Draw a circuit diagram of a full-wave rectifier using p-n junction diodes. Explain its working and show the input-output waveforms.
Show Hint
Uses both half cycles
Output frequency doubles
Can be centre-tapped or bridge type
Approach Solution - 1
Concept: A full-wave rectifier converts both halves of an AC input into pulsating DC output using p–n junction diodes. It provides higher efficiency than a half-wave rectifier.
Circuit Diagram (Centre-tapped full-wave rectifier): Components:
Centre-tapped transformer
Two diodes \( D_1, D_2 \)
Load resistor \( R_L \)
Connections:
Anodes of diodes connected to the ends of secondary winding
Cathodes joined together and connected to load
Centre tap connected to other end of load
Working: Positive half cycle:
Upper end of secondary becomes positive
Diode \( D_1 \) forward biased → conducts
Diode \( D_2 \) reverse biased → off
Current flows through \( R_L \) in one direction
Negative half cycle:
Lower end of secondary becomes positive
\( D_2 \) conducts, \( D_1 \) off
Current again flows through load in same direction
Thus, both halves of AC are rectified → full-wave rectification. Input–Output Waveforms: Input waveform:
Sinusoidal AC wave
Positive and negative halves symmetric
Output waveform:
Both halves appear positive
Pulsating DC with double frequency of input
Graph description:
Input: sine wave about zero axis
Output: series of positive humps (no negative portion)
Key Features:
Output frequency = \( 2f \)
Higher efficiency than half-wave rectifier
Less ripple
Approach Solution -2
A full-wave rectifier is required to convert both the positive and the negative half-cycles of an AC input into a one-directional (though pulsating) output. Here we look at the bridge configuration and focus on the quantities that distinguish it electrically from a half-wave circuit: peak inverse voltage and rectification efficiency.
Circuit arrangement:
Four p-n junction diodes \( D_1, D_2, D_3, D_4 \) are connected in a bridge, with the AC source applied across one diagonal of the bridge and the load resistor \( R_L \) connected across the other diagonal. No centre-tapped transformer is needed.
Working:
During one half-cycle, one diagonal pair of diodes conducts (forward biased) while the other pair is reverse biased and blocks; during the other half-cycle, the roles swap and the second diagonal pair conducts. In both cases, current is steered through \( R_L \) in the same direction, so both halves of the AC cycle contribute a unipolar output.
Peak Inverse Voltage (PIV):
The non-conducting diode pair in a bridge rectifier must withstand a reverse voltage equal only to the peak of the input, \( \text{PIV} = V_m \), which is half of what a centre-tapped rectifier's diodes must withstand for the same output. This makes the bridge design more efficient in its use of diode ratings.
Rectification efficiency and output frequency:
Since current flows through the load during both halves of the input cycle, the output waveform repeats twice per input cycle, so the output ripple frequency is \( 2f \) (double the AC supply frequency). This doubling of pulses per cycle is what gives a full-wave rectifier its characteristically higher efficiency (theoretical maximum about 81.2%) and lower ripple compared to a half-wave rectifier, whose efficiency tops out near 40.6%.
Input-output waveforms:
The input is a smooth sinusoid oscillating symmetrically about zero. The output consists of a continuous series of positive humps with no gaps and no negative excursions -- each hump is a full half-sine derived alternately from the positive and the (inverted) negative half of the input, occurring at twice the input frequency.
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