7-787.6.5 PC(PLC) link Response TimeThe maximum value for the transmission time (T) of one cycle can be calculated using the followingformula.The various items in the formula are calculated as described below.{ Ts (transmission time per station)Ts = scan time + Tpc (PC(PLC) link sending time)Tpc = Ttx (sending time per byte) x Pcm (PC(PLC) link sending size)Ttx = 1/(baud rate x 1000) x 11 ms …. Approx. 0.096 ms at 115.2 kbpsPcm = 23 + (number of relay words + number of register words) x 4| Tlt (link table sending time)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 kbpsLtm = 13 + 2 x n (n = number of stations being added)} Tso (master station scan time)This should be confirmed using the programming tool.~ 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 forcommand 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 kbpsTwt = 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 kbpsTso = Master station scan timeCalculation example 1When all stations have been added to a 16-unit link, the largest station number is 16, relays andregisters 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 msEach Ts = 1 + 6.82 = 7.82 ms Tlt = 0.096 x (13 + 2 x 16) = 4.32 msGiven 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 msCalculation example 2When all stations have been added to a 16-unit link, the largest station number is 16, relays andregisters have been evenly allocated, and the scan time for each PLC is 5 msTtx = 0.096 Each Pcm = 23 + (4 + 8) x 4 = 71 bytes Tpc = Ttx x Pcm = 0.096 x 71 ≒ 6.82 msEach Ts = 5 + 6.82 = 11.82 ms Tlt = 0.096 x (13 + 2 x 16) = 4.32 msGiven 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
