This article discusses the kind of matching network in the following figure.

A common variant shows no capacitor… but for most loads, the capacitance is essential to its operation, even if it is incidental to the inductor or as often the case, supplied by the mounting arrangement of a vertical radiator tube to the mast.

A similar arrangement works for vertical radiators around a λ/2, around means from say 3λ/8 to 5λ/8. All of these present an impedance of some fairly high resistance in series with a significant reactance. Because the current at the feed point is relatively low, these are often used as so-called ground independent antennas.

Above is a Simsmith model of such an arrangement. The load L is from an NEC-4.2 model of a 50MHz 5λ/8 vertical over four λ/4 radials in free space.

The tapped coils is a uniform air cored solenoid of radius 10mm, pitch 3mm, 6t tapped at 33% (ie 2t).

C1 represents the equivalent capacitance of self resonance, equivalent mount capacitance, and possibly some additional capacitance.

Though it seems every ham can explain how this circuit works, there is nothing intuitive about the design of the tapped inductor.

The inductor design was assisted by a little python script which found the tap point and c1 for a given inductor, load, etc.

zl: (59.49-175.4j) pitch: 0.00300 m, radius: 0.01000 m, turns: 6.00, length: 0.01800 m, Ql: 300, Qc: 1000 VSWR=1.00009, tap=32.994 (%), c1=4.91618 (pF)

From that starting point, you could play with Simsmith interactively changing the coil parameters and adjusting things for a new match.

#!/usr/bin/env python # coding: utf-8 import math import cmath from scipy.optimize import minimize from scipy import constants f=50e6 zl=59.49-175.4j #network qc=1000 ql=300 #tapped coil p=0.003 r=0.01 n=6 def wcf(n,r,l): return constants.mu_0*n**2*r*(math.log(1+math.pi*r/l)+1/(2.3004+1.622*l/r+0.4409*(l/r)**2)) def vswr(args): tap,c1=args lt=wcf(n,r,l) l1=wcf(n*tap,r,l*tap) l2=wcf(n*(1-tap),r,l*(1-tap)) m=-(lt-l1-l2)/2 rlt=2*math.pi*f*lt/ql y=1/zl y=y+2*math.pi*f*c1*(1/qc+1j) zl2p=rlt*(1-tap)+2*math.pi*f*(l2-m)*1j z=1/y+zl2p y=1/z yl1p=1/(rlt*tap+2*math.pi*f*(l1-m)*1j) y=y+yl1p z=1/y zm=2*math.pi*f*m*1j z=z+zm y=1/z gamma=(z-50)/(z+50) rho=abs(gamma) vswr=(1+rho)/(1-rho) return vswr l=n*p #starting guess tap=0.5 c1=20e-12 x0=[tap,c1] res=minimize(vswr,x0,method='Nelder-Mead') print() print("zl:",zl) print('pitch: {:0.5f} m, radius: {:0.5f} m, turns: {:0.2f}, length: {:0.5f} m, Ql: {:0.0f}, Qc: {:0.0f}'.format(p,r,n,p*n,ql,qc)) print() print('VSWR={:0.5f}, tap={:0.3f} (%), c1={:0.5f} (pF)'.format(res.fun,res.x[0]*100,res.x[1]*1e12))

Above is the code, it elaborates the solution in small steps so that it is easier to understand. The script does not do much error checking, it will produce valid results on sane input.

If you play with the Simsmith model, you may find that there is a solution with c1=0 for some load impedances, but in reality, the incidental capacitances mean that c1 is unlikely to ever be exactly zero.

For the enquiring mind, the relevant files are attached: EFmatch02.zip.