iR ip +. Vy = Vm sin(or) R Yo Figure 2. Half-wave rectifier with a filter capacitor.

Introductory Circuit Analysis (13th Edition)
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ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
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A half-wave rectifier with an R = 750 N load has a filter capacitor C as
shown in Figure 2 on the second page. The source is 120 V rms, 60 Hz.
Determine (a) the value of the filter capacitor to keep the peak-to-peak
ripple across the load to less than 2.0 V, and (b) the peak and average
values of the diode current ip using the filter capacitor value in part (a)
and assuming that the load current ir can be approximated as ir
Vm/R where Vm is the peak value of the source voltage.
Transcribed Image Text:A half-wave rectifier with an R = 750 N load has a filter capacitor C as shown in Figure 2 on the second page. The source is 120 V rms, 60 Hz. Determine (a) the value of the filter capacitor to keep the peak-to-peak ripple across the load to less than 2.0 V, and (b) the peak and average values of the diode current ip using the filter capacitor value in part (a) and assuming that the load current ir can be approximated as ir Vm/R where Vm is the peak value of the source voltage.
iR
ip
+.
v = Vm sin(@r)
R 'o
Figure 2. Half-wave rectifier with a filter capacitor.
Transcribed Image Text:iR ip +. v = Vm sin(@r) R 'o Figure 2. Half-wave rectifier with a filter capacitor.
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