Q.Explain full wave rectifying action of $p$-$n$ junction diode with the help of circuit diagram. OR Draw the diffraction pattern of light due to single slit and write the expression for position of minima. On what factors does the width of central maxima depend?

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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