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The PC takes longest to respond when it receives the input signal just after the
input refresh phase of the cycle. In this case the CPU does not recognize the
input signal until the end of the next cycle. The maximum response time is thus
one cycle longer than the minimum I/O response time.
Cycle time
Input
signal
Output
signal
Cycle
CPU reads
input signal
Output ON delay
Input ON delay
Program execution
Program execution
I/O refresh
Maximum I/O response time =
input ON delay + (cycle time x 2) + output ON delay
Maximum I/O response time = 1.5 + (20 x 2) +15 = 56.5 ms
6-5-2 I/O Response Times in a SYSMAC BUS System
Here, we’ll compute the minimum and maximum I/O response times for a PC
controlling a SYSMAC BUS System. Both the input and output are on I/O Units
connected to Slave Racks. SYSMAC BUS refreshing is carried out just after I/O
refreshing as one phase of the PC cycle.
The transmission time for a Master is the sum total of the transmission times for
all Slaves connected to it. The transmission time for Slave Racks is 1.4 + (0.2
×
a)
ms, where a is the number of I/O words on the Slave. The transmission time for
I/O Terminals is 2
×
b ms, where b is the number of I/O words on the I/O Terminals.
The data in the following table is used to produce the minimum and maximum
cycle times shown calculated below.
Input ON delay
1.5 ms
Cycle time
20 ms
Output ON delay
15 ms
Slave Rack transmission time
1.4 + (0.2
×
4) = 2.2 ms
I/O Terminal transmission time
2
×
3 = 6 ms
Master transmission time (T
RM
)
2.2 ms + 6 ms = 8.2 ms
Maximum I/O Response
Time
I/O Response Time
Section 6-5
Summary of Contents for CVM1D
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