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SN65HVDA1050A-Q1: Current flowing out of TxD terminal of CAN transceiver

Part Number: SN65HVDA1050A-Q1

Hello.

SN65HVDA1050AQDRQ1 is directly connected to the TxD terminal of the MCU.
The MCU is 3.3V.
The output of the MCU is CMOS output.
The TxD pin of SN65HVDA1050AQDRQ is internally pulled up to 5V.
Therefore, when the output of the MCU is High, it seems that the current will flow from the CAN transceiver to the MCU.

Since resistance is contained in the output stage, I think that it is minute, how much current wll flow?

Regards,
Dice-K

  • Hi Dice-K

    I'll support you on your questions.

    Yes indeed the TxD pin is pulled up to Vcc. Due to the fact that your MCU is operating at a deviant voltage level there will be a minute current flow. 

    Please check the section 6.5 Electrical Characteristics on page 5 of the datasheet. Your MCU is represented by VI (input voltage at pin 1 of the SN65HVDA1050A-Q1)

    • IIH - High-level input current, TXD input is the the current that will flow between (VI at VCC) during the High-level input. The min/max-ratings are -2 µA to 2 µA.
    • IIL - Low-level input current, TXD input is the the current that will flow between (VI at 0 V) during the Low-level input. The min/max-ratings are -50 µA to -10 µA. 
    • For a VI range from 0 V to VCC The current range is -50 µA and 2 µA 
    • In your case VI is between 0 V and 3.3 V -> IIH @ 3.3 V you can expect a negative current below -2 µA and above -50 µA 

    If the current flows out of the SN65HVDA1050A-Q1 the current polarity is determined as negative and vice versa.

    Does this help to answer your question?

    Kind regards 

    Dierk