Is the bandwidth of an LCR resonant circuit independent of the capacitance?

Is the bandwidth of an LCR resonant circuit independent of the capacitance, if the inductance and its series resistance are kept the same, and the Q is high?

This question and the calculation associated with it were originally posted by a radio amateur Anthony G3LAA, who deserves credit for spotting it.

At resonance the reactances of L and C are the same and Q is equal to that reactance divided by the series R of the inductor. If we take Q equal to f0/B where B is the bandwidth, and substitute the usual formulae for Q and the resonant frequency f0 in terms of L, C and R,then C cancels out and we are left with B proportional to 1/L. This is true, for all practical purposes (fapp) at moderate frequencies for both series and parallel circuits, and implies that changing the C in the LCR circuit, although this changes the resonant frequency, the bandwidth remains constant.

I realise that there are a lot of simplifying assumptions in the formulae, both electrical and mathematical, but, fapp, this looks correct to me. Please, am I right?

It has an obvious practical advantage in, say, a TRF receiver, with fixed inductance and a variable capacitor to do the tuning, keeping the bandwidth constant over the tuning range.

(The following is added in response to @Phil Frost's comment, thanks.)

Here is the maths — I assume that the formulae for $Q$, $\omega_0 = 2\pi f_0$ and bandwidth, are familiar enough not to need explanation, except that I use $B$ for bandwidth and brevity. I also accept that the formulae depend on many simplifying assumptions, but, fapp, they are good enough, for high $Q$, above about 10.

$$Q = \frac{1}{\omega_0RC} = \frac{\omega_0L}{R}$$

Reactances equal at resonance, $Q$ is reactance over $R$.

$$\omega_0=\frac{1}{\sqrt{LC}} \tag{1}$$

Substituting for $\omega_0$ in each $Q$ formula we get:

$$Q=\frac{\sqrt{LC}}{RC} = \frac{L}{R} \cdot \frac{1}{\sqrt{LC}}$$

Both of these simplify to

$$Q=\frac{1}{R}\sqrt{\frac{L}{C}} \tag{2}$$

For bandwidth $B$ (frequency difference at half power level) we have $Q=\omega_0/B$ or

$$B=\frac{\omega_0}{Q} \tag{3}$$

Substituting in (3) for $\omega_0$ from (1) and $Q$ from (2) we finally get

$$B = \frac{1}{\sqrt{LC}}\cdot R\cdot\sqrt{\frac{C}{L}}$$

which simplifies to $B=R/L$ i.e. Bandwidth is independent of the capacitor value over the range of resonant frequencies for the inductor.

Is this correct? I now think so and will give that as an answer

• Where have the other comments gone? – Harry Weston Feb 26 '14 at 16:36

1 Answer

I am now convinced that the maths is correct,

Thank you @Kevin, for tidying it all up. I was going to transfer the maths here as an answer, but I will leave it in the question