ledpnptransistors
On the diagram below, there are 4 PNP transistors.
When RA0 is set to HIGH, then current flows from +5V through RA0 to GND and from +5V through 7 segment display, then through (for example) RB0 (if it's set to HIGH) to GND. Is that right? Is setting a processor pin to HIGH state means the pin will allow current flow from the voltage source to GND through itself?
Pretty close.
You should give the PNP a lot more base current to make sure it saturates. Use a forced beta of 10-20, which means that the resistor R2 should be more like 470R.
The value of R3 is not critical, 10K would be fine, so would 20K.
You don't need a Darlington to drive the PNP transistor, a regular NPN will be fine. Use a base resistor of more like 10K. The base resistor limits the current- it is determined by the desired base current, not by the gain of the transistor- you do always need it.
Keep in mind these Darlington things won't switch instantly and if you don't allow some dead time in the drive firmware you'll get 'ghosting'.
You need to do something like this:
simulate this circuit – Schematic created using CircuitLab
The digit select line must be High to select the digit - thia may require a software change. Otherwise, you would need another NPN to invert the signal.
Best Answer
One of RA0 - RA3 must be set low, to select the digit, with the other RAx's set high.
RB0 - RB7 must be set low for the segments you want lit in that digit, and set high for the segments that should be off.