Thursday, August 16, 2018

Notes on Antenna Tuners: Drake MN-4 Matching Network

At a recent swap-meet I purchased a Drake MN-4 "Matching Network" from a friend.  This post describes notes of mine on the tuner and measurements of its performance.

MN-4 Specifications:

From the manual, the MN-4's specifications are:

Interestingly, although the manual specifies that it can match any load-impedance with a VSWR of 5:1 or less, it does not specify to what minimum SWR these loads can be tuned.  Is it 1:1 SWR?  I'll take a look at this further, below...

MN-4 Schematic:

And here is the MN-4 schematic, also from the manual:

Inside the MN-4:

I was curious what the MN-4 circuitry looked like.  After removing the black outer covers, I was surprised to find that the tuner was also well shielded:

Below is a picture of the inside of the tuner with the shield removed.  (Note that the meter is also shielded).

Let's now look at some measurements:

Tuner Performance:  Measuring a Tuner's "Match Space":

Any Antenna Tuner worth its salt should transform a wide range of impedances connected to its Output Port (i.e. its Antenna Port) to 50 ohms (i.e. 1:1 SWR at its Input (Transmitter) Port).

An obvious question is -- what is the actual range of impedances that a specific tuner can match?

It is relatively straight forward to measure these impedances (if one has a Vector Network Analyzer) and to plot them on a Smith Chart.  The resulting region of matchable impedances I call a tuner's "Match Space".

In other words, the "Match Space" is an area on a Smith Chart encompassing the range of impedances that a tuner can transform to 50 + j0 ohms at its input port.

How do we determine what these impedances are?  Do I need a box full of resistors, capacitors, and inductors to connect to the tuner's output and test each combination?

A little network theory...

Assume a two-port network has an impedance of unknown value, Z(unknown), attached to its output port, and assume that the network has been "tuned" so that Z(unknown) is transformed to 50 + j0 ohms when measured at the network's input port.

If I then conect a 50 + j0 load to the network's input port and remove Z(unknown) from the output port, the impedance I measure at the network's output port will be the complex-conjugate of the original Z(unknown) impedance.

So, to determine a tuner's "Match Space", all I need to do is to terminate the tuner's input with 50 + j0 ohms and then, while continuously adjusting the tuner's controls, plot the complex-conjugates of the impedances I measure at the tuner's output.

An ideal task for a Vector Network Analyzer!

Equipment and Software:

To make these measurements I use the following equipment and software:
  1. An HP 8753C Vector Network Analyzer (VNA) and an HP 85046A S-Parameter Test Set.
  2. An Agilent 82357B USB/GPIB Interface Module, to interface a computer to the VNA.
  3. A recent version of MATLAB, with the following toolboxes:  1) RF Toolbox, 2) Instrument Control Toolbox.
In addition, I use a MATLAB script (vna_s11_tracks.m) written by Dick Benson, W1QG, to plot the match-space on a Smith Chart.

Measurement Procedure:

I have discovered that if I first calibrate my VNA at the frequency I want to measure impedance and then launch and execute the MATLAB script vna_s11_tracks.m, the calibration data is lost.  So my procedure for making the match-space measurements is to first launch the script and then perform the calibration:
  1. Connect an accurate 50 ohm reference load to the Tuner's input port (i.e. Transmitter port).
  2. Launch the MATLAB script vna_s11_tracks.m, but do not yet click its RUN_STOP button on its GUI.
  3. At the VNA set Center Frequency to the desired measurement frequency.
  4. At the VNA set Span to 0 Hz.
  5. At the VNA set Number of Points to 3 (under MENU button in the "Stimulus" button group) for the 8753C).  Default is 201 points.
  6. Perform an S11 open-short-load calibration on the end of the coax that will attach to the Tuner's output port (i.e. antenna port).  Note that the other end of this coax should already be connected to Port 1 of the S-Parameter Test Set.
  7. After calibrating, re-attach the "open" reference load to the same end of the coax that had been calibrated in step 4, above.
  8. Click on the RUN-STOP button.  (The STATUS box on the GUI should say "RUN".)
  9. Verify that a small red square outline appears exactly at the 3 o'clock position of the GUI's Smith Chart.  If it is there, then the VNA has been calibrated correctly.  If it is somewhere else on the Smith Chart, you have done something wrong.  Repeat the calibration steps, above.
  10. Click on the GUI's RUN_STOP button to stop the script.  Remove the "open" reference load from the coax and instead attach the coax to the Tuner's output port.
  11. Click RUN_STOP again and start capturing data, per either the first or second method of plotting, discussed below...
Plotting the Match Space:

For a Tuner (or other two-port network) having two variable controls, I can use the MATLAB script to plot the "Match Space" one of two ways.

