Chegg Solutions for Microelectronic Circuits (Adel S Sedra, Kenneth C Smith) (Z-Library)_parte_1872

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Step 1 of 4 17.043P Consider the LCR resonator circuit diagram of Figure 17.17(a) in the textbook. The pair of complex conjugate poles characterized with frequency is, = Substitute 10⁵ rad/s for in equation. 10⁵ = JLC 1 (1) Step 2 of 4 Determine the capacitor C. Consider the pole quality factor Q is, Substitute 10⁵ rad/s for 5 for Q and 10 kΩ for R in equation. = = nF Therefore, the value of capacitor is, 5 nF - Step 3 of 4 Determine the inductor L. Substitute 5 nF for in the equation (1). 10⁵ = = 1 L = 1 = 50 =0.02 = 20 mH Therefore, the value of inductor L is, 20 mH Step 4 of 4 The circuit design of the second order parallel LCR resonator is shown in Figure 1. 10 kΩ R 5 nF C L 20 mH Figure 1 Thus, the design of LCR resonator is done.