Let's first start with an easy topic -- what are common-mode currents?

__1.0 What does "Common-Mode" Current Mean?__If you have done work to mitigate unwanted EM radiation and susceptibility, your usage of the term "common-mode" is probably different from how the term is typically used in amateur radio circles.

In the EMI world, when currents are called common-mode, it refers to their presence, in equal magnitude and phase (i.e direction), on two or more wires. In other words, common-mode currents are identical currents that are

*common*on two or more wires. (Reference [3], pages 300 and 402, and Reference [1], page 96)

Common-mode currents differ from differential-mode currents in that differential-mode currents, while also having equal magnitudes, flow in opposite directions on a pair of wires, while the common-mode currents are equal and flow in the same direction.

For analytical purposes,

*any*combination of currents appearing on the two wires can be decomposed into this definition of common-mode and differential mode elements, even the extreme case of current flowing on one wire but not on the other.

Below is an example of how two unequal currents, I1 and I2, can be decomposed into common-mode and differential mode currents:

Note that in this model the common-mode and differential currents are defined to be:

(1.0-1) Ic = (I1 + I2)/2

(1.0-2) Id = (I2 - I1)/2

Therefore:

(1.0-3) I1 = Ic - Id

(1.0-4) I2 = Ic + Id

In amateur radio it is common to call current traveling on the

*outside*of a coax cable "common-mode", where the outer-surface of a coax cable is considered to be the cable's common-mode path and the coax cable's center-conductor and its shield's inner-surface form the differential-current path (References [4] and [5]).

To distinguish between these two usages of "common-mode", let me call the common-mode currents in which the common-mode current is

*the*

**equal on all lines****"standard definition"**of common-mode, and a

*flowing as the sole common-mode current, the*

**single-current**,**"single-ended"**definition of common-mode.

So, for the "

**single-ended**" definition of common-mode current, we can define the currents through the two coupled inductors. In this case:

(1.0-5) I1 = Id

(1.0-6) I2 = Ise - Id

Note that I am calling the single "single-ended" common-mode current "Ise" to differentiate from the two "standard definition" common-mode currents that are each "Ic".

For analysis, the amount of feedline radiation should be the same whether or not the common-mode current is defined to be "single-ended" versus "standard definition". That is, the total common-mode current should be the same in either case.

In other words, for the "standard definition" of common-mode, the total amount of common-mode current is 2*Ic (because Ic flows on each of the two lines). And for the "single-ended"definition of common-mode, the total amount of common-mode current is Ise.

And therefore, for equivalent amounts of feedline radiation, we can say that:

(1.0-7) Ise = 2*Ic

**2.0 First, Some Notes on Coupled-Inductors and Ideal 1:1 Transformers:**
A common-mode choke (in the lumped-element case) can be modeled as a pair of coupled inductors. The standard model of two

*lossless*coupled-inductors is shown below...This model has the following mathematical relationships, assuming current directions and voltage polarities are per the "dots" in the drawing above:

(2.0-1a) V1 = I1*jωL1 + I2*jωM

(2.0-1b) V2 = I1*jωM + I2*jωL2

(2.0-2) M = k*sqrt(L1*L2)

If the windings of two coupled-inductors are identical, so that their inductances are the same, and if they are tightly coupled together so that their coupling factor is 1, then we can model them as an

**ideal 1:1 transformer**.

Let's derive the "ideal 1:1 transformer" equations.

**2.1 Ideal 1:1 Transformer, Derivation of Voltage Relationship:**First, I'll derive the voltage relationship for an ideal transformer (Reference [12]):

Recapitulating the voltage versus current equations for coupled inductors (equations 2.0-1a and 2.0-1b):

(2.1-1a) V1 = I1*jωL1 + I2*jωM

(2.1-1b) V2 = I1*jωM + I2*jωL2

Solving (10a) for I1 and (10b) for I2:

(2.1-2) I1 = V1/jωL1 - I2*M/L1

(2.1-3) I2 = V2/jωL2 - I1*M/L2

Solving (10a) for I1 and (10b) for I2:

(2.1-2) I1 = V1/jωL1 - I2*M/L1

(2.1-3) I2 = V2/jωL2 - I1*M/L2

Substituting equation 2.1-3 into I2 of equation 2.1-2 and multiplying the equation by jωL1:

(2.1-4) I1*jωL1 = V1 - jωM*(V2/jωL2 - I1*M/L2)

Rearranging:

(2.1-5) V1 = I1*jωL1 + jωM*(V2/jωL2 - I1*M/L2)

Therefore:

(2.1-6) V1 = V2*(M/L2) + I1*(L1 - (M^2)/L2)

If we assume that the windings are equal and the coupling factor is 1, then M = L1 = L2, and this equation reduces to:

(2.1-7)

**V1 = V2**
(Note, a different method of derivation, but with the same result, can be found here).

To derive the ideal 1:1 transformer's current relationship, I'll take a slightly different approach.

**2.2 Ideal 1:1 Transformer, Derivation of Current Relationship:**

Let's assume that a load of impedance "Z" is attached across L2. And let's define I2 so that it

*exits*L2's "dotted" terminal. Like this:
Because I2 exits, rather than enters, L2's dotted terminal, we can replace "I2" in equation 2.1-1b with "-I2". Like so:

(2.2-1) V2 = I1*jωM - I2*jωL2

Because I2 generates a voltage drop across Z, this voltage drop is, by definition, V2:

(2.2-2) V2 = I2*Z

Equating equations 2.2-1 and(2.2-2 through V2 and rearranging to solve for I1:

(2.2-3) I1 = I2*(Z + jωL2)/jωM

Assume perfect coupling between coils. Therefore M = L2. Rearranging again (very slightly):

(2.2-4) I1 = I2*[(Z/(jωL2) + 1 ]

If we now assume that jωL2 is much greater than Z (jωL2 >> Z), then Z/jωL2 goes to 0 and...

