Example 7: Three-line section from C and L matrices

Netlist: examples/three-line-CL.cir

Goal: use the general C/L input, which is what a 2-D field solver gives you, for an inhomogeneous three-conductor section. Then read the modal analysis and build a symmetric coupler with the centre line as input.

The matrices come from comb-over's field solver (manual §4.3) for three 0.8 mm microstrip lines with 0.15 mm gaps on 0.508 mm RO4350B (ε_r 3.66) with 1 oz (34.79 µm) copper:

node solver/cli.mjs --W 0.8,0.8,0.8 --S 0.15 --h 0.508 --er 3.66 --t 0.03479

#Input

cline TL3
  C [103.32058p -24.996611p -1.0254662p; -24.996611p 113.18589p -24.996611p; -1.0254662p -24.996611p 103.32058p]   # F/m (Maxwell matrix)
  L [308.1564n 103.46107n 43.654707n; 103.46107n 294.99833n 103.46107n; 43.654707n 103.46107n 308.1564n]   # H/m
  len 90deg
  A l_cpl c_in  r_cpl
  B l_iso c_out r_iso
end

Requirements on the matrices:

• C is the Maxwell matrix. The diagonal is each line's total capacitance with the others grounded; the off-diagonals are negative. If your solver reports ground and mutual capacitances instead, convert with C_ii = Cg_i + ΣCm_ij and C_ij = −Cm_ij.
• L is the inductance matrix. All entries are positive.
• Both are symmetric, and C must be positive definite.

The tool checks all three.

#The modal analysis

The Line sections panel shows:

How to read it:

• Number of modes. Three conductors give three quasi-TEM modes with three different velocities, which is the hallmark of an inhomogeneous medium.
• Mode 1 is the all-in-phase "even" mode. It has the most field in the substrate, so the highest ε_eff, and the highest impedance.
• Mode 2 is the antisymmetric outer-line mode. The centre line carries no voltage, because by symmetry it cannot couple to this mode.
• Mode 3 drives the centre line against the outer lines.

len 90deg is referenced to the mean of √ε_eff over the three modes, which gives 15.54 mm at 3 GHz. Each mode therefore sees a slightly different electrical length.

The characteristic admittance matrix Y_c = W diag(v) Wᵀ (§9 of the manual) is, in mS,

[ 19.473  −5.689  −0.389 ]
[ −5.689  21.946  −5.689 ]
[ −0.389  −5.689  19.473 ]

Note the small but non-zero coupling between the outer lines (−0.389 mS). The Cg/Cm nearest-neighbour model in the filter examples ignores this kind of term unless you add it with Cm(1,3).

#The coupler

Port 1 drives the centre line at end A. Port 2 is the centre line's other end. Ports 3/5 are the outer lines at the input end (coupled), and ports 4/6 are the outer lines at the far end (isolated).

• Symmetry. db(S31/S51) is zero to 2e−13 dB, because the structure is symmetric and the centre line feeds both outer lines equally.
• Directivity. It is only about 12 dB at f0, the same inhomogeneous-medium problem as in Example 2. Lumped compensation capacitors between the lines at each end would improve it.
• Match. S11 ≈ −27 dB at f0. The widths were picked from a coarse search for a match better than −20 dB with about −9 dB total coupling; they were not optimised.

#The same section from its geometry

The section can also be given by its geometry. The field solver then runs inside comb-over:

.substrate jlc-ro4350b-0.508
cline TL3
  geom msl
  W [0.8mm 0.8mm 0.8mm]
  S 0.15mm
  len 90deg
  A l_cpl c_in  r_cpl
  B l_iso c_out r_iso
end

This gives the same matrices: the substrate preset carries the copper thickness, so the solver gets the same rectangular strips. The geometry form also adds conductor and dielectric loss. At 3 GHz the modal Q is 168 for the in-phase mode, 127 for the outer-line mode and 102 for the centre-against-outer mode, which crowds current into both gaps. S21 goes from −0.76 dB to −0.82 dB.

#Try

• Change the widths in the geometry form, for example W [0.8mm 1.0mm 0.8mm], and watch the match and the coupling change. The centre-line width sets the input impedance.
• Multiply every entry of C by 1.2 and divide every entry of L by 1.2. The ε_eff values are unchanged and every Z_mode drops by 1.2. Watch S11 change.
• Set L = μ0ε0·C0⁻¹ for some air-filled C0 to make your own physically consistent pair. A field solver gives you C and C0 (C with the dielectric removed), and L follows exactly that way.
• Replace the end connections to make a three-line comb or interdigital resonator array. For example, A 0 0 0 and B r1 r2 r3 with loading capacitors gives an inhomogeneous comb-line, which the synthesiser does not yet cover.