L+L EH measurements.

Unlike an L+T network, where a cable shield is directly connected to the bottom radiator of the EH dipole, thus forming only two input terminals, we have three input terminals for the L+L network. Using the same Wheatstone bridge as we have used already for the L+T network, we can measure an impedance between each pair of terminals then we can calculate components of a star or delta equivalent network. First, define these three antenna terminals as shown on the picture below.

Now we have a three terminal "Black box" to be examined and reconstructed as an RLC equivalent. Both star and delta networks are good for this, we start with the star network since we hope to find a fourth terminal hidden inside our black box. This terminal should represent a true ground point. Since we have three unknown impedances inside our circuit, we need at least three different measurements taken on its external terminals. The figure below show this.

This equivalent circuit is already good enough for practical usage: three capacitors and a single resistor only. However, recalling that our cable was close to a quarter of wavelength, we can split our "zc" impedance. Its real part (37) can represent a common mode radiation impedance of the feeder, while its imaginary part may be converted from star to delta with two other impedances, namely "za" and "zb". This is shown on the figure below.

All the calculations are made in Mathcad file  LL_mod0.mcd . The resulting equivalent circuit is intuitively clear, and its internal node "G" is close to ground potential. Each one of C_top and C_bot capacitors has actually a very small radiation resistance (say 0.04 Ohm)  in series, but it can be safely ignored. Now we are ready to connect this circuit with an L+L network and "tweak" it in SPICE model. The SPICE testbench schematic is shown below.

All SPICE node names are marked, a ground node (node 0 in SPICE) is a top end of the coax shield. The RL1 and RL2 resistors represent losses in the coils L1 and L2 respectively, while RV is set to 50 Ohm when we need a frequency response curve (or efficiency figure), and it is set to zero when we need a return loss plot (a voltage source V1 is actually connected between nodes 0 and NS).
Since we have measured all our figures at 16 MHz frequency, it would be a good idea to run our model not far from it. However, if necessary, the figures can be scaled to other frequency with reasonable accuracy. For our 16 MHz "band" we need L1 and L2 close to 3.4 uH (assuming also Q=200), we can sweep C1 and C2 in our model from 5 pF to 35 pF, we set our "frequency of interest" to 16100 kHz and we are looking for a maximum power going into our radiation resistance RC (37 Ohm). The model LL_pwr_0.sp is ready and running it we get a good peaking for C1 at 27 pF:

Using C1=27 pF we can now sweep C2  LL_pwr_1.sp :

This gives us C2=21.5 pF. Running two more iterations we arrive at C1=25.7 pF, C2=21.5 pF. Looking at overall frequency response  LL_pwr_2.sp we see a not too bad figure -2.22 dB at our frequency of interest 16100 kHz:

Removing a source internal resistor (RV) from our model we can plot a reflection losses ( LL_pwr_3.sp ):

For lower radiation resistance value losses in the L+L network become higher: setting RC=3.7 Ohm (instead of 37 Ohm) gives some -10 dB after retuning the circuit.
Now we can measure what we get if we try to use a two terminal parallel equivalent instead of the full three terminal one. Removing CT, CB, CM and RC and replacing them for RP and CP connected across the NT and NB nodes, we get another SPICE model:  LL_sbs_0.sp . Running it for RP=2 kOhm and CP swept from 1 pF to 4 pF we easily get CP=2.13 pF. Sweeping the RP now ( LL_sbs_1.sp ) we note two points, where the S11 value goes to -25 dB: 2.5 kOhm and 3.2 kOhm (with a dip at 2.84 kOhm). Looking at S11 plots ( LL_sbs_2.sp ) for the RP values close to 2.5 kOhm we get almost the same S11 dip (RP=2.55 kOhm), but more bandwidth:

After all the "magnitude" plots, lets look at phase. From  LL_pwr_2.sp (don't forget to set an "acout=0" option for Star-HSPICE) we get it:

A yellow curve on the plot above belongs to the differential voltage across a dipole, a red curve - to the common mode voltage. A phase difference at 16100 kHz frequency is about 40 degrees.

To be continued...