Question

Draw a circuit diagram of a full-wave rectifier using p-n junction diodes. Explain its working and show the input-output waveforms.

Hint:

Full-wave rectifier facts:

Uses both half cycles
Output frequency doubles
Can be centre-tapped or bridge type

Answer 1

Full-Wave Rectifier Using P-N Junction Diodes
A full-wave rectifier converts the entire alternating current (AC) input into pulsating direct current (DC), allowing both halves of the AC waveform to contribute to the output. It typically uses four p-n junction diodes arranged in a bridge configuration. 
- D1, D2, D3, D4: P-N junction diodes - AC: Alternating current source - RL: Load resistor across which DC is obtained 

Working:
1. Positive Half-Cycle of AC Input:
- During the positive half of the AC input, the current flows through diode D1 → Load resistor RL → D3. - Diodes D2 and D4 remain reverse-biased and do not conduct. - A positive voltage appears across RL.
2. Negative Half-Cycle of AC Input:
- During the negative half of the AC input, the current flows through D2 → Load resistor RL → D4. - Diodes D1 and D3 remain reverse-biased. - The current through RL remains in the same direction, so the output is still positive.
Thus, both halves of the AC input contribute to the output voltage, doubling the frequency of the pulsating DC compared to a half-wave rectifier. 
Input-Output Waveforms:
- Input AC waveform:
- Output DC waveform (pulsating):
Summary:
- A full-wave rectifier uses four p-n junction diodes in a bridge configuration. - Both halves of the AC waveform are used, producing a pulsating DC output. - The output frequency is twice the input AC frequency. - Smoother DC can be obtained by adding a filter capacitor across the load resistor RL.
 

Answer 2

Step 1: Circuit description.
A centre-tapped transformer feeds two diodes \( D_1 \) and \( D_2 \), whose cathodes are joined at the load \( R_L \); the other end of \( R_L \) connects to the centre tap of the secondary winding, which is treated as the reference (zero) point.
Step 2: Express the input mathematically.
Let the voltage on each half of the secondary, relative to the centre tap, be \[ v(t) = V_m \sin(\omega t) \] where the two halves are always equal in magnitude but opposite in phase relative to the centre tap.
Step 3: Track conduction over one cycle.
For \( 0 < \omega t < \pi \) (positive half-cycle), the upper half of the winding drives \( D_1 \) into forward conduction while \( D_2 \) stays reverse biased, so the current through \( R_L \) follows \( v(t) = V_m \sin(\omega t) \). For \( \pi < \omega t < 2\pi \), the lower half of the winding drives \( D_2 \) into conduction while \( D_1 \) is off, and because the geometry is mirrored, the current again flows through \( R_L \) in the same direction, following \( |V_m \sin(\omega t)| \).
Step 4: Output expression.
Combining both halves, the load sees \[ v_{\text{out}}(t) = V_m |\sin(\omega t)| \] for every cycle. Since \( |\sin(\omega t)| \) repeats itself every \( \pi \) radians instead of every \( 2\pi \) radians, the output has exactly twice the frequency of the input: \[ f_{\text{out}} = 2f_{\text{in}} \]
Step 5: Waveform picture.
The input trace is a symmetric sine wave crossing zero twice per cycle. The output trace is that same sine wave with every negative lobe flipped upward, so it consists of unbroken positive humps of value \( V_m|\sin(\omega t)| \), never touching zero for an extended interval and repeating at double the input rate. This full-wave output is easier to smooth into steady DC with a filter than a half-wave output, since the gaps between pulses are shorter.