Losses in a helically-wound small transmitting loop antenna

In 2014, K4HKX published measurements showing that a helically-wound small transmitting loop (using 3-inch wide by 0.012-inch thick copper strap wound 40 times around an octagonal support frame) had poorer performance, as evidenced by lower signal strengths on the Reverse Beacon Network, than equivalently-sized loops made of 3-inch diameter copper pipe and 1-inch diameter copper pipe. Specifically, the helically-wound loop's signals were 2 dB below those of the 3-inch pipe loop, and 1 dB below those of the 1-inch pipe loop. Measurements were made on the 40 meter band (7.148-7.150 MHz). See https://www.qrz.com/db/k4hkx for details.

K4HKX's results established without doubt that in this context, a 3-inch strap, when helically wound, is worse than 3-inch diameter pipe or 1-inch diameter pipe. The purpose of this post is to further show, based on further analyses of K4HKX's data, that a helically-wound 3-inch strap is worse than a non-helically-wound 3-inch strap. All measurements below are assumed to have been performed at 7.148 MHz (the K4HKX page mentions both 7.15 MHz in the text and 7.148 MHz on the EZNEC graph).

1. K4HKX measured that an 18-foot perimeter octagon with 3-inch diameter pipe (his so-called "FatLoop") yields signal strength 6 dB (median value) below that of a dipole (at 70 feet). See https://www.qrz.com/db/k4hkx, "Section 7:  Magnetic Loop Comparisons vs My Reference 40M Dipole".
   
2. K4HKX simulated an 18-foot perimeter octagon with 3-inch diameter pipe, which yielded signal strength 6 dB below that of a dipole (at 70 feet) at 45 degrees above the horizon. See the EZNEC graph at https://www.qrz.com/db/k4hkx, "Section 6:  Magloop Version VI:  40 Meter FatLoop".
   
   I could reproduce these simulation results in 4nec2. For the dipole, I modeled an inverted-V using 0.3-mm diameter copper wire, with the antenna apex at 21.336 meters (70 feet), and with arms sloping down at 23 degrees, simulated above real ground with ground conditions specified as "Rocky, steep hills". The resulting far-field graph for the dipole matches very closely the K4HKX simulation, with a maximum gain of about 7.11 dBi (compared with K4HKX's simulated maximum gain of 7.08 dBi), and minimum gain about 7.2 dB below the maximum gain (about the same as K4HKX's simulated minimum gain). The gain of the dipole at 45 degrees above the horizon is -0.77 dB. We will use this value for comparison later.
   
   
   
   
   
   
   For the loop antenna, I used the same ground conditions, used a 3-inch diameter copper conductor, formed an octagon with eight 27-inch segments, and added 15 milliohms of loss for the tuning capacitor, located at the bottom of the loop. The bottom of the loop was 8 feet above the ground, corresponding to K4HKX's report of his original loop height of 8 feet. Again, the resulting graph closely matches K4HKX's graph. Compare the below graph, which overlays my simulated dipole and fat loop, with K4HKX's EZNEC graph linked above (direct link: https://cdn-bio.qrz.com/x/k4hkx/20140926_FatLoop_vs_Dipole.jpg). My simulated loop gains at 15, 30, 45, and 60 degrees above the horizon are 7.11 dB, 7.81 dB, 6.19 dB, and 3.28 dB below the dipole's gain, compared to K4HKX's simulated results of 7 dB, 8 dB, 6 dB, and 2.5 dB below the dipole's gain. So my simulation setup is reasonably close to K4HKX's simulation setup. At 45 degrees, the loop gain of -6.96 dB is 6.19 dB below the dipole gain at 45 degrees of -0.77 dB; the 6.19 dB poorer performance of the loop matches K4HKX's calculated and simulated FatLoop performance of 6 dB below that of the dipole.
   
   
   
   
   
   
   
3. Next, let us consider helical winding. K4HKX measured that an 18-foot perimeter octagon with 3-inch wide x 0.012-inch thick strap wound helically with 40 turns yields signal strength 8 dB (median value) below that of a dipole (at 70 feet height). See https://www.qrz.com/db/k4hkx, "Section 1:  Helically Wound Magloop", which includes a photograph of the helical loop.
4. I modeled an equivalently-performing non-helical loop (which I call the "equivalent-helical" loop below) in 4nec2 as follows. An 18-foot perimeter octagon with 0.43-inch diameter pipe (routed non-helically in the shortest direct path around the octagon's perimeter) yields signal strength 8.23 dB below that of a dipole (at 70 feet height) at 45 degrees above the horizon. This simulated performance is essentially identical to the K4HKX-measured performance of the helically-wound octagon (8 dB below the dipole, and 2 dB below the FatLoop). Therefore, we can say that the helically-wound octagon constructed with 3-inch wide strap (0.012 inch thickness) has the same performance as an identically-sized and non-helically-wound octagon constructed with 0.43-inch diameter pipe.
   
   The simulation results for the equivalent-helical octagonal loop (constructed with a non-helically-wound 0.43-inch diameter pipe) are shown below. The gain at 45 degrees above the horizon of the equivalent-helical loop, shown in blue, is -9 dB.
   
   
   
   
   
   Recall that the dipole had gain of -0.77 dB at 45 degrees above the horizon.
   
   
   
   
   Therefore, the equivalent-helical loop has gain at 45 degrees that is 8.23 dB below that of the dipole at 70 feet, which corresponds with K4HKX's measured result that the helically-wound loop had signal strength 8 dB below that of the dipole at 70 feet.
   
   Also, the equivalent-helical loop has gain at 45 degrees (with a gain value of -9 dB relative to the dipole) that is 2.04 dB below that of the 3-inch pipe FatLoop (which has a gain value of -6.96 dB relative to the dipole, as shown above). Again, this corresponds to the actual K4HKX measurement that the helically-wound loop had signal strength 2 dB below that of the 3-inch pipe FatLoop.
   
5. Having now established that the loop constructed with helically-wound 3-inch strap (0.012 inch thickness) has equivalent performance to a non-helically-wound loop constructed with 0.43-inch diameter pipe, next we can extrapolate a similar result for a non-helically-wound strap. Using W9CF's formulas for the RF resistance of rectangular cross-section conductors, we can calculate that a 0.43-inch diameter pipe with round cross-section has equivalent RF resistance to a 1.44-inch wide strap having rectangular cross section and 0.012-inch thickness. If we denote the strap's width as w=1.44 inches and its thickness as t=0.012 inches, then the equivalent diameter is found by the following W9CF formula, which yields a 0.43-inch diameter for a round-cross-section pipe with equivalent RF resistance.
   
   
   
   
   
6. From #5, we can say that an 18-foot perimeter octagon with 0.43-inch diameter pipe has equivalent RF resistance to an 18-foot perimeter octagon with 1.44-inch wide strap (0.012-inch thickness) routed non-helically in the shortest direct path around the octagon's perimeter.
7. From #6 we can say that an 18-foot perimeter octagon with 1.44-inch wide strap (0.012-inch thickness), routed non-helically in the shortest direct path around the octagon's perimeter, will achieve the same signal strength as the same-sized loop constructed with 0.43-inch diameter pipe, which from #4 we computed to be 8.23 dB below that of a dipole (at 70 feet height) at 45 degrees above the horizon.
8. From #7 and #3, we can say that the K4HKX 18-foot perimeter octagonal loop with 3-inch strap wound in a long helical path has the same performance as an identically-sized loop constructed with 1.44-inch wide strap that is routed in a shorter and non-helical path straight around the loop's perimeter. This is the first important result.
   
9. Next, consider K4HKX's helically-wound loop using 3-inch wide (0.012 inch thick) strap, and let us see what happens if we undo the long and helical winding of the 3-inch strap, and instead route the same 3-inch strap in a shorter, straight, and direct path around the octagon's perimeter. Again using W9CF's formulas for the RF resistance of rectangular cross-section conductors, we can compute that a 3-inch wide strap (with 0.012-inch thickness) has equivalent RF resistance to a 0.809-inch diameter pipe.
10. Adjusting the conductor diameter in the simulation from #4, we find that an 18-foot perimeter octagon with 0.809-inch diameter pipe (routed non-helically in a straight path around the octagon's perimeter) yields signal strength 7.24 dB below a dipole (at 70 feet) at 45 degrees above the horizon.
    
