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.
Note that the terminating resistors do DC-load the signal.
Having a terminating resistor matching the characteristic impedance is for AC functionality - making sure there are no signal bounces.
For a very short cable, a single 120 ohm resistor would normally work. But may not be needed depending on propagation delay compared to your transfer rate.
For short cables, you can often get by without any termination unless the baudrate is very high.
How about your engineers verifies explicitly what reflections they have in the cable? No need to have extra loads if there aren't really any problems with reflections - less load is less power wasted.
Close, but not quite correct. Whether working without a terminator is possible at all depends primarily on which of several physical layers for CAN is in use; speed only comes into play because different PHYs have different ranges of allowed baud rates.
There's the single-wire PHY used by GM that has no termination at all. There's the "low-speed", fault-tolerant CAN PHY which usually doesn't need a terminator, but maxes out at 125 kBit/s. And then there's "high-speed" differential-only PHY, which will not work at all without some kind of termination, but goes up to 1 MBit/s in return. Not to mention CAN-over-fiber.
But when the debate is about 120 ohm or 60 om termination, I think we can rule out your CAN-over-fiber or sing-ewire PHY from the debate.
Many thanks for the helps.
"verifies explicitly what reflections they have in the cable" Can this be done by an oscilloscope? I don't think we know how to do this correctly.
Our baudrate is 500 kBit/s. Does one 60-Ohm resistor provide the same funtionality as two 120-Ohm termination resistors?
Does one 60-Ohm resistor provide the same funtionality as two 120-Ohm termination resistors?
No. Because you'll have reflections on your cable that way. On short cables you may not notice any difference, but if you use a sufficiently long run of cable (> 50 meters, roughly), and use a scope fast enough to measure timing differences below 100 ns, you'll see the difference.
In a nutshell, a single termination resistor is always wrong --- the exact gravity of the inevitable consequences depends on your network's propagation delay compared to the bit time.
now it's CAN, it used to be RS485 these questions were about. However there is only one and the same answer:
The only way to ensure proper transmission is proper termination
Ever so many have "gotten by" without proper termination till, some day, someone intalled the equipment in a different place than the testbench.
what value does (mainly for testing purpose) have, if you are not testing under the conditions that exist for shipped equipment.
Erik
Depends on what is about to be tested.
If testing some software, it may be ok to have just a 10 cm cable between two devices.
====================> en.wikipedia.org/.../CAN_bus
ISO11898-2 describes the electrical implementation formed from a multi-dropped single-ended balanced line configuration with resistor termination at each end of the bus. In this configuration a dominant state is asserted by one or more transmitters switching the CAN- to supply 0 V and (simultaneously) switching CAN+ to the +5 V bus voltage thereby forming a current path through the resistors that terminate the bus. As such the terminating resistors form an essential component of the signalling system and are included not just to limit wave reflection at high frequency.
Does the "the terminating resistors form an essential component of the signalling system and are included not just to limit wave reflection at high frequency" apply to the terminating resistors of CAN-Bus particularly? or it applies to every kind of terminating resistors? <====================
====================> In a nutshell, a single termination resistor is always wrong --- the exact gravity of the inevitable consequences depends on your network's propagation delay compared to the bit time.
So, if
the configuration of "propagation delay compared to the bit time" is not proper, and only one single termination resistor is used,
then even on a low baudrate CAN-Bus, the signals could be damaged? (Am I right?) <====================
Many thanks for your experiences and wisdoms.
_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.
then even on a low baudrate CAN-Bus, the signals could be damaged? (Am I right?)
You're asking the wrong question. Why even bother with what exactly will go wrong if you don't terminate properly, when it's so enormously much cheaper just to terminate correctly and be done with it? You've wasted more time (and thus money) on writing here already than proper termination could possibly cost.
That said, yes, not only could they be, they will be damaged. CAN's MAC layer relies on any node on the bus being able to overpower each recessive bit on the bus into a dominant bit, within that same bit time, and all nodes on the bus seeing that dominant state instead of the recessive one. Improper termination causes reflections, which effectively shortens the usable bit time, because a new bit cannot generally start successfully until the reflections from the previous edge, a sort of ringing effect, have died out. So sooner or later these reflections will keep one node on the bus from detecting the following bit correctly, and thus signals will indeed be damaged.
