I've been looking at input protection circuits and I came across to some circuits that look like this one. Which is just based on the series resistor R1 to limit the current that flows through the MCU pin and also uses the Diodes D1 and D2 (D1 in this case) to release the excessive voltage to the power rail V1.

The question is about permanent voltage levels at 24V not transient effects or more higher than MCU pin levels.

Wouldn't that hurt the power rail (V1)? Even if diodes were omitted it would somehow hurt the power rail through MCU internal diodes. Wouldn't that make LDOs or DC to DC power supplies stop supplying power to the system due to the excessive voltage level and cause damage to the circuit?

I would lower the input voltage at a normal level and wouldn't use such circuit as a voltage translation.

Many other designs just suggest the same, a series resistor with a pair of diodes to both power rails. Some others just add a pull-up to MCU positive rail. And my understanding is that circuits like that are not meant to be used for voltage translation but only for ESD protection or transient responses. Am I missing something regarding to similar designs that could be found online?


simulate this circuit – Schematic created using CircuitLab

Image 1

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    \$\begingroup\$ This is how I do it: electronics.stackexchange.com/questions/334188/… \$\endgroup\$ – Ron Beyer Feb 5 '20 at 23:47
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    \$\begingroup\$ We would need to know a lot more about the 24 V input source, among other things (such as the processing associated with it.) I don't believe there is a single, fits-all answer. \$\endgroup\$ – jonk Feb 6 '20 at 0:22
  • \$\begingroup\$ @jonk this is a general question not a specific issue \$\endgroup\$ – MrBit Feb 6 '20 at 7:48
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    \$\begingroup\$ @MrBit Gathered. As I said, I don't believe there is a single, fits-all answer. I still believe that. But I'm also happy that you have an answer you like. \$\endgroup\$ – jonk Feb 6 '20 at 7:50
  • \$\begingroup\$ @jonk I do believe that too, requirements are very important \$\endgroup\$ – MrBit Feb 6 '20 at 9:21

As for using a circuit like that, it's mostly about the power dissipation. If the series resistor and the diode can handle the power, then it'll be fine.

Usually, from my experience a more common method would be to use two resistors in a voltage divider, to go from 24v down to 3.3v. Again, you need to make sure that the resistors are sized to handle the power.

They each have their tradeoffs. The voltage divider method is sensitive to current draw. The diode clipper circuit, running full time might require a good sized diode, depending on how much current the signal needs.

As long as your power supply has adequate capacitance, and there is enough series resistance to keep the current flow low, then using the diode clipper will not harm the power supply. At 24V and 10k you'll have about 2mA of current flow. Does the rest of the circuit use more than 2mA? If so, it's not an issue. If it is an issue, use a 100k.

To address this concern:

...due to the excessive voltage level and cause damage to the circuit?

Your schematic is missing capacitors. Think about if there were 10uF in parallel with the 3.3v source. What happens when you apply 24V to a cap? At \$t_0\$ nothing happens because you can't change the voltage on a capacitor instantaneously. You have to add or remove charge from a capacitor to change its voltage. In your example circuit you have about 2mA flowing through the diode to the supply, so if your circuit pulls more than 2mA, then the 3.3v supply will provide the rest and the voltage will stay at 3.3v.

  • \$\begingroup\$ Thanks for the answer, my schematic is just a general concept that's trying to capture the main idea behind that kind of protection. But I think you answered the question. So if no more than 24V-3.3V / R is drawn then it's okay, otherwise there's a problem. \$\endgroup\$ – MrBit Feb 6 '20 at 7:46

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