    The image below shows the gain of the octagonal loop constructed with 0.809-inch diameter pipe, equivalent to the non-helically-wound 3-inch strap, in blue. The gain at 45 degrees above the horizon is -8.01 dB.
    
    
    
    
    
    Again, recall that the dipole had gain of -0.77 dB at 45 degrees above the horizon.
    
    
    
    
    Therefore, the non-helically-wound strap loop has gain at 45 degrees that is 7.24 dB below that of the dipole at 70 feet. The loop constructed with a non-helically wound strap has higher gain (having simulated gain 7.24 dB below that of the dipole) than the loop constructed with a helically wound strap (having measured gain 8 dB below that of the dipole from K4HKX's report). This is the second important result.
    

To summarize the results:

• From #3 and #8 we see that helically winding the 3-inch strap, in a long and helical path around the support tubes of the loop perimeter, has reduced its performance to be on par with that of a narrower 1.44-inch strap routed non-helically in the shortest direct path around the octagon's perimeter.
• From #3 and #10 we see that removing the helical winding and routing the same 3-inch strap in a shorter and non-helical direct path around the loop perimeter will increase signal strength by about 1 dB.

In conclusion, there is no performance benefit to be gained by helically winding the conductor in a small transmitting loop antenna; on the contrary, this technique incurs additional resistive loss. This has been stated over and over again since the idea was first introduced 2011, but unfortunately many persons still mistakenly believe that there is some performance benefit to be gained by helically winding the conductor of a small transmitting loop. There is no such performance benefit. 

For the best performance, it is better to run the loop conductor in the shortest direct and non-helical path around the loop's perimeter, minimizing the conductor length while maximizing the loop's enclosed area. 

As with all engineering, the antenna designer must decide whether or not the performance loss caused by helical winding is outweighed by other factors such as ease of construction or mechanical sturdiness.

Fixed image problems with recent posts

Due to some changes in the blogger platform, some images in my recent posts may have been broken. I re-uploaded the images, and they should be working now. 

If you had problems viewing the images in my recent posts on small transmitting loops, please check the articles again to see the images. The updated images include simulations of current density on a loop antenna conductor; evaluations of the effect of crumpled, deformed, and separated cross-sectional conductor shapes; conductor resistance in multi-turn loops; and more.

A 1-meter-diameter small transmitting loop for 7 MHz: part 8

The effect of thin, crumpled, separate copper strips on antenna gain

A small loop antenna with the following parameters was simulated in 4nec2:

1. Dimensions of 1 meter square
2. Conductor diameter of 10 cm (radius of 5 cm)
   
3. 1.5 meters above real ground with "Moderate" conditions
4. Resonated with 276 pF capacitance for zero reactance at 7 MHz
5. 8.2 milliohms of capacitor loss, corresponding to a Q of 10,000 
   

Antenna gain when using a perfectly-shaped round copper conductor

If the conductor loss resistance is computed by 4nec2, the gain at 15 degrees above the horizon is 6.4 dB below that of an isotropic radiator. The same gain value is obtained if the round conductor's loss resistance is analytically computed and specified as 8.72 milliohms (400 cm conductor length / pi*10 cm conductor width = 12.73 copper squares, with 0.685 milliohms RF resistance at 7 MHz for a copper square of 5 skin depths thickness, giving 8.72 milliohms).

Antenna gain when using thin, crumpled, separate strips to form a distorted approximation of a square-cross-section conductor

If the conductor loss resistance is explicitly specified as 21.4 milliohms (located at the highest-current segment of the antenna), to correspond to the values computed in the last article for thin, crumpled, separate strips of copper tape, then the gain at 15 degrees above the horizon reduces to 7.45 dB below that of an isotropic radiator. 

Conclusion

The effect of using thin, crumpled, separate strips of copper tape is to reduce the antenna gain, at 15 degrees above the horizon, by 1.05 dB compared to that of a perfectly round and smooth conductor, giving a gain of -7.45 dBi.

The conductor's RF resistance increases above that of a smooth, round conductor (8.72 milliohms) by 12.68 milliohms, to give 21.4 milliohms RF resistance for the thin, crumpled, separate strips of copper tape formed into an approximately-shaped square-cross-section conductor.

Other antenna simulations were performed in 4nec2 for similarly-sized loop antennas with different configurations, such as multiple turns wound solenoidally, or multiple turns would in a concentric, spiral configuration. When including ground loss, capacitor loss, and the conductor loss as predicted by FEMM, all simulations yielded roughly the same result: around -7 dBi at 15 degrees above the horizon. With the constraint of 1-meter diameter, no multiple-turn configuration of the conductor could be found that significantly exceeds this performance.

The conclusion is that using readily-available copper tape (50 mm wide, 0.030 mm thick), applied to a reasonably but not perfectly-shaped square-cross-section plastic tube, is a feasible method -- and probably the easiest method -- for constructing a low-loss conductor at 7 MHz for use in this application.

A 1-meter-diameter small transmitting loop for 7 MHz: part 7

RF resistance of thin and crumpled copper tape

One feasible approach for constructing a 10-cm-diameter conducting copper tube is to use commercially-available copper tape. Local suppliers sell copper tape of 50 mm width and 0.030 mm thickness. This tape can then be applied to a square-cross-section plastic tube.

This post examines simulations of RF resistance at 7 MHz for four scenarios:

1. Perfectly flat 30 micron tape, arranged to form a perfectly-aligned square-cross-section conductor.
2. Crumpled 30 micron tape, arranged to form a perfectly-aligned square-cross-section conductor.
3. Crumpled 30 micron tape, arranged to form a slightly misaligned (gaps < 5 mm) square-cross-section conductor.
4. Crumpled 30 micron tape, arranged to form to a severely misaligned (gaps > 5 mm) square-cross-section conductor.

General test conditions

Each of the four faces of the 10 cm square-cross-section conductor is formed by two 5 cm-wide copper strips, laid side by side, for a total of 8 copper strips. The strips were separated by small air gaps of about 1 mm (or more in the distorted test cases), so there is no cross-wise conduction of current from one strip to another.

Each 5 cm-wide copper strip was modeled as a vertical cross-sectional strip consisting of 129 sections (0.39 mm/section). Assuming the strip is oriented vertically, tape crumpling was simulated by displacing each of the 129 vertical sections randomly by 0-1 mm in the horizontal direction, leading to a slightly uneven profile to the strip. 

 

During the simulation, because the conductor cross-section is in effect rotated about the vertical axis, the effect of crumpling the cross-section is to form uneven grooves running along the length of the loop conductor. 

The unevenness of the conductor is expected to interfere with the ideal and uniform current distribution, and to increase the RF resistance.

Sharp corners (90 degree angles) of the cross-sectional conductor shape were left as-is in the model; no corner rounding was done. This may lead to unnaturally high current concentrations around the sharp corners, so the overall results should be checked for plausibility.

For the simulation in the FEMM software, we will conceptually trace the 10 cm square cross-section of the conductor along a circular path with 65 cm radius, giving a simulated conductor length of 4.084 m. The reason for choosing this 4.084 m conductor length is that the results can then be related to a 1-meter square loop antenna (see below).

Specific test conditions

 

Case 1: Thin (30 micron), perfectly-shaped conductor

For this case, 8 perfectly flat strips (having a perfectly straight cross-section) of 30 micron thickness (slightly more than 1 skin depth at 7 MHz) were arranged into a perfect square with sides of 10.4 cm. The goal is to simulate the resistance of a 10 cm-wide square conductor, and the extra 0.4 cm of length is to allow small air gaps between the individual strips.

