Problem 63-4
Source: Problem elaborated by the author of the site.
A full-wave, bridged circuit rectifier must supply a load that draws an average current IDC = 1.2 A. At the output of the transformer secondary we have a voltage of 24 Vrms and a frequency of f = 60 Hz.
Find the value of the capacitor so that the ripple factor does not exceed rpp = 8%.
Make a comparison between the values found using the approximation
VCmax ≅ VDC and without using the approximation.
Solution of the Problem 63-4
As we know, we must subtract the voltage drop across the rectifier diodes that are in the conduction state. As for each cycle, in the full-wave rectifier, there are always two diodes in conduction, it means that we have to subtract 1.4 V from the peak value of the input voltage ( V< sub>peak = √2 x 24 = 34 V ), to get the correct value of the voltage at the output of the rectifier. So, after rectification, the value of the maximum voltage across the capacitor is
VCmax = Vpico - 1.4 = 32.6 volts.
And since we want a ripple factor of 8%, so ΔVr = 0.08 x 32.6 = 2.61 V. With this data we can calculate the value of VDC using the eq. 63-01, reproduced below:
Now we can calculate the value of RL, or
And to calculate the value of C, we use eq. 63-13 reproduced below.