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In this datasheet (and in other similar) pins J4, J5, J7, J8 are connected to the ground with a resistor and cap. For 10/100Base-TX pinout it seems they are not used (at least for modern of the standard. for old is stated as "bidirectional"..?).

So why are these 75 ohm resistors there and why is the cap there? Why not just leave it unconnected? Has this something to do with EMC? or does it has something to do with this "bidriectional"?

pinout datasheet 10/100Base-TX pinout

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    \$\begingroup\$ Looks like it is just to avoid having a floating signal. It also may assist test equipment to measure length of cables. \$\endgroup\$
    – Ben Voigt
    May 13 at 20:17
  • \$\begingroup\$ But then why the cap and resistor? Why just not ground it? "measure length of cables" how?.. \$\endgroup\$
    – exzb
    May 13 at 20:21
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    \$\begingroup\$ Also "Bidirectional" comes from the fact your table includes 1000Base-T. There really should have been two separate description columns, one for 10/100 and one for Gigabit. \$\endgroup\$
    – Ben Voigt
    May 13 at 20:56

3 Answers 3

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It's called Bob Smith termination.

As there are multiple pairs, a pair also forms transmission lines with other pairs and may pick up or radiate EMI. The termination network reduces common mode currents. It keeps the voltages between pairs nearly zero by connecting the pairs with resistors to bias them.

The capacitor is required because Ethernet is an isolated interface. There must not be a DC path from connector to ground, so the capacitor is used for AC termination. At high or RF frequencies the capacitor impedance is a short circuit and at DC it is infinite. So any high frequencies see 75 ohms from pair to ground so they are terminated and don't cause large voltage swings, while at low frequencies the voltage can slowly float.

So that is why there must be a capacitor and why it cannot be connected to ground.

It has nothing to do with bidirectionality, the term does not apply in any meaningful way why capacitor and resistors are in the circuit.

All pairs, even unused pairs must be terminated, because all pairs in the cable pick up electromagnetic interference, and all four pairs couple with all four pairs as they run near each other for long distances. The used data pairs can also emit electromagnetic interference to the unused pairs which could act as antennas.

So in short, the Bob Smith termination reduces crosstalk between pairs, makes them less susceptible to receive interference, and makes them emit less interference, while keeping galvanic isolation from any potential such as earth or device ground.

10Base-T and 100Base-TX both use two pairs: one to receive and one to transmit.

1000Base-T uses all four pairs bidirectionally for both reception and transmission simultaneously.

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A slightly simpler answer.

Other answers correctly describe this as Bob Smith termination. Here is the patent that describes how it works. Here is an article that throws some scepticism on it.

However, here's a more simple perspective. The thing being terminated this way is the blue/white pair and the brown/white pair (with your wiring colours). This plays no direct part in the signal for 100Base-T or 10Base-T (as opposed to 1000Base-T which uses all four pairs directly), which is why "ethernet doublers" (mostly) work, and can pack two ethernet connections (10Base-T or 100Base-T) on one cable. I say "no direct part" because the transmission line effects mean that having 2 pairs terminated in this way reduces crosstalk, emissions and interference as Justme wrote. But it Ethernet (mostly) works without it - just not as well.

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It's called "Bob Smith termination" and it's for providing a path for common mode EMI. There's ongoing discussion on "which is better 75 ohms vs 52.3 ohms" and "is it even necessary" which you can make up your own mind on, but to answer your question yes, it's for EMI.

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    \$\begingroup\$ But it is not even connected to anything... these pins are not used in the ethernet cable. Where would the EMI even come from?.. \$\endgroup\$
    – exzb
    May 13 at 20:47
  • \$\begingroup\$ @exzb: If it's not ancient CAT3 cable (and even most CAT3 cable is 4 pair) then yes, these conductors are present in the cable and connected at the jack to the socket. Your own illustration in the question shows 8P8C. \$\endgroup\$
    – Ben Voigt
    May 13 at 20:53
  • \$\begingroup\$ It is not clear how well impedance is controlled on Cat4,5,6 especially CM impedance. Nor is it clear how well Zcm is raised with the CM part of the hybrid at both ends, possibly making Zterm redundant. But make no mistake. EMI is a bidirectional feature ,so do not ignore signal integrity from ingress. \$\endgroup\$ May 14 at 5:24
  • \$\begingroup\$ @exzb The EMI comes from electromagnetic coupling to the lines that are connected. It's similar to light. If you look at the interior of a camera, everything is black, even places that are away from the main light path. That's because some light gets scattered away from its intended path, and you want to absorb that light rather than allow it to reach the focal plane and mess up your image. In an electrical cable, resistive termination is the equivalent of black paint: it absorbs electromagnetic energy. \$\endgroup\$
    – John Doty
    May 14 at 20:11

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