The analytically expected resistance is as follows. At 7 MHz, the surface resistivity per square of copper with 1 skin depth thickness is 1.08 milliohms/square. A 10 cm-diameter circular conductor of 408.4 cm length has 408.4/(pi*10) = 13 squares of copper, for a total expected RF resistance of 13 * 1.08 = 14.0 milliohms. The 10 cm round-cross-section conductor should then have the same RF resistance as that of a 10 cm square-cross-section conductor.

The FEMM simulation result yields 12.7 milliohms.

 

Case 2: Thin, perfectly-shaped, crumpled conductor

This test case is the same as the previous test case, except the cross sections of the strips have been crumpled as described above.

We expect an increase in RF resistance over the previous test case, which yielded 12.7 milliohms.

The FEMM simulation result yields 16.8 milliohms, which as per expectation exceeds that of the previous test case.

Case 3: Thin, slightly distorted, crumpled, conductor

This test case is the same as the previous test case, except that the arrangement of the crumpled strips has been slightly misaligned to form an imperfect square cross-section, with gaps less than 5 mm between the strips.

We expect an increase in RF resistance over the previous test case, which yielded 16.8 milliohms.

The FEMM simulation result yields 20.0 milliohms, which as per expectation exceeds that of the previous test case.

Case 4: Thin, severely distorted, crumpled, conductor

This test case is the same as the previous test case, except that the arrangement of the crumpled strips has been severely misaligned to form a quite distorted square cross-section, with gaps greater than 5 mm between the strips.

We expect an increase in RF resistance over the previous test case, which yielded 20.0 milliohms.

The FEMM simulation result yields 21.4 milliohms, which as per expectation exceeds that of the previous test case.

Summary of results

1. Thin (30 micron), perfectly-shaped conductor: 12.7 milliohms
2. Thin, perfectly-shaped, crumpled conductor: 16.8 milliohms
3. Thin, slightly distorted, crumpled, conductor: 20.0 milliohms
4. Thin, severely distorted, crumpled, conductor: 21.4 milliohms
   

Applying these same results to a square loop antenna

The above simulation results were computed for a 10 cm square-cross-section conductor formed into a loop conductor (loop antenna) of 65 cm radius. This 65 cm radius is measured from from the loop antenna’s center out to the center point of the wide, 10 cm square conductor.

If we instead measure from the loop antenna’s center out to the outer-most extent of the 10 cm square conductor, the outer-most loop antenna radius will be 65 + half of the conductor width, which is 65+5=70 cm. Similarly, the radius of the loop antenna measured up into the inner-most extent of the conductor will be 65-5=60 cm.

Therefore, for the circular loop antenna of 65 cm radius using a 10 cm square conductor, the outer-most flat face of the square conductor traces a circular path of 2*pi*70 cm, or about 4.4 meters. The inner-most face of the square conductor traces a circular path of 2*pi*60 cm, or about 3.77 meters.

We can compare these results with a square loop antenna as follows. A square loop antenna with dimensions of 1 meter per side (measured from the centers of the 10 cm square conductors) uses a similar conductor length. Since the conductor width is 10 cm, the outer-most span of the square loop antenna is 1.1 meters, and the interior span of the square loop antenna is 0.9 meters. The outer-most conductor length is 1.1*4 = 4.4 meters, and the inner-most conductor length is 0.9*4=3.60 meters.

Comparing the outer-most conductor length and the inner-most conductor length, we can see that the conductor length of the 1-meter square loop antenna is equal to or slightly less than the conductor length of the 65-cm radius circular loop antenna.

Therefore, the conductor resistance of the circular loop antenna (which we could simulate directly in the FEMM software) provides a reasonable estimate of the conductor resistance of the square loop antenna (which we could not directly simulate in the FEMM software).

 

Conclusion and next steps

We determined the simulated RF resistance at 7 MHz of various configurations of copper tape having 50 mm width and 0.030 mm thickness. In the worst case, we expect about 21.4 milliohms of conductor loss, based on the types of distortion that could be simulated in the FEMM software. Other unaccounted-for loss mechanisms (such as surface oxides or more complex surface roughness) may push the loss even higher.

We also determined that this result is applicable to a 1-meter square loop antenna.

The next step is to determine the antenna gain of a 1-meter square loop antenna including the 21.4 milliohms of expected conductor loss. To give a more complete picture of the expected performance, the antenna simulation should include not only conductor loss, but also ground loss and capacitor loss.

A 1-meter-diameter small transmitting loop for 7 MHz: part 6

Evaluating conductor loss in the total antenna context

This post provides more context to the conductor loss values computed in part 4 of this series (http://qrp-gaijin.blogspot.com/2022/02/a-1-meter-diameter-small-transmitting.html).

As a review of the simulation results in part 4, we determined that:

1. The radiation resistance of a 1-meter-diameter loop antenna at 7 MHz is 5.7 milliohms.
2. The conductor loss of an ideal round conductor (a hollow 10 cm round conductor with 200 micron shell) is 7.18 milliohms.
3. The conductor loss of a distorted conductor (a hollow 10 cm square conductor with disconnected and misaligned 100 micron shell) is 12.27 milliohms.
4. The effect of the conductor's shape distortion in (3) is therefore 12.27-7.18=5.09 milliohms.
5. Compared to the radiation resistance of 5.7 milliohms, the effect of the conductor's shape distortion was considered "significant", leading to investigation of an alternative configuration, the multi-turn loop, in part 5 of this series (http://qrp-gaijin.blogspot.com/2022/02/a-1-meter-diameter-small-transmitting_13.html).

However, this conclusion, that the 5.09 milliohms of loss due to conductor shape distortion was "significant", fails to consider the magnitude of other losses in the system. In particular, vacuum variable capacitor losses and ground losses may be large enough so that the relative effect of the conductor shape distortion is small.

A more complete analysis should:

1. Specify the entire system including all losses.
2. Define the target performance.
3. Evaluate the effect, on the whole system, of the additional 5.09 milliohms of loss due to conductor shape distortion.

As a starting point we can specify for example:

1. The lowest part of the antenna will be 1.5 meters above NEC-2 real ground, specified as having "Moderate" ground conditions. The capacitor will be assumed to have a Q of 10,000 at the operating frequency, which then introduces an Equivalent Series Resistance (ESR) of ESR=X/Q, where X is the reactance of the capacitor at resonance.
   
2. The antenna gain at 15 degrees above the horizon should be -7 dBi or higher.

Then, given this environment and target gain, we can evaluate the antenna gain (in a NEC-2 based simulator like 4nec2) without and with the 5.09 milliohms of added loss due to conductor shape distortion, and determine the effect on the antenna gain of these shape-distortion-related losses.

The preliminary result of such an analysis (a detailed example will be added later in another post) is that the added 5.09 milliohms causes less than 1 dB difference in antenna gain at 15 degrees above the horizon. The other system losses (ground loss and capacitor ESR) have already degraded the total system performance to such a degree that the added loss of conductor shape distortion is small -- less than 1 dB.

Having quantified the effect of the conductor shape distortion on overall system loss, the revised conclusion is that the loss may be acceptable.

A 1-meter-diameter small transmitting loop for 7 MHz: part 5

Moving to a multi-turn loop

I have been hesitant to implement a multi-turn transmitting loop due to the reported measurements and experience of W8JI, who worked on the commercially-available MFJ loop antenna. His comments can be found here: https://forums.qrz.com/index.php?threads/magnetic-loops.241400/page-3#post-1961567

However, when evaluating such assertions as "multi-turn loops increase loss", it is always important to consider the assumptions underlying those assertions, and whether those assumptions hold for a specific case.

In my case, a multi-turn loop is beginning to look like the best option. My assumptions are as follows:

1. The loop antenna should be 1 meter in diameter and should operate at 7 MHz.
   
2. The loop antenna should be feasible for me to construct physically. For the loop conductor, thick copper tube (10 cm diameter) is not available for purchase locally. Even if it were available, I lack the equipment to solder or braze these tubes together.
3. The loop antenna should be as lightweight as possible to enable easy setup and dismantling, and possible transport for field use. This means that the loop antenna should not be made of heavy, rigid materials. This also implies that the loop antenna may be subject to small shape deformations over time due to transport and repeated setup and dismantling.
   
