Page 32 - Electrician - TT (Volume 2)

ELECTRICIAN - CITS

Neglecting the small forward voltage across D1 we have, 2Vs(peak) = PIV across D2 + 0 = 0
or PIV across D = 2V s(peak)
2
From the above it can be seen that each diode in a fullwave rectifier must have PIV rating greater than the peak
value of the full secondary voltage. 2V s(peak)
In the example considered earlier, the PIV of diodes should be 2 V .
s(peak)

Current rating of diodes in a full wave rectifier : If the load, RL connected in the fullwave rectifier is, say 10W
the DC current through it will be,

In the example considered above, Vdc = 21.6 volts
Therefore,
It is interesting to note this current Idc is shared by the two diodes D1 and D2. This is because each diode
conducts only for one half cycle. Therefore, the DC current through each diode is half the total DC load current Idc.
Hence, the maximum current through each diode with 10W load will be 2.16/2 = 1.08 amps. From this it follows
that the current rating (If(max)) of each diode need only be half the maximum/rated load current.
NOTE: In a halfwave rectifier, since there is only one diode, the current rating of the diode used
should be the maximum current through the load unlike in the case of a full wave rectifier in which
the current rating of the diodes used is only half the maximum current through the load.
Example: In a two diode full wave rectifier, with a load current requirement of 1.8 amps, what should be the current
ratings of the diodes used?
Since it is a two diode full wave rectifier, the current rating of each diode should be = 1/2 the total load current.
Therefore If(max) of diodes should be = 1.8 amps/2 = 0.9 amps.
It is fine if a diode of 1 amp current rating is used for this rectifier circuit.
Disadvantages of TWO DIODE full wave rectifier : The full wave rectifier using two diodes and centre tap
transformer has the following disadvantages
• A centre-tapped transformer that produces equal voltages on each half of the secondary winding is difficult to
manufacturer and, hence, expensive.
• Centre-tapped transformers are generally bulkier than ordinary transformers, and, hence, occupy larger space.
• In a two diode full wave rectifier, only half of the secondary voltage is made use at a time although it works in
both +ve and -ve half cycles.
Bridge rectifier : It is a full-wave rectifier. The circuit is in Fig 5a. In the bridge rectifier four diodes are used.
There is no centre tap on the secondary of the transformer.
During the positive half of the secondary voltage, diodes D2 and D3 are forward-biased. Hence current flows
through diode D2 load resistance RL and D3 to the other end of the secondary. This is illustrated in Fig 5b. During
the negative half of the secondary voltage, diodes D1 and D4 are conducting . The current flows through diode
D4, resistor RL and diode D1 to the other end of the secondary. This is illustrated in Fig 5c.
In both cases the current flows through the load resistor in the same direction. Hence, a fluctuating DC is developed
across the load resistor RL. This is shown in Fig 5d.

19

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