How to find lossy parallel LC tank Q factor, bandwidth, and half-power-frequency?

Question

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.

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).

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

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. Is it possible to find half-power angular frequency from the crazy equetion? Or I have made some mistake in some steps?

Answer 1

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: -

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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: -

Answer 2

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

Explanation:

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;

Evaluation:

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.