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A very basic question:
Must each IO pin of a microcontroller that is serving as an input from a switch or jumper have its own pull-up/down resistor?
In this example, each pin is pulled down by a resistor, then pulled up by a 1P3T switch to VCC.
It seems straightforward, but as more IO pins are used, are more resistors absolutely required? Is there some clever way to keep the resistor count down?
Related question: Sharing a pull-up resistor
As per the schematic provided, if all 3 inputs were sharing a resistor, then any of the lines being pulled high via the switch would raise all 3 lines to high, countering the purpose of the design - the MCU would not know which switch position is selected.
A common way of reducing part count, not resistor count, for such designs is to use a common bus resistor network or array:
(from here)
These are available as through-hole SIP/DIP as well as SMD, in a variety of resistor counts, depending on your needs. The bus pin is connected to ground, and the other pins are connected to the respective MCU inputs as in your schematic.
(from here)
(from here)
Most MCUs have optional pull-up (not down) resistors built in for each pin, so it is normal to pull the pins DOWN with the switch (and deal with the polarity inversion in software).
So - no resistors necessary.
As to why pullup, not down, it's a habit left over from the 1970's TTL logic circuitry, where it took much less current to pull an input up than down - a pulldown resistor would waste more power. This no longer applies with today's CMOS logic, but the tradition of pull-ups has persisted, so that early 5V CMOS chips were compatible with older TTL logic.
You never want to leave an input to logic open on the assumption that it would pull itself up or down. if an input is left open, it is a little antenna and also subject to the currents within the logic device. So you pull up or down to ensure you have a clean and predictable input. I learned this rule while I worked at Fairchild Semiconductor in the 1980s.
