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Considering 74AUP1G07, I was wondering why this design, which I recently came across, supposed to be correct.

since I'm not allowed to share the schematic, I will describe it:

  1. on separate part of the schematic we have a signal "DDR_VTT_CTRL" that starts from an off-page connector, and passes through 0 Ohm resistor and ends with a wire named VDDQ_VTT_MEM_EN
  2. on another separate part, we have "DDR_VTT_CTRL" coming from an off-page connector, going through "A" input in 74AUP1G07. while from the Y input we have a pull up resistor where its node is connected to 0 Ohm resistor and ends with a wire called VDDQ_VTT_MEM_EN.
  3. Note: 74AUP1G07 is marked as EMPTY.

looks like DDR_VTT_CTRL and VDDQ_VTT_MEM_EN edges are shorting the whole device (while R10501 is 0 Ohm). what I am missing here?

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  • \$\begingroup\$ Why do you think it is incorrect? \$\endgroup\$
    – Andy aka
    Apr 6, 2021 at 13:01
  • \$\begingroup\$ because at the beginning I thought that 74AUP1G07 is used but the way it was connected had shorted it. \$\endgroup\$
    – Xhero39
    Apr 7, 2021 at 6:37

1 Answer 1

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If you look at the schematic, many parts such as the 1G07 are marked "EMPTY", so this option is not present on the board, and the option to just short the nodes with a 0 ohm resistor is used.

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