Ideal parallel LC tank has infinite impedance at resonance frequency. However, real components, especially inductors, have unignorable ESR, causing ideally infinite impedance dropping to a finite value. This also cause the LC tank resonant frequency to be lower.

Below is my calculation.

\$\omega_0\$ is resonant angular frequency. \$Z_{in0}\$ is impedance(resistance) when resonance. finding Z0 and w0

According to my calculation, the resonant angular frequency is

$$ \omega_0 = \sqrt{ \frac{L-(R^2)\ C}{(L^2)\ C} } $$

The resonant impedance is

$$ Z_{in0} = \frac{L}{RC} $$

According to my simulation, the result seems correct.

I use 1 V AC signal. The "I(V1)" is current provide by the voltage source. There is 180° phase shift because the current is measured from voltage source positive to negative (from LC tank to voltage source).

$$ \mathrm{R = 40 \Omega , \ L = 10 \ mH , \ C = 100 \ nF. \\ \omega_0 = 31.369\ krad/s \Rightarrow f_0 = 4.992 kHz\\ R0 = 2.5 \ k\Omega } $$

simulation I use these values because I have these components. Maybe I will do the experiment in real circuit.

My question is, how to find out the Q factor, bandwidth, and half-power frequency of the circuit? I have tried to calculate by myself, but it ended up I have no idea what to do next. finding w1 and w2 Is it possible to find half-power angular frequency from the crazy equetion? Or I have made some mistake in some steps?

  • \$\begingroup\$ Go here at this subsection. \$\endgroup\$ Jun 30, 2023 at 13:18
  • \$\begingroup\$ @periblepsis I have ready read this. At first I didn't know that Q factor of the inductor is Q factor of the whole circuit. Thanks to Andy aka telling me that. \$\endgroup\$
    – Willis Lin
    Jun 30, 2023 at 18:23

2 Answers 2


My question is, how to find out the Q factor, bandwidth, and half-power frequency of the circuit?

For your tuned circuit (with inductive ESR), the Q factor of the whole circuit is exactly the same as the Q factor for the lossy inductor. In other words, the Q factor of the circuit is defined by the Q factor of the lossy inductor: -

$$Q = \dfrac{\omega L}{R_{ESR}}$$

For \$\omega\$, you use the natural resonant frequency: \$\dfrac{1}{\sqrt{LC}}\$

If you think about what your circuit looks like "stand alone", all three components (R, L and C) are in series so, you treat the relationship as you would with a series tuned circuit for Q factor. Extract from wiki RLC circuits: -

enter image description here

Given that the Q factor equals resonant frequency divided by bandwidth, you can calculate upper and lower 3 dB points. This is also covered on the same wiki page linked to above: -

enter image description here

  • \$\begingroup\$ Just one comment (for clarification): It is correct that the Q-factor of the circuit is identical to the Q-factor of the inductor - to be clear: This is the pole-Q (Qp). With other words, it the quality factor of the pole (left part of the s-plane). However, the Qp is identical to the relative bandwidth (Q=wo/delta w) for a parallel loss-resistor only. This is because only in in such a case the numerator of the circuits function is pure imaginary (bandpass function). \$\endgroup\$
    – LvW
    Jun 30, 2023 at 10:52
  • \$\begingroup\$ Thank you very much. According to wiki, seem like Q = w0*L/R, not Q = 2pi*w0*L/R? And could you please briefly explain why w0 = 1/sqrt(LC) ? I see this en.wikipedia.org/wiki/RLC_circuit#Other_configurations here. The real resonance frequency is not the natural resonant frequency. Or this is the definition? \$\endgroup\$
    – Willis Lin
    Jun 30, 2023 at 11:01
  • \$\begingroup\$ @WillisLin (1) \$2\pi f\$ is equal to \$\omega\$. (2) \$2\pi\omega\$ is meaningless. (3) There are several definitions of resonant frequency and, one of them (the natural undamped resonant frequency) is \$\frac{1}{\sqrt{LC}}\$. (4) I think you have calculated the frequency where impedance is at a maximum and that is slightly different to the natural resonant frequency. (5) Or, you could have calculated the frequency at which the input impedance is purely resistive. All of them are close in a fairly resonant (high-Q) tuned circuit. \$\endgroup\$
    – Andy aka
    Jun 30, 2023 at 11:16
  • \$\begingroup\$ @Andy aka (1) I know 2pif = w. (2) I mean, seems like you mistyped 2piw in your above answer, at first equation :) (3) I got it. (4)(5) The impedance I calculated was the purely resistive impedance. \$\endgroup\$
    – Willis Lin
    Jun 30, 2023 at 13:08
  • \$\begingroup\$ @WillisLin oops so I did. I thought you were talking about the wiki quote. On its way to being fixed. Thanks. \$\endgroup\$
    – Andy aka
    Jun 30, 2023 at 13:32

When driven with an ideal current source, the shown circuit resembles the superposition of a 2nd-order lowpass and a 2nd-order bandpass. Both functions have the same denominator and, thus, the same pole position. This is confirmed by evaluating the circuits transfer function

Z(s)=(R+sL)/(1+sRC+s²LC)=N(s)/(1 + s/wpQp + s²/wp²).

Therefore, we cannot automatically apply the formalism which applies for an ideal bandpass (tank) circuit. For demonstrating the particular properties of the circuit I have simulated and calculated some typical parameters for two different loss resistors :

1) R=40 Ohm : fmax=5.03kHz; fp=5.03kHz; fo=4.99kHz; B=0.636 kHz; Qc=Qp=Q=7.91

2) R=200 Ohm: fmax=4.89kHz; fp=5.03kHz; fo=3.89kHz; B=3.2kHz; Qc=Qp=Q=1.58


fmax=f for max amplitude; fp=pole frequency from Z(s); fo=f for zero phase (resonance frequency); B=3dB bandwidth; Qp (pole Q from Z(s)); Qc=fmax/B;


1.) The difference between fmax and the pole frequency fp becomes visible for the larger loss resistor only. This shows the influence of the lowpass function on the bandpass response.

2.) In principle, this applies also to the resonance frequency fo. This frequency is smaller than fmax and fp - however, a remarkable difference can be observed for R=200 Ohm only.

3.) As expected, there is a remarkable difference in bandwidth B between both cases - and the same applies to the quality factor Q. It is, however, interesting to note that in both cases, the circuits Q-value equals with sufficient accuracy the pole-Q (Qp) which can be derived from the transfer function very easily.

4.) Hence, to find the circuits bandwidth B it is not necessary to calculate the 3-dB points from the transfer function. The bandwidth can be calculated using the pole-Q and the pole frequency fp from the transfer function: B=fp/Qp

Comment: For R=40 Ohm the difference in magnitude ("dynamic") between 0 Hz and fmax is app. 36 dB. For R=200 Ohm this dynamic is reduced to 10 dB and for R=400 Ohm it is only 0.5dB.

That means: Only for small values for the loss resistor R the circuit can be regarded as a bandpass (dynamic more than app. 30dB). For larger R values the lowpass influence gets larger and larger - and the circuit looses its bandpass character.


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