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I have been trying to find the quality factor of this RLC circuit for some time now. My method is trying to find the equivalent series RLC circuit and calculate the resonant circuit. When I work through the algebra and plug the values in I get values in the order of 10^9, so clearly my working is wrong.

It is an unfamiliar combination as the inductor and capacitor are kind of series and parallel at the same time and it is very confusing.

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    \$\begingroup\$ As Jan points out in an answer, the usual approach to answering your question is to develop the transfer function. (Preferably, into standard form, which Jan did NOT do. Had Jan done so I'd have +1 the answer.) Once in standard form, the answer just drops out as \$Q=R_1\sqrt{\frac{C_1}{L_1}}\$. But I can't point out where your thinking went wrong without any information about your thinking. \$\endgroup\$
    – jonk
    Feb 13, 2022 at 23:45

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Well, the transfer function of the circuit is given by:

$$\mathscr{H}\left(\text{s}\right):=\frac{\text{V}_\text{o}\left(\text{s}\right)}{\text{V}_\text{i}\left(\text{s}\right)}=\frac{\text{R}\space\text{||}\space\frac{1}{\text{sC}}}{\left(\text{R}\space\text{||}\space\frac{1}{\text{sC}}\right)+\text{sL}}=\frac{\text{R}}{\text{CLRs}^2+\text{Ls}+\text{R}}\tag1$$

Where \$\alpha\space\text{||}\space\beta:=\frac{\alpha\beta}{\alpha+\beta}\$.

Now, when working with sinusoidal signals we can use \$\text{s}:=\text{j}\omega\$ (where \$\text{j}^2=-1\$ and \$\omega=2\pi\text{f}\$ with \$\text{f}\$ is the frequency of the input signal in Hertz). So, we get:

$$\underline{\mathscr{H}}\left(\text{j}\omega\right)=\frac{\text{R}}{\text{CLR}\left(\text{j}\omega\right)^2+\text{L}\text{j}\omega+\text{R}}=\frac{\text{R}}{\text{R}\left(1-\text{CL}\omega^2\right)+\text{L}\omega\text{j}}\tag2$$

So, the absolute value if given by:

$$\left|\underline{\mathscr{H}}\left(\text{j}\omega\right)\right|=\frac{\text{R}}{\sqrt{\left(\text{R}\left(1-\text{CL}\omega^2\right)\right)^2+\left(\text{L}\omega\right)^2}}\tag3$$

The quality factor is given by:

$$\mathcal{Q}:=\frac{\hat{\omega}}{\mathcal{B}}\tag4$$

Where \$\hat{\omega}\$ is the frequency (in rad/sec) where \$\left|\underline{\mathscr{H}}\left(\text{j}\omega\right)\right|\$ is at the maximum and \$\mathcal{B}\$ is the bandwidth and is given by \$\mathcal{B}=\left|\omega_--\omega_+\right|\$ with \$\$ and \$\$ are the upper and lower cut-off frequencies (in rad/sec).


Now, we can solve for the unkowns using your values, we get:

$$\frac{\partial\left|\underline{\mathscr{H}}\left(\text{j}\omega\right)\right|}{\partial\omega}=0\space\Longrightarrow\space\omega:=\hat{\omega}=\frac{5000 \sqrt{1999838}}{9}\tag5$$

And:

$$\left|\underline{\mathscr{H}}\left(\text{j}\omega\right)\right|=\left|\underline{\mathscr{H}}\left(\text{j}\hat{\omega}\right)\right|\cdot\frac{1}{\sqrt{2}}\space\Longrightarrow\space\omega=\frac{5000}{9} \sqrt{2 \left(999919\pm9 \sqrt{1999919}\right)}\tag6$$

So, the quality factor is:

$$\mathcal{Q}=\frac{1}{9} \sqrt{\frac{999919 \left(\sqrt{999676013122}+999919\right)}{3999838}}\approx78.5611\tag7$$

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  • \$\begingroup\$ I understand the concept of the transfer function, but how did you calculate Vout and Vin, I can't seem to decipher it from equation (1) \$\endgroup\$ Feb 14, 2022 at 8:26
  • \$\begingroup\$ Also how did you go about doing the derivative of the transfer function with respect to the frequency. It makes sense since at resonance the curve flattens, but how did you go about doing this? \$\endgroup\$ Feb 14, 2022 at 8:27
  • \$\begingroup\$ @BorisvanGriensven Both questions have fairly common answers. If you take an EE course you will learn those things (at least I did). \$\endgroup\$ Feb 14, 2022 at 9:55

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