I've been convinced that AEA's specification of 72% efficiency for the Isoloop at 14 MHz is too high, certainly too high for the MFJ antennas.  So I re-did the calculations using the 59% efficiency figure calculated below.  The new results can be downloaded here:

https://www.dropbox.com/s/ve1v49b3gjvmt64/MFJ-1786-1788_2.pdf?dl=0

If you don't want to download the (1-page) document with the pretty graph, here's a synopsis of the results:

Freq      Eff.     Gain with respect to a 1/2-wavelength dipole
MHz      dB      dBd
7.0         -9.5    -9.9
10.1       -5.1    -5.5
14.0       -2.3    -2.7
18.068   -1.1    -1.5
21.0       -0.7    -1.1
24.89     -0.4    -.8
28.0       -0.26  -0.65

My basic conclusions still stand.  With almost minus 10 dBd of gain on 7 MHz, the 40 meter coverage of the MFJ-1788 doesn't seem very useful.  That is confirmed by some of the reviews I have seen.  I think you'd get better results by just loading up the coax feedline as a random-wire antenna with a tuner.

The 10 MHz performance is a little better.  Good enough to at least allow you to get on the 30 meter band.

On the higher bands, the gain is within less than 3 dB of a full-sized dipole, which seems a useful trade-off for its small size and wide-band continuous coverage.

Disclaimer:  Again, I have never seen one of these things so this is all based on theory and on the many reviews I have read.  Even if my figures are off a bit, at least this gives an idea of the relative performance on the various bands.

Alan N1AL



On 1/18/2021 5:38 PM, Alan Bloom wrote:
Well let's see...

Radiation resistance of a small loop is 31,171 * (Area / wavelength^2)^2

For a loop with a 91cm diameter at 14 MHz, I believe that comes out to 0.064 ohms.

Assuming the loss is due to the RF resistance of the loop:

From the internet I get the volume resistivity and skin depth for 6063 aluminum is 0.03 microohms-meter and 23.3 micrometers respectively, so the surface resistivity is 0.03/23.3 = 0.0013 ohms per square.  The outside circumference of the tubing is PI * 1.05" = 3.3" and the loop length is PI * 36" = 113" so the loss resistance is .0013 * 113/3.3 = 0.045 ohms.

So I calculate an efficiency of 0.064 / (0.064 + 0.045) = 59%

So worse than AEA claimed, but in the ballpark.

Alan N1AL




On 1/18/2021 3:39 PM, Wayne Burdick wrote:
Hi Alan,

72% sounds a bit high. Is this number based on loop size alone ("in theory")? Or are they taking conductor geometry and other losses into account?

Wayne
N6KR


On Jan 18, 2021, at 2:05 PM, Alan Bloom <[email protected]> wrote:

MFJ makes a pair of small, remotely-tuned loop antennas, the MFJ-1786 that covers 10-30 MHz and the MFJ-1788 that covers 7 to 21+ MHz.  As far as I can tell, the two antennas are identical except for the size of the tuning capacitor.  Each consists of a 3 foot (91 cm) diameter loop made of aluminum tubing and a plastic housing that contains the tuning capacitor, motor, and coupling loop.  No control cable is required since the control voltage is sent from the control box in the shack to the motor in the antenna via the coaxial cable.

Before I purchase one of these I wanted to get an idea of the efficiency of such a small loop.  MFJ is silent on the subject so I did my own calculations.  The calculations and results are on a 1-page document that I uploaded to Dropbox and can be downloaded here:

https://www.dropbox.com/s/l8mv67cjrck2ssn/MFJ-1786-1788.pdf?dl=0

My calculations are based on the assumption that the efficiency of the MFJ antennas is similar to the (no longer manufactured) AEA Isoloop (my reasoning for that is in the document) and that AEA's specification of 72% efficiency at 14 MHz is correct.  From that number I can calculate the efficiency and gain on all the other bands.

If you don't want to download the document, here is a summary of the results:

Freq      Eff      Gain with respect to a half-wave dipole
MHz      dB      dBd
7.0        -7.3    -7.7
10.1      -3.5    -3.9
14.0      -1.4    -1.8
18.068  -0.6    -1.0
21.0      -0.4    -0.8
24.89    -0.2    -0.6
28.0      -0.15  -0.5

I'd be interested in any comments people may have on the accuracy of
my assumptions and calculations in the document.

Alan N1AL

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