MX6UL - GPIO State in shutdown

[samuelfriedman]

Is there a default state (pulled-high, pulled-low, high-z) for pins setup as gpios on the mx6ul? Looking through the reference manual, I found (sect 26.4.2.1) that the gpio input: 

   "Receiver is tri-stated when I/O supply (OVDD) is powered down. (Keeper at receiver

   output keeps its previous state)."

which would seem to indicate that the state is defined, but I realize that that the I/O supply may not necessarily be off in shutdown. 

The real goal is to use standard momentary switch + pull-up resistor to drive both a gpio pin and separate, discrete logic. If the default state is open drain or as an input, I should be all set. If it is a driven output, some type of buffer would be needed in shutdown. 

thanks!

Sam 

MX6ul running Debian Jessie R01-3.14.52. and Linux version 3.14.52

[art]

The default out of reset state of all processor's pins can be found in the Tables 95 and 98 (Table 94 for Automotive data sheet) of the i.MX6UL Data Sheet Rev.1 document. Further state of the pins depends on particular pin's function during operation. When the processor goes to low power state without shutting down the power supplies, the pins keep their last state.

Have a great day,
Artur

-----------------------------------------------------------------------------------------------------------------------
Note: If this post answers your question, please click the Correct Answer button. Thank you!
-----------------------------------------------------------------------------------------------------------------------

[samuelfriedman]

Thanks for the information. I don't know if those tables provide what I am looking for however. Do those tables not define the default state on reboot?

For example, the pin in question is NAND_READY_B, which according to the tables you mentioned should be an input out of the reset state. If it were an input in the shutdown state, then I should see some relatively high input impedance looking into the pin and would expect logic high on that pin with an external pull-up resistor to a supply. That is how I intended the circuit to function.

The odd part is that I measure around 0.8V on the pin when the device has shutdown (using shutdown -h now). Current must be flowing somewhere but all that is connected to that pin are the inputs to two cmos logic gates and a pull-up resistor. Also, if I set the pin up as an output, I cannot drive it low. In fact, echoing 1 or 0 causes no distinct change on the output. I know that the hardware is at least connected correctly since I can watch the state change from high to low when the pin is setup as an input.   

Don't know if you have any thoughts, but thanks for your help!

[art]

Actually, when you issue the 'shutdown -h now' command, after completion of all shutdown procedures in software, the power is finally removed from the processor by the external power management circuit. When the power is removed, all GPIO buffers of the processor become non-functional and can behave randomly. Moreover, it is strictly not allowed to externally drive any GPIO singnal when the power is removed from the processor.

Best Regards,

Artur

[lemon_bear]

Dear Artur,

Could you, please, point the document where it is said that externally driving of an IO during device shutdown is dangerous? Is it still dangerous if I pull down it with 100k?   

Best regards,

Aleksandr