5V tolerant pin as output with pullup

[mmensch]

Posted on April 19, 2013 at 16:09

Hi,

I need to replace a 5V-uC by a 3.3V type and maybe it will be a STM32F0xxx. Now there is one question concerning 5V tolerance: If the 3.3V-uC drives an output that has a pullup resistor to 5V, is that any problem because of current flowing into the output pin?

Thank you very much

Martin

[dthedens23]

Posted on April 19, 2013 at 18:12

yes.  You do not drive the pin high.

logic low:   You make the pin an output and drive it low

logic high:  You make the pin an input, then the 5v pull up resistor drives it high

to toggle, just change the pin from input, to output, to input, .... 

[emalund]

Posted on April 19, 2013 at 19:29

if you are replacing a push-pull output with an output pulled up by a resisitor you will see greatly increased risetimes.  sometimes a problem, sometimes not.

Erik

[zzdz2]

Posted on April 20, 2013 at 09:15

You can configure pin as output open-drain so only N output transistor is activated when output is low.

[mmensch]

Posted on April 22, 2013 at 09:55

Hi,

thank you for your answers.

I was asking this question also to the ST online support and got no answer for quite a while, so I decided to post it here. As things are going I received an answer right after that. Just to complete it I also post it here:

You should configure the GPIO as open drain and disable the internal pull-up (the external one only shall be used).

When the FT pin is then output-low, there is max allowed current 25mA through the pin. In output-high guarantees the FT pin circuitry zero injection current when the external voltage applied to the pin is above VDD (and below VDD+4V).

As a conclusion I understand that I should not use the high side FET of the output. Just to improve my knowledge I would like to know what would or could happen if I configured the output as pushpull. Any hints on this would be welcome

Thanks

Martin

[klauskvik]

Posted on April 24, 2013 at 10:26