(2.2-5)

**I1 = I2****Therefore, for an ideal 1:1 transformer, the following is assumed:**

**The coupling factor of the coupled-inductors, k, is 1.****The impedance of of the coupled-inductors (jωL1 or jωL2) is much greater than any load impedance connected across either inductor.**

**With these assumptions, the following are givens for an ideal 1:1 transformer:**

**The voltage across one winding equals the voltage across the other.****If a current goes "into" the "dotted" lead of one winding, a current of equal value leaves the "dotted" lead of the other winding.**

Like this:

Note that the currents entering and exiting the transformer can be considered to be differential currents, per our usage, above.

__3.0 Common-mode Chokes:__Common-mode currents are a common source of unwanted EM radiation and susceptibility, and a goal in product design is to reduce them.

The role of a common-mode choke is to impede (choke) the unwanted common-mode currents while allowing the (desired) differential mode currents to pass unimpeded. And the common-mode choke impedes these unwanted currents by creating a

*series-impedance to common-mode currents*in the choke's lines.

This series-impedance, which can have both resistive (loss) and reactive (e.g. inductive) elements, ideally reduces the common-mode current by simple Ohm's law -- the greater the overall impedance in a loop driven by a signal, the lower the loop's current will be, and thus the lower will be the radiated magnetic field (note that I am ignoring the series-resonance case, in which a choke's inductance can resonate with an external series-capacitance,

*reducing*a loop's impedance and having the opposite of the desired effect).

Therefore, if one inserts a common-mode choke into a two-wire circuit, Ic should be reduced and Id unaffected.

Let's now analyze how common-mode chokes operate. I will look at both lossless and lossy models. Also, to keep the analysis simple and focus on the primary function of a common-mode choke (attenuate common-mode signals but not differential mode signals), I am going to ignore parasitic elements that exist in every choke.

In other words, these will be simple models.

**3.1.0 Analysis, Common-mode Choke, without Loss (Lossless):**
The "lossless" equivalent-circuit of a common-mode choke consists of two coupled-inductors -- there are no resistive (lossy) elements:

Being a pair of coupled-inductors, this common-mode choke will adhere to the coupled-inductor equations stated earlier in this post (equations 2.1-1a and 2.1-1b). Below I've added the voltages and currents, as defined earlier in the coupled-inductor model:

**3.1.1 Common-mode Choke and "Standard Definition" Common-mode Currents:**Let's analyze the common-mode choke's behavior when both differential-mode and "

**standard definition**" common-mode currents are present.

For this analysis, I1 = Ic + Id, and I2 = Ic - Id.

Substituting these two equations for I1 and I2 into equations 2.0-1a and 2.0-1b:

(3.1.1-1a) V1 = (Ic + Id)*jωL1 + (Ic - Id)*jωM

(3.1.1-1b) V2 = (Ic + Id)*jωM + (Ic - Id)*jωL2

If we assume that L1 and L2 are wound similarly, then we can simplify the equations by setting L1 and L2 equal to a common-inductance value, L (i.e. L = L1 = L2).

(3.1.1-2a) V1 = (Ic + Id)*jωL + (Ic - Id)*jωM

(3.1.1-2v) V2 = (Ic + Id)*jωM + (Ic - Id)*jωL

Per the coupled-inductor model M = k*sqrt(L1*L2). Given L1 = L2 = L, then M becomes: M = k*L and our two voltages are:

(3.1.1-3a) V1 = (Ic + Id)*jωL + (Ic - Id)*k*jωL

(3.1.1-3b) V2 = (Ic + Id)*k*jωL + (Ic - Id)*jωL

Rearranging:

(3.1.1-4a) V1 = Ic*(1 + k)*jωL + Id*(1 - k)*jωL

(3.1.1-4b) V2 = Ic*(1 + k)*jωL + Id*(k - 1)*jωL

If we now assume that the two coils are tightly wound together so that k is, essentially, 1, then:

(3.1.1-5a) V1 = Ic*2*jωL

(3.1.1-5b) V2 = Ic*2*jωL

Note that, in the perfect-coupling case, Id does not contribute to voltages V1 or V2. From this we can conclude that the differential-mode currents pass unimpeded through the common-mode choke, while the "standard-definition" common-mode currents are impeded by an impedance of 2*jωL.

Thus, if the coupling factor 'k' is essentially 1, for differential-mode currents with

We can analyze differential-mode currents combined with a "

For this analysis, I1 = Ise + Id and I2 = -Id. In other words, the common-mode current is only present in I1.

(3.1.1-5b) V2 = Ic*2*jωL

Note that, in the perfect-coupling case, Id does not contribute to voltages V1 or V2. From this we can conclude that the differential-mode currents pass unimpeded through the common-mode choke, while the "standard-definition" common-mode currents are impeded by an impedance of 2*jωL.

Thus, if the coupling factor 'k' is essentially 1, for differential-mode currents with

*common-mode currents, we can model a***"standard definition"***lossless*common-mode choke as an ideal 1:1 transformer paralleled with two uncoupled inductors of value 2*L.

**3.1.2 Common-mode Choke and "Single-ended" Common-mode Current:**We can analyze differential-mode currents combined with a "

**single-ended**" common-mode current in a similar fashion.

Let's first return to our common-mode choke with its currents and voltages:

For this analysis, I1 = Ise + Id and I2 = -Id. In other words, the common-mode current is only present in I1.

