# Reverse current into DC/DC or LDO

Suppose we have a generic scheme like this in figure, in order to protect the inputs of a microcontroller or logic gates from overvoltage essentially. For instance now we're applying 10 V at the input, when I expect 3.3 V; assuming that the Schottky forward voltage is 0.3 V we have about 14 mA that goes into the supply rail 3.3 V, where the supply rail is provided by a DC/DC (24->3.3 V). We know the circuit of a DC/DC (figure below), and now my question is, how does the DC/DC react to this reverse current? Isn't the DCDC made for source current and not sink? And instead a LDO? How would it behaves? The question is valid also for higher current.

• How will your 24->3.3 V DC/DC see negative output current? Nov 29, 2023 at 10:38
• What do you mean? I'm wondering about current that goes into the ouput of DCDC Nov 29, 2023 at 10:48
• You could use a schottky which goes to a supply net provided directly from the raw 24V through a high ohm current limiting resistor, then attach a separate zener there to keep the voltage at a low level. Why involve the regulator at all? Keep in mind that in case the schottky Vfwd is 0.3V then that zener should be 3.0V. Nov 29, 2023 at 10:48
• Do you mean that your 10 V external control signal would supply current into your 3.3 V rail? If yes, then you have a point. With zero to little load on the 3.3 V rail and regular diode rectification on your DC/DC output, it would try to push up the 3.3 V to 10 - 0.3 V. If you have more than (10-(3.3+0.3)/470 = 13.6 mA load on your 3.3 V, you are safe. Add a Zener clamp at say 3.6 V on your 3.3 V rail? Nov 29, 2023 at 10:53
• ok, i got it, thank you. The zener clamp is a interesting solution, is intended to keep the voltage from rising too high when i have a little load, right? But But forcing a voltage higher than 3.3 would not cause problems to DCDC in someway? Nov 29, 2023 at 11:19