
Back to category: Arts Limited version  please login or register to view the entire paper.  In figs. 1, 2, and 3 the current rating if the transformer secondary is that flowing through the load since there are no ramifications of the circuits other than the non conducting diode(s) where no current flows, so the current rating can be calculated by simply dividing VL by RL. So the current rating in the secondary is the maximum that can be born by the diodes and that is Vmax/RL. First, w can see that VDC in circuit 1 is approximately half of that in fig.2, which is approximately half of that in fig.3. First, fig.1 is a halfwave rectifier, that is the signal passed by the diode that reached the load is only the positive part of the 7.5Vrms signal so the mean is reasonably half that of fullwave rectifier of the same input voltage, which will pass the absolute value of the signal giving a mean of 2*(mean of positive part). Second, the ripple voltage is the same in figs.1 and 2, 10.5 V, since we are using the same voltage supply which is approximately 7.5 and the vo... Posted by: Ryan Wilkins Limited version  please login or register to view the entire paper. 
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