With a neat circuit diagram, explain the working of a full wave rectifier. Draw input–output waveforms.

Answer: A full wave rectifier (bridge type) uses four diodes so that current flows through the load in the same direction during both half cycles of the a.c. input, giving a pulsating d.c. output.

Circuit (described in words): The a.c. supply is connected to the primary of a transformer. The two ends A and B of the secondary are connected to two opposite corners of a bridge of four diodes D₁, D₂, D₃, D₄. The load resistance \(R_L\) is connected between the other two corners C (where the cathodes of D₁ and D₂ meet, the positive output) and D (where the anodes of D₃ and D₄ meet). D₁ goes from A to C, D₂ from B to C, D₃ from D to B and D₄ from D to A.

Working:

• Positive half cycle (A positive w.r.t. B): D₁ and D₃ are forward biased, D₂ and D₄ are reverse biased. Current flows A → D₁ → C → \(R_L\) → D → D₃ → B.
• Negative half cycle (B positive w.r.t. A): D₂ and D₄ conduct, D₁ and D₃ are off. Current flows B → D₂ → C → \(R_L\) → D → D₄ → A.

In both half cycles the current through \(R_L\) flows from C to D, i.e. in the same direction, so both halves of the input are used. The output is unidirectional but pulsating; its ripple frequency is twice the input frequency. A filter capacitor across \(R_L\) smooths it.

Waveforms (described): Input — a sine wave with positive and negative half cycles about the time axis. Output — a series of positive half-sine humps, one for every half cycle of the input, with no gaps (the negative halves of the input appear flipped above the axis).

(A centre-tap full wave rectifier with two diodes and a centre-tapped transformer is also acceptable: each diode conducts in alternate half cycles and the load current is in the same direction.)