6. COMMUNICATION TIME
MELSEC-A
6-3
The calculation example of the bus cycle time is shown below.
Slave station
(Station No.1)
AJ95TB2-16T
0 points
16 points
Input:
Output:
Slave station
(Station No.2)
AJ95TB3-16D
16 points
0 points
Input:
Output:
Slave station
(Station No.3)
A1SJ71PB93D
1 word
2 words
Input:
Output:
Master station (Station No.0)
AJ71PB92D/A1SJ71PB92D
Transmission speed: 1.5Mbps
Number of slave stations: 3 stations
PROFIBUS-DP
Output data size [byte]
Input data size [byte]
AJ95TB2-16T
AJ95TB3-16D
A1SJ71PB93D
2
0
4
0
2
2
(a) MSI[s] value
MSI[s] = 20 100 10
-6
= 2.0 10
-3
(b) Pt
(i)
[s] value
Slave station
Item
AJ95TB2-16T (Station No.1)
AJ95TB3-16D (Station No.2)
A1SJ71PB93D (Station No.3)
1) Treq
(i)
[s]
{(2 + 9) 11} / (1.5 10
6
)
= 0.081
10
-3
{(0 + 9) 11} / (1.5 10
6
)
= 0.066
10
-3
{(4 + 9) 11} / (1.5 10
6
)
= 0.095
10
-3
Response time [T
Bit
]
of No. i station
150
150
150
2) Max_Tsdr
(i)
[s]
150 / (1.5
10
6
) = 0.1
10
-3
150 / (1.5
10
6
) = 0.1
10
-3
150 / (1.5
10
6
) = 0.1
10
-3
3) Tres
(i)
[s]
{(0 + 9) 11} / (1.5 10
6
)
= 0.066
10
-3
{(2 + 9) 11} / (1.5 10
6
)
= 0.081
10
-3
{(2 + 9) 11} / (1.5 10
6
)
= 0.081
10
-3
Pt
(i)
[s]
0.081
10
-3
+ 0.1
10
-3
+ 0.066
10
-
3
= 0.247
10
-3
0.1
10
-3
+ 0.066
10
-3
+ 0.081
10
-
3
= 0.247
10
-3
0.095
10
-3
+ 0.1
10
-3
+ 0.081
10
-
3
= 0.276
10
-3
(c) Tsdi
(M)
[s] value
Request/response processing time [T
Bit
] of master station (AJ71PB92D/A1SJ71PB92D)
Tsdi
(M)
[s] = 150
/
(1.5 10
6
) = 0.1 10
-3
(d) Lr[s] value
Lr[s] = 1.2 10
-3
+ (3 1.2 10
-6
) = 1.20 10
-3
(e) From/To_time[s] value
Number of FROM/TO instruction execution times within one sequence scan = 2
From/To_time[s] = 2 3.3 10
-3
= 6.6 10
-3
From the values in (b) to (e)
(Pt
(i)
+ Tsdi
(M)
) + Lr + From/To_time = {(Pt
(1)
+ Tsdi
(M)
) + (Pt
(2)
+ Tsdi
(M)
) + (Pt
(3)
+ Tsdi
(M)
)} + Lr + From/To_time
i =1
3
= {(0.347 10
-3
) + (0.347 10
-3
) + (0.376 10
-3
)} + 1.20 10
-3
+ 6.6 10
-3
= 1.07 10
-3
+ 1.20 10
-3
+ 6.6 10
-3
= 8.87 10
-3
Hence, the bus cycle time (Bc) value is as follows.