The question the OP ask is "what happens if I do not do it right" what an utterly stupid question, what excuse does any engineer have for not doing it right (at least when the cost difference, like here, is $0.01). the answer is "if you do not do it right it will fail on the unit farthest away from you at the most inconvenient time" (Murphy's law).
The question "what happens if I do not do it right" can also be seen as an expectation that someone, knowing that things must be done right, spent hours, upon hours analyzing something he was not going to do in the first place.
Why even bother with what exactly will go wrong if you don't terminate properly, when it's so enormously much cheaper just to terminate correctly and be done with it? You've wasted more time (and thus money) on writing here already than proper termination could possibly cost.
The question the OP ask is "what happens if I do not do it right" what an utterly stupid question, what excuse does any engineer have for not doing it right (at least when the cost difference, like here, is .01). the answer is "if you do not do it right it will fail on the unit farthest away from you at the most inconvenient time" (Murphy's law).
I am a totally and completely software guy, don't know anything about hardware and signals. In this company, only 2 or 3 persons believe "one 60-Ohm instead of two 120-Ohm resistors" is wrong. And I need to convince the hardware engineers to do something for me.
Many thanks for your experiences and wisdom.
Hans-Bernhard and Per,
Many thanks for sharing your experiences and wisdom.
Both of you emphasizing the "overpower each recessive bit on the bus into a dominant bit", so I believe this is one of the key-points.
I did some more quick reading, cache.freescale.com/.../AN1798.pdf
2.3 Propagation Segment
The existence of the propagation delay segment, PROP_SEG, is due to the fact that the CAN protocol allows for non-destructive arbitration between nodes contending for access to the bus, and the requirement for in-frame acknowledgement. In the case of non-destructive arbitration, more than one node may be transmitting during the arbitration field. Each transmitting node samples data from the bus in order to determine whether it has won the arbitration, and also to receive the arbitration field in case it loses arbitration. When each node samples each bit, the value sampled must be the logical superposition of the bit values transmitted by each of the nodes arbitrating for bus access. In the case of the acknowledge field, the transmitting node transmits a recessive bit but expects to receive a dominant bit, i.e. a dominant value must be sampled at the sample point(s). The propagation delay segment, PROP_SEG, exists to delay the earliest possible sample of the bit by a node until the transmitted bit values from all the transmitting nodes have reached all of the nodes.
2.4 Synchronisation
Hard synchronisation is performed only at the beginning of a message frame, when every CAN node aligns the SYNC_SEG of its current bit time to the recessive to dominant edge of the transmitted Start-Of-Frame bit. Re-synchronisation is subsequently performed during the remainder of the message frame, whenever a change of bit value from recessive to dominant occurs outside of the expected SYNC_SEG segment.
www.softing.com/.../signal-propagation.php
You have to add up the following delays depending on the selected components: CAN controller (50 ns to 62 ns), optocoupler (40 ns to 140 ns), transceiver (120 ns to 250 ns), and cable (about 5 ns/m).
Improper termination causes reflections, which effectively shortens the usable bit time, because a new bit cannot generally start successfully until the reflections from the previous edge
I don't think I have already understood the above texts completely and correctly. Could you please provide some more hints?
Sorry for my very limited English ability and Technical ability.
If one listener is close and another is far away on the cable, then the message takes different time to reach them.
And the buffers used means that from the transmitter wanting to send something, it takes some while until the signal have jumped through the transceiver or opto-isolator and even reached the bus.
So the transmitter must be patient and make sure that all listeners have had time to get the bit and sample it - and potentially respond and have this response reach back to the transmitter - before the transmitter can check if cable is in dominant or recessive state.
But if the cable is improperly terminated, or there are long stubs cables from the bus cable out to some listeners, or someone have splitted the bus, then there will be reflections in the cable. These reflections are added to the actual transmitted signal. The reflection will also move with the same propagation speed of the cable, so for a longer cable, the reflection will manage to destroy a longer part of the initial part of a transmitted bit. That is why you need to switch to slower transfer speeds when you increase the cable length - so your initial sampling happens after the reflections no longer affects the correct detection of low/high signal state.
Several of the links I posted do show oscilloscope images showing the result on the signal from these reflections. As can be seen, it is vital to minimize reflections if you want to run a high bitrate on a long cable.