NAiS FP Series, Руководство пользователя

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FP
Σ
10.5 PLC Link Response Time
10 - 23
Calculation example 2
When there are no stations that have not been added to a 16-unit link, the largest
station number is 16, relays and registers have been evenly allocated, and the scan
time for each PLC is 5 ms
Ttx = 0.096 Each Pcm = 23 + (4 + 8) x 4 = 71 Tpc = Ttx x Pcm = 0.096 x 71 6.82 ms
Each Ts = 5 + 6.82 = 11.82 ms Tlt = 0.096 x (13 + 2 x 16) = 4.32 ms
Given the above conditions, the maximum value for the transmission time (T) of one cycle will be:
T max. = 11.82 x 16 + 4.32 + 5 = 198.44 ms
Calculation example 3
When there is one station that has not been added to a 16-unit link, the largest station
number is 16, relays and registers have been allocated evenly, and the scan time for
each PLC is 5 ms
Ttx = 0.096 Each Ts = 5 + 6.82 = 11.82 ms Tlt = 0.096 x (13 + 2 x 15) 4.31 ms
Tlk = 0.96 + 400 + 0.67 + 5 407 ms
Note: The default value for the addition waiting time is 400 ms.
Given the above conditions, the maximum value for the transmission time (T) of one cycle will be:
T max. = 11.82 x 15 + 4.13 + 5 + 407 = 593.43 ms
Calculation example 4
When there are no stations that have not been added to an 8-unit link, the largest
station number is 9, relays and registers have been evenly allocated, and the scan time
for each PLC is 5 ms
Ttx = 0.096 Each Pcm = 23 + (8 + 16) x 4 = 119 Tpc = Ttx x Pcm = 0.096 x 119 11.43 ms
Each Ts = 5 + 11.43 = 16.43 ms Tlt = 0.096 x (13 + 2 x 8) 2.79 ms
Given the above conditions, the maximum value for the transmission time (T) of one cycle will be:
T max. = 16.43 x 8 + 2.79 + 5 = 139.23 ms
Calculation example 5
When there are no stations that have not been added to a 2-unit link, the largest station
number is 2, relays and registers have been evenly allocated, and the scan time for
each PLC is 5 ms
Ttx = 0.096 Each Pcm = 23 + (32 + 64) x 4 = 407 Tpc = Ttx x Pcm = 0.096 x 407 39.072 ms
Each Ts = 5 + 39.072 = 44.072 ms Tlt = 0.096 x (13 + 2 x 2) 1.632 ms
Given the above conditions, the maximum value for the transmission time (T) of one cycle will be:
T max. = 44.072 x 2 + 1.632 + 5 = 94.776 ms
Calculation example 6
When there are no stations that have not been added to a 2-unit link, the largest station
number is 2, 32 relays and register 2 words have been evenly allocated, and the scan
time for each PLC is 1 ms
Ttx = 0.096 Each Pcm = 23 + (1 + 1) x 4 = 31 Tpc = Ttx x Pcm = 0.096 x 31
≒
2.976 ms
Each Ts = 1 + 2.976 = 3.976 ms Tlt = 0.096 x (13 + 2 x 2) 1.632 ms
Given the above conditions, the maximum value for the transmission time (T) of one cycle will be:
T max. = 3.976 x 2 + 1.632 + 1 = 10.584 ms
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