Kerala Plus Two Electronics (DHSE): Power Supplies and Rectifiers

Power Supplies and Rectifiers tests the chain from mains AC to a steady DC rail: transformer, rectifier, filter and regulator. Expect comparison of half-wave, centre-tap full-wave and bridge rectifiers, ripple calculations and fixed-voltage regulator pinouts.

Core concepts

• Block diagram: transformer steps down and isolates; rectifier converts AC to pulsating DC; filter smooths; regulator holds the output constant against load and mains changes.
• Half-wave rectifier: one diode, PIV = Vm, ripple factor 1.21, maximum efficiency 40.6%, output ripple frequency equal to supply frequency.
• Full-wave centre-tap: two diodes, PIV = 2Vm, ripple factor 0.482, efficiency 81.2%, ripple frequency 2f.
• Bridge rectifier: four diodes, PIV = Vm, same ripple factor and efficiency as centre-tap, no centre-tapped transformer needed.
• Capacitor filter: charges near the peak and discharges into the load. Larger C and larger RL reduce ripple; for full wave, r ≈ 1/(4√3 f C RL), where f is the supply frequency.
• Regulators: Zener shunt is simple but inefficient; series-pass and 78xx/79xx fixed regulators are common. A 7805 gives +5 V with pin order input–ground–output; dropout voltage must be respected.

Worked example

A bridge rectifier uses a 12 V rms secondary at 50 Hz. Peak voltage is Vm = 12 × 1.414 = 16.97 V. Each diode’s PIV is approximately Vm = 17 V. With two diodes conducting, each dropping 0.7 V, the no-load peak is about 16.97 − 1.4 = 15.57 V. The ripple frequency is 2 × 50 = 100 Hz, not 50 Hz.

Traps that cost marks

• Using 2Vm for bridge PIV or Vm for centre-tap PIV.
• Forgetting that full-wave ripple frequency is double the supply frequency.
• Substituting rms voltage where peak voltage is needed.
• Reversing an electrolytic filter capacitor or ignoring its polarity.
• Assuming a 7805 pinout starts with the output pin; it does not.
• Ignoring regulator dropout: input must be sufficiently above the regulated output.