[SI-LIST] Re: Canbus EMC/SI questions?

  • From: steve weir <weirsp@xxxxxxxxxx>
  • To: javodv@xxxxxxxx, <si-list@xxxxxxxxxxxxx>
  • Date: Mon, 03 Mar 2003 11:29:54 -0800

Javier,

CanBus is a double terminated, multidrop bus.  Each driver / receiver will 
appear to be open when it is receiving.  The 60 ohms arises from the 
parallel 120 ohm terminations on either side of the bus.   Ideally, each 
point along the bus sees 120 ohms looking East to West, and 120 ohms 
looking West to East.  So you want 120 ohm twisted pair.

The AC termination scheme terminates the common mode part of the waveform.

There are three voltages that you are dealing with:

The + side to common
The - side to common
and the difference between + and -.

There are a couple of ways to express these, but they are all pretty much 
different ways of saying the same thing.  It all just depends what you want 
to emphasize.  The bus is DC coupled, so the common mode voltages have to 
remain within reasonable limits, or the receivers will not work.  All those 
specs mean is that when + is high and - is low, there is a 2 volt 
difference, 3.5V +  - 1.5V - = 2V. When the signal switches to the opposite 
state, it is -2V difference, 1.5V +  - 3.5V- = -2V.  Aside from distortion 
the common mode voltage, remains unchanged at Vref:  ( 3.5V + 1.5V ) / 2 = 
( 1.5V + 3.5V ) / 2 = 2.5V.

You definitely want to put the + and - on one pair.  If you need to carry 
extra power, carry that in other pairs.  If you put the + and - in 
different pairs, you will not get the right impedance, have large area 
loops, bad signal quality, and lots of EMI.

As long as the bus is properly terminated, the cable run is not going to 
present a problem.  Personally, I use 1ns / inch as the criteria for determin

You can get the CanBus standard from Bosch.

Regards,


Steve.


At 07:15 PM 3/3/2003 +0100, Javier del Valle wrote:
>Hello to all,
>
>I am a bit confused with the termination of the Canbus. I was wondering if
>you experts could help me because I think that the voltage
>pattern is a bit similar to LVDS.
>
>A common Canbus voltage pattern is like this:
>
>VOLTAGE
>                                   _________ <------------------   3.5Volts
>                                  /                   \
>---------------------/                      \------------------- CAN_H
>---------------------                        -------CAN_L <--  2.5 Volts
>                                 \                      /
>                                  \__________/<---------------- 1.5 Volts
>
>_____________________________________________  TIME
>
>  RECESSIVE          DOMINANT      RECESSIVE
>
>
>   ____________             ___________
>   |   Node 1       |             |   Node 2       |
>   | __________|             |___________|
>        |        |                          |         |
>        |        |                          |         |
>|----x---- |-----------------x------|--------|
>|               |                                    |            |
>Ro           |                                    |           Ro
>|               |                                    |            |
>|----------x-----------------------x--------|
>
>
>Ro is 120 Oms. I have also seen instead of using 120 Ohms the circuit below
>with R1= 60 Ohms and C1=47nF.
>
>x
>|
>R1
>|
>|----C1----GND
>|
>|
>R1
>|
>x
>
>
>"ISO 11898 specifies a bus cable including the two signal wires CAN_H and
>CAN_L with a nominal characteristic impedance of 120 Ohms.. each bus node
>shall be capable of providing a differential output voltage between 1.5 and
>3.0 at a resistive load of 60 Ohms".
>
>I understand, correct me if I am wrong, that the differential impedance that
>the transceiver see at its output is 60 Ohms  (Because each node sees at its
>output the parallel of 120 Ohms and 120 Ohms, then it sees the differential
>impedance of 60 Ohms). Then the CANBUS cable should have a "differential"
>impedance of 120 Ohms. So I am a bit confused because I think that I
>should choose a Canbus cable that provides me with this 120 Ohms
>differential impedance. This is different from choosing a cable in which the
>CAN_H and CAN_L  lines have a Zo=120 Ohms. In fact, if I want a
>differential impedance of 120Ohms then I should choose a cable that has
>CAN_HZo= 60Ohms and CAN_LZo = 60 Ohms aprox.
>
>Zdiff=2.Zo(1-k)
>
>Am I wrong or forgetting something?
>
>- Another thing that is confusing me is that the voltage pattern applied to
>the differential pair in CANBUS is a mix of an odd and even patter.
>(Vref=2.5Volts and X=1Volts)
>
>      CAN_H = Vref + X
>      CAN_L = Vref  - X
>
>So if I use a cable with a differential impedance of 120 Ohms (Zo of each
>one near 60Ohms) the odd voltage pattern is mached but what happen with the
>even voltage??. Is it better to use a  two pairs cable (one pair for the
>CAN_L and CAN_H and the other pair for the GND) and also shielding or
>it is the same if I use a single pair for CAN_L and CAN_H and the shielding
>of this two pair cable to GND. In my case the cables are used inside a
>  machine that generate much EMI. Any suggestion or recommendation
>for Canbus connection?
>
>The texas canbus transceiver SN65HVD230 for example has a the minimun
>differential rise time of 25ns. Then if I use the delay rule of thumb of
>5ns/m of cable, and that a line is a transmission line when Tline is > 20% of
>rise time. Then when the line is more than one meter I should have in mind
>trasmission line concepts. Should I have in mind this kind of things for a
>canbus communication at 125Kbps-1Mbps or am I being too pesimistic?.
>
>Sorry for hitting you with so many questions,
>
>Thanks a lot and Best regards,
>
>Javier
>
>
>
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