# Q-factor in an LRC series circuit

Last weekend BINSIC drove me mad in a desperate rush to finish. This weekend I have been obsessed by this problem, but have finally worked it out.

The Q factor measures the width of a signal at half the resonant power (it is a dimensionless quantity and measures lots of things, but we’ll stick with this for now).

For an AC LRC series circuit, at half power the impedance equals $\sqrt{2}R$ and so the reactances equal R or minus R, i.e.,

$X_L - X_C = R$ at the high frequency end and (1)

$X_L - X_C = -R$ at the low frequency end. (2)

Looking at the angular frequencies, this gives for (1)

$\omega_hL - \frac{1}{\omega_hC} = R$

and for (2)

$\omega_lL - \frac{1}{\omega_lC} = -R$

Subtract (2) from (1) and we get

$2R = L(\omega_h - \omega_l) + \frac{1}{C}(\frac{1}{\omega_l} - \frac{1}{\omega_h})$

Now $\omega_h - \omega_l$ is the bandwidth and we will label this $\Delta$ for ease of exposition.

And we can see that $\frac{1}{\omega_l}-\frac{1}{\omega_h}$ can be expanded to:

$\frac{\omega_h-\omega_l}{\omega_h\omega_l} = \frac{\Delta}{\omega_h\omega_l}$

And (here’s one of the parts I stumbled on), $\omega_h\omega_l$ is the square of the resonant frequency $\omega_0$ (as this is a geometric and not arithmetic mean).

So now we have:

$2R = L\Delta + \frac{\Delta}{C{\omega_0}^2}$

and so $2R = \Delta(L + \frac{1}{C{\omega_0}^2})$ and $\Delta = \frac{2RC{\omega_0}^2}{LC{\omega_0}^2 + 1}$

Now, $Q = \frac{\omega_0}{\Delta}$, so…

$Q = \omega_0\frac{LC{\omega_0}^2 + 1}{2C{\omega_0}^2R}$

Now, ${\omega_0}^2 = \frac{1}{LC}$ (from the resonant frequency definition, see the previous blog)

And so we get $Q = \frac{1}{\sqrt{LC}}\frac{2}{\frac{2CR}{LC}} = \frac{1}{R}\sqrt{\frac{L}{C}}$, the definition you’ll see in a textbook.