Plugging these values for I1 and I2 into the lossless coupled-inductor equations 2.0-1a and 2.0-1b:

(3.1.2-1a) V1 = (Ise + Id)*jωL1 + (- Id)* jωM

(3.1.2-1b) V2 = (Ise + Id)* jωM + (- Id)*jωL2

As we did above, let's set L1 = L2 = L and M = k*L. The voltage equations become:

(3.1.2-2a) V1 = (Ise + Id)*jωL - Id*k*jωM

(3.1.2-2b) V2 = (Ise + Id)*k*jωM - Id*jωL

Rearranging:

(3.1.2-3a) V1 = Ise*jωL + Id*(1 - k)*jωL

(3.1.2-3b) V2 = Ise*k*jωL + Id*(k - 1)*jωL

(3.1.2-1b) V2 = (Ise + Id)* jωM + (- Id)*jωL2

As we did above, let's set L1 = L2 = L and M = k*L. The voltage equations become:

(3.1.2-2a) V1 = (Ise + Id)*jωL - Id*k*jωM

(3.1.2-2b) V2 = (Ise + Id)*k*jωM - Id*jωL

Rearranging:

(3.1.2-3a) V1 = Ise*jωL + Id*(1 - k)*jωL

(3.1.2-3b) V2 = Ise*k*jωL + Id*(k - 1)*jωL

If we now assume that the two coils are tightly wound together so that k is, essentially, 1, then:

(3.1.2-4a) V1 = Ise*jωL

(3.1.2-4b) V2 = Ise*jωL

(3.1.2-4b) V2 = Ise*jωL

Again, note that the differential current does not affect the voltages. Also, although Ise only flows through L1, the voltage drop it causes also appears across L2.

Thus, if the coupling factor 'k' is essentially equal to 1, for this combination of differential-mode currents and

*common-mode current, we can model the***"single-ended"***lossless*common-mode choke as an ideal 1:1 transformer paralleled with a single inductor of value L:

**3.2 Analysis, Common-mode Choke, with Loss:**

Often common-mode chokes are specifically designed with loss (resistance) in the common-mode path. This loss will guarantee that there is still a "choking" action even if the choke is terminated with a capacitive impedance (the latter could series-resonate with a choke's inductance, causing more common-mode current to flow -- thus the desire for a series resistance to limit this possibility).

An example of such a common-mode choke is an antenna-feedline balun in which the feedline coax is wrapped around a ferrite core. The "equal and opposite" differential-mode currents pass through this choke unimpeded, but the "single-ended" common-mode current passes through an impedance with a loss element (the latter being the loss in the ferrite core).

Let's return to the model of coupled inductors:

I'm going to add a loss element to these equations such that, for differential-mode currents, the loss element will cancel out, but it will be in-circuit for common-mode currents:

(Note: I created this "lossy" model to ensure that, for differential currents, there would be no loss. Loss only exists for common-mode currents, as you will see in the analysis below. Thus, this model would seem to duplicate the behavior of a typical antenna balun (a.k.a. common-mode choke), which is often implemented simply as the feedline coax wound around a ferrite core. This model gives me the behavior that I expect (no loss if no common-mode current(s), but better models might exist. If you know of one, please let me know!)

**3.2.1 Lossy Common-mode Choke with "Standard Definition" Common-mode current:**Let's analyze this choke's behavior in the presence of

**differential-mode and "standard definition" common-mode currents**.

Again, recall the voltages and currents associated with our common-mode choke:

In this example I1 = Ic + Id and I2 = Ic - Id.

Substituting these two equations into the "coupled-inductor with loss" model's equations:

(3.2.1-1a) V1 = (Ic + Id)*jωL1 + (Ic - Id)* jωM + ((Ic + Id) + (Ic - Id))*Rc

(3.2.1-1b) V2 = (Ic + Id)* jωM + (Ic - Id)*jωL2 + ((Ic + Id) + (Ic - Id))*Rc

If we assume that L1 and L2 are wound similarly, then we can simplify the equations by setting L1 and L2 to a common-inductance value, L (where L = L1 = L2).

(3.2.1-2a) V1 = (Ic + Id)*jωL + (Ic - Id)* jωM + Ic*2*Rc

(3.2.1-2b) V2 = (Ic + Id)* jωM + (Ic - Id)*jωL + Ic*2*Rc

Recall that M = k*sqrt(L1*L2). Given L1 = L2 = L, then M becomes: M = k*L and our two voltages are:

(3.2.1-3a) V1 = (Ic + Id)*jωL + (Ic - Id)*k*jωL + Ic*2*Rc

(3.2.1-3b) V2 = (Ic + Id)*k*jωL + (Ic - Id)*jωL + Ic*2*Rc

Rearranging terms:

(3.2.1-4a) V1 = Ic*(jωL*(1 + k) + 2*Rc) + Id*jωL*(1 - k)

(3.2.1-4b) V2 = Ic*j(ωL*(1 + k) + 2*Rc) + Id*jωL*(k - 1)

(3.2.1-1b) V2 = (Ic + Id)* jωM + (Ic - Id)*jωL2 + ((Ic + Id) + (Ic - Id))*Rc

If we assume that L1 and L2 are wound similarly, then we can simplify the equations by setting L1 and L2 to a common-inductance value, L (where L = L1 = L2).

(3.2.1-2a) V1 = (Ic + Id)*jωL + (Ic - Id)* jωM + Ic*2*Rc

(3.2.1-2b) V2 = (Ic + Id)* jωM + (Ic - Id)*jωL + Ic*2*Rc

Recall that M = k*sqrt(L1*L2). Given L1 = L2 = L, then M becomes: M = k*L and our two voltages are:

(3.2.1-3a) V1 = (Ic + Id)*jωL + (Ic - Id)*k*jωL + Ic*2*Rc

(3.2.1-3b) V2 = (Ic + Id)*k*jωL + (Ic - Id)*jωL + Ic*2*Rc

(3.2.1-4a) V1 = Ic*(jωL*(1 + k) + 2*Rc) + Id*jωL*(1 - k)

(3.2.1-4b) V2 = Ic*j(ωL*(1 + k) + 2*Rc) + Id*jωL*(k - 1)

For chokes with coupling factors that are essentially equal to 1, these two equations reduce to:

(3.2.1-5a) V1 = Ic*2*(jωL + Rc)

(3.2.1-5b) V2 = Ic*2*(jωL + Rc)

(3.2.1-5b) V2 = Ic*2*(jωL + Rc)

Note that voltages V1 and V2 are solely functions of the common-mode currents and not the differential-mode currents.

