Can anyone tell me what the purpose of split termination is in a differential pair?

If you look at the LVDS article on Wikipedia, you'll see the 100 Ohm termination resistor in parallel across the two pairs. But then if you look in some datasheets for some differential interfaces, you'll see an alternative method with a connection to a reference voltage (see: https://www.semiconductorstore.com/blog/2017/From-Silicon-Labs-Timing-101-The-Case-of-the-Split-Termination/2918/)

enter image description here

EDIT: There is also this scheme for Stratix GX FPGAs (https://www.intel.cn/content/www/cn/zh/programmable/support/support-resources/support-centers/signal-power-integrity/sgl-general.html)

enter image description here

According to the above, this is to provide common mode noise suppression. To me the explanation makes sense. My questions about this are:

  1. What other benefit is there to using split termination? My thought is that the split termination scheme also provides some skew compensation by allowing some current to flow into/out of the receiver.
  2. If there is no skew, then wouldn't the two terminations be equivalent? I would assume the answer is yes, otherwise there would be no benefit in using split termination instead of the 100 Ohm resistor.
  3. Are either of these used outside of LVDS? This is where I've looked but can't get any solid answers. I know it's not used in Ethernet, just not sure about other protocols like USB or PCIe.
  • 1
    \$\begingroup\$ Isn't what you call skew compensation rather inherent to many forms of termination? \$\endgroup\$
    – DKNguyen
    Oct 17, 2021 at 2:56
  • \$\begingroup\$ @DKNguyen I meant to say something to the effect of "allows the two signals to have some skew while still being able to detect the signal". Also, I don't know, is it inherent to many forms of termination? That was my sense with the case above, where the cap to the reference voltage either charges or discharges to provide the required current, but its response is limited in time, thus the limits on allowed skew that can be "compensated". If this is wrong please let me know as I'd like to better understand why this particular termination is used. \$\endgroup\$
    – user224284
    Oct 17, 2021 at 3:03
  • \$\begingroup\$ Sorry, I was thinking slew, not skew. If by "skew" you mean differences in arrival time then I don't see how it can do that at all. \$\endgroup\$
    – DKNguyen
    Oct 17, 2021 at 3:05
  • \$\begingroup\$ 'drift' is probably a better word than 'skew', in this usage. \$\endgroup\$
    – scottbb
    Oct 17, 2021 at 3:15
  • 1
    \$\begingroup\$ @sparaps Well technically the receive circuitry should be detecting the voltage differences between the two lines, but unless it's isolated and floating (which it often isn't) it's probably anchored to some third reference and it certainly helps if that is stable. You don't want everything straying too far from center anyways since it could leave the bounds of valid operation for your absolute input range. I agree about preserving common-mode level at mid-range, regardless of the cause rather than compensating for skew (which would entail shifting arrival times so they line up). \$\endgroup\$
    – DKNguyen
    Oct 17, 2021 at 6:06

1 Answer 1


The advantage of the split termination is that it provides a low impedance path to ground for common mode noise.

Differential signals see the termination resistor as the input impedance of the amplifier. Common mode signals (usually noise) do not create any voltage difference across the single termination resistor in the unsplit version. Thus, the impedance seen by common mode signals in the unsplit version is the terminal impedances of the amplifier, which are typically quite high.

In the split termination version, each line carrying a common mode signal see the resistance in series with twice the capacitative reactance as the impedance. (Twice the reactance because the two lines share the capacitor, so each line sees half the capacitance, hence twice the reactance).

By providing this low impedance path to ground for common mode signals in the form of an RC network, noise of sufficiently high frequency will be attenuated. This will help improve the common mode rejection ratio (CMRR) of the overall system at higher frequencies. At least that is the theory. Note that in the example precision resistors are used. This is important, because if the resistors are not well matched, worse CMRR might result.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.