What is the limiting factor for a CAN bus to exceed 1Mbps bandwidth?What is the maximum bitrate supported in...

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What is the limiting factor for a CAN bus to exceed 1Mbps bandwidth?


What is the maximum bitrate supported in the Can Bus?CAN bus bit timing with 16 MHz crystalSending CAN protocol data(1Mbps) via serial portCAN bus troubleshooting. How?Kvaser Leaf Light CAN BUS Simulator problemsCAN bus TVS diode/EMI for motorcycleCAN Bus Debug MCP25625 CAN to SPI to USB (MCP2210)Interoperability between regular and single-wire CAN busHigh speed CAN configurationHow to calculate bus load of CAN bus?CAN bus bit timing STM32F7 (bxCAN)






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$begingroup$


Why can't CAN baud rate increase beyond 1Mbps










share|improve this question







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  • $begingroup$
    Depends on standards and drivers used. There are faster versions but there are other differences,
    $endgroup$
    – Sunnyskyguy EE75
    6 hours ago












  • $begingroup$
    My question is why can't we achieve higher speeds like upto 100Mbps using CAN?
    $endgroup$
    – Vinay Veeramaneni
    6 hours ago










  • $begingroup$
    Ignition and RF Immunity and delay contention on the bus
    $endgroup$
    – Sunnyskyguy EE75
    6 hours ago












  • $begingroup$
    Check: What is the maximum bitrate supported in the Can Bus
    $endgroup$
    – Huisman
    5 hours ago










  • $begingroup$
    the CAN bus has no overt synchronization, except for detecting collisions.
    $endgroup$
    – analogsystemsrf
    15 mins ago


















3












$begingroup$


Why can't CAN baud rate increase beyond 1Mbps










share|improve this question







New contributor




Vinay Veeramaneni is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$












  • $begingroup$
    Depends on standards and drivers used. There are faster versions but there are other differences,
    $endgroup$
    – Sunnyskyguy EE75
    6 hours ago












  • $begingroup$
    My question is why can't we achieve higher speeds like upto 100Mbps using CAN?
    $endgroup$
    – Vinay Veeramaneni
    6 hours ago










  • $begingroup$
    Ignition and RF Immunity and delay contention on the bus
    $endgroup$
    – Sunnyskyguy EE75
    6 hours ago












  • $begingroup$
    Check: What is the maximum bitrate supported in the Can Bus
    $endgroup$
    – Huisman
    5 hours ago










  • $begingroup$
    the CAN bus has no overt synchronization, except for detecting collisions.
    $endgroup$
    – analogsystemsrf
    15 mins ago














3












3








3





$begingroup$


Why can't CAN baud rate increase beyond 1Mbps










share|improve this question







New contributor




Vinay Veeramaneni is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.







$endgroup$




Why can't CAN baud rate increase beyond 1Mbps







can






share|improve this question







New contributor




Vinay Veeramaneni is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











share|improve this question







New contributor




Vinay Veeramaneni is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









share|improve this question




share|improve this question






New contributor




Vinay Veeramaneni is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









asked 6 hours ago









Vinay VeeramaneniVinay Veeramaneni

162




162




New contributor




Vinay Veeramaneni is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.





New contributor





Vinay Veeramaneni is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






Vinay Veeramaneni is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.












  • $begingroup$
    Depends on standards and drivers used. There are faster versions but there are other differences,
    $endgroup$
    – Sunnyskyguy EE75
    6 hours ago












  • $begingroup$
    My question is why can't we achieve higher speeds like upto 100Mbps using CAN?
    $endgroup$
    – Vinay Veeramaneni
    6 hours ago










  • $begingroup$
    Ignition and RF Immunity and delay contention on the bus
    $endgroup$
    – Sunnyskyguy EE75
    6 hours ago












  • $begingroup$
    Check: What is the maximum bitrate supported in the Can Bus
    $endgroup$
    – Huisman
    5 hours ago










  • $begingroup$
    the CAN bus has no overt synchronization, except for detecting collisions.
    $endgroup$
    – analogsystemsrf
    15 mins ago


