The first method is the fastest, and it simply traces the outline of the match-space by sequentially holding one of the two controls fixed while rotating the other control.

In other words, say a tuner has two controls (control A and control B), this method's steps would be:
  1. Place both controls at their maximum counter-clockwise (CCW) position.
  2. With control B fixed at its counter-clockwise position and rotate control A from CCW to its maximum clockwise (CW) position.
  3. With control A now at its maximum CW position, rotate control B from its CCW position to its maximum CW position.
  4. With both controls now at their maximum CW positions, rotate control A from its CW position to its maximum CCW position.
  5. And finally, rotate control B from its CW position to its maximum CCW position.
An example of an outline-plot of a match-space, made using this method, is shown below:

(Note that Dick's MATLAB script allows the user to assign a different color to each of the steps in steps 2 through 5.  I've defined the color assignments in the comment-box at the lower left of the figure.)

An Important Caveat!  This procedure will accurately trace the outline of a Tuner's match space in most cases.  But not always!  If in doubt, then one can use a second, more time-consuming method:

Again, assuming a tuner has two controls (A and B), this second method's steps would be:
  1. Place both controls at their maximum counter-clockwise (CCW) position.
  2. With control B fixed at its maximum counter-clockwise position and rotate control A from CCW to its maximum clockwise (CW) position and back again.
  3. Rotate control B very slightly clockwise.
  4. Again, rotate control A from its CCW position to its maximum clockwise (CW) position and back again.
  5. Repeat steps 3 and 4 until control B is finally at its maximum CW position and control A has been rotated CCW and back again.
Here is an example of a plot made using this method:

In the example above (using an Icom AT-500 Tuner), I stepped one control through 14 positions, and at each step of the first control position I rotated the other control fully clockwise and then counter-clockwise back again.

This second method is much more time consuming to perform, but it can reveal information that might be missed if  one simply traced the match-space outline using the first method.  For example, below is the match-space outline (using the first technique) overlayed on the second method's plot:

Comparing the two plots, you can see that method 1 missed some of the match-space:

Also, if you examine method 2's plots, you will see that the same match-space area missed by method 1 actually has more than one unique Tuner setting for the load impedances in that region.

For example, in the figure, below, note the intersection of two different curves.  This means that the impedance represented by the Smith Chart at this intersection can be matched with two different tuner settings.

And therefore, because the entire region, above (missed by method 1) actually consists of overlapping curves, the impedances in that entire region can be each matched by two different tuner settings.

Note:  This result does not mean that, in the general case, any area missed by the method 1 (and caught by method 2) can actually be matched by multiple tuner setting.  But for this specific example, it does.

MN-4 Match-Space Measurements:

This section contains match-space plots I've made of my MN-4 tuner.  For these plots I have defined Smith Chart SWR circles representing four SWRs:  1.5:1, 3:1, 5:1, and 10:1.

80 Meter Match Space:

The MN-4 has two band-switch positions for 80 meters.  They are  80A and 80B.

First, outline-plots at 3.5 MHz for the 80A and 80B band-switch positions:

...and then, outline-plots at 4.0MHz for the 80A and 80B band-switch positions:

Note that at 3.5 MHz the Drake specification of matching loads with a VSWR of 5:1 (or less) at any impedance angle is not met, if we assume that the specification is for a match to provide a resulting SWR of 1:1.

At 4 MHz the match-space is closer to the specified 5:1 SWR, but still not quite there.

40 Meter Match Space:

An outline-plot at 7.0 MHz:

...and an outline-plot at 7.3 MHz:

Repeating the 7.3 MHz plot using method 2:

20 Meter Match Space:

The 20 meter match space outline-plot at 14.175 MHz:

15 Meter Match Space:

The 15 Meter match-space outline-plot at 21.0 MHz:

A bit of a strange plot.  How to interpret?

