Differences
This shows you the differences between two versions of the page.
| Both sides previous revisionPrevious revisionNext revision | Previous revision | ||
| can_cabling [2012/06/11 22:51] – heinz | can_cabling [Unknown date] (current) – removed - external edit (Unknown date) 127.0.0.1 | ||
|---|---|---|---|
| Line 1: | Line 1: | ||
| - | 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: | ||
| - | |||
| - | * 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/ | ||
| - | | 800Kbit/ | ||
| - | | 500Kbit/ | ||
| - | | 250Kbit/ | ||
| - | | 125Kbit/ | ||
| - | | | ||
| - | |||
| - | (Table from CiA DS 301 4.02 page 20) | ||
| - | |||
| - | 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. | ||
