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Step 4.021E (a) Refer to Figure 4.23 (a) in the textbook for bridge rectifier During the positive half-cycle of input, the diodes, and D2 are ON and the diodes, and are OFF. During the negative cycle of input, the diodes, D, and are ON and the diodes, are OFF Replace the diodes which are conducting with their constant voltage -drop diode Consider the following circuit for positive half-cycle of the signal: + + ac line voltage R Figure Step Apply voltage law to the circuit in Figure Simplify the expression for the output Consider that the input to the bridge rectifier is, The diodes conduct during both the cycles. Draw the input/output waveforms together 2VD 2V, Step In the waveform Figure 2, the angle at which the input signal reaches and the diode starts conducting The diode conduction terminates at and hence the total conduction Therefore, The average value (dc component) of the Observe from the waveform that the time period, The limits of the integration are from to since the output exists from Step Calculate Here, in the interval, to and to evaluate the integral over the first interval and make twice to get the required value Assume that, then and 2V, Thus, the average value (dc component) of the Step (b) The peak diode current occurs the peak diode voltage, Draw the diode circuit to find the peak diode current + + ac line voltage R Figure 3 Apply Kirchhoff's voltage law to the circuit Figure 3. Therefore, the peak diode current R Step Determine the average value of the output voltage, Consider that the sinusoidal input is (rms) Find peak value =16.97 V From the waveform in Figure Therefore, 16.97 V Recall the expression for the voltage, Substitute and 16.97 for in the expression for =(0.637)(16.97)-1.4 =10.8-1.4 Therefore, the average value of output voltage, Step Calculate the value peak diode current, i, Recall the expression for the peak diode R Substitute V for and 16.97 V for in the expression for in 16.97-2(0.7) 100 A mA Thus, the value 156 mA Determine the peak inverse voltage, PIV. The peak inverse voltage of bridge rectifier circui Substitute 0.7 for and 16.97 V for the expression for PIV. =16.27 the peak inverse 16.3 V