And so, assuming a coupling factor 'k' that is essentially equal to 1, then for the "

**standard definition**" of common-mode currents combined with differential-mode currents, we can model a "lossy" common-mode choke as an ideal 1:1 transformer and the following parallel components:

**3.2.2 Lossy Common-mode Choke and "Single-ended" Common-mode Current:**And finally, let's analyze this choke's behavior in the presence of

**differential-mode and a "single-ended" common-mode current**.

Again, we define the voltages and currents as follows:

And let's define I1 = Ise + Id, and I2 = -Id.

Plugging these values for I1 and I2 into the

*lossy*coupled-inductor equations:
(3.2.2-1a) V1 = (Ise + Id)*jωL1 + (- Id)* jωM + ((Ise + Id) + (-Id))*Rc

(3.2.2-1b) V2 = (Ise + Id)* jωM + (- Id)*jωL2 + ((Ise + Id) + (- Id))*Rc

(3.2.2-1b) V2 = (Ise + Id)* jωM + (- Id)*jωL2 + ((Ise + Id) + (- Id))*Rc

If we assume that L1 and L2 are wound similarly, then we can simplify the equations by setting L1 and L2 to a common-inductance value, L (where L = L1 = L2).

(3.2.2-2a) V1 = (Ise + Id)*jωL + (- Id)* jωM + Ise*Rc

(3.2.2-2b) V2 = (Ise + Id)* jωM + (- Id)*jωL + Ise*Rc

(3.2.2-2a) V1 = (Ise + Id)*jωL + (- Id)* jωM + Ise*Rc

(3.2.2-2b) V2 = (Ise + Id)* jωM + (- Id)*jωL + Ise*Rc

Recall that M = k*sqrt(L1*L2). Given L1 = L2 = L, then M becomes: M = k*L and our two voltages are:

(3.2.2-3a) V1 = (Ise + Id)*jωL + (- Id)* k*jωL + Ise*Rc

(3.2.2-3b) V2 = (Ise + Id)* k*jωM + (- Id)*jωL + Ise*Rc

(3.2.2-3a) V1 = (Ise + Id)*jωL + (- Id)* k*jωL + Ise*Rc

(3.2.2-3b) V2 = (Ise + Id)* k*jωM + (- Id)*jωL + Ise*Rc

Rearranging terms:

(3.2.2-4a) V1 = Ise*(jωL + Rc) + Id*(1-k)*jωL

(3.2.2-4b) V2 = Ise*(k*jωL + Rc) + Id*(k - 1)*jωL

(3.2.2-4b) V2 = Ise*(k*jωL + Rc) + Id*(k - 1)*jωL

For chokes with coupling factors that are essentially equal to 1, these two equations reduce to:

(3.2.2-5a) V1 = Ise*(jωL + Rc)

(3.2.2-5b) V2 = Ise*(jωL + Rc)

(3.2.2-5b) V2 = Ise*(jωL + Rc)

And so, assuming a coupling factor 'k' that is essentially equal to 1, then for the "

**single-ended**" common-mode current (Ise) combined with differential-mode currents, we can model a "lossy" common-mode choke as an ideal 1:1 transformer and the following parallel components:

__4.0 Return-Currents and Common-mode Chokes:__Let's analyze some examples of common-mode chokes with different load configurations:

**4.1 Load grounded at Bottom:**Given the circuit, below, with current "I1" driven to the load via the common-mode choke, what path will the return current take? Will it be via path I2 or via path Ig? Note that any current taking the "Ig" path can be considered to be common-mode current.

Z2 is the impedance of the return-path to the choke and Zg is the impedance between the left-hand and right-hand ground points (Reference [1]).

We can write three equations:

(4.1-1) Ig = I1 - I2

(4.1-2) 0 = V - I2*Z2 + Ig*Rg

Let's also assume that the windings of the common-mode choke are 1:1 and tightly coupled. Therefore, M = L.

Substituting equations 1 and 3 into 2 and replacing M with L:

(4.1-4) 0 = -I2* jωL + I1* jωL - I2*Z2 + (I1 - I2)*Rg

Rearranging:

(4.1-5) I2*(jωL + Z2 + Zg) = I1*(jωL + Rg)

From this equation we can derive the ratio of I2/I1:

(4.1-6) I2/I1 = [jω/(jω + (Z2+Zg)/L)] + [(Zg/L)/(jω + (Z2+Zg)/L)]

As ω increases:

- [jω/(jω + (Z2+Zg)/L)] goes to 1
- [(Zg/L)/(jω + (Z2+Zg)/L)] goes to 0

**I2 essentially equals I1**(i.e. no common-mode current) when ω is appreciably greater than (Z2+Zg)/Ls. That is, when:

**ω > 5*(Z2+Zg)/L**

And thus, when this condition is satisfied,

**the common-mode choke forces the current to return via itself and not via a different path**.

**4.2 Common-mode current with a "T-network" Load:**Let's look at the operation of the common-mode choke for a more general load that incorporates both differential and common-mode impedances. I will use the W9CF's "T-Network" model, shown below (Reference [6]).

Here is the equivalent circuit, with the T-network attached as the load:

For analysis, let'd define voltages and current loops:

Note that I have cunningly defined I1 to be a differential-mode current and I2 to be a common-mode (single-ended definition) current.

