Transformer coupling coefficients
Don Foreman
coefficients (k) might be in ferrite switchmode transformers. I'm
interested in forward converter power range of 50 to several hundred
watts.
I'd be interested in typical values for suitably-sized toroids, pot
cores, E-E cores and U-U cores. Bifilar and/or surrounded windings
(e.g. parallel windings above and below another winding) may be used.
Anybody have any experience with this?
Tim Wescott
Really close to one, in my experience -- which has a lot more to do with
RF than power supplies.
I often see the leakage inductance of a ferrite quoted as being equal to
one turn of wire around the outside of the core. I _don't_ know if this
is really some fundamental constant due to the geometry of the thing, or
if it just works out that way, or if it's all BS. But if it's true you
can back out from leakage inductance to coupling coefficients.
--
Tim Wescott
Wescott Design Services
http://www.wescottdesign.com
Posting from Google? See http://cfaj.freeshell.org/google/
"Applied Control Theory for Embedded Systems" came out in April.
See details at http://www.wescottdesign.com/actfes/actfes.html
Don Foreman
wrote:
>Don Foreman wrote:
>
>> I'd like to know what realistic or reasonable values for coupling
>> coefficients (k) might be in ferrite switchmode transformers. I'm
>> interested in forward converter power range of 50 to several hundred
>> watts.
>>
>> I'd be interested in typical values for suitably-sized toroids, pot
>> cores, E-E cores and U-U cores. Bifilar and/or surrounded windings
>> (e.g. parallel windings above and below another winding) may be used.
>>
>> Anybody have any experience with this?
>
>Really close to one, in my experience -- which has a lot more to do with
>RF than power supplies.
>
>I often see the leakage inductance of a ferrite quoted as being equal to
>one turn of wire around the outside of the core. I _don't_ know if this
>is really some fundamental constant due to the geometry of the thing, or
>if it just works out that way, or if it's all BS. But if it's true you
>can back out from leakage inductance to coupling coefficients.
Yes. Leakage inductance is really what I'm after in the first place.
It really is determined by geometry and the relative permeability of
the core material. Windings that occupy exactly the same space would
have perfect coupling. Leakage happens to the extent that this
impossibility is not realized.
I suppose I could use FEMM before winding up some test samples, but
for k > 0.99 I'm probably in the simulation error and noise.
legg
wrote:
>Don Foreman wrote:
>
>> I'd like to know what realistic or reasonable values for coupling
>> coefficients (k) might be in ferrite switchmode transformers. I'm
>> interested in forward converter power range of 50 to several hundred
>> watts.
>>
>> I'd be interested in typical values for suitably-sized toroids, pot
>> cores, E-E cores and U-U cores. Bifilar and/or surrounded windings
>> (e.g. parallel windings above and below another winding) may be used.
>>
>> Anybody have any experience with this?
>
>Really close to one, in my experience -- which has a lot more to do with
>RF than power supplies.
>
>I often see the leakage inductance of a ferrite quoted as being equal to
>one turn of wire around the outside of the core. I _don't_ know if this
>is really some fundamental constant due to the geometry of the thing, or
>if it just works out that way, or if it's all BS. But if it's true you
>can back out from leakage inductance to coupling coefficients.
Because leakage inductance can vary with the winding structure and
turns ratio, a fixed value based on core structure alone is wishfull
thinking.
RL
Don Foreman
Understood. I'm not looking for a silver bullet or a "simple answer
to a complex issue". I'd just like to know some typical values
that people who use such things routinely may have seen in practice.
That's also why I asked about k rather than leakage inductance. k
does not depend on turns count, just on geometry and relative
permeabilities.
I was surprised to find k's on the order of .998 in a couple of quick
FEMM simulations. Maybe the question is moot.
John Woodgate
Aug 2006, Don Foreman <dfor...@NOSPAMgoldengate.net> writes
>I was surprised to find k's on the order of .998 in a couple of quick
>FEMM simulations. Maybe the question is moot.
I'm not sure what you mean by 'moot'. It can mean 'debatable', but I
suspect you mean 'pointless'.
It's true that for normal constructions, k is so close to 1 that it can
be assumed to be 1. It is more helpful to consider leakage inductance,
which is easy to measure, at least if there are only two windings. You
measure the inductance of one winding with the other winding
short-circuited. You can also include leakage inductance as a discrete
component in simulations.
--
OOO - Own Opinions Only. Try www.jmwa.demon.co.uk and www.isce.org.uk
2006 is YMMVI- Your mileage may vary immensely.
John Woodgate, J M Woodgate and Associates, Rayleigh, Essex UK
J.A. Legris
John Woodgate wrote:
> In message <0os9f29sek6n859dp...@4ax.com>, dated Tue, 29
> Aug 2006, Don Foreman <dfor...@NOSPAMgoldengate.net> writes
>
> >I was surprised to find k's on the order of .998 in a couple of quick
> >FEMM simulations. Maybe the question is moot.
>
> I'm not sure what you mean by 'moot'. It can mean 'debatable', but I
> suspect you mean 'pointless'.
>
It looks like an Anglo/Americano distinction. Merriam-Webster shows a
second sense: deprived of practical significance - made abstract or
purely academic. I've never heard it used in the original sense. Maybe
the distinction on the west side of the pond is moot.
--
Joe Legris
John Woodgate
Wed, 30 Aug 2006, J.A. Legris <jale...@sympatico.ca> writes
this case, because it seemed simple enough to get a result without too
much confusion. The transition in meaning is rather easy, but it doesn't
seem to have made it into English dictionaries (no, I haven't looked at
ALL of them!).
debatable -> purely academic -> pointless
Tim Wescott
You'll still need snubbers on the primary, though.
Don Foreman
<j...@jmwa.demon.co.uk> wrote:
>In message <0os9f29sek6n859dp...@4ax.com>, dated Tue, 29
>Aug 2006, Don Foreman <dfor...@NOSPAMgoldengate.net> writes
>
>>I was surprised to find k's on the order of .998 in a couple of quick
>>FEMM simulations. Maybe the question is moot.
>
>I'm not sure what you mean by 'moot'. It can mean 'debatable', but I
>suspect you mean 'pointless'.
How about "don't amount to a hill of beans", or "hit don't make no
nevermind, ol' Son!"
>
>It's true that for normal constructions, k is so close to 1 that it can
>be assumed to be 1. It is more helpful to consider leakage inductance,
>which is easy to measure, at least if there are only two windings. You
>measure the inductance of one winding with the other winding
>short-circuited. You can also include leakage inductance as a discrete
>component in simulations.
Right. k is generic to a geometry, leakage inductance is specific
to a particular set of windings and the k that exists between them.
For given geometry and k, leakage inductance will increase as
self-inductance increases. Things are clearer for me if I separate
the variables of geometry and numbers of turns because they are
somewhat separate parameters.
Don Foreman
wrote:
Usually so, except for H-bridge or half-bridge configurations.
Question is how much snubbing. Snubbing is dissipative
so less is better -- as long as it's enough!
John Woodgate
>For given geometry and k, leakage inductance will increase as
>self-inductance increases.
But it's reasonably nearly proportional to self-inductance, provided the
windings are not very different in 'window-fill'.
>Things are clearer for me if I separate the variables of geometry and
>numbers of turns because they are somewhat separate parameters.
See above.
Don Foreman
<j...@jmwa.demon.co.uk> wrote:
>In message <nsfbf2lq29gt1nv72...@4ax.com>, dated Wed, 30
>Aug 2006, Don Foreman <dfor...@NOSPAMgoldengate.net> writes
>
>>For given geometry and k, leakage inductance will increase as
>>self-inductance increases.
>
>But it's reasonably nearly proportional to self-inductance, provided the
>windings are not very different in 'window-fill'.
Yes, and the proportionality constant is k -- which is what I'm
asking about. I guess I'll have to do some experiments.
An overnight soak of an inductor and a transformer in some lacquer
thinner did a very nice job of freeing up the little ferrite cores.
Man, the one out of a fluorescent ballast has a 2.61mm gap!
John Woodgate
>n Wed, 30 Aug 2006 18:12:18 +0100, John Woodgate <j...@jmwa.demon.co.uk>
>wrote:
>
>>In message <nsfbf2lq29gt1nv72...@4ax.com>, dated Wed, 30
>>Aug 2006, Don Foreman <dfor...@NOSPAMgoldengate.net> writes
>>
>>>For given geometry and k, leakage inductance will increase as
>>>self-inductance increases.
>>
>>But it's reasonably nearly proportional to self-inductance, provided the
>>windings are not very different in 'window-fill'.
>
>Yes, and the proportionality constant is k --
k is very nearly 1. The leakage inductance is very much smaller than the
winding inductance, not k (or 1/k !) times it. You may say it's (1-k)
times, if you define k to make it true.
Harry Dellamano
"John Woodgate" <j...@jmwa.demon.co.uk> wrote in message
news:p3F8Rxqa...@jmwa.demon.co.uk...
You guys are just killing me! Please read;
http://www.onsemi.com/pub/Collateral/AN1679-D.PDF
regards,
harry
legg
I believe that the coefficient k is an abstraction introduced to
remind the simulator user that something is missing in the model.
It has no physical derivation.
RL
Don Foreman
It is defined in fundamental physical terms: It is a ratio of
flux linkages. Transformer theory was well-developed long before
simulators or even computers existed. Many or most texts
express k as lambda sub ij / lambda sub jj where lambda ij is flux
linking winding i resulting from excitation of winding j.
