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MELSEC-A
3 SPECIFICATIONS
(b) Reception time
An example for calculating the reception time when receiving 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 reception time = 20
×
10/9600
=
0.0208s (20.8ms)
Reception time = 20 + 11.1
×
3 + (11.1
×
2)
*
1
+ 20.8 *
2
+ {(32 + 16)/16
×
11.1} *
3
+ {(512 + 16)/16
×
11.1} *
4
= 96.3 + 3
×
11.1 + 33
×
11.1 = 495.9ms
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 reception time = 20
×
10/9600
=
0.0208s (20.8ms)
Reception time = 20 + 10.9
×
2 + (10.9
×
2)
*
1
+ 20.8 *
2
+ {(32 + 16)/16
×
10.9} *
3
+ {(512 + 16)/16
×
10.9} *
4
= 84.4 + 3
×
10.9 + 33
×
10.9 = 476.8ms
*1 RS (R2 internal processing time)
*2 Data transmission time
*3 Status storage area request/response scan (20
H
(32 words)
worth)
*4 Reception area request/response scan (200
H
(512 words)
worth)