Omron C200HX Operation Manual page 404

I/O Response Time
Maximum Response Time
Cycle time
PC with
Unit 0
Buffer in Unit 0
PC Link Unit
transmissions
Buffer in Unit 7
PC with
Unit 7
Input
Output
386
Output ON delay:
Cycle time for PC of Unit 0:
Cycle time for PC of Unit 7:
Minimum transmission time:
The following diagram illustrates the data flow that will produce the maximum
response time. Delays occur because signals or data is received just after they
would be processed or because data is sent during processing. In either case,
processing must wait until the next scan/polling cycle.
First output to the buffer in the polling unit is delayed by the setting of the number
of LR bits to be refreshed each scan. A similar delay is present when the LR data
reaches Unit 7. The polling delay is the result of the LR data in its PC being up-
dated immediately after the previous was sent to the buffer in the PC Link Unit,
cause a delay until the next polling cycle. One more polling cycle is then required
before the data reaches the buffer in PC Link Unit 7.
I/O refresh
Polling delay
Maximum
transmission
time
The equation for maximum I/O response time is thus as follows:
Response time = Input ON delay + [Cycle time of PC of Unit 0 × (Number of LR
transfer bits ÷ I/O refresh bits)] + α + (PC Link polling time +
Induction sequence processing time) + {Cycle time of PC of
Unit 7 × [(Number of LR transfer bits ÷ I/O refresh bits) × 2 + 1]}
+ β + Output ON delay
If cycle time of PC of Unit 0 > PC Link polling time, α = cycle time of PC of Unit 0. If
cycle time of PC of Unit 0 < PC Link polling time, α = PC Link polling time.
If cycle time of PC of Unit 7 > PC Link polling time, β = cycle time of PC of Unit 7. If
cycle time of PC of Unit 7 < PC Link polling time, β = PC Link polling time.
Inserting the following values into this equation produces a maximum I/O re-
sponse time of 661.3 ms.
Input ON delay:
Output ON delay:
Cycle time for PC of Unit 0:
Cycle time for PC of Unit 7:
PC Link polling time:
Induction sequence processing: 15 ms × (8 PCs – 8 PCs) = 0 ms
15 ms
20 ms
50 ms
2.8 ms+10 ms=12.8 ms
PC Link
polling time
Induction sequence
processing time
Cycle time
1.5 ms
15 ms
20 ms
50 ms
2.8 ms × 8 PCs + 10 ms = 32.4 ms
Section 6-4