The reason to have a 'low' value of pull-down resistor is to ensure the PIC reads the line as low when it should be low. So the value is determined by whatever might wrongly pull the line up.

Leakage from the input pin on the PIC. There should be a worst-case specification on this leakage current (perhaps 1uA or 2uA). You must ensure that the resistor can maintain logic low under these circumstances. Leakage from anywhere else.

This can be a problem if humidity is very high, and you start to get condensation on the pcb. It can be helped by using solder-resist coatings, or by having 'guard' tracks around your sensitive net, connected to 0V.

Injection of electrical noise (by capacitive coupling). Even if your circuit has nothing electrically noisy in it, the environment is noisy. Your operator might have a static charge of 1000V and he pokes his podgy digit right onto the switch! There's only a few picofarads, but it can couple through. When I have done ESD tests on a switch with a (supposedly isolated) metal cap, I have had a spark jump from the cap onto the circuit board! The fix can either be to use a low-value resistor to discharge the spike too quickly for your debounce software* to respond), or (where you don't have inductance problems with long tracks between the PIC and the switch) to use a capacitor, say 10nF, between your input and 0V to short out the spike. *You do have debounce software, don't you.

When things are not critical, I might use 10K - 100K and not worry. For power-critical systems built in 4000-series CMOS logic in a hermetically-sealed enclosure (extremely low leakage) I have used 10Megohm and 10nF. The switch response is fairly slow, but still faster than the human on the other end! Of course there is also plan B: Have a resistor between the PIC and the switch, and put a capacitor across the switch (the other end of which is ground). When idle, have the PIC pin as output low.

There's no volts across the switch, so no current. When you will soon read the switch, have the PIC pin as output high. If the switch is open, the capacitor will charge up. If the switch is closed, the capacitor will stay discharged, and the current will be set by that resistor.

But that current is not very important because you're only in this state for a short time (long enough to reliably charge the capacitor through the resistor and to fit in with the scheduling requirements of your program). To read the switch, set the PIC pin to input. It will read the volts on the capacitor; high means the switch stayed open since the last precharge.

This is quite complicated in software, but if you're only looking at the switch once per second (or whatever) it can get quiescent current very low. And you can arrange the system to 'catch' any presses that were made between key scans. Hope this helps, Danish. Ric, Danish, Thanks very much for your information. To answer your question Ric, please see the attached schematic snapshot - this is the one for the rfPIC project. Pin GP3 on the rfPIC is an input only, and also the MCLR ICD programmer input. Konica Minolta Pi3500 Driver- Download Last Version on this page.