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CANopen Network Interface
1606218 02 08/2006 41
Example The following illustration shows the calculation of the length of a cable located
between two tap junction boxes.
The table below describes the components of a CANopen network:
In this example, we have two tap junction boxes and 6 devices. We start by
calculating the sum of the lengths of cables for each tap junction box, and we obtain
5m (16 ft) and 7 m (23 ft). We keep the longest length, i.e. 7 m (23 ft). The minimum
length of the cable between the two tap junction boxes is equal to 60% of 7 m, i.e.
4.2 m (13.8 ft).
Number of
Connected
Devices
In addition to the length limitations over the whole of the CANopen bus, the following
limitations apply:
Whatever the case, no more than 64 devices may be connected on the same
segment.
Number Description
1 Connected CANopen devices
2 Drop cables (tap junction box / device)
3 Tap junction boxes
4 Connection cables (tap junction box / tap junction box)
ΣL=5 m (16 ft)
Min interval > 0.6 * 7 m (23 ft)
1
2
4
3
1 m
(3 ft)
1 m
(3 ft)
3 m
(10 ft)
ΣL=7 m (23 ft)
1 m
(3 ft)
3 m
(10 ft)
3 m
(10 ft)
Min interval > 4.2 m (13.8 ft)