Go Down

Topic: I2C between 8MHZ & 16 MHz arduinos (Read 839 times) previous topic - next topic

magnethead794

Sep 01, 2012, 04:47 am Last Edit: Sep 01, 2012, 04:54 am by magnethead794 Reason: 1
Any issues with doing it? Just wondering. Need to link a FIO and a mega over I2C.
Yes I know level conversion will be needed.

(the other end of the xBee link is a FIO & teensy2.0++ pairing..same deal there)
KF5RVR

frank26080115

there shouldn't be any issues

level conversion will not be needed, just disable internal pull-up resistors and use external pull-up resistors that "pull" to 3.3V
Freelance engineer, consultant, contractor. Graduated from UW in 2013.

magnethead794


there shouldn't be any issues

level conversion will not be needed, just disable internal pull-up resistors and use external pull-up resistors that "pull" to 3.3V


So the 5V signals coming from the 5V board won't overdrive the 3V3 pins?
KF5RVR

James C4S


So the 5V signals coming from the 5V board won't overdrive the 3V3 pins?

That's why you use external pull-up resistors.
Capacitor Expert By Day, Enginerd by night.  ||  Personal Blog: www.baldengineer.com  || Electronics Tutorials for Beginners:  www.addohms.com

magnethead794



So the 5V signals coming from the 5V board won't overdrive the 3V3 pins?

That's why you use external pull-up resistors.


Doesn't make sense to me, but I'll go with it.
KF5RVR

James C4S

Research "open collector".

The I2C pins are open-collector.  They need to be pulled-up so if you external resistors and pull them up to 3.3V they become 3.3V pins.
Capacitor Expert By Day, Enginerd by night.  ||  Personal Blog: www.baldengineer.com  || Electronics Tutorials for Beginners:  www.addohms.com

magnethead794


Research "open collector".

The I2C pins are open-collector.  They need to be pulled-up so if you external resistors and pull them up to 3.3V they become 3.3V pins.



works for me.
KF5RVR

Go Up