A beginner problem with the latch

Here is the ladder diagram I wrote

please help

Can you label each instruction or symbol and comment what each network is supposed to do please?

I love the comments on the rungs and bits!

/criticism off.

Seriously, comments help.

alyelhefni, have you tried to just program the states like I suggested before? Again, do not worry about real IO at this point. Just program to track the states. You have mulitple rungs going to M1.1, for example.

I suggest you start over, do one thing at a time, program your machine state first. Write out in plain English (or whatever language you are good at) a program for state transition. Come back when you get this done.

I'm just trying to understand what you actually want the program to do.

Click to expand...

Could you gimme a download link for your version of the program?

Click to expand...

I think everyone is confused with trying to figure out WHAT IS the PROJECT ASSIGNMENT? What is the goal? What is supposed to happen? What is the action?

Putting together all the clues, all the hints, and making some guesses, here is my interpretation. Alyelhefni, maybe you would be so kind to say if this picture describes your project, or if it is different?

Lancie
yes that """ is "" my exact problem
we don't have any textbooks or references in this subject except the lectures gives by our dr .

so know I found the place where the dr. copies his problems
""" would you please tell me what book is it ??? """

Yes this is my problem exactly that must be answered in ladder diagram and I finished it on paper
I wanted to know weather it's right or not
so I went to the siemens s7 and started simulating the program
unfortunately I'm facing the problems stated above and I can't get to grips why

or I see you've written what I've scanned on paper

anyway this is the problem

alyelhefni said:

or I see you've written what I've scanned on paper. Anyway this is the problem

Click to expand...

That is a giant step toward your problem. Knowing what the problem is, now it is fairly easy to write PLC logic to solve it.

Think of it in terms of your 3 Outputs Conveyor Motor #1, Flapper, and Conveyor #2. These are the only 3 things that the PLC must control to solve your problem. So why, when, and for how long do we turn on Q1.1. Q1.2, Q1.3? Figure that out and your assignment is 95% finished!

1. Output Q1.1: Turns on to move product to Sensor S1. When S1 = 1, Q1.1 goes off for 2 seconds (use Timer T0). Then if S2 = 1, Q1.1 is turned on again. Instead if S2 = 0, then start Timer T1 AND Output 1.3, wait until T1 is DONE, then restart Q1.1 IF S4 = 1 and Q1.1 is not ON.

2. Output Q1.2: Turns on to move a full box of 24 products off to another place. Assume that Q1.3 runs for 5 seconds (Timer T2) when S4 = 1 AND S3 has triggered a Counter C0 until C0 = 24. If S4 goes off when the full box leaves, then S4 = 1 AGAIN when a new empty box arrives, then turn off Q1.3 and wait until the new box is full.

3. Output Q1.3: Turn on when S1 = 1 AND S2 = 0. Run for 2 seconds (measured by Timer T1), then turn off again, wait for next product.

Those are the 3 Outputs and what each must do, when it goes ON, and when it goes OFF (the only things this PLC can do to any of the 3 devices that control the work).

That is all I have to say about that right now.

I will have mercy on another poor student and show the first 2 ladder networks out of about 8 required.

Here is the program I wrote

well it works except for the counter and I don't know why yet

I discovered an extra annoying fact that to avoid contradictions that happen between different states you should set value for your new state and then reset value of the state you've just left which makes the program much longer

the following is the program I wrote and the last paper is a scanned diagram and table of states

oh sorry I've uploaded the images of my old answer

here is the new one

the problem seems to be with the counter now

if keep checking and unchecking the sensor I 1.3 for 24 times in the simulator but it never counts down
its always giving one at the output