Electronic – Are the currents that travel on the outside of coax really “common mode”

coaxcommon-modedifferential

Consider a ladder-line transmission line. Given any currents on the two conductors, we can describe those currents as the current on each conductor individually, or:

  1. currents that flow one way on one conductor, matched by exactly equal and opposite currents on the other (differential mode)

  2. currents that flow one way on one conductor, matched by exactly equal and like currents on the other (common mode)

That is, the common-mode currents are common to both conductors. This is neat, because I can reject the common-mode currents and measure just the differential-mode signal with a transformer or a differential amplifier:

schematic

simulate this circuit – Schematic created using CircuitLab

But, consider an ideal coaxial transmission line. The currents are:

  1. currents flowing on the inner conductor, matched exactly by equal and opposite currents on the inner side of the shield (differential mode)

  2. currents that flow only on the outside of the shield

I've heard it said that "common mode" currents flow on the outside of the shield. But are they really common? They aren't flowing on both conductors. Therefore, I can't reject these currents with a differential amplifier or transformer:

schematic

simulate this circuit

Therefore, is it really proper to call these currents "common mode"?

Best Answer

When we define the common mode and differential mode currents, we get some relations like

\$ i_{\mathrm{comm}} = \dfrac{1}{2}(i_+ + i_-)\$

\$ i_{\mathrm{diff}} = i_+ - i_-\$.

Now you defined a different way to split up the currents on a coax, recognizing only these two types of current:

  1. currents flowing on the inner conductor, matched exactly by equal and opposite currents on the inner side of the shield (differential mode)

  2. currents that flow only on the outside of the shield

This implies that there's no current on the inner conductor that isn't matched by an equal and opposite current on the outer conductor.

Let's call the current on the center conductor \$i_c\$ and the current on the outer conducter \$i_s\$. And we'll define a positive current on either conductor to be going away from the source.

Now we must have \$i_s = -i_c + i_2\$ where \$i_2\$ is your 2nd category of current.

So, \$i_2 = i_s + i_c\$. And we see that, aside from a scaling factor of 2x, \$i_2\$ is just the same as the common mode current as usually defined.

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