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Parallel Resonant Circuits

Discussion in 'Resonance' started by Tony R. Kuphaldt, Jun 24, 2010.

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  1. Tony R. Kuphaldt

    Tony R. Kuphaldt

    Jun 21, 2010
    A parallel resonant circuit is resistive at the resonant frequency. (Figure below) At resonance XL=XC, the reactive components cancel.

    The impedance is maximum at resonance. (Figure below) Below the resonant frequency, the series resonant circuit looks inductive since the impedance of the inductor is lower, drawing the larger proportion of current. Above resonance, the capacitive rectance decreases, drawing the larger current, thus, taking on a capacitive characteristic.


    A parallel resonant circuit is resistive at resonance, inductive below resonance, capacitive above resonance.

    Impedance is maximum at resonance in a parallel resonant circuit, but decreases above or below resonance. Voltage is at a peak at resonance since voltage is proportional to impedance (E=IZ). (Figure below)


    Parallel resonant circuit: Impedance peaks at resonance.

    A low Q due to a high resistance in series with the inductor prodces a low peak on a broad response curve for a parallel resonant circuit. (Figure below) conversely, a high Q is due to a low resistance in series with the inductor. This produces a higher peak in the narrower response curve. The high Q is achieved by winding the inductor with larger diameter (smaller gague), lower resistance wire.


    Parallel resonant response varies with Q.

    The bandwidth of the parallel resonant response curve is measured between the half power points. This corresponds to the 70.7% voltage points since power is proportional to E2. ((0.707)2=0.50) Since voltage is proportional to impedance, we may use the impedance curve. (Figure below)


    Bandwidth, Δf is measured between the 70.7% impedance points of a parallel resonant circuit.

    In Figure above, the 100% impedance point is 500 Ω. The 70.7% level is 0707(500)=354 Ω. The upper and lower band edges read from the curve are 281 Hz for fl and 343 Hz for fh. The bandwidth is 62 Hz, and the half power points are ± 31 Hz of the center resonant frequency:

               BW = Δf = f[SUB]h[/SUB]-f[SUB]l[/SUB]  = 343-281 = 62 
               f[SUB]l[/SUB] = f[SUB]c[/SUB] - Δf/2 = 312-31 = 281 
               f[SUB]h[/SUB] = f[SUB]c[/SUB] + Δf/2 = 312+31 = 343 
               Q = f[SUB]c[/SUB]/BW = (312 Hz)/(62 Hz) = 5
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