Cycle and reaction times
5.6 Sample calculations
CPU 31xC and CPU 31x, Technical data
5-24 Manual, Edition 08/2004, A5E00105475-05
5.6 Sample calculations
5.6.1 Example of cycle time calculation
Installation
You have configured an S7300 and equipped it with following modules in rack "0":
• a CPU 314C-2
• 2 digital input modules SM 321; DI 32 x 24 VDC (4 bytes each in the PI)
• 2 digital output modules SM 322; DO 32 x 24 VDC/0.5 A (4 bytes each in the PI)
User program
According to the Instruction List, the user program runtime is 5 ms. There is no active
communication.
Calculating the cycle time
In this example, the cycle time is equivalent to the sum of the following times:
• User program execution time:
approx. 5 ms x CPU-specific factor 1.10 = approx. 5.5 ms
• Process image transfer time
Process image of inputs: 100 μs + 8 Byte x 37 μs = approx. 0.4 ms
Process image of outputs: 100 μs + 8 Byte x 37 μs = approx. 0.4 ms
• Operating system runtime at the scan cycle checkpoint:
approx. 0.5 ms
Cycle time = 5.5 ms + 0.4 ms + 0.4 ms + 0.5 ms = 6.8 ms.
Calculating the physical cycle time
• There is no active communication.
• Interrupts are not processed.
Hence, the physical cycle time is 6 ms.
To Next Page
Loading...
Need help?
General
