7-46
Calculation example 3
When all but one station have 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)
≒
4.13 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 all stations have been added to an 8-unit link, the largest station number is 8, relays and
register 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 bytes
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 all stations have 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 bytes
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 all stations have been added to a 2-unit link, the largest station number is 2, 32 relays and
2 register 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 bytes
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
Note:
-
In the description, “stations that have been added” refers to stations which are connected between
station no. 1 and the largest station number and for which the power supply has been turned on.
-
Comparing examples 2 and 3, the transmission cycle time is longer if there is one station that has not
been added to the link. As a result the PC(PLC) link response time is longer.
-
The SYS1 instruction can be used to minimize the transmission cycle time even if there are one or
more stations that have not been added to the link.
Summary of Contents for FP0R Series
Page 1: ......
Page 6: ...iv ...
Page 14: ...xii ...
Page 15: ...Chapter 1 Functions and Restrictions of the Unit ...
Page 24: ...1 10 ...
Page 25: ...Chapter 2 Specifications and Functions of Control Unit ...
Page 38: ...2 14 ...
Page 39: ...Chapter 3 Expansion ...
Page 45: ...3 7 3 4 Terminal layout diagram Model No Terminal layout diagrams E8RS E8RM E16RS E16RM E8YRS ...
Page 48: ...3 10 ...
Page 49: ...Chapter 4 I O Allocation ...
Page 53: ...Chapter 5 Installation and Wiring ...
Page 73: ...Chapter 6 Preparation of USB Port ...
Page 77: ...6 5 5 Click Finish on the following screen to be displayed ...
Page 79: ...6 7 4 Double click on FP0R 5 Click Update Driver ...
Page 84: ...6 12 ...
Page 85: ...Chapter 7 Communication ...
Page 139: ...7 55 Sample program For Type II Use a program as below to directly specify a MODBUS address ...
Page 141: ...Chapter 8 High speed Counter Pulse Output and PWM Output Functions ...
Page 142: ...8 2 ...
Page 199: ...Chapter 9 Security Functions ...
Page 211: ...Chapter 10 Other Functions ...
Page 217: ...Chapter 11 Self Diagnostic and Troubleshooting ...
Page 227: ...Chapter 12 Precautions During Programming ...
Page 242: ...12 16 ...
Page 243: ...Chapter 13 Specifications ...
Page 254: ...13 12 ...
Page 255: ...Chapter 14 Dimensions and Others ...
Page 262: ...14 8 ...
Page 263: ...Chapter 15 Appendix ...
Page 344: ...15 82 15 7 ASCII Codes ...
Page 346: ......
Page 347: ......
Page 348: ......