E-14
FX Series PLC User's Manual - Data Communication Edition
Inverter Communication
2 Specifications
2.4 Execution Times in Inverter Communication Instructions
2.4.3
Calculation example
This is a calculation example for the following communication settings and scan time when communicating
with an inverter.
Communication speed = 19200[bps]
Length of 1 character = 10[bit]
Scan time = 10[ms]
1. Calculation example 1
Calculation of required time when Pr. 3 is read by the IVRD instruction
T
inv
= T
1
+ T
2
+ T
3
= 114[ms]
T
1
= 1[ms], T
3
= 1[ms]
Calculate "T
2
" as follows because Pr.3 does not require change of the 2nd parameter.
2. Calculation example 2
Calculation of required time when Pr.902 is read by the IVRD instruction
T
inv
= T
1
+ T
2
+ T
3
= 155[ms]
T
1
= 1[ms], T
3
= 1[ms]
Calculate "T
2
" as follows because Pr.902 requires change of the 2nd parameter.
T
6
[1] = ( INT( ) + 1 )
10 = ( INT( ) + 1 )
10 = 20[ms]
10
19.8
T
6
[2] = ( INT( ) + 1 )
10 = ( INT( ) + 1 )
10 = 50[ms]
10
40.4
19200
1
T
5
= 1[ms]
T
inv
= T
1
+ T
2
+ T
3
= 1+112+1 = 114[ms]
T
7
[1] + T
9
[1] = ( ( )
(11+4)
10 )
1000 = 7.8[ms]
T
7
[1] + T
8
[1] + T
9
[1] = 7.8 + 12 = 19.8[ms]
T
8
[1] = 12[ms]
19200
1
T
7
[2] + T
9
[2] = ( ( )
(9+11)
10 )
1000 = 10.4[ms]
T
7
[2] + T
8
[2] + T
9
[2] = 10.4 + 30 = 40.4[ms]
T
8
[2] = 30[ms]
T
4
= ( INT( ) + 1 )
10 = 20[ms]
T
2
= 2
(T
4
+T
5
) + T
6
[1] + T
6
[2] = 2
(20+1) + 20 + 50 = 112[ms]
Sending and
receiving
frequency
The first
sending and
receiving
The second
sending and
receiving
10
15
10
T
7
[1] + T
8
[1] + T
9
[1]
10
T
7
[2] + T
8
[2] + T
9
[2]
T
6
[1] = ( INT( ) + 1 )
10 = ( INT( ) + 1 )
10 = 20[ms]
10
19.8
19200
1
T
7
[1] + T
9
[1] = ( ( )
(11+4)
10 )
1000 = 7.8[ms]
T
7
[1] + T
8
[1] + T
9
[1] = 7.8 + 12 = 19.8[ms]
T
8
[1] = 12[ms]
T
6
[2] = ( INT( ) + 1 )
10 = ( INT( ) + 1 )
10 = 20[ms]
10
19.8
T
6
[3] = ( INT( ) + 1 )
10 = ( INT( ) + 1 )
10 = 50[ms]
10
40.4
19200
1
T
5
= 1[ms]
T
inv
= T
1
+ T
2
+ T
3
= 1+153+1 = 155[ms]
T
7
[2] + T
9
[2] = ( ( )
(11+4)
10 )
1000 = 7.8[ms]
T
7
[2] + T
8
[2] + T
9
[2] = 7.8 + 12 = 19.8[ms]
T
8
[2] = 12[ms]
19200
1
T
7
[3] + T
9
[3] = ( ( )
(9+11)
10 )
1000 = 10.4[ms]
T
7
[3] + T
8
[3] + T
9
[3] = 10.4 + 30 = 40.4[ms]
T
8
[3] = 30[ms]
T
4
= ( INT( ) + 1 )
10 = 20[ms]
T
2
= 3
(T
4
+T
5
) + T
6
[1] + T
6
[2] + T
6
[3] = 3
(20+1) + 20 + 20 + 50 = 153[ms]
Sending and
receiving
frequency
The first
sending and
receiving
The second
sending and
receiving
The third
sending and
receiving
10
15
10
T
7
[1] + T
8
[1] + T
9
[1]
10
T
7
[2] + T
8
[2] + T
9
[2]
10
T
7
[3] + T
8
[3] + T
9
[3]
Summary of Contents for FX-485ADP
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Page 344: ...E 2 FX Series PLC User s Manual Data Communication Edition Inverter Communication ...
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