Don Foreman
<har...@tdsystems.org> wrote:
>
>"John Woodgate" <j...@jmwa.demon.co.uk> wrote in message
>news:p3F8Rxqa...@jmwa.demon.co.uk...
>> In message <qcnbf2l59la6q55jt...@4ax.com>, dated Wed, 30 Aug
>> 2006, Don Foreman <dfor...@NOSPAMgoldengate.net> writes
>>>n Wed, 30 Aug 2006 18:12:18 +0100, John Woodgate <j...@jmwa.demon.co.uk>
>>>wrote:
>>>
>>>>In message <nsfbf2lq29gt1nv72...@4ax.com>, dated Wed, 30
>>>>Aug 2006, Don Foreman <dfor...@NOSPAMgoldengate.net> writes
>>>>
>>>>>For given geometry and k, leakage inductance will increase as
>>>>>self-inductance increases.
>>>>
>>>>But it's reasonably nearly proportional to self-inductance, provided the
>>>>windings are not very different in 'window-fill'.
>>>
>>>Yes, and the proportionality constant is k --
>>
>> k is very nearly 1. The leakage inductance is very much smaller than the
>> winding inductance, not k (or 1/k !) times it. You may say it's (1-k)
>> times, if you define k to make it true.
k is defined as a ratio of flux linkages, not something I define.
Leakage inductance will depend on turns ratio as well as k, but you're
right in that it is a 1-nk sort of relationship.
Don Foreman
<har...@tdsystems.org> wrote:
I saw nothing in that appnote dealing with values of k found in common
practice.
So far all I'm hearing is "close to 1", which isn't very helpful. The
difference between .990 and .995 can be significant, and .999 works
even better. Are these realistic values? I guess nobody here knows
either, oh well! Thanks anyway and nevermind.
John Woodgate
Dellamano <har...@tdsystems.org> writes
What astonishing insight does this give? To try to introduce all that in
a news article to someone more or less beginning to understand
transformers is simply counter-productive.
John Woodgate
>So far all I'm hearing is "close to 1", which isn't very helpful. The
>difference between .990 and .995 can be significant, and .999 works
>even better. Are these realistic values? I guess nobody here knows
>either, oh well! Thanks anyway and nevermind.
I HAVE addressed that. Unless you have an unusual construction, k is so
close to 1 that you can assume it IS 1. Yes, values above 0.998 are
realistic. Anything lower requires an unusual construction.
Don Foreman
<j...@jmwa.demon.co.uk> wrote:
>In message <pjvcf2l4raocajc2i...@4ax.com>, dated Thu, 31
>Aug 2006, Don Foreman <dfor...@NOSPAMgoldengate.net> writes
>
>>So far all I'm hearing is "close to 1", which isn't very helpful. The
>>difference between .990 and .995 can be significant, and .999 works
>>even better. Are these realistic values? I guess nobody here knows
>>either, oh well! Thanks anyway and nevermind.
>
>I HAVE addressed that. Unless you have an unusual construction, k is so
>close to 1 that you can assume it IS 1. Yes, values above 0.998 are
>realistic. Anything lower requires an unusual construction.
Ah! NOW you have! That's exactly what I was after. .998 is
considerably more informative than "close to 1". Thanks!
Don Foreman
<j...@jmwa.demon.co.uk> wrote:
>In message <ANmJg.3761$N84.3740@trnddc08>, dated Wed, 30 Aug 2006, Harry
>Dellamano <har...@tdsystems.org> writes
>> You guys are just killing me! Please read;
>>http://www.onsemi.com/pub/Collateral/AN1679-D.PDF
>
>What astonishing insight does this give? To try to introduce all that in
>a news article to someone more or less beginning to understand
>transformers is simply counter-productive.
I'm not exactly just beginning to understand transformers, I was just
looking for a "typical number", like John's .998, from an experienced
practicioner.
All that mishmash in the appnote is an attempt to boil things down to
pat turn-crank formulae. Ugh! It is a LOT easier to deal with and
understand transformers by using matrix notation and MathCAD. This
is particularly true when dealing with multiply-wound loosely-coupled
xfmrs as rod-core ignition transformers where k's of 0.7 and less are
not only typical but desirable to produce the desired high-output-Z
transfer function. See, e.g., US patents 5,521,444 and 6,191,956.
legg
<dfor...@NOSPAMgoldengate.net> wrote:
Fundamental physical terms - ie a mathematical model.
RL
Don Foreman
Yup. For every pair of windings i and j, there is a k sub ij and a k
sub ji. These are not necessarily equal. This can be seen either
with bench tests or in a FEMM simulation.
In the simplified expression for mutual inductance derived from the
general matrix equations, M = k * sqrt (L1*L2), that k is the
geometric mean of k sub ji and k sub ij for that pair of windings.
SPICE uses that k. The geometric mean happens when the system of
equations is solved for a particular value.
The Phantom
>sub ji. These are not necessarily equal. This can be seen either
>with bench tests or in a FEMM simulation.
>
> In the simplified expression for mutual inductance derived from the
>general matrix equations, M = k * sqrt (L1*L2), that k is the
>geometric mean of k sub ji and k sub ij for that pair of windings.
>SPICE uses that k. The geometric mean happens when the system of
>equations is solved for a particular value.
In a linear, passive, bilateral, time-invariant network of several
coupled inductances, L1, L2,...,Ln we can write down an impedance matrix
for the network, representing the self inductance of each inductor as L11,
L22, L33,...,Lnn and the mutual inductances as L12, L21; L13, L31; L23,
L32; Lij, Lji; etc. It is an experimental fact discovered by Faraday that
Lij = Lji, and this fact can also be demonstrated by energy considerations.
I've never seen k defined in such a way as to depend on which inductor is
being excited, but if it were then presumably the definitions for a pair of
inductors would be kij = Lij/SQRT(Lii*Ljj) and kji = Lji/SQRT(Lii*Ljj).
Since Lij = Lji, then kij = kji. This is also evidenced by the application
of the reciprocity theorem to the impedance matrix for the network.
I suppose that real-world lab measurements may give slightly different
results for kij and kji due to measurement error and non-linearities, but
in the ideal case, there is only one k between a pair of inductors.
To Legg, who says:
"I believe that the coefficient k is an abstraction introduced to
remind the simulator user that something is missing in the model.
It has no physical derivation."
A simple physical interpretation of coupling coefficient would be
something like this:
Imagine you have a pair of inductors wound with an equal number of turns
in each winding, and with superconducting wire so that IR drops don't
figure in; ignore parasitic capacitance, etc., etc. Apply an AC voltage to
the first inductor and measure the induced voltage at the terminals of the
second inductor. If the inductors were perfectly coupled, the voltage at
the second would be equal to the voltage applied to the first. If the
inductors are not perfectly coupled the voltage at the second inductor will
be some fraction (less than unity) of the voltage applied to the first
inductor. That fraction *is* the coupling coefficient; hence its name. It
is the ratio of flux linkages. The first inductor links all of its own
flux, but the second inductor only links a fraction of the flux produced by
the first inductor; that fraction is k.
If the inductors are wound with different numbers of turns, just make the
obvious adjustment to the voltage expected at the terminals of the second
inductor.
John Woodgate
Sep 2006, The Phantom <pha...@aol.com> writes
>I've never seen k defined in such a way as to depend on which inductor
>is being excited, but if it were then presumably the definitions for a
>pair of inductors would be kij = Lij/SQRT(Lii*Ljj) and kji =
>Lji/SQRT(Lii*Ljj). Since Lij = Lji, then kij = kji. This is also
>evidenced by the application of the reciprocity theorem to the
>impedance matrix for the network.
>
> I suppose that real-world lab measurements may give slightly
>different results for kij and kji due to measurement error and
>non-linearities, but in the ideal case, there is only one k between a
>pair of inductors.
I'm not sure about that. Consider that leakage inductance in a
two-winding transformer is a result of flux from one winding not linking
with the other winding. Now think of a rectangular core with two 'legs'.
On one is a coil which extends over the full width available to it, but
the winding depth is just one layer. Very nearly all of the flux is in
the core. Now consider the other winding to be very narrow, but with
many layers. The pattern of flux generated by this winding is very
different from that generated by the first winding. Can we be sure that
the non-linking flux is the same in both cases?
For one winding energized, it is certainly true that the two k's are
equal, because ether is only a single flux pattern that causes both.
legg
wrote:
<snip>
> I suppose that real-world lab measurements may give slightly different
>results for kij and kji due to measurement error and non-linearities, but
>in the ideal case, there is only one k between a pair of inductors.
>
>To Legg, who says:
>
>"I believe that the coefficient k is an abstraction introduced to
>remind the simulator user that something is missing in the model.
>
>It has no physical derivation."
>
> A simple physical interpretation of coupling coefficient would be
>something like this:
>
> Imagine you have a pair of inductors wound with an equal number of turns
>in each winding, and with superconducting wire so that IR drops don't
>figure in; ignore parasitic capacitance, etc., etc. Apply an AC voltage to
>the first inductor and measure the induced voltage at the terminals of the
>second inductor. If the inductors were perfectly coupled, the voltage at
>the second would be equal to the voltage applied to the first. If the
>inductors are not perfectly coupled the voltage at the second inductor will
>be some fraction (less than unity) of the voltage applied to the first
>inductor. That fraction *is* the coupling coefficient; hence its name. It
>is the ratio of flux linkages. The first inductor links all of its own
>flux, but the second inductor only links a fraction of the flux produced by
>the first inductor; that fraction is k.
