Question

How does the p-n junction diode is used as the half wave rectifier ? Explain its working by drawing simple circuit.

Hint:

For the rectifier, remember that the key is the diode's one-way-street behavior for current. The output is "half" of the input wave.

Answer 1

Step 1: Understanding the Concept:
Rectification is the process of converting an alternating current (AC) into a direct current (DC). A half-wave rectifier uses a single p-n junction diode to achieve this. The basic principle is that a p-n diode allows current to flow in only one direction—when it is forward-biased—and blocks current flow in the opposite direction—when it is reverse-biased.

Step 2: Simple Circuit Diagram:
The circuit consists of an AC input source (usually connected via a step-down transformer), a p-n junction diode (D), and a load resistor (\(R_L\)) across which the DC output is obtained. \begin{center} Circuit Diagram: An AC voltage source is connected to the primary coil of a transformer. The secondary coil is connected in series with a p-n diode (D) and a load resistor (\(R_L\)). The output voltage is measured across \(R_L\). \end{center}

Step 3: Working:
The working can be explained by considering the two half-cycles of the input AC voltage. \begin{enumerate} \item During the Positive Half-Cycle of AC Input: The upper end of the transformer's secondary coil becomes positive with respect to the lower end. This makes the p-side of the diode positive relative to the n-side, putting the diode in forward bias. The diode conducts current, and a current flows through the load resistor \(R_L\). Consequently, a voltage (output) is developed across \(R_L\), which follows the shape of the positive half-cycle of the input. \item During the Negative Half-Cycle of AC Input: The polarity of the secondary coil reverses. The upper end becomes negative with respect to the lower end. This makes the p-side of the diode negative relative to the n-side, putting the diode in reverse bias. In this state, the diode offers very high resistance and (ideally) does not conduct any current. Therefore, no current flows through \(R_L\), and the output voltage across it is zero. \end{enumerate} This process repeats for every cycle, resulting in an output voltage that consists only of the positive half-cycles of the original AC input. This output is a pulsating, unidirectional (DC) voltage.