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In my recent endevours in switching regulators I came to understand the need of a flyback diode in non-synchrounos regulators. From my experience with relays I know the flyback diode is placed in parallel with the coil.

What I don't understand is that in the regulators datasheets that I've looked so far, the flyback diode is placed in parallel with the inductor on the GND side, where there is a capacitor in parallel with the diode, instead of the diode being in parallel with the inductor 'directly', like shown below:

enter image description here

Doesn't the capacitor change the discharge behaviour? Why does this work at all considering that the two plates of the capacitors don't physically touch, and so can't allow current to flow and thus dissipate?

Thanks.

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  • \$\begingroup\$ In your second schematic, the energy stored in the inductor would not be transferred to the output capacitor. Have you tried to simulate it? \$\endgroup\$
    – winny
    Dec 19, 2021 at 9:50
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    \$\begingroup\$ In this converter, you want to exploit the flyback current, so the parallel-to-coil diode would ruin it \$\endgroup\$
    – tobalt
    Dec 19, 2021 at 10:40
  • \$\begingroup\$ You want to use an accumulated inductor energy, not waist it. However, both are an option. \$\endgroup\$
    – user208862
    Dec 19, 2021 at 18:05

4 Answers 4

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Diodes are used with inductors, but for different reasons in different circuits.

When suppressing a relay coil, the relay EMF generated when the current is switched off is unwanted, and all we want to do is channel it somewhere so that the energy won't do damage to the driving transistor.

When used in a buck converter, the coil EMF is wanted, and we want to route it to the output filter capacitor, where the energy can be usefully used to drive a load.

With the diode in your position, the energy is harmlessly dissipated in the diode and coil resistance. Safe, but useless as a power supply.

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  • \$\begingroup\$ Perfect! That's it thanks! \$\endgroup\$
    – StefanoN
    Dec 19, 2021 at 10:14
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When the main power switch \$SW\$ turns on, it magnetizes the inductor and current flows from the source to feed the load with the output capacitor. When the switch turns off, you need to offer a path to this inductive current through the so-called freewheel diode which, in conduction, brings the common point SW/K to \$-V_f\$. The below picture is excerpted from my APEC 2019 seminar on the buck converter:

enter image description here

With this scheme in mind, it is convenient to represent the buck converter as a low-impedance square-wave generator toggling between ground and \$V_{in}\$ then supplying a low-pass \$LC\$ filter. Should you instead connect the diode across the inductor as you thought, you would defeat this scheme by preventing the off-time inductive current to correctly flow in the output capacitor.

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The whole idea is to pass current from the inductor to the capacitor. With your scheme current just loops through the inductor and wouldn't get to the capacitor.

With the recommended circuit current flows in the loop L1, Cout, D1 resulting in Cout being charged and raising its voltage. The capacitor then supplies the load while L1 is recharged.

Why does this work at all considering that the two plates of the capacitors don't physically touch, and so can't allow current to flow and thus dissipate?

Capacitors don't pass DC but do "pass" AC by charging and discharging.

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The capacitor is not in parallel with the capacitor or inductor.

When the buck converter switch is off, the diode, capacitor and inductor are all in series, allowing the energy stored in the inductor to be transferred to the capacitor.

What you propose by putting the diode in parallel with the inductor means that when the buck converter switch is off, energy stored into inductor is only dissipated in the diode and only energy stored in the capacitor flows to the output.

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