  • $begingroup$
    Depends on standards and drivers used. There are faster versions but there are other differences,
    $endgroup$
    – Sunnyskyguy EE75
    6 hours ago












  • $begingroup$
    My question is why can't we achieve higher speeds like upto 100Mbps using CAN?
    $endgroup$
    – Vinay Veeramaneni
    6 hours ago










  • $begingroup$
    Ignition and RF Immunity and delay contention on the bus
    $endgroup$
    – Sunnyskyguy EE75
    6 hours ago












  • $begingroup$
    Check: What is the maximum bitrate supported in the Can Bus
    $endgroup$
    – Huisman
    5 hours ago










  • $begingroup$
    the CAN bus has no overt synchronization, except for detecting collisions.
    $endgroup$
    – analogsystemsrf
    15 mins ago
















$begingroup$
Depends on standards and drivers used. There are faster versions but there are other differences,
$endgroup$
– Sunnyskyguy EE75
6 hours ago






$begingroup$
Depends on standards and drivers used. There are faster versions but there are other differences,
$endgroup$
– Sunnyskyguy EE75
6 hours ago














$begingroup$
My question is why can't we achieve higher speeds like upto 100Mbps using CAN?
$endgroup$
– Vinay Veeramaneni
6 hours ago




$begingroup$
My question is why can't we achieve higher speeds like upto 100Mbps using CAN?
$endgroup$
– Vinay Veeramaneni
6 hours ago












$begingroup$
Ignition and RF Immunity and delay contention on the bus
$endgroup$
– Sunnyskyguy EE75
6 hours ago






$begingroup$
Ignition and RF Immunity and delay contention on the bus
$endgroup$
– Sunnyskyguy EE75
6 hours ago














$begingroup$
Check: What is the maximum bitrate supported in the Can Bus
$endgroup$
– Huisman
5 hours ago




$begingroup$
Check: What is the maximum bitrate supported in the Can Bus
$endgroup$
– Huisman
5 hours ago












$begingroup$
the CAN bus has no overt synchronization, except for detecting collisions.
$endgroup$
– analogsystemsrf
15 mins ago




$begingroup$
the CAN bus has no overt synchronization, except for detecting collisions.
$endgroup$
– analogsystemsrf
15 mins ago










3 Answers
3






active

oldest

votes


















3












$begingroup$

It's because the CAN 2.0B standard did not specify any higher in order to reduce hardware costs and still meet the various requirements of the standard (like distance and noise immunity). It's not a technical barrier.



The standard was written that way probably since they deemed the extra speed unnecessary for the intended application, and specifying a higher speed needlessly would increase the cost of all the hardware supporting the standard when the capability would be underutilized.



If the standard is written that way, few IC manufacturers will bother trying to exceed it since there's no point. It's not really a technical barrier.






share|improve this answer









$endgroup$





















    3












    $begingroup$

    From Controller Area Network Physical Layer Requirements



    CAN termination




    CAN is an open collector technology – the protocol could not work otherwise. This means that the recessive state of a CAN transceiver is not actively driven. The termination resistors together with transceiver input capacitance and cable capacitance create an RC time-constant discharge when an actively-driven dominant bit on the bus transitions to an un-driven recessive bit. For signaling rates greater than CAN's 1Mbps, a technology that actively drives the bus in both states such as RS-485 is required to facilitate the bus transitions required for high-speed signaling rates.




    Ultimately, the answer to the question is how the CAN protocol is implemented at a physical level. Change that protocol and a higher data rate can be used.



    From Understanding Microchip’s CAN Module Bit Timing




    ... the CAN protocol implements a non-destructive bitwise arbitration scheme that allows multiple nodes to arbitrate for control of the bus.
    Therefore, it is necessary for all the nodes to detect/ sample the bits within the same bit time. The relationship between propagation delay and oscillator tolerance effect both the CAN data rate and the bus length.




    enter image description here



    Two masters on either end of the CAN bus must be able to communicate and arbitrate which one has the bus, while each are on the bus at the same time.



