Power Supplies and Rectifiers
Power Supplies and Rectifiers tests conversion from AC mains to regulated DC: transformer selection, rectifier configuration, filtering, ripple, and regulator choice. Expect waveform reasoning, average-voltage formulas, and safety-aware component ratings.
Core rules
- Half-wave rectifier: Vdc ≈ 0.318 Vm; ripple factor ≈ 1.21. Full-wave and bridge: Vdc ≈ 0.636 Vm; ripple factor ≈ 0.482.
- A bridge conducts in pairs, so two diode drops (≈1.4 V for silicon) appear in the loop.
- Capacitor-filter ripple for full-wave is approximately Vr ≈ Iload/(2fC), where f is the mains frequency; the 2 accounts for two charging pulses per cycle.
- PIV for a bridge is about Vm; for a centre-tapped full-wave rectifier it is about 2Vm.
- A zener shunt regulator needs a series resistor to drop the excess voltage and limit current. A linear series regulator wastes (Vin − Vout) × Iload as heat.
Worked example
A 12 V RMS transformer feeds a bridge with 1.4 V total diode drop and a 4700 µF filter. Peak secondary = 12 × 1.414 = 16.97 V; DC peak after bridge ≈ 15.57 V. For 0.5 A load at 50 Hz, Vr ≈ 0.5/(2 × 50 × 0.0047) = 1.06 V. Average output ≈ 15.57 − 1.06/2 ≈ 15.0 V before regulation.
Traps
Using 12 V RMS as if it were peak, forgetting the bridge drop, and using f instead of 2f in the ripple formula are frequent. A zener without a series resistor will overheat. Filter capacitors need a voltage rating above the unloaded peak, not the RMS value. Never treat a transformer as ideal when selecting fuse and diode ratings.