Where:

Emax = the peak value of the load voltage pulse.

Eavg = 0.637 x Emax (the average load voltage).

Imax = the peak value of the load current pulse.

Iavg = 0.637 x Imax (the average load current).

supply a full-wave rectifier is 300 volts. Find the average load voltage (ignore the drop

across the diode).

one-half of this value, or 150 volts. Since the secondary voltage is an RMS value, the

formula for the peak load voltage is as follows:

Emax = 1.414 x Es

Emax = 1.414 x 150

Emax = 212 volts

The average load voltage is computed as follows:

Eavg = 0.637 x Emax

Eavg = 0.637 x 212

Eavg = 135 volts

NOTE*: *If you have problems with this equation, review the portion of

TC 9-60 that pertains to this subject.

4-22. Remember, every circuit has advantages and disadvantages. The full-wave rectifier

is no exception. In studying about 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

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