Quality Factor (AKA: Q) Explained (059)

[@thed8229]

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

[@eie_for_you]

Woohoo! Good news! 🙂

[@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! 🙂

[@W1RMD]

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

[@eie_for_you]

Thank you and you are welcome! 🙂

[@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! 🙂

[@nathanherling9836]

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

[@eie_for_you]

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

[@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. 🙂

[@acestudioscouk-Ace-G0ACE]

Thanks very much for this explanation of Q. You presented it in an easy to understand way and I'm grateful.

[@eie_for_you]

You are very welcome, my friend! 🙂

[@SpinStar1956]

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

[@eie_for_you]

Thanks! 🙂

[@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. 🙂

[@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?

[@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. 🙂

[@eemtech1606]

Excellent explanation, Sir. Hat off to you!

[@eie_for_you]

Thank you so much! :-)

[@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! 🙂

[@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.

[@yt4krist0f]

Simple and perfect explanation..! Thanks!

[@eie_for_you]

Thanks!

[@jashimuddin612]

Very helpful, thanks for your video

[@eie_for_you]

You are very welcome, my friend! 🙂

[@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

[@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. 😄

[@garysewell5465]

Great 🙂

[@eie_for_you]

Thanks! 🙂

[@mrhahmad1]

Tqvm...

[@eie_for_you]

???????????

[@Matt-pd2cq]

Isn't wider bandwidth good?

[@eie_for_you]

It depends entirely on what you are trying to do, actually. Like receiving CW (Morse Code), you want a very narrow bandwidth so that adjacent signals do not interfere with what you are trying to receive. On the other hand, listening to an A.M. signal, you want much wider bandwidth because the signal is wider. Notice, too, that the difference is the height of the resonance, not the width of the base. 🙂

[@tahirsongelen7781]

tnc ur expl and efford BR fm istanbul

[@eie_for_you]

Thanks!

[@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]

❤

[@allennebu7478]

@@eie_for_you yes

[@eie_for_you]

@ True that! 🙂

[@factify2022]

Thank you for reminding me this verse