This link clearly explains how high impedance nodes are susceptible to capacitive coupling noise. Is the same true for inductive coupling?

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As per my understanding, inductive coupling would induce an emf ( mutually induced emf) in a nearby circuit and this would cause a current flow ( if 'circuit is closed') which would oppose the change of flux.

Hence if a circuit has lower impedance, it can result in higher current flow (due to noise) and oppose the change of flux and hence reduce the resultant emf generated ( Resultant emf is a result of net flux change). ie, resultant effect of noise will be lower (provided voltage is what matters to us!)

Why we don't hear about inductive coupling often? Everybody seems to talk about only capacitive coupling!

  • 1
    \$\begingroup\$ Your reasoning sounds okay to me but you'll have to get someone else to confirm. I think you hear more about inductive coupling in high current circuits like motor runs than low power circuits. I think inductive coupling is more through the magnetic field like in a transformer so you need higher currents to notice it whereas a lot of microelectronic circuits are very low current (and high voltage, at least relative to the current levels involved). But I'm not quite sure about that either. \$\endgroup\$
    – DKNguyen
    Jan 28, 2022 at 5:14

2 Answers 2


Capacitive coupling produces currents. If these currents flow through a high impedance, you will see big voltages. (The source of the current is a voltage swing somewhere.)

Inductive coupling produces a voltage across things. If this voltage is applied to high impedance, not much current will flow. If it is applied to a low impedance, lots of current will flow. ( The source of the induced voltage is a current flowing somewhere.)

So, you are correct.

  • \$\begingroup\$ "Capacitive coupling produces currents"- Capacitive coupling doesn't necessarily induce currents. It just induces a voltage at the other end of parasitic capacitance. Remember a capacitor cannot change the voltage 'across' it unless there is a current through it. If one end of a parasitic capacitance is experiencing a voltage change, the other end would also be forced to have same voltage changes unless there is a current flow. If the impedance is lower ie, higher current through parasitic capacitance, lesser would be the impact on the receiving end. \$\endgroup\$
    – Divya K.S
    Jan 28, 2022 at 5:55

Consider the following circuit:


simulate this circuit – Schematic created using CircuitLab

It is a transimpedance amplifier with a very high input impedance node "IN". The impedance is given by Rsrc||R1

If you capacitively couple to this node, you change its voltage and the output will react to compensate. If Rsrc is small, then only a small amount of the offender's voltage will actually materialize in "IN", the rest being shunted away through Rsrc. Therefore, high impedance (large Rsrc) circuits are more vulnerable to capacitive coupling than low impedance circuits.

If you couple inductively, you will generate a voltage between the ends of the "IN" node, and the voltages at both ends will try to float away in opposite directions. The op-amp will keep its end of the line grounded, for which it needs to send a reverse current through R1, creating an output voltage. If the left end has a very high Rsrc (>> R1), it will float away and be fine. But if the left side has an impedance much lower than R1, its end will be forced to some voltage, too and that end will also send a reverse current through through "IN", through R1 and generate an output voltage, as well. So low impedance circuits are indeed more susceptible to inductive coupling, but even high impedance circuits will react appreciably to inductive coupling.


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