7/4/2022 7:58 AM | |
Joined: 9/2/2021 Last visit: 8/31/2024 Posts: 26 Rating:
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Dear Mr.dchartier, Thank you very much for your answer. But the problem is that when a segment of the program is activated, the PLC output is not activated. For example, I have made screen shots to the problem and insert them in the file "The my problem.docx". At the Fig.1 in the program block "Regimes_Changes", Network 4 is activated the program block "MAN" (Fig. 1). At the Fig.2 in the program block "MAN", Network 29 is activated the output D44 Q - (Address %Q12.2) . End finally the output D44Q (Address %Q12.2) - is not active. This is shown at Fig.3 "PLC tags". I have attached the file The my problem.docx with Fig.1, Fig.2 and Fig.3 inside. Please, loock at figures attached in docx format. Thank you Yordan Chervenkov AttachmentThe my problem.docx (393 Downloads) |
7/4/2022 8:23 AM | |
Joined: 9/2/2021 Last visit: 8/31/2024 Posts: 26 Rating:
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Dear bit7, I've tried without success. I'll try again, but it's straight Thank you very much Yordan Chervenkov |
7/4/2022 8:31 AM | |
Joined: 1/21/2013 Last visit: 9/11/2024 Posts: 3093 Rating:
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Hi Yordan Chervenkov, as Daniel Chartier has already mentioned, the last write access to the PLC output is crucial. Once each PLC cycle, the process image of the outputs will be written to the hardware. This happens at the beginning of a cycle. When monitoring a program block, the values written in the seen network / program line are shown. Since it is possible to write the same variable later again, the process image can be different to your expectation. When monitoring a PLC tag list, the value of the process image at cycle start should be shown. And as I mentioned above, these are the values written to the hardware. To be absolutely sure, you can use a watch table in extended mode and select "Permanently, at end of scan cycle" for the trigger. You should take a look at the cross-references for your output tag. In the worst case, you have to take a look at all write accesses for the tag. In your screenshots, the data blocks "MAN_DB" and "MAN_DB_1" are shown. This looks like you have two instances of your function block "MAN". If so, you must ensure, only one of the instances is active. Kind regards |
7/4/2022 9:58 AM | |
Joined: 6/19/2019 Last visit: 9/5/2024 Posts: 697 Rating:
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Try press directly on output to see cross reference (F11) and maybe somethere else it's written in other blocks |
7/5/2022 10:03 AM | |
Joined: 9/2/2021 Last visit: 8/31/2024 Posts: 26 Rating:
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Dear Mr.V, I used the advice above but to no avail. With kind regards Yordan Chervenkov |
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