Posted 10 Nov 2021 According to picture 1, ''Piston A moves to the right. After the set time, K = 4, the normally closed timer T450 contacts open and the normally open timer T450 contacts close.'' But according to picture 2 After the power flow from Left power rail to right power rail, it should directly go to 2nd rung and cause the normally open timer T450 contacts to close only right? but why it will go back 1st rung and cause the normally closed timer T450 contacts to open? Anybody can help explain me regarding this. Thank You in advance Share this post Link to post Share on other sites
Posted 10 Nov 2021 Continuous Scanning is the reason. If we look at a timing chart starting at zero absolute time. (NOTE - the time intervals are purely arbitrary for illustration purposes) 0.000 - Read Inputs to Memory 0.002 - Start Scanning Rung 1 0.004 - Start Scanning Rung 2 0.010 - Scan Last Rung 0.012 - Write Output from Memory to Real World 0.014 - Overhead to display status on your PC 0.016 - Read Inputs to Memory 0.018 - Start Scanning Rung 1 0.020 - Start Scanning Rung 2 0.026 - Scan Last Rung 0.028 - Write Output from Memory to Real World 0.030 - Overhead to display status on your PC This continues for 250 cycles until 4.000 - Read Inputs to Memory 4.002 - Start Scanning Rung 1 - T450 is True 4.004 - Start Scanning Rung 2 4.010 - Scan Last Rung 4.012 - Write Output from Memory to Real World 4.014 - Overhead to display status on your PC and I could continue this for some time, but I hope you get the idea. A Program in a PLC is never a single pass. 1 person likes this Share this post Link to post Share on other sites