• Tags: rectifier, half-wave, full-wave
• Difficulty: Moderate

A rectifier converts AC to DC using the one-way property of diodes. Half-wave rectifier: single diode passes only the positive half of the AC cycle. Output frequency = input frequency. $V_{DC}$ = $\frac{V_0}{pi}$. Ripple factor = 1.21 (poor). Full-wave rectifier (center-tapped transformer): two diodes, each conducting during alternate halves. Output frequency = 2 x input frequency. $V_{DC}$ = 2*$V_0$/pi. Ripple factor = 0.48 (better). Bridge rectifier: four diodes, no center tap needed. Same output as full-wave. A capacitor filter across the output smooths the ripple: larger C and larger $R_L$ give smoother DC. With filter: $V_{DC}$ ≈ $V_0$ - $\frac{I_DC}{2fC}$. JEE primarily asks conceptual questions: sketch output waveforms, identify circuit type, and compare efficiencies.