uint8 Pin_Write(void)

[Anonymous]

 Hi all, 

   

If a port is connected to a signal generator, can I still use uint8 Pin_Write(void) function to set the value of that port to the other value for some certain time period, like 0. Thank you so much!

[ETRO_SSN583]

Yes, the pin logic is not a f() of the signal injected into

   

it, but you could get unexpected results. For example,

   

if you are driving the pin with 50 ohm generator, and you

   

config pin as strong drive output, you could get high currents

   

flowing that PSOC I/O supply routing on die cannot handle.

   

Hence logic  could fail, part could get damaged. Case in point

   

you drive output to logic low and generator to supply rail.

       

Regards, Dana.

[ETRO_SSN583]

Yes, the pin logic is not a f() of the signal injected into

   

it, but you could get unexpected results.

       

Clarify this that when pin is subjected to inappropriate conditions

   

then pin logic can fail, stated another way pin logic then would be a

   

function of the way you drive it.

       

Regards, Dana.

[Bob_Marlowe]

First: the function is void Pin_Write(uint8).

   

You cannot safely run two signals against each other when their type does not match,

   

When the output of your function generator is open drain or a resistive output there is a good chance to get it to work as you like to. When the generator's output has a phase of high impendancy you may drive the signal by the pin, there is a output enable signal to swithch a pin on and off.

   

Plenty of possibilities, but some combinations may blow your pin or generator.

       

Bob

[Anonymous]

 Thank you for your help! Yup! I can't feed two conflicting signal to a pin.