Quality Factor (AKA: Q) Explained (059)

aaah yes, there is Q which stands for Quality Factor!
In this video I am going to explain what Q is and point out the basic things that can cause the Q to be less than stellar.
What is Q?
How does it relate to components?
How does it relate to resonant circuits?
How do I calculate it?
=========================
Time Markers for Your Convenience
----------------------------
00:05 Introductory Comments
01:05 Q on the Component Level
01:06 What is it?
02:44 Calculating it
04:44 Q and the Resonant Circuit
04:48 What is it?
05:11 The Series Resonant Circuit
06:43 The Parallel Resonant Circuit
08:05 Calculating it
08:17 Finding the Bandwidth of a Resonant Circuit
08:20 EXAMPLE #1
08:23 The Bandwidth
09:47 Calculate the Q
10:29 EXAMPLE #2
10:43 The Bandwidth
11:21 Calculate the Q
12:00 Concluding Comments and Toodle-Oots

[@thed8229]

I understand Q better than I did 12mins ago. Mission accomplished!

[@eie_for_you]

Woohoo! Good news! 🙂

[@RobLudwick]

I love your scene transitions, btw. :) "Meet me at camera 2"

[@eie_for_you]

Thanks! I'm working at trying to develop some more variety for some of my future videos. 🙂

[@34u90vutjwifjklSFJKJKSDjs9]

Thank you for the simple explanation! Q finally makes sense.
The only thing I am unsure of is whether or not the two definitions for Q are mathematically identical, ie. is X/R = f0 / BW ? Or is this an approximation or generalization.

[@eie_for_you]

Well...just thinking about it ... one is used for components and the other for a resonant circuit. At resonance the total effective value for X is zero. So, this raises the question, "What about the 3 dB points? I just did a mathamatical exercise to answer this. A parallel tuned circuit with a Q of 625 per the fo/BW method (fo=50MHz; BW=0.08MHz). If we calculate the Q using X/R at the 3dB point we get a Q of 1.24 (Z=9963.56+12373.65j). Good question! But it doesn't work out that way.

[@nathanherling9836]

I appreciate your 'turn to the camera' segues. .. .

[@eie_for_you]

Thanks! (as he turns to the camera and smiles)🙂

[@N6WLM]

I was thinking about making some of the NVARC “Ugly” Filters DIY bandpass filters. Sadly, the capacitors they recommend in the build are no longer sold. The instructions often referenced component Q when picking parts, so I was very confused how to determine the Q of a part to find replacements. This is very useful information, thank you! The Ugly Filters encompass many topics you've covered on your channel, it would make for a great video!

[@eie_for_you]

One way that I got a feel for the relative Q of various capacitors was to set up a quicky representative resonant circuit. I put this series resonant circuit (made with a coil I wound on some PVC pipe + my test capacitor) in series between port 1 and port 2 of my VNA and did a through (S21) measurement. This is not a way to actually measure the Q of the capacitor, but you can try various capacitors and definitively see which one provides the best, high Q, response relative to its cousins. With this you should see a peak in the response as the series resonant circuit becomes a short at resonance.
This feels like an interesting possibility for a short video ... you think?

[@NickFrom1228]

So many things make sense now. I learn more in one video on this channel than multiple other sources. Huge thanks!

[@eie_for_you]

You are very welcome, Nick! 🙂

[@W1RMD]

Best description of 'Q" yet. Thanks for making this so simple and straight forward.

[@eie_for_you]

Thank you and you are welcome! 🙂

[@davidluther3955]

AS USUAL AN EXCELLENT EXPLANATION!ONE WAY OF CALCULATING THE RESONANT FREQUENCY IS TAKING THE GEOMETRIC MEAN OF THE LOW AND HIGH FREQUENCIES OF THE BANDWIDTH.

[@eie_for_you]

In the simplest cases, yes. I have also seen nearby (downstream or upstream) resonant circuits or circuitry "pull" the bandwidth edge of a resonant circuit so this didn't apply. 🙂

[@tsharathchandra7074]

This is by far one of the best practical explanations I have ever seen. Thank you

[@eie_for_you]

Thank you and you are welcome! 🙂

[@d46512]

For God so loved the world that he gave his only begotten son, that whosoever believes in him should not perish, but have eternal life. ✝

[@eie_for_you]

Best news in the whole world!

[@factify2022]

❤

[@SpinStar1956]

Great video!
Great use of graphics and clear explanation!
Thanks Ralph!
73...

[@eie_for_you]

Thanks! 🙂

[@aasm-f4l]

Thank you so much it is very easy to understand with practical explanation hoping for more such content 🙏🏻

[@eie_for_you]

Thanks! I'm glad it was helpful. 🙂

[@SeyedAgent47]

Perfect job... Please continue thanks👍

[@eie_for_you]

Thanks! That is my plan!😁

[@abderraouflalla]

Thank you for the clear explaining sir.

[@eie_for_you]

You are very welcome! 🙂

[@anthonyselby8337]

thank you very good explanation as a novice hobbyist trying to rap my head around electronics this video actually explains it very well

[@eie_for_you]

Thank you! I am so glad that this helped you understand what Q is. 🙂

[@yt4krist0f]

Simple and perfect explanation..! Thanks!

[@eie_for_you]

Thanks!

[@tahirsongelen7781]

tnc ur expl and efford BR fm istanbul

[@eie_for_you]

Thanks!

[@oldblokeh]

Thanks, Ralph. I'd be grateful if you could say a few words on loaded Q, which is a term often heard.

[@eie_for_you]

I've been thinking about this very thing myself, knowing that I did not address it in my video. There is the Q of a resonant circuit when it is just sitting there all by itself, enjoying the influx of energy and minding its own business. But, generally speaking, resonant circuits don't exist floating out in space by themselves. Something is drawing energy from them to, say, and antenna or another circuit. This very action changes the Q of the circuit. This new Q is referred to as "loaded Q." Hope this makes sense.

[@oldblokeh]

@@eie_for_you Yes, it does make sense, thanks.

[@eie_for_you]

@@oldblokeh I am glad to hear that. 😄

[@Festus2022]

Nice video. I have a dumb question. Why are you decibel numbers negative? Thanks1

[@eie_for_you]

Glad you found the video helpful. The dB values are all negative because I was talking about a passive circuit where the resonant circuit is the output impedance. Being a voltage divider, the voltage gain will always be a negative dB number. Hope this makes sense. :-)

[@Festus2022]

Thanks. I will dwell on that answer.
@@eie_for_you

[@eie_for_you]

@@Festus2022 Just to give you food for thought... Vout/Vin = Rout/(Rout+Rin);
Rout/(Rout+Rin) is always less than 1.
dB = 20*log(Vout/Vin)
if Vout/Vin is less than 1, then the log (Vout/Vin) will be negative.
Hope that helps the noodling. 🙂

[@Festus2022]

Thanks again!! Appreciated.@@eie_for_you

[@mrhahmad1]

Tqvm...

[@eie_for_you]

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