I regularly see the use of Schottky diodes in order to protect microcontroller inputs from overvoltages (>5V). The Schottky diodes are connected to the 5V power supply (cathode) and the microcontroller input (anode).

I was wondering what happens when current flows through these diodes due to overvoltage? Where does this current flow into? It probably flows into the power supply unit...but what happens there? Isn't it so, in fact, that we are applying an overvoltage to the power supply output and that we expect the power supply to take care of it...?

Anyone with bad experiences with this solution? Are there any better solutions?

(In fact, the ESD protection diodes in the microcontroller itself apply the same trick for short spikes...the spike current flows into the power supply...)

  • \$\begingroup\$ In addition to clipping diode there is always extra resistor whose function is to dissipate energy and voltage of the overvoltage. \$\endgroup\$
    – Al Bundy
    Jun 10, 2015 at 8:21
  • \$\begingroup\$ @Jon'sanswer is good. Note that while the ESD protection diodes clamp controller pins (in most cases) at about 0.6V above or below supply rails, any current that they conduct has the potential to cause problems. This is covered by other SE EE answers. What happens is that the clamp diodes are between pin and the IC substrate and the point of injection and the path it then takes are not defined. Even extremely small currents can end up charging isolated nodes which are not designed to ever be charged and spurious FETs can be formed or existing ones can be controlled in unintended manners. ... \$\endgroup\$
    – Russell McMahon
    Jun 10, 2015 at 8:46
  • 1
    \$\begingroup\$ ... I have seen such charges cause IC misoperation and ICs take minutes to hours after power down for such charges to leak away and return normal operation. Any situation where body diodes conduct falls at best in the "absolute maximum" datasheet area and outside the guaranteed operating conditions area. \$\endgroup\$
    – Russell McMahon
    Jun 10, 2015 at 8:47
  • \$\begingroup\$ Pascalm, see my answer here, and see that the input series resistor limits current in the condition that there is a constant/"slow" over-voltage condition on that pin: electronics.stackexchange.com/questions/163464/… \$\endgroup\$
    – KyranF
    Jun 10, 2015 at 17:00

3 Answers 3


Your analysis is correct. The idea is that the current flows through the diode and into the power supply. What happens to it in the power supply depends on the design of that part of the circuit.

If the current is a transient pulse (from EMI or something like that) then the circuit's decoupling capacitance and the output capacitance of the power supply will easily absorb it. If it is a small constant current then it will be used by the chips in the circuit instead of drawing current from the power supply. If it is excessive, it may cause the power rail to rise and destroy the rest of the circuit as most power supplies are not bi-directional.

The key thing is to ensure that the last situation cannot happen, and this is one of the reasons you normally see a series resistor before the protection diodes. If the potential current is still too high it is better to use a disconnection FET or shunt clamp design instead.


The internal diodes are only there for ESD protection while handling the device (while it is not fitted on a PCB). They are not rated for human body discharge model 8kV/16kV. They only satisfy some JEDEC standard.

In your final product, if your microcontroller's pins are exposed, external connectors for example, you fit ESD protection diodes to protect against more severe discharge events. They can not protect for over-voltage! you will need some series resistors if that what you are trying to achieve. Also remember, regulators do not sink! So if you increase the supply rail from another source, your 5V WILL go up. 99.9% of the regulator designs do nothing about this situation. An ESD event is a short pulse and gets absorbed by the decoupling capacitors (shorted), not by the regulator.

I hope this helps.



If I understand correctly:

1) In most cases, the power supply will go up, due to current leaking through the schottky diode.

2) This means that, even with a series resistor at the input, power supply will go up. How fast it goes up depends on the power supply design and the series resistor used.

3) So, if it (slowly) goes up, then not only all circuitry powered by this supply is in danger, but also the input that I wanted to protect, as it stays only at about 0.6V below power supply at that moment (power supply minus schottky forward voltage)...

So, if inputs at the microcontroller input could rise @ x2 the allowed input voltage and very slowly, one would advise not to use this kind of solution at all?


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