Needin help to verify this program

when the start button is pushed the conveyor will start running.
After a 20 second the part is dropped onto the loading station depressing a mechanical limit switch , when the switch is actuated the part is ready to be pushed onto the conveyor.
1 second after the part touches the limit switch the part is pushed onto the conveyor.
The part travels down the conveyor and is detected by a photo electric sensor , the part is counted and continues down the conveyor where is it stamped at the stamping station. The part continues down the conveyor , where it is manually removed from the conveyor.

Programming Details

There is a 20 second time delay between parts being dropped on the loading station.

Input 0 is a stop button
Input 1 is a start button
Input 2 is the part detect limit switch
Input 3 is the Photo Electric Sensor

Output 0 is the conveyor
Output 1 is the drop solenoid
Output 2 is the pusher solenoid
Output 3 is the Stamp solenoid


Does this program work

Wow , I`ve never had a post that NOBODY commented on.

Then let me be the first...

Something hit me almost immediately: the STOP button in rung zero. Is that a normally open pushbutton or normally closed? It will work if the button is normally closed, and, thinking about it, that's probably the correct failure mode for this start-stop logic, too.

I'll continue looking at this, and I'll probably have some more questions as we go. If no one else chimes in, I'll edit this post instead of doubling up...

[EDIT] You might also consider that you could use something like CutePDF to output your ladder as a PDF file (CutePDF acts as a "virtual printer" device and outputs a PDF file to a file name you select - it's free, too. http://www.cutepdf.com/) instead of doing screenshots? Might help get you an answer quicker - my old eyes are having some trouble as I get further down the ladder...

[EDIT2] In rung 7 is there a reason you're not driving the stamp solenoid latch "directly" from an XIC addressed to the photo-eye? No modification occurs in rung 6 other than just passing the state of the photo-eye.

I'll chew on this some more, but I've had a long day today and need to take some time to really "get" what you're trying to do (I need to visualize the conveyor layout for it to make sense to me...) so let me put this aside for now, please.

Later On,
D

One thing I noted on the first two pdf files was that your outputs to your pusher and drop solenoids MAY need to be on timers that limit the time they are on. I didn't read the other two pdfs yet. Horus is correct, viewing multiple pdfs is kinda cumbersome.

Where's the 20 second delay?

I haven't verified everything else, but I agree about the stop button being normally closed.

[edit] The pusher delay is set to .1 second right now instead of 1 sec.

BernardWright said:

Wow , I`ve never had a post that NOBODY commented on.

Click to expand...

It's the weekend and many are absent from this board on weekends. There are many people here that will jump in to help, just give them a little time. You might want to consider zipping and posting the .rss file in .zip format. Even posting a file in .pdf form as Rube and Horus pointed out would be easier to follow.

horus said:

Then let me be the first...

Something hit me almost immediately: the STOP button in rung zero. Is that a normally open pushbutton or normally closed? It will work if the button is normally closed, and, thinking about it, that's probably the correct failure mode for this start-stop logic, too.

Click to expand...

Stop buttons in PLC logic is always N/O as a standard stop pushbutton in the physical world is N/C, therefore your input will always be ON unless the PB is pressed.

file re-submitted.

Horus thanks for the advice on using pdf files. I was taught that the stop button should be programed XIC because it`s gonna be hardwired normally closed..



when the start button is pushed the conveyor will start running.
After a 20 second the part is dropped onto the loading station depressing a mechanical limit switch , when the switch is actuated the part is ready to be pushed onto the conveyor.
1 second after the part touches the limit switch the part is pushed onto the conveyor.
The part travels down the conveyor and is detected by a photo electric sensor , the part is counted and continues down the conveyor where is it stamped at the stamping station. The part continues down the conveyor , where it is manually removed from the conveyor.

Programming Details

There is a 20 second time delay between parts being dropped on the loading station.

Input 0 is a stop button
Input 1 is a start button
Input 2 is the part detect limit switch
Input 3 is the Photo Electric Sensor

Output 0 is the conveyor
Output 1 is the drop solenoid
Output 2 is the pusher solenoid
Output 3 is the Stamp solenoid


Does this program work

In rung 0 The stop I/0 and start I/1, is as I would expect it to be, with a seal in B3/0 added to hold B3/0 in.

Now this is only a comment from me,
If you do not fully understand TOF operations it pays to avoid them,
in fact, I very seldom use then, I think I could count on my fingers the number I have ever used in 10 years.

Another suggestion from me,
in your case is replace the Latch Unlatch operation with an OTE,
using a seal in contact as you did in rung 0.
I feel this would allow you to simplify your program and use TON instead of TOF.

PeterW said:

Stop buttons in PLC logic is always N/O as a standard stop pushbutton in the physical world is N/C, therefore your input will always be ON unless the PB is pressed.

Click to expand...

Yeah, I think I even mentioned in my previous that this thinking is driven by the failure mode desired (i.e., if a NC pushbutton fails, it usually fails OPEN).

The important thing is to make sure that was the conscious choice in the design of the system...