Inversion of the digital IO pins
 

I was hoping to use an external EEPROM to store data for our autonomous routine, it uses an I2C interface, but I have code written to bit-bang, so that is ok, but here is my question. I know the pins are inverted as an input, since when they are low, you get a 1 in software, and when they are high, you get a 0. But are they inverted as outputs too?

Thanks

Re: Inversion of the digital IO pins
 

They're not inverted in my experience, as either input or output. You might be thinking of the fact that they're internally pulled up to 5 volts, so a switch to ground reads as a logic 1 when it's open and a logic 0 when it's closed.

Re: Inversion of the digital IO pins
 

I thought that a pull up floated high which meant that it will always read as a logic 1 unless you ground the signal pin

Re: Inversion of the digital IO pins
 

You thought correctly. As I said, closing the switch to ground yields a logic 0.

Re: Inversion of the digital IO pins
 

A pullup does not imply inversion. It just means that when you leave it floating, you will get a logic one on the input. If you have an input connected to it that will pull the input down, then you will get a logic zero as the input.

Matt

Re: Inversion of the digital IO pins
 

Nevermind, limit switches, use OPEN and CLOSED, not a bit test... bah sorry
Its been a long 6 weeks... (*mutters on about being designer, team leader, and programmer...*)

I thought that they made it so that when the switch was closed, the bit read 1, and when it was open it read 0.

in user_routines.h
I guess I won't have to modify my programs after all...