The circuit in figure 3 addresses the second of these problems since at no time is the output voltage 0V. This time four diodes are arranged so that both the positive and negative parts of the AC waveform are converted to DC. The resulting waveform is shown in figure 4.
When the AC input is positive, diodes A and B are forward-biased, while diodes C and D are reverse-biased. When the AC input is negative, the opposite is true - diodes C and D are forward-biased, while diodes A and B are reverse-biased.
One disadvantage of the full-wave rectifier is that there is a voltage loss of 1.4V across the diodes. Why not 2.8V as there are four diodes, that is because there is only two of the diodes are passing current at any one time.
While the full-wave rectifier is an improvement on the half-wave rectifier, its output still is not suitable as a power supply for most circuits since the output voltage still varies between 0V and Vs-1.4V. So, if you put 12V AC in, you will 10.6V DC out.
The formula Bridge full wave Rectifier is,
Full wave
Bridge
|
||
Number of diodes
|
4
|
|
PIV of diodes
|
Vm
|
|
D.C output voltage
|
2Vm/phi
|
|
Vdc,at
no-load
|
0.636Vm
|
|
Ripple factor
|
0.482
|
|
Ripple
frequency
|
2f
|
|
Rectification
efficiency
|
0.812
|
|
Transformer
Utilization
Factor(TUF)
|
0.812
|
|
RMS voltage Vrms
|
Vm/√2
|
|
A bridge rectifier makes use of four diodes in a bridge arrangement to achieve full-wave rectification. This is a widely used configuration, both with individual diodes wired as shown and with single component bridges where the diode bridge is wired internally. So, in this project it will use 4 x 1N4001 diode (as picture below).
Diode type 1N4001
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