If I play around with the tuning controls (using plotting method 2), I discover that the match-space outline should actually encompass more impedances...

Repeating the same procedure at 21.45 MHz, the match-space outline is:

10 Meter Match Space:

Outline-plot at 28.0 MHz:

Looks pretty lousy!

Here's the match-space outline-plot at 29.7 MHz:

It's clear that on 10 meters the MN-4 does not come close to meeting the specification of matching any load with a 5:1 VSWR.

17 Meter Match Space:

I was curious how the tuner might work on the WARC bands.  So as an example, I thought I would plot the match-space for 17 meters (measured at 18.130 MHz), setting the MN-4's band-switch first to 20 meters and plotting the match-space, and then setting the band-switch to 15 meters and again plotting the match-space.

Here is an outline-plot of the 17 Meter match-space with the band-switch set to 20 Meters:

Hmmm...looks strange.  How should I interpret the big yellow circle?

Let's see what happens if I plot the match-space using the second plotting method (described earlier):

Ah-ha!  There's a big hole in the match-space!

And below is the 17 Meter match-space (plotted using the second method) with the band-switch set to 15 Meters:

In either case, the MN-4 leaves something to be desired if attempting to tune impedances on 17 meters.

Power Loss Measurements:

Tuner power loss measurements are made using the procedure described in this blog post:

At the moment I only have two reference loads for testing power loss.  They are 5 and 500 ohms.  In other words, they lie on the 10:1 Smith Chart SWR circle and thus are well outside the specified load range for the MN-4 of being loads with an SWR of 5:1 or less.

Never the less, on 10 meters the MN-4 can match a 5 ohm load.  I made a loss measurement at 29.7 MHz, and the MN-4, tuned to match a 5 ohm load, has an internal loss of 0.5954 dB, or 12.8% of the power applied.

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.

Sunday, August 12, 2018

Additional Notes on Measuring Antenna Tuner Power Loss

This post updates the technique I have used for measuring antenna tuner power loss.  As such, it supersedes the techniques I described a few years ago in this blog post:

This current method of measuring tuner power loss is, at its essence, based upon the fifth technique (Technique 5) described in this earlier blog post (above).  But I have simplified its calculations to focus on "Operating Power Gain" (Gp) (although I do also calculate "Transducer Gain" (Gt), but this is primarily for comparison purposes).

Operating Power Gain, Gp:

Gp is simply the ratio of the power delivered to a network's load divided by the power delivered to the network (which in this case is a two-port network).  Any difference in these two powers must be due to power lost within the network.

Gp can be calculated using a network's 2-port S-parameters (S11, S12, S21, S22) and its load's gamma using the following equations:

(Derivation of Gp can be found here: UC Berkeley Notes)

Transducer Gain, Gt:

Gt is slightly different from Gp in that it is the ratio of the power delivered to a network's load divided by the power available from the generator.

Note that the Gt formula now includes terms representing the source's gamma.  So Gt is also a function of the match between the source's impedance and the network's input impedance.  Thus, if the source is not matched to the network's input impedance (i.e. not conjugately-matched), not all of the power that is available from the generator will be delivered to the network.

So, for the purposes of calculating Antenna Tuner loss, Gt can be thought of as comprising two sources of power loss.  The first source of loss being the loss within the network (i.e. the loss calculated with Gp), and the second loss being a loss due to the mismatch between source and network-input impedances (i.e. the loss due to a non-conjugate match between source and network-input impedances).

For antenna tuners adjusted for low SWR's (measured at the network's input port), Gt and Gp should be (essentially) identical.  But as SWR increases, Gt loss will increase compared to Gp loss because of the the Gt loss's increasing source-to-network mismatch loss.  Thus, Gt is not nearly as useful for determining Tuner power loss because it includes this additional "external to the tuner" loss.

(Derivation of Gt can be found here: UC Berkeley Notes)

Determining Gp:

From the formulas above, to calculate Gp we need to measure a tuner's two-port S-parameters (S11, S21, S12, S22) when the tuner has been tuned to match (as best it can) a load, and we should also have the load's gamma (calculated from its own S11 parameters).

