# Matching antenna at 4.2 GHz using an LC network

I want to match an antenna using an LC matching network:

Operation frequency: 4.2 GHz
Vacuum wavelength: 7.2 cm

Cable: Source Impedance: 50 Ω\

The antenna impedance is unknown, but it is approximately a full wavelength and I think it should be around 100 Ω at 4.2 GHz.

My mathematica code (below) tells me I would need a variable capacitor and a variable inductor that i can tune around 80pF and 3 nH. This is very low and I am having trouble finding parts that can be tuned around these values.

What is the best way to do this? Is the LC network a good idea?
Can I just put enough capacitors in series and enough inductors in parallel to lower the effective capacitances and inductions? Does that work, or will I run into some unforeseen issue? In the end the signal going to the antenna should be on the order of a Watt.\ Do you know of a vendor where I can find the right parts?

freq=4200*10^6;
\[Omega]=2\[Pi]*freq;
sourceImpedance=50;
antennaImpedance=75-I*100;
LImpedance[\[Omega]_,L_]:=I*L*\[Omega];
CImpedance[\[Omega]_,C_]:=1/(I*C*\[Omega])

LParallelCSeriesImpedance[\[Omega]_,antennaImpedance_,L_,C_]:=(1/(antennaImpedance+CImpedance[\[Omega],C])+1/LImpedance[\[Omega],L])^-1;
LSeriesCParallelImpedance[\[Omega]_,antennaImpedance_,L_,C_]:=(1/(antennaImpedance+LImpedance[\[Omega],L])+1/CImpedance[\[Omega],C])^-1;
LParallelCSeriesImpedanceAlt[\[Omega]_,antennaImpedance_,L_,C_]:=(1/antennaImpedance+1/LImpedance[\[Omega],L])^-1+CImpedance[\[Omega],C];
LSeriesCParallelImpedanceAlt[\[Omega]_,antennaImpedance_,L_,C_]:=(1/antennaImpedance+1/CImpedance[\[Omega],C])^-1+LImpedance[\[Omega],L];
ReflectionRatio[sourceImpedance_,finalImpedance_]:=(finalImpedance-sourceImpedance)/(finalImpedance+sourceImpedance);

solutions=NMinimize[{Norm[ReflectionRatio[sourceImpedance,LParallelCSeriesImpedance[\[Omega],antennaImpedance,L,C]]],L>0,C>0,L<10^-7,C<10^-8},{L\[Element]Reals,C\[Element]Reals}];
Plot[Norm[ReflectionRatio[sourceImpedance,LParallelCSeriesImpedance[omega*(2\[Pi])*10^9,antennaImpedance,L/. solutions[[2]],C/. solutions[[2]]]]],{omega,4,4.4},AxesLabel->{"\[Omega] (GHz)","|\[CapitalGamma](\[Omega])|"},PlotRange->All,PlotLabel->StringForm["Subscript[C, opt]=pF \n Subscript[L, opt] = nH",(C/. solutions[[2]])/10^-12,(L/. solutions[[2]])/10^-9],ImageSize->Large]
solutions=NMinimize[{Norm[ReflectionRatio[sourceImpedance,LSeriesCParallelImpedance[\[Omega],antennaImpedance,L,C]]],L>0,C>0,L<10^-7,C<10^-8},{L\[Element]Reals,C\[Element]Reals}];
Plot[Norm[ReflectionRatio[sourceImpedance,LSeriesCParallelImpedance[omega*(2\[Pi])*10^9,antennaImpedance,L/. solutions[[2]],C/. solutions[[2]]]]],{omega,4,4.4},AxesLabel->{"\[Omega] (GHz)","|\[CapitalGamma](\[Omega])|"},PlotRange->All,PlotLabel->StringForm["Subscript[C, opt]=pF \n Subscript[L, opt] = nH",(C/. solutions[[2]])/10^-12,(L/. solutions[[2]])/10^-9],ImageSize->Large]
solutions=NMinimize[{Norm[ReflectionRatio[sourceImpedance,LSeriesCParallelImpedanceAlt[\[Omega],antennaImpedance,L,C]]],L>0,C>0,L<10^-7,C<10^-8},{L\[Element]Reals,C\[Element]Reals}];
Plot[Norm[ReflectionRatio[sourceImpedance,LSeriesCParallelImpedanceAlt[omega*(2\[Pi])*10^9,antennaImpedance,L/. solutions[[2]],C/. solutions[[2]]]]],{omega,4,4.4},AxesLabel->{"\[Omega] (GHz)","|\[CapitalGamma](\[Omega])|"},PlotRange->All,PlotLabel->StringForm["Subscript[C, opt]=pF \n Subscript[L, opt] = nH",(C/. solutions[[2]])/10^-12,(L/. solutions[[2]])/10^-9],ImageSize->Large]
solutions=NMinimize[{Norm[ReflectionRatio[sourceImpedance,LParallelCSeriesImpedanceAlt[\[Omega],antennaImpedance,L,C]]],L>0,C>0,L<10^-7,C<10^-8},{L\[Element]Reals,C\[Element]Reals}];
Plot[Norm[ReflectionRatio[sourceImpedance,LParallelCSeriesImpedanceAlt[omega*(2\[Pi])*10^9,antennaImpedance,L/. solutions[[2]],C/. solutions[[2]]]]],{omega,4,4.4},AxesLabel->{"\[Omega] (GHz)","|\[CapitalGamma](\[Omega])|"},PlotRange->All,PlotLabel->StringForm["Subscript[C, opt]=pF \n Subscript[L, opt] = nH",(C/. solutions[[2]])/10^-12,(L/. solutions[[2]])/10^-9],ImageSize->Large]

• You don't find 3nH inductors because a few mm of wire in a straight line has more than 3nH of inductance. Commented Jul 2 at 18:50