how low speed can iso1050 at

Zhao,

I have spoken to one of our CAN experts, Bryce Sinclair, and he provided the following explanation:

1. The ISO1050 has a feature called Dominant Timeout (tDOM) that prevents a node from monopolizing the bus in the event of a fault.  Unfortunately it kicks in around at a datasheet minimum = 300us or 3.33kbps and would be the absolute minimum speed for one dominant bit.  In the bit stuffing algorithm from the CAN protocol, the longest sequence of 1’s or 0's is five, so the minimum frequency would be 16.66kbps.

2. In CAN, the Dominant-to-Recessive timing is achieved by the RC decay of the network (Passive).  70 nodes on 600m of a double terminated CAT5E 100 Ohm Differential Bus, the Rdiff =~47.9 ohms (50 ohms || [80k/70] ohms) and Cdiff = ~31.41nF (3pF * 70 + 52pF * 600).  This would give a total theoretically 2.3 RC (90% transitioned) and 5RC (99.3% transitioned) Dominant to Recessive system timing [{2.3|5} * 47.9 ohms * 31.4nF] = ~3.46us | ~7.52us respectively. In actuality, these numbers will be significantly larger due to internal device charging.

3. Another important consideration is the VOD attenuation loss at 600 meters of the cable.  600m of CAT5E cable models as a 56.4 series resistors.  The differential load a driver sees under this long cable is ~64.2 ohms (100 ohms || 212.8 ohms || [80k/70] ohms) so VOD = ~1.55V. The VOD over the longest run length a receiver will have is ~730mV which is below the VIT(+) trip level of this device [900mV].

Conclusion: We assumed CAT5E cable because the values were readily available, but it gives you a proxy and equations to perform your own modeling.  A CAN device is not ideal for your 1000m system, you might want to look into our ISO RS-485 devices [ISO3082 might be a good choice] which has faster transitions with double the output amplitude, and a +/-200mV Input Threshold (VIT’s) at the receiver.

Regards, Thomas