S-parameters are ideally measured with a Vector Network Analyzer (VNA), and the resulting data can then be used to calculate Gp.  Although this calculation can be performed by hand, MATLAB software provides a convenient way to both capture the S-parameter data from the VNA (via MATLAB's "Instrument Control Box") and to perform the Gp calculation (via MATLAB's powergain() function in its RF Toolbox).  Note that a single MATLAB statement will calculate Gp:

Gp = powergain(s_params,z0,zl,'Gp')

What could be simpler?

S-parameter Capture and Gp Calculation Using MATLAB:

The rest of this blog post describes how I measure Tuner power loss using my VNA and MATLAB.

I use the following equipment and software:
  1. An HP 8753C Vector Network Analyzer (VNA) and an HP 85046A S-Parameter Test Set.
  2. An Agilent 82357B USB/GPIB Interface Module, to interface a computer to the VNA.
  3. A recent version of MATLAB, with the following toolboxes:  1) RF Toolbox, 2) Instrument Control Toolbox.
(A note regarding MATLAB:  their "Home" version is, in my opinion, a great value, and it is this version that I personally use.)

Previously I have measured Tuner Loss with the tuner terminated (and tuned for) either a 5 ohm (resistive) or 500 ohm (resistive) load (when measuring loss in the Elecraft KAT-500 and also in my own tuner design).  Both of these loads represent an SWR of 10:1.  But other loads could be used, too.

As an example, I will go through the steps of measuring the loss within a Drake MN-4 "Matching Network" when it has been terminated with a 5 ohm load and tuned to a 1:1 SWR at 29 MHz.

(Note that the Drake MN-4 is only spec'd to tune loads with a maximum SWR of 5:1.  So a 5 ohm load is well outside its specified load range!  Never the less, the tuner will match 5 ohms to 50 ohms on 29 MHz, and this is a load I have at hand, so I will use it as an example for measuring Tuner Loss).

Procedure to Calculate Tuner Loss:

Step 1.  Measure the Load's S11 characteristics:

Strictly speaking, if the impedance of the Test Load is known (e.g. 5 ohms), this step can be skipped and the Test Load's impedance can simply be inserted into the "zl" field of MATLAB's powergain function: powergain(s_params,z0,zl,'Gp').

But I prefer to use the VNA to measure the load's actual S11 characteristics for the frequency range of interest.  Below is my 5 ohm Test Load attached to Port 1 of the S-Parameter Test Set:

(Note that the 5-ohm load consists of two 10 ohm 1% resistors connected in parallel across the back of a female SMA connector).

The frequency range will be 28 to 30 MHz (because, from previous experiments, I know that the MN-4 can match a 5-ohm load at 29 MHz).

After setting up the VNA for the appropriate frequency range and performing a full two-port calibration using the cables I will use to connect to the Tuner's ports (do not skip this step!), I use a MATLAB routine (vna_s11.m) written by Dick Benson, W1QG, to capture the load's S11 data:

Step 2:  Store this S11 Data in S1P File Format:

This same routine is then used to store the captured S11 data in S1P file format.

In this example, I have named the file: 180810_8753C_S11_5ohm_Load_28-30MHz.s1p .

Step 3:  Attach the Test-Load to Tuner and Adjust for the Tuner for Minimum SWR:

Now I need to tune the Tuner to match the 5 ohm Test Load.  To do this, I first:
  • Attach the 5-ohm load to the Tuner's "Antenna" port.
  • Attach Port 1 of the S-Parameter Test Set to the Tuner's "Transmitter" port.
With the VNA still in S11 mode, I then tune the Tuner for minimum SWR at the frequency of interest (e.g. 29.0 MHz).

Step 4:  Measure S11 with the Tuned Tuner and Load:

This next step is optional (because it is not needed for calculating Gp).  I do it to validate my measurements, as I will show, later.

With the Tuner tuned for minimum SWR, I then make another S11 capture using the VNA and the vna_s11.m MATLAB routine.

The results are stored in a new file, again in S1P format.

