# Designing a high-$Q$ RLC circuit

I would like to make a series RLC circuit which will serve as RF receiver. I would like very high frequency selectivity, i.e. narrow bandwidth, i.e. $\pm$200 Hz.

The sharpness of a resonance is described by $Q$ which itself is equal to

$$Q = \frac{\omega_{0}L}{R} = \frac{1}{R \omega_0 C}$$ but also $$Q = \frac{\text{Energy Stored}}{\text{Energy Dissipated}}$$ but also $$Q = \frac{\omega_0}{B}$$ where $B$ is the bandwidth and $\omega_0$ is the resonance frequency.

I am confused which one to use for designing process. If I take the first formula, it looks simple at first - big $L$ small $R$ and that's all. However big inductors has also bigger resistance i.e my 1mH inductor has 50 Ohm resistance.

How should I design high selective RLC circuit which can be use as RF selector for example?

PS: I would apreciate any answer as well as resources recommendation :)

• Why do you want such a narrow bandwidth? – Farcher Mar 15 '18 at 14:35
• You're missing an R in the numerator for your Q expression I believe; it should be $R/(\omega_0\,C)$. Aside from this, you've gotten your thoughts straight; you've rightly identified of the problems of lumped circuit technology. Also, as @Farcher says: that is a curiously narrow bandwidth; what's your application and that is your RF carrier frequency? – Selene Routley Mar 15 '18 at 14:43
• The long history of amateur radio suggests that you have an impossible task. 200Hz roofing filters are definitely not lumped RLC elements. You might consider asking for suggestions over on the Ham SE. – Jon Custer Mar 15 '18 at 15:14
• @Farcher, I want as narrow response as possible +/- 200 Hz was just an example. – DannyS Mar 15 '18 at 18:10
• Yeah, get a lower resistance inductor. $50 \, \Omega$ is a lot. – DanielSank Mar 15 '18 at 18:44

The formula $$Q = \frac{\omega_0 L}{R}$$ is a good design formula. Say we want $Q \geq 10$ and a receiver frequency of $3.5 \, \text{kHz}$. Let's arbitrarily say we want the characteristic impedance of the tuned circuit to be $50 \, \Omega$. The characteristic impedance is given by $$Z = \sqrt{\frac{L}{C}}$$ and the resonance frequency is $$\omega_0 = \frac{1}{\sqrt{LC}} \, ,$$ so $$L = \frac{Z}{\omega_0} \qquad C = \frac{1}{\omega_0 Z}$$ with values $$L = 2.2 \, \text{mH} \qquad C = 9 \times 10^{-7} \text{F} \, .$$
Now using the very first formula, we have $$R = \frac{\omega_0 L}{Q} = \sqrt{\frac{L}{C}} \frac{1}{Q} < \sqrt{\frac{2.2 \, \text{mH}}{0.9 \, \mu\text{F}}} \frac{1}{10} \approx 5 \, \Omega \, .$$
Can we find a $\sim 2.2\, \text{mH}$ inductor with resistance less than or equal to $5 \, \Omega$? We surely can. Try the usual online electronics suppliers such as Digi-Key and Mouser.
• Thanks for very comprehensive answer. However I have couple of questions. First, where this equation $$Z = \sqrt{\frac{L}{C}}$$ comes from? Second, similary, how did you get equations for L and C. Using only resonant frequency equation I got $$L = \frac{1}{C \cdot \omega^{1/2}}$$ – DannyS Mar 15 '18 at 19:35
• @DannyS For your first question, look up characteristic impedance. The Wiki article is for distributed transmission lines, but the same principles work for lumped resonators. For your second question, just try evaluating $Z/\omega_0$ and $1/\omega_0 Z$ using the first two fomulas in the answer. Also, if you like an answer on this site, upvote it, and when you think an answer is acceptable, accept it. – DanielSank Mar 15 '18 at 19:38