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 I_DC = 1.2 A. At the output of the transformer secondary we have a voltage of 24 V_rms and a frequency of f = 60 Hz. Find the value of the capacitor so that the ripple factor does not exceed r_pp = 8%. Make a comparison between the values found using the approximation V_Cmax ≅ V_DC 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 V_Cmax = V_pico - 1.4 = 32.6 volts.

And since we want a ripple factor of 8%, so ΔV_r = 0.08 x 32.6 = 2.61 V. With this data we can calculate the value of V_DC using the eq. 63-01, reproduced below:

Now we can calculate the value of R_L, or

And to calculate the value of C, we use eq. 63-13 reproduced below.