If my CAN baud rate is 500Kbps. So 1 bit time will take this amount of time
1/(500Kbps) =0.000002 bits so if the CAN frame contains 128bits then the total time to transmit is 0.000002*128 = 128uS. Am i correct in these calculations? Please advise. 128uS seems to be very high number.

 

0.000002*128 = 128uS

T = 0.000002*128 = 256 uS.

Please advise. 128uS seems to be very high number.

How many CAN controllers (Node ?) that you trying to connected together?

What kind of device that you want to connecting them?

 

T = 0.000002*128 = 256 uS.


How many CAN controllers (Node ?) that you trying to connected together?

3 Nodes

T = 0.000002*128 = 256 uS.
What kind of device that you want to connecting them?

It is an automotive application. A Body Control Module.

 

When the speed raise up to 500 Kbps and the distance probably only 100 meters, is that distance suit for the application?

If the 3 Nodes just used in one car then the distance isn't an issue, and still have chance to raise up to 1 Mbps.

 

My doubt is can i send the CAN message at the rate of 100 uS? This message is only for debugging purpose.

 

* High Speed CAN offers baud rates from 40 Kbit/s to 1 Mbit/sec, depending on cable length. This is the most popular standard for the physical layer, since it allows for simple cable connection between devices. This is the physical standard used in the DeviceNet and CANopen specifications. High speed CAN networks are terminated with 120 ohm resistors on each end of the network.

CAN Physical Layer Standards: High-Speed vs. Low-Speed/Fault-Tolerant CAN.

 

The length of a CAN message is not fixed. The data portion can be any length from 0 to 8 bytes. Don't forget to allow for bit stuffing, the identifier field, the length field, and the CRC field. At 500 kbps your budget is 50 bits. The fixed overhead is 44 bits, so there is no room in a standard frame for even 1 byte of data. A one byte data frame will take 52 bits plus bit stuffing.