ladder logic question

It seems to me that You ought to implement flip flop type D.

SOR XIC I:002/4 ONS B3/10 OTE B3/0 EOR SOR BST XIC B3/0 NXB XIC B3/1 BND BST XIO B3/1 NXB XIO B3/0 BND OTE B3/1 EOR

That should do it for you.

Stanislaw,

Don't you mean set-reset?

bakerjw2000,

Does the device repertoire not include a toggle switch? If not, you can do what you want with two pushbuttons and either a set-reset flipflop or a latching relay. (You can make a latching relay from a single two-pole relay a normally-closed, and a normally-open pushbutton.)

Jerry

No, one input, output divide by 2. For set-reset You must use two inputs of PLC.

...

Thanks for explaining. There are toggle flip-flops that aren't D types. With your arrangement, I would demand an indicator output to make the state of the toggle known. Working blind is not good practice.

Jerry

I arrange my branches differently, but the effect is the same (corrected for SLC addressing & instructions). The trick is remembering that ladder logic executes instructions one at a time. SOR XIC I:1.0/0 OSR B3:0/10 OTE B3:0/11 EOR SOR BST XIC B3:0/11 XIO B3:0/12 NXB XIO B3:0/11 XIC B3:0/12 BND OTE B3:0/12 EOR

SOR XIO B3:0/13 OTE B3:0/13 EOR alternates on every scan.

Mike

With the advice of a flip flop, I was able to locate a forum that addressed multiple methods of performing the operation that I was tring to perform. i.e. Press a button and a contact turns on. Press it again and the contact goes off.

Really need a monospace font to view the ladder. PB Target R1

--] [-+-]/[--+--------------------( )- | | R1 | PB |

--] [-+-]/[--+

PB R1 Target

--]/[-+-] [--+--------------------( )- | | Target| PB |

--] [-+-] [--+

Thanks for all of the advice. It always helps to see multiple ways of doing things.

That's a good way to do it.

I thought that your method might save a word or two of memory, but both your example and mine compile down to 13 instruction words (in a SLC500, anyway.)

Actually, doing it the following way uses only 10 words of memory, but by the time you add a rung to examine C5:0.ACC/0 you are back up to 13 words, and then you are using 3 rungs, so I generally avoid it.

SOR XIC B3:0/0 OSR B3:0/10 CTU C5:0 2 0 EOR SOR XIC C5:0/DN RES C5:0 EOR

Thanks for the tip.

I see you found one of the

threads on flip flops. Search for tidbits and toggles too. There are many threads about this on because unlike you, the rookies can't search and think they are the first to have this problem. BTW, beat this.

Output = Output xor ( Input and not LastInput ) LastInput = Input

This would translate into 4 instructions. You can get it down to 3 if you change LastInput to NotLastInput

Output = Output xor ( Input and NotLastInput ) NotLastInput = not Input

This takes advantage of the fact that the not instruction can negate and move at the same time. The downside is that NotLastInput must be initialized at first scan. Perhaps initializing is necessary anyway.

Rockwell does not provide a bitwise xor or alternate function except on the 5000 family. I use word or dword functions to process 16 or 32 bits at a time anyway instead of processing the bits one by one.

Huh? Just pick an address (some internal bit) like %V115.3 or what ever your application uses.

The pushbutton address will be something like X22.3 depending on input config.

Your kidding right?

Look it up on the net

Daveb

Sorry, I missed the point of the question

Daveb

...

I often find that what at first seems like a dumb question is in fact a deficient reading of the question. I'd embarass myself more often if I didn't realize that.

Jerry

Funny thing is, everytime I do exactly what he was asking I always say hmm how did I do it last time.

Regards Daveb