Ad d0-06 ladder seq.

[nykommling]

Hi, Could someone explain to me why this code will not set output Y1 as it is written. I placed rung 9(set Y1) after rung 10(reset) and it worked fine. How can the reset happen if Y1 in rung 10 isn't true?

Thank You

 

[Eric Nelson]

How can the reset happen if Y1 in rung 10 isn't true?

Y1 is true in rung 10. The state of Y1 is updated immediately, although it won't update the 'physical' output until the end of the scan.

EDIT: It looks like you want Y1 to alternate each time the counter reaches preset. Are you trying to make a flip-flop?



-Eric

 

[Tom Jenkins]

First off, the outputs are only written to the actual hardware output at the end of each scan, but the status of the output as stored in the memory location is updated as each rung is scanned.

That means that in your example when rung 9 is solved, Y1 is OFF and if CT0 and C12 is ON the logic will set - turn ON - Y1. This stores a "1" or "ON" in the memory location of the PLC for Y1.

Then when rung 10 is solved, Y1 is seen as ON. That's because the previous rung stored a "1" in Y1. Since Y1 is ON, if C10 is also on Y1 is reset - turned OFF and a 0 or OFF stored in the memory location for Y1.

At the end of the scan Y1 has a 0 in it's memory, so when the output table is written to the physical outputs Y1 is OFF and so is any load connected to it. When the program scan repeats, it gets to rung 9, and Y1 is off, so .................

This logic is an example of why I personally hate and despise SET and RESET commands. It is not obvious what is going on, and trouble shooting is much harder than it needs to be.

Incidentally, I also hate one shots (PD coils) for the same reason.