PIC18F2550 Limits

diodespic

I want to use the PIC18F2550 with this dual power circuit but I am afraid about what can happen with the PIC inputs pins.

If Vself is 5V, the diode drop (let's say it's a Shottky) will make the Vdd of the PIC be something around 4.7V. The problem is I am reading some inputs coming from other systems, and this systems operates in 5V.

How will the PIC pins react to that? Could it damage the PIC? I read the datasheet, and things appear to be pretty tight, so will my PIC be less reliable?

Best Answer

The two key specifications you need to consider are (taken from the datasheet):

  • Voltage on any pin with respect to VSS: -0.3V to VDD + 0.3V
  • Input clamp current, IIK (VI < 0 or VI > VDD): ±20 mA

A simple way of protecting your device from exceeding either one is with a voltage divider that drops from 5V to 4.7V.

That being said, you can actually get schottky diodes that only drop 0.2 V @ 500 mA, so you have a bit more margin. Also consider that the outputs of your other system may not be rail to rail, and will have a slight voltage drop.

Therefore it will probably be enough to put a small resistor in series at the input pin, to prevent any significant current flowing through the clamping diode.

In case your external signal comes from a pullup to 5V, you probably don't need to do anything at all, because the pullup resistor will prevent any significant current from flowing through the clamping diode, and the clamping diode itself will do its job and prevent over-voltage. If you'd rather not have the clamping diode draw any dc current, you could put a large resistor to ground and "complete" the voltage divider to 4.7V.