0
\$\begingroup\$

I am trying to calculate the quality factor for the response curve of a series RLC circuit. I measured the voltage across the capacitor at several frequencies and calculated its ratio to the forcing voltage. The curve I obtained is as follows: enter image description here

How is it possible to find quality factor when there no points where the curve reach -3db on the left side of the peak?

\$\endgroup\$
3
  • \$\begingroup\$ The Q factor of a series resonant circuit is simply XL / R (at resonance). \$\endgroup\$ Commented Feb 23, 2019 at 14:54
  • \$\begingroup\$ You have a VERY LOW Q. \$\endgroup\$ Commented Feb 23, 2019 at 15:14
  • 2
    \$\begingroup\$ You might try measuring voltage across the resistor rather than capacitor. I get component values of \$ C \approx 1.6nf \$ and \$ L \approx 0.011 H \$ for \$R=2193 \$ \$\endgroup\$
    – glen_geek
    Commented Feb 23, 2019 at 15:23

2 Answers 2

2
\$\begingroup\$

\$Q=\dfrac{1}{2\zeta} = \sqrt{A_{pk}^2 + 1}\$ for peak gain Apk

When Q is the quality factor of resonance gain approaches 1 or less, we switch to damping factor as resonance becomes dampening factor is more relevant.

Rule of Thumb: We approximate high Q to be just the resonant gain for Q>>1.

But in your case, Q is very low, and the peak/flat gain = 1.25

If gain, Apk=1.25 then Q = 1.6 , or ζ = 1/3.2 This is your answer from reading graph.

Other useful formulae for 2nd order RLC filters depend if in series or parallel resonant mode.

\$X_C(\omega_o)=\dfrac{1}{\omega_oC}\$ and \$X_L(\omega_o)=\omega_oL\$

Consider a series LP filter.

schematic

simulate this circuit – Schematic created using CircuitLab

\$\omega_o=\dfrac{1}{\sqrt{LC}}=2\pi*f\$

Parallel resonant circuits for high Q, Rp >> X(ωo) the impedance of both reactances: are low and equal so the current cycles between these and very little loss current in Rp. (underdamped)

  • e.g. L=10uH, C=10uF Rp=100 , thus Q=100 at ωo=1e-5 f=√2*10kHz=14.14kHz

For series resonant circuits: \$Q = \frac{1}{Rs} \sqrt{\frac{L}{C}}\$
\$Q=\dfrac{\omega_o}{BW_{-3dB}}\$ for Bandpass filters (BPF)

More math background on 2nd order systems

enter image description here

Normal for high Q we simplify this to just voltage or current peaking linear gain which you can convert to dB.

Another way is lookup on any "RLC Impedance Nomograph chart" and instantly find the unknown for any of the other knowns for L, C, Q, f, Z(f) using the log-log-log 3D chart like a slide rule. For examples see here.

When Q=ζ=0.707, it is the common Butterworth filter, which also has a critically damped response with no overshoot or "unity" Voltage gain as shown above.

For a more mathematical explanation read here. from U of Carlton.

Useful Q benchmarks and for fun

Q's of < 50 are easy in LC or active filters and >200 hard due to tolerances and thermal drift. However for Q > 10k it is easily possible with any piezo-crystal resonator. (MEMs, tuning fork and AT cuts) But with some Quartz SC-cut the Q => 100k only at one temperature regulated by an oven, (OCXO). Cesium clocks it is even higher, the phase noise is poor so they lock SCt OCXO to the Cesium clock.

This Q factor thus is related to noise BW, jitter, phase shift and the crystal has equivalent RLC paramters with L in the mH to Henries and C in the fempt-farads with ESR [Ohms] and a load capaitance [pF].

64k$ question

What is the Q of an atom? ( hint it is the only lossless thing in nature)

\$\endgroup\$
12
  • \$\begingroup\$ How did you come up with this equation for Q? \$\endgroup\$ Commented Feb 23, 2019 at 17:44
  • \$\begingroup\$ I did it from memory, but this may help you cnx.org/contents/Cy752CuW@4/Second-Order-Filters \$\endgroup\$ Commented Feb 23, 2019 at 17:49
  • \$\begingroup\$ make sense yet? \$\endgroup\$ Commented Feb 23, 2019 at 17:59
  • \$\begingroup\$ Sim falstad.com/afilter/… \$\endgroup\$ Commented Feb 23, 2019 at 22:52
  • \$\begingroup\$ Tony Stewart: I have some problems with the first line (Q=f(Apk). Example 1: Apk=1 (Butterworth) gives Q=1.414 (wrong); Example 2: Apk=1.414 (Chebyshev with 3dB ripple) gives Q=1.732 (wrong). \$\endgroup\$
    – LvW
    Commented Oct 18, 2020 at 11:52
1
\$\begingroup\$

It all depends on the DEFINITION of the Q-value.

  • For second-order LOWPASS functions (as in the described example) there is only one single definition: The Q-value is the POLE QUALITY FACTOR Q=Qp, which is defined based on the pole location sp in the s-plane.

Hence, the 3dB points of the magnitude plot are not relevant for for the Q-value of a lowpass.

  • The same applies to second-order BANDPASS functions - however, this pole quality Qp is idenbtical to the Q value defined as Q=fo/BW (mid frequency devided by 3dB-bandwidth).

Example: Obviously, the problem description shows a second-order lowpass and the Q-value is the pole-Q which can be found using the expression: 2Qp²=wp²/(wp²-wm²) with pole frequency wp and frequency wm at peak magnitude.

\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.