S5 Help please !!!!

Join Date
Mar 2017
Location
6 October
Posts
69
Hi to all ,
I don't have a good experience in PLC ( you can say a beginner )
I had a problem with a PLC S5 135U /155U CPU 928B
The problem is i can't understand a network where it's written in STL and CSF
The Flag is F 15.2 and it's related to stopping the machine ( F15.2 is on randomly when the machine is running or stopped ) and the only way to run it back is to reset PLC
here is a picture of the PLC when it's Running or ready to run
154yqh1.jpg

2ng4mxl.jpg

2dryijk.jpg

and when it's stopped
2ev6zk8.jpg

16920kx.jpg

2yzgaht.jpg
 
Hello

IF one or more bits in FW124 OR FW126 OR FW128 are ON OR Input 18.1=OFF OR input 6.4=OFF then F15.2 goes to ON state.

In your screenshot FW124 = 8 so not equal to zere and thats why F15.2 goes to ON state.

FW124 HEX = 0 0 0 8

FW124 binary = 0000 0000 0000 1000<=this bit is F125.0
________ F124.7_F124.0______<=this bit is F125.3
_________________F125.7___F125.0
So F125.3=ON

Hope this helps you?

Kind regards
Henny
 
Hello

IF one or more bits in FW124 OR FW126 OR FW128 are ON OR Input 18.1=OFF OR input 6.4=OFF then F15.2 goes to ON state.

In your screenshot FW124 = 8 so not equal to zere and thats why F15.2 goes to ON state.

FW124 HEX = 0 0 0 8

FW124 binary = 0000 0000 0000 1000<=this bit is F125.0
________ F124.7_F124.0______<=this bit is F125.3
_________________F125.7___F125.0
So F125.3=ON

Hope this helps you?

Kind regards
Henny

Thanks Henny ,

yes it was F125.3 and i follow it and at last was a lubrication level switch :cry:

the point i learned how you determined the problem

FW 124 = Flag Word 124 = Flag ( Word = two bytes ) 124 & 125 = Flag ( one byte = 8 bit ) = F124.7 < F124.0 & F125.7 < F125.0
Change 0 0 0 8 from HEX to Binary
0 = 0000
8/2 0
4/2 0
2/2 0
1 1
8 = 1000
this lead to
0000 0000 0000 1000
compare the signal 1 to F in same place = F125.3

Hello mitakka ,

Hope i could learn more in PLC all Ver. but the way is too long and am just started
 

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