# Obtaining capacitor value for ripple voltage

Discussion in 'Homework Help' started by frisbee4all, Feb 23, 2009.

1. ### frisbee4all Thread Starter Member

Feb 22, 2009
13
0
Im trying to solve this problem:

Given a bridge rectifier circuit with a filter capacitor C placed across the load resistor R, for the case in which the transformer secondary delivers a sinusoid of 12 V (rms) having a 60-Hz frequency, and assuming V_DO = .8V and a load resistance R = 100Ω. Find the value that results in a ripple voltage no larger than 1 V peak to peak. I am using the equation C = V_p / (2*f*V_r*)R, where f is the frequency, and V_r is the ripple voltage. I am getting an answer that is off from the 1283 μF that the book is giving me. Also, the next step of the problem says to find the dc voltage at the output. I am assuming this would be the same as the peak-peak voltage if we assume that CR is much greater than T (the period). I don't fully understand how the book obtains dc values for its output, the only thing I can think of is that after the rectifier circuit and capacitor the signal is now consider a DC signal thust the output voltage across the load resistor is the DC voltage output. Is this about right? Thanks!!

2. ### ecb123 New Member

Mar 6, 2008
7
0
frisbee4all:

The peak AC voltage out of the transformer is:
12.0*(2^0.5)=16.970 Volts Peak AC.

This passes through the rectifier converting it to a full-wave DC pulse voltage seen at the cap and load is:

Minus 2 diode forward voltage drops, 0.8*2=1.6 Volts DC:

16.970-1.6=15.370 Volts peak DC

We are given the load resistance of 100 ohms, so can calculate the needed load current that the cap must supply b/t pulses.