How is electromagnetic interference possible when measuring voltage difference?

Question

This question is primarily concerned with audio but it's probably applicable to other areas as well.

As far as I understand, EMI is caused by electromagnetic waves adding power to wires as they transmit a signal. Let's say you are transmitting an audio signal and the reference voltage wire is at zero watts and the signal wire is at one watt.

Now imagine you have the world's worst EMI and it causes an extra watt to be added to the signal wire and the reference voltage wire. The difference between the wires is still two so the signal hasn't changed. What gives? Why does EMI cause nasty noises at the other end when it would presumably be changing both wires' power equally?

Answer 1

Now imagine you have the world's worst EMI and it causes an extra watt to be added to the signal wire and the reference voltage wire. The difference between the wires is still two so the signal hasn't changed.

This is why differential wiring is often chosen. It does help to reject radiated interference.

What gives? Why does EMI cause nasty noises at the other end when it would presumably be changing both wires' power equally?

It won't change both wires' power (or voltage or current) exactly equally.

If one wire is slightly further from the source than the other (even by millimeters) then it will see a slightly lower E field from that source. At high frequencies, the signal reaching the further wire will be significantly out of phase with the one reaching the closer wire.

Furthermore, any AC magnetic field in the space between the two wires will induce an EMF in one direction on one wire, and in the opposite direction on the other, exactly a differential current signal. To combat this we can twist the wires so that the EMFs induced in one section of the wire pair oppose those produced in another section. But the twist is never perfectly uniform, and the field is never exactly uniform over the length of the wire pair.

Finally, your differential receiver can not perfectly reject common mode signals, so in your example, where a extremely large common mode interference signal is injected on the wire, imperfections in the receiver will cause this to confound your attempts at a differential mode measurement.

Answer 2

You are on the track of a valid idea, which is the basis of why differential signaling is often used - for example in the balanced signals on XLR connectors in professional audio which do things like help keep interference out of the microphone signals in a live show.

But common single ended systems do not evaluate a signal in comparison to its return, but rather in comparison to a local ground, which would not contain the same interference.

And even in a differential setup, the idea only works to the extent that interference is coupled into both the non-inverted and inverted sides of the differential pair equally and to the degree that the differential receiver successfully rejects common mode interference. For example, if the induced voltage exceeds the input range of the differential receiver, it may cause problems regardless of it being equal on both sides of the pair.

Additionally, radio frequency noise may bypass the intended behavior of the differential amplifier, or may find paths of coupling in to the receiving circuit which bypass the differential amplifier itself. So typically gear is also designed with elements which try to reject this higher frequency energy.

Answer 3

Suppose you have a ower wire 1 meter away from, and running parallel to, a twin-lead TV cable.

Assume the wire has 100 amps peak current surges, as the diodes rectify the power; assume the diodes turn on in 10 microSeconds, that speed of turnon defined by the inductance of the power system and the slewrate of the power voltage, and the q/kt of the silicon rectifier. This produces 10,000,000 amps per second dI/dT.

Now back to the twin-lead. Assume the twin-lead wires are 1cm apart, and the twin-lead runs parallel to the power wiring for 10 meters.

Can we predict the induced interferer? Yes. Use a formula for magnetically-induced voltage into a rectangular loop of wire, from an interfering wire which is in the same plane as the loop and (for easy math) parallel to one axis of the rectangular loop.

Vinduce = [ MU0 * MUr * Area / (2 * pi * Distance) ] *dI/dT

For air and copper (non-magnetic materials), this simplifies to

Vinduce = [2e-7 * Area / Distance ] * dI/dT

We ignore a small logarithm coefficient, so the causality becomes obvious

Now insert values

Vinduce = 2e-7 * (10 meters * 1cm )/1meter * 1e+7 amps/second

Vinduce = 2e-7 * (10 * 0.01 ) * 1e+7 = 2e-7 * 0.1 * 1e+7 = 0.2 volts

Can your system tolerate 0.2 volts on top of your sensor signal?

Answer 4

It is the pesky slow speed of light that causes these kinds of problems. It causes the physical lengths and shapes of the conductors to form antennas allowing radio frequency (RF) energy to be absorbed into and radiated out of the circuit. In the case of RF energy absorption, the components intended to protect against electrostatic discharge rectifies the RF energy forming an AM receiver. This effect starts to be an issue at frequencies above 100 KHz. As a first order approximation use 15 nH/inch for the inductance of a wire as long as the physical length of the wire is less than a twenth of a wavelength. The wavelength is the speed of light divided by the frequency.