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I am using a chip whose part number is ADC3444 in my design.

The outputs are LVDS signals. In the datasheet, it is recommended that a 100 ohm resistor needs to be placed as termination between the LVDS signal pair and an external load capacitance needs to be placed from each output pin to Ground.

The below image is from the datasheet where the above point has been mentioned.

datasheet picture

In my application, these LVDS signals will be going to FPGA. Do i need to place the resistors and capacitors as recommended? If i don't, what might be the issues?

Kindly give a solution to my query. Thank-you

Link for datasheet:- https://www.ti.com/lit/ds/symlink/adc3444.pdf?ts=1731156418751&ref_url=https%253A%252F%252Fwww.ti.com%252Fproduct%252FADC3444

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  • \$\begingroup\$ do not place the capacitors, they're an indication of how much it can drive \$\endgroup\$
    – Neil_UK
    Commented Nov 9 at 13:51
  • \$\begingroup\$ It is common for fpgas to integrate termination for lvds, make sure that you don't already have that before adding a second resistor. \$\endgroup\$
    – user1850479
    Commented Nov 9 at 14:50

1 Answer 1

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LVDS requires a termination resistor directly at the receiver. If the FPGA does not have it built in, then you must add it.

You are not supposed to add any capacitors. The output works with a load of about 3.3 pF, but the typical trace capacitance is roughly 1 pF/cm, and that the receiver's input pins also have a certain capacitance. And there are no specified minimum/maximum limits; you should not try to match this value.

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  • \$\begingroup\$ Would it be correct to say that the capacitance here is more of a metric of how well-matched the transmission line on the output can be, considering that a perfectly impedance-matched and terminated transmission line would appear as a perfectly resistive impedance? \$\endgroup\$
    – Hearth
    Commented Nov 10 at 19:16
  • \$\begingroup\$ This specification does not appear to be related to the characteristic impedance. Even if the line's characteristic impedance is perfectly matched, the 'real' capacitance still needs to be (dis)charged when switching. \$\endgroup\$
    – CL.
    Commented Nov 11 at 14:52

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