>
> If the inductors are wound with different numbers of turns, just make the
>obvious adjustment to the voltage expected at the terminals of the second
>inductor.
I should love to ignore IR drops, stray capacty, core loss, leakage,
source impedances, fringing, crowding, eddying, self-resonance and the
like. It would really be quite interesting to see what was left in
transformer action.
It's no wonder I think of 'k' as an abstraction, if all that it
identifies is an amplitude error caused by factors that are not being
conveniently ignored..
As a Doubting Thomas, I expect that this factor wouldn't repeat, given
the simplest turns ratio adjustment, when scaled, unless another
factor was identified for removal.
RL
Don Foreman
wrote:
>On Fri, 01 Sep 2006 12:50:52 -0500, Don Foreman
>>Yup. For every pair of windings i and j, there is a k sub ij and a k
>>sub ji. These are not necessarily equal. This can be seen either
>>with bench tests or in a FEMM simulation.
>>
>> In the simplified expression for mutual inductance derived from the
>>general matrix equations, M = k * sqrt (L1*L2), that k is the
>>geometric mean of k sub ji and k sub ij for that pair of windings.
>>SPICE uses that k. The geometric mean happens when the system of
>>equations is solved for a particular value.
>
> In a linear, passive, bilateral, time-invariant network of several
>coupled inductances, L1, L2,...,Ln we can write down an impedance matrix
>for the network, representing the self inductance of each inductor as L11,
>L22, L33,...,Lnn and the mutual inductances as L12, L21; L13, L31; L23,
>L32; Lij, Lji; etc. It is an experimental fact discovered by Faraday that
>Lij = Lji, and this fact can also be demonstrated by energy considerations.
>I've never seen k defined in such a way as to depend on which inductor is
>being excited, but if it were then presumably the definitions for a pair of
>inductors would be kij = Lij/SQRT(Lii*Ljj) and kji = Lji/SQRT(Lii*Ljj).
>Since Lij = Lji, then kij = kji. This is also evidenced by the application
>of the reciprocity theorem to the impedance matrix for the network.
Circular argument. You redefined k in terms of mutual inductance. I
said above that this k will be the geometric mean of kij and kji, so
by (your) definition kij = kji =k. If kji is the percentage of
flux produced by winding i that links winding j (as originally
defined), kij and kji need not and in general are not identical.
The way to measure these experimentally is not by inductance
measurements but rather by open-circuit voltage measurements corrected
for turns ratio. V2 = (N2/N1) * k * V1 because V2 = N2 dphi2/dt
>
> I suppose that real-world lab measurements may give slightly different
>results for kij and kji due to measurement error and non-linearities, but
>in the ideal case, there is only one k between a pair of inductors.
Depends on how you define k. If it is defined in terms of flux
linkages, there can be and in general are differences between kji and
kij.
>
>To Legg, who says:
>
>"I believe that the coefficient k is an abstraction introduced to
>remind the simulator user that something is missing in the model.
>
>It has no physical derivation."
>
> A simple physical interpretation of coupling coefficient would be
>something like this:
>
> Imagine you have a pair of inductors wound with an equal number of turns
>in each winding, and with superconducting wire so that IR drops don't
>figure in; ignore parasitic capacitance, etc., etc. Apply an AC voltage to
>the first inductor and measure the induced voltage at the terminals of the
>second inductor. If the inductors were perfectly coupled, the voltage at
>the second would be equal to the voltage applied to the first. If the
>inductors are not perfectly coupled the voltage at the second inductor will
>be some fraction (less than unity) of the voltage applied to the first
>inductor. That fraction *is* the coupling coefficient; hence its name. It
>is the ratio of flux linkages. The first inductor links all of its own
>flux, but the second inductor only links a fraction of the flux produced by
>the first inductor; that fraction is k.
You're having a little problem with consistent definition of k here.
Don Foreman
Regarding flux linkages in a transformer as an abstraction is an
interesting perspective!
The Phantom
wrote:
>In message <2c2hf2hud1rg8s5dp...@4ax.com>, dated Fri, 1
>Sep 2006, The Phantom <pha...@aol.com> writes
>
>>I've never seen k defined in such a way as to depend on which inductor
>>is being excited, but if it were then presumably the definitions for a
>>pair of inductors would be kij = Lij/SQRT(Lii*Ljj) and kji =
>>Lji/SQRT(Lii*Ljj). Since Lij = Lji, then kij = kji. This is also
>>evidenced by the application of the reciprocity theorem to the
>>impedance matrix for the network.
>>
>> I suppose that real-world lab measurements may give slightly
>>different results for kij and kji due to measurement error and
>>non-linearities, but in the ideal case, there is only one k between a
>>pair of inductors.
>
>I'm not sure about that. Consider that leakage inductance in a
>two-winding transformer is a result of flux from one winding not linking
>with the other winding. Now think of a rectangular core with two 'legs'.
>On one is a coil which extends over the full width available to it, but
>the winding depth is just one layer. Very nearly all of the flux is in
>the core. Now consider the other winding to be very narrow, but with
>many layers. The pattern of flux generated by this winding is very
>different from that generated by the first winding. Can we be sure that
>the non-linking flux is the same in both cases?
I have a long single layer coil of 180 turns of 26 gauge wire on a
porcelain cylinder, .75" in diameter. I wound about 120 additional turns
as a fairly compact bundle near the middle. That's fairly close to what
you are describing. I posted a picture of it over on abse.
I denote the single layer coil as L1 and call it the primary, and the
compact bundle as L2, the secondary. The measurements are as follows:
Self inductance of L1 = 93.6 uH
Self inductance of L2 = 450 uH
Inductance of L1 + L2, connected series aiding = 704 uH
Inductance of L1 + L2, connected series opposing = 385 uH
From this we can determine the mutual inductance to be
(704-385)/4 = 79.75 uH
and the coefficient of coupling to be
k = 79.75/SQRT(93.6 * 450) = .3886
When coils are as loosely coupled as this, it isn't appropriate to use
turns ratio to determine the voltage induced in a second winding by the
voltage applied to a first winding. The appropriate ratio is the square
root of the ratio of the primary self inductance to the secondary self
inductance. Note that for coils wound on ungapped ferromagnetic cores
(regular old transformers, in other words), this ratio is very nearly the
same as the turns ratio. I'm going to call this ratio ~N because it's
analogous to a transformer turns ratio. From the measurements above,
~N = .456
So, if the flux generated by applying V1 volts to the primary (L1) were
completely linked by the secondary, we would expect to measure V1/~N volts
at the secondary. But since not all the primary flux is linked by the
secondary, we need to include the coupling coefficient in the expression.
Thus, the voltage we should expect to measure at the secondary would be
(k / ~N) * V1 volts. If we apply V2 volts to the secondary (L2), we should
expect to measure (k * ~N) * V2 volts at the primary.
In the case of the coil assembly whose measurements are given above, we
have:
For V1 applied to L1, the voltage V2 at L2 should be
V2 = (k / ~N) * V1 = .8522 * V1
For V2 applied to L2, the voltage V1 at L1 should be
V1 = (k * ~N) * V2 = .1772 * V2
Now for the measured 25 KHz voltages:
With .4639 volts applied to L1, I measured .3893 volts at L2; I should have
expected to measure .8522 * .4639 = .3953 volts.
With 2.19 volts applied to L2, I measured .3859 volts at L1; I should have
expected to measure .1772 * 2.19 = .3881 volts.
These measurements don't really justify 4 digits; at 25 KHz with wires
draped over the bench, readings are a little jumpy, but those were the
numbers I wrote down.
It's pretty clear that k is the same in both directions, even for these
two coils where the flux distributions are very different depending on
which winding is energized.
This is the rather counterintuitive fact which Faraday discovered.
I should have mentioned in my first post in this thread that these
relationships I've been discussing are voltage transfer ratios.
Reciprocity only applies as well as it does in this hardware example if the
inductances have rather high Q. I implied it when I said that the example
would use superconducting wire.
The reciprocity theorem only says strictly that the transfer admittances
and transfer impedances are equal. In other words, if we apply a voltage
to L1 and measure the short-circuit current in L2 it will be the same as if
we apply the same voltage to L2 and measure the short-circuit current in
L1. But if the resistances in this circuit (a transformer, albeit loosely
coupled) are negligible, then the voltage transfer ratios are equal when
adjusted for turns (square root of self-inductance) ratios, because k is
the same in both directions.
Don Foreman says:
"...kij and kji need not and in general are not identical.
The way to measure these experimentally is not by inductance
measurements but rather by open-circuit voltage measurements corrected
for turns ratio. V2 = (N2/N1) * k * V1 because V2 = N2 dphi2/dt"
Actually, this is not the way to measure k experimentally. One would not
expect the open circuit voltage transfer ratios to be equal; that's not
what the reciprocity theorem says. Perhaps this is why Don thinks he has
experimental evidence that k is different in the two directions; he
measures different voltage transfer ratios in the two directions and
believes this means that k is different in the two directions.
"V2 = (N2/N1) * k * V1" is not the fundamental relationship, but rather
V2 = SQRT(L2/L1) * k * V1 as I explain above. The N2/N1 ratio is an
approximation which is only acceptably accurate for closely coupled
inductors such as found in transformers.
Note that for the (loosely coupled) transformer experiment described
above, we would expect the voltage V2 to be
(N2/N1) * k * V1 = 120/180 * .3886 * .4639 = .12 volts if it were true
that V2 = (N2/N1) * k * V1. But the measured voltage V2 was .3893 volts,
considerably different. The expression V2 = SQRT(L2/L1) * k * V1 gives the
correct result.
