Page 31 - Electrician - TT (Volume 2)

ELECTRICIAN - CITS

Fig 3

The load current is in the same direction during both the half-cycles of the AC input. The output of the full-wave
rectifier is shown in Fig 3d.
DC output : Since a full wave rectifier is nothing but a combination of two half-wave rectifiers, the average or DC
value of a full wave rectifier is naturally twice the output of a half wave rectifier driven by the same secondary voltage.
From Fig 3 it i s evident that the average of DC value of a full wave rectified output is
V = 0.318 V s(peak) + 0.318 V s(peak)
dc
V = 0.636 V s(peak)
dc
where, V is the equal peak voltage between the centre-tap and any one end A or B of the transformer secondary.
s(peak)
In terms of V s(rms) V of full wave rectifier is given by,
dc
V s(rms) = 0.707 V s(peak)

Example
Suppose the secondary voltage of the transformer is 24-0-24V(rms), the Dc output voltage of a full wave rectifier
using this transformer will be,

For a two diode full wave rectifier
Vdc = 0.9 Vs(rms)
Therefore, in the given example
V = 0.9 x V s(rms) = 0.9 x 24 = 21.6 volts
dc
Ripple frequency in a full wave rectifier: From

Fig 3c it can be seen that two cycles of output occur for each input cycle of AC voltage. This is because, the full
wave rectifier has inverted the negative half cycle of the input voltage. As a result, the output of a full wave rectifier
has frequency double the input AC frequency. If mains AC supply is used as input to a full wave rectifier, the mains
frequency is 50 Hz, the output frequency of the pulsating DC will be 100 Hz.

Note: This increased ripple frequency has certain advantages when the pulsating DC is smoothed.
This will be dealt with in further lesson.
Peak inverse voltage: Fig 4 shows the full wave rectifier at the instant the secondary voltage reaches its maximum
positive value.
Applying Kirchhoff’s law around the outside loop, we get, 2V s(peak) - Reverse voltage(PIV)
across D + Forward voltage across D = 0
1
2
18

CITS : Power - Electrician & Wireman - Lesson 60-69