In this example, the file is: 180810_8753C_S11_MN-4_5ohm_load_28-30MHz.s1p 

Step 5:  Remove the Test-Load and Measure the Tuner's S11, S12, S21, and S22 characteristics:

Without touching the Tuner's settings (!!!), I remove the 5-ohm Test Load from the Tuner's "Antenna Port" and connect this port instead to Port 2 of the S-Parameter Test Set, as shown below (The tuner's Transmitter port is on the left, and its Antenna port is on the right):

Using the VNA, I capture S11, S12, S21, and S22 with a different MATLAB routine written by Dick Benson, W1QG to perform two-port S-Parameter capture (vna_gui_1.m):

Step 6: Store this Data in S2P File Format:

Via this same program (vna_gui_1.m), I then store this data in S2P file format.

In this example, the file is: 180810_8753C_ s2p_MN-4_tuned_for_5_ohms_28-30MHz.s2p.

Step 7:  Calculate the "Operating Power Gain" (Gp):

To calculate Gp (and thus Tuner Power Loss)) I will use the following S-parameter files that I have created (above):

  • The Load's S1P file: 180810_8753C_S11_5ohm_Load_28-30MHz.s1p 
  • The Tuner's S2P file: 180810_8753C_ s2p_MN-4_tuned_for_5_ohms_28-30MHz.s2p 
The result will be the Tuner's power loss when it has been adjusted to match a 5 ohm load at 29 MHz.

To perform this calculation I use the MATLAB script below (Plotting_Network_Gp_from_S2P.m).
Note:  I usually only write MATLAB routines once or twice a year, and in between each session I typically forget what I had learned previously (the hazards of age!).  Please forgive any awkward MATLAB code that might offend the more experienced programmer!
% Plotting_Network_Gp_from_S2P.m
% Predicting Network Loss with MATLAB
% Date:  180811
% By Jeff Anderson, K6JCA,
% from code originally written by Dick Benson, W1QG.

close all;


% Note: objects must have the SAME number of points (e.g.
% 201 or 401)
% HP 8753C generates S2P files with 201 points, while the
% HP 3577A generates S1P and S2P files with 401 points.

% Get the Load-only S1P file.  This is simply the S11 data
% of the load, itself.
s= '180810_8753C_S11_5ohm_Load_28-30MHz.s1p';
[Load_Obj,Load_Notes,Load_State] = spar_read(path,s);

% Get Network-only S2P file -- this is S11, S12, S21, S22
% of the network "tuned" to an SWR of 1:1 when the load
% in the S1P file (above) was attached, but captured via
% the VNA *without* the load attached.
s= '180810_8753C_ s2p_MN-4_tuned_for_5_ohms_28-30MHz.s2p';
                         = spar_read(path,s);

% As an optional check, get the S1P file of the Network
% with Load attached. This will then be compared to the
% gamma data generated from the Network-only S2P file (above)
% and the Load-only S1P file (also above).
% In other words, this is the S11 data for the network
% tuned to an SWR of 1:1 WITH the load still attached.
s= '180810_8753C_S11_MN-4_5ohm_load_28-30MHz.s1p';
                        = spar_read(path,s);

Zo = Network_Obj.Z0;  % 50 ohms for all s-parameter files.
FreqMHz= Network_Obj.Freq*1e-6; % Freq Vector.

% Convert the load's S11 data to Impedance (Z):
Z_Load = gamma2z(Load_Obj.S_Parameters,Zo);

% Calculate the Gamma at the network's input from the
% network's S2P file data (with network tuned for
% appropriate load) and that load's impedance
% (also known as Z_Load).
gamma = gammain(Network_Obj.S_Parameters,Zo,Z_Load);

% And for the Gt calculation, set Zs to equal Zo.
%   Note that if Zs is set to equal the
%   complex conjugate of the impedance seen at the
%   tuner's input port, then Gt, when
%   calculated, will equal Gp.
%   To see this, set Zs to be:
%   Zs = conj(gamma2z(gamma));
%   instead of the next statement:
Zs = Zo;      % So Zs now equals 50 ohms.

