4
\$\begingroup\$

I am studying about CAN buses, and there is one thing I just can't find an explanation for. I understand that the idea of a wired-and connection is, that if any node is driving the bus to the dominant state, the bus will get to the dominant state regardless of the number of nodes transmitting a recessive state.

However, I find that in CAN the dominant and recessive states are as shown in the image below. enter image description here

I could easily image an implementation like this, if it was the other way around, and the dominant state was the one where the wires are on the same voltage level:enter image description here

But this implementation would result in the states being swapped. So how is it possible for the dominant state to be the state with the voltage difference?

\$\endgroup\$
3
\$\begingroup\$

Because CAN is not driven in the way you're imagining.

Instead, the termination resistor(s) are connected between the lines (in the position of your transistors), and each driver has two transistors, connected between one line and either Vcc or Gnd.

This makes sure that the wires are impedance-balanced and terminated properly for maximum signal integrity.

\$\endgroup\$
2
\$\begingroup\$

CAN drivers are like this

schematic

simulate this circuit – Schematic created using CircuitLab

For example.

\$\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.