    If the bus length is 30m, the time it takes to propagate the signal over the bus is: $$t_{BUS} = 30m @ 5.5 ns/m = 165ns$$



    Assuming the input comparator delay is $t_{CMP}$ = 40ns and the output driver delay is $t_{DRV}$ = 60ns for all devices.



    The round trip time for a bit on the physical bus will be:



    $$t_{PROP} = 2(t_{BUS} + t_{CMP} + t_{DRV}) = 2 (165ns + 40ns + 60ns) = 530ns$$
    $$TQ = 530ns/6 = 88.33ns $$
    $$t_{BIT} = 10times TQ = 883.3ns $$
    $$ f = 1/t_{BIT} = 1 / 883.3ns = 1.13MHz $$



    The maximum rate is governed by bus length, line capacitance, connected nodes and the drivers selected by the protocol. In principle at 30m, CAN could do 1.13Mz if everything was perfect.



    Longer the bus, the slower the data rate. But a shorter bus would mean a higher rate. CAN bit rate vs Bus length:



    CAN BIT RATE VS. BUS LENGTH



    Both referenced documents go into this at greater length.






    share|improve this answer









    $endgroup$





















      2












      $begingroup$

      It can. Meet CAN-FD.



      Why was a new protocol needed? CAN is a multi-master bus with arbitration and error reporting. These features limit the data rate based on the cable length, since it takes a certain amount of time for the signal to make a round trip between the two farthest nodes. That, along with backwards compatibility requirements, led to CAN-FD.



      Classic CAN at 1 Mbps is limited to a 40-meter bus length. (In practice, I think it's lower due to stray capacitance.) At 100 Mbps, you'd be lucky to have even half a meter of usable bus length, which is not enough for automotive and industrial applications.






      share|improve this answer









      $endgroup$














        Your Answer






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        3 Answers
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        active

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        3 Answers
        3






        active

        oldest

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        active

        oldest

        votes






        active

        oldest

        votes









        3












        $begingroup$

        It's because the CAN 2.0B standard did not specify any higher in order to reduce hardware costs and still meet the various requirements of the standard (like distance and noise immunity). It's not a technical barrier.



        The standard was written that way probably since they deemed the extra speed unnecessary for the intended application, and specifying a higher speed needlessly would increase the cost of all the hardware supporting the standard when the capability would be underutilized.



        If the standard is written that way, few IC manufacturers will bother trying to exceed it since there's no point. It's not really a technical barrier.






        share|improve this answer









        $endgroup$


















          3












          $begingroup$

          It's because the CAN 2.0B standard did not specify any higher in order to reduce hardware costs and still meet the various requirements of the standard (like distance and noise immunity). It's not a technical barrier.



          The standard was written that way probably since they deemed the extra speed unnecessary for the intended application, and specifying a higher speed needlessly would increase the cost of all the hardware supporting the standard when the capability would be underutilized.



          If the standard is written that way, few IC manufacturers will bother trying to exceed it since there's no point. It's not really a technical barrier.






          share|improve this answer









          $endgroup$
















            3












            3








            3





            $begingroup$

            It's because the CAN 2.0B standard did not specify any higher in order to reduce hardware costs and still meet the various requirements of the standard (like distance and noise immunity). It's not a technical barrier.



            The standard was written that way probably since they deemed the extra speed unnecessary for the intended application, and specifying a higher speed needlessly would increase the cost of all the hardware supporting the standard when the capability would be underutilized.



            If the standard is written that way, few IC manufacturers will bother trying to exceed it since there's no point. It's not really a technical barrier.






            share|improve this answer









            $endgroup$



            It's because the CAN 2.0B standard did not specify any higher in order to reduce hardware costs and still meet the various requirements of the standard (like distance and noise immunity). It's not a technical barrier.



            The standard was written that way probably since they deemed the extra speed unnecessary for the intended application, and specifying a higher speed needlessly would increase the cost of all the hardware supporting the standard when the capability would be underutilized.



