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:

A series resonant circuit looks like a resistance at the resonant frequency. (Figure below) Since the definition of resonance is X_{L}=X_{C}, 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.

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

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 = f_{c}/Q Where f_{c}= resonant frquency Q = quality factor

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

BW = Δf = f_{h}-f_{l}= f_{c}/Q Where f_{h}= high band edge, f_{l}= low band edge f_{l}= f_{c}- Δf/2 f_{h}= f_{c}+ Δf/2 Where f_{c}= 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 f_{l} and 355 Hz for f_{h}. The bandwidth is 64 Hz, and the half power points are ± 32 Hz of the center resonant frequency:

BW = Δf = f_{h}-f_{l}= 355-291 = 64 f_{l}= f_{c}- Δf/2 = 323-32 = 291 f_{h}= f_{c}+ Δf/2 = 323+32 = 355

Since BW = f_{c}/Q:

Q = f_{c}/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.