Wiring; Wiring Procedure - Mitsubishi MELSEC iQ-F FX5 User Manual

■Calculation example 3
Calculation of required time when Pr. 10 to Pr. 14 [(s2) = 5] are written by the IVBWR instruction
Tinv=T +T +T =851[ms]
1 2 3
T =10×(s2)=50[ms], T
1
Calculate "T " as follows because Pr.10 to Pr. 14 do not require change of the 2nd parameter and the time required for writing
2
is same in each parameter.
T = ( 2 × (T +T ) + T
2 4 5
Time required to write Pr. 10
= 5 × ( 2 × (T +T
4 5
(s2)
1
T = ( INT ( +1 ) × 10 ) + ( INT (
4
10
1
T = INT ( +1 ) × 10 = 10 [ms]
5
10
T [1]+T
7
T [1] = ( INT (
6
T [1] + T [1] + T
7 8
T [1] + T [1] = ( (
7 9
T [1] = 12 [ms]
8
T [2]+T
7
T [2] = ( INT (
6
T [2] + T [2] + T
7 8
T [2] + T [2] = ( (
7 9
T [2] = 30 [ms]
8
Tinv = T + T + T = 50 + 800 + 1 = 851 [ms]
1 2 3
5.5

Wiring

This section explains about the wiring.

Wiring procedure

1.
Select the connection method
Confirm the inverter connection method. (Page 101 Connection method)
2.
Make arrangements for wiring
Prepare cables (Page 103 Cable), distributors (Page 105 Connection devices (RJ45 connector and distributor)) and
termination resistors (Page 106 Termination resistor setting) required for wiring.
3.
Turn OFF the PLC power
Before wiring, make sure that the PLC power is OFF.
4.
Wire the communication equipment.
Connect RS-485 communication equipment of PLC with the communication port of the inverter. (Page 107 Connection
diagram)
5.
Set or connect termination resistors.
Set or connect termination resistor of the inverter farthest from the PLC. (Page 106 Termination resistor setting)
6.
Connect (Ground) a shielding wire (Class-D grounding)
When using a twisted pair cable, connect a shielding wire. (Page 107 Shielded wiring)
5 INVERTER COMMUNICATION
100
5.5 Wiring
=1[ms]
3
[1]+T [2] ) + ( 2 × (T +T
6 6 4
Time required to write Pr. 11
) + T [1] + T [2] ) = 5 × ( 2 × (30+10) + 30 + 50 ) = 800 [ms]
6 6
11
+1 ) × 10 ) = 30 [ms]
10
[1]+T [1]
8 9
) +1 ) × 10 + ( INT (
10
[1] = 7.8 + 12 = 19.8 [ms]
9
1
) × (11+4) × 10 ) × 1000 = 7.8 [ms]
19200
[2]+T [2]
8 9
) +1 ) × 10 + ( INT (
10
[2] = 8.9 + 30 = 38.9 [ms]
9
1
) × (13+4) × 10 ) × 1000 = 8.9 [ms]
19200
) + T [3]+T [4] ) + •••
5 6 6
1
+1 ) × 10 ) = ( INT (
10
1
+1 ) × 10 ) = ( INT (
10
19.8
+1 ) ×10 +10 = 30 [ms]
10
38.9
+1 ) ×10 +10 = 50 [ms]
10