microcontroller
I can't tell how many IO pins that are available on the C8051F975.
SiLabs website, Farnell and the datasheet p. 42 says 19 IO pins, so I guess that is correct, but the according to the datasheet p. 38 there is 21 IO pins.
I'm trying to use this MCU in a schematic, but I cannot get it finished before I know which of the pins are actually available.
Further to Keelan's answer:
The official name for the technology is "peripheral pin select" or PPS.
The 4th column in table 1-1 is the PPS column. You can see that the UART pins are all listed as 'Yes' which tells you that you need to use PPS to get these functions physically connected to pins.
Table 4-16 lists the input register map: file RPINR18 concerns the UART receive function. The value you write to RPOR18's low byte will determine which pin acts as UART receive.
Table 10-2 lists the output registers for PPS. UART transmit is 0x03; you would write 0x03 to the RPOR register corresponding to your desired output pin to connect it to UART transmit.
A colleague pointed out that each of port A and B only have 8 pins each. This can be seen by referencing Figure 3-1 of the datasheet. Unfortunately it means that there are in fact only 16 pins with ADC, not 32 as I had previously thought. Similarly, since the timers are on ports C, D, E, and F, and also are limited to 8 pins there must only be 16 pins capable of PWM generation.
Best Answer
There are 24 pins (+ GND pad) on the QFN-24 package on page 38.
2 pins are VDD and GND -> 22pins.
1 pin is RST/C2CK -> 21.
If you want to use a hardware debugger, then do not use
1 pin P5.2/C2D as an I/O -> 20 I/O pins.