About leakage inductance in transformers

[John Larkin]

Either you are being obtuse or you are being stupid.

Of course you might be being both.

DNA

I said "I think", not that I was sure. Of course, some people will
always object to thinking, or taking a risk of being wrong.

What about a non-classic, non-concentric transformer? Like a toroid
with windings on opposite sides, or a super-CE-insulated transformer
with windings in different places, not on top each other? With no
core, primary flux will barely brush the secondary... for the toroid,
maybe 5% will couple. Add the core, and a lot more primary flux is
directed through the secondary, and how much depends on the
permeability.

Feel free to call me stupid again. You don't matter, you know.

John

 

[Paul Mathews]

I said "I think", not that I was sure. Of course, some people will
always object to thinking, or taking a risk of being wrong.

What about a non-classic, non-concentric transformer? Like a toroid
with windings on opposite sides, or a super-CE-insulated transformer
with windings in different places, not on top each other? With no
core, primary flux will barely brush the secondary... for the toroid,
maybe 5% will couple. Add the core, and a lot more primary flux is
directed through the secondary, and how much depends on the
permeability.

Feel free to call me stupid again. You don't matter, you know.

John- Hide quoted text -

- Show quoted text -

FWIW, I can confirm that the case of the toroidal core produces much
different results: leakage flux is much lower than for air core coils
in the same geometry.
I occasionally use ring-decay and voltage step methods to measure
leakage inductance. In the latter, you simply switch a voltage step
and measure current slope, solving L = V/(di/dt). The ring decay
method has been discussed extensively in this group. In both cases,
secondary is shorted. In some cases, it may make more sense to measure
secondary leakage inductance and calculate it for the primary based on
turns ratio.
Using a bridge, use must use an excitation frequency high enough that
the impedance of the leakage is significant compared to secondary
resistance multiplied by the square of the turns ratio. This makes 1
kHz bridges useless for many high freq transformers. It also means
that the method of shorting the secondary can become critical. Short
the secondary with as short and stout a link as possible.
Paul Mathews

 

[Jim Thompson]

I said "I think", not that I was sure. Of course, some people will
always object to thinking, or taking a risk of being wrong.

What about a non-classic, non-concentric transformer? Like a toroid
with windings on opposite sides, or a super-CE-insulated transformer
with windings in different places, not on top each other? With no
core, primary flux will barely brush the secondary... for the toroid,
maybe 5% will couple. Add the core, and a lot more primary flux is
directed through the secondary, and how much depends on the
permeability.

Feel free to call me stupid again. You don't matter, you know.

John

Some great sage said, "Stupid is as stupid does!" ;-)

...Jim Thompson

 

[Rich Grise, Plainclothes Hippie]

Either you are being obtuse or you are being stupid.

Of course you might be being both.

Damn! Wiki is my friend:
http://en.wikipedia.org/wiki/Leakage_inductance

It even has a diagram of the stuff, which is great for ignorami
like me; it was like an "Ah!" moment, like popping a big mental
zit. :-D

I don't think John is being obtuse at all... more like scalene. %-}

Cheers!
Rich

 

[John Larkin]

Damn! Wiki is my friend:
http://en.wikipedia.org/wiki/Leakage_inductance

It even has a diagram of the stuff, which is great for ignorami
like me; it was like an "Ah!" moment, like popping a big mental
zit. :-D

I don't think John is being obtuse at all... more like scalene. %-}

Cheers!
Rich

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

 

[Rich Grise]

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?

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

Cheers!
Rich

 

[MassiveProng]

It probably will, but not much, since the shorted secondary is doing its
damnedest to keep flux out of the core.

In order for ANY energy to pass from the primary to the secondary,
it MUST pass through the core. Perhaps you were making some vague
reference to stored energy?

 

[The Phantom]

Damn! Wiki is my friend:
http://en.wikipedia.org/wiki/Leakage_inductance

It even has a diagram of the stuff, which is great for ignorami
like me; it was like an "Ah!" moment, like popping a big mental
zit. :-D

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.

 

[Terry Given]

In order for ANY energy to pass from the primary to the secondary,
it MUST pass through the core. Perhaps you were making some vague
reference to stored energy?

nope. If they are wound concentric (and possibly interleaved) then there
is still plenty of coupling, sans core.

