Andy aka
    Commented Jan 19, 2021 at 14:48

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