RSlogix Emulate 500

[showshocka]

Possibly OT, but I can't understand the function of your timer as it is written in your program.

i agree that this rung is defeating a purpose, that's why i am investigating it. the last IT designed this rung and from my understanding it never worked. All i want to do it use that timer to read longer then what it usually would, while the conveyor is running, and when that happens the timers DN bit sends a 1 to B3:6/2(L) for a fault condition.

 

[MarkNightingale]

The JSR instruction needs to come before the END instruction, but as for where it comes in your ladder we wouldn't be able to tell you without seeing LAD 2 because we can't assume that there are aren't any jumps within that ladder.

To begin with try putting it just before the END instruction and see if it works.

If not post the full code (zip it first) and we will have a look for you.

Mark

 

[dcooper33]

i agree that this rung is defeating a purpose, that's why i am investigating it. the last IT designed this rung and from my understanding it never worked. All i want to do it use that timer to read longer then what it usually would, while the conveyor is running, and when that happens the timers DN bit sends a 1 to B3:6/2(L) for a fault condition.

Yes, my thoughts exactly.

i see that mark has only rung 0000 and 0001 (end) where would i be putting that new rung? before the end at the end i have already? i have 0095 rungs in lad2 where would i put the new JSR instruction and "end" instruction exactly?

It depends on when in the sequence of your scan you want that ladder file executed. Normally, all of your JSR instructions will be at the bottom of Ladder 2 in sequence, i.e.:

---------------------------------------------------[JSR U:3]

---------------------------------------------------[JSR U:4]

---------------------------------------------------[JSR U:5]

        |||||||||||||||||||||||||||||||||||||

---------------------------------------------------[JSR U:68]

---------------------------------------------------[END]

If Ladder 68 is the last subroutine in your program.

That's normally how I've seen it done, but there may be a reason in your program why the sequence is conditional, or ordered differently, post your file and let us have a look.

Dustin

 

[showshocka]

The JSR instruction needs to come before the END instruction

I put the JSR at the end and this is what it keeps giving me, also not letting me go online. it's looping me around to this box.

 

[jrwb4gbm]

Go to the RSLogix Emulate 500, Click on the HLT button, then Download your program, this should break the looping. You also need to verify the entire project, you have some shorting branches in file 20. Shorting branches are not allowed. If you want to do away with the contact that has the shorting branch, first, Click on the lower left corner of the shorting branch, then, Right Click-> Delete. Next , click on the contact, Right Click-> Delete. Repeat as necessary.
Forcing the input won't work unless the Forces Enabled is on, try Right Click->, then, Toggle Bit instruction instead.

PS:
Please note that all your JSR rungs are at the Top of LAD2 not at the Bottom just before the END instruction in the file you posted.

 

[ddnnis]

I did a toggle on I:6/11, and timer T4:15 starts timing.

 

[ddnnis]

I think you need to make sure you select the enable all forces option after you force it on.

 

[showshocka]

with all you guys help it should have been no reason the program was not working. i simply rebooted my computer and reloaded everything and worked on the first try, maybe because the computer has been on for weeks without rebooting. thanks for all your help guys!!!!! I'm coming far from be being totally lost. i'll be posting the new change when i redesign the rung.

 

[showshocka]

I did a toggle on I:6/11, and timer T4:15 starts timing.

When i toggle I:6/11. my timer starts timing. i have my T4:15DN bit sending a 1 to B3:6/2(L). once i forced off my I:6/11... over in run 15 my B60:1/0 when toggled, it will not reset B3:6/2(U). why is this? not until i added a test input bit I:12/0, if i toggle I:12/0 then my B60:1/0 will reset. i mean in regular operations B60:1/0 push button bit will reset those other bits that needs unlatching. why not now...?

 

[showshocka]

why will 0:3/5 not force on also?

 

[MarkNightingale]

When you are toggling B60:1/0, there is an OTE instruction which is turning this bit off as soon as you toggle it.

Try making the rung with the OTE true and you will see the bit unlatch.

Mark

And I believe the same thing is happening for your output. Go to the OTE for the output and force it on there.

 

[showshocka]

When you are toggling B60:1/0, there is an OTE instruction which is turning this bit off as soon as you toggle it.

Try making the rung with the OTE true

ok, that worked for both, for the output i had to got to the 2 timer instructions and make them 0.