Transformers (coils) wired in parallel

Here's a question. Say I have a transformer and instead of wiring the primary and secondary in series (like all the textbooks do) I want to
wire them in parallel. What total inductance would I measure for the two windings both with adding and subtracting mutual coupling?
But the real problem I'm interested in would be three close spaced wire loops in parallel. Each loop has it's own inductance L and a given mutual inductance M to each of the other loops. What is the value of the total inductance measured at the terminals. I suspect it's one over the sum of all the 1/L and 1/M terms. [None of the Ls and Ms have the same values] But I haven't been able to prove it. Any hints?
Thanks.
Benj
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Excuse me- what texts wire transformer primaries and secondaries in series? If you are referring to "equivalent circuits" which often present primary and secondary leakage inductances in series as seen from one side-ignoring the relatively high mutual inductance- then go back to basic principles to see the basis for this model and why it is a simple and useful approximation to a better approximation, for many, in particular power, transformers. The problem of additive/subtractive mutuals is well covered in any of the texts that I have seen. I don't know what you have seen but it appears that a crucial step has been missed.
If you have a multiwinding situation -total inductance as seen from one set of terminals is not as you suspect. It does depend on what is connected to each winding. If winding 2 and 3 are open circuited then the inductance of winding 1 is that which would occur if windings 2,3 weren't even there. If, as you are considering, a 3 or more winding transformer, you are getting into a matrix model -are you ready for this at the present time? It is a bit messier and your intuitive guess (based on parallel resistances?) is not correct.
--

Don Kelly snipped-for-privacy@shawcross.ca
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Don Kelly wrote:

OK. I've just looked in a few old books I have here. What I'm talking about is they start with the basic equations: You know V = L di/dt + M di/dt etc. and then show that if you wire those two mutually coupled coils in series you can get an inductance that is the sum of each coil's L and either PLUS twice the mutual inductance or MINUS twice the mutual if the mutual coupling has the dots the other way. That's an easy problem and as you say, well-covered.

Let me clarify again. The problem I'm proposing is to take a multi- winding transformer (or more exactly mutually coupled coils as above each with inductance and a coupling to the others and this time hook them all in PARALLEL! No windings are open! All windings are hooked in parallel with all other windings. The whole thing ends up as a TWO terminal device with a certain inductance (Just as in the series case above). The question is what is the formula for that inductance? I've made a guess but I haven't been able to prove it. I keep thinking that like the series case the parallel case can't be that hard to do, but I still can't seem to find the way to do it! Is that more clear?
Here's a practical example: Say I've got this "IF can" type transformer. It has 3 windings. If I wire those three windings in parallel and hook that circuit to an inductance meter, what inductance would I read? It's obvious the self-inductance of each coil add together like typical parallel inductors 1/( 1/L +1/L +1/L) [I'm ignoring subscripts here for ASCII reasons], but the serious question is what role does the mutual inductances between the coils play in the final value? IF M = 0 it's obvious the above formula gives the answer. But what happens when the various Ms are not zero and you actually have coupling between the coils? OK?
If anybody has a reference to a book that has this worked out that would be great! So far, I haven't found one.
Thanks in advance!
Benj
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Benj wrote:

Why do you make this so convoluted? Just google "inductors in parallel" Here's the first hit: http://www.play-hookey.com/dc_theory/parallel_inductors.html
Ed
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------snip-------
----------------- Unfortunately your reference is incorrect, having made a fundamental error. It has assumed that the inductances can be represented by a parallel combination of L1+M and L2+M. This would be like writing KVL as v1=v=(L1+M)di1/dt v2=v =(L2+M)di2/dt where in fact: v1=v=L1di1/dt +Mdi2/dt v2=v=Mdi1/dt +L2di2/dt is correct.
This makes quite a difference.
--

Don Kelly snipped-for-privacy@shawcross.ca
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ehsjr wrote:

Well, DUH!
Man I HATE it when I've neglected to do the obvious thing first!!!!!
OK, NOW, I've done it!
But sorry, no cigar! Your reference has a nifty formula and I presume since it's on the Internet is has to be true....Only it's not!
They assume that the mutual Inductance simply adds on to the self- inductance as it does in a series case. In series inductors the current through one inductor has to be the same as the current through any other inductor. But in parallel that is no longer true. This means if you look at the equations, that you cannot group L + M as a factor.
I looked through all the crap in the search and while there is a lot of good stuff there that I should have examined first <sigh>, nevertheless a derivation of parallel inductors with mutual coupling from basic equations: I = I1+I2 and V=L1 dI1/dt + M dI2/dt = L2 dI2/dt +MdI1/dt. that ends up with L total = ? (expression with Ls and M) is not among the search that I could find.
Thanks for getting my mind right, though!
Benj
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--------------------- Ok, the way I would approach it is to write the equations for each coil v1=L11pi1 +L12pi2 +L13pi3.... v2= L21pi1 +L22pi2 +L23pi3.... as a family of simultaneous equations where L11, L22 etc are the self inductances and L12, L21 etc are the mutuals (The Lij values include the polarity sign ) and pi =did/t Since the coils are in parallel, v1 =v2 =v3 ... and you can solve for pi1, pi2, pi3 etc Easy to see the form in matrix notation. The sum of the currents is the total current so you know know pi total and v so assuming a single Lequivalent. v/(pi total) =Lequivalent
for two inductors this would give Leq ={L11 +L22 -2M}/{L11*L22 -M^2} where M=L12=L21
For three it will be a quite a bit messier. Usually in cases of multiple windings it is easiest to use the basic equations as above -in matrix form- as that would be quite general. All you are adding is the constraint that each coil has the same voltage. It usually is not worth the bother of working out the parallel equivalent for more than 2 or 3 coils unless you have numerical data and a program to crunch the matrix.
--

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Correction: 1/Leq =={L11 +L22 -2M}/{L11*L22 -M^2}
so Leq={L11*L22 -M^2}/=={L11 +L22 -2M}
--

My apologies-past bedtime.

Don Kelly snipped-for-privacy@shawcross.ca
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----------------------------

Really past bedtime!
Leq={L11*L22 -M^2}/{L11 +L22 -2M}
--

Don Kelly snipped-for-privacy@shawcross.ca
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Uh, oh! I see that none of my recent posts are appearing here. Gosh, ain't Google Groups great! Guess I'll have to try later and try to re- write what I said. <sigh> I HATE this work-eating feature of computers and half-baked software!
Just a moment.
Benj (testing)
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OK. My investigation to this point shows that Google is actually posting my efforts but is not showing it as posted. In other words you can't read anything I've said IN Google. However, if you have a "real" newsreader, the recent stuff is there.
The bad news is that I haven't got my "free" news service working properly yet and therefore cannot post there with a "spoof" email address. The bottom line is I'm sort of cut off right now.
But I will make a comment or two. Ahem. There is a copy of Terman right on my shelf here and it was one of the FIRST places I looked. Blanket statements about how "easy" it is to derive all this or that Terman "talks about" mutual coupling are total BS. If it's so damn easy then how about YOU just whip it out for us all, OK? We will be forever grateful! As you go on in life, you will one day discover that the most difficult questions to answer are the ones that seem the simplest to ask! I would hardly call the question of parallel coil connections "irrelevant"! I guess all electrical questions are irrelevant. You will recall that the FIRST thing I did was ask if anyone had a reference to a book that had this problem worked out. (That would be "worked out" not "talks about mutual inductance"!) Zero references from Salmon Egg or anyone else!
I have indeed derived the equations shown in Wikipedia (I believe the signs are wrong) but as I've indicated it's not the MATH that is the problem it is the reality of it! The problem has to do with an equivalent circuit for the situation. Any math can be defined valid but if the answers don't match reality it isn't of much electrical value. And the Wikipedia result does appear to agree at least in part with measured data. (The Wikipedia circuit assumes that the changing current in one leg results in a series voltage induced in the other leg)
By the way, the other comment someone made about a power transformer with two different windings is VERY interesting! Yes, that would be similar to what we are talking about although I'm not exactly sure how two different voltage windings would fight each other. I mean if you have a 10 volt winding and you hook it to a 15 volt winding that essentially makes the 10 volt winding have a 5 volt input, right? Which would be reflected back to the primary which then alters the over all output! I'm not sure it actually would smoke the thing. But it might. It's not wise to bet against smoke! If we hooked tow outputs in parallel OUT of phase, it's clear that this would cause the primary to appear with no inductance or as a piece of wire. This is a result apparent in the parallel inductor thing. And yeah, that might result in some heavy currents.
Since 'm sort between a rock and a hard place right now, I'll have to back off until my posts are no longer banned on Google or I fix my terranews setup. But I will keep an eye out for comments.
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On 4/25/07 10:11 AM, in article snipped-for-privacy@r30g2000prh.googlegroups.com, "Benj"

