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.