Full-Wave Rectifier
Bridge of 4 diodes — V_out = |V_in| − 1.4 V with optional capacitor filter.
Key Notes
A full-wave rectifier inverts the negative half of the AC cycle, producing pulses for BOTH halves.
Two common topologies: (i) Centre-tapped transformer with 2 diodes; (ii) bridge rectifier with 4 diodes (no centre tap).
Output frequency = 2× input frequency (100 Hz from 50 Hz mains).
Average (DC) output: V_dc = 2V_m/π ≈ 0.637 V_m. DOUBLE that of half-wave.
RMS: V_rms = V_m/√2.
Ripple factor: γ ≈ 0.482 — much better than half-wave's 1.21.
Efficiency: ~81.2% — twice half-wave.
PIV: 2V_m for centre-tap; V_m for bridge. Bridge is preferred because diode rating is half — but uses 4 diodes.
Combined with a filter capacitor and (optionally) a voltage regulator, you get a usable DC supply.
Formulas
DC output
Average over both rectified halves.
RMS output
Same as RMS of original sine wave (entire cycle utilised).
Ripple factor
Much better than half-wave.
Efficiency
Maximum theoretical for ideal diodes, resistive load.
PIV
Bridge wins on diode stress.
Important Points
Bridge rectifier is the standard topology in modern power supplies — needs no center tap.
Output ripple at 2f (100 Hz in India) is easier to filter than 50 Hz half-wave ripple.
Adding a smoothing capacitor: output approaches V_m with small ripple proportional to 1/(fRC).
For Indian mains, 220 V_rms ⇒ V_m = 311 V ⇒ V_dc(no filter) ≈ 198 V, V_dc(capacitor-filtered) ≈ 310 V.
Diode forward drop: ~0.7 V per diode in conduction. Bridge has 2 diodes in series each half-cycle ⇒ 1.4 V drop.
Switching power supplies use high-frequency rectification — much smaller filter components.
Full-Wave Rectifier notes from sciphylab (also known as SciPhy, SciPhy Lab, SciPhy Labs, Physics Lab). Class 12 physics revision for JEE Mains, JEE Advanced, NEET UG, AP Physics 1/2/C, SAT, and CUET-UG.