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I’m trying to make the question clear by simplifying a single ended circuit as a constant voltage source is coupled to a measuring device such as a scope or an ADC ac below:

enter image description here

The source V1 above has an output resistance Rs we can vary. And the source is coupled to a measuring unit which has 100Meg input resistance. And as in the drawing the wires form a rectangle |ABDC|. I also draw x y z coordinates where x is horizontal axis of rectangle y is the vertical axis and z is the axis going out of screen which penetrates the rectangle loop.

Now It I assume there is 60Hz magnetic and electric fields around this circuit.

1- If we only focus on magnetic field’s effect on this can we say the following?

According to Faraday’s law, the varying magnetic field component through |ABDC| (along with z axis) will induce voltage across the loop. Can we say that the more Rs is increased the more the 60Hz noise will be observed at B? How can we model this coupling as a lumped circuit element? A series 60Hz voltage source?

2- If we only focus on electric field’s effect, regarding Maxwell equations what is the 60Hz electric field’s path which induces 60Hz displacement current through the circuit? Is it the electric field looping through |ABDC|? (x an y axis). And how is this type of coupling modelled? Will increasing Rs increase the 60Hz current?

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  • \$\begingroup\$ Read my answer to your last question which has the same error as this. B field in mA/mm or A/m times loop impedance in that field and loop area follows what law? \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 May 5 '19 at 20:35
  • \$\begingroup\$ Which of Maxwell equations is related to E field coupling? \$\endgroup\$ – cm64 May 5 '19 at 20:37
  • \$\begingroup\$ How does the 60Hz E field induce current through the rectangle? \$\endgroup\$ – cm64 May 5 '19 at 20:38
  • \$\begingroup\$ Read my Reply to last question again. E fields are V/m induce into dielectrics to conductors. B fields are A/m induce current into loop area conductors not capacitors \$\endgroup\$ – Tony Stewart Sunnyskyguy EE75 May 5 '19 at 20:48
  • \$\begingroup\$ How can we visualize E field coupling? Can u provide an image like E field vectors coupling to the rectangle? My problem is visulasing what is happening:( \$\endgroup\$ – cm64 May 5 '19 at 20:59
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Here is two models for (1) magnetic coupling, and for (2) electric coupling.

(1) assume the source of the magnetic field is a long straight wire, in the plane of a rectangular loop; you will encounter lots of these topologies (a wire and a loop) in circuitry.

To compute the induced voltage, which we model as a source inserted into the loop, I use this formula (which ignores the Integral needed for accuracy when the wire is near the loop)

Vinduce = [MUo * MUr * LoopArea / (2 * PI * Distance_wire_to_Loop)] * dI/dT

For MUo = 4 * PI * 1e-7, for MUr = 1 (air, copper, Fr-4, aluminum), the formula becomes

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

Example of use: switching regulator, I = 1 amp, switching is 100 nanosecond, Distance between Switcher wire and the circuit loop being 1cm, and the circuit has 1cm * 1cm loop.

Vinduce = [2e-7 ( 1cm * 1cm) / 1cm ] * 10million amps per second dI/dT

Vinduce = 2e-7 * 1cm * 1e+7 = 0.02 volts

Thus tis difficult to have analog circuits NEAR a switching power supply. Also the operating of a crystal oscillator near a switching power supply, or near ANY switching current (such as from MCUs) has risk of edge jitter in the clock timing.

Notice the wire is NEAR the loop, thus this equation needs the Integral (Using natural log) for complete accuracy.

(2) I will provide example of Efield interference modeling, later in the day.

schematic

simulate this circuit – Schematic created using CircuitLab

We need the capacitance between Transmitter and Receiver. To get started, pick the smaller area of the two, and use the parallel-plate capacitive model:

C = E0 * Er * Area / Distance

Suppose trace is 30mm by 2mm, and is 30mm from a 117VAC (160v peak) at 60Hz power wire, that has NO SPIKES ON THE POWER; if pure sin, we know the slew rate exactly; if power has motor spikes, assume 100X faster spikes; if next to MCU Dataline, assume 5 volts/1nanosecond dV/dt.

What is the capacitance, between Transmitter and Receiver?

C = 9e-12 farad/meter * (er=1 for air) * 30mm * 2mm/ 30mm

C = 9e-12 farad/meter * 0.002 meter

C = 9e-12 * 2e-3 = 18e-15 ~~ 20 femtofarads

That forms a voltage divider with the 20picoFarads I assumed the Receiver PCB trace to look like: metal over ground + ESD diodes of the OPAMP + FET gate capacitance of the OpAmp's diffpairs + capacitance of the 10MegOhm resistors. Thus 20 pF is a useful starting point.

Notice this is capacitive divider, just like a scope probe. We have

18 femtofarad / 20picoFarad ==== 60dB drop, or 1,000:1 smaller.

Thus the 160 volts at 60Hz will appear, on the opamp feedback path, at 0.16 volts. But there is more.

Unless the High_Pass_Filter dominates. Does it? What is the time constant of 20pF and 5MegOhms? 100 microSeconds, or 1,600Hertz HPF corner.

The ratio between 60Hz and 1,600Hz is 25:1, thus the HPF further reduces the 60Hz injection by 25:1, or 0.16/25 = 6 milliVolts.

Notice we are assuming a CLEAN PURE SIN, no spikes, at 60Hz.

If the Transmitter has 10uS spikes, the HPF would have no effect.

And an MCU dataline, with 1nanoSecond edges, would ignore the HPF.

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  • \$\begingroup\$ Please add some drawing or visualization formulas alone no help. I asked this question to learn how does the electric field couple and induce noise Details dont help if I dont get the basic picture. I hope you get what I mean \$\endgroup\$ – cm64 May 5 '19 at 22:32
  • \$\begingroup\$ yes----I will also provide diagrams for both Hfields and Efields. Can you imagine, on a PCB, many situations where a trace is in the same plane as loop? Thus the math can become rather exact, especially if Natural Log (and skin effect) are included. \$\endgroup\$ – analogsystemsrf May 5 '19 at 23:25
  • \$\begingroup\$ Your second part I was looking for. Is there any resource in this topic I can learn more but not advanced. \$\endgroup\$ – cm64 May 6 '19 at 17:20
  • \$\begingroup\$ check out Signal Chain Explorer, when the "gargoyles" are enabled. \$\endgroup\$ – analogsystemsrf May 7 '19 at 5:23
  • \$\begingroup\$ cm64 ---- what additional discussion, or illustrations, might assist you? \$\endgroup\$ – analogsystemsrf May 11 '19 at 6:38
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1)

If the output of source V1 is loaded by so high resistance like 100Meg, the small Rs affects the loop almost nothing.

According to electrical model, I would simulate it as coupled inductors:

enter image description here

2) Its the electric field generated from probably power line wire, which has some little inductance.

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Michal podmanicky is correct for the magnetic coupling. It appears as a voltage in series with the circuit. And the electric field coupling can be modeled as a current injecting into the signal wire. The magnetic coupling ( a series voltage) arises from some currents flowing around a loop somewhere. The electric coupling ( a current injected) arises from a changing voltage on some conductor somewhere.

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