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enter image description here

I don't understand why second way of implementing is better than the first one? I have my own basic logic regarding this topic, but i am not confident if what i think is correct! so i was looking forward to have some general view on this topic!

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Ask yourself which cable is likely to have the lowest loop area:

cable

A large loop area has greater inductance and can emit more EM interference. It can also receive more EM interference.

If each forward conductor has its own return wire then this potentially minimizes each circuits loop area.

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    \$\begingroup\$ Oww !! just watching at the image made everything clear !! Thank you! \$\endgroup\$ – Hilton Khadka Apr 26 '16 at 9:32
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    \$\begingroup\$ A decent compromise if you're low on pins / wires is to have 2 wires share a GND -- still better than the first example where the only return is far far away. \$\endgroup\$ – Krunal Desai Apr 26 '16 at 23:33
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If you were transmitting something of the order of a 10s to 100s of kHz, number 1 may be perfectly adequate, but number 2 does have the advantage that each signal has a return path that is closer and therefore reduces the total current loop. This is useful in minimising radiated emissions (and radiated susceptibility too), to say nothing of crosstalk.

I would not say they are perfectly balanced in number 2 (Signal 1 has only return 1, signal 2 has return 1 and return 2). That said, the signals are shielded more effectively in this configuration.

In number 1, the single return would have to carry all the return currents and would need some careful checking to ensure the path is capable of the current loading.

In a high speed world, things would have balanced return paths for both impedance control (the distance to the return path is an important component of the track impedance) and to separate the return currents for various reasons.

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  1. The second alternative provides far better shielding. Since the current in signal N and return N flows in opposite directions, their EM fields cancel each other out. So in the first example, when signal 1 is active, signal 2 is more exposed to crosstalk as it is in the second example.

  2. As the picture title says, a single return path creates unequal (and much larger) loops than dedicated return paths. If the cable in example 1 is exposed to external EMI source, signal 1 will receive four times as much interference as signal 4 will. In the second example, both signals will receive the same, minimal amount of interference.

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  • \$\begingroup\$ OK, this makes more sense now! \$\endgroup\$ – Hilton Khadka Apr 26 '16 at 9:37
  • \$\begingroup\$ but could you explain me, why signal 1 would receive four times larger interference than signal 4 ? \$\endgroup\$ – Hilton Khadka Apr 26 '16 at 9:46
  • \$\begingroup\$ Because the distance between signal 1 and return is 4 times larger than the distance between signal 4 and return. \$\endgroup\$ – Dmitry Grigoryev Apr 26 '16 at 9:53
  • \$\begingroup\$ But wouldn't the signal 4 be affected by the EMI produced by other 3 signal lines? \$\endgroup\$ – Hilton Khadka Apr 26 '16 at 9:55
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    \$\begingroup\$ @HiltonKhadka On/off DC is fairly easy, but you rarely would use multi-conductor cabling for that. (Two wires is enough.) Of course, you have interference during rise and fall periods when turning the source on or off respectively, but other than that there's no big issue. However, when you do that hundreds of thousands or millions of times per second, the effect adds up! Look at a "square" wave in detail on an oscilloscope some time; it is an enlightening experience. \$\endgroup\$ – a CVn Apr 26 '16 at 14:51
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Basically, when the path that current takes encloses an area, this creates inductance. Inductance is like electrical inertia, it prevents the current from changing quickly (like how regular inertia slows down speed changes), this electrical inertia rounds off square pulses and attenuate fast signals, worse still the current still want's to flow even when the load is removed, creating voltage spikes. Loops also act as good antennas and will broadcast their contents all over the place. These effects get worse the larger the loop area (larger inductance). Having a return right next to each signal keeps the loop area really small and keeps the signal lines isolated from each other - each return is essentially a shield.

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  • \$\begingroup\$ Well, I can't guarantee it'd be 4 times more, but it's got a bigger loop area (more or less 4 times bigger as it's about 4 times further from the return), loops act like antennas, larger antennas collect more energy, that energy has to somewhere and ends up on your signal lines. An antenna 4 times the size will roughly pick up 4 times as much interference. \$\endgroup\$ – Sam Apr 26 '16 at 9:59
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    \$\begingroup\$ Also good to see: In the first case, the 4 loops overlap. Not only do they act as antenna's, they act as paired emitters and receivers. \$\endgroup\$ – MSalters Apr 26 '16 at 14:07
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I'm surprised no one has mentioned differential signalling. https://en.wikipedia.org/wiki/Differential_signaling

If you want fast clear signals, you use this technique which requires the "return" line be used in an active manner. In this case, you wouldn't want multiple drivers of the single return line.

From source: "The technique minimizes electronic crosstalk and electromagnetic interference, both noise emission and noise acceptance, and can achieve a constant or known characteristic impedance, allowing impedance matching techniques important in a high-speed signal transmission line or high quality balanced line and balanced circuit audio signal path."

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  • \$\begingroup\$ I think this is more related to single-ended signals, where there is just a common ground for the signals. Perhaps they are low-speed. The return wire for each single-ended wire can still help with ground loop/noise immunity, without having to be fully differential. With differential you'd still want a reference ground wire somewhere, shared by all signals, in addition to each differential pair. \$\endgroup\$ – KyranF Apr 26 '16 at 17:23
  • \$\begingroup\$ @KyranF Good points. \$\endgroup\$ – horta Apr 26 '16 at 17:51
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In addition to the EMI issues, wires which run parallel to each other will have a certain amount of capacitive coupling. Capacitively coupling signal wires to ground will increase the amount of energy lost to the resistance in the cable or signal source, but that can be dealt with much more easily than crosstalk caused by capacitive coupling of signal wires to each other. In clocked parallel protocols which are slow enough that data wires will have time to stabilize before their clock arrives, it may not be necessary to worry too much about crosstalk between data wires, but modern protocols are more prone to send lots of data quickly over a few wires than to send a smaller amount of data more slowly over each of a large number of wires.

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This gets a bit involved, if it makes you feel better, many electronics designers are not perfectly clear on this either. Hence the "black magic" of EMI.

Here's an article which attempts a simple explanation of what's going on: http://learnemc.com/identifying-current-paths

So for "high" frequencies (larger than a few kHz), inductance starts to dominate and the "easiest" return path for the current is to follow the driving current as closely as possible in the reverse direction. This would be on one of the signal lines in your example! On low frequencies the shortest way wins so it would probably mostly flow through the single ground wire.

The higher the frequency and the worse this gets, not to mention your signal will become wonky.

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