The DC resistance and the skin and proximity effect (equivalent)
resistances in the windings will generally cause unequal (adjusted for
turns ratio if closely coupled, otherwise adjusted for square root of
self-inductance ratio) voltage transfer ratios.
You can see why the voltage transfer ratio isn't necessarily the same
(adjusted for turns ratio) in both directions by a simple thought
experiment. Imagine you have a two winding transformer on a low loss
ferromagnetic core, with the two (equal turns) windings closely coupled
(bifilar wound). Since the self-inductances will be finite, there will be
some magnetizing current; let's say that for an applied voltage of 100
volts, the magnetizing current is .1 amps (a nearly pure reactance of 1000
ohms), for either winding. If we apply the 100 volt excitation to the
primary of this transformer, we will measure nearly 100 volts at the
secondary. And if we apply 100 volts to the secondary, we will measure
nearly 100 volts at the primary; the same voltage transfer ratio in both
directions. Now put a 1000 ohm resistor in series with the primary (and
consider it part of the primary), and apply 141 volts to the combination.
We will measure about 100 volts at the secondary, so we might think the
coupling coefficient is about .707. But if we apply 100 volts to the
secondary, we will measure nearly 100 volts at the primary (with the 1000
ohm resistor in series), because our measurement is made open circuit and
no current passes through the 1000 ohm resistor, and we would think the
coupling coefficient is very near 1.
The resistor has caused the voltage transfer ratio to be different in the
two directions. If the external 1000 ohm resistor bothers you, just
imagine the primary to be wound with resistance wire of 1000 ohms total DC
resistance, and you get the same result. This is how (probably different)
losses in the different windings can cause the voltage transfer ratio to be
different in the two directions.
But if we apply 100 volts to the primary and short the secondary, the
secondary current will be .1 amp. If we apply 100 volts to the secondary,
the short-circuit primary current will be .1 amp. The transfer
*impedances* are equal as the reciprocity theorem requires.
The Phantom
What is left is most of what a transformer does. All of those things are
generally not necessary to consider for a first cut at a transformer
design. It's up to the designer to know when to take them into account.
When you analyze a circuit made up of R, L and C, don't you ignore the L
and C in the R, the R and L in the C, and the R and C in the L? Even
though you know that each element is, in the real world, contaminated by
the other two, don't you do a lot of analysis of your schematic assuming
most of them are pure?
In the example I gave, the coupling coefficient would behave just as I
described. And it is a necessary component of the expression (formulated
without mutual inductance, as may be convenient) for the behavior of
coupled inductors, as I explain in my long post in reply to John Woodgate.
The Phantom
Can you give us a citation of a book (with page number) or other
authority that explains how it is that kij # kji?
Harry Dellamano
"The Phantom" <pha...@aol.com> wrote in message
news:silhf2dms73faa8u7...@4ax.com...
> wrote:
>
Dear Phantom,
My questions below pertain to two windings, loosely coupled, K<0.95. In the
above you calculated mutual inductance by dividing the difference of aiding
and opposing inductance by 4. Four is not intuitive to me, why not two?
Does shorting one winding and measuring inductance on the other yield any
useful information at all? Is any measurement useful with a shorted winding?
This app. note; http://www.onsemi.com/pub/Collateral/AN1679-D.PDF on page
four they show a two winding transformer and ways to make a SPICE model.
Let's say the coupling is very poor, K< 0.95, how would you correct his 8
steps so the model is correct. Multiplying Np by K would certainly help.
Thanks for your insight,
Harry
Don Foreman
wrote:
>
> Can you give us a citation of a book (with page number) or other
>authority that explains how it is that kij # kji?
>
"Electric Machinery" Fitzgerald & Kingsley, 1961, McGraw Hill, page
26.
Jim Thompson
Kingsley... what a horse's ass... just about the worst professor I
ever had.
...Jim Thompson
--
| James E.Thompson, P.E. | mens |
| Analog Innovations, Inc. | et |
| Analog/Mixed-Signal ASIC's and Discrete Systems | manus |
| Phoenix, Arizona Voice:(480)460-2350 | |
| E-mail Address at Website Fax:(480)460-2142 | Brass Rat |
| http://www.analog-innovations.com | 1962 |
I love to cook with wine. Sometimes I even put it in the food.
The Phantom
wrote:
<SNIP>
>Dear Phantom,
> My questions below pertain to two windings, loosely coupled, K<0.95. In the
>above you calculated mutual inductance by dividing the difference of aiding
>and opposing inductance by 4. Four is not intuitive to me, why not two?
I won't go into all the details; any circuit theory text should give
them. What's going on is that when the two windings are connected series
aiding, the measured inductance is L1 + L2 + 2*m, and when they're
connected series opposing, the measured inductance is L1 + L2 - 2*m. So,
when we subtract, we get (L1 + L2 + 2*m)-(L1 + L2 - 2*m) = 4*m. Notice
that when the coils are closely coupled, m will be small, and using this
method will require us to subtract two nearly identical numbers (because
the L's are much bigger than m), causing severe numerical cancellation.
This will happen when k is some number like .998, very close to 1. For
coils not so tightly coupled, the method works ok.
> Does shorting one winding and measuring inductance on the other yield any
>useful information at all? Is any measurement useful with a shorted winding?
Yes. When you measure the inductance of one winding of a two winding
transformer with the other winding shorted, you are measuring the leakage
inductance of the primary plus the leakage inductance of the secondary
referred to the primary. One caveat is that the frequency used to measure
the inductance should be high enough that the measured reactance dominates
the value. In other words, the imaginary part of the impedance should be
much greater than the real part.
Consider a two coil transformer having a primary inductance of L1, with
resistance R1, and a secondary having inductance of L2, with resistance R2.
If we set R1 and R2 equal to zero and calculate the impedance seen at the
primary with the secondary shorted, we get (s is the complex frequency
variable, jw):
Zin = s*(L1 - m^2/L2) = s*[L1 * (1-k^2)]
If R1 and R2 are not zero, the impedance is more complicated:
s^2*(L1*L2 - m^2) + s*(R1*L2 + R2*L1) + R1*R2
Zin = ---------------------------------------------
s*L2 + R2
In each case, to get the equivalent inductance, divide by s. In the
first case, the calculated inductance is L1*(1-k^2). In the second case,
it's more complicated. But, the thing to note is that the presence of
non-zero resistances changes the apparent (leakage) inductance measured
with the secondary shorted. For the two coil inductor for which I gave all
the measurements in the earlier post, if we calculate (with the frequency f
set to 1000 Hz) the inductance using the complicated formula with R1 =
1.858 ohms and R2 = 2.256 ohms (these are the measured DC values), we get
84.96 uH. But, if R1 and R2 are set to zero, we get 79.46 uH, the same as
the simple formula.
Now set the frequency to 10,000 Hz and re-evaluate the formulas. With R1
and R2 set to zero, the result is the same, 79.46 uH. But with their
non-zero values, the calculated apparent inductance is 79.55 uH. If you
increase the frequency to 100,000 Hz, the value with non-zero resistance
gets even closer to the value with zero resistance.
I measured the inductance of L1 with L2 shorted on a GenRad Digibridge
using a test frequency of 1000 Hz and got 85.0 uH, mighty close to the
calculated value of 84.96 from the paragraph above. Theory triumphs
again!, as my former EE professor Don Reynolds used to say.
The lesson here is to measure the inductance of a transformer's primary
with the secondary shorted at a frequency near the operating frequency.
This is especially important for transformers with ferrite cores, because
ferrite can have a fairly low permeability, giving a winding reactance not
enough higher than the resistance to avoid error. Most low cost meters use
a 1000 Hz (120 Hz is also sometimes available; do not use this frequency
with a ferrite cored transformer!), and may not give good results. This
may be why you aren't getting good results with shorted winding
measurements.
(Try this: Get a 60 Hz (or 50 Hz) power transformer and connect a 25 ohm
rheostat across the secondary. Measure the primary inductance with a meter
using a test frequency of 1000 Hz. Start with the rheostat set to zero
ohms (thereby providing a short on the secondary), and mote the indicated
inductance. Now increase the resistance of the rheostat and watch the
indicated inductance change. How can just adding resistance to the
secondary cause the *inductance* to change so much? The answer is in the
complicated formula above. If you can, change the test frequency to a
higher one, and you will see that the indicated inductance doesn't change
as much as you increase the rheostat's resistance.)
It should be possible to manipulate the network algebra to get an
expression for the true leakage inductance, corrected for the error caused
by the non-zero winding resistances when the measurement is made with a low
test frequency. I'll look at this possibility later.
And, of course, as shown in step 4 of the AN1679 procedure, you can
compute k from a good short-circuit measurement.
> This app. note; http://www.onsemi.com/pub/Collateral/AN1679-D.PDF on page
>four they show a two winding transformer and ways to make a SPICE model.
>Let's say the coupling is very poor, K< 0.95, how would you correct his 8
>steps so the model is correct. Multiplying Np by K would certainly help.
> Thanks for your insight,
> Harry
>
In the App note, the author uses a symbol, Lps, apparently standing for
open circuit inductance, primary side. I will use Lss for the secondary
side open circuit inductance.
In step 1, you must change the computation for N. Instead of computing
N = Np/Ns, compute N (in my long post I called this value ~N, because it's
different from the turns ratio when k is substantially less than 1) as
N = SQRT(Lps/Lss). Use this value for N in later calculations. This is
the change that will give you good results when the coupling is loose. You
will have to *measure* Lss because you can't calculate it from measurements
made only on the primary side.
