CAN Transceiver Transmit Mode
I try to understand the transmit functionality of a CAN transceiver. I found this post a good starting point, but it doesn't completely answer my question.
Let's say we have the following CAN transceiver block diagram (VDD = 5 V / VSS = GND = 0 V).
CAN Bus Basics
In my understanding there are a few simple basics for an ISO-11898-4 "complient" CAN-Bus.
- In the recessive state: CANH and CANL should be at the same voltage level (≈ 2.5 V).
=> CANH ≈ 2.5 V & CANL ≈ 2.5 V
- In the dominant state: CANH and CANL should have a high voltage difference (≈ 2 V).
=> CANH ≈ 3.5 V & CANL ≈ 1.5 V
- There must be at least one termination resistor of ≈ 120 Ω.
- A logical TxD LOW or 0 bit is transformed to a dominant state.
- A logical TxD HIGH or 1 bit is transformed to a recessive state.
This should be enough for the first understanding!
- CANH and CANL output transistors are open (high impedant).
- The recessive voltage level of CANH and CANL = VDD / 2 ≈ 2.5 V is created by the following circuit part.
- CANH and CANL output transistors are closed.
- Let's say:
Transistor (collector emitter voltage): Uce = 0.1 V
Didode (diode drop voltage) UD = 0.5 V.
=> CANH = VDD - (Uce + UD) = 5 V - (0.1 V + 0.5 V) = 4.4 V.
=> CANL = Uce + UD = 0.1 V + 0.5 V = 0.6 V.
=> UT (Rtermination voltage) = VDD - (2 * (Uce + UD)) = 5 V - (2 * (0.1 V + 0.5 V)) = 3.8 V.
=> IT (Rtermination current) = UT / 120 Ω = 3.8 V / 120 Ω ≈ 32 mA
- Did I miss something or is my calculation valid (just for basic understanding :) )?
- Can anyone explain how the recessive voltage generator in picture 2 works in detail? How big are the resistors to CANH and CANL?
- Will Itermination double if there is a second termination resistor?