Short circuit protection of a controller for flyback converter

Question

I would like to have your opinions about this short circuit protection. Here is the schematic associated given into the datasheet.

And you can find into the same datasheet how their short circuit protection works.

Here are my questions:

1 - " When the output of a flyback power supply is shorted, the primary VDD decreases due to the coupling polarity between the aux winding and the secondary winding of a transformer"

It means that the current flowing through the secondary windings which is high produces a flux (according to the Ampere's theorem) which can cross the inductance of the auxilliary winding and then have an influence on the auxilliary voltage? Theoritically, the mutual inductance between the secondary winding and the auxilliary inductance should be null? I.e. coupling factor = 0? So the datasheet assumes that the transformer is not well designed.

2:nd question: "However, it is possible that the VDD voltage remains higher than the UVLO level even if the output is shorted. This happens when the coupling between the aux and the primary winding is too good." Is it asking to reduce the mutual inductance between primary and auxilliary, ie to increase the leakage inductance on the primary side? What it really asking is to have a lower coupling between the primary and the aux and a better coupling between the output and the aux. Imagine I follow the instructions, as the current sink by the auxilliary winding is low, even if the coupling factor is poor, the flux which will not be tranfered to the auxilliary side and so will remain to the primary would not be able to produce a high leakage inductance? As Lleakage = remaining_flux/I_primary (I’m not sure of this last definition)

Answer 1

This is a classic in switching power supplies. The auxiliary \$V_{cc}\$ is an image of \$V_{out}\$ and is dependent upon the transformer turns ratio linking the auxiliary and power windings. That is to say, if you have a 12-V output and a turns ratio of 1:1 on the auxiliary winding, then the auxiliary will be 12 V also. This is a theoretical approach because the transformer hosts many parasitics such as leakage inductances, ohmic drops and capacitances properly modeled via a cantilever model (see this paper).

When you create an output short circuit, the peak current goes to a maximum value (clamped by the PWM controller but dependent upon \$V_{in}\$ and the propagation delay) and the energy stored in the leakage terms is maximum. A peak appears on the auxiliary winding which is peak-rectified by the diode and \$V_{cc}\$ increases even if the "clean" plateau is closer to \$V_{out}\$ which is theoretically 0 V in short circuit (you still reflect a bit of voltage made of the power diode drop and the drops in the PCB traces). The picture below is an excerpt from the book I wrote on power supplies:

Despite an output short circuit, the voltage cannot collapse to trigger the controller under voltage lockout (UVLO) and there is not much you can do beside trying to better couple the two windings. The peak lasts until the leakage inductance is reset. To appease its effects, you can add a small resistance in series with the diode (not really effective with a badly-coupled transformer), insert a small \$LC\$ filter to get rid of the peak (damp the filter properly) or better, resort to a more modern controller which monitors the feedback pin rather than the \$V_{cc}\$ for fault detection. If the FB pin goes to the max, it means that the loop is no longer closed (like in a short circuit where the LED bias disappears) and a timer starts. At the end of the timer (usually around 30-50 ms), all pulses are stopped and the IC either latches off or, more commonly, goes into an auto-recovery hiccup mode (you can hear the tic-tic noise).