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I have two sytems, one is a master system and the other is a slave system. Both systems are galvanic isolated and share no common potential. The two systems need to communicate over CAN (around 1m appart). I would like to prevent ground loops, so the galvanic isolation is also necessary for CAN.

Now there are many CAN transceivers, that come with galvanic isolation, but the problem is, that here are two different CAN systems, two nodes on the master system and one node on the slave system. In the future, there might come up to 3 slave devices, so this should already be considered in the design.

First Question: How can I galvanic isolate the two systems and still communicate over CAN?

Second Question: What about the termination with galvanic isolation? Should there be added two 120 Ohm on each network separately or is one termination resistor already enough?

Third Question: Since all systems have already a galvanic isolation, there should be no currents in between the systems, right? If so, there is no isolation necessary for CAN and I could directly connect the GND on each node together over the CAN connectors? But I doubt this is correct.

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First Question: How can I galvanic isolate the two systems and still communicate over CAN?

You first need to identify galvanically isolated from what? From the utility or from each other? Assuming they do need to be isolated from each other, then you require isolated CAN since the physical layer of CAN needs a common reference.

The ISO-1042-Q1 chipset ( https://www.ti.com/lit/ds/symlink/iso1042-q1.pdf ) can help solve this issue. You need to provide an isolated rail (a simple flyback or an ISOPOWER chipset). Your CANbus now becomes a 3-wire system as both sets of transievers need a common reference

The ISOW1044 ( https://www.ti.com/lit/ds/symlink/isow1044.pdf ) incorporates a DCDC converter within the transceiver chip, simplifying the needed physical layer

Second Question: What about the termination with galvanic isolation? Should there be added two 120 Ohm on each network separately or is one termination resistor already enough?

The physical bus of a CANBus network needs termination at both ends, 120R between CANH and CANL. Whether this is floating or referenced to some node does not change this

Third Question: Since all systems have already a galvanic isolation, there should be no currents in between the systems, right? If so, there is no isolation necessary for CAN and I could directly connect the GND on each node together over the CAN connectors? But I doubt this is correct.

There will be some current flowing between the CANBus nodes as each line is dominant or recessive. The majority will be via the terminating resistors but there will be a tiny amount via the reference

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  • \$\begingroup\$ There is no direct need for the isolation, but the power supply already have internal galvanic isolation. What I want to prevent is a GND loop between the devices, thats why I tought about isolation the CAN. The problem with all these isolated CAN transceiver is, that the are only for one node in the system, but one system has already two nodes in it, that share the same GND reference on the same PCB \$\endgroup\$
    – HansPeterLoft
    Commented Aug 26, 2021 at 10:09
  • \$\begingroup\$ If the concern is todo with "ground loop" then going to the extent of a fully isolated CAN might be too much and you can provide a 3rd wire to reference the two units together. However, without a full grounding scheme or information with respect to how the two units will behave under some form of reference perturbation (ground bounce), it isn't easy to definitively state. Isolated CAN will obviously remove the possibility. \$\endgroup\$
    – user16222
    Commented Aug 26, 2021 at 10:14
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A standard two-channel high-speed bi-directional digital isolator can be used.

Connect the CANH and CANL signals from the Master System to one side of the isolator IC and connect CANH and CANL from Slave System to the other side of the isolator IC.

Provide the 3.3V connection from the Master and Slave system to power the respective side of the isolator IC.

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  • \$\begingroup\$ Well, the signals are differential, so there must be a special isolator for this, that does not pull the signals to GND or V3.3. I will have a look, thanks. \$\endgroup\$
    – HansPeterLoft
    Commented Aug 26, 2021 at 9:44
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First Question: How can I galvanic isolate the two systems and still communicate over CAN?

If you want to isolate two parts of a CAN system, you need an isolated CAN bus repeater. This has impact on the maximum bus length due to the latency of the repeater and the speed of the bus. Eg: TIDA-01487

Second Question: What about the termination with galvanic isolation? Should there be added two 120 Ohm on each network separately or is one termination resistor already enough?

Since it's a repeater, you will have two CAN systems and 4 terminations.

Third Question: Since all systems have already a galvanic isolation, there should be no currents in between the systems, right? If so, there is no isolation necessary for CAN and I could directly connect the GND on each node together over the CAN connectors? But I doubt this is correct.

There may be AC leakage currents trough the isolation barriers and Y-caps. And there will be tiny current running trough the reference line from the tranceivers.
When your nodes are isolated (not ground referenced) you need a CAN-GND line since the common mode of CAN-H and CAN-L is limited.

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  • \$\begingroup\$ How do you mean I need a CAN_GND line when I use a isolation repeater? The grounds are connected to the isolation repeater and internally connected over a capacitor? \$\endgroup\$
    – HansPeterLoft
    Commented Aug 26, 2021 at 10:04
  • \$\begingroup\$ @HansPeterLoft CAN bus requires 3 wires to operate. \$\endgroup\$
    – Jeroen3
    Commented Aug 26, 2021 at 10:38

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