1786-RPFRXLB |FORCE| Internal Relay

Description

1786-RPFRXLB |FORCE| Internal Relay

1786-RPFRXLB (c) shows the second scanning cycle after pressing the SB1 button. ① Input scanning, as the input state remains unchanged, the input image area remains unchanged. ② During the execution of the program, on the first line, the X0 contact is closed. Since Y1 was in the ON state in the previous cycle, the Y1 contact is also closed and Y0 is also in a conductive state; In the second line, Y1 is still in a conductive state. The states of Y0 and Y1 are both “1”. ③ During the output refresh process, both lights are on. Attention: Due to the short scanning cycle of PLC, the phenomenon we can see with the naked eye may be that both lights are on simultaneously. If the button remains unchanged, the internal relay and output device status remain unchanged.

This 1786-RPFRXLB has no practical significance, but it is very beneficial for analyzing the program execution process. The system input only needs to be connected to one button without output, as shown in Figure 5-2, only X0 is connected. Analyze the program execution process by observing the current value of the counter in the three scenarios of (a), (b), and (c) in Figure 5-5.
In the program, 1786-RPFRXLB is a special auxiliary relay. As long as the PLC is in operation, it will continuously emit a 10ms pulse signal (5ms on, 5ms off). In the program, T0 is timed for 1 second, and 1 second after X0 is closed, T0 conducts. C0 is an increase counter, which records the number of pulses emitted by M8011 while X0 is closed and T0 is not closed. In theory, when T0 is conducting and the C0 counter stops counting, the current value of the counter should be 100 pulses (1s/10ms=100 pulses). In the three segment program, only the position before and after execution was changed, but the results were different. Analyze the reasons based on the corresponding time sequence diagram.