Problem 63-5    Source: Problem elaborated by the author of the site.

In the secondary of a transformer the voltage is V_s = 15 Vrms. Using a tap-central full-wave rectifier, calculate the capacitor value for a 4% ripple and the load value if it draws a current of 1 A (assume f = 60 Hz). Make a comparison between the values found using the approximation V_Cmax ≅ V_DC and without using the approximation.

Solution of the Problema 63-5   

As the secondary voltage was given in rms values, we must transform to peak voltage, that is, V_p = √ 2 x 15 = 21.21  V. Therefore, in a full-wave rectifier, to calculate the maximum voltage across the capacitor, we must subtract from the peak voltage the value of the voltage drop across the diode. So the maximum voltage across the capacitor will be V_Cmax = 21.21 - 0.7 = 20.51  V.

Since a 4% ripple is desired, this results in ΔV_r = 0.04 x 20.51 = 0.82  V.

Now let's calculate the value of C using the approximation mentioned in the statement. For that, let's use the eq. 63-14, reproduced below.

So, substituting the numerical values, we get