4. Transmitter power will be 5 watts.
   

Until now, the strategy for this project was to use a single-turn loop, and to construct a thick copper tube piecewise, by: (1) changing the tube cross section from circular to square; (2) conceptually separating the four sides of the square-cross-section tube with a small air gap, yielding four flat copper sheets; (3) purchasing flat copper sheets and gluing them onto the surface of a square-cross-section plastic former; and (4) electrically joining several such tubes together to form a large loop (an octagon, or a square), by soldering the flat copper sheets together.

However, as described in the last post (http://qrp-gaijin.blogspot.com/2022/02/a-1-meter-diameter-small-transmitting.html), small deformations in the cross-sectional shape of a 10-cm-diameter tube can cause unacceptably high losses in a single-turn, 1-meter-diameter loop at 7 MHz. Hand-assembly of such a copper tube will not yield a perfectly-shaped cross-section, and will contain small distortions, which may increase over time. Removing such distortions, and preventing distortions over time, would require more rigid and heavy materials, and more thorough and solid construction techniques, all of which go against the project assumptions for my case.

Overcoming objections to a multi-turn loop

An alternative is to construct a multi-turn loop. W8JI's objections to a multi-turn loop are as follows (https://forums.qrz.com/index.php?threads/magnetic-loops.241400/page-3#post-1961567):

Three things kill the small loop's efficiency when it is multiple turns. The first is turn-to-turn capacitance. The turn-to-turn capacitance adds circulating currents across each turn of the loop, and this seriously de-Q's the system by increasing I^2 R losses.

The second problem was air. In order to not arc, the turns needed spaced. This means for a given volume of area more of it is useless air. The same amount of material in a thin single tube occupying the same area would have much more surface area and less resistance.

The final problem was current bunching or pushing, just like occurs in an inductor or even in a single conductor (skin effect). Current tends to push out from the most concentrated area of magnetic fields, and a larger amount of current flows in a smaller area of surface. There was a lot less of this effect in a round tube compared to a flat strip (like the AEA loop), and multiple turns made it act more like a flat strip.

 Examining these objections in turn:

1. Yes, coil self-capacitance will cause circulating currents "across" the turns instead of "along" the turns as desired. It is open for debate whether or not the concept of "turn-to-turn capacitance" is actually valid, because attempts to explain coil self-capacitance that are based on "turn-to-turn capacitance" generally fail (see e.g. https://www.g3ynh.info/zdocs/magnetics/appendix/self-res.html, which asserts that "The inter-turn capacitance approach is found to have no predictive power").
   
   If some form of inter-turn capacitance actually exists and can be explained by "capacitors" formed by opposing faces of adjacent turns, then this capacitance can likely be reduced by (1) increasing the turn spacing, hence increasing the distance between the "plates" of each "inter-turn capacitor", and hence reducing the "inter-turn capacitance"; and (2) using a thin strip conductor to minimize the conductor surface area between turns, so that any "capacitor plates" formed by adjacent turns have very small surface area (along the thin edge of the strip conductor).
   
   Conclusion: This concern is valid, and can be mitigated.
   
   
2. Yes, it is true that turns must be spaced, introducing "useless air" into the volume of the loop. It is also true that "The same amount of material in a thin single tube occupying the same area would have much more surface area and less resistance."
   
   However, as described above, when the tube diameter approaches 10 cm, then avoiding deformations become critical to ensure low RF resistance along the tube. A perfect 10-cm-diameter single tube would be ideal, but is difficult to obtain and more difficult to construct.
   
   While useless air does increase the total volume required for the loop antenna, it is not a critical requirement in my case. For a commercially-sold antenna like the MFJ loop, which must be shipped to customers via post, it is understandable that minimising volume was a concern.
   
   Furthermore, the RF resistance -- including skin effect and proximity effect -- can be computed and analyzed by Finite-Element Methods, using the FEMM software. It is not necessary to rely on general statements about conductor geometry and resistance; it is instead possible, and much better, to compute and visualise the current distribution and the RF resistance, to evaluate a specific conductor configuration and also to distort the shape of that conductor configuration to test the effects of the distortion.
   
   Conclusion: Useless space caused by multiple turns can be tolerated. The specific RF resistance can be computed in FEMM.
   
   
3. Yes, multiple turns will cause a skin-effect-like proximity effect; the current in the turns will not be exactly equal.
   
   But we can easily compute the actual current distribution in either FEMM (which accounts only for magnetic effects, like proximity effects and skin effect) or in NEC-based Method-of-Moments simulators (which account for full electromagnetic field effects between wires, including both electric-field and magnetic-field effects, but which does not include proximity effects and skin effects on the surface of a single conductor).
   
   Conclusion: The actual distribution of RF current can be computed, and its effect on efficiency evaluated, in FEMM or NEC.

Simulation results of RF resistance of multiple turns in series

A future post may examine the simulations in detail, but here are some preliminary impressions:

1. Analytically, a square-shaped loop antenna with 4 turns (in series) has a radiation resistance at 7 MHz of 148 milliohms. For an intuitive explanation of why series turns increase the radiation resistance, see https://owenduffy.net/blog/?p=16986 -- each turn carries the same current (since all are in series), and because the turns are physically arranged to drive this same current around the perimeter of the antenna, the result is increased current flowing around the loop antenna's perimeter, and hence increased radiation. Turns in parallel do not (contrary to some mistaken information online) cause a similar increase in radiation resistance due to the available current dividing simply itself among the turns, leading to no increase in current around the loop; again, see https://owenduffy.net/blog/?p=16986 for details.
2. 4nec2 simulations of a 4-turn square loop with 1-meter sides yield a radiation resistance of 170 milliohms. This is in reasonable agreement with the analytical value. This was measured as the real part of feedpoint impedance at the highest-current point in the antenna, in free space, with wire loss set to zero, so that the only resistive component of the feedpoint impedance is the radiation resistance. The reactive part of the feedpoint impedance was zero, since the antenna was adjusted with a capacitance to be resonant at 7 MHz.
   
   
   
   
   
   Due to the potential for calculation inaccuracies of the NEC-2 engine with small loops, the simulation was re-run with a double-precision port (by VK3DIP; see http://www.yagicad.com/Projects/mininec.htm) of the MININEC simulator, which uses a completely different implementation of the Method-of-Moments simulation and which therefore provides a valuable second opinion for sanity-checking the NEC-2 results. MININEC is also reported to be able to measure much smaller loops with high accuracy; the report at https://archive.org/details/DTIC_ADA181682, p. 51, Section 3.2  asserts that MININEC with single-precision floating point calculations can accurately model loops as small as 0.01 wavelengths in circumference -- which is much smaller than loop under investigation, which is about 0.10 wavelengths in circumference. The MININEC results, run with shorter segment lengths (46 segments per meter, or about 2.17 cm per segment), yielded a radiation resistance of 167.78 milliohms (measured at the feedpoint which was located at the wire segment with maximum current), which agrees well with the NEC-2 reported result of 170 milliiohms. The following is an excerpt of the MININEC output, run in free space with no resistive losses.
   
   
   ********************    SOURCE DATA     ********************
   PULSE 359       VOLTAGE = (1,0J)
                  CURRENT = (5.960077 , -0.000552131 J)
                  IMPEDANCE = (0.1677831 , 1.554313E-05 J)
                  POWER = 2.980038 WATTS
    
   ---------- POWER BUDGET ---------
   TOTAL POWER = 2.980038 WATTS
   RADIATED POWER= 2.980038 WATTS
   TOTAL LOSS= 0 WATTS
   
    EFFICIENCY = 100.000 %
   
   
   The following is another excerpt from the MININEC output, showing that the feedpoint location (pulse 359) is the segment of maximum-current, which is where the radiation resistance should be measured as the real part of the feedpoint impedance.
   
