Page 30 - Electrician - TT (Volume 2)

ELECTRICIAN - CITS

V = 0.318 V
dc p

= 0.45 V rms

For example if the input AC voltage is 24 volts the output DC of the half wave rectifier will be Vdc = 0.45 x 24= 10.8 V
The DC load current is I
dc
Ripple frequency: From Fig 1 it is evident that the frequency of the rectified pulsating DC is same as the frequency
of the input AC signal. This is true for all half-wave rectifiers.
Peak inverse voltage: Fig 1(a) shows the half-wave rectifier at the instant the secondary voltage is at its maximum
negative peak.

In this condition, since the diode is reverse biased, it behaves as an open switch as in Fig 2b. Since the diode is
reverse biased, there is no voltage across the load RL. Therefore, from Kirchhoff’s Voltage law, all the secondary
voltage appears across the diode as shown in Fig 2a. This is the maximum reverse voltage that appears across
the diode in the reverse biased condition. This voltage is called the peak reverse voltage or more commonly as
the peak inverse voltage (PIV). Therefore, in a half-wave rectifier the peak inverse voltage across the diode is
equal to the -ve peak value of the secondary voltage V s(peak) . Since the -ve peak voltage and +ve peak voltage in
a sinusoidal wave is same in magnitude, the peak inverse voltage (PIV) across the diode in a halfwave rectifier
can be taken as a V s(peak) .
In the example considered earlier, the PIV across the diode will be,

Fig 2

To avoid break down of the diode used, the PIV appearing across the diode of the designed HW rectifier must be
less than the PIV rating of the diode. For instance, in the above example to avoid break down of the diode, the
PIV rating of the diode should be greater than 34 volts.

However this condition changes when a filter capacitor is used in the output DC circuit.
Full wave rectifier (FW): A full wave rectifier circuit is in Fig 3. The secondary winding of the transformer is centre-
tapped. The secondary voltage is divided equally into two halves, one end of the load RL is connected to the
centre tap and the other end of RL to the diodes.
It is seen that two half-wave rectifiers are conducting on alternate half cycles of the input Ac.
During the positive half cycle of the secondary voltage, diode D1 is forward-biased and diode D2 is reverse-biased.
(Fig 3b) The current flows through the load resistor RL, diode D1 and the upper half of the secondary winding.
During the negative half cycle of secondary voltage, diode D2 is forward-biased and diode D1 is reverse-biased.
Therefore, current flows through the load resistor RL diode D2 and the lower half of the secondary winding. (Fig 3c)

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