Q.With the help of a circuit diagram, explain the working of a p-n junction diode as a full wave rectifier. Draw its input and output waveforms.

Full-Wave Rectifier Using P-N Junction Diodes

A p-n junction diode conducts only when forward-biased, so it acts as a one-way valve for current. A full-wave rectifier uses two diodes with a centre-tapped transformer so that both halves of the AC input drive current through the load in the same direction.

Circuit: the AC mains feed the primary of a transformer whose secondary is centre-tapped. The two ends A and B of the secondary connect through diodes $D_1$ and $D_2$ respectively to one terminal of the load $R_L$; the other terminal of $R_L$ returns to the centre tap. The voltages of A and B relative to the centre tap are always equal and opposite ($180^\circ$ out of phase).

Operation in the two half-cycles:

1. Positive half-cycle: end A is positive with respect to the centre tap, so $D_1$ is forward-biased (conducts) while $D_2$ is reverse-biased (blocks). Current flows A → $D_1$ → $R_L$ → centre tap.
2. Negative half-cycle: end B is now positive, so $D_2$ conducts and $D_1$ blocks. Current flows B → $D_2$ → $R_L$ → centre tap — through the load in the same direction as before. …

The Concept: Why Full-Wave Rectification?

A p-n junction diode conducts only when forward-biased (p-side positive, n-side negative) and blocks current when reverse-biased. This one-way valve property makes it perfect for rectification — converting alternating current into direct current.

In half-wave rectification, we waste half the AC cycle by blocking it entirely. Full-wave rectification is smarter: it makes the negative half-cycle also contribute, so both halves of the AC input deliver current to the load. The result is a higher average output voltage, less ripple, and better utilisation of the transformer.

The centre-tapped transformer design achieves this elegantly by providing two secondary voltages that are $180^\circ$ out of phase with each other. While one diode conducts during the positive half-cycle of its input, the other conducts during what would have been the negative half-cycle — which, from its own end of the winding, is a positive half-cycle.

The Circuit

The circuit consists of:

• A transformer whose primary coil is connected to the AC mains.
• A centre-tapped secondary with three terminals: the two ends A (top) and B (bottom), and the centre tap C.
• Two diodes: the anode of $D_1$ is connected to end A and the anode of $D_2$ to end B. Their cathodes are joined together and connected to one terminal of the load resistor $R_L$.
• The other terminal of $R_L$ returns to the centre tap C, which serves as the common reference.

Because C is the midpoint of the winding, the instantaneous voltages of A and B measured from C are always equal in magnitude and opposite in sign: if $V_{AC} = +V_m\sin(\omega t)$, then $V_{BC} = -V_m\sin(\omega t)$.

Working Principle: Step-by-Step

1. The centre tap creates two anti-phase voltages

The centre-tapped secondary effectively gives us two AC sources of equal magnitude and opposite polarity, sharing the common terminal C.

2. During the positive half-cycle of the input (0 to $\pi$)

When terminal A is positive with respect to C:

• Diode $D_1$ is forward-biased (its anode at A is positive) and conducts.
• Diode $D_2$ is reverse-biased (its anode at B is negative relative to C) and blocks.
• Current flows along the path A → $D_1$ → load $R_L$ → C.
• The load receives a positive half-sine pulse of voltage.

3. During the negative half-cycle of the input ($\pi$ to $2\pi$)

When terminal A swings negative with respect to C, terminal B becomes positive:

• Diode $D_1$ is now reverse-biased and blocks.
• Diode $D_2$ is forward-biased and conducts.
• Current flows along the path B → $D_2$ → load $R_L$ → C.
• The load still receives current in the same direction — from the joined cathodes toward the centre tap.

4. The key result: unidirectional current

In both half-cycles, current through the load flows in the same direction. The load simply receives one positive half-sine pulse after another — pulsating DC.

5. Output characteristics

• Frequency: the output ripple frequency is $2f$, where $f$ is the input AC frequency — each input cycle produces two output pulses.
• Average DC voltage: $V_{dc} = \frac{2V_m}{\pi}$, where $V_m$ is the peak secondary voltage measured from the centre tap to either end. …