% Calculating the power gains...
% Note that Gp has no dependence on the source impedance
% (Zs).
% It only depends on the Networks' S-Parameters (which
% are Zo referenced) and the Load impedance.
% Gt is also calculated, but as SWR diverges from 1:1,
% Gt will worsen (because it also depends upon the
% source impedance's match to the Tuner's input Z).
% Gp is thus the preferred value to use because of its
% independence from the source impedance.
GpSystem = powergain(Network_Obj.S_Parameters,...
GtSystem = powergain(Network_Obj.S_Parameters,...

% Calculate the SWR from the measured S11 data of the
% Network with Load attached.
SWR_System = vswr( squeeze(System_Obj.S_Parameters));

% Plots:

% Figure 1:
% Plot Gp and Gt losses

h_lines= plot(FreqMHz,10*log10(abs(GpSystem)),...
xlabel('Freq in MHz');
ylabel('Loss in dB');
legend('Gp Loss','Gt Loss','Location','east');
grid on
grid minor
ax = gca;
ax.GridAlpha = 0.4;  % Make grid less transparent.
ax.GridColor = [0,0,0]; %
ax.MinorGridAlpha = 0.5;  % Make grid less transparent.
ax.MinorGridColor = [0,0,0]; %
title(['Power Loss, Drake MN-4 Matching Network',...
    ' Tuned for a 5 ohm Load']);

% Figure 2:
% Plot Smith Chart and compare the measured gamma
% (from S11 measured at the network's input with the
% load attached) versus the gamma derived from the
% Network-only S2P file and the Load-only S1P file.

hsm1 = smithchart(squeeze(System_Obj.S_Parameters));

hsm2 = smithchart(gamma);  %

legend('Measured gamma',...
    'Calculated gamma');
tx=text(-1.8,1.1,['Drake MN-4 Matching Network,',char(10),...
     '           5 ohm Load,',char(10),...
     '            28-30 MHz']);
tx.FontSize = 16;
tx.FontWeight = 'bold';
Published with MATLAB® R2017a
Note that this code has file names and other descriptors specific to the measurement of the Drake MN-4 Tuner when terminated with a 5 ohm load and measured at 29 MHz.

To use for other networks, loads, or frequencies, the following changes should be made:
  • Change the three file names (one S2P file and two S1P files).
  • Change the Title of Figure 1 (Gp and Gt losses),
  • Change the Text of Figure 2 (Smith Chart)

Below are the resultant plots:

First, the Power Loss plot:

Note that Gp is -0.5954 dB at 29 MHz.  This represents a power loss of 12.8%, calculated using this formula:

Power Loss (percent) = 100 * ( 1 - 10^(Gp/10) )

(Note, too, that Gp must be negative when calculating a loss with this formula).

Gt is also plotted, but it is not as useful for determining Tuner loss because Gt doesn't just represent Tuner loss, it also includes power loss due to any mismatch between the source's impedance and the tuner's input impedance.  So if the SWR, after tuning the Tuner, is low (i.e. the tuner's input appears to be Zo (typically 50 ohms)), then Gt will essentially equal Gp.  But as SWR increases, Gt will diverge from Gp and thus Gp will no longer accurately reflect only Tuner loss.

As an aside, note that if the source impedance, Zs, in the Gt calculation is defined to be the complex-conjugate of the tuner's input impedance when the tuner is terminated with the Test-load, then Gt will equal Gp.  You can see this using the MATLAB code, above, by changing the definition of Zs from "Zs = Zo" (i.e. Zs = 50 ohms) to:

Zs = conj(gamma2z(gamma))

where "gamma" is calculated earlier in the code and represents the gamma of the Tuner's input when the Tuner is tuned to minimize SWR with the Test-load attached to the Tuner's Antenna port.

As an additional (and optional) verification step (per Step 4, above), I also compare the measured gamma versus the gamma predicted from the independently captured Network S2P and Load S1P files...

You can see that the two curves are quite close, as they should be.

One final note:  as I mentioned above, this 5 ohm load is well outside the MN-4's range of spec'd loads (it is spec'd for loads having up to a 5:1 SWR).  Please do not take this example of a 12.8% loss to represent what MN-4 loss would be with less extreme loads.

Obtaining the MATLAB scripts:

If you would like copies of any of the MATLAB scripts or GUIs listed above, please email me.

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.

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:

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:

[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], 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.