   
   WIRE NO. 9:
   PULSE         REAL          IMAGINARY     MAGNITUDE     PHASE
    NO.          (AMPS)        (AMPS)        (AMPS)        (DEGREES)
   J            5.945504       -0.0007043754 5.945504      -0.006787942
   345          5.94675        -0.0006978156  5.94675      -0.006723317
   346          5.94794        -0.0006914593  5.94794      -0.006660743
   347          5.949186       -0.0006847983  5.949186     -0.006595197
   348          5.950457       -0.0006779634  5.950457     -0.006527976
   349          5.951738       -0.0006709474  5.951738     -0.006459031
   350          5.953006       -0.0006637438  5.953006     -0.006388322
   351          5.954234       -0.000656331   5.954234     -0.006315673
   352          5.955411       -0.0006486758  5.955411     -0.006240776
   353          5.95651        -0.0006407233  5.95651      -0.006163129
   354          5.957508       -0.0006323904  5.957508     -0.006081956
   355          5.95838        -0.0006235364  5.95838      -0.005995925
   356          5.959102       -0.0006139791  5.959102     -0.005903307
   357          5.959646       -0.0006028565  5.959646     -0.005795837
   358          5.959982       -0.0005922992  5.959982     -0.005694018
   359          5.960077       -0.000552131   5.960077     -0.00530778
   360          5.95989        -0.0005934233  5.95989      -0.005704913
   361          5.959375       -0.0006051083  5.959375     -0.00581775
   362          5.958478       -0.0006173661  5.958478     -0.005936494
   363          5.957143       -0.0006280719  5.957143     -0.006040793
   364          5.95532        -0.0006380953  5.95532      -0.006139076
   365          5.953004       -0.0006476296  5.953004     -0.00623323
   366          5.950294       -0.0006568312  5.950294     -0.006324671
   367          5.947456       -0.0006658013  5.947456     -0.006414105
   368          5.944841       -0.0006746035  5.944841     -0.006501761
   
   
   
3. FEMM simulations of the RF resistance of 4 turns in series, with 1 meter diameter, yielded 130 milliohms for a perfectly-aligned series of turns, or 137 milliohms for a slightly-misaligned (by a few millimeters) series of turns.
   
   The following image shows a conceptual diagram (not to scale) of the perfectly-aligned loop geometry. Four exactly identical loops are placed exactly above one another at exactly the same spacing, and the RF resistance is computed assuming the turns are in series.
   
   
   
   The following image shows the computed RF resistance and the flux density for a perfectly-aligned, 4-turn loop (four turns of copper in series, each 0.1mm thick and 100mm wide, placed exactly vertically above one another, with 15mm spacing between turns).
   
   
   
   
   The following image shows a conceptual diagram (not to scale) of the misaligned loop geometry. Four loops with slightly differing tilted cross-sectional profiles are approximately above one another at approximately the same spacing, and the RF resistance is computed assuming the turns are in series. This misalignment can capture some of the effects of distorted turns that are not exactly spaced and not exactly in line with one another. (Due to restrictions in the FEMM software, it is not possible to simulate distortions that alter the tilt within one turn, because FEMM only allows definition of a single cross-sectional profile for each turn, that is then uniformly rotated about the vertical axis.)
   
   
   
   
   
   The following image shows the computed RF resistance and the flux density for misaligned conductors. The top and bottom ends of each conductor's cross-sectional profile have been slightly displaced in the x direction by a few millimeters, so that the turns are not exactly aligned with one another.
   
   
   
   
   
   This simulation assumes a round loop shape -- a circle of 1 meter diameter, not a square with 1-meter sides -- and so a square loop shape will have slightly higher conductor resistance. This could be approximated by making the round loop diameter slightly larger in FEMM, so that the total conductor length of the 4 round loops corresponds to the total conductor length of 4 square loops.
4. From the above point 3, we see that small distortions in the geometry of the multi-turn loop do cause variations in RF resistance of several milliohms. However, unlike the single-turn loop, the multi-turn loop has higher radiation resistance. Therefore, the effect of several added milliohms of Rloss is insignificant. For example, using the figures above, the perfectly-aligned loop has an efficiency of Rrad/(Rrad+Rloss) = 148/(148+130) = ~53%, and the misaligned loop has an efficiency of 148/(148+137) = ~52%.
   
   As stated above, the Rloss is actually higher due to the longer conductor path required for a square loop perimeter, but nevertheless it can be seen that the higher radiation resistance of the multi-turn loop makes it much more forgiving of milliohm-level increases in loss resistance caused by distortions or misalignments of the loop geometry.
   
   The price we pay for this higher radiation resistance is the increased volume of the multi-turn antenna, needed due to the inter-turn spacing.

A rough sketch of how the multi-turn loop might be implemented looks as follows.

A tuning capacitor will be connected between the open ends of the loop. 4nec2 simulations of a similar loop indicate that, depending on the turn spacing and conductor diameter, the loop may become self-resonant below 7 MHz, which would make operation at or above 7 MHz impossible. 4nec2 simulations (using the NEC-2 engine) can only serve as a guideline here, since NEC-2 only models round conductors, but the actual antenna will use a flat strip conductor. The actual self-resonance of the structure as constructed with a flat strip will need to be measured in practice.

Spiral winding: a possible (but lossy) alternative

An alternative to winding the turns in a solenoidal coil is to wind the turns into a flat spiral coil, with all the turns lying in a plane and with later turns being nested inside of earlier turns. This has the advantage of reducing the physical volume required for the antenna. The disadvantage is increased losses due to proximity effect and possibly due to inter-turn capacitance.

The current distribution and RF resistance of spiral-wound (concentrically-wound) turns can be simulated in FEMM by modeling each turn to lie inside of the previous turns. Conceptually, this would look like the following diagram, where 4 strip conductors are placed concentrically.

FEMM can then simulate the RF resistance, assuming the turns are connected in series. The result is an RF resistance of 250.57 milliohms -- almost twice as high as the 130-137 milliohms of the solenoidally-wound loop. Note that the simulation shown below uses perfectly-aligned conductors. No test was performed with misaligned conductors.

The reason for this high RF resistance is the proximity effect: with turns placed in proximity and carrying identical current flowing in the same direction, the current flows influence each other and cause redistribution of current, under-utilization of the conductor, and an increase in the RF resistance.

Furthermore, if we accept that the opposing faces of adjacent turns form "capacitors" that contribute to "inter-turn capacitance", then the spiral-wound configuration -- in stark contrast to the solenoidally-wound configuration -- has very large surface areas from adjacent turns that face one another, potentially increasing the "inter-turn capacitance" (if the concept itself is in fact valid), and hence potentially increasing unwanted current flow "across" turns (via capacitance, i.e. displacement current) instead of the desired current flow "along" the turns. Unfortunately, when FEMM simulates proximity effect and RF resistance, it cannot account for electric field effects like capacitance. Therefore, the effect of "inter-turn capacitance" cannot be seen in FEMM. It can, however, be seen in NEC-2-based simulations, but in NEC-2, proximity effects and skin effect are not modeled, and all wires are assumed to be round, not flat. Synthesizing results from both simulators will give the best insight -- but the preliminary result is clear: spiral winding is lossier than solenoidal winding.

Self-resonance, capacitor current, and capacitor loss

An interesting aspect of a nearly-self-resonant antenna is that the required resonating capacitance is very small. This makes it feasible to construct a high-quality variable capacitor rather easily. Furthermore, the current flowing through the capacitor is quite low, making losses in the capacitor far less significant. Most of the current flows through the nearly-self-resonant structure, and little of the current flows through the capacitor. In the limiting case of self-resonance, no current flows through the capacitor, and all of the current flows through interior of the self-resonant structure. However, there is a disadvantage -- at or near self-resonance, we can expect that the entire structure will be very sensitive to detuning by its surroundings. Environmental changes such as people walking near the loop; the addition of moisture due to rain; or strong winds that change the loop's orientation relative to its surroundings or that cause fluttering of the individual turns or of the feedline may change the loop's self-capacitance and change the loop's tuning.

Future work

Good agreement was established between the MININEC results and the NEC-2 results for a loop approximately 0.10-wavelengths in circumference. This is very encouraging, because the single-precision MININEC was validated to return valid results for much smaller loops down to 0.01 wavelengths; the double-precision port by VK3DIP can likely simulate even smaller loops. Future work can conduct further MININEC simulations of the radiation resistance of multi-turn loops with smaller loop diameters and with more turns.

