Is there any derived equation/formula to calculate the bus load?
I need transmit below frames(Assume Standard frame format) at baud rate 500 kbps on the bus
5ms ----- 50 frames.
10ms ----- 10 frames.
100ms ---- 30 frmaes.
Event messages ---- 10 frames.
How to calculate bus load, does anyone knows kindly illustrate the same?
Bus load = Used capacity / Max capacity. In practice this means how much of the possible bandwidth that was used over a certain period of time.
Assuming standard identifier, a CAN frame consists of:
So the amount of bits is variable depending on how much data there is and also on how many stuffing bits you end up with. Ignoring stuffing, you have an overhead of 47 bits/frame. The maximum frame size with 8 bit data with worst-case stuffing is something like 47+64+19=130.
Lets say you use 250kbps and measure bus load over 10ms intervals. The theoretical maximum amount of bits is then 10ms / (1/250kbps) = 2500 bits. But this is of course not necessarily divisible by full CAN frames.
If you send 10 frames with 250kbps during 10ms, each frame being 11 bit identifier, no data and no stuffing, then you end up with 47*10 = 470 bits, out of the theoretical maximum 2500. 470/2500 = 18.8% bus load.
Whereas 10 of my standard id rough worst-case 130 bits/frame, would give 1300/2500 = 52% bus load.
In practice, tools that measure bus load just clock the time after the end of each intermission field, until the next dominant start bit of another frame appears, then sum all such time periods.
Afaik, bus load is a percentage.
bus load = #bytes send / speed
e.g. 1000 bytes (to be send in a second) * 8 (bits/byte / 500,000 (baudrate bps) * 100% = 1.6 %
In your case you are sending per second:
note: 76 is assumed the average bits per message (32 bit data), this depends on what you define, so it can vary. See the CAN bus message in Wikipedia.
Total: 760,000 + 76,000 + 22,800 = 858,800
Bus load = 858,000 / 500,000 * 100% = 171,76%
This already is too much.
ID allocation
In the early 1990s, the choice of IDs for messages was done simply on the basis of identifying the type of data and the sending node; however, as the ID is also used as the message priority, this led to poor real-time performance. In those scenarios, a low CAN bus use of circa 30% was commonly required to ensure that all messages would meet their deadlines. However, if IDs are instead determined based on the deadline of the message, the lower the numerical ID and hence the higher the message priority, then bus use of 70 to 80% can typically be achieved before any message deadlines are missed.
So if you are using IDs based on priority, 70-80% is ok, and you are well below that number.
Load in CSMA/CA system like CAN is totally application dependant and need to be evaluated case by case, given the quality of service required. In theory, as already stated, you just need to pick the maximum frame length (assuming average stuffing), put some recessive bits between frames and every so often a dozen more to resync joining station. So, from a strictly theorical point, you could reach some 99.9% bus utilization.
The real question is? your 100ms frames how late can arrive? how much jitter is permissible on repetition rate? Do your frames have all 8 bytes payload or is there some compression scheme? Are maybe some event more important than your 5ms frames?
Process bus scheduling can be extremely tricky. There are CAN based time slot protocols, too (TT CAN, for example), the analysis is completely different in that case.
