TIP: S7-1500 - cycle and response times

Pegaia

I corrected the open text.

Dear Pegaia,

First of all, very big thank you for your big effort to make people understand how a "response" time calculated on PN network. I could see its very old TIP, unfortunately now only am came across it. I was trying to understand my self how the calculation is done. Now its very clear. I have small doubt in below calculation.

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