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Robustness of SN65LVDS93A Output Circuit

Patrick Rayero
Intellectual 290 points
Other Parts Discussed in Thread: SN65LVDS93A

Hi guys,

We are using a SN65LVDS93A to serialise camera data for a FPGA. During power-up of the FPGA, we found that the FPGA is driving 1.8V through internal 25kOhm resistors to the SN65LVDS93A outputs. These 1.8V are present for ca. 7 seconds until the FPGA is fully configured. During this phase, the SN65LVDS93A is still unpowered.

According the datasheet, no voltages greater than Vcc+0.5V are allowed at the outputs, which would be in our case 0.5V. But before we spend some extra circuitry to protect the SN65LVDS93A outputs, we would like to know if these 1.8V pulled through a 25kOhm resistors are really an issue? Because 1) current into pins is very low and 2) other TI serializers with LVDS output have a similar output circuit but maximum output voltage is not Vcc referenced.

Thanks,

Patrick

  • 0
    TI__Genius 11280 points
    Hello Patrick,

    Obviously the better solution is apply power to the LVDS93 when the FPGA is configuring. Unfortunately if you apply signals above the absolute maximum ratings, the circuit could have problems. It's a better idea put some protection for the LVDS93, in my opinion.

    Please provide a schematic for a double check.

    Regards,
    Gerardo
  • 0 in reply to Gerardo
    Intellectual 290 points

    Hi Gerardo,

    unfortunately I cannot share schematics due to confidentially. But you would anyhow don't see much since it is a direct connection between SN65LVDS93A outputs and 1.8V I/O bank of FPGA (the 25kOhm pull-ups are internally).

    But I still wonder that our case could really cause problems on the SN65LVDS93A because when you check output circuitry, current could only flow through the resistors and then current would be even lower (1.8V/(25k+10k)).

    Regards,

    Patrick

  • 0 in reply to Patrick Rayero
    TI__Genius 11280 points

    Hello Patrick,

    I've discussed it with other coworkers and we don't think that the 1.8V applied to digital input will cause damage to the device before power supply is ready. Customer should do their own test to see if it causes any functionality issue at system level though.

    Regards,

    Gerardo

  • 0 in reply to Gerardo
    Intellectual 290 points

    Hi Gerardo,

    great, thank you very much for double-checking. Sure, we will intensively test.

    Have a nice week and thanks again,

    Patrick