dahnuguy
Back in post #24 (page 2) you wrote....
"You need AN unlatch somewhere"
True.
And a false OTE will not unlatch a latched bit in my experience..........
Sorry to have to correct you here, but a false OTE will certainly unlatch a "latched" bit.
Whether it stays off is entirely dependant on the rung condition(s) driving the OTL instuction when it is next scanned. It may get "latched" on again if the rung is true.
I believe you have made the assumption that the bit actually knows that it was latched, or unlatched, and will not respond to an OTE instruction. That simply is a false assumption. Each bit is written to by the execution of the ladder at the time it is encountered during the program scan. Whether a bit is written or not is dependant upon the instruction it is attached to.
I deliberately put speech marks around the word "latched" above, to emphasise the point that the bit
does not know that it was latched, it is simply written to by the logic processor according to the rules of the instructions.
OTE - writes 1
or 0 dependant on the "logic-continuity" of the rung - True or False respectively
OTL - writes 1 only, and only if the rung is true
OTU - writes 0 only, and only if the rung is
true
Here is an example by way of explanation....
Say rung 50 has an OTL, and the rung is true, then the ladder processor will write a "1" to the data bit. If the rung goes false, the OTL instruction does nothing to the bit.
If absolutely nothing else addresses that bit (as a rung output), then it will stay latched forever, even if the rung with the OTL goes false. In other words, an OTL instruction only writes "1"s.
Now, say, rung 60 has an OTU of the same bit, and that rung is also true, then the ladder processor will write a "0" to the bit. Likewise as with the OTL, an OTU instruction can only write "0"s to the data bit.
Now, imagine both rungs 50 and 60 are true, then what will happen is that rung 50 will turn ON the bit, then rung 60 will turn it off again. If you are inspecting the bit asynchronous to the program scan (eg. RSLinx) you will see it both on and off, at different times.
Oh and finally, as i said, a
false OTE will "unlatch" a bit, but so will a
true OTU, and the latter is cheaper in terms of memory usage and execution time, since you need another instruction on the rung to make the OTE false.
Sorry to rant, but I want everyone to understand the priciples of how the processor scans the logic, writing to the data tags as it goes.