You are a lost soul an nobody appears to be inclined to rescue you. Life is tough.
Bill -- Fermez le Bush--about two years to go.
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-----------
You were told, by me and by "donotreply", how to do this. Have you bothered to go back to the basic equations? By the way, look up autotransformers. You seem to have some big holes in your education.
--

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Don Kelly wrote:

Yes, this is what I've been trying to do only I'm not seeing the form in matrix notation! In fact I wasn't trying matrix notation, but it's a good idea especially when going to more coils.

This is the sort of thing I was looking for! Having the answer should make my efforts to make the derivation a bit easier! I do believe, however that your formula for Lequiv. is upside down, right?

Thanks. This is a great hint. Only I wish there were some place it were all worked out for 3 or more coils (in matrix notation!), so I wouldn't have to deal with the math (which always drives me nuts!) But I do feel like I'm getting closer!
Benj
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Benj wrote:

I want to bounce this idea off you guys. I can't figure out how to get the above formula! I wonder what the equivalent circuit is? So consider this.
Take a peer at the OSU lesson below.
http://www.physics.ohio-state.edu/~gan/teaching/summer04/Lec2.pdf
It's a very simple and straight-forward derivation of inductors in parallel easily extended to N inductors in parallel. Now consider this. I'm proposing an equivalent circuit for two inductors with mutual coupling as two inductors in parallel AND where both are in parallel with an ideal transformer (with no self-inductance in the legs) of coupling M. If we look at the inductance of this device we find that the same derivation holds even though the voltage in each leg is due to the current in the OPPOSITE leg. But that doesn't matter to the derivation! What I've done is separate the self-inductance part out from the mutual inductance part in a parallel model! The bottom line is I end up with the formula for total parallel inductance that I had "guessed" at before!
In other words 1/L = 1/L1 +1/L2 + 1/M + 1/M
The idea is easily extended to many coils where it becomes a sum of 1/ L terms and a sum of all 1/M terms between all coils!
OK. Why is this wrong? Is it wrong? Why can't I make an equivalent circuit like this separating out the mutual inductance from the self- inductance? Inquiring minds want to know!
Thanks.
Benj
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----------------------------
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Oh man! I can't believe this! I mean people have been building and measuring and fooling around with coils since the 19th freakin' CENTURY and here in the 21 century it seems nobody has bothered to figure out the case of parallel inductors with mutual coupling and put it in a textbook somewhere!!!
So far we have three candidates. The Don Kelly formula below, the Wikipedia formula which Don and I agreed was wrong, and my "guess" as a sum of one over all Ls and Ms.
Don Kelly wrote:

So the question is which formula does God approve of? In other words which one fits reality?
Lets look at Don's in the case of two identical closely coupled coils. In that case L1=L2 and M~= L. We see Don's formula has two cases of M. One where the denominator is zero and the other where it equals 4L. However in both cases the numerator is equal to zero!!! This means that we have one case with zero inductance and another indeterminate case.
A zero case is always expected because when mutual inductance subtracts from the self-inductance we have a non-inductive coil with zero inductance. But the 0/0 is not able to make sense in this case.
Lets next look at "my" formula 1/L = 1/L1+1/L2+ 1/M+1/M Here if L1=L2 and M=L we find that L eq. = L/4
Finally lets look at the wikipedia formula:
1/L = 1/(L1+M) + 1/(L2+M) In this case, 1/L = 2/2L =1/L or L eq = L In other words the inductance of the parallel combination of two closely coupled coils should pretty much equal the inductance of one of the single coils before being hooked in parallel with a second one!
So what does God say, or as close as we can get to God which is Grover's book on inductance! We look at stranded wire and lo: We find that 10 meters of 2mm diameter wire has an inductance of 18.307 microHenries. And TWO parallel wires like that have an inductance of 16.454 microHenries! Not 1/4 not 1/2 but very nearly the SAME value!
Whoa! It seems the Wikipedia is actually the one giving the answer that agrees with reality! It seems that the correct equivalent circuit is where an ideal mutually coupled transformer is inserted in series with the bottom of the two inductors. This is as opposed to my circuit where I put the transformer coils across the input voltage and Kelly's circuit as a T network with an M inductance in the bottom leg.
Hmmmm.
Comments?
Benj
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On 4/24/07 6:15 PM, in article snipped-for-privacy@n35g2000prd.googlegroups.com, "Benj"

