In Fig. 12.69, RL=2.2kΩ and νS=VS[cos(ω1t)+cos(ω2t)] with f1=100kHz and f2=200 kHz. The circuit is intended to block the 200 kHz component and pass the 100 kHz component of the source voltage. You have available a selection of 200 and 100 kHz coils with inductances ranging from 10 to 200

Question

In Fig. 12.69, [latex]R_L=2.2 k \Omega[/latex] and [latex] u_S=V_S\left[\cos \left(\omega_1 t ight)+\cos \left(\omega_2 t ight) ight][/latex] with [latex]f_1=100 kHz[/latex] and [latex]f_2=200 kHz[/latex]. The circuit is intended to block the 200 kHz component and pass the 100 kHz component of the source voltage. You have available a selection of 200 and 100 kHz coils with inductances ranging from 10 to 200 mH in steps of 10 mH . All have quality factors of about 25 at their intended operating frequencies. You also have available capacitors ranging from 10 to 91 pF and a wide selection of trimmer capacitors in the same range, both having quality factors in excess of 100. Specify the circuit components such that the 100 kHz component of the load voltage is as large as possible and the 200 kHz component is as small as possible. Evaluate your design (with the help of a computer).

Accepted Answer

The impedance of a series RLC circuit is minimum at resonance and the impedance of a parallel R L C circuit is maximum at resonance. The idea here is to make the series RLC section resonant at 100 kHz and the parallel R L C section resonant at 200 kHz . For the series circuit, we must choose a 100 kHz coil and a capacitor such that

[latex] \frac{1}{\sqrt{L_1 C_1}}=\omega_1=200 \pi imes 10^3 s ^{-1} . [/latex]

Given the large assortment of components and in absence of other specifications, there are a great many solutions. Choosing [latex] C_1=15 pF [/latex] gives

[latex] L_1=\frac{1}{\omega_1^2 C_1}=169 mH, [/latex]

For the parallel section, we require

[latex] \frac{1}{\sqrt{L_2 C_2}}=\omega_2=400 \pi imes 10^3 s ^{-1}[/latex]

We choose [latex] C_2=62 pF [/latex] and calculate

[latex] L_2=\frac{1}{\omega_2^2 C_2}=10.2 mH ext {. }[/latex]

For frequencies near resonance, we can express the impedance of an inductor as

[latex] \begin{aligned} Z_L &=R_L+j \omega L=\frac{\omega L}{Q_L}+j \omega L \\ &=j \omega L\left(1-j Q_L^{-1} ight) . \end{aligned}[/latex]

Similarly, for a capacitor,

[latex] \begin{aligned} Z_C &=R_C+\frac{1}{j \omega C}=\frac{1}{Q_C \omega C}+\frac{1}{j \omega C} \\ &=\frac{1}{j \omega C}\left(1+j Q_C{ }^{-1} ight) . \end{aligned} [/latex]

The impedance of the series section in Fig. 12.69 is given by

[latex] \begin{aligned} Z_1 &=Z_{L 1}+Z_{C 1} \\ & \cong j \omega L_1\left(1 - \frac{j}{Q_L} ight)+\frac{1}{j \omega C_1}\left(1+\frac{j}{Q_C} ight) . \end{aligned}[/latex]

Similarly, the impedance of the parallel section is given by

[latex] \begin{aligned} Z_2 &=\frac{Z_{L 2} Z_{C 2}}{Z_{L 2}+Z_{C 2}} \\ &=\frac{\left[j \omega L_2\left(1-j Q_L^{-1} ight) ight]\left[\left(1+j Q_C^{-1} ight) ight]}{\left(1+j Q_C^{-1} ight)-\omega^2 L_2 C_2\left(1-j Q_L^{-1} ight)} . \end{aligned} [/latex]

The load voltage is given by

[latex] ilde{V}_L(\omega)=\frac{R_L ilde{V}_S}{R_L+Z_1(\omega)+Z_2(\omega)} [/latex]

Figure 12.70 shows a graph of the normalized load voltage (or voltage gain) in dB , given by

[latex] A_ u=20 \log \left|\frac{ ilde{V}_L}{ ilde{V}_S} ight|, [/latex]

where [latex] ilde{V}_S=1 V [/latex]. The resonant peak at 100 kHz and the resonant valley at 200 kHz are evident. Direct calculation gives [latex] A_ν\left(f_1 ight) \cong-14.5 dB, A_ν\left(f_2 ight) \cong-42.9 dB [/latex], so the 200 kHz component of the load voltage is 28.4 dB below the 100 kHz component, if the components have their specified values.

Question 12.47: In Fig. 12.69, RL=2.2kΩ and νS=VS[cos(ω1t)+cos(ω2t)] with f...

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