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Series Resonant Circuit in a 50 Ohm System

chiques

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Hello Forum Peeps,

I have a series resonant circuit which I have simulated and built on an actual grounded coplanar waveguide. My question is regarding the Q factor of the circuit. In the case of the LTSpice simulation, I used generic (ideal) lumped resistor, inductor and capacitor. I added this resistor only because my signal generator has a 50 Ohm impedance (not sure if this is correct) although nonexistent on the physical circuit.

1683095796660.png


I’m trying to wrap my head around the fact that even in an ideal world, my circuit Q will only be as high as 9.1.

1683095840885.png


1683095848550.png


Am I using the correct R or is this more like the DCR of the inductor?

For example: CoilCrafts 1812CS-472XJR is specified to have a DCR of 5.4 Ohm and a Q of 63.

1812CS-472XJR_ 4.7 @ 7.9 MHz 5 63 @ 50 MHz 115 5.4 230

https://www.coilcraft.com/getmedia/677afbbb-f368-4f0b-b59e-ab9befcb1f0e/1812cs.pdf

What about the capacitor, won’t that affect the Q factor the circuit as well?

I’m asking this because I’m currently using a high current pulse inductor PA4548.472NLT (seen on my board) but the datasheet does not show Q. I noticed many high current inductors such as this do not have a specified Q. Is this because they are not intended to be used in RF? I'm guessing the Q is so bad they don't want to publish it..???


My goal is to get a better return loss than -12.8dB

1683095890998.png
 
Changed .net syntax, works with old LTspice version too. Need series-parallel circuit to match 5.4 ohm resistance. Unfortunately matching is much more sensitive to parameter variations due to higher Q.

1683630623599.png
 

Attachments

  • rlc-series-parallel.zip
    826 bytes · Views: 73
Apparently a bug in LTspice .net statement calculation. The simulation setup follows LTspice examples, but it gives completely wrong results.

Present at least in Version 4 bis 17.0.35 (July 2022). Fixed in newest versions. I should have followed my intuition that the parallel resonant circuit can't directly match 50 ohms (post #9). I'll come back if I have time to redesign the matching.

Sorry for the confusion, I must confess I never experienced a serious LTspice bug like this.
--- Updated ---

Turns out that the bug was detected by LTspice users in 2012 https://groups.io/g/LTspice/topic/50200398, but at that time Mike Engelhardt claimed it's not a bug, altough the setup follows LTspice directions. Now it's fixed in V17.1
I'm using 17.1.8. This does look like a complicated bug. Thanks for sharing.
--- Updated ---

Both seem good now.
1683655540860.png

1683655561271.png

This format of .param and .step was separated on mine. I switched over to this syntax
1683655604053.png


I also updated C1 to be the variable {Cs}
 

Attachments

  • rlc-series_1.zip
    326.1 KB · Views: 64
  • 1683655525310.png
    1683655525310.png
    172.5 KB · Views: 67
Last edited:
Final question: does suggested matching circuit what you want?
It does except for a couple of items I couldn't figure out.
  • Why was a shunt matching component chosen instead of a series?
  • Why am I getting a different voltage out of the source in the simulation than my calculation?
Here’s a summary of what I’ve done. I hope it’s correct :/

I started with this:
1683870848195.png


I removed the resistor and used the 50 Ohm series feature in V1. This increases the current across the elements but has a horrible match where ~ 50% of my signal is reflected (on my bench test).


1683870868753.png



I also needed to add a resistor that represents the inductor DCR and the ESR.

This version has a bad match between the source and load:
1683870899036.png


The impedance looking into the RLC circuit is ~30Ohm
1683870931128.png

Visual separation of the source and load
1683870951713.png

To match the 30 Ohm to the 50 Ohm source, I needed a matching component. To find that matching component the impedances should be in phasor format. For simplicity, the circuit had some slight adjustments. Change C1 to 30.5pF and C2 to 674.5pF (this is shown in the phasor calculation below)

Calculating the load using phasors (all in series):
1683870969683.png

The slight circuit modifications change the load impedance to 13.3 Ohms
1683870988740.png

A 674.5pF capacitor has a conjugate impedance of ~-17.5i which seems good shown by the –66dB of return loss in the simulation. Adding C2 has a source impedance of ~-17.5i Ohms in shunt matches the load and source. 1) Is there a logical reason why a shunt C2 was chosen instead a series C2?
1683871016880.png

1683871026491.png

Good return loss:
1683871049997.png


2) I’m trying to wrap my head around the voltage source calculation vs the simulation. If I perform a transient simulation of the matched circuit, I measure the source putting out ~1.9A/15.4V even though I have it set to 100V amplitude. I interpret the ‘Amplitude’ of a signal being the peak voltage/Irms.


This should be ~100V, not 15.4V.

{I hit my screenshot limit here so no more graphics}

The calculations of supply that is capable of providing 90W has ~95V/1.9A (pk)
 
i am not sure i would call that "coplanar waveguide". the two ground planes are a mile away from the center conductor. it is likely acting more like microstrip.

but if it WERE coplanar waveguide, you would need to connect the ground side of capacitor C1 to BOTH top ground planes!

also, depending on the frequency, those transmission line lengths really start to matter!
 

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