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The CAN bus has CAN-High and CAN-Low lines. Normally, in the recessive state the voltages are each 2.5 V and in the dominant state the CAN-High line goes to 3.5 V and the CAN-Low line goes to 1.5 V.

  • How are these voltages actually achieved in the transceiver? What would be the circuit for the sake of understanding (using FETs or transistors or something else)?

  • In dominant, is that 1 V drop and rise happening in the CAN-High and CAN-Low lines?

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There's no rocket science in CAN transeivers.

Normally, reference voltages are achieved using a simple voltage drop circuit or a linear volatage regulator.

Note, as there may be a large voltage drop between transmitter and receiver grounds (80 V AFAIK) there's no need for precisely 2.5 V. Low-voltage transceivers use a lower common-mode voltage.

So only a differential voltage of ±0.5 V is actually regulated, and the common mode voltage can be set by a simple voltage divider.

There are some interesting schematics in the article Isolated CAN Transceiver Assures Robust Fieldbus Design.

Note the diodes feeding CAN-Low and CAN-High. Together with output transistors they give about a 1.5 V drop. (They also protect the transistors from reverse current.)

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