Q.With the help of a diagram, explain the working of a full-wave rectifier, using a p-n junction diode. ($1+\tfrac12+\tfrac12+1=3$) OR Draw the forward characteristic curve of a p-n junction diode. Explain the terms depletion region and barrier potential or junction potential. (1+1+1=3)

Full-Wave Rectifier: From Intuition to Precision

Imagine you have a tap that gives water in alternating bursts — forward, then backward, then forward again. You want a steady stream that always flows in one direction. A half-wave rectifier is like a one-way flap that only lets the forward bursts through, wasting the backward ones. A full-wave rectifier is smarter: it flips the backward bursts around so they also flow forward. Nothing is wasted.

That is the core idea. The AC voltage swings positive and negative. A full-wave rectifier takes both halves of the cycle and makes them contribute to a unidirectional (DC) output.

How It Works (The Intuition)

A full-wave rectifier uses either:

• Two diodes and a centre-tapped transformer, or
• Four diodes in a bridge configuration (the bridge rectifier — far more common).

In both cases, the diodes are arranged so that during the positive half-cycle, one pair conducts, and during the negative half-cycle, the other pair conducts — but the current through the load always flows the same way.

The Precise Statement

A full-wave rectifier is a circuit that converts the entire input AC waveform (both positive and negative half-cycles) into a pulsating DC output. The output voltage is always of the same polarity, and its ripple frequency is twice the input AC frequency.

For a sinusoidal input $V_i = V_m \sin(\omega t)$, the output voltage (ideal diodes) is:

That absolute value is the mathematical signature of full-wave rectification.

Key Result: Ripple Frequency

If the input AC has frequency $f$ (e.g., 50 Hz), the output ripple frequency is:

Why? Because each input cycle gives two output pulses — one from the positive half and one from the inverted negative half. So for 50 Hz mains, you get 100 pulses per second. This makes filtering much easier than in a half-wave rectifier (where ripple frequency equals $f$).

Average (DC) Output Voltage

For a full-wave rectifier with a sinusoidal input of peak voltage $V_m$, the average (DC) output voltage is:

Compare this to a half-wave rectifier, where $V_{\text{dc}} = V_m/\pi$. The full-wave gives double the average output for the same peak input — another reason it is more efficient.

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