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I am reading data sheets on the LIN bus, but apparently the same terminology is used for CAN.

What does "recessive" and "dominant" stand for, e.g http://hw-server.com/products/rs_optika/prevodniky_optika.html ?

How relate these terms to "low" and "high", or 0 and 1?

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In order to understand this, you need to understand CSMA/CA. –  Lundin Jul 1 at 7:56
    
thanks for the pointer! –  poseid Jul 1 at 8:36
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2 Answers 2

up vote 6 down vote accepted

Dominant is 0. Recessive is 1. Dominant applies to 0 because if two arbitration ID's are being transmitted at the same time and the first 4 bits are the same and the fifth is 0 for one of them and 1 for the other, the ID with the 0 will end up being transmitted. Transmission of the message with the larger arbitration ID will be tried again after the other message is done.

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In addition to the perfectly accurate answer given elsewhere, it may also be useful to consider the lower level makings of the phrases dominant and recessive. In both CAN and LIN at the physical layer the bus "floats" to a particular state when no nodes are communicating. This is the recessive state. Any node which drives a dominant bit will override this state (hence the word "dominant").

This is used within the protocol to allow non-destructive arbitration to occur, where the node with the lowest id "wins". This is due to the election to signify a logic low by a dominant bit. But the standards could have been written the other way around. There's nothing that demands a dominant bit is interpreted as a zero in any other context.

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Also, CAN bus is Wire-ANDed... And in AND operation, 0 will "dominate" the 1. –  Swanand May 28 at 11:48
    
that is nice shortcut to remember –  poseid May 28 at 12:52
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