In step 8, he uses the DC values for Rp and Rs. I suspect that at the
high frequencies typically used for flyback operation, these values will be
too low.
Finally, he made a major boo-boo in Figure 8. The resistance Rp should
be in series with LI1, not Lm. There should be another resistance in
parallel with Lm to represent the core loss.
The Phantom
<To-Email-Use-Th...@My-Web-Site.com> wrote:
>On Sat, 02 Sep 2006 16:42:38 -0500, Don Foreman
><dfor...@NOSPAMgoldengate.net> wrote:
>
>>On Fri, 01 Sep 2006 20:58:42 -0700, The Phantom <pha...@aol.com>
>>wrote:
>>
>>>
>>> Can you give us a citation of a book (with page number) or other
>>>authority that explains how it is that kij # kji?
>>>
>>"Electric Machinery" Fitzgerald & Kingsley, 1961, McGraw Hill, page
>>26.
>
>Kingsley... what a horse's ass... just about the worst professor I
>ever had.
Did you ever take a class from Guillemin? In his book, "Introductory
Circuit Theory" on page 380-381, he discusses coupling coefficients, and
explicitly says L12=L21 in an example. Nowhere does he suggest that k is
different in different directions. I am unable to find any reference to
that effect.
>
> ...Jim Thompson
Jim Thompson
wrote:
>On Sat, 02 Sep 2006 15:23:06 -0700, Jim Thompson
><To-Email-Use-Th...@My-Web-Site.com> wrote:
>
>>On Sat, 02 Sep 2006 16:42:38 -0500, Don Foreman
>><dfor...@NOSPAMgoldengate.net> wrote:
>>
>>>On Fri, 01 Sep 2006 20:58:42 -0700, The Phantom <pha...@aol.com>
>>>wrote:
>>>
>>>>
>>>> Can you give us a citation of a book (with page number) or other
>>>>authority that explains how it is that kij # kji?
>>>>
>>>"Electric Machinery" Fitzgerald & Kingsley, 1961, McGraw Hill, page
>>>26.
>>
>>Kingsley... what a horse's ass... just about the worst professor I
>>ever had.
>
> Did you ever take a class from Guillemin? In his book, "Introductory
>Circuit Theory" on page 380-381, he discusses coupling coefficients, and
>explicitly says L12=L21 in an example. Nowhere does he suggest that k is
>different in different directions. I am unable to find any reference to
>that effect.
>>
>> ...Jim Thompson
Nope.
I was in course 6B... honors EE.
We had Harry B. Lee for network theory.
He was fabulous... taught from his own hand-outs... didn't use
Guillemin's book.
That's why I'm still so good at nodal and loop analysis after all
these years.
Homer J Simpson
> Did you ever take a class from Guillemin? In his book, "Introductory
> Circuit Theory" on page 380-381, he discusses coupling coefficients, and
> explicitly says L12=L21 in an example. Nowhere does he suggest that k is
> different in different directions. I am unable to find any reference to
> that effect.
If they were different, wouldn't you have invented a free energy machine?
Don Foreman
wrote:
>On Sat, 02 Sep 2006 15:23:06 -0700, Jim Thompson
><To-Email-Use-Th...@My-Web-Site.com> wrote:
>
>>On Sat, 02 Sep 2006 16:42:38 -0500, Don Foreman
>><dfor...@NOSPAMgoldengate.net> wrote:
>>
>>>On Fri, 01 Sep 2006 20:58:42 -0700, The Phantom <pha...@aol.com>
>>>wrote:
>>>
>>>>
>>>> Can you give us a citation of a book (with page number) or other
>>>>authority that explains how it is that kij # kji?
>>>>
>>>"Electric Machinery" Fitzgerald & Kingsley, 1961, McGraw Hill, page
>>>26.
>>
>>Kingsley... what a horse's ass... just about the worst professor I
>>ever had.
>
> Did you ever take a class from Guillemin? In his book, "Introductory
>Circuit Theory" on page 380-381, he discusses coupling coefficients, and
>explicitly says L12=L21 in an example. Nowhere does he suggest that k is
>different in different directions. I am unable to find any reference to
>that effect.
I don't dispute that. When working with L, it's correct to use k
which is the geometric mean of k ij and k ji, and when working with
circuits one uses L. Flux linkages are certainly not an abstraction
to a transformer, though they may well be to a circuit designer. When
solving the matrix equations for L's, the K ji's and K ij's get
cross-multiplied so once in the realm of L it doesn't matter if
they're alike or different since they've disappeared as entities.
I didn't find much help in textbooks or in the literature when I was
developing my theory some years ago. Seems like it must be somewhere,
but I never found it. I've never published it and don't plan to. I
can assure you that it works, y'all can decide whether or not you
care to believe that. There was a TDK appnote at the time (1995 or
so) that stated something like "rod core transformers cannot be
designed analytically, they must be developed empirically." Wrong! I
wish I still had a copy of that app note! Maybe using FEMM is
empirical, but the results from it wouldn't be of any use without the
theory.
Let's try a heuristic argument. By reciprocity, the permance of the
flux path linking two windings is the same regardless of which
winding is excited. But the permeance of the total flux paths for
the windings may be quite different, particularly in loosely coupled
situations if the windings are geometrically rather different
relative to one another and/or in how they're disposed on a core
that doesn't comprise a closed path. Therefore, the ratio of
flux linking a pair of windings to total flux for a given winding
(k ij) can be quite different. FEMM finds these flux linkages as
space integrals so flux partially linking a winding (not the whole
winding) gets "partial credit".
Let's look at a simple familiar model. This one is limited to 2
windings while the general case is n windings, but it may help
clarify. Consider the classic three-terminal model of a transformer:
------ Z1 - M -----+------ Z2-M-------
|
M
|
--------------------------------------------
L1 is primary inductance, L2 is secondary inductance, M is mutual
inductance. In conventional mesh notation, L11 = L1, L22 = L2,
L21 = L12 = M. Mr. Guillemin smiles. But the ratios of mutual to
self, call them k21 and k12, are M/L1 from the left and M/ L2
from the right. Even though L21 = L12 = M, the ratios are not
necessarily the same. From any text, M = k * sqrt (L1 * L2). By
simple algebra, k21 = k sqrt (L2/L1) and k12 = k sqrt (L1/L2). The
geometric mean, sqrt (K21*K12) = k. QED.
If one is working with an existing transformer, it's definitely
easiest to measure and think in terms of inductances. If one is
designing loosely-coupled transformers, it's easier to use FEMM to see
what flux linkages occur with a given geometry. It's vert easy to
get k ji and k ij from a FEMM simulation. One can examine variants
that way a lot faster than winding, vacuum-potting and testing actual
specimens -- and at 30KV in a small xfmr vacuum-potting is not
optional though somewhat messy. Once the flux linkage data is
converted to inductances in a math model, Lij = Lji in every case
as Guillemin and others assert.
Both approaches are correct, and both work. Transformers I've built
accurately to a given geometry usually exhibited circuit parameters
within 5% or so of what I expected from FEMM modelling and math
modelling. The math models were used to predict the effects of
distributed capacitances and hence self-resonances, and they did
account for winding resistances. Fringing and all that are non-issues
because FEMM analyzes and integrates fluxes as they are determined to
actually exist. The resulting values of L ij and k were then used in
SPICE to predict circuit behavior with associated electronics. Once
actual circuits and transformers were made, they performed at the
bench as expected.
The application here was very low cost 30KV 30-watt high-frequency
oil-ignition transformers operating at resonance with a capacitor in a
self-excited power oscillator. Loose coupling is useful here
because it enables the desired high output impedance without requiring
additional inductances.
The circuit designer could care less about k ij and k ji, since all he
can see and all he need care about is inductances. For
closely-coupled designs as used in SMPS, the difference between k ji
and k ij must be miniscule in any case since k is "very close to 1".
I now know, thanks to the experience and helpful post of Mr Woodgate,
that values of .998 are quite achievable so I guess I need not worry
about k or k ij or such claptrap for my little SMPS project. I'll
just wind the suckuh and snub as necessary. I apologize for the
diversion.
The Phantom
Funny you should mention that. While I was going through all my texts, I
noticed in Skilling's circuit theory book a section about "equality of mutual
inductances". In a footnote on that page he gives a brief outline of a proof
that if they aren't equal, you would be able to extract energy from an inductor.
>
>
>
The Phantom
<snip>
>
> Consider a two coil transformer having a primary inductance of L1, with
>resistance R1, and a secondary having inductance of L2, with resistance R2.
>If we set R1 and R2 equal to zero and calculate the impedance seen at the
>primary with the secondary shorted, we get (s is the complex frequency
>variable, jw):
>
>Zin = s*(L1 - m^2/L2) = s*[L1 * (1-k^2)]
>
> If R1 and R2 are not zero, the impedance is more complicated:
>
> s^2*(L1*L2 - m^2) + s*(R1*L2 + R2*L1) + R1*R2
>Zin = ---------------------------------------------
> s*L2 + R2
>
> In each case, to get the equivalent inductance, divide by s.
I should have said, "...to get the equivalent inductance, divide the imaginary
part of the impedance by s.
Homer J Simpson
>>If they were different, wouldn't you have invented a free energy machine?
>
> Funny you should mention that. While I was going through all my texts, I
> noticed in Skilling's circuit theory book a section about "equality of
> mutual
> inductances". In a footnote on that page he gives a brief outline of a
> proof
> that if they aren't equal, you would be able to extract energy from an
> inductor.
