Concept:By rectifying both half-cycles, a full-wave rectifier doubles the DC output component (\( 2V_m/\pi \) vs \( V_m/\pi \)) and shifts ripple energy to higher harmonic frequencies, significantly reducing the ratio of AC ripple to DC output.
Formula:$$r = \sqrt{\left(\frac{V_{\text{rms}}}{V_{\text{dc}}}\right)^2 - 1} \implies r_{\text{full-wave}} = 0.482 < r_{\text{half-wave}} = 1.21$$
Solution:- Half-wave rectifiers have dead-time gaps during alternate half-cycles, leaving large AC ripple content (\( r = 1.21 \)).
- Full-wave rectifiers deliver continuous output pulses, doubling the average DC voltage (\( V_{\text{dc}} = 0.637 V_m \)) and lowering the ripple factor to \( 0.482 \).
Why other options are incorrect:- Option A: Full-wave rectifiers conduct over the full \( 360^\circ \) period.
- Option B: Full-wave rectifiers shift ripple to even harmonics (\( 2f, 4f \)), but do not eliminate them completely.
- Option C: Half-wave rectifiers produce a non-zero DC output of \( V_m / \pi \).
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