What happens to PIC when pin is using too much current?

just curious what happens on the hardware level when a i/o pin is using more than its rated current. For example: direct driving an IR led with only a 50ohm resistor in series. VDD is 5V, led Vf is 1.7 so 66mA used by the LED.

I was experiencing erratic chip behavior where the device would end up in states that were not explicitly in the code, or device would need a power cycle to resume correct behavior. The erratic behavior wasn't always repeatable and seemed to happen randomly.

I wasn't sure what was going on until I realized I recently made a resistor change from 1Kohm to 50ohm which would make the LED pin use too much current.

My hypothesis is that using too much current on a pin may create brown out conditions where the program counter may jump to wrong parts of the code or incorrectly execute certain processes.

Can anyone confirm my hypothesis that this is indeed what is causing the issue?

How a PIC behaves under overload conditions is unpredicatable but the usual failure is the output driver transistor in the top or bottom of the pin driver. It shorts and as a result the pin becomes unusable but the remainder of the IC generally carries on working. Drawing the extra current may have caused a dip in overall supply voltage though, that depends on how much current is available. Note that many PICs have brown-out protection that creates a graceful reset rather than random code execution. You also have to consider the total current drawn from all the pins, if the extra current to your LED pushed the total pin current over the limit there would be risk of permanent damage to the silicon. PICs are pretty tough though, I've seen them run too hot to touch and yet still work OK when the fault has been cleared and they have cooled down.

Brian.