Ah, sorry, I missed that the problem was now only the bitrate.
If we convert your CNF registers to mathematical variables, then:
becomes:
As you said, you are using a sample point of 75%. But the MCP2515 datasheet (4th paragraph in 5.3 Programming Time Segments) says:
You will land in the middle-ish of the recommended sample point percentage.
When you only have 8 TQs to work with and the sample point is placed too close or too far away, you can get synchronization problems, which could make the MCP2515 send the bits at a higher bitrate.
In closing, try using these CNF values instead:
If we convert your CNF registers to mathematical variables, then:
Code:
CNF1 = 0x00CNF2 = 0x91CNF3 = 0x81Code:
sjw = 1t_bit = t_sync_seg + t_prop_seg + t_ps1 + t_ps2 = 8t_sync_seg = 1t_prop_seg = 2t_ps1 = 3t_ps2 = 2sample_point = (t_sync_seg + t_prop_seg + t_ps1) / t_bit = 75%If you instead put the sample point at ~62%, using these variables:Typically, the sampling of the bit should take place at about 60-70% of the bit time, depending on the system parameters.
Code:
sjw = 1t_bit = t_sync_seg + t_prop_seg + t_ps1 + t_ps2 = 8t_sync_seg = 1t_prop_seg = 1t_ps1 = 3t_ps2 = 3sample_point = (t_sync_seg + t_prop_seg + t_ps1) / t_bit = 62.5%When you only have 8 TQs to work with and the sample point is placed too close or too far away, you can get synchronization problems, which could make the MCP2515 send the bits at a higher bitrate.
In closing, try using these CNF values instead:
Code:
CNF1 = 0x00CNF2 = 0x90CNF3 = 0x82Statistics: Posted by rzetterberg — Thu May 28, 2026 8:18 pm
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