3/23/2013 10:21 AM | |
Joined: 9/23/2005 Last visit: 1/14/2025 Posts: 4392 Rating: (1463)
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Hi, to better unterstand the theory, here there is a practical example: Buttom -> DI module (ET200S)-> IM151-3PN (ET200S)-> PROFINET->S7-1511->program (on open contact + one coil)->DO (central rack) -> main contactor 1x Buttom delay (0 ms - as definition) 1x DI Input delay (4DI DC24V HF 6ES7131-4BD01-0AB0 http://support.automation.siemens.com/WW/view/en/25388101 page 10) = set as 0.1 ms => worst case = 0.15 ms 2x Time for IM read the signal (IM151-3 PN HIGH FEATURE 6ES7151-3BA23-0AB0 http://support.automation.siemens.com/WW/view/en/30609285, page 55): Response time [μs]: 390 + Maximum of (380 + 9m + 11do) or (24m + 40ai + 80t) + Maximum of (120 + 9m) or (24 + 9do + 40ao + 80t) In our case: m=2 (PM-E+DI), do=0, ao=0, ai=0, t=0 => 390 +max [(380+2*9+0), (24*2+40*0+80*0)]+max[(120+9*2),24+9*0+40*0+80*0)= 390+max[380,48]+max[138,24] =390+380+138=908 μs = 0,908ms 2x PROFINET IO time =2ms (defined by programmer) 2x CPU internal time (“SIMATIC S7-1500, ET 200MP, ET 200SP Cycle and response times” http://support.automation.siemens.com/WW/view/en/59193558) as page 25: Internal response time = 2 x transfer time of the process image inputs + 2 x processing of the user program + 2 x transfer time of the process image outputs Transfer time of the process image inputs, as page 16: 35us*1+0,5μs*1 = 25,5μs= 0,0255ms Response time for program: 1 open contact + 1 coil = 2 binary instruction = 2 *60ns = 120ns=0,12μs=0,00012ms Transfer time of the process image inputs, as page 16: 35us*1+9 μs*1 = 42 μs = 0,042ms Total CPU internal time without communication last = 0,0255ms + 0,00012ms + 0,042ms = 0,06462 ms Communication load = 15% (defined by programmer) Total CPU internal time with communication last 0,06462 ms *100/(100-15) = 0,076 ms 1x DO Output delay (6ES7522-1BF00-0AB0DQ 8X24VDC/2A HF http://support.automation.siemens.com/WW/view/en/59193089 pg22) = 500 μs = 0,5ms 1 x main contactor (?... I define as 5ms) Total response time = 1*(0+0,15)+2*(0,908+2+0,076)+1*(0,5+5)= 11,618 ms |
Last edited by: Pegaia at: 3/23/2013 1:08 PMDenilson Pegaia |
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This contribution was helpful to4 thankful Users |
3/25/2013 10:18 AM | |
Posts: 14 Rating: (0) |
Thank you for clarifying me a few important things. I'll address my further questions to correct conference.
matus |
4/9/2013 10:07 PM | |
Joined: 9/23/2005 Last visit: 1/14/2025 Posts: 4392 Rating: (1463)
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Hi, The internal response time for ET200MP could be found in the Apendix B of the manual "ET 200MP IM 155-5 PN ST interface module (6ES7155-5AA00-0AB0)" at http://support.automation.siemens.com/WW/view/de/59193106/0/en PS: Thanks for Mr Andreas Kraft for the tip. |
Denilson Pegaia |
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