For my projects next revision, I'd like to drop the ULN2803A transistor array I currently use to drive eight optocouplers.
Using optocouplers with a lower current requirement means that I'd only need to sink / source 5 mA of current for each of the eight pins.
Unfortunately, neither the datasheet for the MCU nor the specs of the development board include a rating I'm used to – like the 40 mA per I/O pin limit on the ATmega328P.
Table 12 of the datasheet (1) contains a formula for some Imax;
Imax = N x C x V x (0.5 x F)
where F is the "Data change rate" in Hz – but I believe that would imply I couldn't supply any current if the outputs were static, which is probably not the case.
What currents can this MCU sink / source? How do I calculate that? Can I (safely) drive eight 5 mA optocouplers directly via the MCU?
Best Answer
I think you are misinterpreting the datasheet. That Imax is how much current the GPIO is consuming. Because it's CMOS based DC current is negligible. Higher data rates increase current consumption.
I looked through the datasheet and did not find an absolute max GPIO output current. For what it's worth tables 26/27 mentions GPIO output impedance.