A beginner problem with the latch

alyelhefni · Apr 24, 2013

Hi guys

It's my first time to write a ladder diagram program

I faced a problem that the latch doesn't work and I don't know why

please view the attached image that explains the problem

thanks in advance

Lancie1 · Apr 24, 2013

Alyelhefni,

The logic that you show does not use a Latch instruction, but instead trys to use a seal-in method to keep M6.0 ON or sealed in, once it has been turned on by I0.0

However the dashed lines at | |--- seem to indicate that there is a problem with your logic. It could be that the dashed line indicates that there is a missing connection at this point. Maybe you need to draw in that line to complete the circuit.

On the other hand, coud the program error handler be trying to show that M6.0 is being turned OFF somewhere else?

alyelhefni · Apr 24, 2013

I've kept trying various things and I came up with a partial solution to the problem :

The program works when I use memory bit of 124.1 124.2 124.3 .... etc

I don't know why 6.0,1,2,...etc and 1.0,1,2....etc are "bad" and 124.number is accepted

There must be a method where I assign or declare memory bits numbers or names I want

but I cant find that >>> I'm using siemens symatic manager s7 if anyone could help

Lancie1 · Apr 24, 2013

I don't know why 6.0,1,2,...etc and 1.0,1,2....etc are "bad" and 124.number is accepted.

Different models of PLCs have different amounts of memory in different locations. You cannot AssUMe that all memory locations are available for all instructions in all models of the Siemens S7 PLCs.

What model of Siemens PLC do you have selected for your program to control?

In general, the "I" memory type is reserved for Siemens PLC Inputs, which are very limited in number, and depend totally on the model of PLC. Some have fewer, some have more Inputs. The "M" type of memory is usually for internal relays and flag bits. In Siemens, Outputs are often the "Q" memory type. Check your manual for your model number to see what memory areas are available for each type. It may be difficult to find the right Siemens manual that will show what you need to know.

cbuysse · Apr 24, 2013

I only have limited experience with Siemens S7 programming myself, but my first thought is that maybe you've used the 6.0 memory address elsewhere in the program. Don't forget to check for MW 6 or MB 6.

JOLTRON · Apr 24, 2013

Right click on the tag and select "Goto" -> "Location" this will pull up the cross reference data for that memory bit. In the lower left corner of the box I believe there is a check box for "Overlapping memory locations" this will show you if this bit is used in another location as a byte / word / dword etc..

alyelhefni · Apr 25, 2013

Well this haven't solved the problem yet

I made a simple exercise to spot out the problem
please check the 2 cases in the following pictures
the whole program is those 3 branches

figure 1

figure 2

it's obvious that the memory bit here turns zero if it's located as an output in other branches and it doesn't get a one , it kinda resets it self to zero again

thats really frustrating because like this its impossible to get back to a certain state after leaving it

how can I solve this ?

bernie_carlton · Apr 25, 2013

It looks like the double output syndrome. DO NOT use as an output ( ) an address in more than one place. The last rung which has the bit as an output 'wins'. Its logical truth is the last one applied before the rung which uses a contact of that bit. If you feel you need to use complex rungs combine the logic to drive just one instance of the output bit.

alyelhefni · Apr 25, 2013

I've understood that now that for a single address output the last always wins
and therefore I shouldn't include same address as an output in more than one place
That really explains why the program wasn't working well

But I really can't figure another way to get back to old states
meaning
for example : I push the start button and motor1 runs while motor 2 is off.... so this is state 1
and then when a sensor gives 1 motor 1 stops and motor 2 starts working .... and that is state 2

when sensor 2 gives one stop motor 2 and run motor 1 ..... and that is the original first state (state 1)

so how can I write such a ladder diagram without repeating same addressed outputs ?

harryting · Apr 25, 2013

You are jumping over the states all together.

Let's think about this. Say you want to water the lawn on Monday and prune the tree on Tuesday. Do you have any problem doing that? No. Because your calendar tells you what day it is.

You need something to keep track of the day, or in this case, the state. As always i would suggest new programmer to write in pseudo code rather jump right into ladder logic. In pseudo code:

If State1=true then turn on pump1 AND turn off pump2
If State2=true then turn off pump1 AND turn on pump2

Then you have separate logic (calendar?) to keep track of the state (day):

If Button=true then state1=false, state2=true

Now, there's a hundred different way you can keep track of states. You can find some of those by search for "sequencer". You can use an interger, for example, State=1,2,3.4.... and so on

jrwb4gbm · Apr 25, 2013

alyelhefni said:
I've understood that now that for a single address output the last always wins
and therefore I shouldn't include same address as an output in more than one place
That really explains why the program wasn't working well

But I really can't figure another way to get back to old states
meaning
for example : I push the start button and motor1 runs while motor 2 is off.... so this is state 1
and then when a sensor gives 1 motor 1 stops and motor 2 starts working .... and that is state 2

when sensor 2 gives one stop motor 2 and run motor 1 ..... and that is the original first state (state 1)

so how can I write such a ladder diagram without repeating same addressed outputs ?

I will be happy to help but right now, I'm just trying to understand what you actually want the program to do.
Here is a short list of questions.
What causes the sensors to change state?
When do the sensors change state?
Are they floats on a tank or something else?
Is motor 1 and motor 2 pump motors or something else?
What condition do you want to stop the motors?
After both have run or just at any time?

alyelhefni · Apr 25, 2013

Sure

the whole thing is I'm a last year student and I'm asked to write PLC programs in ladder diagram

so after finishing I wanted to check if it really works
I went to the software and wrote it and faced problems that I'm discussing on this thread

here is the problem jrwb4

alyelhefni · Apr 25, 2013

bloodredsoul · Apr 25, 2013

could you gimme a download link for your version of the program? i will come up with a simple ladder diagram for you as reference for how yours should look like

jrwb4gbm · Apr 25, 2013

I dont know about Siemens software. Here is just something in LogixPro I threw together, mostly just some ideas based on the thread posts. Maybe it will help get you started in a direction.

A beginner problem with the latch

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