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I want to use the ADS58B18 from Texas Instruments. This ADC is differential and the outputs can set to either LVDS or LVCMOS with programming an internal register.

I have a question regarding the clock inputs and outputs. Lets say I am programming the ADC to use LVDS outputs. Is it still possible to clock the ADC with a LVCMOS signal or does it have to be a LVDS clock now?

Thanks

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  • \$\begingroup\$ The datasheet will answer that. \$\endgroup\$
    – user16324
    Commented Oct 12, 2020 at 19:06
  • \$\begingroup\$ Welcome to EE.SE. It's a good idea to include the link to the datasheet that you're using. It saves us a search and it means that we're all using the same one. Hit the edit link. \$\endgroup\$
    – Transistor
    Commented Oct 12, 2020 at 19:09

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Per the datasheet, the pin functionality of the clock input does not change regardless of which mode you are using:

Pinout is same in both CMOS and LVDS modes

The clock signal is always a differential input. However it is possible to drive it from an LVCMOS signal by AC coupling and applying a DC offset as shown later in the datasheet. The LVDS/CMOS mode selection has no bearing on this ability.

Driving Clock input

In this configuration (single-ended input) you are actually making a pseudo-differential signal with a common mode offset that keeps changing, though the performance of this will be worse in terms of jitter performance and is not recommended for high speed sampling.

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  • \$\begingroup\$ Thats the answer I was looking for, thanks! \$\endgroup\$
    – electrococuk
    Commented Oct 12, 2020 at 19:20
  • \$\begingroup\$ Hey Tom, I have another question regarding the input clock. The data sheet only says that Vpp = 700mV, which is +-350mV, Vcm = 1.7V. Is it possible to use a LVDS clock with Vpp = 700mV and Vcm = 1V? \$\endgroup\$
    – electrococuk
    Commented Oct 13, 2020 at 8:33

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