Overunity in transformer flyback mode

DC to DC Power Supply including an Energy Transferring Snubber Circuit - Bond US 4,899,270

Why is maximum flux attained in transformer when off load?

Why in transformer flux remains constant no load to full load?

Why transformer is called constant flux machine?

Flyback Combined with Induced Generator Coils into Cap and Load

Joel Lagace

Master Ivo

Most or all here (hopefully) understand how transformer works, primary magnetizes the core, if secondary is unloaded this magnetizing current is small and almost 90° out of phase with voltage, those few degrees offset from 90° are due to losses which are circa 1-5% of full load power. Of course when secondary is loaded it generates the counter flux due to lenz which demagnetizes the core and thus inductance of the primary suddenly drops and consequently inductive reactance across the primary which makes current in the primary to skyrocket 'trying' to get the flux back to previous value which it never fully manages so no load flux is maximum and full load flux is slightly lower. This skyrocketing of current in the primary due to lenz of the secondary is expressed as Ia=Ve-Eb/Zs Ia is of course current in the primary, Ve source voltage, Eb voltage across the inductance of the primary and Zs impedance aka complex resistance.

This was just a little recap on working of a transformer, conventional stuff. Now, with that out of the way, consider this...some food for thought.

Let's establish the facts we can all agree on, if someone disagrees, feel free to say it.

So called magnetizing current is tiny compared to load current.

Circa 1-5% of full load power is enough to establish max flux in the core.

Point is energy stored in the core is proportional to flux.

(Also power transferred through a transformer is flux times frequency but that's not the point now)

In other words, 1-5% of full load power input creates full potential energy in the core.

It is quite obvious form this that every transformer operated in flyback mode should be OU.

But practice shows it's usually not as simple as that. It is important to understand exactly why.

Let make it absolutely clear to remove any doubt that

1) Flux in the core is is max in no load state

2) Max flux at same frequency means max potential energy

BackEMF is often claimed to be source of OU.

Also, with high perm cores even smaller current (and power) is needed to generate large flux so those cores operated in flyback mode should be many many times OU and, of course, they can be tamed that way, but it's usually not as simple

This should be cleared up. If someone sees a flaw in my logic, i'd like to hear, but i don't see any.

There is absolutely no doubt that max flux in the core at same frequency means max potential energy.

And this max potential energy is generated with 1-5% of full load power.

Why then, if we magnetize the core and then turn off the source and use the collapsing field to run the load, should not that collapsing field energy be what it is, a full-load energy.

Nix

TRANSFORMER EXPLAINED

When transformer is unloaded (secondary open) only magnetizing current flows through the primary which is tiny compared to load current and is always (almost) 90° outta phase with the driving voltage, almost but not 90° due to various losses, namely, eddy currents, hysteresis, magnetostriction and copper losses (P = I²R). Average flux in the core is maximum in no-load state and slightly smaller in the full-load state.

When secondary is loaded, counterflux developed by the secondary demagnetizes the core and this makes the voltage across the primary to drop since this voltage is directly proportional to rate of change of flux, we all know Faraday's (or should i say Henry's) law V=--N*dF/dt.

Voltage across an inductor can also be expressed as V = L(di/dt)

And current through an inductor I = (V-E)/Z where V is voltage of the source driving the primary and E is voltage drop across primary's inductance. Clearly, when secondary demagnetizes the core and flux through the primary drops, so does it's inductive reactance Z and voltage across it E, V remaining the same means current must rise and so it does trying to bring the flux back to the original value but it never fully manages to do so, so, as said before, max load flux is slightly less than no load flux.

To the circuit driving the primary, it appears as if a resistor appears in parallel with the inductance of the primary, bigger the load smaller the resistor appears, obviously.

As said above magnetizing current is always almost 90° out of phase with the driving voltage while the load current is always in-phase. At least when the load is purely resistive.

If load is inductive then picture is not so clear, but extending what happens with the resistive load, we can assume this inductive load will also appear as an inductor and resistor in parallel and larger the work done larger the virtual resistor in parallel will appear again bringing IV in phase.

