Why is UPDI pin having Vcc as output?

[tobivan]

Hi there,

simple circuit: LED at PA1 :-)
Trying some blinky-stuff, and I wonder why the UPDI pin is showing +Vcc
To save some energy, I took this function and I thought that PIN0 was the UPDI pin:

void disableAllPeripherals(void) {
  PORTA.DIRCLR = PIN0_bm; //  set port A0 to input
  //PORTA.DIRCLR = PIN1_bm; //  LED
  PORTA.DIRCLR = PIN2_bm; //  set port A2 to input
  PORTA.DIRCLR = PIN3_bm; //  set port A3 to input
  PORTA.DIRCLR = PIN6_bm; //  set port A4 to input
  PORTA.DIRCLR = PIN7_bm; //  set port A5 to input        
        
  PORTA.PIN0CTRL = PORT_ISC_INPUT_DISABLE_gc; // disable input buffers
  //PORTA.PIN1CTRL = PORT_ISC_INPUT_DISABLE_gc;
  PORTA.PIN2CTRL = PORT_ISC_INPUT_DISABLE_gc;
  PORTA.PIN3CTRL = PORT_ISC_INPUT_DISABLE_gc;
  PORTA.PIN6CTRL = PORT_ISC_INPUT_DISABLE_gc;
  PORTA.PIN7CTRL = PORT_ISC_INPUT_DISABLE_gc;
}

What am I missing? Do I need to connect UPDI pin if not needed?
Thanks a lot!

[hmeijdam]

You need to first set the UPDI pin as GPIO pin to be able to use it as GPIO pin.
After doing so, you will need to have a HV programmer to ever set it back to UPDI and reprogram the chip. Which is the reason why this choice is hidden/commented-out as "expert-mode" in the boards.txt file.

[tobivan]

Thanks for your answer.
I don't want to use it as a GPIO pin. But why can I read VCC voltage on the UPDI pin? If it is a "programming pin" I would have expected it to be either input or connected to GND, but in no case to Vcc.

[hmeijdam]

Because the UPDI is HIGH in it's idle state and it has a internal pullup resistor. That's what you are measuring. If you connect a led to it, you will see it lights up a little (current flowing through the pullup resistor).

If you want to read more about the half-duplex protocol over one wire, this Microchip application note may be of interest to you.

[tobivan]

Thanks, that explains it very well!
But unless not connected, it cannot draw any current, can it?

[hmeijdam]

But unless not connected, it cannot draw any current, can it?

"Unless not connected" means it is connected.

If not connected it cannot draw current.

[SpenceKonde]

Which uses more power - assuming nothing other than dangling wires connected to the pins, no external loads, powerdown sleep?
An ATtiny with all pins set to OUTPUT and HIGH?
An ATtiny with all pins set to OUTPUT and LOW?
An ATtiny with all pins set INPUT_PULLUP?
An ATtiny with all pins set INPUT?
An ATtiny with all pins set INPUT, but the PINnCTRL's input sense configuration set to disable input buffer?

(remember, this is with nothing actually connected that would draw current from that pin)

All except the fourth are very similar. The floating inputs can dramatically increase power consumption and must never be permitted in a battery operated design.

As an aside, there are several ways (depending on what chip and die rev) where actions involving the UPDI pin can cause increased power consumption! First, driving it low fires up an oscillator briefly. Holding it low keeps that on, and destroys any hope of low power operation.

I think also If you enter UPDI mode with a programmer (possibly on some early modern parts, it would get in this state even if you didn't enter UPDI mode, not sure), but it fails uncleanly, that can leave the UPDI peripheral and it's oscillator running and sucking power until next POR or UPDI reset. So yeah.

The UPDI pin is high. That's good. let it stay that way except as part of programming