A 1-meter-diameter small transmitting loop for 7 MHz: part 4

Abstract

This post describes the results of Finite-Element-Method simulations (using the software package FEMM) of the RF resistance at 7 MHz of various conductor shapes for use in a small transmitting loop antenna. The conclusion – already known from past theoretical literature, and verified here by simulation – is that a large-diameter, round-cross-section tubular conductor can be replaced with a square-cross-section conductor whose cross-sectional width and height correspond to the tube’s diameter, with negligible change in RF resistance. A new finding is that electrically isolating the four flat faces of the square-cross-section conductor from each other has negligible effect only if the four faces are perfectly aligned to form a perfectly-square cross-section; if, however, the four sides are slightly misaligned to form a distorted approximation of a square, then the RF resistance significantly increases due to uneven distribution among the isolated faces of the RF current which cannot redistribute itself due to the inter-face isolation. Electrically connecting the four flat faces of the square-cross-section conductor may, depending on the conductivity of the connecting material, reduce the RF resistance, by allowing current redistribution and compensating for an imperfect shape of the square cross-section. However, using tin (solder) as a connecting material was found to offer no benefit over leaving the faces disconnected.

Background: Round-cross-section vs. Square-cross-section conductors

As mentioned in a previous post (http://qrp-gaijin.blogspot.com/2017/09/a-1-meter-diameter-small-transmitting.html), H. A. Wheeler’s 1955 article  "Skin Resistance of a Transmission-Line Conductor of Polygon Cross Section" already established that:

If a conductor cross section is any straight-sided polygon that can be circumscribed on a circle, it is found to have the same skin resistance as a conductor whose cross section is this circle. For example, a square wire has the same resistance as a round wire of the same radius, though the square perimeter is 4/pi times as great.

To verify this, a number of Finite-Element-Method computer simulations were performed with the FEMM software.

Analytically computed, the RF resistance Rloss at 7 MHz of a 3.14 meter length (corresponding to a loop antenna of 1 meter diameter) of round copper conductor having 10 cm diameter is 6.81 milliohms, as can be verified by the formulas and calculator at https://chemandy.com/calculators/round-wire-ac-resistance-calculator.htm. The identical result can also be found by assuming an equal current distribution along the entire conductor surface area (as will be the case if the conductor has a round cross-section and is not influenced by the magnetic fields of nearby conductors) and computing the RF sheet resistance of the conductor (https://www.microwaves101.com/encyclopedias/sheet-resistance) using the spreadsheet at https://www.microwaves101.com/uploads/RF-Sheet-Resistance-Rev-7.xls. Using that spreadsheet, the skin depth is computed as 24.60 microns, the surface resistivity per square for a conductor of 5 skin depths thickness is 0.685 milliohms, and for a sheet conductor of 3.14 meters length and 0.314 meters width, the number of squares on that sheet is 10 (3.14m / 0.314m), so the total RF sheet resistance is 10 squares * 0.685 milliohms/square = 6.85 milliohms, which is essentially identical to the previously-computed result of 6.81 milliohms.

Analytically computed, the radiation resistance Rrad of a single-turn, 1 meter diameter loop antenna is approximately 5.7 milliohms, using the approximate formula 31171*(A/lambda^2)^2 (reference:  https://www.ece.mcmaster.ca/faculty/nikolova/antenna_dload/current_lectures/L12_Loop.pdf). Various small-loop calculators also return this same value of radiation resistance for a 1 meter diameter loop (reference: https://owenduffy.net/blog/?p=1693).

Therefore, assuming no other losses, the efficiency is Rrad / (Rrad + Rloss) = 5.7 / (5.7 + 6.85) = ~45%. However, the very low Rrad means that milliohm-level increases in Rloss can have a large effect on efficiency. For example, if Rloss increases by 5 milliohms, then the efficiency drops to 5.7 / (5.7 + 11.85) = ~32%. If Rloss increases by 10 milliohms, the efficiency drops to 5.7 / (5.7 + 16.85) = ~25%.

Given this background, the following FEM simulations were performed.

Simulated RF resistance of a solid, round-cross-section copper conductor

For this first simulation, a solid copper conductor was modeled. FEMM allows modeling a cross-sectional slice of the conductor, which is then effectively rotated about the Z axis to form a loop conductor. The complex impedance of the conductor can be computed at a given frequency, and the real part of the impedance gives the resistive losses. It is also possible to plot the flux density and current density, although this data is not extensively used in this article. We focus mainly on the computed resistive loss at 7 MHz, which includes skin effects and also proximity effects.

Conditions: Mesh resolution was set to 4 microns in a 200-micron-thickness shell at the outer edge of the copper conductor. Higher resolution was not possible due to memory limitations. Mesh resolution was set to automatic (coarser resolution) in the interior of the copper conductor, and in the air region exterior to the conductor. Simulation domain consisted of 3955864 Nodes and 7911456 Elements.

The following image shows the entire simulation domain. The circle in the middle represents the round cross-section of the conductor, which is placed at a distance (radius) of 50 cm from the central vertical axis. The simulation then effectively rotates this cross section around the vertical axis, forming a loop conductor 1 meter in diameter. The several semi-circles at the edge of the problem domain represent the boundary conditions, which were automatically generated by FEMM.

The following images zoom in successively closer to the copper conductor, showing how the problem has been configured with the highest simulation density near the surface of the conductor, to capture skin effects.

 

 

 

 

Result: RF resistance of 527.96 milliohms was computed. This is an implausible result and is too high. Examination of the simulated current density shows very high current density in the interior of the conductor, which we know analytically to be incorrect. This incorrectly-computed interior current flow, probably computed inaccurately due to the coarse mesh resolution in the interior conductor region, reduces the computed current flow in the outer shell region (which should actually be carrying all of the current), which ultimately leads to a wrong overall current density and a wrong RF resistance value. 

Simulated RF resistance of a hollow, round-cross-section copper conductor

For this next simulation, the interior region of the conductor was replaced with non-conducting air, to prevent any current flow in this region. The only conductive region is the 200-micron copper shell, forming a thin-wall conductor. Since the skin depth at 7 MHz is ~25 microns, the 200-micron shell thickness covers more than 5 skin depths, which is generally considered sufficient for practically minimum RF resistance. 

Conditions: Mesh resolution was set to 5 microns in the 200-micron-thick copper shell. Mesh resolution was set to automatic (coarser resolution) in the air region exterior to the conductor. Simulation domain consisted of 2546258 Nodes and 5092242 Elements.

Result: RF resistance of 7.18 milliohms computed. This is in good agreement with the analytical result of 6.85 milliohms. The RF current is concentrated in the outer-most layers of the thin shell, as expected.

Simulated RF resistance of a thinner-shelled, round-cross-section copper conductor

For this next simulation, the copper shell thickness was reduced from 200 microns to 100 microns. This corresponds to only four skin depths at 7 MHz, not the typically-recommended 5 skin depths.

Conditions: Mesh resolution was set to 3 microns in the 100-micron-thick copper shell. Mesh resolution was set to automatic (coarser resolution) in the air region exterior to the conductor. Simulation domain consisted of 3553075 Nodes and 7105877 Elements.

Result: RF resistance of 7.19 milliohms was computed. This is again in good agreement with the analytical result of 6.85 milliohms. The reduction in shell thickness from 200 microns to 100 microns results in only a negligible change in RF resistance.

Simulated RF resistance of a thin-shelled, almost perfectly square-cross-section copper conductor

For this next simulation, the copper shell shape was changed from round to almost perfectly square. Width was 100.1 mm and height was 100.4 mm. The extra height is to accommodate inserting an isolating air gap between the flat faces, for the next simulation. Shell thickness was 100 microns. To avoid computation errors and unrealistically high current crowding at sharp corners, the 90-degree corners  of the square cross section were changed into rounded curves to smooth out the current flow to more realistic values (since perfectly sharp corners do not exist in physical conductors).

