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calculating saturation current of transformer

I have designed a 1:1 transformer that will be used in a low voltage DC:DC converter. I have tested the current at which the transformer saturates by decreasing the frequency of the primary voltage while keeping the secondary open loop.

I would like to calculate the theoretical value of the current at which the transformer will saturate. I tried using the formula in Flanagan's which relates Imax to Bmax and I then used the Bmax of ferrites but that doesn't correspond to the current I measure on the oscilloscope.

Also, please clarify the following for me since I find it very confusing: is magnetizing inductance the same as primary inductance? How do I correctly calculate magnetizing inductance?

Reply to
johanwagener
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I forgot to mention that my method of calculating the magnetizing inductance was with the formula: Lprim = (Uo x Ua x N^2 x Ae) /Ie. Is this correct? Is the effective path length the length that the flux must travel in the core or the length that the electrical current must travel in the winding?

Reply to
johanwagener

What's Ie in the above formula? Not a current, I hope. Inductance isn't proportional to current (well, it _is_ due to saturation, but for the linear region, no).

If Ie is the core mean path length, then this looks OK. But I could be leading the witness.

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Paul Hovnanian	paul@hovnanian.com
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Have gnu, will travel.
Reply to
Paul Hovnanian P.E.

that is l for length yes. So that explains my other question as well, it is the length the flux travels.

Using that formula the calculation doesn't work out on the same value as the value simulated with Ansoft Pexprt.

Reply to
johanwagener

That formula is really only good far away from saturation and doesn't account for hysteresis or core losses. So I'd trust the Ansoft modle a little better in these cases.

How do either of these models agree with measured values?

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Paul Hovnanian     mailto:Paul@Hovnanian.com
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Stop Continental Drift! Re-unite Gondwanaland!
Reply to
Paul Hovnanian P.E.

I have designed a 1:1 transformer that will be used in a low voltage

** How is that finding a "current" ??

What you have likely found is a combination of primary voltage and frequency that (just) fully magnetises the core.

I suppose you realise that the applied voltage and frequency that produce any given level of core magnetisation are in inverse proportion ??

** A verbal nonsense.

Seems you have transformers all mixed up with inductors.

What are you REALLY trying to find ??

..... Phil

Reply to
Phil Allison

To those interested I have found the solution to my problem:

Amperes Law: A current carrying conductor produces a magnetic field of intensity H whose SI units is in (A/m). The line integral of the magnetic field intensity H equals the total enclosed current:

$B-s(BHdl = $B-t(Bi

The above equation can be rewritten as for most practical circuits

$B-t(B HkIk = $B-t(BNmim

Further simplifying the equation you can use:

ISAT = HSAT x l

Looking at the B-H curves for the transformer material it can be seen that the core is saturated at a H value of around 200 A/m.

The effective path length of two E18 cores is 24mm.

Therefore Isat = 200 x 0.024 = 4.8A

Looking at my measured graphs I can see saturation starting to occur at 90kHz with an Ip-p = 8.8A. Decreasing the frequency further the unmistakable saturation knee starts to form and the base of the knee stays at just below 10A p-p.

snipped-for-privacy@gmail.com wrote:

Reply to
johanwagener

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