Equations:

(4.2-1a) Vs = V1 + (I1+I2)*Z1 + I1*Z2 - V2

(4.2-1b) Vs = V1 + (I1+I2)*Z1 + I2*Z3

Equating these two equations via Vs and then reducing, we get:

(4.2-2) I1*(Z2 + jω(L2 - M) = I2*(Z3 + jωM)

A useful ratio is the ratio of I2 (common-mode current) to I1 (differential-mode current). This is a ratio we would like to drive to zero.

From the equation above, this ratio is:

(4.2-3) I2/I1 = (Z2 + jω(L2 - M))/(Z3 + jωM)

If we assume that the common-mode choke is tightly coupled and wound 1:1, then L1 should equal L2, which should equal M (i.e. the coupling-factor "k" is 1). Equation 4.2-3 reduces and...

Given a T-Network load, the

**ratio of common-mode to differential-mode current**for a

**common-mode choke becomes:**

*lossless*(4.2-4)

**I2/I1 = Z2/(Z3 + jωL)**

Our goal is to minimize common-mode current. This means that we want the ratio of I2/I1 to go to zero. There are three ways to do this:

- Make Z3 larger.
- Make Z2 smaller.
- Make L larger.

Suppose the choke is

**lossy**(i.e. coax wound on a ferrite core)...

**4.3 T-Network Load with a***Lossy*Common-mode Choke:Finally, if we wanted to add resistive loss into the "single-ended" common-mode current path, in the "ideal-transformer" model we would add a resistor in series with L, as shown below:

If I go through a similar derivation of I2/I1, I find:

Given a T-Network Load, the

**ratio of common-mode to differential-mode current**for a

**common-mode choke becomes:**

*lossy*(4.3-1)

**I2/I1 = Z2/(Z3 + (Rc + jωL))**

Note that Rc is in the denominator. To reduce the common-mode current I2 (which is "single-ended") with respect to the differential current I1, we now have four options, instead of three:

- Make Z3 larger.
- Make Z2 smaller.
- Make L larger.
- Make Rc larger.

*worsen*common-mode current. For this reason, if Z3 is unknown, Rc (e.g. a ferrite core's loss) is the primary method to minimize the single-ended common-mode current.

__5.0 Coax as a Common-mode Choke:__Like the common-mode choke analyzed above, a coax cable, driven by a differential source connected between the coax center conductor and its shield, also naturally forces the source's "return" current to return to the source via the coax shield, even if the cable is connected to ground at both ends. (Note that this behavior, though, is frequency dependent).

An "intuitive" way to understand this behavior is to recognize that the return current, given two parallel return paths of different impedances (e.g. via coax shield or via ground), will want to return via the path of lowest impedance. And the coax-shield return path is the lowest impedance path for two reasons: first, it encompasses the smallest loop area, therefore its inductance should be lowest, and second, the coax center-conductor and the shield act as mutually coupled inductors, which, to equal and opposite differential currents, appear to have no inductive reactance at all: they appear simply as wires to the differential signal.

For this analysis I will replace the transmission line with a "lumped-element" equivalent circuit of two coupled inductors (References [1], [2] (pages 29, 51), [7]). Note that to do this, I am assuming the length the coax is significantly shorter than the wavelength of the operating frequency.

Let's analyze this behavior when the coax is grounded at both ends (References [1] and [7] ):

With both the load and source ends tied to ground, return current has two path choices -- to return via the shield of the coax or via ground.

In the diagram above, "Zg" is the impedance of the ground between Vs and RLoad, "I1" is the source current to the load (on the center-conductor of the coax cable), "Is" is the shield current, and "Ig" is the ground current.

Below is the equivalent lumped-element circuit (References [1], [2] (pages 29, 51), [7]).

Through basic circuit analysis techniques (Reference [1]) similar to those used above, the relationship between I1 and Is is:

(5.0-1) Is = I1*((jωM/(jω + (Rs+Zg)/Ls)) + ((Zg/Ls)/(jω + (Rs+Zg)/Ls)))

For coax cable, M = Ls (References [2] (equation 2-21), or [7] (Appendix I)). Making this substitution and rearranging the terms of the equation, the ratio of Is/I1 can be expressed as:

(5.0-2)

**Is/I1 = [jω/(jω + (Rs+Zg)/Ls)] + [(Zg/Ls)/(jω + (Rs+Zg)/Ls)]**For frequencies well above

**ω**(e.g. 5 times ω

_{c}= (Rs + Zg)/Ls_{c}), this ratio becomes, essentially, equal to 1 (because [jω/(jω + (Rs+Zg)/Ls)] goes to 1 and [(Zg/Ls)/(jω + (Rs+Zg)/Ls)] goes to 0).

**And thus at this frequency and above, virtually all of the return-current**

*flows via the shield of the coax*, not via ground.In other words, as the frequency of operation increases, the currents on the coax are forced to be differential, with essentially no common-mode component.

This frequency of 5*ω

_{c}is typically in the KHz range for a variety of coax cables (Reference [2] (page 47).

Note, too, that if Zg were very large (i.e. the coax shield (and load) at the load-end of the coax are isolated from ground), then the right-hand term of the Is/I1 equation dominates and the ratio reduces to:

(5.0-3) Is/I1 = (Zg/Ls)/(Zg/Ls) = 1

Not too surprising -- if the coax and load were both truly isolated from ground, all of the return current should be via the shield!

Let's look at a more general ;psf with both a differential-mode component and a common-mode component: the W9CF T-Network (Reference [9]).

Again, here is that T-Network:

Terminating the coax with it:

Let's look at the equivalent circuit:

Note that Ls is the coax cable's shield inductance and Li is the inductance of the coax cable's center-conductor (Reference [7]).

I'll also add that the current paths I choose do not need to represent the actual paths the current take. The analysis will work out either way (Reference [10]).

