Tuesday, July 10, 2018

Thoughts and Notes: Common-mode Current, Chokes, and Coax

Thoughts-had and notes-made while investigating chokes and baluns for my 80 meter loop (the latter here and here).

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


    (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 equal on all lines the "standard definition" of common-mode, and a single-current, flowing as the sole common-mode current, the "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

Where I1 is the coax center-conductor current, and I2 is the total current on the coax shield (both inner and outer surfaces).

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:

For reference, let me assign equations numbers to each of these three equations:

    (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

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)


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


   (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 impede's 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, reduces the common-mode current by simple Ohm's law -- the greater the 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


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 "standard definition" common-mode currents, we can model a 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


    (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

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 "single-ended" common-mode current, we can model the 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)

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)

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

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

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

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

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)

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 

    (4.1-3)  V = -I2* jωL + I1* jωM
Equation 4.1-2 is the sum of the voltages around the lower loop, and equation 4.1-3 is simply equation 2.0-1b, but with the minus sign reflecting that I2 is flowing in the opposite direction from its definition in the model.

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


    (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
This ratio shows us that 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.


    (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 lossless common-mode choke becomes:

    (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:
  1. Make Z3 larger.
  2. Make Z2 smaller.
  3. 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 lossy common-mode choke becomes:

    (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:
  1. Make Z3 larger.
  2. Make Z2 smaller.
  3. Make L larger.
  4. Make Rc larger.
Note that if Z3 is capacitive, it could series-resonate with the inductor, which could 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 ωc = (Rs + Zg)/Ls (e.g. 5 times ωc), 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*ωc 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!

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:

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


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:

And the equations for V1 and V2 reduce to:

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

   (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 not 0, so a magnetic field 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:

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?

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, the shield has become a radiator (draw an Amperian loop around it -- the current piercing the loop's surface sums to Iy).

Below might be a more general representation of current flow.  Note that I've drawn the dipole's field, coupling as it should between dipole elements (Ix), is 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:

[3]  C.R. Paul, Introduction to Electromagnetic Compatibility, Wiley-Interscience, 1992

[7]  R.J. Mohr, Coupling Between Open and Shielded Wire Lines Over a Ground Plane, IEEE Transactions on Electromagnetic Compatibility, September, 1967

[10]  Desoer & Kuh, 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

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.

Thursday, June 28, 2018

Common-mode Chokes (Baluns), Power Dissipation versus Effectiveness

Apropos my 80 meter loop, some notes I've made while thinking about common-mode chokes and feedline radiation...

One question I had -- how can I use EZNEC to model feedline radiation of a coax-cable feeding an antenna?

Let start by looking at a dipole fed with coax...

Feedline Radiation, Example 1: Coax-fed 80 Meter Dipole.

I'm going to start with an explanation that is also discussed elsewhere in excellent articles by the following authors (click on a callsign to be routed to their link): W0IVJ, W7EL, W2DU.

Let's take a look a a standard half-wave dipole up in the air, fed via coax from a grounded transmitter.  Like this:

(Note: dipole and coax drawn sideways for convenience)

The tight coupling between coax shield and coax center conductor forces the currents to and from the source to be equal and opposite while they are within the coax, with the current returning to the source flowing on the inner surface of the shield.

In other words, the coax itself acts as a very good common-mode choke (in the sense that it forces differential currents to return via the coax, rather than via a less desirable path). (For more information, please refer to: Noise Reduction Techniques in Electronic Systems, 2nd Ed., by Henry W Ott, page 51.)

But this common-mode choke behavior of coax stops at the ends of the coax, outside of which we can then have currents flowing on the outside surface of the coax, as shown below.

Note that I1 on the inner-surface of the coax (returning to the source) is composed of I2, the desired current-path in the dipole element, and I3, the undesired path on the coax shield's outer surface.  And note, too, that I1 flowing on the inner-surface of the coax shield is equal and opposite I1 flowing on the coax center conductor.

Given the currents, as drawn, above no current flows from ground directly to the source.  This lets us redraw the model without the source tied directly to ground:

Now the source only connects to the coax, and I have added a conductor (representing the outer-surface of the coax shield, with the shield's length and dimensions) that runs from the antenna-feedpoint end of the coax to back to ground near the source.

