# Selectivity Curve for Parallel Resonant Circuits

The $Z_T$ versus frequency curve of Fig. 1 clearly reveals that a parallel resonant circuit exhibits maximum impedance at resonance ( fp), unlike the series resonant circuit, which experiences minimum resistance levels at resonance. Note also that $Z_T$ is approximately $Rl$ at $f = 0 Hz$ since $Z_T=Rs || Rl \appro Rl$.
Fig. 1: ZT versus frequency for the parallel resonant circuit.
Since the current I of the current source is constant for any value of ZT or frequency, the voltage across the parallel circuit will have the same shape as the total impedance $Z_T$, as shown in Fig. 2.
Fig. 2: Defining the shape of the $V_p (f)$ curve.
For the parallel circuit, the resonance curve of interest is that of the voltage VC across the capacitor. The reason for this interest in $V_C$ derives from electronic considerations that often place the capacitor at the input to another stage of a network.
Since the voltage across parallel elements is the same, $$\bbox[10px,border:1px solid grey]{V_C = V_p = IZ_T}$$ The resonant value of VC is therefore determined by the value of ZTm and the magnitude of the current source I. The quality factor of the parallel resonant circuit continues to be determined by the ratio of the reactive power to the real power. That is, $$Q_p = { V^2/X_{L_{p}} \over V^2/R}$$ where $R = Rs || Rp$$, and Vp is the voltage across the parallel branches. The result is$$ \bbox[10px,border:1px solid grey]{Q_p = {R \over X_{L_{p}}} = {R_s||R_p \over X_{L_{p}}}}$$or since X_{L_{p}} = X_C at resonance,$$ \bbox[10px,border:1px solid grey]{Q_p = {R_s||R_p \over X_{C}}}$$For the ideal current source (R_s = \infty Ω) or when R_s is sufficiently large compared to R_p, we can make the following approximation:$$R = R_s || R_p$$and$$ Q_p = {R_s||R_p \over X_{L_{p}}} = { R_p \over X_{L_{p}}} \\ = { (R^2_l + X^2_L) / R_l \over (R^2_l + X^2_L) /X_{L}}$$so that$$ \bbox[10px,border:1px solid grey]{Q_p = { X_L \over R_l} = Q_l} $$which is simply the quality factor Q_l of the coil. In general, the bandwidth is still related to the resonant frequency and the quality factor by$$ \bbox[10px,border:1px solid grey]{BW = f_2 - f_1 = { f_r \over Q_p}} \tag{a}$$The cutoff frequencies f1 and f2 can be determined using the equivalent network of Fig. 20.25 and the unity power condition for resonance. The half-power frequencies are defined by the condition that the output voltage is 0.707 times the maximum value. However, for parallel resonance with a current source driving the network, the frequency response for the driving point impedance is the same as that for the output voltage. This similarity permits defining each cutoff frequency as the frequency at which the input impedance is 0.707 times its maximum value. Since the maximum value is the equivalent resistance R of Fig. 3 , the cutoff frequencies will be associated with an impedance equal to 0.707R or (1/\sqrt{2})R. Fig. 3: Setting the input impedance for the network of Fig. 3 equal to this value will result in the following relationship:$$ Z = { 1 \over {1 \over R} + j ( wC - { 1 \over wL})} = 0.707 R$$which can be written as$$ Z = { 1 \over {1 \over R}[1 + j R( wC - { 1 \over wL})]} = {R \over \sqrt{2}} $$or$${ R \over 1 + j R( wC - { 1 \over wL})} = {R \over \sqrt{2}} $$and finally$${ 1 \over 1 + j R( wC - { 1 \over wL})} = {1 \over \sqrt{2}} $$The only way the equality can be satisfied is if the magnitude of the imaginary term on the bottom left is equal to 1 because the magnitude of 1 + j 1 must be equal to \sqrt{2}. The following relationship, therefore, defines the cutoff frequencies for the system:$$ R( wC - { 1 \over wL}) = 1$$Substituting w = 2\pi f and rearranging will result in the following quadratic equation:$$ f^2 - {f \over 2\pi RC} - { 1 \over 4 \pi^2 LC} = 0$$having the form$$af^2 + bf + c = 0$$with a=1, b= -{1 \over 2\pi RC} and c = - { 1 \over 4 \pi^2 LC} Substituting into the equation:$$ f = { -b \pm \sqrt{b^2 - 4ac} \over 2a}$$will result in the following after a series of careful mathematical manipulations:$$ \bbox[10px,border:1px solid grey]{f_1 = { 1 \over 4 \pi C}[ { 1\over R} - \sqrt{{1 \over R^2} + { 4C \over L}}]} \tag{1} \bbox[10px,border:1px solid grey]{f_2 = { 1 \over 4 \pi C}[ { 1\over R} + \sqrt{{1 \over R^2} + { 4C \over L}}]} \tag{2}$$Since the term in the brackets of Eq. (1) will always be negative, simply associate$f_1$with the magnitude of the result. The effect of Rl, L, and C on the shape of the parallel resonance curve, as shown in Fig. 4 for the input impedance, is quite similar to their effect on the series resonance curve. Fig. 4: Effect of Rl, L, and C on the parallel resonance curve. Whether or not$R_l$is zero, the parallel resonant circuit will frequently appear in a network schematic as shown in Fig. 4. At resonance, an increase in Rl or a decrease in the ratio L/C will result in a decrease in the resonant impedance, with a corresponding increase in the current. The bandwidth of the resonance curves is given by Eq. (a). For increasing$R_l$or decreasing L (or L/C for constant C), the bandwidth will increase as shown in Fig. 4. At low frequencies, the capacitive reactance is quite high, and the inductive reactance is low. Since the elements are in parallel, the total impedance at low frequencies will therefore be inductive. At high frequencies, the reverse is true, and the network is capacitive. At resonance ($f_p\$), the network appears resistive. These facts lead to the phase plot of Fig. 5. Note that it is the inverse of that appearing for the series resonant circuit because at low frequencies the series resonant circuit was capacitive and at high frequencies it was inductive.
Fig. 5: Phase plot for the parallel resonant circuit.