Approximation of the Bus Load
A short example for the calculation of the bus load to check if an application is possible.
Example:
- 1 CAN message: 8 byte of data + CAN ID… = 108 bits
- CAN baud rate: 1000000
- Send cycle: 1 ms
- Bus load: 70% → = 700000 bit/s
700000 bit/s / (108 bit * 1000 msg/s ) = ~6.5 messages
To not exceed a busload of 70%, 6 messages with 1 kHz and 1 MBaud can be transferred!!
→ using 500 kBaud: max. 3 messages are possible (@ 1 kHz, bus load ~70 %)
Example 2:
- 20 channels with 4 Byte each should be transmitted
- CAN baud rate: 1000000
- 1 CAN message: 8 byte of data + CAN ID… = 108 bits
- 2 Channels per message, results in 10 Messages and 1080 Bit per cycle
- Send cycle: to be determined
- Bus load: 70% → = 700000 bit/s
700000 bit/s / 1080 bit = ~650 Transfers of all 20 channels per second
To not exceed a busload of 70%, 650x 20 channels can be transfered per second at 1 MBaud
→ using 500 kBaud: max. 325 messages are possible (bus load ~70 %)