Mitsubishi Electric FX5 User Manual page 78

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]
■Calculation example 1
Calculation of required time when Pr. 3 is read by the IVRD instruction
Tinv=T +T +T =181[ms]
1 2 3
T =10[ms], T =1[ms]
1 3
Calculate "T " as follows because Pr.3 does not require change of the 2nd parameter.
2
T = 2 × (T + T ) + T
2 4 5
Number of
send/receives send/receive
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 = 10 + 170 + 1 = 181 [ms]
1 2 3
■Calculation example 2
Calculation of required time when Pr.902 is read by the IVRD instruction
Tinv=T +T +T =251[ms]
1 2 3
T =10[ms], T =1[ms]
1 3
Calculate "T " as follows because Pr.902 does not require change of the 2nd parameter.
2
T = 3 × (T +T ) +
2 4 5
Number of
send/receives send/receive
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] = 12 [ms]
8
T [3]+T
7
T [3] = ( INT (
6
T [3] + T [3] + T
7 8
T [3] + T [3] = ( (
7 9
T [3] = 30[ms]
8
Tinv = T + T + T = 10 + 240 + 1 = 251 [ms]
1 2 3
4 INVERTER COMMUNICATION
76
4.4 Specifications
[1] + T [2] = 2 × (30 + 10) + 30 + 60 = 170 [ms]
6 6
The first The second
send/receive
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] = 10.4 + 30 = 40.4 [ms]
9
1
) × (9+11) × 10 ) × 1000 = 10.4 [ms]
19200
T [1] + T [2] +
6 6
The first The second
send/receive
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] = 7.8 + 12 = 19.8 [ms]
9
1
) × (11+4) × 10 ) × 1000 = 7.8 [ms]
19200
[3]+T [3]
8 9
) +1 ) × 10 + ( INT (
10
[3] = 10.4 + 30 = 40.4 [ms]
9
1
) × (9+11) × 10 ) × 1000 = 10.4 [ms]
19200
1
+1 ) × 10 ) = ( INT (
10
1
+1 ) × 10 ) = ( INT (
10
T [3] = 3 × (30+10) + 30 + 30 + 60 = 240 [ms]
6
The third
send/receive
1
+1 ) × 10 ) = ( INT (
10
1
+1 ) × 10 ) = ( INT (
10
1
+1 ) × 10 ) = ( INT (
10
19.8
+1 ) ×10 ) +10 = 30 [ms]
10
40.4
+1 ) ×10 ) +10 = 60 [ms]
10
19.8
+1 ) ×10 ) +10 = 30 [ms]
10
19.8
+1 ) ×10 ) + 10 = 30 [ms]
10
40.4
+1 ) ×10 +10 = 60 [ms]
10