About leakage inductance in transformers

In the figure on the wiki page, it sure looks to me like the permeability of the core will affect the leakage inductance. Am I wrong about that? Genome?

John

Yeah, like John said, without the core, it's almost _all_ leakage inductance!

Cheers! Rich

The diagram on Wikipedia is incomplete, and therefore misleading. Nobody builds transformers like that except for extreme high voltage or some such special use. All the leakage flux is first shown as being in the core inside the primary winding. Actually, most of the leakage flux never enters the core.

See the .pdf I posted over on ABSE showing the true state of affairs.

This is almost right. Look at the expression for measured Lpri posted by Tony Williams. What you want is for the self reactance of the secondary (shorted winding) to be large compared to the resistance of that winding. It would be possible to have a high leakage transformer where the leakage reactance was high, but the self reactance of the shorted winding was comparable to its resistance. Then you would get a bad result.

I should have been more specific in my first post in this thread. I had in mind the classic concentric transformer.

Other geometries are certainly worth considering, but I wanted to call attention to an aspect of making shorted-winding measurements that is often overlooked, and I wanted to use the classic transformer geometry for the discussion.

I liked your 3-legged transformer in another thread. As I read that I was thinking that the ultimate extension of that would be to dunk a transformer in a bucket of high permeability ferrofluid. If the air space where the leakage flux normally passes in a classic concentric wound transformer is replaced with some high permeability substance, then of course the leakage *inductance* will increase very much.

I wish we could refrain from name calling. I don't do it, and I don't see any benefit to it.

See my posting over on ABSE. I'm also posting some measurements on a small transformer taken with an LCR meter.

So..... stop being stupid and start thinking. What does your magnetic circuit look like... where is the energy stored... where does it go.

I don't know either but I can take a few guesses that are better than some of the 'statements' that you are coming up with.

I would explain it like but its toooooo hard. Plus if you are going to be stupid and say yahbut what if I put rice crispies on it you'll be wanting milk as well.

Next thing you'll be talking about ad hominem attacks.

FFS, did someone kidnap Larkin?

DNA

No you are probably not wrong. If I wanted to play the same game then I would tell you that the B field depends on the core.

I don't know this stuff myself but some of it might leak into my head.

BTW, I know it doesn't work for some people but.... in as much as you are not bothered I'm not bothered either and you insult yourself by suggesting I might be.

DNA

Oh, and just to prove it there is a rubbery face-sitting movie being posted in the latex group by rix500 at the moment.

Hope the sunrise looks good in AmericaLand this Sunday Morning.

DNA

Sorry about the delay in replying to this post, piddling about with sums on scraps of paper. That effective Xpri is not the leakage inductance and I can now see where the frequency-sensitive values for Rpri and Xpri come from.

The problem starts with the fact the the only definition of leakage inductance is based on a transformer without any winding resistance.

Lp(leak) +---+ +-----+ +--///---+---+ +-----+ | M | | | | ) ( 2 ) )|( Ratio = Lp ) ( Ls ----> k.Lp ) )|( 2 ) ( ) )|( k .Lp/Ls | | | | | +----+ +-----+ +---------+---+ +----+

The Lp-M-Ls transformer can be represented by an equivalent circuit of an uncoupled inductor, Lp(leak), value being given as Lp.(1-k^2), a shunt inductor of value (k^2.Lp) and a perfect transformer with inductance ratio k^2.Lp/Ls.

The coupling, k is of course defined by k^2 = M^2/Lp.Ls.

Ok, now add the winding resistances, Rp and Rs to the equivalent circuit and the short circuit on the secondary.

Rp Lp.(1-k^2) Rs +---/\\/\\--///---+---+ +---------/\\/\\--+ | | | | ) )|( Ratio = | k^2.Lp ) )|( | ) )|( k^2 .Lp/Ls | | | | | +----------------+---+ +---------------+

Now transform Rs over to the primary side.

Rp Lp.(1-k^2) R= Rs.k^2.Lp/Ls +--/\\/\\---///---+----/\\/\\---+ | | ) | k^2.Lp ) | ) | | | +---------------+-----------+

Now convert the parallel L//R into the series-equivalents to make it easy to see Effective Rpri and Lpri.

(w.k)^2.Rs.Lp.Ls Rp Lp.(1-k^2) Ra Ra = --------------- +--/\\/\\---///------/\\/\\---+ Rs^2 + (w.Ls)^2 | ) La ) (w.k.Ls)^2.Lp ) La = ---------------- | Rs^2 + (w.Ls)^2 +-------------------------+ Effective Rpri = Rp + Ra.

Effective Lpri = Lp.(1-k^2) + La.

But Ra and La are both frequency sensitive.

In fact if you expand-out Ra and La in Rpri and Lpri you get back to Wm Fraser's originals, complete with the required minus sign in the Lpri calculation.

As above. Lp(leak) = Lp.(1-k^2), where k^2 = M^2/Lp.Ls.

The rest is a red herring, swimming up a blind alley. :)

Typo correction. I read off the wrong scrap of paper.

The numerator in La should be (k.Rs)^2.Lp.

That and similarly leaky constructions are common nowadays...

because of safety isolation standards.

John

Look at the aspect ratio of the Prem xfmrs and compare to the Wikipedia drawing. The only time I've ever seen a transformer with the secondary as far from the primary as in the Wiki drawing is the horizontal output (flyback) transformer in a television.

But the effect of the core permeability on leakage inductance will be severe for such a non-concentric structure.

John

perhaps the word "necessarily" should be inserted here; at a high enough measurement frequency it's pretty close. We just need to pick w so that (w.Ls)^2 >> (Rs)^2 in the denominator above.

Excellent. This is what I was trying to bring to the attention of the group. That the secondary resistance causes the measured Lpri to increase at low frequencies seems to not be commonly understood. The plots I posted over on ABSE show the effect clearly. To get an accurate result, one needs to either make the measurement at a high enough frequency that (w.Ls)^2 >> (Rs)^2, or calculate it as L(leak) = Lp.(1-k^2). By the way, we've all heard of the term "coupling coefficient" for the quantity k = m^2/(Lp*Ls). Additionally, I saw in the 1943 book I mention elsewhere that the expression (1-k^2) is called the "leakage coefficient".

Quite right. Using this relationship, one can adjust the mearurement made at a too low frequency and thus obtain a much better value for the leakage inductance. It is not a perfect compensation, but it's a lot better than no compensation at all. Given the difficulty of making accurate measurements on iron cored transformers, it's probably as good as you're going to get.

This should be understood by every engineer who ever measures leakage inductance.

In the book I mention elsewhere is an interesting tidbit. The term:

(w^2.M^2) ------------------- in the expression for Effective Xpri is equal to (Rs)^2 + (w.Ls)^2

the square of the open circuit voltage ratio; that is, (Eoc1/V2)^2, where Eoc1 is the voltage measured at the open circuit primary when a voltage V2 is applied to the secondary.

I don't think it's that bad, but anyway it's an acceptable tradeoff. When the output of the secondary is 30 kV, you can't let said secondary get very close to anything.

I have one of those lying around; I'll have to make some measurments on it.