Asymmetry of things like this is usually a big red flag.
The Phantom
wrote:
>I'd like to know what realistic or reasonable values for coupling
>coefficients (k) might be in ferrite switchmode transformers. I'm
>interested in forward converter power range of 50 to several hundred
>watts.
>
>I'd be interested in typical values for suitably-sized toroids, pot
>cores, E-E cores and U-U cores. Bifilar and/or surrounded windings
>(e.g. parallel windings above and below another winding) may be used.
>
>Anybody have any experience with this?
Remembering that you did start this thread with a request for some typical
numbers, I found an old EC41 core set in the bin with some wire already on it,
and I took some measurements.
The first thing on the bobbin is a single layer, bifilar wound with a pretty
red and green pair of wires. The k between these two, calculated from
measurements at 1 KHz is .9991.
Then there is some Nomex, about 3 wraps, and another winding. The k between
one of the first layer windings, and the isolated winding on top of the Nomex is
.9962.
legg
wrote:
>On Fri, 01 Sep 2006 12:50:52 -0500, Don Foreman
I guess I've had my head up my butt on this issue, having spent too
much time in conventional power conversion. The only time k ever
showed signifigant effect was with pluggable, rfid or transcutaneous
coupling (and even this was mostly on paper).
At 3 to 5KW, I tended to take the heating of adjacent hardware as the
effect of leakage, rather than as the product of accidental coupling.
Even there, amplitude errors in the main coupler wasn't a measurable
or signifigant symptom.
RL
Don Foreman
Wow! When Woodgate said "close to 1", he wasn't kidding! Thank
you for that data, Phantom. I have some of those EC cores also --
somewhere. I think I'll hunt for them today.
Harry Dellamano
Thanks Phantom, great stuff.
Now let's say that the two windings are on two different cores, but
different sizes and material and in a fixed position. Does any of the above
change?
Regards,
Harry
Harry Dellamano
Phantom, one problem!
I measure my DUT for: Lpo, Lps, Lso, Lss, Rs, and Rp at the operating
freq (130KHz) on a precision LCR meter. Then do the indicated calculations
and make my SPICE model and run it. Simulations are right on. Then, being a
doubting Thomas, I interchange secondary and primary readings and do
indicated calculations, which are all different. Lm winds up on secondary
instead of primary so I do the obvious T^2 trick to transform it to other
side. Everything gets transformed into a perfect mirror image but the K
factor (0.111) has not changed!
Just kidding, this looks great. I will call it "Phantom's Reversible
Coupling Factor".
Highest Regards,
Harry
The Phantom
wrote:
<snip>
>
> Thanks Phantom, great stuff.
> Now let's say that the two windings are on two different cores, but
>different sizes and material and in a fixed position. Does any of the above
>change?
If the core material had really low permeability, like molypermalloy with a
perm of 30, or something like that, you might have watch out for that error that
occurs when you measure the leakage inductance with one winding shorted.
Other than that, it should all work. As a general rule, you can *always* use
the SQRT(L1/L2) formulation rather than N1/N2. You can even use it when the
coils are tightly coupled, because it's the fundamental relation for coupled
coils. The N1/N2 ratio is only an approximation, although a quite good
approximation when L1 and L2 are tightly coupled.
There is another technique which I'll describe later tonight which I think may
avoid the error caused by winding resistance..
Also, I want to do further analysis and see if it isn't possible to compensate
for the error caused by the winding resistance.
> Regards,
> Harry
>
The Phantom
wrote:
<snip>
>
> Phantom, one problem!
> I measure my DUT for: Lpo, Lps, Lso, Lss, Rs, and Rp at the operating
>freq (130KHz) on a precision LCR meter.
As a matter of curiosity, what make and model is your LCR meter?
>Then do the indicated calculations
>and make my SPICE model and run it. Simulations are right on. Then, being a
>doubting Thomas, I interchange secondary and primary readings and do
>indicated calculations, which are all different. Lm winds up on secondary
>instead of primary so I do the obvious T^2 trick to transform it to other
>side. Everything gets transformed into a perfect mirror image but the K
>factor (0.111) has not changed!
Are you telling me that you have a transformer with a coupling coefficient of
only .111? No wonder you couldn't get the model in the App note to work!
The Phantom
<snip>
>
> There is another technique which I'll describe later tonight which I think may
>avoid the error caused by winding resistance..
>
Hz power transformer by applying a voltage to the primary and then measuring the
open circuit voltage at the secondary, you don't get the reciprocal of the value
you get when you do the measurement in the other direction? It's close for a 60
Hz power transformer, but if you try it with a loosely coupled transformer, it
isn't even close. That's because the expression, V2 = V1 * N2 / N1 is only an
approximation. The correct expression is V2 = V1 * k * SQRT(L2 / L1); the
simpler expression is as good as it is because for a power transformer, k is
usually very nearly 1, and N2/N1 is very nearly SQRT(L2 / L1). You could think
of SQRT(L2 / L1) as a pseudo turns ratio.
Let ~N be SQRT(L1 / L2). The notation, V1' and V2' will mean that V1 or V2 is
the excitation voltage, respectively. Then V2 = V1' * k / ~N, or k / ~N =
V2/V1' = V12 (note that L1 was excited for this measurement). Also, V1 = V2' *
k * ~N, or k * ~N = V1/V2' = V21 (L2 was excited for this measurement). The two
quantities V1/V2' and V2/V1' are the result of open circuit measurements, and
*note well*, V2/V1' is *not* equal to 1/(V1/V2').
Let V12 be the result of applying excitation to L1 and dividing it into the
open circuit voltage at L2 (that is, V2/V1'). Let V21 be the result of applying
excitation to L2 and dividing it into the open circuit voltage at L1 (that is,
V1/V2').
Thus, we have V12 = k / ~N, and V21 = k * ~N.
Finally, V21 * V12 = (k * ~N) * (k / ~N) = k^2
And, V21 / V12 = (k * ~N) / (k / ~N) = ~N^2
So here is a way to determine k and ~N with only open ciruit measurements, and
the winding resistances shouldn't cause any error. We don't even need to be
able to measure inductance to get k and ~N.
Let's see how it works out with the loosely coupled inductors I have been
using as an example.
I applied .4639 VAC to L1, and measured .3893 VAC at L2; this gives V12 =
.8392. I also applied 2.19 VAC to L2 and measured .3859 VAC at L1; this gives
V21 = .1762.
Then V21 * V12 = k^2 = .14787, and SQRT(.14787) = .3845 = k.
And V21 / V12 = ~N^2 = .20997, and SQRT(.20997) = .458 = ~N.
Looking back and comparing to the results I got with the other method
(measuring short circuit inductance), this method did quite well.
The winding resistances shouldn't have much effect on the results, but the
flux distribution in the windings will be slightly different for different
measurement frequencies, giving rise to slightly different values for k and ~N.
Harry, it would be most instructive if you have time to make these
measurements on your transformer and compare results to those made with the
other method.
I proof read all this several times, but Murphy has probably inserted several
dumb mistakes.
Harry Dellamano
It's a Quad Tech Model 1920 and gives AC resistance and inductance at the
measuring frequency range of 20Hz to 1.0MHz. Nice feature when Rac/Rdc>>1.0
My windings are on different cores, separated by up to 1.00", hence 0.100"
< K< 0.60" even if measured from either direction.
Just to answer Don question, I have wound many transformers for power
electronics and K<0.995 is only acceptable if the large leakage inductance
is used to store energy and cause some sort of zero voltage switching. Most
Ks in hard switching units are greater than 0.997.
Cheers,
Harry
Harry Dellamano
Hey Ph. will do it tomorrow, Got to go get my G+T, mostly gin and cook
pizza on the Barbie.
Harry
wendell...@comcast.net
Harry Dellamano
Phantom,
By inspection the above will be true:
If: V2 = V1*K*SQRT(L2/L1)
And: K= SQRT(V21*V12)
And: N= SQRT(V21/V12)
And: N= SQRT(L2/L1)
Then: V2=V1*V21 which is obviously true and tell us to measure V21 and use
it in place of N because K, L2 and L1 are not known.
Here are my new measurements at 130KHz:
V12= 0.0678/9.720 = 6.97mV
V21= 4.920/3.870 = 1.27V and
K= SQRT(1.27*6.97m)= .094 compared to 0.111 using inductor measurement
method.
N= SQRT(1.27/6.97m)= 13.5 compared to 9.4 using inductor measurement
method.
Close but the inductor method yields all parasitic values and can be used
in a SPICE model. Also, Lpo, Lps, Lso, Lss, Rp and Rs can be easily measured
at the operating frequency by a transformer fab house on one setup, a LCR
meter.
What are your thoughts?
Harry
wendell...@comcast.net
Hi Don
I just found this site so in this case am late in replying but can help
if you want me to.
I have been designing switch mode power supplies since 1980. Am
currently employed and have held top engineering positions including
Senior Staff Scientist for Emerson Electric in 1980 thru 1989. In my
earlier years, designed pulse transformers, now designing switch mode
transformers, so am very experienced in calculating leakage inductance,
ac conductor losses both skin and proximity effect and effective
capacity of stepup HV transformers.
I can send you the appropriate formulas which are quite easy to use
that will allow you to calculate leakage inductance referred to any
winding (usually the driven primary winding). The only rules for the
equations to valid for are: (1) all windings must be of the same
traverse or length-of-winding. (2) the length of the windings must be
at leat 30 times the thickness of the insulation between the windings.
