microcontrollerprotection
I'm revisiting a design I made.
The circuit is powered by 24V with polarity protection.
I placed a power loss detection before the polarity protection:
In case the power pins were replaced, the 3v3 zener diode will have forward voltage of ~1V.
will this pin go through R32 to the "unpowered microcontroller" and destroy its IO pin?
The absolute maximum is -0.3V in the mcu:
It's only 100uA limited by R32, but I can't be sure.
Is there such detailed information about the IO pin?
The easiest way to deal with a digital signal like that is to feed it to a BJT or MOSFET with a pullup from the processor voltage. For example:
simulate this circuit – Schematic created using CircuitLab
The diode allows the input to go negative as well without damage to the processor or transistor. Brief spikes of +/-100V won't hurt anything, regardless of the processor power state. Cost is pennies, at most.
For over-voltage protection of signal lines a zener diode is generally a poor choice. You haven't given enough details of your application for me to give many specific reasons, but you should consider the following:
A BZX84C 3.3V zener has up to 450pF capacitance. You have a 1MHz signal, which will probably be attenuated due to a RC filter effect in your circuit.
Zeners do not have a sharply defined conduction voltage. They start to conduct at a much lower voltage than the zener voltage Vz. This becomes a problem when your signal has a high source impedance, or if you have introduced some series resistance to try to limit the effects of over-voltage. To counter this you may need to use a higher voltage zener, but this then defeats purpose of including a zener.
If you are concerned about high energy over-voltage transients then a zener is of no use at all. It reacts too slowly, and cannot absorb much energy.
Generally, for signal lines, a Transient voltage suppressor is a much better part.
Best Answer
Usually the datasheets, in the absmax section, give an information like "-0.3 V or -10 mA, whichever occurs first", but unfortunately not in this case; therefore a proper answer is not possible, but I can give a bit of insight.
The -0.3V limit is there because for each pin there is a diode from ground to the pin, and from the pin to vdd. With 10 kOhm in series, this is a non issue; I would expect almost any IC to be able to handle at least a few mA in the protection diodes, and here you have a 10 kOhm resistor in series. You can even bump the resistor up to 100 kOhm, if I understand your circuit correctly.
To sum it up, with your series resistor you are not violating the absmax because the diode will conduct and block the voltage around -0.3 V, probably even less. With 100 uA you can sleep safe.