Quality Factor and Bandwidth of Series Resonant Circuits

In this article, you’ll learn about Q factor and bandwidth of series resonant circuits. This article is linked with 7 other articles you might be interested in reading:

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A series resonant circuit looks like a resistance at the resonant frequency. (Figure below) Since the definition of resonance is XL=XC, the reactive components cancel, leaving only the resistance to contribute to the impedance. The impedance is also at a minimum at resonance. (Figure below) Below the resonant frequency, the series resonant circuit looks capacitive since the impedance of the capacitor increases to a value greater than the decreasing inducitve reactance, leaving a net capacitive value. Above resonance, the inductive reactance increases, capacitive reactance decreases, leaving a net inductive component.

At resonance the series resonant circuit appears purely resistive. Below resonance it looks capacitive. Above resonance it appears inductive.

Current is maximum at resonance, impedance at a minumum. Current is set by the value of the resistance. Above or below resonance, impedance increases.

Impedance is at a minumum at resonance in a series resonant circuit.

The resonant current peak may be changed by varying the series resistor, which changes the Q. (Figure below) This also affects the broadness of the curve. A low resistance, high Q circuit has a narrow bandwidth, as compared to a high resistance, low Q circuit. Bandwidth in terms of Q and resonant frequency:

          BW = fc/Q
          Where  fc = resonant frquency
                 Q = quality factor
A high Q resonant circuit has a narrow bandwidth as compared to a low Q

Bandwidth is measured between the 0.707 current amplitude points. The 0.707 current points correspond to the half power points since P = I2R, (0.707)2 = (0.5). (Figure below)

Bandwidth, Δf is measured between the 70.7% amplitude points of series resonant circuit.
BW = Δf = fh-fl = fc/Q 
           Where fh = high band edge,  fl = low band edge 
           fl = fc - Δf/2 
           fh = fc + Δf/2
           Where  fc = center frequency (resonant frequency)

In Figure above, the 100% current point is 50 mA. The 70.7% level is 0.707(50 mA)=35.4 mA. The upper and lower band edges read from the curve are 291 Hz for fl and 355 Hz for fh. The bandwidth is 64 Hz, and the half power points are ± 32 Hz of the center resonant frequency:

           BW = Δf = fh-fl  = 355-291 = 64 
           fl = fc - Δf/2 = 323-32 = 291 
           fh = fc + Δf/2 = 323+32 = 355 

Since BW = fc/Q:

           Q = fc/BW = (323 Hz)/(64 Hz) = 5

Article extracted from Lesson in Electric Circuits AC Volume Tony R Kuphaldt under Design Science License. Heading and title are modified/added. Additionally, the content is visually edited and rearranged for purpose of ease in skimming.

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