I can't believe this disbelief! I do not understand how someone can finish a as junior sophomore year, let alone a junior year, in an electrical engineering program without being able to solve such (relatively meaningless) problems. I m giving away my age by referring to "Elementary Electric-Circuit Theory" by Richard Frazier. If that is not good enough, Frederick Terman has described magnetic coupling in his books on radio engineering.
Believe it or not, a technical library can be an interesting place.
Bill -- Fermez le Bush--about two years to go.
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------- Maybe because it is a relatively trivial problem of little interest or use. In the case of parallel wires there is some interest but there is an easy way to deal with it. ----------

---------- Huh? Something missing in your background Sneak up on it. L1=M+e =L2 -plug into the expression and see what happens as e goes to 0. Result is Leq =M. Taking limits is a standard approach in such cases. If M is -ve then the equivalent L =0.
Note that using the T equivalent circuit also gives this without having to sneak up on the problem. What's the big deal? Result is Leq =M =L or Leq=0 Using the equivalent circuit also gives this without having to sneak up on the problem .
Your "formula" apparently ignores Kirchoff's Laws (without which circuit theory is meaningless) and is incorrect because of this.

------- Note that the incorrect Wikipedia approach is actually correct for two coils when L1=L2 for any M It doesn't give the correct answer when L1< >L2 except for M=0 and cannot be extended to more than 2 coils. -----------

--------- No kidding, Have you looked at the calculations of inductance for finite or infinite wires and the concept of GMR or approximations needed? >

--- Only because you have L1=L2. By the way, reality is not L1=L2=M.

---------- However the T circuit does give the expected result -so again what is the problem. In the case of your transformer - just what is being coupled in the two windings? Your description is questionable. The mutually coupled windings already provide a non-ideal transformer. You can use an ideal transformer but note that the ideal transformer has L1=L2 =M all approaching infinity so the transformer without any load will be an open circuit. Is this what you want? You need a load across one side of it- I suggest M. Hell, you can have a black box with a gremlin in it and a pair of hand crank generators and ammeters if you want -AS LONG AS THE BASIC CIRCUIT EQUATIONS ARE SATISFIED. I suggest that you look at the these equations, given in damn near any text, for coupled windings and make sure that they are satisfied and also that current in coil 1 is only i1 and current in coil 2 is only i2 as in real life- then go for an equivalent circuit 1 and 2 coupled by an ideal transformer, or a T or pi model- lots of ways to skin a cat but make sure you have the right cat to start with.
Generally they all have some form of
v1=L11di1/dt +L12di2/dt v2=L21di1/dt +L22di2/dt
where, (with exceptions in some rotating machines) L12 =L21 =M Look, in the library for some texts such as Krause "Analysis of Electric Machinery" or even Fitzgerald Higgenbotham, Grabel "Basic Electrical Engineering" (a once very popular book, updated many times-used for non-EE students. These are now old but newer circuits and energy conversion texts deal with this stuff. It would also be good for you to look at the magnetic situation involved -i.e. what leads to L11 L22 and L12. The flux relationships leading to these KVL equations are also given in a wide variety of texts.
If you believe, as the preponderance of authors do, that these equations are valid, then both the T circuit and the expression for equivalent inductance follow. If you disagree with these equations- then why do you disagree? I would like to know. If you have a problem with the approach I gave, then question it- I welcome questioning, but do your homework- it is a trivial problem.
--

Don Kelly snipped-for-privacy@shawcross.ca
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(snip)

(snip)
You'll have, effectively, a lot of shorted turns and thus no inductance. A power transformer has very tight coupling between windings. If you actually did what you said (BTW, dont!), the "two terminal" result would be close to a short circuit. Cheers,. Roger
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