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On a past project, I've used the LM2576 buck converter with success. However, I've never understood how the catch diode works (D1 on the diagram).

Typical application diagram

The datasheet explains the importance of the diode, and why it has to be a Schottky. TI LM2576 Datasheet

My doubt is around the workings of this diode on the example circuit. Being a DC output, when the IC is shut down, the inductor will release charge flowing in all directions. IC output will prevent sinking that current, so it must take the load way. Is the voltage drop of the diode what is important here, maintaining the orientation of current? Does this means that current will flow to the load, after shutdown, and return via GND multiple times, using that voltage drop to slowly drop to zero? Does this solution prevent damage to the load, or the voltage source?

Thanks in advance for your time.

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    \$\begingroup\$ Simply the diode serves as a return current path for the load when the chip is shut off. \$\endgroup\$ – vini_i Feb 25 at 17:47
  • \$\begingroup\$ Watch the first 4 mins of this video, that'll answer your question: youtube.com/watch?v=CEhBN5_fO5o \$\endgroup\$ – Big6 Feb 25 at 17:56
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    \$\begingroup\$ "The inductor will release charge flowing in all directions" - that is not how inductors work. \$\endgroup\$ – user253751 Feb 25 at 19:07
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enter image description here

Figure 1. Buck converter.

Inductors don't like you to change the current suddenly. When you try large voltages are induced.

  • In the on-state current flows through the inductor to the load and the filter capacitor.
  • In the off state there is no current supplied by the PSU. The inductor tries to keep the current flowing and since the right side of the inductor is held at the output voltage the left side of the inductor is driven below zero volts until the diode starts to conduct. This will maintain the current for a short period of time.
  • As the inductor current decays the feedback (in your circuit) will trigger another pulse from the converter and the next cycle begins.

A few points:

Being a DC output, when the IC is shut down,

The IC isn't shutting down. It's just turning off its output.

... the inductor will release charge flowing in all directions.

Not really. It will flow in the same direction as when the converter output was on.

IC output will prevent sinking that current, ...

Yes, the output is switched off so no current flows back into the converter.

... so it must take the load way.

Correct.

Is the voltage drop of the diode what is important here, ...

The voltage drop of the diode is important to keep the efficiency high.

... maintaining the orientation of current?

The orientation of the diode ensures that current flows in the correct direction.

Does this means that current will flow to the load, after shutdown, and return via GND multiple times, using that voltage drop to slowly drop to zero?

  • In the on-state the power supply provides energy to the load and to charge the inductor. Current will flow around the red loop.
  • In the off-state energy is released from the inductor to power the load. Current will flow around the red loop.

Does this solution prevent damage to the load, or the voltage source?

A successful design has to do both. The voltage is usually the critical parameter for most electrical loads. The maximum current may be a critical factor for the power supply feeding the converter.

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  • \$\begingroup\$ One more point: A Schottky diode also has a fast recovery time (delay from being positively biased, to positive current flow). There are fast recovery PN junction diodes, but they aren't as fast as Schottky's. And if you put a common 1N400X diode in the circuit, you will get a double hit on your efficiency, high voltage drop and long recovery time. \$\endgroup\$ – Mattman944 Feb 25 at 17:58
  • \$\begingroup\$ Thank you very much for your detailed answer. \$\endgroup\$ – Damián González Feb 25 at 19:59
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D1 is not best described as a 'catch' diode. That's a better description of the diode across a relay coil, which 'catches' the high voltage transient when its driver turns off.

In a buck converter, best to call it a 'freewheel' diode.

During the on phase of the IC, the output is taken up to the input voltage. Current flows from left to right in L1, increasing as L1 sees the excess of the input voltage over the load voltage across it.

When the IC switches off, the output pin goes high impedance. Current is still flowing from left to right through L1, but it can't be sourced from the output pin. For a very brief time, L1's current is sourced by the stray capacitance of the output pin and D1, and the voltage there falls rapidly. Eventually, the voltage drops to -0.4v, and D1 starts to conduct, sourcing L1's current. This allows the current to continue flowing (it freewheels) through L1 to the load, although the current is now falling as L1 sees the load voltage plus the D1 voltage drop across it.

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It is a switch mode regulator. It first turns on to allow current to flow into the coil, and when it turns off, current must keep flowing to output to release the energy to output, so the diode allows it. The less voltage drop over diode, the more efficient the buck converter is.

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  • \$\begingroup\$ even more efficient would be to not have a junction drop at all... \$\endgroup\$ – dandavis Feb 25 at 18:44
  • \$\begingroup\$ Yes and that is why in synchronous converters there is a FET in place of the diode. \$\endgroup\$ – Justme Feb 25 at 19:30

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