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Wattage Requirements for CAN-Bus Terminating Resistor with SN65HVD233

Other Parts Discussed in Thread: SN65HVD233, SN65HVD255

This post concerns the SN65HVD233 CAN transceiver device.

We use a Device Net cable with 5 conductors.  Two of the conductors carry the CANH and CANL signals, and two conductors carry 24V and GND.  We have had several instances in the field where a 'bad' cable causes 24V to be shorted to CANH or CANL.  In that situation, what is the wattage requirement of the 120 ohm CAN-bus terminating resistor?  See my attached pdf drawing that helps explain the situation.

The SN65HVD233 datasheet indicates that the "Short Circuit Output Current (Ios)" can be 250mA at 12V.  In our case, the voltage is 24V instead of 12V.

The calculations in the attached document indicate that the resistor needs to be at least a 7.5W resistor, which seems very large in comparison to the 0.25W value that is recommended in most CAN-bus applicaiton notes.

Perhaps I am miss-interpreting the datasheet?


CAN-Bus Terminating Resistor.pdf
  • Hi Eric,

    Very good question and I will try to expaling the missing points about how CAN and the system work and why many of these app notes got down to 0.25W.  In your case it will not get down to 0.25W and I would normally recommend at least 0.5W in a 12V system and at least 1W in a 24V system as a more conservative assumption as will be outlined.

    1.  CAN is a switching device with data and will normally be at recessive and thus not really a pure DC value
    1. The IOS (short circuit current) you show is only for a dominant (logic low) when the CAN driver is trying to drive the bus, and is the short straigh into the device. Per the picture not much current will flow through the resistor into CANH at it will not try to sink a high positive voltage but a the short circuit protection amount from the datasheet will flow from the short into CANL.   
    2. A proper CAN system will not have a DC dominant with the high DC current unless you have 2 faults:  D pin of the HVD233 is driven constantly low and the CANL pin is shorted to the 24V.  In a normal CAN system if you short the bus you may have high current during the dominant portions of CAN messages as it tries to send dominant bits but eventually the uP's CAN controller will go to bus off and doesn't transmit anymore.  Depending on how many faults and what level of "worst case" you want to see the CAN data it would be all domianants for the frame except for the stuff bits and recessive bits pre-set in the protocol.  This is very unlikely but if you run with it you could get 5/6 bits domianant in a row.  Most likely in these cases you will hit the thermal shut down of the CAN transceiver which will shut the driver off and then only leakage currents will flow.  The real worst case for the resistor is actually a short of one bus line to the supply and the other to GND.  
    3. At 24V across 120 ohm the resistor would current limit to 200mA (i=V/R=24/120=0.2A) which is below what the CAN short circuit will limit to so worst case double fault math would lower your power to 0.2^2 x 120 = 4.8W.  If you scale for 5/6 dominant due to bit stuffing you get to 4W.  But a more realistic case is really looking at most networks are designed for only about 50-60% bus loading and then 50% dominant vs recessive bits would give you closer to 1.2W as a sort of average power in the resistor under the short circuit.  This 1.2W scales pretty well to the normal automotive assumption for 12V where they ask for 0.5W resistors. 
    4. The CAN pin short circuit current depends on which pin and what level the short its.  A short of +24V to CANH will not have much current as it will not try to sink it.  CANL is where the higher short circuit currents will be for positive voltages. CANH will try to source a lot of current if it is about 2.5V or lower (ground and going negative voltages). 
    • Your resistor power choice needs to depend on how many faults you want to design for at the same time.  The most conservative approach would be 4.8W which will allow for a CANH or L short to the power supply and the other bus pin to GND.  This also covers a similiar bus line short to 24V and a software or PCB fault driving D low permanently.  If you can de-rate it based on the CAN system aspects then you could derive to the lower power termination resistances.

    Below is an application section out of our SN65HVD255/6/7 datasheet on this topic which I actually working on adding into the HVD23x datasheets plus some other system level hints.  You may also want to look at page 15 of the SN65HVD255 datasheet for a short description of split termination which may also be helpful to you (this is also being added to the HVD233 datasheet shortly).  Hopefully these all help you come to the decision you want to take for your system assumptions. 

     Best Regards, Scott

    CAN Bus Short Circuit Current Limiting

    The device has several protection features that limit the short circuit current when a CAN bus line is shorted. These include CAN driver current limiting (dominant and recessive). The device has TXD dominant state time out which prevents permanently having the higher short circuit current of dominant state in case of a system fault. During CAN communication the bus switches between dominant and recessive states, thus the short circuit current may be viewed either as the current during each bus state or as a DC average current. For system current and power considerations in the termination resistors and common mode choke ratings the average short circuit current should be used. The percentage dominant is limited by the TXD dominant time out which prevents permanently driving dominant and CAN protocol has forced state changes and recessive bits such as bit stuffing, control fields, and interframe space. These ensure there is a minimum recessive amount of time on the bus even if the data field contains a high percentage of dominant bits.

     APPLICATION NOTE: The short circuit current of the bus depends on the ratio of recessive to dominant bits and their respective short circuit currents. The average short circuit current may be calculated with the following formula:

           IOS(AVG) = %Transmit * [(%REC_Bits * IOS(SS)_REC) + (%DOM_Bits * IOS(SS)_DOM)] + [%Receive * IOS(SS)_REC]

     Where IOS(AVG) is the average short circuit current, %Transmit is the percentage the node is transmitting CAN messages, %Receive is the percentage the node is receiving CAN messages, %REC_Bits is the percentage of recessive bits in the transmitted CAN messages, %DOM_Bits is the percentage of dominant bits in the transmitted CAN messages, IOS(SS)_REC is the recessive steady state short circuit current and IOS(SS)_DOM is the dominant steady state short circuit current.

    APPLICATION NOTE: The short circuit current and possible fault cases of the network should be taken into consideration when sizing the power ratings of the termination resistance and other network components.  

    Thermal Shutdown

    If the junction temperature of the device exceeds the thermal shut down threshold the device will turn off the CAN driver circuits thus blocking the D to bus transmission path. The shutdown condition is cleared once the temperature drops below the thermal shutdown temperature of the device.

    APPLICATION NOTE: During thermal shutdown the CAN bus drivers will be turned off thus no transmission is possible from D to the bus.  The CAN bus pins will be biased to recessive level during a thermal shutdown and the receiver to R path will remain operational.

  • Scott,

    Thank you for the incredible detail in your response.  I could not have imagined such a thorough answer to my question.  Your response was VERY helpful.  Thanks.