Multiple CAN bus devices broadcasting at the same time...possible conflict?

Question

I've worked with CAN bus recently and started to wonder what would happen if multiple devices on a network suddenly happen to broadcast at the same time.

I mean every device on the CAN-bus network receives each other's messages. But since they are on the same bus, wouldn't there be a conflict resulting in some devices not receiving some of the other's messages? I see it as a one-way street where you have two cars trying to go each other's opposite direction resulting in a crash and not reaching their destination.

I couldn't find anything about this on the Internet....plus this question probably relates to I²C as well or any other similar communication protocols.

Answer 1

As you stated, every node receives the others' messages. This allows the nodes to prevent any conflicts on the bus by listening the bus. Each node listens the bus before transmitting a message. If the bus is not busy then the node transmits its message.

There's also a priority mechanism to avoid conflicts: If a message has a higher priority (i.e. lower ID) then it gets the bus access. An example:

Let's assume there are two transmitters trying to transmit a message. One having an ID of 0x7FA and the other 0x7FF. They started to transmit right at the same time by transmitting the SOF (Start of frame) bit.

A: ¯¯¯¯¯¯¯¯¯|___|¯ SOF

B: ¯¯¯¯¯¯¯¯¯|___|¯ SOF

There's no conflict here. Then they send the ID bit-by-bit but MSB first.

A: ¯¯¯¯¯¯¯¯¯|___|¯¯¯|¯¯¯|¯¯¯|¯¯¯|¯¯¯|¯¯¯|¯¯¯|¯¯¯|___|¯¯¯|___| SOF 1 1 1 1 1 1 1 1 0 1 0 | 7 | F | A | | | conflict here | B: ¯¯¯¯¯¯¯¯¯|___|¯¯¯|¯¯¯|¯¯¯|¯¯¯|¯¯¯|¯¯¯|¯¯¯|¯¯¯|¯¯¯|¯¯¯|¯¯¯| SOF 1 1 1 1 1 1 1 1 1 1 1 | 7 | F | F |

The conflict occurs when the 9th bit (i.e. 3rd bit of the ID) is sent. The 2nd device loses arbitration (because it has lower priority due to higher ID) and leaves out the bus thus the 1st device wins and continues transmitting its message.