Why does LVDS (or any differential signaling for that matter) use a common mode voltage of 1.2V instead of 0V?

Is there an easy way to shift the common mode voltage to 0V?


1 Answer 1


Because, then the signal would have to be negative also. The trick is to use a positive single power supply. Now the differential signal is centered arround 1.2V and the applied differential signal is +/-175mV.

Using a 0v common mode, the signal would be +/-175mV meaning that you need dual power supply - negative and positive at both sides, which is more expensive.

  • \$\begingroup\$ I see. I wonder why they picked 1.2V instead of 0.175V (or a similarly low) common mode voltage. That would still let you use a positive supply and the output can swing between 0V and 0.350V. \$\endgroup\$
    – Navin
    Oct 3, 2018 at 9:30
  • 2
    \$\begingroup\$ Because the output buffer can't go such low (0V) or such high as rail supply. It can also elliminate the noise, since the noise would have a common mode voltage of 0V. \$\endgroup\$ Oct 3, 2018 at 9:39
  • 1
    \$\begingroup\$ ... and 1.2v is sufficiently high that it will ensure the transistors at the receiving end are appropriately biased so the signal can be easily amplified to whatever internal logic level is required. \$\endgroup\$
    – Jules
    Oct 3, 2018 at 11:20
  • 1
    \$\begingroup\$ Some similar interface standards such as MIPI_DSI (used for cell-phone displays) do indeed use 200mv common mode. \$\endgroup\$ Oct 3, 2018 at 14:22
  • \$\begingroup\$ The little tiny transistors have to pull down and to pull up. They need headroom in each case. Thus some voltage about the middle of 3.3 volts was chosen. \$\endgroup\$ Oct 4, 2018 at 3:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.