1
\$\begingroup\$

I have two CAN busses which are each terminated on each end by a 120Ω resistor, as shown in the diagram below. The CAN bus consists of a CAN high, CAN low and ground connection. To terminate it, a 120Ω resistor connects CAN high and CAN low.

CAN bus setup

Sometimes I want to use both CAN busses completely separate but in some situations, I want to connect the devices D3 and D4 together, making a single bigger CAN bus.

To connect the two, I need to remove the two resistors R2 and R3 and then connect the CAN bus together. My CAN connection currently terminates in a D-SUB connector X1 (at each end D1, D3, D4, D6) to easily connect other CAN devices. At the same time I have a switch (S1) that will enable or disable the CAN bus termination, as can be seen in the schematic below.

Simplified schematic

What I want to do is to remove the manual operation of the switch S1 by including something in the D-Sub connector X1. When nothing is connected to it, the terminating resistor R1 will connect CAN high and CAN low. But when I connect another device to the connector X1, the resistor R1 will not connect CAN high and CAN low anymore.

I would appreciate any tips that point me towards a possible solution. I should also mention that connecting and disconnecting any device will happen when everything is powered off and does not need to happen during operation.

\$\endgroup\$
1
\$\begingroup\$

The first thing you must do is to fix your incorrect connector pin-out and turn it into CAN. DB9 connectors for CAN are widely standardized and professional engineers follow industry standards. You should have CANHI=7, CANLO=2, GND=3.

Once you have turned your pin-out into CAN, the simplest solution might be to buy pre-made DB9 terminators and plug those in whenever you don't connect the buses together. For example these: https://www.kvaser.com/product/kvaser-d-sub-9-pin-120-ohm-termination-adapter-2/ (though Kvaser specifically are ridiculously expensive - there's plenty of 2nd source doing the same thing for a 10th of the price).

Alternatively, you could also connect two pins of the DB9 connector together at each end (to once again violate the standard pin-out...) and then have that signal activate an analog switch, which in turn enables the resistor on that board only. Larger BOM but cheaper external adapters.

\$\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.