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MELSEC-A
3 SPECIFICATIONS
(2) Calculation example
(a) Transmission time
An example of calculating the transmission time for transmitting 10 words
(20 bytes) of data is shown below.
Item Setting details
Transmission size of each area Default value
Transmission speed 156kbps
No. of connected modules Only one R2 module
Master station sequence program scan time 20ms (Hypothetical)
Transmission speed 9600bps
Data bit length 8
Stop bit length 1
Parity bit Even
1) When the master station is the A/QnA series
LS = 51.2 {29.4 + (8
×
4.8) + (8
×
9.6) + (1
×
32.4) + (1
×
4.8) + (1
×
9.6)}
+ 1300
=
11100
µ
s (11.1ms)
Data transmission time = 20
×
10/9600
=
0.0208s (20.8ms)
Transmission time = 20
×
2 + 11.1
×
6 + (11.1
×
2) *
1
+ 20.8 *
2
+ {(136 + 16)/72
×
11.1} *
3
+ {(512 + 16)/72
×
11.1} *
4
+ {(32 + 16)/16
×
11.1} *
5
= 149.6 + 3
×
11.1 + 8
×
11.1 + 3
×
11.1
= 305ms
2) When the master station is the Q series
LS = 51.2 {27 + (8
×
4.8) + (8
×
9.6) + (1
×
30) + (1
×
4.8) + (1
×
9.6)}
+ 1300 +0 +0
=
10854
µ
s (10.9ms)
Data transmission time = 20
×
10/9600
=
0.0208s (20.8ms)
Transmission time = 20
×
2 + 10.9
×
4 + (10.9
×
2) *
1
+ 20.8 *
2
+ {(136 + 16)/72
×
10.9} *
3
+ {(512 + 16)/72
×
10.9} *
4
+ {(32 + 16)/16
×
10.9} *
5
= 126.2 + 3
×
10.9 + 8
×
10.9 + 3
×
10.9
= 278.8ms
*
1
R2 (R2 internal processing time)
*
2
Data transmission time
*
3
Transmission area 1 request/response scan (88
H
(136 words)
worth)
*
4
Transmission area 2 request/response scan (200
H
(512 words)
worth)
*
5
Status storage area request/response scan (20
H
(32 words)
worth)