Filter
Top Products
3 - 25
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 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
+ {20 + 11.1 + 11.1 + 11.1 + 20 + 11.1
×
2 + (10
+ 16)/16
×
11.1 + 11.1 + 11.1}
×
1 *
3
= 96.3 + (20 + 11.1 + 11.1 + 11.1 + 20 + 22.2 + 11.1
+ 11.1)
= 96.3 + 139.9
= 236.2ms
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 (10.9 ms)
Data transmission time = 20
×
10/9600
=
0.0208s (20.8ms)
Transmission time = 20
×
2 + 10.9
×
2 + (10.9
×
2) *
1
+ 20.8 *
2
+ 1 + 10.9
×
[6 + {(10 + 16) / 16}
×
1.067] *
3
= 84.4 + 89.6606
= 174.0606
=
174.1ms
*1 R2 (R2 internal processing time)
*2 Data transmission time
*3 Transient transmission time (10 words worth)