2
\$\begingroup\$

What is the purpose of the capacitors (C4 and C5, 56 pF) used on the CANH and CANL lines?

And how do I calculate the values?

As shown in the attached image:

Enter image description here

\$\endgroup\$
3
\$\begingroup\$

These capacitors are to suppress CAN bus line transients high-frequency noise by conducting it to earth.

Their values will be a function of the expected noise and the line imedance. Is this circuit a recommended application circuit from a reputable source? If so, I would stay with the recommended values unless you have the equipment for diagnostic EMC testing readily to hand. Otherwise you can revise these values during initial EMC testing later.

\$\endgroup\$
  • \$\begingroup\$ I thought that the CAN bus was not sensitive to noise (due to CANH and CANL) \$\endgroup\$ – M.Ferru May 22 '17 at 9:21
  • 3
    \$\begingroup\$ @M.Ferru Just because the CAN data link is working doesn't mean the system will pass the required EMC tests. \$\endgroup\$ – Andrew May 22 '17 at 9:36
  • 1
    \$\begingroup\$ @M.Ferru, all signalling systems are sensitive to noise to different degrees, CAN is just less sensitive than others. But reducing EMC isn't to keep your system working, it's to stop your equipment interfering with other equipment around it. \$\endgroup\$ – TonyM May 22 '17 at 9:36
  • 1
    \$\begingroup\$ Thanks to all, @TonyM its already proven design so i would like to reuse it before my doubt cleared. Thanks for your help \$\endgroup\$ – ramesh6663 May 22 '17 at 9:57
  • 1
    \$\begingroup\$ the can bus lines are probably terminated at either end by 130 ohms or so. The RC value of these capacitors times the line impedance defines a risetime which is probably picked to be some fraction of a bit time. That way there is no big effect on the signal but some attenuation of higher frequency noise. So the vaiue of the capacitors depends on what data rate you are operating at. If you want to decide if you are using the right capacitance, put a scope on your bus lines and swap in and out a few capacitors. \$\endgroup\$ – electrogas Jun 20 '17 at 4:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.