Question

Discuss the working method of a half-wave rectifier using a p-n junction diode.

Hint:

A diode conducts only when forward biased: it passes current during one half of the AC cycle and blocks the other, giving a pulsating DC output across the load.

Answer 1

Step 1: Key property used.
A p-n junction diode acts like a one-way valve for current: it passes current when the p-region is at higher potential (forward bias) and stops it when the n-region is at higher potential (reverse bias). This directional conduction is what makes rectification possible.

Step 2: Connections.
An input transformer steps the mains AC to the required level. One end of the secondary goes to the anode (p-side) of the diode; the cathode (n-side) connects through the load resistor \(R_L\) back to the other end of the secondary. Output is measured across \(R_L\).

Step 3: Follow one full input cycle.
For the first half of the input cycle the anode is positive, the junction is forward biased, and a current \(i\) flows through \(R_L\) producing an output voltage \(V_{out} = iR_L\).

Step 4: The blocked half.
In the second half the anode goes negative, the junction is reverse biased, only a negligible leakage current flows, and \(V_{out} \approx 0\).

Step 5: Nature of the output.
The result is a series of one-directional voltage pulses, one per input cycle, i.e. pulsating DC. Because output exists for only one of the two half cycles, the arrangement is a half-wave rectifier; a filter (capacitor) can later smooth the ripple.
\[\boxed{\text{Only positive half cycles appear at the output } (V_{out}=iR_L)}\]