We had agreed some weeks ago, that we shouldn't talk about 'modes'.
Say instead: "If port is emitting logic low state." Or: "If port output is logic low." Or something similar.


This negative current means, it's flowing outside of package, right what you need to turn-on a NPN transistor. Unfortunately this output current is not strong enough to fully turn-on a NPN, especially if its load current is rather high. Assume NPN must turn-on a load of 20mA. Then, as base current should be 1/20 of load current, about 1mA should flow into the base of NPN. A standard 80C51 output cannot deliver this current. An additional pull-up is needed.


It simply means, what it states: If you force the port pin, which was programmed as input by writing '1s' into according SFR, to logic low, then up to 50µA will flow out of the pin.


This means, that if you want to force a port pin, which was programmed as input by writing '1s' into according SFR, to logic low, that you must take care that source impedance is not too high. Here, less than 2V / 650µA = ca. 3kOhm. Otherwise the potential at port pin will never fall down to less than 2V, which is needed to detect logic low state.

You should read the according 'bible' chapters, to learn how the port topology with its three active pull-ups works.

Kai