            If the standard is written that way, few IC manufacturers will bother trying to exceed it since there's no point. It's not really a technical barrier.







            share|improve this answer












            share|improve this answer



            share|improve this answer










            answered 5 hours ago









            ToorToor

            2,046215




            2,046215

























                3












                $begingroup$

                From Controller Area Network Physical Layer Requirements



                CAN termination




                CAN is an open collector technology – the protocol could not work otherwise. This means that the recessive state of a CAN transceiver is not actively driven. The termination resistors together with transceiver input capacitance and cable capacitance create an RC time-constant discharge when an actively-driven dominant bit on the bus transitions to an un-driven recessive bit. For signaling rates greater than CAN's 1Mbps, a technology that actively drives the bus in both states such as RS-485 is required to facilitate the bus transitions required for high-speed signaling rates.




                Ultimately, the answer to the question is how the CAN protocol is implemented at a physical level. Change that protocol and a higher data rate can be used.



                From Understanding Microchip’s CAN Module Bit Timing




                ... the CAN protocol implements a non-destructive bitwise arbitration scheme that allows multiple nodes to arbitrate for control of the bus.
                Therefore, it is necessary for all the nodes to detect/ sample the bits within the same bit time. The relationship between propagation delay and oscillator tolerance effect both the CAN data rate and the bus length.




                enter image description here



                Two masters on either end of the CAN bus must be able to communicate and arbitrate which one has the bus, while each are on the bus at the same time.



                If the bus length is 30m, the time it takes to propagate the signal over the bus is: $$t_{BUS} = 30m @ 5.5 ns/m = 165ns$$



                Assuming the input comparator delay is $t_{CMP}$ = 40ns and the output driver delay is $t_{DRV}$ = 60ns for all devices.



                The round trip time for a bit on the physical bus will be:



                $$t_{PROP} = 2(t_{BUS} + t_{CMP} + t_{DRV}) = 2 (165ns + 40ns + 60ns) = 530ns$$
                $$TQ = 530ns/6 = 88.33ns $$
                $$t_{BIT} = 10times TQ = 883.3ns $$
                $$ f = 1/t_{BIT} = 1 / 883.3ns = 1.13MHz $$



                The maximum rate is governed by bus length, line capacitance, connected nodes and the drivers selected by the protocol. In principle at 30m, CAN could do 1.13Mz if everything was perfect.



                Longer the bus, the slower the data rate. But a shorter bus would mean a higher rate. CAN bit rate vs Bus length:



                CAN BIT RATE VS. BUS LENGTH



                Both referenced documents go into this at greater length.






                share|improve this answer









                $endgroup$


















                  3












                  $begingroup$

                  From Controller Area Network Physical Layer Requirements



                  CAN termination




                  CAN is an open collector technology – the protocol could not work otherwise. This means that the recessive state of a CAN transceiver is not actively driven. The termination resistors together with transceiver input capacitance and cable capacitance create an RC time-constant discharge when an actively-driven dominant bit on the bus transitions to an un-driven recessive bit. For signaling rates greater than CAN's 1Mbps, a technology that actively drives the bus in both states such as RS-485 is required to facilitate the bus transitions required for high-speed signaling rates.




                  Ultimately, the answer to the question is how the CAN protocol is implemented at a physical level. Change that protocol and a higher data rate can be used.



                  From Understanding Microchip’s CAN Module Bit Timing




                  ... the CAN protocol implements a non-destructive bitwise arbitration scheme that allows multiple nodes to arbitrate for control of the bus.
                  Therefore, it is necessary for all the nodes to detect/ sample the bits within the same bit time. The relationship between propagation delay and oscillator tolerance effect both the CAN data rate and the bus length.




                  enter image description here



                  Two masters on either end of the CAN bus must be able to communicate and arbitrate which one has the bus, while each are on the bus at the same time.



                  If the bus length is 30m, the time it takes to propagate the signal over the bus is: $$t_{BUS} = 30m @ 5.5 ns/m = 165ns$$



                  Assuming the input comparator delay is $t_{CMP}$ = 40ns and the output driver delay is $t_{DRV}$ = 60ns for all devices.



