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DS15BR401: DS15BR401 Consulting

Part Number: DS15BR401
Other Parts Discussed in Thread: DS15BR400,

Hello

using DS15BR400 as LVDS Driver,In my system,the external teminate resistor of Receiver side was easy fall off, so did you have any way to detect this situation, to detect if the terminate resistor was soldered well or existed ?

  • Hi Fandy,
    For LVDS interface, when 100-ohm termination is not connected, circuit is not closed. This means there is no current path. LVDS interface drives 10mA into this 100-ohm creating 300mV peak to peak swing across this 100-ohm. So if you use a voltmeter across DS15BR400 you would see a very low voltage due to 10mA going through voltmeter resistance.
    Regards,,nasser
  • Hi Nasser

    I am confused for your answer because I am not familiar with LVDS product ,so could you explain more detail ? thanks 

    "For LVDS interface, when 100-ohm termination is not connected, circuit is not closed" because DS15BR400's output has internal 100 ohm resistor, so I didn't think the circuit is not closed, did I think right ?

    "LVDS interface drives 10mA into this 100-ohm creating 300mV peak to peak swing across this 100-ohm"  10mA through 100 ohm resistor should be create 1000mV across 100 ohm ?

  • Hi Fandy,

    The Ds15BR401 uses a voltage mode output driver.  It will drive the correct output voltage with or without a Rx termination on the far end.  Still there should always be a 100 Ohm termination on the Rx side to match the transmission line impedance.  This termination may be internal or external to the Rx device.

    Regards,

    Lee

  • Hi Lee
    Yeah, I agree that it should always be a 100 ohm termination on the Rx side to match the transmision line impedance , actually I want to ask if this 100 ohms on the Rx side is not connected,what will hapan ? and I really want to add some design to detect this 100 ohm is exist or not, do you have some way advises ?
  • Hi Fandy,
    Without a 100 Ohm termination resistor the DS15BR40x device will have an elevated VOD reading. This may be difficult to detect since the output common mode will be the same in either case.

    Regards,
    Lee