194 Electrical Power Systems Technology
The ideal filtered DC voltage would be one with no AC ripple and a
value equal to the peak voltage (Vmax) from the rectifier output. Note that
in Figure 7-28, the value of Vdc is approaching that of Vmax, and compare
this to the full-wave rectified voltage of Figure 7-28. There are two time in-
tervals shown in Figure 7-28. Time period t 1 represents diode conduction
that charges the filter capacitor (C) to the peak rectified voltage (Vmax).
Time period t 2 is the time required for the capacitor to discharge through
the load (RL).
If a different value of filter capacitor were put into the circuit, a
change in the rate of discharge would result. If capacitor C discharged a
very small amount, the value of Vdc would be closer to the value of Vmax.
With light loads (high resistance), the capacitor filter will supply a high DC
voltage with little ripple. However, with a heavy (low resistance) load con-
nected, the DC voltage would drop, because of a greater ripple. The in-
creased ripple is caused by the lower resistance discharge path for the
filter capacitor. The effect of an increased load on the filter capacitor is
shown in Figure 7-28.
By utilizing the value indicated on the waveforms of Figure 7-28, it
is possible to express Vdc and Vr(rms) as:
Vr(p–p)
Vdc = Vmax – ———
2
and
Vr(p–p)
Vr(rms) = ———
2 × √3
Sample Problem:
Given: the measured AC ripple voltage at the output of a filtered
power supply is 0.8V (rms), and the DC output voltage is 15.0 volts.
Find: the percent ripple of the power supply.
Solution:
0.8V × 100
%r = —————
15.0 V
= 5.33%
Sample Problem:
Given: a power supply has a p-p ripple of 1.2 volts at its output.
Find: the rms ripple voltage of the power supply.