Below are straight and twisted pair transmission lines:

enter image description here

From the above illustration the induced noise current due to the magnetic coupling are opposite direction in each lines.

It seems like magnetic coupling causes differential noise rather than common mode noise. Because the currents above are not flowing the same direction.

On the other hand, I read many places that STP wiring minimizes the induced noise and differential ended inputs are used to subtract the same noise on both lines.

But this illustration contradicts that. For diff-input amplifier to cancel out equal noises the currents due to the magnetic coupling must flow in same direction.

Where am I thinking wrong here?


In below diagrams there are two independent scenarios where the only source of noise around is assumed to be magnetic field such as 60/50Hz or so. In both scenarios the transmission line is perfectly balanced and twisted pair is used.

The only difference is that in Scenario I single-ended inputs used; and in the Scenario II differential-inputs used.


simulate this circuit – Schematic created using CircuitLab

In this case, would the noise due to magnetic coupling be differential? And would using single ended or diff-ended inputs will not make any difference?


It seems like magnetic coupling causes differential noise rather than common mode noise.

In the illustration, that is what is going on.

On the other hand, I read many places that STP wiring minimizes the induced noise and differential ended inputs are used to subtract the same noise on both lines.

Suppose there's a third wire (ground, for example) connecting the two boxes.

And there is a changing magnetic flux in the loop between the data wires and the ground wire.

This will cause a common-mode (same direction) current on the two data wires. And using a differential receiver means this current won't (ideally) be detected as part of the signal.

If the loop between the ground wire and the data lines is bigger (for example, the ground wire loop may actually go through the building mains wiring) than the loop between the two data lines, the common mode current may be much larger than the current shown in your illustration, so it's a good thing we can ignore it at our receiver.

  • \$\begingroup\$ Current isn't induced. \$\endgroup\$ – Andy aka May 15 '18 at 17:26
  • \$\begingroup\$ @Andyaka it is if there's a path for the induced emf to drive it around. \$\endgroup\$ – The Photon May 15 '18 at 17:26
  • \$\begingroup\$ Then it is the voltage that is induced and the current serves to reduce the magnetic field. \$\endgroup\$ – Andy aka May 15 '18 at 17:27
  • 1
    \$\begingroup\$ @Andyaka, what term do you prefer? "generated current"? \$\endgroup\$ – The Photon May 15 '18 at 17:28
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    \$\begingroup\$ "The current produced by the induced voltage" is a bit unwieldy. \$\endgroup\$ – The Photon May 15 '18 at 17:29

The whole point of twisted pair is that the voltage induced on one wire equals the voltage induced on the other. However, if the wires were run side by side with no twisting, the wire nearest to the interfering magnetic field source would receive slightly more induced voltage than the wire slightly further away. That then becomes a differential interference.

What was the context of the picture you embedded? It may well be explaining something different to what you think it implies.

Scenario 1 shorts one of the twisted wires to ground and there can be no induced voltage on that line but, there is still an induced voltage on the other line hence this circuit won’t get rid of common mode induced voltages.

Scenario 2 is better but the impedance at both diff amp inputs may well be high but still unbalanced hence, this will work if a light impedance to local ground is applied to both wires. To balance out induced noise both induced voltages have to be the same and both impedances to ground also need to be the same. If V2 was grounded and presented a low output impedance to R1 then this would be a reasonably good circuit but, with V2 shown as a floating source it isn’t good for maintaining earth impedance balance (the technical term used in telecoms).

Please don’t edit your question any more because I don’t like editing answers to suit an evolving question where the goal posts are shifting bit by bit. If you don’t understand something raise it as a comment.

  • \$\begingroup\$ Okay I will draw a schematics \$\endgroup\$ – user1245 May 15 '18 at 17:54
  • \$\begingroup\$ I edited my question, hopefully more clear what I'm after. \$\endgroup\$ – user1245 May 15 '18 at 18:09
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    \$\begingroup\$ I thought the whole point of twisted pair is to create a controlled transmission line impedance. I can think of tons of applications where the cables are not twisted and still have common mode rejection and are differential. For instance, in professional audio applications you have differential signals with normal cables. Twisting are not needed because the low frequencies cause negligible reflections and transmission impedance doesn’t need to be controlled. \$\endgroup\$ – PDuarte May 15 '18 at 18:30
  • \$\begingroup\$ @PDuarte well telephone companies in the olde days always ensured that the overhead telephone wires were alternated pseudo randomly between poles to prevent differential pick up. Balanced Microphone wires are twisted as far as I know. \$\endgroup\$ – Andy aka May 15 '18 at 18:41
  • \$\begingroup\$ @Andy Aka I see but the point of my question is about whether there is common mode induced voltage or not. You wrote for Scenario I "this circuit won’t get rid of common mode induced voltages." Is there Common Mode induced voltage in Scenario I? Isn't the induced voltage noise due to magnetic field is differential in Scenario I? \$\endgroup\$ – user1245 May 15 '18 at 18:42

Wires near a source of electromagnetic interference are loosing coupled between signal and interference dependent on the mutual orientation, gap, shielding, quality of ground and quality balance in the wire pair's impedance load.

Since each wire radiates a magnetic field , B , the return current flowing in the oppposite direction at the same time will tend to cancel out which is why twisted pairs offer good immunity.

We can readily understand that two current loops will have crosstalk when they are parallel and close togther and minimal when the loop is at right angles to each other. The intensity of the induced current is proportional to the length and 1/gap² while the voltage created depends on the impedance of the signal. Being just air gap and a large gap the current is very small but with high impedance inputs and low voltage can cause trouble. The mutual inductance coupling for V=L*dI/dt should tell you that more voltage is induced dependent on the current slew rate and thus frequency spectrum.

The amount of current induced that is common mode and differential mode depends on the impedance (f) of each wire. Therefore balanced pair impedance is good and balanced high impedance is best. This is what CM chokes do is raise the CM impedance (e.g. of both ground and signal) so that the DM voltage induced from the magnetic field, B is minimize.

Low frequency crosstalk occurs in large loops while high RF occurs in small loops. A shielded cable offers no protection for low frequency magnetic currents but does for electric fields. So a shielded twisted pair gets the benefit of twisted pairs for magnetic fields and the shield for electric fields at low frequency. At high frequency , the common mode inductance and "transfer impedance" is affected by many different methods where solid coax is far better than braid or foil.

Electric field crosstalk is different in that it couples by capacitance of surface area gap, but also transfers more with increasing frequency as as Zc reduces.

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