252
7.3.2 Transmission delay time calculation example
This section provides a calculation example of the cyclic transmission delay.
(1) Transmission delay time in a single network system
(a) System configuration and conditions
(b) Link scan time ( Page 244, Section 7.1)
(c) Link refresh time ( Page 245, Section 7.2)
Item Description
CPU module Q06HCPU
Total number of stations per network 8 stations
Total link device points
LB/LW 1024 points for each
LX/LY 0 point
SB/SW 512 points for each
Sequence scan time 1ms
File register None
Interlink transmission None
Transient transmission None
Station-based block data assurance Assured
CC-Link IE Controller Network module Installed to slot 0 of main base unit
Faulty station None
LS = [KB + (N × 56) + {LB + LY + (LW × 16) 8 × 0.016 + (NT × T × 30)] 1000 + Nc [ms]
= [1100 + (8 × 56) + {1024 + 0 + (1024 × 16)} 8 × 0.016 + (0 × 2 × 30)] 1000
= 1.58 [ms]
(1) Sending-side link refresh time, receiving-side link refresh time
T, R
= 130 × 10
-3
+ 0.41 × 10
-3
× {(1024 + 0 + 0 + 512) 16 + 1024 + 512} + 0 + 0
= 0.80 [ms]
To Next Page
Loading...