As long as we do not change the length of this conductor (representing the coax shield), we can move the source towards or away from the antenna feedpoint with no change in performance (assuming ideal, lossless, coax).  Like this:

Reducing the length of the coax in the above illustration to zero, our model reduces to:

Let's add our currents back in...

To minimize feedline radiation, we want to minimize I3, and the best way to do this is to add an impedance in the coax-shield-to-ground path at the source, which is actually the antenna feedpoint, not the original location of the source (at the other end of the coax), as shown below:

Let's assume we are inserting a large resistance as a choke (rather than a reactance, which could worsen I3 current if it series-resonates with the impedance of the coax-shield-to-ground).  How large does this resistance need to be to reduce I3 to an acceptable level?

Let's look at a worst-case model using EZNEC...

Feedline Radiation, EZNEC Model, 80-meter Dipole Fed with 1/2 Lambda Length of Coax:

This dipole's elements has been tuned to resonate at 3.5 MHz.  I will examine its currents and choking requirements at resonance (3.5 MHz) and at the other end of the 80-meter band (4.0 MHz).

The feedline length is 144 feet which, when represented by a conductor connecting one end of the dipole's feedpoint to ground (wire 5 in the EZNEC model, representing the coax shield's outer surface), results in maximum current flowing on this conductor.

(I did mention that this was a worst-case model, didn't I?)

Let's look more closely at the model.  Here's how the antenna is driven, at ground level:

...and here is how its feedpoint looks:

  1. The transmission line (ideal, Zo = 50 ohms, Vf = 0.66) runs from the center of wire 4 to the center of wire 2).
  2. Note that one side of wire 4 connects to ground (via wire 7).  This will let me check if any currents flow from ground back to the source.
  3. Wire 5 (and wire 6, which is in series with it down at the ground-end) represents the outer-surface of the coax shield (and thus the unwanted radiating path).
  4. The little red square in wire 5 (up near wire 2) represents a choke in wire 5 at the dipole's feedpoint whose impedance I can vary to model its effect upon current in wire 5, as well as its power dissipation.

Here are the antenna's currents:

Note that very little current flows on the dipole arm represented by wire 3; it instead flows on the coax shield.

Not good!

But we can improve upon this by inserting a choke (represented by a resistance-only impedance) into wire 5 up at the antenna's feedpoint (near wire 2).

Here's the impedance I will insert:

And here is its effect on the antenna system's currents:

Much better!

(A quick sidebar...

What would happen if we were to place a choke at the source-end of a feedline, when the feedline is 1/4 lambda long?

First, let's examine the currents with no choke: 

Looks good!  (Note that the current for wire 7 (from one end of the source to ground) is shown to demonstrate that, within the limits of the EZNEC model, essentially no current flows from ground to the source)

Now let's add the choke (5000 ohms in wire 6)...

Now we have feedline radiation!

So the point is:  When in doubt, with an antenna system in which the antenna is fed with coax, choke at the antenna's feedpoint, NOT at the transmitter.

...end of sidebar).

Let's look at how different values of choking resistance affect wire 5's current, and how much power is dissipated by this resistance.

If we use G3TXQ's criteria of reducing the unwanted current by 30 dB, then a choke with 800 ohms resistance is sufficient to achieve this goal (note that, given the EZNEC source amplitude of 1.0 Amps, a 30 dB reduction would be 0.03 amps).

Here is a table summarizing the results of different choke impedances at 3.5 MHz:

Note that if I have an ineffective choke (e.g. 20 ohms, resistive), it could dissipate quite a bit of power!

What happens to currents if we tune our transmitter to the other band edge at 4.0 MHz?

Without any CM choke inserted, here are the antenna system's currents at 4 MHz:

There is some feedline radiation, but not as bad as the antenna system at 3.5 MHz.  But this isn't surprising, because the feedline length is no longer 1/2 lambda long, so the impedance it presents at the antenna's feedpoint will be higher.

But, my goal is to look at currents in the worst-case scenario.  So let's say that we were to add, at the dipole feedpoint, a choke whose reactance series-resonates with wire 5's reactance.