Cheers
Terry

 

[The Phantom]

FWIW, I can confirm that the case of the toroidal core produces much
different results: leakage flux is much lower than for air core coils
in the same geometry.
I occasionally use ring-decay and voltage step methods to measure
leakage inductance. In the latter, you simply switch a voltage step
and measure current slope, solving L = V/(di/dt). The ring decay
method has been discussed extensively in this group. In both cases,
secondary is shorted. In some cases, it may make more sense to measure
secondary leakage inductance and calculate it for the primary based on
turns ratio.
Using a bridge, use must use an excitation frequency high enough that
the impedance of the leakage is significant compared to secondary
resistance multiplied by the square of the turns ratio.

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.

 

[The Phantom]

I said "I think", not that I was sure. Of course, some people will
always object to thinking, or taking a risk of being wrong.

What about a non-classic, non-concentric transformer?

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.

Like a toroid
with windings on opposite sides, or a super-CE-insulated transformer
with windings in different places, not on top each other? With no
core, primary flux will barely brush the secondary... for the toroid,
maybe 5% will couple. Add the core, and a lot more primary flux is
directed through the secondary, and how much depends on the
permeability.

Feel free to call me stupid again. You don't matter, you know.

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

 

[The Phantom]

In the figure on the wiki page, it sure looks to me like the
permeability of the core will affect the leakage inductance.

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

 

[MassiveProng]

nope. If they are wound concentric (and possibly interleaved) then there
is still plenty of coupling, sans core.

That depends on the frequency. The lower it gets, the less coupling
will occur without a core.

 

[Genome]

I said "I think", not that I was sure. Of course, some people will
always object to thinking, or taking a risk of being wrong.


Feel free to call me stupid again. You don't matter, you know.

John

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

 

[Terry Given]

That depends on the frequency. The lower it gets, the less coupling
will occur without a core.

buggered if I know what happened to my post, so here it is again (kinda):


That is completely wrong.

Sans core, coupling is defined solely by geometry.

And it certainly doesnt depend on frequency, high or low.

If you dont believe me, go read Grover. Not that you'll understand it.
Or you could try Magnetic Circuits and Transformers, MIT rad lab series.
Given your demonstrated lack of understanding, Smit & Wijn will confuse
the hell out of you, and of course Perry is so far out of your league it
isnt even worth mentioning

(NB: at very low frequencies, the currents associated with low
inductances will eventually exceed the capability of the source, thereby
convincing the gullible that coupling has changed; at high frequencies
capacitive coupling will swamp inductive coupling, again leading the
foolish to the same erroneous conclusion; at really high frequencies
things get quite interesting, but for practical transformers its fair to
neglect displacement current)


This makes two MassivelyWrong posts re. magnetics, in relatively short
order. Give up.

Cheers
Terry

 

[Genome]

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

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.

http://focus.ti.com/lit/ml/slup123/slup123.pdf
http://focus.ti.com/lit/ml/slup105/slup105.pdf
http://focus.ti.com/lit/ml/slup088/slup088.pdf

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

 

[Genome]

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.

http://focus.ti.com/lit/ml/slup123/slup123.pdf
http://focus.ti.com/lit/ml/slup105/slup105.pdf
http://focus.ti.com/lit/ml/slup088/slup088.pdf

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

 

[Tony Williams]

w^2.M^2.(Rl+Rs)
Effective Rpri = Rp + --------------------
(Rs+Rl)^2 + (w.Ls)^2


( (w^2.M^2).Ls )
Effective Xpri = j.w.( Lp - -------------------- )
( (Rs+Rl)^2 + (w.Ls)^2 )

[/QUOTE]

It's even stranger than that. If the secondary resistance
(Rs+Rl, or just Rs if you set Rl to zero) is less than w*Ls, then
the Effective Rpri will decrease with decreases of Rs. If the
secondary resistance is greater than w*Ls, then Rpri will
*increase* with decreases of Rs.

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.

L1 = 12.5 mH
R1 = 32.7 ohms
L2 = 938 uH
R2 = 1.72 ohms
m = 2.8 mH

and plot the Effective Lpri vs frequency from 60 Hz to 10 kHz.

Then answer my question:

"Is this really the leakage inductance? Under what conditions
might it not be a good value for the leakage inductance?"

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.

 

[Tony Williams]

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

La ) (w.k.Ls)^2.Lp
) La = ----------------
| Rs^2 + (w.Ls)^2

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

 

[John Larkin]

The diagram on Wikipedia is incomplete, and therefore misleading. Nobody
builds transformers like that except for extreme high voltage or some such
special use.

That and similarly leaky constructions are common nowadays...

http://www.premmagnetics.com/SPW050d.htm

because of safety isolation standards.

John