Chegg Solutions for Microelectronic Circuits (Adel S Sedra, Kenneth C Smith) (Z-Library)_parte_483

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4.023E Consider following circuit for rectifier with placed load resistor, + ac line voltage D, Step 2 of 10 Consider that peak to-peak ripple Determine the value of using the following expression ripple voltage Consider that (rms) Find peak =16.97 V The expression The output voltage end the discharge Step 3 of 10 The discharge occurs almost time Use following approximation since the time constant very large compared discharge time Use equation The expression Step 4 of 10 Simplify expression get expression fo the Substitute 16.97 V for 1 V for V, 60 Hz for and 100 R 16.97-2(0.8) 2(1)(60)(100) 15.37 Thus, the value capacitor required is Step 5 of 10 =16.97-2(0.8) voltage the 15.37 V Consider the voltage calculate the voltage - =15.37-0.5 =14.87 V voltage output for better 14.87 V Step 6 of 10 Calculate considering the ripple voltage Substitute 15.37 and for R 15.37 V =0.1537 A to find the conduction angle The effect o voltage the peak voltage V, Substitute 16.97 V for 2(1) (16.97-2(0.8) =0.36 rad =20.7° Thus, Step 8 of 10 The average diode current is Here, he load current and peak voltage the ripple voltage and =14.87V 14.87 100 =0.149 16.97-2(0.8) 2(1) =(0.149)(9.7) average curren Step 9 of 10 the cad current and the peak voltage are affected the ripple voltage and hence =14.87V 14.87 100 =0.149 16.97-2(0.8) 2(3.14) 2(1) =(0.149)(18.4) peak Step 10 of 10 voltage of the Substitute 16.97 V for V, =16.97-0.8 =16.17 the peak inverse voltage is 16.17 V Thus select 20