Transformers (coils) wired in parallel

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.

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

It will be good if you clarify what type of transformer you are referring to , only in Auto transformer there is electrical connection between primary & secondary windings, in normal transformer the coils are not electrically connected, in the power system study transformer is modeled as such but I guess this is not the subject which you are looking for. In the Steady state condition of your circuit when your coils are connected in parallel to the AC source with fixed frequency following should be valid Leq=(L1*L2*L3)/(L2*L3+L1*L3+L1*L2) in your measurements you should distinguish between reactance Xl & inductance L, in regards to the mutual inductance, this is depend if there is any considerable magnetic coupling between them for example if they are using the same core or if you are transferring any power signal from one coil to the other , as an example if you connect all three in the same active source, you could ignore M.

Regards Mansour

You guys are trying to make this problem much more complex than I am asking! No, it's NOT an autotransformer. No, it's NOT a power transformer. And for now, I'm not assuming ANY ferromagnetic core, not iron, not powered iron, not ferrite, NUTTIN'! Consider it an RF transformer (untuned) as in a radio. Cores can be dealt with later once I figure out what is going on here.

Simple parallel inductance is no problem. It's that MUTUAL inductance that is the fly in the ointment!

Thanks for the reply,

Benj

---------------------------------- mjalali wrote:

Why do you make this so convoluted? Just google "inductors in parallel" Here's the first hit:

--------------------- 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.

------snip-------

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.

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

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

Really past bedtime!

Leq={L11*L22 -M^2}/{L11 +L22 -2M}

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 , 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

Thanks for getting my mind right, though!

Benj

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

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.

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

----------------------------- Yes, I did have it upside down and corrected it in a later post. You can get an expression for coupled inductors by finding the inverse of the L matrix and then noting that pi/v = the row sums of the inverse matrix and as the total pi = the sum of the individual currents, the sum of all the elements in the inverse of the inductance matrix is the inverse of the equivalent inductance. In the case of the two element case: L=| L11 M | | M L22| It's inverse is |L22 -M | |-M L11| divided by L11L22-M^2

Summing the terms gives (L11+L22 -2M) /(L11L22-M^2) =1/Leq or Leq=(L11L22-M^2)/(L11+L22 -2M)

This can be extended to higher order cases

Reasoning : For any coil j KVL gives: vj=v=Lj1*pi1 +Lj2*pi2 +..... for each j where p=d/dt leads to a family of simultaneous equations, one for each coil, which can be expressed as [V]=[L][pi] so we get [pi]= [P] [v] where P is the inverse of the L matrix [v] is a column [v v v v ...]=v*[1 1 1 1..] in particular for row k pi(k)/v =Pk1+Pk2 +Pk3 +..... and the total pi/v is the sum of all the pi(k) values which is the sum of all the elements of the inverse matrix. so Leq = 1/{sum of all the elements of the inverse of the inductance matrix} It is not necessary to solve for the currents

The only reason this can be done is that all voltages v1,v2, etc are the same.

The OSU expression deals with inductors which are NOT mutually coupled. Then the equivalent inductance "formula" is simply that for resistances in parallel- substituting L for R. It assumes that v=L1di1/dt v=L2di2/dt for independent windings. so that 1/Leq=1/l1 +1/L2 to extend this by adding v=M di?/dt implies a separate current through each of the M's which is NOT true so adding 1/M terms to the 1/Leq is not correct.

The actual equations are

v=L1di1/dt +Mdi2/dt v=Mdi1/dt +L2di2/dt

We don't know the voltage across the individual self and mutual inductances and cannot assume that the voltage across L1=voltage across L2 =voltage across each M and cannot then treat L1, L2 and the two M's as being in parallel. That is why the OSU expression is NOT correct for mutually coupled inductances.

(to verify my solution above, solve for di1/dtand di2/dt in terms of v and note that the total pi =pi1+pi2 to get an expression for total pi in terms of v- you are then done).

For the two coil case here is an diagram based on the common 3 terminal T equivalent circuit with terminals 1,2 tied together. ___1__________2________ | | L1 -M L2-M | | ----------------- | M ________|___________

That is, representing L1-M in parallel with L2-M and the combination in series with M as done in the T equivalent and tying terminals 1,2 together. This satisfies the correct KVL and KCL equations. The equivalent inductance obtained from this is the same as I calculated above. Note that my expression reduces to the OSI form if the mutuals are ignored. It will reduce to the form given by ejhsr(?) IF the inductances are equal (in the 2 coil case-haven't checked further).

Don Kelly snipped-for-privacy@shawcross.ca remove the X to answer

I think you should think of the product of current and the number of windings, "Ampere-Turns" if you like that terminology. You know have the inductance of one coil with the others open circuit. If all the other windings are identical then they have the same number of windings. All you are doing is increasing the amount of parallel conductors, but not the number of windings - thus it will have the SAME inductance as a single winding. In the case of a real set of coils it will have a lower resistance and thus a greater current capacity.

That's if I have understood what you mean by: "*I* am wiring the three coils in parallel!"

Robert

Benj wrote:

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 D>

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:

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

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

Yes, that is the idea and the general expected result in a case where M is larger and all coils are identical! However, we are struggling here with a case where the self-inductance (size) of each coil of different and the mutual coupling of each coil to the other two is also variably and not necessarily large. Of course as I pointed out in a recent post, once you go to identical coils that are very close coupled, one expects the inductance to be nearly the same as for a single winding! (I used the idea of using stranded wire) So what you say is right but still not the general solution I'm looking for!

Benj

------------------- snipped-for-privacy@reply.n> I think you should think of the product of current and the number

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.

OK I accept that it's no so simple: it's a matrix calcualtion, as follows:

Imagine you have three separate circuit segments, coupled by the transformer, the self and mutual inductances are: M11, M12, M13, M22, M23, M33 (M11=L1, etc.) the circuits have currents, with the derivatives (note the convenient Newtonian dot on the top ;-) ): i1, i2, i3 and voltage across the inductor: V1, V1, V3 so you get the matrix relation:

/V1\ /M11 M12 M13\ /i1\ |V2| = |M12 M22 M23| |i2| \V3/ \M13 M23 M33/ \i3/

or, more succinctly (just like the single circuit):

V = M . i

solve for the currents:

i = M^ . V

(where ^ denotes inverse)

Now we connect them all together, in parallel, so the voltages are identical and the the individual currents in the branches may be calculated. Of course they will all be in phase because the voltage is the same for each, so it is easy to sum them to get the overall current. Then, clearly, the overall inductance is the reciprocal of the sum of all the elements of the inverse of the inductance matrix.

Is that better?

Robert

snipped-for-privacy@reply.n> I think you should think of the product of current and the number

Which works out to: (L1.L2.L3 - L3.M12^2 - L2.M13^2 - L1.M23^2 + 2.M12.M13.M23)/ (L1.L2 + L1.L3 + L2.L3 -2(L1.M23 + L2.M13 + L3.M12) + 2(M12.M13 + M12.M23 + M13.M23) - M12^2 - M13^2 - M23^2)

... Simple really ;-)