On Tue, 9 Sep 1997 23:48:38 CET "Marc 'Nepomuk' Heuler" writes: >Hi Adolfo (Adolfo Cobo (Ingenieria Fotonica)), in ><3.0.3.32.1997090911240= >8.00789824@193.144.187.40> on Sep 9 you wrote: > >> And now the problem: the current from the pin of parallel port at >log= >ic >> level high seems to feed the entire PIC circuit, because it runs >even >> without power. > >The current you apply to the port pin flows through a protection diode >to= > [...] >Otherwise you should isolate the port pin by transistor, or change >polari= >ty >of your signal so you can use a pulldown instead of a pullup resistor. If the signal direction is always from the PC to the PIC, put a large (100K~1M) resistor in series to isolate the two. If the PIC doesn't have power, the resistor will prevent a heavy current from flowing through the protection diode and trying to power the PIC circuit. It also gives a much greater degree of protection against static and excess voltage than just connecting directly to the PIC. This also works to receive RS-232 signals directly into a PIC. It is hard to damage a PIC this way anyway. Generally PIC inputs don't mind being left floating, unless you need low power consumption or the input to assume a known state when not connected. In that case put the pull-up/pull-down resistor on the PC side of the isolation resistor. The internal pull-ups can't be used with an isolation resistor.