How it works
During the positive half-cycle, the diode forward biases and current flows through RL. During the negative half-cycle, the diode reverse biases and current is zero. The output is a series of positive pulses at the supply frequency f. DC output voltage is Vdc = Vm/π, where Vm is the peak of the secondary voltage. RMS output is Vrms = Vm/2. Ripple factor γ = √((Vrms/Vdc)² − 1) = 1.21, meaning the AC component in the output is 121% of the DC value — very high. Peak inverse voltage (PIV), which the diode must withstand during negative half-cycles, equals Vm. Rectifier efficiency is just 40.6%, the worst of all rectifier types.
Key points to remember
Key numbers to memorise: Vdc = Vm/π, ripple factor = 1.21, efficiency = 40.6%, and PIV = Vm. The ripple frequency equals the supply frequency (50 Hz in India), unlike the full wave rectifier where it doubles to 100 Hz. Adding a shunt capacitor filter reduces ripple; for light loads the output approaches Vm but the diode peak current increases sharply. Form factor (Vrms/Vdc) = π/2 ≈ 1.57. The transformer utilisation factor (TUF) is only 0.287, meaning the transformer is severely under-used — this is why half wave rectifiers are rarely used in practice beyond very low power applications.
Exam tip
Every Anna University question on rectifiers expects you to state ripple factor = 1.21, efficiency = 40.6%, and PIV = Vm for the half wave rectifier — write all three even if only one is asked, since partial credit is common.