The transmission line need to be multiple of wavelength.2m is for dipole This sensor acts as a antenna, but almost no gain.
I found web site to calculate coplanar capacitor.
www.daycounter.com/Calculators/Plate-Capacitor-Calculator.phtml
Equation:
C= K*Eo*A/D, where Eo= 8.854x10-12
where:
K is the dielectric constant of the material,
A is the overlapping surface area of the plates,
d is the distance between the plates, and
C is capacitance
A (plate area) 20 (mm2)
d (distance) 0.1 (mm)
K(dielectric constant inside 4layer pcb) 4.2
Capacitance 7.44 (pF)
distance 0.05mm capacitance = 14.9pF
The closer the strips the long the range of electromagnectic field.
I also calculated the capacitive reactance impedance of this strips 10mm*2mm = 20mm2
http://www.sengpielaudio.com/calculator-RC.htm
For example 10pF at 100MHz has an resistance of 159ohm.
This two strips on pcb are used on vh400 sensor at 80MHz . Kind of 80MHz antenna with receiver do DC.
The two strips have 10pF equals to ~160ohm at 100MHz and AD8307 power detector has internal resistence 1,1Kohm.
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Oscillator output cmos can drive 500 ohm at 100MHz 3V rail to rail output.
The AD8307 has an resistence of 1,1Kohm
probe capacitance.
A (plate area) 20 (mm2)
d (distance) 0.1 (mm)
K(dielectric constant inside 4layer pcb) 4.2
Capacitance 7.44 (pF)
picture: electromagnetic field
How can I configure external R , C or L needed?
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This circuit is a power meter with AD8307
How can I redesign this RLC network to mach two pcb strips at 100MHz?
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LC ressonance calculator
www.daycounter.com/Calculators/LC-Resonance-Calculator.phtml
If frequence 100MHz and capacitance from probe 10pF inductance will be 253nH