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10,000pF Capacitor Tank Circuit

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chiques

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Hello,
I need to build a tank circuit using Muratas GCM55D7U3A103JX01 at 13MHz. I noticed the capacitor has ~1.2 Ohm of impedance at this frequency.

If I put this in my simulator I get a 'blah' resonance.
1654562525996.png


Any tips on how to overcome this and get it to resonate with at least 20-30dB of return loss?
 

Have you never heard that \[ Q=R\sqrt{\frac{C}{L}} \] for Parallel Tank Circuit ?
And this \[ BW=\frac{f_{o}}{Q} \]
 

Have you never heard that \[ Q=R\sqrt{\frac{C}{L}} \] for Parallel Tank Circuit ?
And this \[ BW=\frac{f_{o}}{Q} \]
Yeah but how does that overcome a 1.2 Ohm load?
 

Yeah but how does that overcome a 1.2 Ohm load?
This is not Parallel Resistance to the Capacitor, it's Magnitude of the Impedance. You have to click [R] tab to able to know the "Series Resistance" that is 0.004 Ohm@13MHz.
 

Using the s-parameters GCM55D7U3A103JX01_series.s2p and Coil Crafts 08HQ16N.S2P I get a slight resonance. Below are what I believe are my calculations. I don't see how they correlate.

1654576367457.png

1654576379184.png
 

In order to find exact Resonance Frequency and Quality Factor, you should use S21.View attachment Quality Factor Testbench.png
View attachment Q Factor Testbench Results.png

As you can see, the Resonance Factors are almost same but the Quality Factor varies depending on Loading Condition. Since your circuit will be loaded by a charge, you have to consider Loaded Quality Factor.
There are 4 types of Quality Factor Definitions here.
First- Unloaded Quality Factor that is compatible with the Theory
Second- Similar one but obtained by S11
Third- The Real Quality Factor loaded by 50 Ohm
Fourth- The Quality Factor by Impedance ( Useless but gives an insight )
 

* Try different L:C ratios. It's common for the inductor to be 1,000 to 1 million times the capacitor value.

* Steepness of response curve is affected by level of current going through the LC network.
 

Hello,
I need to build a tank circuit using Muratas GCM55D7U3A103JX01 at 13MHz. I noticed the capacitor has ~1.2 Ohm of impedance at this frequency.

If I put this in my simulator I get a 'blah' resonance.
View attachment 176684

Any tips on how to overcome this and get it to resonate with at least 20-30dB of return loss?
so you are deliberately using a junky capacitor?
what is the question? Do you expect to make a high Q resonator with that?
Use two of these 5000 pf caps:

what is the q of your inductor?
 

The parallel LC resonator is open circuit at resonance while each reactance cancels the other at what 1.2 Ohms If you want a high Q amplifier or a passive BPF , the requirements are different but the load reistance is the same which determines the Q by the R/X ratio. You must choose the load R or Q and L to produce the gain and resonace at 13 MHz.

from my simulation , L = 50 nH which is an air coil of about 5 cm of magnet wire in a helical shape around a coarse threaded bolt which you can tune gently. use rigid heavy copper wire so that it's resistance is << 1% of 1.2 Ohms which is the Q of the inductor and must be much higher than the Q of the filter for best results and put inside a square tin shield grounded but at least the radius away from the wire. Thats < 10 mOhm and you can look up AWG tables to estimate it's resistance. The Cap must be metal film which are stable but have inductance in the wire so the SRF must be >> 13 MHz to work well. Obviously at ~ 0.8 nH/mm to 1nH/mm short means 1 mm lead length.

For a 50 Ohms in series from a voltage source the Q = 50/1.2 for 7% ohm , it's higher,. Trying to go much higher poses limitations on your ability to all the above and control temperature drift of wire inductance but metal film caps are pretty stable.

I would try the 10 nH coil with a PN2222A or better transistor in common emitter mode to a 5V supply with Ic =50 mA and R=100, L= 50 nH +/- the tolerance of the capacitor (10%?) Then you can AC couple some narrow 50 mA pulses and see it resonate at 50 mA* 100 ohms= 5Vpk or 10Vpp. This may elevate temperature (Vrms^2/R* 200'C/W )= Vrms^2 *2 for a TO92 http://www.falstad.com/afilter/circuitjs.html?cct=$+0+0.000005+5+50+5+50 %+0+180609918.59465617 l+560+368+560+304+0+1.5000000000000002e-8+0 c+512+368+512+304+0+1e-8+0 w+512+304+560+304+0 r+448+304+512+304+0+100 g+512+368+512+384+0 g+560+368+560+384+0 170+448+304+416+304+2+20+4000+5+0.1 O+560+304+608+304+0

The above is the passive filter response and below is the time domain with a 100 kHz pulse of low duty cycle from 3.3 V. You can adjust it anyway you want. https://tinyurl.com/27ufgd58
 
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