Pc(Plc) Link Response Time - Panasonic FP0R User Manual

7.6.7 PC(PLC) Link Response Time

The maximum value for the transmission time (T) of one cycle can be calculated using the following
formula.
The various items in the formula are calculated as described below.
Ts (transmission time per station)
○
1
Ts = scan time + Tpc (PC(PLC) link sending time)
Tpc = Ttx (sending time per byte) x Pcm (PLC link sending size)
Ttx = 1/(baud rate 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
Tlt (link table sending time)
○
2
Tlt = Ttx (sending time per byte) x Ltm (link table sending size)
Ttx = 1/(baud rate x 1000) x 11 ms .... Approx. 0.096 ms at 115.2 kbps
Ltm = 13 + 2 x n (n = number of stations being added)
Tso (master station scan time)
○
3
This should be confirmed using the programming tool.
Tlk (link addition processing time) .... If no stations are being added, Tlk = 0.
○
4
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/(baud rate x 1000) x 11 ms .... Approx. 0.096 ms at 115.2 kbps
Twt = Initial value 400 ms (can be changed using SYS1 system register instruction)
Tls = 7 x Ttx (sending time per byte)
Ttx = 1/(baud rate x 1000) x 11 ms .... Approx. 0.096 ms at 115. 2 kbps
Tso = Master station scan time
Calculation example 1
When all stations have 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 PLCs is 1 ms.
Ttx = 0.096 Each Pcm = 23 + (4 + 8) x 4 = 71 bytes Tpc = Ttx x Pcm = 0.096 x 71 ≒ 6.82 ms
Each Ts = 1 + 6.82 = 7.82 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
Calculation example 2
When all stations have 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 bytes Tpc = Ttx x Pcm = 0.096 x 71 ≒ 6.82 ms
Each Ts = 5 + 6.82 = 11.82 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
Tlt = 0.096 x (13 + 2 x 16) = 4.32 ms
Tlt = 0.096 x (13 + 2 x 16) = 4.32 ms
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