4
\$\begingroup\$

Some devices output differential outputs since both wires carry varying signals which are mirrored. Most devices on the other hand output one signal carrying and one GND wire. There is also differential-ended wiring and single-ended wiring. My question will be about "external" interference type of noise(EMI or RFI or maybe capacitive coupling) impinging upon both wires.

After some reading, if I'm not wrong I came to understand the fact that what we really want is "the same noise should appear upon each line" regardless of it is a differential or single ended wiring. And the noise must be common(common mode) so that the differential amplifier can cancel it out.

As far as I understand, there are two main ways to make the noise common. One of them is twisted pairing. The idea here is the wires are very close to each other because they are twisted and the same magnetic filed or electric field hits upon both wires. The second method to make the external noise common is to make the total output impedance(from source to receiving end) of both wires equal, so if an EMI happens both wires would induce the same voltage at the same time.

Upto here was just a summary of my understanding to show/check to whom read this.

Here is a simple illustration of a single-ended and a differential-ended wiring:

enter image description here

If they are both using twisted wiring and they are all balanced, they would both reject the common mode noise.

So what is the point to send the signals in mirrored fashion if one can balance and twist a single-ended wiring as well? Is that because something is practically easy to establish in differential/mirrored outputs? I really couldn't find a good answer to this.

edit illustration for a comment:

enter image description here

enter image description here

\$\endgroup\$

3 Answers 3

Reset to default
3
\$\begingroup\$

As has been mentioned by peufeu, if you use balanced differential drivers you create twice the p-p voltage and hence the SNR is doubled because noise can be assumed to remain as it was previously. But, there is another subtlety that was alluded to in the question and this relates to impedance balance.

Quite simply, if the balanced driver had an output impedance of 20 ohm (ignoring the output feed resistor R5) you would need to add 20 ohm to the balancing drive resistor (R6) that connects to 0 volts to keep balance.

But it's a bit harder than this because the output impedance of a driver chip will likely vary across its bandwidth so, it's much easier to use a dual inverting driver and equal values for R1 and R2 (right hand diagram).

\$\endgroup\$
11
  • \$\begingroup\$ I see I think it is not practically easy to measure the output impedance of a transducer and size and add a resistor to its signal ground R6. Imagine I have 12 transducers single-ended going to same ADC. If I want balance I would have to measure the exact output impedance of each transducer. But in diff ended since we know they are "both" 20 Ohm we can just add 100Ohm to each line or nothing at all. Did I get you correct? \$\endgroup\$
    – user1245
    Sep 16, 2017 at 13:25
  • \$\begingroup\$ I have no idea what you are saying, sorry. If you can't estimate the output impedance of a device then use a balanced driver to interface it to the line. Is this what you are getting at? \$\endgroup\$
    – Andy aka
    Sep 16, 2017 at 13:29
  • \$\begingroup\$ I mean practically one doesnt have to deal with measuring output impedances of the signal sources in diff ended outputs. Because as you say diff output wires already have same output impedances. But in single ended I have to measure the output impedance or look at data sheet and add resistor to GND. I was looking for a practical benefit of using diff ended output. \$\endgroup\$
    – user1245
    Sep 16, 2017 at 13:33
  • \$\begingroup\$ In a single ended output you have to take account of its output impedance and match that onto the 0 volt wire. You generally don't need to do this when using a balanced driver because of impedance similarity. \$\endgroup\$
    – Andy aka
    Sep 16, 2017 at 13:35
  • \$\begingroup\$ Okay this is the answer and the "practical advantage". \$\endgroup\$
    – user1245
    Sep 16, 2017 at 13:36
3
\$\begingroup\$

Good answer are already posted but I would also add another point.

Single ended systems usually rely on ground as the differential signal. That is fine if the connection is short and you can guarantee the grounds at both ends are the same and you do not have too many of them. Differential transmission removes this reliance and provides some degree of ground isolation and prevents ground loops.

Also, from your question you seem to imply that balancing the impedances and using twisted pairs are alternative solutions. They are not, for best performance you need to do both. In fact, if you don't balance the impedances, using twisted pair is almost pointless.

Also you might find my answer here interesting.