Writing equations:

(5.1-1a) Vs = V1 + (I1 + I2)*Z1 + I1*Z2 - V2

(5.1-1b) Vs = V1 + (I1 + I2)*Z1 + I2*Z3

Equating these two equations via Vs:

(5.1-2) V1 + (I1 + I2)*Z1 + I1*Z2 - V2 = V1 + (I1 + I2)*Z1 + I2*Z3

This reduces to:

(5.1-3) I1*Z2 - V2 = I2*Z3

Using the equations for the standard coupled-inductor model, we can write V2 as:

(5.1-4) V2 = (I1 + I2)*jωM - I1*jωLs

We also know (References [2] (equation 2-21), and [7] (Appendix I)), that M = Ls. Making this substitution for M and reducing equation 5.1-4:

(5.1-5) V2 = I2*jωLs

Substituting equation 5.1-5 into equation 5.1-4:

(5.1-6) I1*Z2 - I2*jωLs = I2*Z3

Therefore:

Given a T-Network load at the end of a coax cable,

(5.1-7)

This equation is exactly the same as equation 4.2-4 for a standard common-mode choke, with the exception that the "L" in equation 4.2-4 is now "Ls", the shield's inductance, in this new equation.

Note that in this analysis I have treated the coax as a two coupled-inductors. Can we replace the coax cable's coupled-inductor lumped-element circuit model with an ideal 1:1 transformer paralleled with an inductor?

Well, we can, but it's a slightly different model because the coupling factor (k) of the coax cable's coupled-inductor equivalent-circuit

Let's look more closely at this...

Mohr (reference [7]) has determined the shield, center-conductor, and cable inductances for various coax cables.

From his data we can calculate k, the mutual coupling between the coax center-conductor and the coax shield.

Recall that, from our coupled-inductor model, M = k*sqrt(L1*L2). Replace L1 with the center-conductor's inductance, Li. Replace L2 with the the shield's inductance, Ls. Also, we know that M = Ls (Reference [2] (equation 2-12)). Substituting these terms into the equation for mutual inductance (M = k*sqrt(L1*L2)) and we get:

And the equations for V1 and V2 reduce to:

(5.3-3a)

(5.3-3b)

You can see, by inspection, that equations 5.3-3a and 5.3-3b are exactly equivalent to our original coax-as-coupled-inductor equations 5.3-1a and 5.3-1b.

Can I move Ls to the other side of the ideal transformer?

Let's check...

By inspection, we can see that V2 = jωLs*(I1 + I2). So equation 5.3-3b is unchanged.

On the left side of the transformer, if we follow the voltage drops around the I1 current loop, then we see that V1 = I1*jω(Li - Ls) + V2. And this resolves to: V1 = jωLi*I1 + jωLs*I2, which is the same as equation 5.3-3a.

Therefore, because equations 5.3-3a and 5.3-3b remain unchanged, this transformation is valid.

Let's use this new model to analyze the earlier circuit where the coax cable was terminated with the W9CF T-network.

If I replace the original coupled-inductors (representing the coax) with this new network and draw in three current-loops, the equivalent circuit becomes:

(Note that I have changed the loop names from I1 and I2 to Ix, Iy, and Iz. Otherwise, relating these currents to the coupled-inductor-as-ideal-transformer model can get very confusing, because the model's currents are also named I1 and I2.)

If we relate the currents in the drawing above to the model's currents I1 and I2, we get:

(5.3-4) I1 = Ix + Iy

(5.3-5) I2 = - Ix

Note that the current into the parallel inductor that is across the transformer's secondary,

(5.3-6) Iz = Iy

Voltage equations around each loop:

(5.3-7) Vs = V1 + (Ix + Iy)*Z1 + Ix*Z2 - V2

(5.3-8) Vs = V1 + (Ix + Iy)*Z1 + Iy*Z3

(5.3-9) V2 = Iz*jωLs

Substitute Iy into Iz and then use this result to replace V2 in equation 5.3-7. If we then equate equation 5.3-7 to equation 5.3-8 via Vs, we get:

(5.3-10) Ix*Z2 - Iy*jωLs = Iy*Z3

In this circuit the differential-mode current is Ix and the common-mode (single-ended) current is Iy, and thus, the ratio of common-mode current to differential-mode current is:

(5.3-11)

This ratio is exactly identical to equation 5.1-7.

Let's recall the equivalent circuit from the previous section:

Does this equivalent-circuit provide a mechanism for feedline radiation?

Yes it does. But, in this "lumped-element" model, it isn't quite the same as the typical explanation, which is the coax acting as an "additional radiating element" to an antenna (Reference [8]).

Let's draw an "Amperian Loop" (Reference [13]) whose surface slices transversely through the coax:

Given that Iz = Iy (equation 5.3-6), if we sum the currents passing through the Amperian surface, the total current is:

(40i) Itotal = Iy + Ix - Ix + Iy - Iy = Iy

Itotal is

But note that this radiation would seem to be due to a current

Continuing with the same model, here's how these currents might be flowing if we replace the T-network with a dipole...

I am assuming that the current on the inside of the shield is equal and opposite to the current on the center conductor. But is it, really? And I am assuming that the current "Iy" returns to the source via ground and not via the coax.

Suppose I draw Ls and its ground connection as shown in the next illustration, and define I2 as traveling from Z3, up Ls, to the lower winding of the ideal 1:1 transformer. Like this:

This circuit could represent an actual antenna and its field, as shown below. Note that, because Iy couples to the coax cable's shield and travels in the

Let's put these paths all together. Below I've drawn the dipole's field coupling, as it should, between dipole elements (Ix), but also coupling to the coax shield (Iw) as well as ground (Iy), and that feedline radiation is therefore due to the "Iy+Iw" current on the shield.

