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?
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.)
