_Pavel_ Posted April 29, 2017 Report Posted April 29, 2017 (edited) Hello All, I used CJ2 PLCs where program capcity is given in Ksteps. After few projects when ROM was unexpectedly over, I have cought, that 5Ksteps is too little and say 20Ksteps is enough. Now I'm selecting NJ PLC where it's given in Mbytes of program memory. So 3MB of program memory - is it much or few? How to compare Ksteps and Mbytes of program memory? Thanx! Edited April 29, 2017 by _Pavel_
Crossbow Posted May 1, 2017 Report Posted May 1, 2017 I'd swear I have seen a guideline on this somewhere but can't locate it at the moment, if I find it I will reply again with details.
Michael Walsh Posted May 2, 2017 Report Posted May 2, 2017 An Omron engineer did some testing on this and determined that 1.5 MB of program memory in an NX1P (same family of controllers) is roughly equivalent to 100k steps on a CJ2H processor. If I can find more info or specifics on NJ testing, I will add to the discussion.
_Pavel_ Posted May 11, 2017 Author Report Posted May 11, 2017 Thanks for the replies! I"ve been searching myself as well and I've just found here https://www.ia.omron.com/products/family/3111/specification.html 5 MB (100 KS)
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