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CAN & CANopen DISCUSSION
I built a small CAN network with four nodes on about 8 m of twisted pair at 500 kbit/s. It works with one 120 Ω resistor, works with two, and even seemed to work for a while with none on a very short test cable. Two of my modules also have a termination jumper fitted by default.
Why is the value 120 Ω and why two of them? How do I check on an installed bus that the termination is correct?
Termination does two jobs on CAN. The first is the usual transmission-line one: a resistor equal to the cable's characteristic impedance absorbs the signal at the end of the line instead of reflecting it. The cable specified for CAN is twisted pair with a nominal 120 Ω impedance, so each physical end gets 120 Ω between CANH and CANL.
The second job is specific to CAN. Transceivers actively drive only the dominant state; to return to recessive they simply let go. The termination resistors are what pull the differential voltage back to zero. With no termination, the recessive edge is a slow discharge through the receivers' input resistance, which is why an unterminated bus fails even on a short cable once the bit rate rises.
Two resistors because a line has two ends, and the transceivers are specified for that load: two 120 Ω resistors in parallel present 60 Ω, and transceiver output levels are characterised for a nominal 60 Ω bus load. One resistor (a 120 Ω load) often works on a short bus since the recessive return still happens, but one end is reflecting.
Three or four terminators (40 Ω or 30 Ω) overload the drivers: the dominant differential voltage shrinks and errors appear at the far nodes. So those default jumpers matter. Only the two nodes at the physical ends should have them fitted.
The multimeter check: power everything down and measure the resistance between CANH and CANL anywhere on the bus. About 60 Ω means two terminators. About 120 Ω means one is missing, or a wire is broken somewhere between the meter and the far terminator. About 40 Ω means a third terminator is fitted. Measure with the power off, because active transceivers and bias voltages distort the reading.
With power on and the bus idle, both lines should read about 2.5 V to ground. With traffic, a meter shows CANH slightly above and CANL slightly below 2.5 V, since it averages the dominant bits.
A refinement worth using on new designs is split termination: replace each 120 Ω resistor with two 60 Ω resistors in series and connect their midpoint to ground through a capacitor, commonly 4.7 nF. The differential termination is unchanged, but common-mode noise now has a low-impedance path to ground, which improves emissions and immunity. The two 60 Ω resistors should be well matched, so use 1 percent parts.
Also think about the resistor power rating on vehicle buses. In the worst case a wiring fault that puts 12 V or 24 V across a 120 Ω resistor dissipates 1.2 W or 4.8 W, far beyond a small chip resistor.
On topology: the terminators belong at the two ends of a single line, with nodes attached by short stubs. The guideline for 1 Mbit/s is to keep each stub under about 0.3 m; at lower bit rates longer stubs are tolerated. A star of several long branches has no two ends to terminate, and it is the usual reason a bus that measures 60 Ω still shows ringing and error frames.
With 8 m and four nodes at 500 kbit/s, a simple daisy chain with the two jumpers fitted at the end nodes is all you need.
