184 Electrical Power Systems Technology
voltage during each alternation of the AC input. The rectified output of the
full-wave rectifier has twice the DC voltage level of the half-wave recti-
fier.
The full-wave rectifier utilizes a center-tapped transformer to transfer
AC source voltage to the diode rectifier circuit. During the positive half
cycle of AC source voltage, the instantaneous charges on the transformer
secondary are as shown in Figure 7-21. The peak voltage (Vmax) is devel-
oped across each half of the transformer secondary. At this time, diode D1
is forward biased, and diode D2 is reverse biased. Therefore, conduction oc-
curs from the center-tap, through the load device, through D1, and back to
the outer terminal of the transformer secondary. The positive half cycle is
developed across the load, as shown.
During the negative half cycle of the AC source voltage, diode D1 is re-
verse biased, and diode D2 is forward biased by the instantaneous charges
shown in Figure 7-21. The current path is from the center-tap, through the
load device, through O 2 , and back to the outer terminal of the transformer
secondary. The negative half cycle is also produced across the load, devel-
oping a full-wave output, as illustrated in Figure 7-21.
Each diode in a full-wave rectifier circuit must have a piv rating of
twice the value of the peak voltage developed at the output, since twice
the peak voltage (2Vmax) is present across the diode when it is reverse bi-
ased. The average voltage for a full-wave rectifier circuit is:
Vdc = 2 (0.318 × Vmax)
= 0.636 × Vmax
Sample Problem:
Given: a single-phase, half-wave rectifier has 120 volts (rms) applied
to its input.
Find: DC output voltage of the rectifier.
Solution:
Vdc = 0.636 × Vmax
= 0.636 × (120 V × 1.41)
Vdc = 107.61 volts
This type of rectifier circuit produces twice the DC voltage output
of a half-wave rectifier circuit. However, it requires a bulky center-tapped
transformer, as well as diodes that have a piv rating of twice the peak val-
ue of applied AC voltage.