EXAMPLE 2-8 FIGURE 2-48 Determine the ripple factor for the filtered bridge rectifier with a load as indicated in Figure 2-48. 120 V 60Hz 10:1 Dy 1000F All diodes are 1N4001, Output w 2200 Solution The transformer turns ratio is n = 0.1. The peak primary voltage is Vi 1.414V 1.414(120 V) - 170 V The peak secondary voltage is VV-0.1(170 V) - 17.0V The unfiltered peak full-wave rectified voltage is V-V-1.4V 17.0V-1.4V 15.6 V The frequency of a full-wave rectified voltage is 120 Hz. The approximate peak-to- peak ripple voltage at the output is JRC (120 Hz)(220 (1) 1000F)) -0.591 V The approximate de value of the output voltage is determined as follows: -(1-2/+C)/m - (1-(240 HD)(220 x 100,F)) 156V-153 <- The resulting ripple factor is Vpp 0.591 V 0.039 Voc 15.3 V The percent ripple is 3.9%. Related Problem Determine the peak-to-peak ripple voltage if the filter capacitor in Figure 2-48 is in- creased to 2200 µF and the load resistance changes to 2.2 k. Problem 1: Solve the given example again from your online lecture (slide 9) to determine the peak-to-peak ripple voltage if the filter capacitor is increased to 2200µF and the load resistance changes to 2.2k2