proworx Nxt

'lo all. I have been programming with A&B for many moons. And now I have my first Modicon project. I am looking for some help to see if I understand this correctly. In the attached pic am I correct by saying that rung 4 will never execute but rung 5 will?

First off, you don't have the timer configured correctly. As shown the timer will NEVER time. You must provide an enable input to the timer as well as a control input. This can be as simple as connecitng together the two inputs with a | short.

Now for the question:
Here is how the network is solved - ie, it is solved by columns. I'm not sure what you mean by saying rung 4 or 5 does or does not execute. Every cell in the network, even blank cells, is solved. Do you mean to ask whether coils 00022 and 00023 will ever be true? The answer is yes, they will be true, but for only one scan during the scan immediately following the timer reset by the NOT 00020 instruction in front of the timer (assuming that you are trying to enable and reset the timer by the 00020 coil).

The way I see it, niether rung 4 or 5 will ever fire the outputs.
Remember, the modicon solves vertically, not horizontal. I set this up in ProWorx and ran it in emulation. 4 nor 5 are turning on the outputs.

I thought that rung 4 would never execute because the XIC 00021 is in column 3, and since the coil is not evaluated/solved (or so I believe) until column 4 is scaned then only rung 5 would ever solve.


I think I see what you are saying. Coil 23 will be true on the same scan as when 21 is true, but coil 22 will be true on the following scan, because 21 will not be reset until column 4 is evaluated on that second scan.

Thanks for your help this is a little confusing.

On the scan where the timer at 2,2 (40001) reaches its preset it will pass power along the top of the rung. When column 3 is solved, cell 3,2 will solve true. 3,4 will solve false (we havent scolved column 7 yet to set 00021) – guaranteeing that 00022 will not turn on this scan. The same holds at cell 4,5. When column 7 is solved then 00021 and 00020 will turn on.



The following scan, cell 1,2 will evaluate false, causing the timer to reset when the second column is solved. It is important to remember that in the bit table bits 00021 and 00020 are still set. They will remain set until column 7 of the network is scanned. Thus when cell 3,4 and cell 4,5 are solved, the solver checks the bit table, sees that bit 00021 is still on, and solves those cells as true. Coils 00022 and 00023 will be set on. Now when column 7 is finally solved, bit 00021 will be reset.



On the next scan cells 3,4 and 4,5 will solve false.



Bits 00022 and 00023 are on for one scan only on the falling edge of the timer enable.

Does it not matter where I place the XIC if I am using the horizontal short like in rung 4 and 5? Does that mean Coil 22 and 23 are actually in the 2nd column? And what I was trying to point out would only be valid if I had occupied the two/three nodes prior to the XIC 21 in rung 4 and 5 with other instructions.

Since the horizontal shorts consumes a node, that stretches out the coil evaluation. A more accurate diagram would be:

A coil is ALWAYS solved in the column in which it is placed - no exceptions, no ifs-ands-or-buts. If you place it in column 7 and have horizontal shorts in columns 1-6 then it is still solved in column 7.

Be aware that there are two options in Proworx to display all coils in column 7 or in the column in which they are placed. If you pick the first option then even though you place a coil in colunn 3 it will be displayed in column 7 with a dotted line ------ from column 3 thru column 6. The coil is however still ALWAYS solved in column 3.

Thanks for your help

Unlike other PLCs, a network in a Modicon processor always has a fixed size. The size of the network does not change with the instructions placed in it. A network occupies 77 words (plus some overhead). Always, whether it has one instuction or 77 instructions in it. When you add a network memory is allocated for a 7 x 11 array. Instruction Op codes are inserted into positons of the array exactly where you put them. When a newtork is solved, each and every square in the array is solved, even if it is empty. The solver looks to the square to the left, if it is false then the square will also be false. If the square to the left is true then the solution result for the square will be based on the instruction contents of the square. If the square to the left has a horizontal short and the square to the left is true, then the current square is true. The horizontal short can be treated as an instruction which is an ALWAYS TRUE instruction.