Theoretically, I can't see a problem if the twisted pair has an end of cable termination that is: -

  • A single resistor (R) that matches the characteristic impedance of the cable placed across the two ends of the pair or,
  • Two resistors (\$\dfrac{R}{2}\$) across the two ends of the pair with the centre-point tied also to the shield/screen.

enter image description here

Practically, when looking through data sheets I tend to see option 2 more than option 1.

Today, I had to use option 2 because option 1 caused a noticeable time lag (about 2 or 3ns) between the two conductors over 50m of cable. This surprised me and I'm wondering why this should be so. The signal I was driving at one end was about 2V logic levels and very balanced in nature (no discernible time difference or noticeable amplitude difference).

Question - why should option 2 be better than option 1 in the set-up I've described and is it possible that there is something theoretically better about option 2?

  • \$\begingroup\$ What is the signal frequency? \$\endgroup\$
    – deadude
    Commented Nov 11, 2013 at 23:24
  • \$\begingroup\$ its 80Mbps with a guaranteed transition every 6 bits \$\endgroup\$
    – Andy aka
    Commented Nov 11, 2013 at 23:36
  • \$\begingroup\$ For option 1, the resistor is in a pull-up configuration or in series ? \$\endgroup\$
    – zeqL
    Commented Nov 12, 2013 at 16:05
  • \$\begingroup\$ @zeqL Neither, the single resistor is across the ends of the two twisted wires. \$\endgroup\$
    – Andy aka
    Commented Nov 12, 2013 at 16:24
  • \$\begingroup\$ Are you sure data is good at the receiver end? (tested with a BERT or similar)? \$\endgroup\$ Commented Nov 12, 2013 at 16:33

2 Answers 2


Scheme #1 is terminating only the differential mode signal, not the common mode.

Scheme #2 is terminating both differential and common mode.

Even with a perfectly symmetrical differential output signal you will have what we call "differential to common mode conversion" in the cable. So at the receiver you will have both common mode and differential mode.

One source of this is the different propagation delay for the two signals of the pair (length mismatch and other effects). You measure this to 2-3ns, so you know it's there.

At the receiver, the common mode signal sees no termination and is reflected 100% (voltage doubling) with scheme #1. With scheme #2 some of that energy is absorbed by the termination resistors (note that the common mode impedance match may not be perfect, but it's definitely better than in scheme #1).

I did a quick simulation to show the effect of the two termination schemes with a 2ns skew in an otherwise perfect setup. See for yourself how much of a difference it makes.

Scheme #1 Scheme #1 with only differential mode termination.

Scheme #2 Scheme #2 with both differential and common mode termination.


There is a bit more details in this blog post I wrote while I was at it:


Update 2:

I swapped the plot for scheme #1 for the correct one. Guess you won't notice the difference, but the simulation was not done correctly.

  • \$\begingroup\$ now that sounds reasonable +1 \$\endgroup\$
    – Andy aka
    Commented Nov 16, 2013 at 15:20
  • \$\begingroup\$ @Andy aka: Thanks. Obviously I can't know for sure what the explanation in your specific case is, but this could explain what you see. BTW: If you want a bit more details I used this problem for a blog post as well while I was at it. Not sure exactly how the rules are for referring to that here though. \$\endgroup\$ Commented Nov 16, 2013 at 16:32
  • \$\begingroup\$ Not a problem. Please link the blog post and I'll read it Rolf \$\endgroup\$
    – Andy aka
    Commented Nov 16, 2013 at 18:00
  • \$\begingroup\$ I've read the blog but your simulation is flawed in scheme 1. I'm not saying this greatly affects the correctness of the answer but maybe you can to a re-look at this. What simulator did you use BTW? I'm quite happy to try it if it is free! Thanks for your effort on this, it is appreciated Rolf. \$\endgroup\$
    – Andy aka
    Commented Dec 3, 2013 at 13:43
  • \$\begingroup\$ I fixed the 50R->100R error in the plot both here and in the blog post. Thanks for spotting the error. Results are not much different - probably why I didn't catch the error myself. The simulator is Cadence SigXplorer (= not free). You should be able to do the same with Spice if you have a lot of time. Full disclosure: Cadence sponsors use of SigXplorer for the SI courses I teach around the world. \$\endgroup\$ Commented Dec 4, 2013 at 9:18

One potential issue is EMI - in option 1 you are essentially creating a magnetic loop antenna which would pick up way more noise than in option 2 where a lot more induced noise is grounded via resistors.

  • \$\begingroup\$ How can that be so with twisted pair shielded cable. And why would that be different for option 2. \$\endgroup\$
    – Andy aka
    Commented Nov 13, 2013 at 15:28
  • \$\begingroup\$ The effectiveness of a lot of EMI filters is determined by how close the ground is. For option 1, even if you have a twisted pair, you have 50m of cable which is bound to pick up some junk. In case 2 most of that junk is taken to ground via resistors. Another issue with option 1, is it carries any sort of signal resulting from imperfect input of the op-amp back along the 50 m cable. \$\endgroup\$
    – Yuriy
    Commented Nov 13, 2013 at 17:42
  • 1
    \$\begingroup\$ I don't understand why you mention EMI filters - this was a lab test with the cable on a reel in a near perfect environment. I'm looking for theoretical reasons why cable might perform better on skew when terminated in options 1 or 2. \$\endgroup\$
    – Andy aka
    Commented Nov 14, 2013 at 8:51

Your Answer

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

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