Hi,

>  Hello to all,
>   I need advice with a "problem" I've found: I'm working in a project with
> a PIC16F84-04/P that will be connected to a Pc using the parallel port. An
> input pin (RA1) of the PIC is connected to an out (D0) of the LPT port,
> and, because the circuit may be connected or not, I decided to add a
> pull-up resistor of *330K* to the pin, forcing a high level when the PC is
> not connected.
>   And now the problem: the current from the pin of parallel port at logic
> level high seems to feed the entire PIC circuit, because it runs even
> without power. The voltage across the pull-up resistor is 1.2V, so I supose
> that the PIC chip is consumming 3.6 microAmps. Of course, I'm not using the
> SLEEP mode and there is not short-circuits in the PCB. May be that the
> current flows internally from the RA1 ping to Vcc??? It seems very strange,
> Any suggestions???
>  Thanks in advance.
>
It is, as other people have already stated, the internal protection diode
that allows current to flow to Vcc of you PIC.  You can use the
transistor circuit already mentioned, or you can use something like this:
                   VCC
                   |
                   \
                   /
                   \
             /|    |
PC --------|< |--------PIC
             \|

The diode should actually be a Schottky diode since the maximum low level
is something like 0.8V and the normal silicon diode has a +-0.7-0.8V
drop.  The Schottky diode only has a 0.4V drop.  Or you could use some of
those germanium glass diodes found so commonly in radio receivers.  They
also have a 0.3-0.4V voltage drop.

Niki