Omron C200HE ユーザーズマニュアル
383
I/O refresh bits for Unit 0
256
I/O refresh bits for Unit 7
256
Reducing Response Time
IORF(97) can be used in programming to shorten the I/O response time greater
than is possible by setting a high number of refresh bits. (Remember, increasing
the number of refresh bits set on the back-panel LED shortens response time,
but increases the cycle time of the PC.)
than is possible by setting a high number of refresh bits. (Remember, increasing
the number of refresh bits set on the back-panel LED shortens response time,
but increases the cycle time of the PC.)
In programming the PCs for PC Link Units #0 and #7, IORF(97) is executed dur-
ing every PC scan for the PC Link Units. The basic equation for the maximum I/O
response time is as follows:
ing every PC scan for the PC Link Units. The basic equation for the maximum I/O
response time is as follows:
Response time = Input ON delay + [Cycle time of PC of Unit 0
×
(Number of LR
transfer bits
÷
Number of I/O refresh bits
÷
2)] +
α
+ PC Link
polling time + Induction sequence processing time + {Cycle
time of PC of Unit 7
time of PC of Unit 7
×
[(Number of LR transfer bits
÷
Number of
I/O refresh bits
÷
2)
×
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.
The required data from the example System configuration is as follows:
Input ON delay
1.5 ms
Output ON delay
15 ms
Cycle time of PC of Unit 0
20 ms + 5.7 ms = 25.7
(5.7 ms required for IORF execution)
Cycle time of PC of Unit 7
50 ms + 5.7 ms = 55.7
(5.7 ms required for IORF execution)
Number of PC Link Units
8
Number of LR bits
1,024
Number of refresh bits for Unit 0
256
Number of refresh bits for Unit 7
256
PC Link polling time
2.8 ms
×
8 PCs + 10 ms = 32.4 ms
Induction sequence processing time
15 ms
×
(8 PCs – 8 PCs) = 0 ms
Placing these values into the equation produces a maximum I/O response time
of 466.9 ms, approximately 200 ms shorter than when IORF is not used.
of 466.9 ms, approximately 200 ms shorter than when IORF is not used.
6-4-5 One-to-one Link I/O Response Time
When two C200HX/HG/HEs are linked one-to-one, the I/O response time is the
time required for an input executed at one of the C200HX/HG/HEs to be output
to the other C200HX/HG/HE by means of one-to-one link communications.
time required for an input executed at one of the C200HX/HG/HEs to be output
to the other C200HX/HG/HE by means of one-to-one link communications.
One-to-one link communications are carried out reciprocally between the mas-
ter and the slave. The respective transmission times are as shown below, de-
pending on the number of LR words used.
ter and the slave. The respective transmission times are as shown below, de-
pending on the number of LR words used.
Number of words used
Transmission time
64 words (LR 00 to LR 63)
39 ms
32 words (LR 00 to LR 31)
20 ms
16 words (LR 00 to LR 15)
10 ms
I/O Response Time
Section 6-4