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I am trying to calculate the inductance for the KA4823-CL Transformer. According to the datasheet it is 28uH. I know that they use a 3C90 RM14 ferrite core and 6 windings on the primary. As I was planning on introducing an airgap into the transformer I went through the calculations as described at How to correctly use the formula for calculating the air-gap in a pulse transformer?, but with the gap being 0 for now.

This leads me to:

Effective length = 70mm = 0.07m

Permeability = 1850

Gap = 0mm = 0m

Effective permeability = 1850 (for obvious reasons)

Now calculating the inducatance:

Crosssection area = 198qmm = 0.000198qm

Turns = 6

which leads to an inductance of 236,72 uH. According to the datasheet that is wrong.

Calculating using a different formula based on AL = 6600 and N = 6 I get 23,76 uH, which is at least close to the value from the datasheet.

Why is in this case the calculation using the formula from the other post not working properly? I was doing it in Excel with the values from the original post and I get the same result as stated in the answer over there (so I assume I entered all formulas correctly) , but why do I get such a different result in my case?

Jenny
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    Which formula are you using? Can you show your calculations? – user28910 Aug 08 '23 at 18:55
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    It's already gapped. Even if Np were 6t, an ungapped transformer wouldn't achieve Isat = 10A. Datasheet: https://www.coilcraft.com/getmedia/bf866ea4-2079-4c16-a5ce-87ab161660fa/ja4635.pdf But I think you are also misreading turns, because only a ratio is given, and the secondary has the factor of 6, not the primary. – Tim Williams Aug 08 '23 at 19:15
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    @TimWilliams the number of turns (and the entire transformer design) is given in Microchip Appnote AN1338 which is mentioned in the transformer's datasheet. This is probably how the OP knows the number of turns. – Rohat Kılıç Aug 08 '23 at 19:33
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    Jenny please provide links to the data sheets you mention. – Andy aka Aug 08 '23 at 20:01
  • @RohatKılıç Haha wow, they actually went through the design detail! That was unexpected. Link for posterity: https://ww1.microchip.com/downloads/en/AppNotes/01338B.pdf – Tim Williams Aug 09 '23 at 06:09

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but why do I get such a different result in my case?

Because you confuse something with something else. Also, your calculations are partly wrong.

A flyback transformer (actually a coupled inductor) has to have an air gap because flyback topology is based on energy storage (in the magnetic field) and transfer. And almost all of the energy is stored in the gap.

The inductance given in the transformer's datasheet is the gapped inductance so you can't use the un-gapped AL or take lg=0 in the formula given in the question topic you linked. Instead, if you are curious, you can calculate the gap length using the simplified formula given in my answer.

Calculating using a different formula based on AL = 6600 and N = 6 I get 23,76 uH, which is at least close to the value from the datasheet.

No, it isn't. You calculated it wrong. Let's have a look at both approaches again:

  • Un-gapped core, using the theoretical formula:

$$ \mathrm{L = \frac{\mu_0 \ \mu_r \ A_e}{\ell_e+\mu_r \ell_g} \ N^2 = \frac{4\pi 10^{-7} \cdot 1850 \cdot 198 \ 10^{-6}}{0.07+0} \ 6^2 = 236.7 \ \mu H} $$

  • Using un-gapped AL value from the datasheet:

$$ \mathrm{L = A_L \ N^2 = 6600 \cdot 36 = 237600 \ nH = 237.6 \ \mu H} $$

The difference is less than 0.1%. Also, you can see what AL equals in the 1st formula.

Basically, your calculations are wrong and also, as already stated, the given value is the gapped inductance.

Rohat Kılıç
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  • Ah, thanks a lot! Indeed the formula I got from https://sciencing.com/how-8293803-calculate-inductance-ferrite-inductor.html appears to be wrong. L = AL/[(100/N)^2] is not correct apparently. In the AppNote there is no mentioning of a gap, but the calculations suggest that they used about 0.3mm as a gap. As I know that they glued the core halves together maybe they used a center gap of 0.24mm and the glue is providing an additional gap, which then results in the 28uH inductance mentioned in the datasheet. – Jenny Aug 08 '23 at 22:57
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    @Jenny AL comes in a variety of units; pay no attention to random articles. The manufacturer is authoritative, or physics if nothing else. Some manufacturers have given values in "mH/100t" (read: "millihenry at 100 turns"---not "per 100" because it goes as the square of turns!), usually labeled properly as such. Use base SI units, evaluate whatever unit multipliers, and there is your answer. Use a unit-aware calculator if you must (e.g. enter an expression into Google). – Tim Williams Aug 09 '23 at 06:29