\$\endgroup\$
26
  • \$\begingroup\$ Thanks I read your answer but I would be glad if you can answer these 1-) Can you illustrate/(simple drawing maybe) how ground is not an issue in diff ended but can be an issue in single ended. I cannot picture what you wrote thats why. 2-) If I use coax cables do I still need twisting? Is it better to use twisted shielded cables for single-ended than using simple coax cables for distances around 15 meters? \$\endgroup\$
    – user1245
    Sep 16, 2017 at 13:53
  • 1
    \$\begingroup\$ @user134429 In differential the receiver is looking for a difference in potential between the two signal lines, not a difference in absolute values. You are effectively using a different reference. Using COAX does not help with common mode noise since the ground shields the signal wire from noise, twisted pair shielded cable is better. \$\endgroup\$
    – Trevor_G
    Sep 16, 2017 at 14:01
  • \$\begingroup\$ Thanks. I plan to balance a single ended system . So first I have to change the coax to shielded twisted pair then. Imagine you yourself want to balance a single ended system. And you know the output total impedance and you concluded to add resistor to 0V wire and you think you are done with balancing. Imagine the wiring is twisted as well so you think the system will create the common mode noise and reject it. Is there a technique to create interference and "test" if the system is really rejecting noise? I mean can I experiment if I balance a single ended system? Is there a way to verify? \$\endgroup\$
    – user1245
    Sep 16, 2017 at 14:10
  • \$\begingroup\$ @user134429 that is a whole other thing and beyond the scope of this answer. You should post that as a new question. \$\endgroup\$
    – Trevor_G
    Sep 16, 2017 at 14:13
  • \$\begingroup\$ I see. But regarding what you've written: "twisted pair shielded cable is better." I now use coax and the shield of the coax is earth grounded in receiving end. And the source is floating. If I change to shielded twisted pair. Should I use one wire of the pair as ground? Or should I use shield as ground? Because there will be 3 wires now. One pair and a shield. Should I just not connect shield any where and use one pair as ground? \$\endgroup\$
    – user1245
    Sep 16, 2017 at 14:17
1
\$\begingroup\$

So what is the point to send the signals in mirrored fashion if one can balance and twist a single-ended wiring as well?

2x more voltage swing, thus dynamic range and SNR doubles.

For high-speed signals, driving the cable differentially also reduces the amount of radiated noise. On a PCB, it keeps return currents out of the ground plane. For logic signals (LVDS) it also reduces the amount of noise and ground bounce at the driver side, by making the power supply current of the driver relatively constant and independent of signal. On the receiver side, equal and opposite signals cancel as they couple through parasitic caps, so this also reduces noise.

Consider USB3: if the drive wasn't differential, you'd need 2x more voltage swing for the same noise margin, also you'd need more elaborate common mode filter (ferrites) to reduce radiated noise.

\$\endgroup\$
5
  • \$\begingroup\$ I thought in mirrored outputs the wires are already balanced(they have the same output impedances from manufacture) and one does not have to spend time on balancing. But in single-ended one has to measure the output impedance of the source's signal and add a corresponding resistor to GND to balance the wires. I was thinking this was the reason. \$\endgroup\$
    – user1245
    Sep 16, 2017 at 13:12
  • \$\begingroup\$ Btw regarding your first argument. One can easily send 2x voltage swing by just doubling the output voltage in single-ended output as well(?). Could you explain this why SNR increases in differential? Thanks \$\endgroup\$
    – user1245
    Sep 16, 2017 at 13:15
  • \$\begingroup\$ Well, you can increase signal level, but not more than the supply voltage... and low voltage is popular these days! Consider an ADC running on 3.3V, can't take input voltage above the supply, so a differential input driven by differential signals doubles dynamic range... \$\endgroup\$
    – bobflux
    Sep 16, 2017 at 13:19
  • 1
    \$\begingroup\$ Also if you want to drive ethernet through 100m of cables, from a 3V3 chip, considering termination resistors eat 50% of the signal, differential drive really helps in feeding as much power as possible into the cable... \$\endgroup\$
    – bobflux
    Sep 16, 2017 at 13:23
  • \$\begingroup\$ Oh okay that explains why diff ended is mostly used for low level voltages (?) \$\endgroup\$
    – user1245
    Sep 16, 2017 at 13:27

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.