                  The round trip time for a bit on the physical bus will be:



                  $$t_{PROP} = 2(t_{BUS} + t_{CMP} + t_{DRV}) = 2 (165ns + 40ns + 60ns) = 530ns$$
                  $$TQ = 530ns/6 = 88.33ns $$
                  $$t_{BIT} = 10times TQ = 883.3ns $$
                  $$ f = 1/t_{BIT} = 1 / 883.3ns = 1.13MHz $$



                  The maximum rate is governed by bus length, line capacitance, connected nodes and the drivers selected by the protocol. In principle at 30m, CAN could do 1.13Mz if everything was perfect.



                  Longer the bus, the slower the data rate. But a shorter bus would mean a higher rate. CAN bit rate vs Bus length:



                  CAN BIT RATE VS. BUS LENGTH



                  Both referenced documents go into this at greater length.






                  share|improve this answer









                  $endgroup$
















                    3












                    3








                    3





                    $begingroup$

                    From Controller Area Network Physical Layer Requirements



                    CAN termination




                    CAN is an open collector technology – the protocol could not work otherwise. This means that the recessive state of a CAN transceiver is not actively driven. The termination resistors together with transceiver input capacitance and cable capacitance create an RC time-constant discharge when an actively-driven dominant bit on the bus transitions to an un-driven recessive bit. For signaling rates greater than CAN's 1Mbps, a technology that actively drives the bus in both states such as RS-485 is required to facilitate the bus transitions required for high-speed signaling rates.




                    Ultimately, the answer to the question is how the CAN protocol is implemented at a physical level. Change that protocol and a higher data rate can be used.



                    From Understanding Microchip’s CAN Module Bit Timing




                    ... the CAN protocol implements a non-destructive bitwise arbitration scheme that allows multiple nodes to arbitrate for control of the bus.
                    Therefore, it is necessary for all the nodes to detect/ sample the bits within the same bit time. The relationship between propagation delay and oscillator tolerance effect both the CAN data rate and the bus length.




                    enter image description here



                    Two masters on either end of the CAN bus must be able to communicate and arbitrate which one has the bus, while each are on the bus at the same time.



                    If the bus length is 30m, the time it takes to propagate the signal over the bus is: $$t_{BUS} = 30m @ 5.5 ns/m = 165ns$$



                    Assuming the input comparator delay is $t_{CMP}$ = 40ns and the output driver delay is $t_{DRV}$ = 60ns for all devices.



                    The round trip time for a bit on the physical bus will be:



                    $$t_{PROP} = 2(t_{BUS} + t_{CMP} + t_{DRV}) = 2 (165ns + 40ns + 60ns) = 530ns$$
                    $$TQ = 530ns/6 = 88.33ns $$
                    $$t_{BIT} = 10times TQ = 883.3ns $$
                    $$ f = 1/t_{BIT} = 1 / 883.3ns = 1.13MHz $$



                    The maximum rate is governed by bus length, line capacitance, connected nodes and the drivers selected by the protocol. In principle at 30m, CAN could do 1.13Mz if everything was perfect.



                    Longer the bus, the slower the data rate. But a shorter bus would mean a higher rate. CAN bit rate vs Bus length:



                    CAN BIT RATE VS. BUS LENGTH



                    Both referenced documents go into this at greater length.






                    share|improve this answer









                    $endgroup$



                    From Controller Area Network Physical Layer Requirements



                    CAN termination




                    CAN is an open collector technology – the protocol could not work otherwise. This means that the recessive state of a CAN transceiver is not actively driven. The termination resistors together with transceiver input capacitance and cable capacitance create an RC time-constant discharge when an actively-driven dominant bit on the bus transitions to an un-driven recessive bit. For signaling rates greater than CAN's 1Mbps, a technology that actively drives the bus in both states such as RS-485 is required to facilitate the bus transitions required for high-speed signaling rates.




                    Ultimately, the answer to the question is how the CAN protocol is implemented at a physical level. Change that protocol and a higher data rate can be used.



                    From Understanding Microchip’s CAN Module Bit Timing




                    ... the CAN protocol implements a non-destructive bitwise arbitration scheme that allows multiple nodes to arbitrate for control of the bus.
                    Therefore, it is necessary for all the nodes to detect/ sample the bits within the same bit time. The relationship between propagation delay and oscillator tolerance effect both the CAN data rate and the bus length.




                    enter image description here



                    Two masters on either end of the CAN bus must be able to communicate and arbitrate which one has the bus, while each are on the bus at the same time.



