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| - | Some rules on how to design a CAN network. Consider reading [[Problem Solving Questions]]. | ||
| - | The rules are general and valid for an CAN network including [[CAN Higher Layer Protocols: | ||
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| - | * Always use twisted pairs for CAN_H and CAN_L | ||
| - | * Maximum cable length is between the two farthermost nodes on the CAN bus line. Remember, in the worst case a signal has to travel from the node at one end to the node on the other and of the line. It is a function of the bit rate as follows: | ||
| - | |||
| - | |bit rate|max cable length (in m)| | ||
| - | | 1Mbit/s| 25| | ||
| - | |800Kbit/s| 50| | ||
| - | |500Kbit/s| 100| | ||
| - | |250Kbit/s| 250| | ||
| - | |125Kbit/s| 500| | ||
| - | |50Kbit/s| 1000| | ||
| - | |||
| - | (Table from CiA DS 301 4.02 page 20) | ||
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| - | If you ask for a formula? There is one rule of thumb: | ||
| - | cable length < 40000 Kbit/s * 1 m / BitRate in Kbit/s | ||
| - | BitRate given in Kbit/s as well. Using this, results in < 40m for 1Mbit/s or | ||
| - | < 320m for 125Kbit/s. In this formula a typical transceiver propagation of 25ns and a typical cable propagation of 5 ns/m is used for calculation. | ||
