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SN65HVD20

Carl Reese
Prodigy 230 points

Other Parts Discussed in Thread: SN65HVD20

My question concerns the SN65HVD20 RS485 transceiver. Please assume that I have a total of 32 nodes on my RS485 link. On the RS485 line side, how will the SN65HVD20 influence the RS485 link if one of the SN65HVD20 devices is powered off ? 

  • 0
    TI__Mastermind 28265 points

    The data sheet shows on page 4 that the driver output current under power-off conditions depends on the common-mode voltage.

    For the normal common-mode range (-7V to +12V) it can be up to 0.5mA and for the extended range (-20V to +25V) it can rise up to 1mA. Depending on how many transceivers you power off at once, you could have a significant impact on the buss.

    regards,

    Thomas

  • 0 in reply to Thomas Kugelstadt
    Prodigy 230 points

    Hey Thomas:

     

    I found, on page 5 of the datasheet, "Output Current with Power OFF" listed at -400uA to 500uA over -7V to 12V.  But I'm a bit confused by this, so feel free to treat me like a 6-year-old.

    I don't understand this.  If my transceiver has no power (but is still hanging on the RS48 bus), how can my transceiver source any current at all, much less 500uA ?  My goal is to understand  how a de-powered SN65HVD08D will look in terms of current draw to an active transmitter on an operational RS485 link.  

  • 0 in reply to Carl Reese
    TI__Mastermind 28265 points

     

    1) The term "output current" is misleading or simply wrong. (our fault, sorry). The main loading of a powered-down transceiver comes from the internal common-mode attenuator (R1 to R3) of both inputs, A and B.

    2) From the data sheet (page 8) take the equivalent circuit diagrams for inputs A and B and convert the resistor network into a single common-mode input resistor as shown below.

    3) Then divide the upper and lower common-mode limits of 12V and -7V by this single resistor value and you receive the corresponding load currents, which are flowing in and out of the transceiver bus terminals, depending on the polarity of the common-mode voltage applied. For the higher common-mode voltages of 25V and -20V you will get higher currents of course.

    Hope this clarifies it.        Regards, Thomas

     

  • 0 in reply to Thomas Kugelstadt
    Prodigy 230 points

    Thanks Thomas.  Excellent explanation.  Thank you.  You may close this item

     

    Carl Reese