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Q&A CAN Bus - Internal Oscillator

CAN is asynchronous and, if you receive a string of consecutive zeros, the UART mechanism that turns raw CAN data into "a byte" might have to wait a great deal longer than 1 bit of data before it c...

posted 4y ago by Andy aka‭  ·  edited 4y ago by Andy aka‭

Answer
#2: Post edited by user avatar Andy aka‭ · 2020-12-19T15:59:26Z (almost 4 years ago)
  • CAN is asynchronous and, if you receive a string of consecutive zeros, the UART mechanism that turns raw CAN data into "a byte" might have to wait a great deal longer than 1 bit of data before it can re-sync.
  • If data is coming 101010101010 then that's great but, if data is 010000000010 then you have to freewheel across all those zeros and still be able to reliably detect the next 1 without over-counting the number of steady bits you have received in the interim period.
  • CAN is asynchronous and, if you receive a string of consecutive zeros, the UART mechanism that turns raw CAN data into "a byte" might have to wait a great deal longer than 1 bit of data before it can re-sync.
  • If data is coming 101010101010 then that's great but, if data is 010000000010 then you have to freewheel across all those zeros and still be able to reliably detect the next 1 without over-counting the number of steady bits you have received in the interim period.
  • By the way, 1 Mbps is a maximum frequency of 500 kHz because there are two symbol slots in one cycle.
#1: Initial revision by user avatar Andy aka‭ · 2020-12-19T15:48:37Z (almost 4 years ago)
CAN is asynchronous and, if you receive a string of consecutive zeros, the UART mechanism that turns raw CAN data into "a byte" might have to wait a great deal longer than 1 bit of data before it can re-sync.

If data is coming 101010101010 then that's great but, if data is 010000000010 then you have to freewheel across all those zeros and still be able to reliably detect the next 1 without over-counting the number of steady bits you have received in the interim period.