Calculation Example - Mitsubishi Electric FX3U-232-BD User Manual

FX Series PLC User's Manual - Data Communication Edition
Inverter Communication
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 = T + T
inv 1
T = 1[ms], T
1
Calculate "T
T = 2(T +T
2 4
Sending and
receiving
frequency
T = ( INT(
4
T = 1[ms]
5
T [1] = ( INT(
6
T [1] + T [1] + T
7 8
T [1] + T
7 9
T [1] = 12[ms]
8
T [2] = ( INT(
6
T [2] + T [2] + T
7 8
T [2] + T
7 9
T [2] = 30[ms]
8
T = T + T
inv 1
2. Calculation example 2
Calculation of required time when Pr.902 is read by the IVRD instruction
T = T + T
inv 1
T = 1[ms], T
1
Calculate "T
T = 3(T +T
2 4
Sending and
receiving
frequency
T = ( INT(
4
T = 1[ms]
5
T [1] = ( INT(
6
T [1] + T [1] + T
7 8
T [1] + T
7 9
T [1] = 12[ms]
8
T [2] = ( INT(
6
T [2] + T [2] + T
7 8
T [2] + T
7 9
T [2] = 12[ms]
8
T [3] = ( INT(
6
T [3] + T [3] + T
7 8
T [3] + T
7 9
T [3] = 30[ms]
8
T = T + T
inv 1
E-14
+ T = 114[ms]
2 3
= 1[ms]
3
" as follows because Pr.3 does not require change of the 2nd parameter.
2
) + T [1] + T [2] = 2(20+1) + 20 + 50 = 112[ms]
5 6 6
The first The second
sending and sending and
receiving receiving
15
) + 1 )  10 = 20[ms]
10
T [1] + T [1] + T [1]
7 8 9
) + 1 )  10 = ( INT(
10
[1] = 7.8 + 12 = 19.8[ms]
9
1
)  (11+4)  10 )  1000 = 7.8[ms]
[1] = ( (
19200
T [2] + T [2] + T [2]
7 8 9
) + 1 )  10 = ( INT(
10
[2] = 10.4 + 30 = 40.4[ms]
9
1
)  (9+11)  10 )  1000 = 10.4[ms]
[2] = ( (
19200
+ T = 1+112+1 = 114[ms]
2 3
+ T = 155[ms]
2 3
= 1[ms]
3
" as follows because Pr.902 requires change of the 2nd parameter.
2
) + T [1] + T [2] + T
5 6 6
The first The second
sending and sending and
receiving receiving
15
) + 1 )  10 = 20[ms]
10
T [1] + T [1] + T [1]
7 8 9
) + 1 )  10 = ( INT(
10
[1] = 7.8 + 12 = 19.8[ms]
9
1
)  (11+4)  10 )  1000 = 7.8[ms]
[1] = ( (
19200
T [2] + T [2] + T [2]
7 8 9
) + 1 )  10 = ( INT(
10
[2] = 7.8 + 12 = 19.8[ms]
9
1
)  (11+4)  10 )  1000 = 7.8[ms]
[2] = ( (
19200
T [3] + T [3] + T [3]
7 8 9
) + 1 )  10 = ( INT(
10
[3] = 10.4 + 30 = 40.4[ms]
9
1
)  (9+11)  10 )  1000 = 10.4[ms]
[3] = ( (
19200
+ T = 1+153+1 = 155[ms]
2 3
19.8
) + 1 )  10 = 20[ms]
10
40.4
) + 1 )  10 = 50[ms]
10
[3] = 3(20+1) + 20 + 20 + 50 = 153[ms]
6
The third
sending and
receiving
19.8
) + 1 )  10 = 20[ms]
10
19.8
) + 1 )  10 = 20[ms]
10
40.4
) + 1 )  10 = 50[ms]
10
2.4 Execution Times in Inverter Communication Instructions
2 Specifications