The following image shows the square-cross-section conductor.

The following image shows how the sharp corners of the conductor have been rounded, to avoid simulation anomalies.

Conditions: Mesh resolution was set to 3 microns in the 100-micron-thick copper shell. Mesh resolution was set to automatic (coarser resolution) in the air region exterior to the conductor. Simulation domain consisted of 4601765 Nodes and 9203200 Elements.

Result: RF resistance of 7.09 milliohms was computed. The change of conductor cross section from round to square results in only a negligible change in RF resistance. The value is still in good agreement with the analytical result of 6.85 milliohms.

Simulated RF resistance of a thin-shelled, almost perfectly square-cross-section copper conductor with isolated walls

For this next simulation, the four faces or walls of the square-cross-section conductor were disconnected from each other by inserting small air gaps at each corner. Because the current ideally should flow longitudinally along each face and along the circumference of the loop, the air gaps – which are in line with the current flow – should theoretically not hinder the current flow and should not change the RF resistance. The air gaps only prevent the current from flowing cross-wise from one face to another adjoining face. However, normally we would expect that current need not flow cross-wise from one face to another face.

Conditions: Mesh resolution was set to 3 microns in the 100-micron-thick copper shell. Mesh resolution was set to automatic (coarser resolution) in the air region exterior to the conductor. Simulation domain consisted of 4595785 Nodes and 9191231 Elements.

The following image shows how an air gap has been inserted between adjoining sides of the square-cross-section conductor.

Result: RF resistance of 7.27 milliohms was computed. The change of isolating the adjoining conductor faces by insertion of air gaps (in line with the direction of the current flow) results in only a negligible change in RF resistance. The value is still in good agreement with the analytical result of 6.85 milliohms.

Simulated RF resistance of a thin-shelled, distorted square-cross-section copper conductor with isolated walls

For this next simulation, the four isolated faces or walls of the square-cross-section conductor were distorted from the perfect square shape by randomly moving each end of each edge by a few millimeters in the x and y directions. A real physical construction of such a square-cross-section conductor may have small imperfections in the square cross-sectional shape, and the purpose of this simulation was to see if these imperfections would cause an increase in RF resistance.

Conditions: Mesh resolution was set to 3 microns in the 100-micron-thick copper shell. Mesh resolution was set to automatic (coarser resolution) in the air region exterior to the conductor. Simulation domain consisted of 4521864 Nodes and 9043451 Elements.

The following image shows the distorted, approximately-square cross-section of the conductor, with the adjacent sides still disconnected by an air gap.

Result: RF resistance of 12.27 milliohms was computed. This is a significant increase in RF resistance. A perfectly square cross-sectional shape leads to perfectly balanced current distribution among four faces, even if those faces are electrically isolated from one another. However, a distorted cross-sectional shape leads to a poorer current distribution among the faces, which, due to the inter-face isolation, can no longer redistribute itself more evenly among the faces. The resulting poor current distribution reduces the amount of the conductor that is used to carry current, hence increasing the RF resistance.

Simulated RF resistance of a thin-shelled, distorted square-cross-section copper conductor with thin copper connectors between adjoining walls

For this next simulation, the isolated faces of the distorted square-cross-section conductor were connected with each other. The hope is that by electrically connecting the isolated and misaligned faces, the poor current distribution – caused by the distorted shape and misaligned faces – can somewhat balance itself out again if the current is allowed to flow cross-wise from one misaligned face onto an adjoining misaligned face. The air gaps between the corners of the faces were bridged with a copper conductor. Again, sharp edges were changed to round edges to prevent simulation anomalies. 

Conditions: Mesh resolution was set to 3 microns in the 100-micron-thick copper shell. Mesh resolution was set to automatic (coarser resolution) in the air region exterior to the conductor. Simulation domain consisted of 4824805 Nodes and 9649327 Elements.

The following image shows how the air gaps at the corners of the misaligned sides have been bridged with a copper conductor.

 

Result: RF resistance of 8.12 milliohms was computed. This is lower than the 12.27 milliohms of the distorted conductor with disconnected faces. Therefore, we can conclude that connecting the misaligned faces to allow cross-wise current flow between adjacent misaligned faces does improve the current distribution and does reduce the RF resistance. However, the resulting RF resistance of this misaligned and connected square-cross-section conductor is still higher than the 7.09 milliohms of the perfectly-aligned and connected square-cross-sectional conductor. So even with inter-face connections, the distortion of the cross-sectional shape still has a small, detrimental effect on the RF resistance.

Simulated RF resistance of a thin-shelled, distorted square-cross-section copper conductor with thin stainless steel connectors between adjoining walls

For this next simulation, the copper conductors that bridge the gaps between the misaligned faces were replaced with stainless steel conductors. The background to this idea is that in a physical construction of a square-cross-section conductor, the four flat faces (flat sheets of copper) can be laid separately on the outside of a square-cross-section plastic tube. This leaves four air gaps at the corners, with each gap running along the length of the square tube, and with each gap separating the the edges of two adjoining flat copper sheets. These air gaps can then be soldered for electrical connectivity. However, the electrical resistance of solder is higher than that of copper. For simulation purposes, it could be appropriate to use tin as a conductor to simulate the higher resistance of solder. However, the FEMM software does not offer tin as part of its material library. As a subsitute, 304 stainless steel (which is provided with the default FEMM material library, with conductivity specified as 1.45 MS/m) was used as a connecting material in the simulation. Steel has a much higher resistance than copper or tin, and so should provide an upper bound on the RF resistance. The expectation was that since the connecting material’s area is so small compared to the main copper conductors’ area, the impact on RF resistance should be negligible. 

Conditions: Mesh resolution was set to 3 microns in the 100-micron-thick copper shell. Mesh resolution was set to automatic (coarser resolution) in the air region exterior to the conductor. Simulation domain consisted of 4825265 Nodes and 9650247 Elements.

The following image shows how the connecting corner conductor, that bridges the gap between the misaligned vertical conductor and misaligned horizontal conductor, has been changed to a steel conductor, while the main conducting areas are still configured as copper.

Result: RF resistance of 32.41 milliohms was computed. This is contrary to expectation and seems unusually high. Introducing the stainless steel connectors has actually increased the RF resistance far above the RF resistance that occurs when the faces are left unconnected. The presence of stainless steel seems to be worse than no connection at all.

Examination of the current density showed extremely high current concentration in the stainless steel regions. The cause appeared to be the insufficient thickness (100 microns) of the stainless steel conductor. 100 micron thickness covers 4 skin depths of copper at 7 MHz, but it covers less than 1 skin depth in stainless steel at 7 MHz, because the skin depth of stainless steel at 7 MHz is approximately 161 microns. To cover 5 skin depths in stainless steel requires approximately 805 microns of thickness, so another simulation was run with thicker stainless steel connectors.

Simulated RF resistance of a thin-shelled, distorted square-cross-section copper conductor with thick stainless steel connectors between adjoining walls

For this next simulation, the thickness of the conductor thickness in the stainless steel connecting areas was increased to approximately 900 microns, to cover 5 skin depths in stainless steel at 7 MHz. After thickening the connectors, the sharp corners were rounded. Copper conductors were left at 100 microns thickness. The expectation was that the current flow in the steel areas should be essentially maximal, due to the 5-skin-depth thickness, and that the overall impact on RF resistance should be small, since the circumferential length (around the conductor’s cross-sectional perimeter) occupied by the steel connectors is small compared to the circumferential length occupied by the copper conducting surfaces.

Conditions: Mesh resolution was set to 3 microns in the 100-micron-thick copper shell. Mesh resolution was set to 30 microns in the approximately 900-micron-thick stainless steel connecting areas, to encompass 4 to 5 skin depths of stainless steel at 7 MHz. Mesh resolution was set to automatic (coarser resolution) in the air region exterior to the conductor. Simulation domain consisted of 4825560 Nodes and 9063542 Elements.

