Is the DC resistance of a transformer winding DIRECTLY related to the volt drop?

[uer166]

That's complete garbage. Leakage inductances represent how well two windings are coupled. If you remove the core there will be almost no coupling between the two windings and the leakage inductance will be much higher.

That is some serious confidence.. You're mixing up terms: there is leakage inductance, and there is coupling coefficient. This isn't just semantics either, it's an important distinction.

The leakage inductance as a number (e.g. 10nH, 1uH, etc) does not change when you increase or decrease the amount of mutual inductance. It is purely a windings geometry thing, and the only way to reduce it is to couple the windings more closely via say interleaving, bifilar windings, or for the absolute lowest leakage coax windings. You can also increase leakage when needed (e.g. LLC transformers) by separating the windings into 2 banks.

When for example designing a flyback, the leakage inductance energy is lost in snubber, and everything else being equal, a core with higher permeability WILL NOT reduce those losses, only a windings geometry change will.

What actually represents how well the pri/secondary is coupled is *coupling coefficient* (e.g. 0.9, 0.999, etc), and that is now a function of both the mutual/magnetizing inductance, and the leakage inductance. You can increase coupling given a fixed leakage inductance by increasing magnetizing inductance, or you can increase it by decreasing leakage inductance as well.

[TizianoHV]

I'm sorry for being rude but I do not  agree with your theory:


"the excitation voltage level should not matter in determining the leakage inductance."   That's true.


"In fact you can remove the core entirely and still get the same result via shorting the secondary and measuring primary inductance."    But I can't agree with that. Maybe we are talking about two different things. Are you telling that the two transformers in the photo have same leakage inductance since it isn't influenced by the core?

[uer166]

I'll have to think about that one. Here you might be right, a magnetic shunt would be a special case since it's equivalent to adding an entire second inductor in series. This isn't really a "transformer" anymore, but rather a series inductor combined with transformer on one core. In this case excitation level would also change the leakage inductance as well, since the magnetic shunt can saturate.

I was talking about normal transformers in general though, and will maintain that the leakage inductance does not change whenever core is removed. That would apply to split bobbin, interleaved, non interleaved etc windings in cases where you're trying to control or minimize leakage. The context was specifically to show that the excitation level doesn't really matter in those cases.

[TizianoHV]

Such cores are used in Neon Signs transformers and Microwave Oven Transformers. I used it as an extreme example to show clearly where I'm pointing.

Before going in deeper I want to show how I convert magnetic circuit to electric circuit: Fig.1, starting from a draw of the transformer, you visualize the fluxes, reluctances and mmf's.
Now, to convert to electric equivalent, you need to swap series/parallel connections (step 3, 4) and rearrange the equivalent electric in a more intuitive disposition(5,6).
Since Xfe is much larger than Xa1, Xa2 we can say that the leakage reactance Xk =~ Xa1 + Xa2.


In Fig.2 I've removed the core. And yes, Xa1 and Xa2 are still there, but how can you measure them when there's Xa (Xa < Xa1,Xa2) shunting them?


Edit. And that's just a theoretical representation, in practice you have to immage leakage fluxes excaping from the core and everywere

[najrao]

As an old school 'power' electrical engineer, I am amused by the extreme views expressed in this thread.

Be that as it may, this forum caters largely to power electronics users, and they would be concerned with how less-than-ideal a practical transformer would work in their circuits. A most common application would be a 'mains' transformer feeding a capacitor 'input' rectfier. In dozens of other posts in our forum, I have read, with amusement,  it expressed that the dc output voltage of such rectifier would be the peak of the applied ac,  less for diode drops. In a practical situation, the average dc voltage would be far lower, perhaps closer to the effective ac itself. And it is just not due to the Idc drawn dropping  voltage in the transformer resistance! Or even leakage 'inductance'.

Yes, the transformer impedance is the culprit, but what multiplies it is not the Idc at all, but the many times larger peak cap charging current.  After the serious drop, the cap will charge to much less than the 'peak' taken for granted. The large fast rising charge current is much more curtailed by the inductance than the resistance of the transformer, because it has much higher frequency components.

Even with 'small' transformers where resistance may predominate over reactance, the latter is not negligible at all in cap input rectifier application. Wise designers use practical 'curves' to substitute rules of thumb. Or end up with having to up the transformer specs after test!