I know the reasons for using terminating resistors on a CAN bus and how important it is.
But why 120 ohm? How did this value come up? Is there any specific reason to use 120 ohm?
Why does the CAN bus use a 120 ohm resistor as the terminating resistor and not any other value
cantermination
Related Solutions
You seem to have your answer from the comments, so this is mainly for posterity purposes.
Your calculations are correct, at a max of 5V differential voltage you would have ~42mA which would dissipate ~210mW. You are safe using a 0.25W resistor, especially since you know exactly what is going to be connected to the bus.
I would like to add that I work with CAN on a regular basis, some of which can be in rather demanding environments, and our typical termination is shown below. We use standard 0805 resistors with a rating of 0.125W and I have yet to see a burnt resistor.
simulate this circuit – Schematic created using CircuitLab
What you have is essentially an unterminated stub. This TI application note mentions on page 9 the maximum length allowed for a stub according to the CAN specification. It is 0.3 m, slightly longer than your 10, therefore I think you should be fine.
If you can change the slew rate of your CAN driver, as they suggest, then you might have longer unterminated stubs, but it does not apply to your case, since your HMI panel probably cannot change its slew rate.
edit: Oops, mixed 10 inches with 10 feet, which is way larger than the maximum allowed stub.
Best Answer
You need to be familiar with Transmission Line Theory to understand the deeper physics in play here. That said, here's the high-level overview:
How important termination is to your system is almost exclusively determined by how long the bus wires are. Here length is determined in terms of wavelengths. If your bus is shorter than one wavelength over 10, the termination is irrelevant (practically) since there is plenty of time for the reflections introduced from an impedance mismatch to die out.
Length defined in wavelengths is a strange unit on first encounter. To convert to standard units you need to know the velocity of the wave and it's frequency. Velocity is a function of the medium it travels through and the environment surrounding the medium. Usually this can be estimated fairly well through the dielectric constant of the material and assuming free-space surrounding that medium.
Frequency is a little more interesting. For digital signals (such as those in CAN), you are concerned with the maximum frequency in the digital signal. That is well approximated by f,max = 1/(2*Tr) where Tr is the rise time (defined 30%-60% of the final voltage level, conservatively).
Why it's 120 is simply a function of the design limited by physical size. It isn't specifically important which value they picked within a broad range (for example, they could have gone with 300 Ohms). However, all devices in the network have to conform to the bus impedance, so once the CAN standard was published there can be no more debate.
Here's a reference to the publication (Thanks @MartinThompson).