These conditions are what you want anyway to reduce the leakage figure.
If you wish I can phone you to discuss how its done and send the
documentation to you. If you are in the USA I can call you for nothing
since I use Vonage as my phone service. My email is
wendell...@comcast.net
Email me and I will call you back.
By the way the coupling coefficient for real practical switch mode
transformers is typically about 1% of the primary open circuit
inductance (OCL) or a figure of 0.99498
It has nothing to do with the OCL but is often close to the figure
above. It is in fact the square root of the value (1-LLP/LP) where LLP
is the leakage inductance referred to the primary winding.
Also a good choice for your forward converter, subject to the
specification details, would be a forward diagonal topology which
clamps the two switches to the DC line voltage and returns any stored
leakage energy to the DC source capacitor.
Best Regards
Wendell Boucher
Harry Dellamano
<wendell...@comcast.net> wrote in message
I think we all would be interested in your formulas, please publish to
this group.
Above you state that coupling coefficient (K) has nothing to do with OCL,
then you state: K=SQRT(1-LLP/LP). But is not OCL the same as LP? Please
explain.
Harry
The Phantom
wrote:
>
> Phantom,
> By inspection the above will be true:
>
>If: V2 = V1*K*SQRT(L2/L1)
>And: K= SQRT(V21*V12)
>And: N= SQRT(V21/V12)
>And: N= SQRT(L2/L1)
>Then: V2=V1*V21 which is obviously true and tell us to measure V21 and use
>it in place of N because K, L2 and L1 are not known.
> Here are my new measurements at 130KHz:
> V12= 0.0678/9.720 = 6.97mV
> V21= 4.920/3.870 = 1.27V and
> K= SQRT(1.27*6.97m)= .094 compared to 0.111 using inductor measurement
>method.
> N= SQRT(1.27/6.97m)= 13.5 compared to 9.4 using inductor measurement
>method.
> Close but the inductor method yields all parasitic values and can be used
>in a SPICE model. Also, Lpo, Lps, Lso, Lss, Rp and Rs can be easily measured
>at the operating frequency by a transformer fab house on one setup, a LCR
>meter.
>What are your thoughts?
> Harry
>
Harry, can you provide the following:
Inductance of L1 and L2 at both 1 KHz and 130 KHz
Resistance of L1 and L2 at both 1 KHz and 130 KHz
Inductance of L1 at 1 KHz and 130 KHz with L2 shorted.
Resistance of L1 at 1 KHz and 130 KHz with L2 shorted.
Inductance with L1 and L2 series aiding at both 1 KHz and 130 KHz.
Inductance with L1 and L2 series opposing at both 1 KHz and 130 KHz.
With all these numbers I should be able to give a full account of what's
going on.
I'm guessing you have most of them already, so a few additional
measurements should provide the rest.
The Phantom
wrote:
>
In my work above, I defined ~N to be SQRT(L1/L2), not SQRT(L2/L1) which
won't be consistent with the rest of the expressions.
>Then: V2=V1*V21 which is obviously true and tell us to measure V21 and use
>it in place of N because K, L2 and L1 are not known.
> Here are my new measurements at 130KHz:
> V12= 0.0678/9.720 = 6.97mV
> V21= 4.920/3.870 = 1.27V and
> K= SQRT(1.27*6.97m)= .094 compared to 0.111 using inductor measurement
>method.
> N= SQRT(1.27/6.97m)= 13.5 compared to 9.4 using inductor measurement
>method.
I'm surprised you didn't get better agreement. Maybe it's the presence
of a core. My little coupled inductor arrangement was air core. With a
winding shorted, the core effect is different than for open-circuit
measurements.
> Close but the inductor method yields all parasitic values and can be used
>in a SPICE model. Also, Lpo, Lps, Lso, Lss, Rp and Rs can be easily measured
>at the operating frequency by a transformer fab house on one setup, a LCR
>meter.
>What are your thoughts?
> Harry
>
This last method will be mainly useful with 60 Hz power transformers
where the series aiding and series opposing method to determine m (and
subsequently, k) may suffer from numerical cancellation. The method is
only for determining ~N and k anyway. If you need anything else, it's back
to the LCR meter.
Harry Dellamano
What is the meaning of "series aiding" when the two inductors are on
different cores?
Also, I just happen to know the author of the app note we are critiquing
(C. Basso) and sent him some of our correspondence. He is interested in your
work and sent me an appendix from his new book with his SPICE coupling
model. I tried forwarding it to you but your e-mail address you show here
does not work for me. My e-mail address is correct so if you e-mail me
directly, I will get you hooked up.
Cheers,
Harry
wendell...@comcast.net
wendell...@comcast.net
calculation itself has nothing to do with the magnetic core. However
the ration of the
leakage inductance figure to the OCL does determine the coupling
coefficient as can
be seen from the equation above. ( my earlier posting was a typo,
cleared up here)
The Phantom
>On Sat, 02 Sep 2006 19:20:29 GMT, "Harry Dellamano" <har...@tdsystems.org>
>wrote:
>
><SNIP>
>
>>Dear Phantom,
>> My questions below pertain to two windings, loosely coupled, K<0.95. In the
>>above you calculated mutual inductance by dividing the difference of aiding
>>and opposing inductance by 4. Four is not intuitive to me, why not two?
>
> I won't go into all the details; any circuit theory text should give
>them. What's going on is that when the two windings are connected series
>aiding, the measured inductance is L1 + L2 + 2*m, and when they're
>connected series opposing, the measured inductance is L1 + L2 - 2*m. So,
>when we subtract, we get (L1 + L2 + 2*m)-(L1 + L2 - 2*m) = 4*m.
I suffered a major brain infarction in what follows:
>Notice
>that when the coils are closely coupled, m will be small, and using this
>method will require us to subtract two nearly identical numbers (because
>the L's are much bigger than m), causing severe numerical cancellation.
>This will happen when k is some number like .998, very close to 1. For
>coils not so tightly coupled, the method works ok.
Replace with:
Notice that when the coils are loosely coupled, m will be small, and using
this method will require us to subtract two nearly identical numbers
(because the L's are much bigger than m), causing severe numerical
cancellation. This will happen when k is some number like .5, not very
close to 1. For coils that are tightly coupled, the method works ok.
>> Does shorting one winding and measuring inductance on the other yield any
Paul
This is a very nice piece of work Phantom, but:
The Phantom wrote:
> "V2 = (N2/N1) * k * V1" is not the fundamental relationship, but rather
> V2 = SQRT(L2/L1) * k * V1 as I explain above. The N2/N1 ratio is an
> approximation which is only acceptably accurate for closely coupled
> inductors such as found in transformers.
The difference between your 2 formulas is N2/N1 vs. SQRT(L2/L1)
Thus the validity condition should be:
The approximation is acceptable if the 2 coils have approximately the same inductance per
turn (A sub l), which is obviously only true if the 2 windings have approximately the same
geometry (same length, same cross-section, same number of turns, etc.)
(L1=N1^2 x Al)
The coupling factor doesn't enter into it. The approximation is acceptable for every
possible coupling value [0...1].
It is very possible that a lot of the confusion around transformers arises from attempts
to model unequal A sub l values of primary and secondary coils by tweaking k
Paul
The Phantom
>Hello all,
>
>This is a very nice piece of work Phantom, but:
>
>The Phantom wrote:
>> "V2 = (N2/N1) * k * V1" is not the fundamental relationship, but rather
>> V2 = SQRT(L2/L1) * k * V1 as I explain above. The N2/N1 ratio is an
>> approximation which is only acceptably accurate for closely coupled
>> inductors such as found in transformers.
>
>The difference between your 2 formulas is N2/N1 vs. SQRT(L2/L1)
>Thus the validity condition should be:
>The approximation is acceptable if the 2 coils have approximately the same inductance per
>turn (A sub l), which is obviously only true if the 2 windings have approximately the same
>geometry (same length, same cross-section, same number of turns, etc.)
>(L1=N1^2 x Al)
>
>The coupling factor doesn't enter into it. The approximation is acceptable for every
>possible coupling value [0...1].
Are you saying that since "The coupling factor doesn't enter into it.",
V2 = (N2/N1) * V1, and k shouldn't appear in this expression?
Paul
> On Wed, 06 Sep 2006 09:33:26 +0200, Paul <pa...@thuis.nl> wrote:
>
>> Hello all,
>>
>> This is a very nice piece of work Phantom, but:
>>
>> The Phantom wrote:
>>> "V2 = (N2/N1) * k * V1" is not the fundamental relationship, but rather
>>> V2 = SQRT(L2/L1) * k * V1 as I explain above. The N2/N1 ratio is an
>>> approximation which is only acceptably accurate for closely coupled
>>> inductors such as found in transformers.
>> The difference between your 2 formulas is N2/N1 vs. SQRT(L2/L1)
>> Thus the validity condition should be:
>> The approximation is acceptable if the 2 coils have approximately the same inductance per
>> turn (A sub l), which is obviously only true if the 2 windings have approximately the same
>> geometry (same length, same cross-section, same number of turns, etc.)
>> (L1=N1^2 x Al)
>>
>> The coupling factor doesn't enter into it. The approximation is acceptable for every
>> possible coupling value [0...1].
>
> Are you saying that since "The coupling factor doesn't enter into it.",
> V2 = (N2/N1) * V1, and k shouldn't appear in this expression?
