10
\$\begingroup\$

Do I need a ground connection when connecting differential signals between two systems?

I was thinking this should not be needed, but how will common mode voltage at the receiving end behaves? Without connected the ground between the two systems, I probe the inputs at the receiving end and observe that the inputs rides on a 10V peak to peak 60Hz(probably interference from the power line). As there is a +/-5V common mode limit on the receiving input, this probably causes the erratic behavior that I'm now experiencing.

Will an earth connection at both systems help? Or a ground connection is still needed for connecting the differential signals?

\$\endgroup\$
3
  • \$\begingroup\$ Unless it is AC-coupled, ground connection is needed. \$\endgroup\$
    – venny
    Commented Sep 10, 2014 at 11:11
  • \$\begingroup\$ How is this normally accomplished in practice? Through the shield connection in a shielded cable? \$\endgroup\$
    – Park
    Commented Sep 10, 2014 at 11:16
  • \$\begingroup\$ That should work, no current should flow in the reference path. Why can't you go with the capacitive coupling? \$\endgroup\$ Commented Sep 10, 2014 at 11:23

3 Answers 3

6
\$\begingroup\$

There are 3 main types of receivers used to detect "differential signals":

DC coupled differential signals

RS-485, RS-422, CANbus, LVDS, USB, SATA, PCI Express, etc. directly connect differential signals to the receiver chip -- "DC-coupled". They require a ground connection to keep the signal at the receiver's end of the bus within the common-mode range of the receiver chip.

Often such systems stop working when the voltage offset is more than a few volts, and can be permanently damaged if the voltage offset ever reaches a few dozen volts. (That is, the voltage offset between the system "ground" at one end of the cable and the system "ground" at the other end of the cable).

Often 2 boxes with a cable between them carrying such a protocol (or a single-ended protocol such as SPI or RS232) seem to work fine in the lab sitting next to each other, but have intermittent communication or stop communicating entirely when placed in the field with long distances between them. When that happens people often end up buying 2 "isolators" that internally use one of the following approaches, and putting the long cable between those isolators.

opto-isolator coupled differential signals

Systems like MIDI connect more-or-less differential signals to the LED of an opto-isolator at the receiver.

With proper design, similar systems can and sometimes do work just fine with kilovolts of offset between the system "ground" at one end of the cable and the system "ground" at the other end of the cable.

transformer-coupled and capacitor-coupled differential signals

Analog audio, LonWorks(a), etc. connect differential signals to DC-blocking capacitors.

Ethernet, etc. connect differential signals to DC-blocking transformers.

Broadband-over-powerline receivers typically have both DC-blocking capacitors and DC-blocking transformers.

With proper design, they can and sometimes do work just fine with kilovolts of offset between the system "ground" at one end of the cable and the system "ground" at the other end of the cable.

These systems block the DC offset with a transformer or capacitors or both to carry the signal across the isolation boundary. (To reduce EMI and protect against cable discharge events, many systems also connect each cable wire with resistors or capacitors or both -- a Bob Smith AC termination -- to the chassis ground (b) (c) (d) (Intel AP-434); often with additional capacitors to support power over ethernet (e). )

Such offset voltages are the main reason behind " 2kV capacitor on ethernet? ".

Differential over a cable

How is this normally accomplished in practice?

When sending Ethernet, LonWorks, opto-isolated data, etc. over a cable, a ground wire is not required. All the wires in the cable can be used for data transmission. (PoE systems often end up pulling the two system grounds close together anyway; non-PoE systems allow the two system grounds to float apart).

When sending RS-485, CANbus, etc. over a cable, typically at least 1 wire in the cable is reserved for the ground wire, which pulls the system ground at one end of the cable and the system ground at the other end of the cable closer -- hopefully close enough to allow communication or at least to prevent permanent damage.

Many people use exactly the same (unshielded) CAT5 cable with standard RJ45 plugs at both ends for both kinds of systems.

When using a shielded cable, some people are very careful to design the system with the socket where that cable plugs in to have a separate "chassis ground"/"frame ground" and connect it to the shielding in the cable, and separate from the "data ground"/"signal ground" on, for example, pin 9 of a DB9 connector carrying RS232 data. Other people simply connect all the grounds together. I'm not going to say more here about that raging controversy.

\$\endgroup\$
3
\$\begingroup\$

If you used a transformer at one end, the ground would be isolated in which case you would not need a ground wire between.

But in theory I think you could also make a circuit without a transformer that just used identical drivers on both terminals with capacitors. Then it would be completely isolated. But in practice the impedance of both output terminals would have to be matched exactly. And the receiver would have to have good CMR. Then I think ground noise would basically be canceled out. And it might work better if signal was transmitted out-of-phase on both wires as opposed to just putting signal on one wire.

But in practice it's probably a dubious thing to do. If you want maximum ground isolation and great CMRR, then transformers rule. Otherwise, you need a 3rd wire so that both circuits are pushing / pulling against a common reference voltage (ground) so that any noise on that wire will not matter (not considering the wire resistance / inductance).

\$\endgroup\$
2
\$\begingroup\$

Even though it's differential, you do need to constrain the common-mode range. That's what the common ground is for; it keeps the references somewhat equal so that the common mode isn't influenced by a floating Tx, Rx, or both.

Of course, by adding that connection, you may then create a ground loop. See here for details on that: Ground Loop for balanced connection

\$\endgroup\$

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.