Constraints on Coupling Constants of Multiple Mutual Inductors

L2 .990;\\n*K5 L1 L3 .995;\\n*K6 L2 L4 .995;

Will transfer the file to my other HD and OS to look at and mess with. Meanwhile, have you tried adding in leakage inductance, winding resistance, and capacitance? See Pg 275 Fig 8-14(b) in Electronic and Radio Enginnering by F. E. Terman (Fourth Edition) for a good equivalent circuit of a transformer.

I recently asked a similar question about 3 coupled inductors in the LTspice Yahoo discussion group and got a very helpful answer.

Here is a clip of the response I got in message:

Begin excerpt: Re: Three coupled inductors (limits on coupling factors)

For a system of coupled inductors, physical realizability requires that the coupling matrix be positive definite. For most structures, the matrix is symmetric (i.e. K12 = K21, etc). The following inequality must be met for a typical three-winding transformer:

(K12*K12)+(K13*K13)+(K23*K23)-(2*K12*K13*K23)

Found another reference to the same paper:

Any combination of 3 inductors have to satisfy this constraint.

It already is "the coupling matrix (must be) positive definite". So dig through your friendly local linear algebra text for the definition of "positive definite" and go to it.

I'll even do it for you -- if you pay me :)

I think you can develop a similar sort of required relationship between the k's of a 3-winding transformer by assuming that each winding can be represented by a perfectly coupled inductor, L1, in series with an uncoupled inductor, L1', plus a mutual inductance, Mnm, between each pair of windings.

eg, (L1 - L1')*(L2 - L2') = M12^2, and M12^2 = L1*L2*k12^2.

Write down the three equations, solve for L1', L2' and L3'.

L1' = L1*( 1 - k12*k13/k23 ).

L2' = L2*( 1 - k23*k12/k31 ).

L3' = L3*( 1 - k13*k23/k12 ).

Now require that every calculated L' must not result in a negative value. ie, All three terms inside the brackets must be => 0.

I'm not 100% certain how valid that is, nor how it compares with the relationship you have quoted from that paper, but it does seem to work.

Are you sure the circuit is supposed to work like this? I get horrible ringing primarily with 400V peaks, and across the D1 and D6 there are 360V reverse. There will also build up some DC-current, I get even missing switch phases. So on your bench the results are as predicted, when it blows up. :-( Your are putting stress outside the operating range, which is not caught by swcad. Also your transformer will saturate, when you don't monitor the current or balance it better, already the first pulse will unbalance the transformer by 50%.

With a minus sign missing ;-)

...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 | |

Yes. Good catch.

I have tried to extend this to 4 inductors. I came up with

K122 + K132 + K14^2 + K232 + K24^2 + K34^2 - 2*K12*K13*K14*K23*K24*K34

Messed up the exponents. Corrected below. Sorry about that.

K12^2 + K13^2 + K14^2 + K23^2 + K24^2 + K34^2 - 2*K12*K13*K14*K23*K24*K34

You guys are great -- thanks for all the replies.

Well the circuit does work in reality -- it even passed the UL torture test where they block all the ventilation holes and let it bake.

The real circuit has very little ringing on the primary but nasty ringing on the leading edge of the power pulse on the secondary. I went to 600V diodes but I realize that is a poor subsitute to understand what is going on, say by modifying the winding of the transformer or tweeking the snubber values. Hence my desire to properly simulate, hence my query about coupling constants.

Not shown on the circuit are the PWM controller which has cycle-by-cycle current limiting and the overall feedback which regulates the output voltage. In the simulation I pre-load the output inductor to try to get something which approaches steady state. I may have left a poor set of K's in the file I attached to my OP. Set all the K's back to .999 and it should look better.

I am a bit rusty doing algebra with inequalities, but the furthest I can get is dealing with every combination of 3 of the 4 inductors:

K12^2 + K13^2 + K23^2 -2*K12*K13*K23

A symmetric matrix with real elements is positive definite if its eigenvalues are all > 0, which is equivalent to saying that the determinant is > 0.

The criterion (K12*K12)+(K13*K13)+(K23*K23)-(2*K12*K13*K23)

should be < 1

There has to be at least one boo boo in a posting like this!

should be < 1

The other part i do not understand, is why the "dead time" between each FET on time and the other FET on time. Isn't it supposed to be a CT drive equivalent to a symmetrical, single primary drive?

Try pre-loading with 1.643; that seems to be very close to "perfect"; verify with longer simulation time (say 200uSec).