Results: RF resistance of 29.65 milliohms was computed. This is only a small improvement over the previous RF resistance (with thinner steel connectors) of 32.41 milliohms. Therefore, the thickness of the stainless steel connectors does not seem to be the primary cause of the increased RF resistance.

One explanation may be that the very presence of the stainless steel itself is causing the increased RF resistance. An anecdotal report that stainless steel greatly increases RF resistance can be found here: https://www.eham.net/community/smf/index.php/topic,83113.0.html.

Another alternative explanation may be that the simulation resolution in the stainless steel area (30 microns) is too coarse to compute an accurate resistance. It is not possible, in the above simulation, to increase the simulation resolution in the thick stainless steel areas, because then the mesh size exceeds the available memory. To investigate this hypothesis further, it should be possible to run a separate simulation with a thin-shelled (5 skin depths) stainless steel conductor, once with 3 micron resolution, and once with 30 micron resolution. The mesh size can be minimized to fit within available memory by reducing the conductor diameter as necessary. Then, the RF resistance can be compared between the 3-micron-resolution run and the 30-micron-resolution run. If the computed RF resistance is significantly different, this indicates that a 30-micron resolution is too coarse.

Simulated RF resistance of a thin-shelled, distorted square-cross-section copper conductor with thin aluminum connectors between adjoining walls

For this next simulation, the connecting areas were changed from stainless steel to aluminum (which is provided with the default FEMM material library, with conductivity specified as 24.59 MS/m). The problem geometry was copied from the previous test case with thin copper connectors, meaning that all conductors and connectors were set to 100 micron thickness and sharp corners were rounded in the same way as in the previous test case with thin copper connectors. 100 micron thickness should be enough to cover 3 skin depths in aluminum at 7 MHz.

Conditions: Mesh resolution was set to 3 microns in the 100-micron-thick copper shell and in the 100-micron-thick aluminum connectors. Mesh resolution was set to automatic (coarser resolution) in the air region exterior to the conductor. Simulation domain consisted of 4825560 Nodes and 9650835 Elements.

Result: RF resistance of 9.71 milliohms was computed. This is higher than the 8.12 milliohms RF resistance when using thin copper connectors, but lower than the 29.65 milliohms RF resistance when using thick stainless steel connectors. It is also lower than the 12.27 milliohms RF resistance when the faces are disconnected from one another.

Simulated RF resistance of a thin-shelled, distorted square-cross-section copper conductor with thin tin connectors between adjoining walls

For this next simulation, the connecting areas were changed from aluminum to tin, to simulate the resistance of ordinary solder. The FEMM default material library does not provide a tin material. Examination of other default FEMM material library entries for copper, aluminum, and stainless steel revealed that the only parameter that is necessary to define a new material is the material's conductivity in MS/m. The conductivity of tin is 8.7 MS/m, and a new FEMM material was created with this conductivity, leaving all other material parameters as-is. The problem geometry was copied from the previous test case with thin copper connectors, meaning that all conductors and connectors were set to 100 micron thickness and sharp corners were rounded in the same way as in the previous test case with thin copper connectors. A 100 micron thickness covers slightly more than one skin depth (64.49 microns) of tin at 7 MHz.

Conditions: Mesh resolution was set to 3 microns in the 100-micron-thick copper shell and in the 100-micron-thick tin connectors. Mesh resolution was set to automatic (coarser resolution) in the air region exterior to the conductor. Simulation domain consisted of 4825560 Nodes and 9650835 Elements.

Result: RF resistance of 12.84 milliohms was computed. This is higher than the RF resistance when using copper or aluminum connectors, but is lower than the RF resistance when using thick stainless steel connectors. However,  the RF resistance is actually higher than the 12.27 milliohms obtained when the faces are disconnected from one another. Even higher total RF resistance would be expected if the resistivity of the solder is actually higher than that of tin,as for example shown in the resistivity table at https://owenduffy.net/antenna/conductors/loss.htm. 

Summary of simulated RF resistance values

The following FEM simulation results were obtained for the RF resistance of a 10 cm diameter conductor of various configurations. These should be compared with the analytical result of 6.85 milliohms RF resistance at 7 MHz for a solid round conductor with 10 cm diameter. 

• Solid 10 cm round conductor: 527.96 milliohms (wrong result)
• Hollow 10 cm round conductor with 200 micron shell: 7.18 milliohms
• Hollow 10 cm round conductor with 100 micron shell: 7.19 milliohms
• Hollow 10 cm square conductor with 100 micron shell: 7.09 milliohms
• Hollow 10 cm square conductor with disconnected 100 micron shell: 7.27 milliohms
• Hollow 10 cm square conductor with disconnected and misaligned 100 micron shell: 12.27 milliohms
• Hollow 10 cm square conductor with misaligned 100 micron shell connected by thin copper connectors: 8.12 milliohms
• Hollow 10 cm square conductor with misaligned 100 micron shell connected by thin stainless steel connectors: 32.41 milliohms
• Hollow 10 cm square conductor with misaligned 100 micron shell connected by thick stainless steel connectors: 29.65 milliohms
• Hollow 10 cm square conductor with misaligned 100 micron shell connected by thin aluminum connectors: 9.71 milliohms 
• Hollow 10 cm square conductor with misaligned 100 micron shell connected by thin tin connectors: 12.84 milliohms

Conclusions

For a 1-meter-diameter loop antenna at 7 MHz, the low radiation resistance of 5.7 milliohms requires a correspondingly low-loss conductor to achieve reasonable efficiency. With a 10 cm diameter conductor, a square-cross-section conductor – which in some respects is easier to fabricate than a round-cross-section conductor – can provide sufficiently low RF resistance.

However, small details of the conductor geometry can cause significant changes in the overall RF resistance. This is important because fabricating a 10 cm-wide square-cross-section conductor by hand, by separately assembling several flat sheets of copper, will inevitably lead to small misalignment among the faces and distortions from the ideal, perfectly-shaped, square cross-section.

If the flat sides of the square cross-section conductor are galvanically isolated from one another, then small misalignment, on the order of a few millimeters, of the flat sides of the square-cross-section conductor can lead to significant increases of RF resistance.

If the misaligned flat sides of the square-cross-section conductor are connected to each other with copper, the RF resistance decreases to acceptable levels (only slightly above that of the perfectly-aligned case) because the current is allowed to flow cross-wise from one misaligned face to an adjoining misaligned face, which will improve current distribution and reduce the conductor’s RF resistance.

However, if the misaligned flat sides of the square-cross-section conductor are connected to each other with low-conductivity stainless steel, the RF resistance significantly increases to unacceptably high levels -- higher than if the faces were left unconnected.

If an aluminum conductor is used to connect the misaligned flat sides, then the RF resistance lies between that of copper connectors and that of stainless steel connectors.

If a tin conductor is used to connect the misaligned flat sides, then the RF resistance is comparably high to the case where the flat sides are left disconnected.

Therefore, the material used to connect the misaligned flat sides of the square-cross-section conductor will have an influence on the overall RF resistance. In practice, tin solder, with a resistivity similar to that of tin, will be the most likely material used to connect the faces. In the above simulations, using tin solder to connect misaligned faces was found to have no benefit over leaving the misaligned faces disconnected. 

Further studies with varying amounts of misalignment would be needed to clarify whether there are any cases where soldering misaligned faces would bring any benefit, but the current results suggest there will be minimal, if any, benefit. 

A likely more beneficial strategy would consist of two points. First, attempt to avoid misalignment of the cross-sectional shape from a perfect square or perfect circle; this can be done with a rigid former and/or with thick copper that is less subject to deformation, but both of these measures increase the weight and the difficulty of working with the materials. Second, avoid separating the faces in the first place, by avoiding the use of four separate copper sheets to form four separated flat faces of one tube, and instead rolling a continuous copper sheet into a large-diameter tube. This again is more difficult to construct (since a single and large copper sheet must be rolled to yield a perfectly circular or square cross-sectional shape), but likely will yield the lowest RF resistance for the conductor. This roll-up construction would still leave one unconnected seam along the tube's length. The effect of this seam is likely minimal, but a simulation should be performed to confirm this.