Since the output waveform is essentially a sine wave with both alternations at the same polarity, the
average current or voltage is 63.7 percent (or .637) of the peak current or voltage.
As an equation:
Eavg (the average load voltage) = .637 × Emax
Emax = The peak value of the load voltage pulse
Iavg (the average load current) = .637 × Imax
Imax = The peak value of the load current pulse
Example: The total voltage across the high-voltage secondary of a transformer used to supply a full-
wave rectifier is 600 volts. Find the average load voltage. (Ignore the drop across the rectifier tube.)
Solution: Since the total secondary voltage is 600 volts, each diode is supplied one-half of this value,
or 300 volts. As the secondary voltage is an rms value, the peak load voltage is:
Emax = 1.414 × Es
Emax = 1.414 × 300
Emax = 424 volts
The average load voltage is:
Eavg = .637 × Emax
Eavg = .637 × 424
Eavg = 270 volts
NOTE: If you have problems with this equation, review NEETS, module 2, pertaining to this area.
As you may recall from your past studies in electricity, there are advantages and disadvantages in
every circuit. The full-wave rectifier is no exception. In studying the full-wave rectifier, you have found
that when the output frequency is doubled, the average voltage is also 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 a full-wave rectifier is only half the peak voltage in a half-wave rectifier. This is because the secondary
of the power transformer in a full-wave rectifier is center tapped; therefore only half the source voltage
goes to each diode.
Fortunately, there is a rectifier that produces the same peak voltage as a half-wave rectifier and the
same ripple frequency as a full-wave rectifier. This circuit, called the BRIDGE RECTIFIER, will be the
chapter of our next discussion.
Q12. What is the ripple frequency of a full-wave rectifier with an input frequency of 60 Hz?