I currently read an application note from ON Semiconductor concerning forward converters. Latter explains why one needs core reset in case of the "single-switch" forward converter.

Consider the following slide, especially the graph of the current in the primary side, i.e. the magnetizing inductance.

enter image description here I know that a transformer can be modeled by an "ideal" transformer with infinite inductance (here X1) plus Lmag on the primary side and some leakage inductance (which was obviously omitted here because the explanations target the primary side effects only?)

I also know that the magnetizing inductance is a model for the energy needed to "allow" a magnetic field to change inside the transformer core (so I suppose it is somehow related to core losses.)

To my question: When Q1 turns off, where is the REAL WORLD path for the current? Since Q1 is open and ILmag cannot be "stopped" immediately, there must be a way for the current to circulate until the magnetizing energy is dissipated. All the slides I found draw this "circulating path" though Lmag and (for the given picture) the primary side of X1.

I don´t understand that, neither conceptually nor for real-world scenarios.

Conceptually, why would the current circulate through the primary of X1 and Lmag? (is there even a current through an infinite inductor? and if there is, should not the voltage across the inductor then be zero, so no voltage anywhere across the inductor and therefore a short?)

For real-world considerations: Since the primary of X1 is just a model for the ideal transformer and no real thing, where does the current circulate that flows in Lmag?


2 Answers 2


With the schematic posted as is, when Q1 turns off, in the ideal case, there is no current path available as you said. In the real world, a large positive voltage would appear at the drain of the Q1. Unless D1 breaks down first, the most likely current path is through Q1 under Vds breakdown. I think the circuit can actually work under this condition because a MOSFET can survive Vds breakdown if the power being dissipated is low enough.

By the way, infinite inductor means infinite impedance. Therefore an infinitesimal amount of current corresponds to a voltage approaching infinity. So the assumption such a (unloaded) transformer as being a short is not right.

In the real world, a common scheme of resetting the magnetizing current is by an extra reset winding on the transformer.

Going back to the diagram post, with the schematic as is, the current would not behave as illustrated. In order for the current to behave as illustrated, the transformer needs to be shorted by something when Q1 is off. One easy way to do something close to that is to put a diode (sometimes called a flyback diode in some applications) across the primary. Such a diode would not produce a working design in this application.

  • \$\begingroup\$ What do you mean by "would not produce a working design" in this application? I know that typically in forward converters, a path with a reset winding and a series diode is provided to reset the core - so the reset current flows through an extra winding, and that is done to feed the energy of the magnetizing current back to the input. But from an theoretical point of view and If I don´t mind the power dissipation: why would "shorting" the primary with a diode not work? \$\endgroup\$
    – Junius
    Nov 23, 2016 at 17:58
  • \$\begingroup\$ \$V_L = L \frac{dI_L}{dt}\$. With a reset winding, the flux energy is usually dumped back to the input voltage (Vin) at chosen turn ratio. With a diode across the primary, the discharge voltage would be the voltage drop across the diode. Therefore the current slope would be shallow, approximating the chart you posted. I suppose I should have said such a diode would not produce a working design with a reasonable duty cycle. \$\endgroup\$
    – rioraxe
    Nov 23, 2016 at 20:56
  • \$\begingroup\$ I understand your explanation and know the formula you posted, but I was wondering why voltage quantity is of concern here since i thought that if a parallel diode would allow current to flow, that would be sufficient for demagnetisation. I now think that is wrong. The current through the primary not just has to decay to zero, it actually has to reverse for demagnetisation, right? (I now think of remanence on the B-H curve). Is that true? \$\endgroup\$
    – Junius
    Nov 23, 2016 at 21:12
  • \$\begingroup\$ If you rely on a diode across the primary to discharge the flux, VL = Vdiode, therefore \$dI_L/dt\$ is small like Vdiode, so it approximates the chart as posted and takes a long long time to discharge back to zero. \$\endgroup\$
    – rioraxe
    Nov 23, 2016 at 21:22
  • \$\begingroup\$ By the way, the sign of VL is determined by the sign of \$\frac{dI_L}{dt}\$. The simple interpretation is that the sign changes depending on whether the magnetizing current is ramping up (positive slope) or ramping down (negative slope). \$\endgroup\$
    – rioraxe
    Nov 23, 2016 at 21:26

Even when Q1 turns off, the core is magnetized, i.e. the electron orbitals are aligned because it's ferromagnetic. That means a sheet current is part of the magnetized core material (probably ferrite). It is that sheet current in the magnetic material that is being modeled as a parasitic inductive element, Lmag. An ideal transformer doesn't magnetize, and for AC it's a different problem, but this is a pulsed DC circuit.

The magnetic polarization of the core stores energy just as an 'ideal' inductor does, though it has no wires that you could put through an ammeter. Without taking that energy OUT of the core somehow, it would build up until the population of electrons no longer includes any un-aligned orbits (and the inductor saturates).

That saturation would be a cause for unintended high currents (Q1 or the battery will blow up, if the printed wiring doesn't melt first). That's why a source of resetting current (or sometimes just a long wait) is required.

  • \$\begingroup\$ I like the explanation though it suggests that things get a bit hard to understand intuitively at this point ;-) You mentioned things to change for AC. What does actualy change and when should this be considered? \$\endgroup\$
    – Junius
    Nov 20, 2016 at 10:34
  • \$\begingroup\$ Another comment on your answer: you say, one cannot "directly" measure the current through Lmag with an ammeter. In fact, though, if i measure the current through the physically availabe transformer connections on the primary, it IS the magnetization current I see, right? And therefore still the question: where is the current loop? ;-) i feel that my initial question still not answered?! \$\endgroup\$
    – Junius
    Nov 20, 2016 at 11:26
  • \$\begingroup\$ A long string of AC cycles has the effect of resetting a transformer core average magnetization (but it has to be AC, using a transformer on a dimmer isn't pure AC and runs into troubles). For high power, a common switching scheme is to only switch at AC zero crossings (minimizes inrush current when the switch is part-on or bouncing), but for a transformer it is better to switch at an AC peak instead, in order to minimize the probability of saturation. \$\endgroup\$
    – Whit3rd
    Nov 20, 2016 at 19:42
  • \$\begingroup\$ The 'sheet current' in a magnetized object is in the coinciding orbitals of the valence electrons. Geometrically, it surrounds the magnetic flux lines (which are intended to fill the core), so is on the surface of the core. Think of a honeycomb, each cell having a clockwise current: at the honeycomb edge, there's a net current, but in each cell-cell boundary two currents in opposite direction (which cancel). \$\endgroup\$
    – Whit3rd
    Nov 20, 2016 at 19:45
  • \$\begingroup\$ I will need some time to think and read about your last explanation before I can accept your answer and therefore close the question.. \$\endgroup\$
    – Junius
    Nov 22, 2016 at 9:59

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