                    If the bus length is 30m, the time it takes to propagate the signal over the bus is: $$t_{BUS} = 30m @ 5.5 ns/m = 165ns$$



                    Assuming the input comparator delay is $t_{CMP}$ = 40ns and the output driver delay is $t_{DRV}$ = 60ns for all devices.



                    The round trip time for a bit on the physical bus will be:



                    $$t_{PROP} = 2(t_{BUS} + t_{CMP} + t_{DRV}) = 2 (165ns + 40ns + 60ns) = 530ns$$
                    $$TQ = 530ns/6 = 88.33ns $$
                    $$t_{BIT} = 10times TQ = 883.3ns $$
                    $$ f = 1/t_{BIT} = 1 / 883.3ns = 1.13MHz $$



                    The maximum rate is governed by bus length, line capacitance, connected nodes and the drivers selected by the protocol. In principle at 30m, CAN could do 1.13Mz if everything was perfect.



                    Longer the bus, the slower the data rate. But a shorter bus would mean a higher rate. CAN bit rate vs Bus length:



                    CAN BIT RATE VS. BUS LENGTH



                    Both referenced documents go into this at greater length.







                    share|improve this answer












                    share|improve this answer



                    share|improve this answer










                    answered 4 hours ago









                    StainlessSteelRatStainlessSteelRat

                    3,956821




                    3,956821























                        2












                        $begingroup$

                        It can. Meet CAN-FD.



                        Why was a new protocol needed? CAN is a multi-master bus with arbitration and error reporting. These features limit the data rate based on the cable length, since it takes a certain amount of time for the signal to make a round trip between the two farthest nodes. That, along with backwards compatibility requirements, led to CAN-FD.



                        Classic CAN at 1 Mbps is limited to a 40-meter bus length. (In practice, I think it's lower due to stray capacitance.) At 100 Mbps, you'd be lucky to have even half a meter of usable bus length, which is not enough for automotive and industrial applications.






                        share|improve this answer









                        $endgroup$


















                          2












                          $begingroup$

                          It can. Meet CAN-FD.



                          Why was a new protocol needed? CAN is a multi-master bus with arbitration and error reporting. These features limit the data rate based on the cable length, since it takes a certain amount of time for the signal to make a round trip between the two farthest nodes. That, along with backwards compatibility requirements, led to CAN-FD.



                          Classic CAN at 1 Mbps is limited to a 40-meter bus length. (In practice, I think it's lower due to stray capacitance.) At 100 Mbps, you'd be lucky to have even half a meter of usable bus length, which is not enough for automotive and industrial applications.






                          share|improve this answer









                          $endgroup$
















                            2












                            2








                            2





                            $begingroup$

                            It can. Meet CAN-FD.



                            Why was a new protocol needed? CAN is a multi-master bus with arbitration and error reporting. These features limit the data rate based on the cable length, since it takes a certain amount of time for the signal to make a round trip between the two farthest nodes. That, along with backwards compatibility requirements, led to CAN-FD.



                            Classic CAN at 1 Mbps is limited to a 40-meter bus length. (In practice, I think it's lower due to stray capacitance.) At 100 Mbps, you'd be lucky to have even half a meter of usable bus length, which is not enough for automotive and industrial applications.






                            share|improve this answer









                            $endgroup$



                            It can. Meet CAN-FD.



                            Why was a new protocol needed? CAN is a multi-master bus with arbitration and error reporting. These features limit the data rate based on the cable length, since it takes a certain amount of time for the signal to make a round trip between the two farthest nodes. That, along with backwards compatibility requirements, led to CAN-FD.



                            Classic CAN at 1 Mbps is limited to a 40-meter bus length. (In practice, I think it's lower due to stray capacitance.) At 100 Mbps, you'd be lucky to have even half a meter of usable bus length, which is not enough for automotive and industrial applications.







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                            answered 5 hours ago









                            Adam HaunAdam Haun

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