Sorry for my limited English ability and Technical ability.
It is very common in our company that, the engineers don't care about (or don't know) the termination resistors on a CAN-Bus. They just simply put one 60-Ohm resistor between two or several CAN-Devices (mainly for testing purpose). I know this is not standard, and I don't like this workaround. I tried to figure out why such a workaround is bad, but failed.
It seems that many engineers use such a workaround. ==============================> www.microchip.com/.../m177894-print.aspx a single 60 ohm termination for lab testing when bus is only a few feet (probably not allowed, but it works).
www.microchip.com/.../m144034-print.aspx If you don't have two 120-ohm resistors, you can accomplish the termination (on a small network) with a single 60 ohm resistor(120 in parallel with 120 is 60) or anything close 55-65 is fine. <==============================
I found a document saying this workaround is a mistake, but it does not provide why. ==============================> www.scribd.com/.../SAE-J1939 The bus is linear and should be terminated with 120-Ohm resistors at either end. It is a common mistake to use one 60-Ohm resistor instead of two 120-Ohm resistors. This does not work correctly, however. <==============================
What is the disadvantage/mistake, if we use one 60-Ohm resistor instead of two 120-Ohm termination resistors on a CAN-Bus, assuming the communication distance is not very long and only 3-5 CAN-Nodes (Devices) involved? In another words, when will such a workaround fail to work?
-- It is very difficult to find another job in southern Taiwan. So I am still struggle with idiotic things, and busy in urgent projects.
_Some_ busses requires the resistors for both DC purposes and for stopping reflection.
_Some_ busses only needs the resistors for stopping reflection.
And for CAN, there exists more than one physical layer standard (just as Ethernet have thin and thick coaxial cables and TP cables). Some variants are very similar to RS-485.
Since the wire have quite low DC resistance, it is easy to believe that two 120 ohm resistors with a low-resistance cable between them is equivalent with a 60 ohm resistor. It is - but only for the DC case.
For the AC case - i.e. taking care of reflections - a 60 ohm resistor is not equivalent. A 60 ohm resistor is much off compared to the characteristic impedance of the cable. And next thing - where to place this single resistor since it can't be at both ends of the cable at the same time.
Anyway, the case of recessive and dominant bits is a big different from many other busses, since all transceivers on the cable are involved. It is a form of an open-collector drive, where multiple transmitters can start to send at the same time - if one transmitter then "sends" a recessive bit while another sends a dominant bit, the guy sending a recessive bit will notice on the cable that the transmission failed. So it will stop sending and wait for a later time to try again. The transmitter(s) that sent a dominant bit at that position will not see any errors, so it/they will continue to send more frame bits.
You can find a bit of information in this application note for a TI chip: www.ti.com/.../slla270.pdf
Note that this specific document is for ISO 11898:1993 and relates to high-speed CAN.
Many other busses do not have any recessive/dominant bits but instead use other means to figure out who may send, to avoid collissions, and what means to use for acknowledge. Maybe a specific master. Maybe some form of token passing. Maybe time slots. In that case, they can make use of a fully symmetric definition of the bits.