• $summary_{type}$: application
• $word_{count}$: 150

Rectifiers convert AC to DC using diode's one-way property. Half-wave: single diode passes one half-cycle. $V_{DC}$ = $\frac{V_0}{pi}$ = 0.318*$V_0$. Output frequency = input frequency. Ripple factor = 1.21 (high ripple). Full-wave (center-tap): two diodes, each conducting alternate halves. $V_{DC}$ = 2$V_0$/pi = 0.637*$V_0$. Output frequency = 2*input frequency. Ripple factor = 0.48. Bridge rectifier: four diodes, no center tap needed. Same performance as full-wave. A capacitor filter across the output smooths the ripple. Larger C gives smoother DC but slower response to load changes. With filter: $V_{DC}$ ≈ $V_0$ - $\frac{I_DC}{2fC}$. The key metric is ripple factor = $\frac{V_ac_rms}{V_DC}$ — lower is better. For JEE: know the circuit diagrams, output waveform sketches, and the doubling of output frequency in full-wave rectification. The diode drops ~0.7 V per junction in the output path.