New to PLC's, First Question

Grover · Oct 4, 2002

Hi everyone.

This is my first post and first question. What a great forum!

I have 20 years exerience in electronics, but I am new to the world of PLC's. Ladder logic is coming along, but not quickly enough for what I need to get done.

My first question is very simple (well, not for me at the moment) and it involves an AB SLC5. Once a timer (T4:4 for example) is done, I understand the DN bit gets set (1). When does the DN bit reset (0) and where did I miss that in the help file?

Grover

jimtech67 · Oct 4, 2002

As long as the rung the timer is on is held true and the preset has been reached the DN bit is high.

Martin T. · Oct 4, 2002

As long as the rung is true and your preset is reached the DN bit will stay set a 1, if you just want to keep cycling to your preset add a XIO T4:4.DN in the rung with the Timer. What are you using the timer to control? Knowing this might help with a better answer.

Grover · Oct 4, 2002

Jim & Martin,

Thanks for replying.

The rung goes true when:
- the motor start button is closed
or
- the T4:4 TT bit is set.

The rung output:

- T4:4 is a three second motor start delay.
and
- O:15/2 gets set to sound an audible alarm

It would appear that once the operator releases the start button the rung will go false, therefore, the T4:4 DN bit will reset. Did I get it right?

Allen Nelson · Oct 4, 2002

I don't think you've quite got it.

First, this is what, from your desrciption, I think your ladder looks like:



      MOTOR

   START BUTTON            +-------- TON ---+

---+---| |---+-------------+ Timer:    T4:0 |

   |         |             | Time Base: 1.0 |

   | T4:0.TT |             | Preset:      3 |

   +---| |---+             | Accum:       0 |

                           +----------------+



                             AUDIBLE ALARM         

     T4:0/DN                    O:15/2 

-------| |-------------------------( )

What's going on is this:

When the operator presses the START button, for a moment or for a long time, the timer starts.

While the timer is timing (counting up to 3.0 sec), the Timer_Timing bit (T4:0.TT) is set (ON). This will keep "power" to the timer, even if the operator lets go of the button.

After 3 seconds, the done bit (T4:0/DN) comes on, sounding the horn.

Now here's the trick part:

When the done bit comes ON, the timer is no longer "timing", and so the TT bit goes OFF (is reset). If the operator has let go of the button, there is now no "power" going to the timer block. Therefore the timer block resets itself, putting the ACCumulated time back to zero, and resetting the done bit.

Therefore, if the operator holds the button, the ALARM will sound 3 seconds after he's pressed it, but will stop as soon as he lets go.

If he let go before it sounds, the done bit will only be on for one scan, and so the ALARM will "chirp" once (i.e., it will be ON for one scan - about 2 msec).

Does that clear things up?

Grover · Oct 4, 2002

Hey Allen,

Thanks for the reply and the template. Here is the actual rung.

MOTOR
START BUTTON +-------- TON ---+
0028---+---| |---+-------------+ Timer: T4:0 |------
| | | | Time Base: 1.0 | |
| T4:0.TT | | | Preset: 3 | |
+---| |---+ | | Accum: 0 | |
| +----------------+ |
| |
| AUDIBLE ALARM |
| O:15/2 |
--------------( )-----------

When the done bit comes ON, the timer is no longer "timing", and so the TT bit goes OFF (is reset). If the operator has let go of the button, there is now no "power" going to the timer block. Therefore the timer block resets itself, putting the ACCumulated time back to zero, and resetting the done bit.

The T4:0.DN is examined during a later rung but suffice to say that until rung 0028 is scanned again (assuming T4:0 is not used elsewhere) the T4:0.DN bit will remain set. When 0028 does get scanned again, both the T4:0.TT and the motor start button will be false, therefore the timer will be completely reset (EN, TT, DN all OFF).

Better?
:c{ )

Grover · Oct 4, 2002

Sorry Allen, I forgot to use the preview feature. Here is the
ladder diagram for my previous post.



      MOTOR

   START BUTTON            +-------- TON ---+

---+---| |---+-------------+ Timer:    T4:0 |

   |         |             | Time Base: 1.0 |

   | T4:0.TT |             | Preset:      3 |

   +---| |---+             | Accum:       0 |

                           +----------------+



                             AUDIBLE ALARM         

     T4:0/DN                    O:15/2 

-------| |-------------------------( )

Grover · Oct 4, 2002

How about this. Ignore my last two posts and go from here.





          MOTOR

       START BUTTON             +-------- TON ---+

0028---+---| |---+--------------+ Timer:    T4:0 |---------

       |         |        |     | Time Base: 1.0 |   |

       | T4:0.TT |        |     | Preset:      3 |   |

       +---| |---+        |     | Accum:       0 |   |

                          |     +----------------+   |

                          |                          |

                          |       AUDIBLE ALARM      |   

                          |          O:15/2          |

                           ------------( )-----------

How does this sound now?

The T4:0.DN is examined during a later rung but suffice to say that until rung 0028 is scanned again (assuming T4:0 is not used elsewhere) the T4:0.DN bit will remain set. When 0028 does get scanned again, both the T4:0.TT and the motor start button will be false, therefore the timer will be completely reset (EN, TT, DN all OFF).

Grover
:c{ )

rsdoran · Oct 4, 2002

What I see is an input (start button pushed in this case) that sounds the alarm horn and starts timer which in turn latches the horn on as long as timer is timing. I would "assume" (since I do this regularly) that the timer DN allows the next sequence to start...ie it starts after horn is done sounding.

BTW here is the Instruction Set Reference for SLC 500

Grover · Oct 4, 2002

Yes, that is exactly what it does. The operator presses start, the horn sounds for three seconds and then the motor starts (I left out a few rungs which check for safety interlocks).

My original question focused on the behavior of the TON instruction and I now see my perception of ladder logic is more the problem. I believe I better understand how TON works and more importantly, how it and other instructions behave in rungs.

Thanks for the link to AB. The wording in that manual was a bit different than the help in RSLogix and I glossed over some key words.

Thanks all! This has been most helpful. I'm looking forward to future conversations.

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