
480 SECTION 8 Solid-State Devices
RECTIFIERS
Diodes can be used to perform many jobs, but their
most common use in industry is to construct a rec-
ti er. A recti er is a device that changes or converts
AC voltage into DC voltage. The simplest type of rec-
ti er is known as the half-wave recti er. Refer
to the circuit shown in Figure 50–5. The half-wave
recti er can be constructed with only one diode, and
gets its name from the fact that it will rectify only half
of the AC waveform applied to it. When the voltage
applied to the anode is positive, the diode is forward
biased and current can ow through the diode, load
resistor, and back to the power supply. When the
voltage applied to the anode becomes negative, the
diode is reverse biased and no current will ow. Since
the diode permits current to ow through the load in
only one direction, the current is DC.
Diodes can be connected to produce full-wave
recti cation, which means both halves of the AC
waveform will be made to ow in the same direc-
tion. One type of full-wave recti er is known as
the bridge recti er and is shown in Figure 50–6.
Notice the bridge recti er requires 4 diodes for
construction.
To understand the operation of the bridge recti er,
assume that point X of the AC source is positive and
point Y is negative. Current will ow to point A of the
OHMMETER
+–
Figure 50–3
Testing a diode. (Source: Delmar/Cengage Learning)
Figure 50–4
A diode connected in the reverse direction.
(Source: Delmar/Cengage Learning)
OHMMETER
+–
Figure 50–5
Half-wave rectifi er. (Source: Delmar/Cengage Learning)
LOAD
recti er. At point A, diode D4 is reverse biased and
D1 is forward biased. The current will ow through
diode D1 to point B of the recti er. At point B,
diode D2 is reverse biased, so the current must ow
through the load resistor to ground. The current
returns through ground to point D of the recti er. At
point D, both diodes D4 and D3 are forward biased,
but current will not ow from positive to positive.
Therefore, the current will ow through diode D3 to
point C of the bridge, and then to point Y of the AC
source, which is negative at this time. Since current
owed through the load resistor during this half
cycle, a voltage is developed across the resistor.
Now assume that point Y of the AC source is
positive and X is negative. Current will ow from
point Y to point C of the recti er. At point C, diode