A designer in our team has a habit of putting freewheeling diodes on the output of every voltage regulator. Buck converters, LDOs, you name it. The loads in our applications are mostly logic gates, FPGAs, DDR3 RAMs and similar. No motors or relays (inductive loads). The reason for this practice he says, is that if there is a sudden negative transient during power off, the freewheeling diode will catch this.

But. I have never seen this in any other application except for systems with inductive loads. Is he just wasting board space?


simulate this circuit – Schematic created using CircuitLab

  • 1
    \$\begingroup\$ circuit diagram? \$\endgroup\$
    – jippie
    Oct 23, 2015 at 17:59
  • \$\begingroup\$ I tend to do this in any battery application with large capacitance, due to the reason that the battery can be disconnected quickly and often the circuit will not have a large cap in parallel with the battery. \$\endgroup\$
    – MadHatter
    Oct 23, 2015 at 18:44
  • \$\begingroup\$ Sorry! Missed to add it. \$\endgroup\$
    – Dejvid_no1
    Oct 23, 2015 at 19:52
  • \$\begingroup\$ Base on your schematic and description, he is wasting board space. The question to him is where does the potential negative transient come from. If a potential source can be identified, take care of that accordingly. For example, a potential source can come in through an external interface, if that is the case, it usually has to be taken care of at the interface (not at the power rail after transient has burnt out the interface pins already). \$\endgroup\$
    – rioraxe
    Oct 24, 2015 at 7:21

1 Answer 1


Assuming you are talking about a topology as can be seen below:


simulate this circuit – Schematic created using CircuitLab

Then yes, there are valid concerns to at least consider doing this with (some) linear regulators (in some situations).

In the event that the input is forced low & especially if there is larger output capacitance than the local input capacitance, you end up in a case where the output voltage is higher than the input with a relatively high source of charge.

This is a valid concern in the automotive industry and you find regulators with such a diode on-die.


When using large capacitors at the output of these regulators, a protection diode connected input to output may be required if the input is shorted to ground. Without the protection diode, an input short will cause the input to rapidly approach ground potential, while the output remains near the initial VOUT because of the stored charge in the large output capacitor.

The capacitor will then discharge through a large internal input to output diode and parasitic transistors. If the energy released by the capacitor is large enough, this diode, low current metal and the regulator will be destroyed. The fast diode in Figure 21 will shunt most of the capacitors discharge current around the regulator. Generally no protection diode is required for values of output capacitance ≤ 10 μF.


  • \$\begingroup\$ Very interesting! However this was not how the diode was connected in our systems. I've added the missing schematic, sorry. \$\endgroup\$
    – Dejvid_no1
    Oct 23, 2015 at 19:53

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