CBSE 2025, Set 1

Question

Differentiate between half-wave and full-wave rectification. With the help of a circuit diagram, explain the working of a full-wave rectifier.

Edvaya · Accepted Answer

The correct answer is Subjective. **Difference between half-wave and full-wave rectification:** | Feature | Half-wave Rectifier | Full-wave Rectifier | |---|---|---| | Output cycles | Only one half-cycle (positive or negative) | Both half-cycles converted to same polarity | | Efficiency | ~40.6% | ~81.2% | | Ripple frequency | Equal to input frequency ($f$) | Twice input frequency ($2f$) | | Number of diodes | 1 | 2 (centre-tap) or 4 (bridge) | | Output ripple | High | Lower | **Half Wave & Full Wave Rectification** ![Wave Diagram](https://www.edvaya.com/images/cbse/physics/2025/set1/q25-wave-diagram.webp) **Full Wave Rectifier** ![Full-wave rectifier circuit diagram](https://www.edvaya.com/images/cbse/physics/2025/set1/q25-full-wave-rectifier-circuit.webp) The circuit uses a **centre-tap transformer** with two diodes $D_1$ and $D_2$, and a load resistor $R_L$. **Working:** **During positive half-cycle** of input AC: Terminal A is at higher potential than centre-tap B. Diode $D_1$ is forward biased and conducts; diode $D_2$ is reverse biased and does not conduct. Current flows through $D_1 	o R_L 	o$ centre-tap. **During negative half-cycle:** Terminal A is at lower potential. Now diode $D_2$ is forward biased and conducts (current flows through lower half); $D_1$ is reverse biased. Current again flows through $R_L$ in the **same direction**. Since current flows through $R_L$ in the same direction during both half-cycles, we get a **unidirectional (pulsating DC)** output at twice the input frequency. > **Key formula:** For a full-wave rectifier, ripple frequency = $2f$ (twice the input AC frequency). The ripple factor $\gamma = 0.48$, and rectification efficiency $\approx 81.2\%$.

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