Finally, let me remind the reader (and myself) that the equivalent-circuits above are "lumped element" circuits. That is, the circuit assumes that the coax length (and all other circuit lengths) are much less than a wavelength of the operating frequency.

**5.1 Coax Terminated with a T-Network:**Let's look at a more general ;psf with both a differential-mode component and a common-mode component: the W9CF T-Network (Reference [9]).

Again, here is that T-Network:

Let's look at the equivalent circuit:

Note that Ls is the coax cable's shield inductance and Li is the inductance of the coax cable's center-conductor (Reference [7]).

I'll also add that the current paths I choose do not need to represent the actual paths the current take. The analysis will work out either way (Reference [10]).

Writing equations:

(5.1-1a) Vs = V1 + (I1 + I2)*Z1 + I1*Z2 - V2

(5.1-1b) Vs = V1 + (I1 + I2)*Z1 + I2*Z3

Equating these two equations via Vs:

(5.1-2) V1 + (I1 + I2)*Z1 + I1*Z2 - V2 = V1 + (I1 + I2)*Z1 + I2*Z3

This reduces to:

(5.1-3) I1*Z2 - V2 = I2*Z3

Using the equations for the standard coupled-inductor model, we can write V2 as:

(5.1-4) V2 = (I1 + I2)*jωM - I1*jωLs

We also know (References [2] (equation 2-21), and [7] (Appendix I)), that M = Ls. Making this substitution for M and reducing equation 5.1-4:

(5.1-5) V2 = I2*jωLs

Substituting equation 5.1-5 into equation 5.1-4:

(5.1-6) I1*Z2 - I2*jωLs = I2*Z3

Therefore:

Given a T-Network load at the end of a coax cable,

**the ratio of common-mode current (I2) to differential current (I1) is:**(5.1-7)

**I2/I1 = Z2/(Z3 + jωLs)**This equation is exactly the same as equation 4.2-4 for a standard common-mode choke, with the exception that the "L" in equation 4.2-4 is now "Ls", the shield's inductance, in this new equation.

Note that in this analysis I have treated the coax as a two coupled-inductors. Can we replace the coax cable's coupled-inductor lumped-element circuit model with an ideal 1:1 transformer paralleled with an inductor?

Well, we can, but it's a slightly different model because the coupling factor (k) of the coax cable's coupled-inductor equivalent-circuit

**.***is not 1*Let's look more closely at this...

**5.2 Coax-cable Coupling Factor (k):**Mohr (reference [7]) has determined the shield, center-conductor, and cable inductances for various coax cables.

From his data we can calculate k, the mutual coupling between the coax center-conductor and the coax shield.

Recall that, from our coupled-inductor model, M = k*sqrt(L1*L2). Replace L1 with the center-conductor's inductance, Li. Replace L2 with the the shield's inductance, Ls. Also, we know that M = Ls (Reference [2] (equation 2-12)). Substituting these terms into the equation for mutual inductance (M = k*sqrt(L1*L2)) and we get:

(5.2-1) Ls = k*sqrt(Li*Ls)

Solving for k:

**(5.2-2)**

**k = sqrt(Ls/Li)**

Using Mohr's coax-cable inductance data, here are the calculated 'k' values for three 50-ohm coax cables. Note that Lcenter is simply Li, renamed. Ditto for Lshield and Ls:

Given that k is not 1 for coax, let's now take a look at what model with an ideal 1:1 transformer would be appropriate as a lumped-element equivalent circuit for coax.

**5.3 An Ideal-Transformer Model for Coax:**
Our original coupled-inductor model of coax is below (References [1]. [2], [7]):

Li is the center-conductor's inductance and Ls is the shield's inductance. The mutual inductance, M, equals Ls.

We know that this model's equations for V1 and V2 are:

(5.3-1a) V1 = I1*jωLi + I2*jωLs

(5.3-1b) V2 = I1*jωLs + I2*jωLs

Because the coupling factor for coax is not 1, I will use a more general model of coupled-inductors that still uses a 1:1 turns ratio "ideal" transformer (Reference [11], page 15, or Reference [12]):

This model's equations for V1 and V2 are:

(5.3-2a) V1 = jωL1*(1-k^2)*I1 + jωL1*(k^2)*(I1 + I2)

(5.3-2b) V2 = jωL1*(k^2)*(I1 + I2)

To use this model for coax, we substitute Li for L1 and note, because M = k*sqrt(Li*Ls) for coax and M = Ls, that therefore k = sqrt(Ls/Li).

If we make these substitutions, this "general" model becomes coax-specific:

Li is the center-conductor's inductance and Ls is the shield's inductance. The mutual inductance, M, equals Ls.

We know that this model's equations for V1 and V2 are:

(5.3-1a) V1 = I1*jωLi + I2*jωLs

(5.3-1b) V2 = I1*jωLs + I2*jωLs

Because the coupling factor for coax is not 1, I will use a more general model of coupled-inductors that still uses a 1:1 turns ratio "ideal" transformer (Reference [11], page 15, or Reference [12]):

This model's equations for V1 and V2 are:

(5.3-2a) V1 = jωL1*(1-k^2)*I1 + jωL1*(k^2)*(I1 + I2)

(5.3-2b) V2 = jωL1*(k^2)*(I1 + I2)

To use this model for coax, we substitute Li for L1 and note, because M = k*sqrt(Li*Ls) for coax and M = Ls, that therefore k = sqrt(Ls/Li).

If we make these substitutions, this "general" model becomes coax-specific:

(5.3-3a)

**V1 = jωLi*I1 + jωLs*I2**(5.3-3b)

**V2 = jωLs*(I1 + I2)**You can see, by inspection, that equations 5.3-3a and 5.3-3b are exactly equivalent to our original coax-as-coupled-inductor equations 5.3-1a and 5.3-1b.