I guess we can extend the last paragraph to capacitive loads too, of course, electric field just like magnetic field can also be used to do work and as you all probably know there are various electrostatic motors, some newer ones of significant power (some even speculate about replacing magnetic ones).

As for flyback, the only difference is, as said before in the thread, induction in the secondary happens with a delay (due to internal diodes blocking the current in one direction) when the primary flux collapses. Ignition coil uses the same principle, store, collapse, get 10x (or more) voltage in the primary and x turns ratio in the secondary.

Nix

Hi Simon,

Do you find anything wrong with Nix's reasoning?

Best Wishes, Hermes

Hi Hermes,

Basically, the back emf isn't a source for OU. The system puts energy into the magnetic field, then takes it out again (with a few losses from resistance and core losses). From a system point of view, it's symmetrical. Your energy is either in current times voltage or in magnetic field. Change the number of turns around that magnetic field, and you change the voltage and current, but not their product.

Still, the main proof is practical. Test it, and you'll find no actual extra energy coming out. If there was, this would have been found a long time ago and we'd be using it for at least a century by now. Of course, with sparks and other high-frequency spikes, your meter may mis-read it and you might think there's more energy coming out, but if you're really wanting to be sure you'd put the output through a resistor and heat some water with it.

If you put a voltage across a coil, then the current increases at a certain rate. The longer you leave the voltage on there, the more the current gets and the larger the magnetic field is. With a perfect voltage source and no resistance in the coil, there's no limit to the current it will take. A 1-henry coil and a 1V source will take an extra amp every second. With a real coil and source, you've got resistances so it will asymptote to a limiting current and probably release the magic smoke that makes it work. If you've got a core in the coil, then that will have a limiting magnetisation, above which the inductance of the coil goes down and the current rises faster. Still, apart from the losses, the energy you put in goes into the magnetic field anyway. You don't get more energy out of that magnetic field than you put into it.

It's maybe a bummer, but coils and magnets working at fairly low frequencies just isn't going to deliver OU. There's no breaking of symmetry, and to get OU you have to break a symmetry.

One other thing that might be interesting is that the energy isn't stored in the core, but in the gap. Doesn't take much energy to magnetise some ferrite, but the gap does take a lot. Thus to store a lot of energy, you increase the gap size. This also makes the core less permeable. Still, you may have wondered why flyback transformers always have a gapped core. Now you know.

There is however a possibility of OU if you can re-route magnetic flux using less energy than you can get out of the local flux-change. That's a difficult symmetry to break, but may be possible using the Meissner effect. Might also be possible using the effect of mechanical stress on permeability, but I need to test that experimentally. Didn't appear to work using a quick and dirty test, but maybe a more-complex double path could work. I don't hold out a lot of hope for the mechanical modulation of permeability, but probably worthwhile testing even if only to knock it on the head.

To find a way of getting OU,think of some plan or technique, then reduce it to the underlying principle. Then look at what people have done before that uses that same principle. Most of the time, you'll find that a lot of people have thought of the same principle, maybe used a different setup to test it, and it wasn't OU. Unless you're doing something significantly different (and can point to that difference and why it is really different), and you've found what they missed, yours won't work, either.

That's the value of the lists of OU inventions - you know that they don't work because if they had they'd be in use by now, so there's no point in replicating them. They tell what not to do. So many other people will have tried them, with variations, to be certain they don't actually work. It might seem harsh, but you have to find something that someone else hasn't tried in order to have any chance of succeeding.

Best regards, Simon

Hi Hermes,

Yeah, that’s the guy. I did not watch all his videos but I noticed some of the guys here also gave some comments on his videos. If people don’t have so many views, usually they respond and do so with sensible replies. Thanks for the page for the backspike material!

My gut feeling is we need some batteries, some electronics for pulsing and of course some measuringequipment. But all that is not enough; you also need communicate with each other. Thats what makes 1 and 1 add up to more like 3.

Kr Hans

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