Skip to Content

Register Login

Industry Online Support

Technical Forum

Navigation

Need explanation of this STL program

Created by: Sasongko at: 6/26/2012 6:42 PM (8 Replies)

Rating (3)

Thanks 7

9 Entries

6/26/2012 6:42 PM	Rate (0)
Sasongko Posts: 87 Rating: (0)	Dear Expert, I don't know to much about S5 programming. Can you explain this STL program :L FW 200 :SRW 3 :FDG :LWD =FAKT :xG :L KG +5000000+00 :+G :GFD :T =AIST : :BE Many thanks
Sasongko Posts: 87 Rating: (0)
Suggestion To thank Quote Answer

6/26/2012 8:51 PM	Rate (1)
BeVee Gold Member Joined: 7/18/2011 Last visit: 10/14/2024 Posts: 521 Rating: (91)	Hi Sasongko, :L FW 200 // Load MW 200(16bits) :SRW 3 // Shift 3 bits to right ( = division by 8) :FDG // convert to floating point format :LWD =FAKT //Load inputparameter in floating point format :xG//Floatingpoint multiplication :L KG +5000000+00// Load .5 floating point constant :+G // and ADDit to result of multiplication (for rounding purpose) :GFD //convert result to DINT format :T =AIST // transfer result to output parameter : :BE // block end Kind regards, BV

Suggestion To thank Quote Answer
This contribution was helpful to 2 thankful Users Sasongko E.Ridy

7/3/2012 9:21 AM	Rate (0)
Sasongko Posts: 87 Rating: (0)	Dear Beevee, Can you help me once again? This is the actual Program/function Block with CPU 6ES5926-3SA12 and AI 6ES5464-8MC11. I think this program to process analog input/output. I don't know what the purpose of this program/function block. PB 121 (with FB 205): : DO DB 161 : : L PW 154 : T FW 200 : : A F 200.7 : = F 255.0 : : AN F200.7 : JC M001 : L FW 200 : CSW : T FW 200 M001 : L FW 200 : SRW 3 : FDG : LWD +1000000+01 : XG : L KG +5000000+00 : +G : GFD : T DW 70 : : BE Thank you very much.
Sasongko Posts: 87 Rating: (0)
Suggestion To thank Quote Answer

7/3/2012 9:58 AM	Rate (1)
fritz Diamond Expert Joined: 10/7/2005 Last visit: 11/16/2024 Posts: 3027 Rating: (1057)	Hello Sasongko it's actually quite simple, but you need to appreciate the structure of the S5 Analogue input value (see also attached pic and grab also a copy of the S5-90U/S5-95U Programmable Controller System Manual which contains your AI card). The mostsignificant bit (bit 15) is the sign bit. The least three significant bits (bit 0, 1 and 2) contain overflow and error bits. Your Block is a simple analogue input "scaling" FB which actually doesn't do all that much useful stuff, but here's a commented version of it anyhow: : DO DB 161 // open DB161 : : L PW 154 // load raw AI value : T FW 200// and store in FW200 : : A F 200.7 // If raw AI value is negative(Sign bit = 1) : = F 255.0 // memorise this : : AN F200.7 // If raw AI is positive(Sign bit = 0) : JC M001 // --> Jump : L FW 200 // Else: (raw AI value is negative) : CSW // do two's complement (= make negative value positive) : T FW 200 // and store it back in FW200 M001 : L FW 200 : SRW 3 // get rid of overflow, error and unused bits 0-2 : FDG // convert to floating point value : L KG +1000000+01 // load 1.0 : XG // multiply raw AI with 1.0 : L KG +5000000+00 // load 0.5 : +G // add 0.5 to above result : GFD // convert to Double Integer : T DW 70 // storevalue in DW70 (of DB161) I hope this helps Attachment S5 AI value representation (95U manual).jpg (841 Downloads)
	Cheers fritz The advice and opinion given in this thread is that of the author and does not necessarily reflect the views of I S Systems Pty Limited. To find out how I S Systems can help you with your automation needs please visit www.issystems.com.au.
Suggestion To thank Quote Answer
This contribution was helpful to 2 thankful Users Sasongko E.Ridy

7/3/2012 11:17 AM	Rate (0)
Sasongko Posts: 87 Rating: (0)	Dear Fritz, I am sorry for my foolish. Is FW 200 = FB 200 + FB 199 ? Many thanks
Sasongko Posts: 87 Rating: (0)
Suggestion To thank Quote Answer

7/3/2012 12:11 PM	Rate (0)
BeVee Gold Member Joined: 7/18/2011 Last visit: 10/14/2024 Posts: 521 Rating: (91)	Hi Sasongko, FW200 = FY200 + FY201 or in German mnenotics MW200 = MB200 + MB201 Kind regards, BV

Suggestion To thank Quote Answer

7/3/2012 12:32 PM	Rate (0)
Sasongko Posts: 87 Rating: (0)	Dear BeVee, So, Why Fritz said: : A F 200.7 // If raw AI value is negative (Sign bit = 1) Based on the attached pict, Sign bit at bit 7 on High Order Byte. Can you give explanation? Many thanks
Sasongko Posts: 87 Rating: (0)
Suggestion To thank Quote Answer

7/3/2012 9:24 PM	Rate (1)
PLCforEver Posts: 99 Rating: (3)	Dear Sasongko, As per bit representation of picture "S5 AI value representation....jpg" in previous answer, in Siemens PLC, the byte equivalent of the word adress is filled from left to right. So High-Order Byte will be FY200, and Low-order Byte willbe FY201. And F200.7will match the position of the bit sign. As an easy reminder,if we make a byte representation of the beginning of the Flag area PLC memory wehave this: FY000, FY001 FY002, FY003 FY004, .... Best regards
PLCforEver Posts: 99 Rating: (3)
Suggestion To thank Quote Answer
This contribution was helpful to 1 thankful Users Sasongko

7/4/2012 3:41 AM	Rate (0)
Sasongko Posts: 87 Rating: (0)	Dear BeVee, Fritz, and PLCforEver Thank you very much for your help and knowledge. I know more about Siemens process now.
Sasongko Posts: 87 Rating: (0)
Suggestion To thank Quote Answer
This contribution was helpful to 2 thankful Users fritz BeVee

|

Share this page:

Share this page on...

Follow us on