Lenzless resonant transformer

[verpies]

I wrote this before but I will repeat it again:
Mutual magnetic coupling between two transformer windings, "consumes" the free inductance of the secondary winding, that otherwise would be available to form an LC Tank with a capacitor connected to this secondary winding.

In an extreme case (when the the mutual coupling coefficient = 1) the entire inductance of the secondary winding becomes "consumed" (converted to an EMF source) and there is no free inductance remaining in it to form an LC Tank with a capacitor connected to this winding.
The frequency response of such secondary circuit is the same as of a lone capacitor (no resonant peaks) - Itsu has witnessed this response when his windings were wound to maximize the mutual coupling coefficient.

[MileHigh]

Itsu:

Thanks and I am not surprised that you also tried crossing the wires.  I realize you can't put everything in your clips.  So you appear to get a much lower resonant frequency but the amplitude is very low.  Something is going on, perhaps your investigations will figure it it.  Of course it's hard to visualize all the nuances associated with these things in your head.  I can only suggest that there is still something funny going on because the the L1 and L2 are seeing clockwise and counter-clockwise flux at the same instant from L3, while at the same time they are magnetically connected by the toroid.  So if L1 is driving a load, the counter-flux induced in L1 also flows into L2, etc, etc.

Verpies:

Mutual magnetic coupling between two transformer windings, "consumes" the free inductance of the secondary winding, that otherwise would be available to form an LC Tank with a capacitor connected to this secondary winding.

Indeed, that means my posting #165 is bogus.  To be honest I had some nagging doubts about it.  I realize that when the capacitor is discharging energy and sending it back to the source, it sees a real load.  The toroid couples the AC power back to the source.  So I suppose the instantaneous impedance the capacitor sees is the output impedance of the signal generator (typically a 50-ohm resistor) plus the impedance caused by the instantaneous output voltage of the signal generator signal.  And of course there is a transformer turns ratio to factor in also.

So in this case the L1 or L2 secondary is not an inductance, it's just part of the "gear system" that either transmits AC power forward into a target load, or it transmits AC power "backwards" into the signal generator source, which also looks like a load to the capacitor.  I use the term "gear system" because a great analogy for a transformer is a set of meshed gears.

In my opinion, at this point what's really happening is Itsu is exploring the phasor relationships in the power flow as he tries different configurations.  There is no pot of gold at the end of the rainbow, but it's still interesting.  There are a couple of ways to do the math behind the way the phasors act and react.  I did it in anther lifetime!  lol

MileHigh

[MileHigh]

Itsu:

I watched clip 16 and I figured it out, but I am too tired to write it up.  Still lots of fighting going on.  If it ain't flux fighting, then it's EMF fighting.  There is your big clue.

MileHigh

[itsu]

It should be so only if you are connecting two separate inductors in parallel (on two separate cores). 
Note, that winding direction is not an issue for completely separate inductors.

However, if there is any flux sharing between two inductors and/or any mutual inductance, then their relative winding direction becomes an issue and formulas for Mutually Coupled Inductors in Parallel should be used instead.
Of course, because L3 affects only the leakage flux that closes outside of the core through the air (bar core saturation).

Ok, great link, it explains exactly what is happening in this situation:

parallel aiding    LT = L1 = L2 = M  meaning 713mH whether they are paralleled or not, L stays the same

parallel opposing  LT = ( L ± M ) ÷ 2 meaning zero with perfect components, and in my real world 260uH (this was how my tests were done yesterday)
(This time, if the two inductances are equal in value and the magnetic coupling is perfect between them,
the equivalent inductance and also the self-induced emf across the inductors will be zero as the two
inductors cancel each other out. )

"Correct" to maximize what parameter?

good question, i leave that to Jack, but my feeling is telling me that we should go for the parallel aiding = 713mH setup,
i mean why should we want to cancel out anything?

Thanks,  regards Itsu

[itsu]

Itsu:

I watched clip 16 and I figured it out, but I am too tired to write it up.  Still lots of fighting going on.  If it ain't flux fighting, then it's EMF fighting.  There is your big clue.

MileHigh

Hmmm,   in "parallel opposing" yes, both flux fighting and EMF fighting going on as mentioned earlier, but when in "parallel aiding", all should be fine and dandy, right?

Regards Itsu