Do Common-mode Currents Split Equally across Circuit Conductors?

This post discusses common-mode currents in circuits, examining the commonly-held belief that a cable's common-mode currents are divided equally among the cable's conductors.  It is a companion piece to my blog post on transmission-line common-mode currents:  https://k6jca.blogspot.com/2024/12/do-common-mode-currents-split-equally.html

Introduction:

When designing electronic circuitry for products, one requirement is to ensure that any unintentionally radiated EMI (Electromagnetic Interference) is below limits set by EMC (Electromagnetic Compatibility) specifications.

Although EMI emissions can sometimes be generated by a voltage (e.g. high voltage nodes in off-line switching power supplies), in my experience EMI emissions are more often created by unintentional currents within the electronic system. These EMI-creating currents are commonly known as "common-mode currents," and they can have current-loop paths (the complete path that a current take, outbound from the source and then back again) whose encompassed-area is large, resulting in significant magnetic-field induced EMI.

On the other hand, "differential-mode" currents are currents of intentionally created signals and, if their routes are well designed, both the signal's path to its load and its return path will be in close proximity.  The loop area enclosed by the entire differential-mode current path should be small and thus magnetic-field EMI radiation from this signal minimized.

Given a two wire cable, it is commonly accepted that the differential-mode currents on such a cable will be of equal magnitude and flow in opposite directions, while common-mode currents, if they exist, will be of equal magnitude but flow in the same direction, as shown in the diagram, below (adapted from Introduction to Electromagnetic Compatibility, C. R. Paul, Wiley Interscience, NY, 1992):

But must common-mode currents across multiple conductors always be equal?

Unequal Common-mode Currents in Electronic Systems:

Below is an image (from a Murata Manufacturing Company, Ltd. document) illustrating how differential-mode and common-mode currents might exist within an electronic system.

Note that the differential-mode currents are equal in magnitude and travel in opposite directions on well-defined circuit paths.  They are driven by the "Signal Source," which is a differential-mode voltage.

In the circuit above, the "Noise Source" is a common-mode voltage.  The combination of this common-mode voltage (Vcm) and the differential-mode voltage (Vdm) would be modeled per the illustration below, such that Vcm attaches to the midpoint of Vdm, making Vdm balanced with respect to Vcm:

The circuit's common-mode currents, in this case generated by an unwanted noise voltage, travel in the same direction on the two conductors along a path that includes parasitic capacitance coupling of the load to ground.  But are these two common-mode currents equal?

In the diagram below I've added parasitic capacitances to ground at the load-end of each wire.  Note that the common-mode currents on the two wires will only be equal if the capacitances are equal -- that is, if the circuit load is balanced with respect to ground.  

Therefore, if the impedance-to-ground seen by Vcm looking into one wire is different from the impedance Vcm sees looking into the other wire, the common-mode currents will not be equal.

Analysis of the Circuit's Currents:

The following images provide a more in-depth analysis of the circuit:

So, we have shown that the equations: I1 = Ic + Id and I2 = Ic - Id (and therefore the equations Ic = (I1 + I2)/2 and Id = (I1 - I2)/2) are not always true, but instead their applicability is a function of circuit component values and the balance-about-ground of the differential-voltage driving the circuit.

Another Interesting Observation:

Recall that R3 is the load for our differential-mode signal, Vdm.  Let's examine the equation for the current through R3:

We do not want Vcm (i.e. a noise source) to corrupt the Vdm signal across R3!  To prevent this from occurring in the circuit above, R2*R4 should equal R1*R5.

SPICE Simulations:

The following SPICE simulations can give us actual values of the currents in each component.  With these currents, we can test if and when Ic = (I1 +I2)/2 and Id = (I1-I2)/2 are true.  

Note that the following values for current calculated with LTSpice are the same values as would be calculated using the equations derived with MATLAB, shown earlier in this post.

LTSpice with Balanced Vdm and Balanced Load:

In the circuit below Vdm equals 1 volt (i.e. Vdm1 = Vdm2 = 0.5V), and it is balanced with respect to ground (e.g. when Vcm is set to 0 for Vdm analysis).  That is, its midpoint is referenced to ground.

Also, the "parasitic loads to ground," R4 and R5, are of equal value and balanced with respect to ground.

To analyze the currents when driven by Vcm alone, Vdm is set to 0 (i.e. the Principle of Superposition) and the circuit simulated.