Also -- to more easily determine when the common-mode current is 30 dB down from the source current, I'm going to move the source up to wire 2 (the antenna's feedpoint) and eliminate the transmission line.  We can do this because EZNEC's transmission line is perfect and does not radiate -- we are using wire 5 to represent the "imperfect" part of the transmission line which radiates.  To illustrate this point, note that the current distribution of the "source-at-top" (i.e. in wire 2) antenna system, shown below, is essentially the same as the current distribution of the "source-at-ground" (i.e. in wire 4) antenna system, shown above.

With the source now at the top (wire 2), the source's current at that point is 1.0 A, rather than some other value with a phase angle.  And so to meet the G3TXQ criteria, I simply need to increase the choke's resistance in wire 5 until the current through the choke is 30 dB down from the source's current in wire 2.

But first, let's force the current in wire 5 (representing the radiating feedline) to be maximum (it's a worst-case analysis, after all).

If I add some reactance (-j350 ohms) as the series-load in wire 5 and adjust its value for maximum feedline current,  the current through the choke is maximized, and the antenna currents become:

Note the increase in feedline (wire 5) current!.

Here's a summary of different choke impedances for the off-resonance dipole:


  1. 90 ohms resistance exhibits max power dissipation.  And it is a LOT of power!
  2. Even at 1000 ohms of resistance, 14 percent of the transmitter power is dissipated in the choke.  That is a lot.
  3. The choke's resistance must be at least 4000 ohms to reduce the unwanted feedline current to 30 dB below the transmit current.  At this level, the power dissipated in the choke is 4.2 percent of the total power.

And again, I want to stress -- I am looking at worst-case scenarios!

Next, let's look at a full-wave coax-fed loop...

Feedline Radiation, Example 2: 80 Meter Full-Wave Loop.

Let's look at a Full-Wave 80 meter loop, cut to be resonant at 3.5 MHz and again (like the dipole) feed from a grounded source with 144 feet of coax.

From my discussion, above, the source itself is effectively isolated from ground due to the common-mode choke behavior of the coax cable, itself.  So I can simplify my model (just as I did when examining dipole currents at 4 MHz, above), and move my source to wire 6 at the loop's feedpoint.

Below is the loop.  Note that wire 5 simulates the coax shield's outer-surface that connects one side of the loop feedpoint to ground, below:

Here's a closeup of the feedpoint, as well as data on the source and data on the series-load (which will be used to mimic a choke) in the first segment of wire 5.

And here are the currents:

Note the huge difference in at-resonance (3.5 MHz) currents from the dipole (earlier, above).  Even though this loop has the same feedline path to ground as the dipole (simulated by the 144 foot conductor to ground), it has very little feedline current.

You can see that this feedline current is already 34 dB below the source's current (compared to 1 Amp), so: no choke is needed when the loop is operated at resonance.

Let's see what happens if we move the transmit frequency to the other side of the band (4 MHz)...

What happens to currents if we tune our transmitter to the other band edge at 4.0 MHz?

Now there are common-mode currents:

The magnitude of the current on the coax shield, near where it attaches to the feedpoint (and at the point where I will insert a choke), is 1.44A.

I can simulate worst-case feedline current by inserting  +j110 ohms (inductive) in series into the feedline common-mode current path (inserted close to the loop's feedpoint).  In this case, the current is 1.716 Amps.

To continue examining this worst-case scenario, let's keep this reactance of +j110 ohms, and add to it different values of resistance while examining how this resistance lowers the common-mode current as well as its power dissipation.

This table summarizes the results:


1.  When operated off-resonance, the choke's resistance must be 10,000 ohms to reduce the worst-case common-mode current down to a level at least 30 dB below the transmit current of 1 amp (per G3TXQ criteria).

2.  Significant power dissipation can occur even at fairly large values of resistance.  Note that with a choke resistance of 4000 ohms, the power dissipation is 10 percent of total power.

So my conclusions, comparing worst-case Common-Mode radiation scenarios on Dipole and Loop coax feedlines, are:

1.  An 80 meter loop, at resonance, requires little common-mode choking.  The 80 meter dipole will require a choke.

2.  A loop, off resonance, can require significantly more common-mode choking (at least twice as much) than a dipole fed with the same length of coax.

3.  More power dissipation will occur in a loop's choke compared to a dipole's choke of similar resistance and worst-case reactance values.

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.