>
No, I mean to say that the value of the coupling factor is of no importance in the
validity of "V2 = (N2/N1) * k * V1"
Paul
The Phantom
I see. You're saying no matter what k is, *if* N2/N1 is a good
approximation to SQRT(L2/L1) because of the particular geometry of a pair
of inductors (...if the 2 windings have approximately the same geometry
(same length, same cross-section, same number of turns, etc.)), then
V2 = (N2/N1) * k * V1 will give approximately the same result as
V2 = SQRT(L2/L1) * k * V1. I would have to agree.
Instead of saying:
"The N2/N1 ratio is an approximation which is only acceptably accurate for
closely coupled inductors such as found in transformers."
I should have said:
"The N2/N1 ratio is an approximation which is only acceptably accurate, in
all cases of inductor geometry, for closely coupled inductors such as found
in ordinary transformers (when they *are* closely coupled, as is the usual
case)."
This would allow for the possibility that N2/N1 might equal SQRT(L2/L1)
for some particular geometry even if k is low.
Consider it said.
I could have also dealt with this particular case by stating my main
point in a different way:
The formula V2 = SQRT(L2/L1) * k * V1 *always* gives a correct result, no
matter what the geometry of the inductors is, or what k is. The formula
V2 = (N2/N1) * k * V1 *doesn't always* give a correct result for all k, and
is especially likely to give an incorrect result when k is substantially
less than 1.
>
>Paul
legg
The Asub l is less sensitive to winding geometry if coupling is
performed by relatively high permeability material ie not by air.
I think you're both saying the same thing.
RL
Paul
'Likely' is probably right, since it is _impossible_ for 2 coils which are closely coupled
to have very different Al values, but when k is low, Al1 and Al2 _can_ differ substantially.
They are subject to the following equations:
k SQRT(Al1/Al2) <= 1
k SQRT(AL2/AL1) <= 1
k <= 1
So if k is approximately 1, Al1 is definately close to Al2
For low coupling k, for example 0.1, Al1/Al2 can be up to 100!
Paul
The Phantom
This is just another way of saying that when k is low, N2/N1 can differ
substantially from SQRT(L2/L1), but when k is nearly 1, N2/N1 scarcely differs
from SQRT(L2/L1). That is the main point I've been making in the larger thread.
Do you see some particular usefulness in restating it in terms of Al1, Al2?
Paul
It helped me understanding the difference between the two. I feel very comfortable when
geometric differences between the coils are reflected by a parameter.
I had some bad trouble simulating low-k transformers (a very effective way of looking at
electromotors) before I realised this.
I think my original point was lost.
You're right of course when you say: all you ever have to know about a transformer is k,
L1 and L2.
If you're happy with an approximation and you know k to be approximately 1, then you can
safely approximate L2 with L1 x(N2/N1)^2.
What I was commenting on was: "only acceptably accurate for closely coupled inductors".
I'm saying: even with low coupling, the approximation _can_ still be valid, just imagine 2
identical air-coils at considerable distance.
The approximation is also valid for the coils of a 3-phase motor. There, the coupling is
also significantly smaller than 1.
Paul
John Woodgate
Paul <pa...@thuis.nl> writes
>I'm saying: even with low coupling, the approximation _can_ still be
>valid, just imagine 2 identical air-coils at considerable distance.
>The approximation is also valid for the coils of a 3-phase motor.
>There, the coupling is also significantly smaller than 1.
Well, of course. I have two physically identical inductors, one with 50
turns and one with 60. One is here and the other 100 m away. k is very
small, but L1/L2 = (N1/N2)^2.
It would be very strange indeed if one inductor at a considerable
distance could affect the inductance of another. You could call it
'quantum entrapment' Oh, you have! (;-)
--
OOO - Own Opinions Only. Try www.jmwa.demon.co.uk and www.isce.org.uk
There are benefits from being irrational - just ask the square root of 2.
John Woodgate, J M Woodgate and Associates, Rayleigh, Essex UK
The Phantom
Your original point wasn't lost at all. I thought it wan't clearly put
in your first post, and I had to ask you for clarification, but when I
realized that you were talking about one small point in the continuum of
possible inductor geometries and values, I replied:
" I see. You're saying no matter what k is, *if* N2/N1 is a good
approximation to SQRT(L2/L1) because of the particular geometry of a pair
of inductors (...if the 2 windings have approximately the same geometry
(same length, same cross-section, same number of turns, etc.)), then
V2 = (N2/N1) * k * V1 will give approximately the same result as
V2 = SQRT(L2/L1) * k * V1. I would have to agree.
Instead of saying:
"The N2/N1 ratio is an approximation which is only acceptably accurate for
closely coupled inductors such as found in transformers."
I should have said:
"The N2/N1 ratio is an approximation which is only acceptably accurate, in
all cases of inductor geometry, for closely coupled inductors such as found
in ordinary transformers (when they *are* closely coupled, as is the usual
case)."
This would allow for the possibility that N2/N1 might equal SQRT(L2/L1)
for some particular geometry even if k is low.
Consider it said."
Your comment about 3-phase motors makes me realize that in that
situation, the geometries and number of turns in windings would tend to be
nearly identical, and the N1:N2:N3 ratios would be nearly the same as
L1:L2:l3. You should have brought this (the motor connection) up in your
original post and we would have seen the relevance of your point. The
thread was about transformers, and I couldn't see what you were getting at.
Since SQRT(L1/L2) always works in the case of transformers I couldn't see
why bother with N1/N2 since in most of the continuum of possibilities with
low k transformers, N1/N2 will be in error.
But I see that for motors the A sub l parameterization could be useful
since the windings will fall on just that point in the continuum where
N1:N2:N3 give the correct result.
Paul
> On Mon, 11 Sep 2006 10:56:38 +0200, Paul <pa...@thuis.nl> wrote:
> I should have said:
>
> "The N2/N1 ratio is an approximation which is only acceptably accurate, in
> all cases of inductor geometry, for closely coupled inductors such as found
> in ordinary transformers (when they *are* closely coupled, as is the usual
> case)."
>
> This would allow for the possibility that N2/N1 might equal SQRT(L2/L1)
> for some particular geometry even if k is low.
>
> Consider it said."
>
I would still like to point out for the interested reader (if anyone is left) that 'close
coupling' is only possible when primary and secondary geometry are the same, which is as
you point out the case in ordinary transformers.
Different coil geometries necessarily result in a lower coupling.
If anyone out there wants to build a closely coupled transformer, they should understand
that this necessarily implies equal geometry for primary and secondary.
Therefore my preferred formulation would have been:
"The N2/N1 ratio is an approximation which is acceptably accurate _only_ in
cases of _equal_ inductor geometry, such as necessarily the case for closely coupled
inductors such as found in ordinary transformers (when they *are* closely coupled, as is
the usual case)."
Another example I have of an unusual transformer is a high voltage (1MV) supply I built 10
years ago.
It consisted of a stack of 25 'decks', each generating 40kV. Each deck was supplied from
its own oscillating LC circuit, the coil being magnetically coupled with the neighbouring
decks, and the whole supplied from 1 coil at the bottom.
The thing could be seen as one 25-coils air-cored transformer, each of the coils coupling
with all the other coils, k decreasing as distance increased.
All the coils were the same (number of turns and geometry), so the L1/L2=N1/N2
approximation was valid although k was 0.7 maximum between direct neighbours.
Paul
The Phantom
>The Phantom wrote:
>> On Mon, 11 Sep 2006 10:56:38 +0200, Paul <pa...@thuis.nl> wrote:
>> I should have said:
>>
>> "The N2/N1 ratio is an approximation which is only acceptably accurate, in
>> all cases of inductor geometry, for closely coupled inductors such as found
>> in ordinary transformers (when they *are* closely coupled, as is the usual
>> case)."
>>
>> This would allow for the possibility that N2/N1 might equal SQRT(L2/L1)
>> for some particular geometry even if k is low.
>>
>> Consider it said."
>>
>
>I would still like to point out for the interested reader (if anyone is left) that 'close
>coupling' is only possible when primary and secondary geometry are the same, which is as
>you point out the case in ordinary transformers.
>Different coil geometries necessarily result in a lower coupling.
>If anyone out there wants to build a closely coupled transformer, they should understand
>that this necessarily implies equal geometry for primary and secondary.
Actually this is not absolutely true.
>
>Therefore my preferred formulation would have been:
>"The N2/N1 ratio is an approximation which is acceptably accurate _only_ in
>cases of _equal_ inductor geometry, such as necessarily the case for closely coupled
>inductors such as found in ordinary transformers (when they *are* closely coupled, as is
>the usual case)."
While reading this I reminded myself (as Legg reminded us) that if the
coils are wound on a high permeability core, the geometries can be quite
different and still have high k. Since I have been using an air core pair
of coils for the measurements I've quoted in the larger thread, air core
has been uppermost on my mind. That's my excuse, and I stand by it. :-)
I took an EC-41 core set and wound two bifilar 14 turn coils in a single
layer on the bobbin. I then cut a couple of cardboard discs and put them
on top of the single layer, with the two discs very close together. Next I
wound 28 turns in a thin disc between the cardboard formers. I then
inserted and clamped the two ferrite core halves.
The coupling coefficient between the two 14 turn bifilar coils measured
.9995. I then wired the two 14 turn bifilar coils in series aiding,
forming a single layer 28 turn winding. The coupling coefficient between
this single layer winding and the pancake, or disc, winding measured .9973.
It would be hard to get any more different geometries than that, but the
two windings are closely coupled nonetheless, due to the (relatively) high
permeability core.