FPΣ
Communication Function 3 PLC Link Function
10 - 22
10.5 PLC Link Response Time
10.5.1 PLC Link Response Time
The maximum value for the transmission time (T) of one cycle can be calculated using
the following formula.
T max. = Ts1 + Ts2 + ----+Tsn+Tlt
+Tso+ Tlk
4
Tlk (link addition processing time)
3
Tso (master station scan time)
2
Tlt (link table sending time)
1
Ts (transmission time per station)
The various items in the formula are calculated as described below.
1
Ts (transmission time per station)
Ts = scan time + Tpc (PLC link sending time)
Tpc = Ttx (sending time per byte) x Pcm (PLC link sending size)
Ttx = 1 / transmission speed x 1000 x 11 ms - - - approx. 0.096 ms at 115.2 kbps
Pcm = 23 + (number of relay words + number of register words) x 4
2
Tlt (link table sending time)
Tlt = Ttx (sending time per byte) x Ltm (link table sending size)
Ttx = 1 / transmission speed x 1000 x 11 ms - - - approx. 0.096 ms at 115.2 kbps
Ltm = 13 + 2 x n (n = number of stations being added)
3
Tso (master station scan time)
This should be confirmed using the programming tool.
4
Tlk (link addition processing time) - - - If no stations are being added, Tlk = 0.
Tlk = Tlc (link addition command sending time) + Twt (addition waiting time) + Tls (sending time for
command to stop transmission if link error occurs) + Tso (master station scan time)
Tlc = 10 x Ttx (sending time per byte)
Ttx = 1 / transmission speed x 1000 x 11 ms - - - approx. 0.096 ms at 115.2 kbps
Twt = Initial value 400 ms (can be changed using SYS2 system register instruction)
Tls = 7 x Ttx (sending time per byte)
Ttx = 1 / transmission speed x 1000 x 11 ms - - - approx. 0.096 ms at 115.2 kbps
Tso = Master station scan time
Calculation example 1
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 1 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 = 1 + 6.82 = 7.82 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. = 7.82 x 16 + 4.32 + 1 = 130.44 ms