4-10 Solution: Since the total secondary voltage (E_{S}) 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: E_{max} = 1.414 E_{S}E_{max} = 1.414 150 E_{max} = 212 volts The average load voltage is: E_{avg} = 0.637 E_{max}E_{avg} = 0.637 212 E_{avg} = 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 (E_{avg}) 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.

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