AB SLC 5/05 Compare

averytc

Member
Join Date
Jul 2002
Posts
114
Ok, I’ve run into something I’ve not seen before. It would seem strait forward, but. I’m comparing two integers N13:10 & N13:11 using the EQU instruction. When they are the same value, the rung should be true and a timer starts. I want the timer to reset when they are not true. This is the way it’s supposed to work, right?? I’m looking at the timer active with the enable bit and done bits set.

I've never seen this before. Am I missing something?

Thanks,
Charlie

Edit, I removed the drawing. How do you all get the line drawings to look right? Mine looked good until I posted it.
 
Last edited:
To be exact, the processor is a 1747-L552B/C CPU 32K Mem. OS501 Series C. I've just tryed the same thing on an identical processor, and it works the way I would expect. Has anyone seen a processor do this before?
 
Try this:

Code:
equ N13:10 N13:11 ton t4:0 .01 0

Obviously, you'll need to change the timer register to whichever timer you are using...
 
Ozee, thanks for your reply. I just found where I was already using N13:10 indirectly... I've changed it and all is well.

It's been that kinda week!!

Thanks again.

Charlie
 

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