Similarly, to analyze the currents when the circuit is driven by Vdm, Vcm is set to 0 and the circuit is simulated.

Total currents are calculated with both Vdm and Vcm set to 1 volt .  Note that the total currents should equal the sum of the Vcm-driven currents plud the Vdm-driven currents.

Recall our equations Ic = (I1 + I2)/2 and Id = (I1 - I2)/2.  I1 is equal to the total current through R1, which is 0.0131 Amps.

Similarly, I2 is equal to the total current through R2, which in the circuit above is -0.0087 Amps (negative because I2 is defined to travel from left-to-right, while the current through R2 is actually traveling right-to-left).

Per the equations for Ic and Id:

     Ic = (I1 + I2)/2 = (0.0131-0.0087)/2 = 0.0022 A

     Id = (I1 - I2)/2 = (0.0131+0.0087)/2 = 0.0109 A

Note that the calculated value for Ic exactly matches the simulated values of the Vcm-driven currents through R1 and R2.

Similarly, the calculated value for Id exactly matches the simulated values of the Vdm-driven currents through R1 and R2.

Therefore, for the circuit above, the equations Ic = (I1+I2)/2 and Id = (I1-I2)/2 are TRUE.

LTSpice with Unbalanced Vdm and Balanced Load:

Now let's examine the same circuit, but with the Vdm source unbalanced (i.e. single-ended drive).  Note that the Vdm drive remains unchanged at 1V.

Calculating Ic and Id, recall that I1 = IR1 and I2 = IR2.  Using the "Total Current" values from the LTSpice simulation:

     Ic = (I1 + I2)/2 = (0.0142-0.0075)/2 = 0.0034 A

     Id = (I1 - I2)/2 = (0.0142+0.0075)/2 = 0.0109 A

If we compare these values to the values generated with LTSpice, we can easily see that the common-mode currents through R1 and R2 ae each 0.0022 A, not the calculated value of Ic = 0.0034 A.

Similarly, the circuit's Vdm-driven currents through R1 and R2 do not equal Id.  In fact, a portion of the differential-mode current travels out through the parasitic resistors R4 and R5 and takes the long way back to Vdm (i.e. via Vcm), rather than returning via R2.

Therefore, for the circuit above, the equations Ic = (I1+I2)/2 and Id = (I1-I2)/2 are FALSE.

LTSpice with Balanced Vdm and Unbalanced Load:

Finally, suppose that, with respect to ground, Vdm is balanced but the load is unbalanced -- in this case, R4 = 500 ohms and R5 = 400 ohms.

Calculating Ic and Id, recall that I1 = IR1 and I2 = IR2.  Using the "Total Current" values from the LTSpice simulation:

     Ic = (I1 + I2)/2 = (0.0128-0.0085)/2 = 2.15 mA

     Id = (I1 - I2)/2 = (0.0128+0.0085)/2 = 10.65 mA

If we compare these values to the Vcm-driven values of IR1 and IR2 generated with LTSpice, we can easily see that the common-mode currents through R1 and R2 are 2.0 and 2.5 mA, respectively, not each 2.15 mA.

Similarly, the Vdm-driven currents through R1 and R2, are also unequal.  In fact, a portion of the differential-mode current travels out through Vcm and takes the long way back to Vdm2 via the parasitic resistor R5.

Also, in this circuit we can see that a small portion of the Vcm-driven current goes through R3.  If R3 is our load for the differential-drive signal Vdm, then, if Vcm is an unwanted noise source, we have corrupted Vdm's signal across R3 with this noise.

Therefore, for the circuit above, the equations Ic = (I1+I2)/2 and Id = (I1-I2)/2 are FALSE.

Conclusions:

Although there is often an assumption that common-mode current in a cable must be equally distributed across the cable's conductors, in fact it is possible for the conductors to have different amounts of common-mode current.  Ditto for the differential-mode current.

In addition, if a circuit is not perfectly "balanced", it is possibly for a common-mode noise voltage to add unwanted noise to a differential-signal's load.

Standard Caveat:

As always, I might have made a mistake in my equations, assumptions, drawings, or interpretations.  If you see anything you believe to be in error or if anything is confusing, please feel free to contact me or comment below.

And so I should add -- this information is distributed in the hope that it will be useful, but WITHOUT ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.