4-10
Solution: Since the total secondary voltage (ES) is 300 volts, each diode is supplied one-half of this
value, or 150 volts. Because the secondary voltage is an rms value, the peak load voltage is:
Emax = 1.414
ES
Emax = 1.414
150
Emax = 212 volts
The average load voltage is:
Eavg = 0.637
Emax
Eavg = 0.637
212
Eavg = 135 volts
NOTE: If you have problems with this equation, review the portion of NEETS, module 2, that
pertain to this subject.
As you may recall from your past studies in electricity, every circuit has advantages and
disadvantages. The full-wave rectifier is no exception. In studying the full-wave rectifier, you may have
found that by doubling the output frequency, the average voltage has doubled, and the resulting signal is
much easier to filter because of the high ripple frequency. The only disadvantage is that the peak voltage
in the full-wave rectifier is only half the peak voltage in the half-wave rectifier. This is because the
secondary of the power transformer in the full-wave rectifier is center tapped; therefore, only half the
source voltage goes to each diode.
Fortunately, there is a rectifier which produces the same peak voltage as a half-wave rectifier and the
same ripple frequency as a full-wave rectifier. This circuit, known as the BRIDGE RECTIFIER, will be
the subject of our next discussion.
Q8. What was the major factor that led to the development of the full-wave rectifier?
Q9. What is the ripple frequency of a full-wave rectifier with an input frequency of 60 Hz?
Q10. What is the average voltage (Eavg) Output of a full-wave rectifier with an output of 100 volts peak?
The Bridge Rectifier
When four diodes are connected as shown in figure 4-8, the circuit is called a BRIDGE RECTIFIER.
The input to the circuit is applied to the diagonally opposite corners of the network, and the output is
taken from the remaining two corners.
