0
\$\begingroup\$

I'm trying to implement MCP25625 CAN controller and tranceiver in my design. I checked the dev board of Microe MCP25625 Click and especially in the the lines CANH and CANL like it is showen in the picture below

enter image description here

Question 1 : there are 2 x 60 Ohm resistors which I guess it is the CAN bus termination ! But if I am plugging this to an OBD for example! there is no need for 120Ohm termination resistor ? Question 2 : Is 4.7nF between the 2 resistors mondatory here! I usually put 120 Ohm and nothing else ?

Question 3 : What's the purpose of the 47pF capacitor on CANH ? in the MCP25625 datasheet they don't talk about it ?

Full schematic link

I would appreciate your inputs and maybe some links to help me understand the role of these components.

\$\endgroup\$
1
  • \$\begingroup\$ The 4n7 cap is straight from the MCP2562 datasheet. That is, the transceiver datasheet, not the controller datasheet. Some transceivers have an optional "split" pin that can be connected between the two 60R terminating resistors. \$\endgroup\$
    – Lundin
    Feb 26, 2020 at 8:42

1 Answer 1

1
\$\begingroup\$

1) If you are plugging this to a bus that is already terminated at both ends, your device in the middle must not have termination, and it must be removed.

2) The 4.7nF capacitor between 2 resistors is not mandatory. When the 120 ohm resistor is split into two 60 ohm resistors with capacitor in the middle, it is called split termination. The capacitor is there for terminating the common mode signal voltages that would otherwise be unterminated. And since most likely you are connecting to a bus that is already terminated at both ends, you must remove the resistors and this capacitor as well.

3) There is a 47pF on both CANH and CANL wires, not just on one of them. It can be there for many reasons, such as limiting slew rate, filtering of noise spikes or just for filtering radiation or reception of electromagnetic interference.

\$\endgroup\$

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.