TCAN1044AV-Q1: Questions when use TCAN1044AV-Q1 and TCAN1043A-Q1

Hi Bryce,

Thanks for bringing your questions to E2E. 

Bryce Xun said:

How to determine whether to use 5V CAN FD or 3.3V CAN FD if MCU supply voltage is 3.3V? Is there any advantages for 5V CAN FD over 3.3V CAN? In most applications, I have seen that 5V CAN FD is used though MCU supply voltage is 3.3V.

5V supplies are standard for CAN transceivers as this is how the ISO 11898-2 CAN specification defines CAN behavior. TCAN1044AV and TCAN1043A are both 5V supplied CAN transceivers that can support 3.3V Vio for MCU interfacing. The advantage to using a 3.3V supplied CAN transceiver such as TCAN334 is that a 5V supply is not required. This allows the transceiver to share a power source with a 3.3V powered MCU. However, the ISO spec currently does not define behavior for a 3.3V supplied CAN transceiver, so such devices can only claim compatibility with devices that follow this specification (not compliance). Let me know if you have more questions regarding the use of 3.3V supplied CAN. 

Bryce Xun said:

What is the advantage of TCAN1043A-Q1 over TCAN1044A-Q1? I think one value is that TCAN1043-Q1 has sleep, silent and standby mode while TCAN1044-Q1 only has standby mode, so that TCAN1043-Q1 is more power saving device. Is there any other values for 1043 compared with 1044? In which application will prefer to use TCAN1043?

Yes, you are mainly right here. The advantage of the Sleep mode and extra pins (INH and WAKE) for TCAN1043A is that it may act as the low-power mode control for a local node. The INH output may be used to control local voltage regulators so the node may be completely unpowered while it is in Sleep mode. The TCAN1043A then remains active in a low power mode supplied by Vsup. In this state, it can monitor the CAN bus and WAKE pin for wake conditions. When a wake condition is detected, it can re-enable the local voltage regulator by asserting INH and turn on power for the rest of the node. 
In contrast, 8-pin CAN transceivers such as TCAN1044A can not control local supplies in this way. For TCAN1044A, a 5V rail will need to remain active in order to recognize bus wake events. When the INH function of a 14-pin CAN transceiver is not needed, simpler 8-pin devices can help simplify the design. 

Bryce Xun said:

In my view, when 1043 works in sleep mode, all system components like MCU will not be powered except 1043. I am not sure whether it is correct. However, in the mandarin version of the datasheet, the description of the device is that in sleep mode of 1043, power will be delivered to all components except 1043. I think these two descriptions are opposite to each other.

Yes, your understanding is correct. In Sleep mode, it is normal for only the TCAN1043A to be powered by the Vsup supply (battery) with Vcc and Vio disabled. This saves power by ensuring the rest of the node does not draw extra current while idle. It sounds like the mandarin description may be inaccurate here as the intention of Sleep mode is to power-off all other components. 

Bryce Xun said:

For the WUP of 1044, once the bus recognize the WUP, the RXD output is driven low. I think this time the node will return to normal mode. After it returns to normal node, why will it still need to wait for twk_filter so that the RXD output can be driven low?

Once the transceiver is in Normal mode (STB = low), the RXD pin will mirror the state of the CAN bus. The twk_filter timer has no impact on the RXD pin while in Normal mode. 

Bryce Xun said:

Why CANH is 3.5V and CANL is 1.5V when bus state is dominant? I am not sure about the relationship about 2V differential voltage with Vcc(5V). Will it be related with the termination resistor? Is there any formula or explanation about this relation?

These values are defined by the ISO 11898-2 CAN specification. Overall, they describe the valid dominant-state differential a CAN transceiver needs to drive on the bus across a 50Ω - 65Ω load. This differential will be smaller than the 5V supply due to losses across the CANH and CANL drivers. It is important that the differential is large enough at the receiving nodes so that the bus is recognized as a dominant. When in a recessive state, the active drivers are off and CANH and CANL decay to a recessive level (2.5V) for a differential of 0V. 

Let me know if any of my response is unclear and if you have any more questions.

Regards,
Eric Schott

Hi Eric,

Thanks a lot for your reply.