Chegg Solutions for Microelectronic Circuits (Adel S Sedra, Kenneth C Smith) (Z-Library)_parte_475

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Step 4.019E (a) Refer to Figure 4.21 (a) in the textbook for half wave rectifier Replace the diode, D the circuit with its piece-wise linear That is replace the diode, D with and ro series. Neglect the effect of diode and draw the equivalent + R3 Figure Step Apply voltage law the circuit in Figure Simplify the expression for the output voltage, Consider that the input to the half wave rectifier is, The diode does not conduct during the negative half cycle Draw the t/output waveforms together Figure 2 Step In Figure the angle at which the input signal reaches and the diode starts conducting. The diode conduction terminates at and hence the total conduction Therefore, Therefore for the -cycles during which the diode conduction begins at angle and terminates for total conduction angle Step Consider that the sinusoidal input is 12V (rms) Find the peak value =16.97 V From the =16.97 V Recall the expression for the Substitute 0 V for and 16.97 0.7 (0.04) =2.4° Therefore, the value 2.4° The total conduction angle is, Therefore, the total conduction 175° Step (b) The average value (dc component) of the Observe from the waveform that the time period, The limits the integration are from to since the output exists from Calculate output the to Assume that, then and Therefore, 2 Thus, the average value component) the Substitute V for and 16.97 V for to calculate the value 16.97 0.7 Thus, value Step of 7 (c) The peak diode current the peak diode voltage, Draw the diode circuit find the peak diode + D + R Figure 3 Step Apply voltage law to the circuit in Figure R Therefore, the peak diode current is, R V for 16.97 V for and 100 for R to calculate the value 16.97-0.7 100 0.1627 A Thus, the value is, 163 mA Calculate the peak inverse voltage (PIV). Substitute 16.97 Therefore, the peak inverse voltage,