- I am guessing a “trap network” is a series resonant circuit, but I’m not sure.
YES, IT IS A SERIES TRAP NOT A PARALLEL TRAP BUT IT IS IN SHUNT WITH THE OUTPUT, CH1. SO THIS IS A S21 MEASUREMENT. IN AFFECT YOU ARE USING THE VNA AS A SPECTRUM ANALYZER-VOLTMETER AND TAKING A MEASUREMENT OF THE NOTCH DEPTH. IF THE SERIES R OF THE LC NETWORK UNDER TEST IS SMALL, SAY ON THE ORDER OF MILLI-OHMS, YOU WOULD BE SEEING Q's ON THE ORDER OF 300 OR GREATER.
THE MEASUREMENT PROCESS IS JUST THAT OF A VOLTAGE DIVIDER.
- I don’t know what constitutes a “high Q variable capacitor” which your method requires.
IF THE Q OF YOUR COIL IS ASSUMED TO BE 300 OR MORE... YOU ARE GOING TO NEED TO FIND A C LARGER THAN THAT! AIR VARIABLES LIKE BC BAND ARE PRETTY GOOD AND THEIR ARE SOME CERAMIC CAPS THAT HAVE Q'S GREATER THAN 500... SEE AVX CAPS
- When X is 0 at resonance, I get that Rs is all that is left of impedance… but why is that in parallel with the termination resistance?
AGAIN, RS IS IN SHUNT WITH THE LOAD, 50 OHMS, BUT ITS VALUE IS SO SMALL, THAT THE NOTCH DEPTH CREATED IS THAT OF A VERY SMALL SHUNT R. THE END RESULT IS THE NOTCH DEPTH IS DIRECTLY CORRELATED TO THE Rs OF THE LC NETWORK AT RESONANCE.
- Indeed, is “termination resistance” simply the built-in resistance of the meter?
YES, 50 OHMS.
- What is a “notch” and why is there an attenuation depth when series resonance provides a lack of attenuation?
SERIES RESONANCE PROVIDES A LACK OF ATTENUATION IN THE SERIES MODE. BUT THIS HAS IT CONFIGURED IN THE PARALLEL MODE!
- Couldn’t I simply measure the series resonant circuit with the VNA and get a Rs reading directly, but would it be any better than what I get off the coil alone?
YES, BUT THE NUMBERS ARE SOME ARE DIFFICULT TO DISCERN. THIS IS AN INDIRECT WAY TO GET TO A NUMBER WITH SOME IMPROVED ACCURACY. KEEP IN MIND, IF THE FIXTURE LOSSES OR R VALUES ARE NOT SMALL, YOU WILL NOT OBTAIN CORRECT ANSWER.
Parallel Q is the same as series Q at a single frequency. Google the concept of a series to parallel conversion. There you will find the proof and the math. It is not hard, but will require your familiarity of complex numbers.
