Reverse current into DC/DC or LDO

Question

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.

Answer 1

A generic regulator only sources current, and will source curtent to keep the output voltage at the setpoint. They do not, generally, sink current to bring voltage down to the setpoint.

So if your device requires 100mA, but you feed in 10mA through protection network to 3.3V node, obviously the consumption is still larger and regulator needs to provide only 90mA to fill the 100mA consumption.

But if your circuit only consumes 5mA, and you feed in 10mA through protection network, there is not enough consumption to consume the 10mA. This will raise the voltage of the supply voltage node until there is a balance.

This means the supply voltage may rise above 3.3V and current consumption in loads may rise too along with the voltage, like for example current in a power-on LED. If the voltage can rise too much then something breaks due to overvoltage, overcurrent, or too much heat due to increased power dissipation.