Target is S32K148:
In the S32k reference manual the attached spread sheet for the IO pins, the default state is marked as disabled.
What is meant by disabled?
if I don't initialize the pin in my code, what would be the current consumption for that specific pin with the default state as "Disabled"?
I know the best way to reduce the current consumption for the unused pin is to configure it as an input with the internal pulls enabled.
I'm just very curious to know what "disabled" state means .
is it totally disconnected with CMOS structure of the pin? is it consuming any power? what state is the
pin's driver gate gate ?
Thanks,
Koorosh Hajiani
Hello Koorosh,
It means that it is disconnected from all the digital peripherals in the port's MUX including GPIO.
Please refer to AN5426, Hardware Design Guidelines for S32K1xx Microcontrollers.
It can be found at the link below under Application Note
Table 7. Used pins configuration
Footnote 2.
Regards,
Daniel