Can I move Ls to the other side of the ideal transformer?

Let's check...

By inspection, we can see that V2 = jωLs*(I1 + I2). So equation 5.3-3b is unchanged.

On the left side of the transformer, if we follow the voltage drops around the I1 current loop, then we see that V1 = I1*jω(Li - Ls) + V2. And this resolves to: V1 = jωLi*I1 + jωLs*I2, which is the same as equation 5.3-3a.

Therefore, because equations 5.3-3a and 5.3-3b remain unchanged, this transformation is valid.

Let's use this new model to analyze the earlier circuit where the coax cable was terminated with the W9CF T-network.

If I replace the original coupled-inductors (representing the coax) with this new network and draw in three current-loops, the equivalent circuit becomes:

(Note that I have changed the loop names from I1 and I2 to Ix, Iy, and Iz. Otherwise, relating these currents to the coupled-inductor-as-ideal-transformer model can get very confusing, because the model's currents are also named I1 and I2.)

If we relate the currents in the drawing above to the model's currents I1 and I2, we get:

(5.3-4) I1 = Ix + Iy

(5.3-5) I2 = - Ix

Note that the current into the parallel inductor that is across the transformer's secondary,

**, is I1 + I2. In the equivalent circuit, above, I have called this current is Iz. Therefore, Iz = I1 + I2 = (Ix + Iy) + (-Ix) = Iy.***per the model*(5.3-6) Iz = Iy

Voltage equations around each loop:

(5.3-7) Vs = V1 + (Ix + Iy)*Z1 + Ix*Z2 - V2

(5.3-8) Vs = V1 + (Ix + Iy)*Z1 + Iy*Z3

(5.3-9) V2 = Iz*jωLs

Substitute Iy into Iz and then use this result to replace V2 in equation 5.3-7. If we then equate equation 5.3-7 to equation 5.3-8 via Vs, we get:

(5.3-10) Ix*Z2 - Iy*jωLs = Iy*Z3

In this circuit the differential-mode current is Ix and the common-mode (single-ended) current is Iy, and thus, the ratio of common-mode current to differential-mode current is:

(5.3-11)

**Iy/Ix = Z2/(Z3 + jωLs)**This ratio is exactly identical to equation 5.1-7.

**6.0 Coax Feedline Radiation:**Let's recall the equivalent circuit from the previous section:

Does this equivalent-circuit provide a mechanism for feedline radiation?

Yes it does. But, in this "lumped-element" model, it isn't quite the same as the typical explanation, which is the coax acting as an "additional radiating element" to an antenna (Reference [8]).

Let's draw an "Amperian Loop" (Reference [13]) whose surface slices transversely through the coax:

Given that Iz = Iy (equation 5.3-6), if we sum the currents passing through the Amperian surface, the total current is:

(40i) Itotal = Iy + Ix - Ix + Iy - Iy = Iy

Itotal is

*0, so a magnetic field***not***external*to the coax exists.**Therefore the feedline radiates.**But note that this radiation would seem to be due to a current

*circulating within*the coax. Around and around, as shown below:Continuing with the same model, here's how these currents might be flowing if we replace the T-network with a dipole...

*But are the above currents truly representative of how current is flowing?*I am assuming that the current on the inside of the shield is equal and opposite to the current on the center conductor. But is it, really? And I am assuming that the current "Iy" returns to the source via ground and not via the coax.

Suppose I draw Ls and its ground connection as shown in the next illustration, and define I2 as traveling from Z3, up Ls, to the lower winding of the ideal 1:1 transformer. Like this:

This circuit could represent an actual antenna and its field, as shown below. Note that, because Iy couples to the coax cable's shield and travels in the

*same direction*as Iy on the center conductor, the shield has become a radiator (draw an Amperian loop around it -- the current piercing the loop's surface sums to Iy).Let's put these paths all together. Below I've drawn the dipole's field coupling, as it should, between dipole elements (Ix), but also coupling to the coax shield (Iw) as well as ground (Iy), and that feedline radiation is therefore due to the "Iy+Iw" current on the shield.

Finally, let me remind the reader (and myself) that the equivalent-circuits above are "lumped element" circuits. That is, the circuit assumes that the coax length (and all other circuit lengths) are much less than a wavelength of the operating frequency.

__6.0 References:__
[2] H.W. Ott,

*Noise Reduction Techniques in Electronic Systems, 2nd Edition*, Wiley-Interscience, 1988
[3] C.R. Paul,

*Introduction to Electromagnetic Compatibility*, Wiley-Interscience, 1992*Coupling Between Open and Shielded Wire Lines Over a Ground Plane*, IEEE Transactions on Electromagnetic Compatibility, September, 1967

*Basic Circuit Theory*, McGraw-Hill, 1969

[11] K.K. Clarke, D.T. Hess,

*Communication Circuits: Analysis and Design*, Addison Wesley, 1971

[12] IntgCkts,

*Coupled Inductors as Transformer*, web post

[13] Brilliant.org,

*Ampere's Law*, web post

__My Balun (and 80-Meter Loop) posts:__
80 Meter Loop, Part 1: http://k6jca.blogspot.com/2018/05/adventures-with-80-meter-loop-antenna.html

80 Meter Loop, Part 2: http://k6jca.blogspot.com/2018/05/adventures-with-80-meter-loop-antenna_30.html

Notes on 1:1 Baluns: http://k6jca.blogspot.com/2018/06/transmit-common-mode-chokes-11-current.html

Notes on Common-Mode Currents: http://k6jca.blogspot.com/2018/07/thoughts-and-notes-common-mode-current.html

__Standard Caveat:__I might have made a mistake in my designs, equations, schematics, models, etc. If anything looks confusing or wrong to you, please feel free to comment below or send me an email.

Also, I will note:

This design and any associated 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.