A half-wave rectifier uses a diode with forward resistance Rf = 20 Ω connected to a load resistance

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Question

A half-wave rectifier uses a diode with forward resistance Rf = 20 Ω connected to a load resistance RL = 980 Ω and an AC source of peak voltage 10 V. The efficiency of rectification is approximately:

Accepted Answer

Correct Answer: A — 40.6%Step-by-step Solution:For a half-wave rectifier:Peak current: Im = Vm/(Rf + RL) = 10/(20 + 980) = 10/1000 = 0.01 ADC (average) current: Idc = Im/πRMS current: Irms = Im/2Efficiency of rectification:η = Pdc/Pinput(AC) = (Idc² × RL) / (Irms² × (Rf + RL))= [(Im/π)² × RL] / [(Im/2)² × (Rf + RL)]= [RL/π²] / [(Rf + RL)/4]= 4RL / [π²(Rf + RL)]= 4 × 980 / [9.87 × 1000] = 3920/9870 ≈ 0.397 ≈ 40%Maximum efficiency of a half-wave rectifier = 40.6% (when Rf → 0).Key Concepts from Semiconductor Electronics (NCERT Class 12, Chapter 14):For a half-wave rectifier: ηmax = 40.6% (when Rf L).For a full-wave rectifier: ηmax = 81.2%.Idc (HWR) = Im/π; Idc (FWR) = 2Im/π.Irms (HWR) = Im/2; Irms (FWR) = Im/√2.Ripple factor (HWR) = 1.21; (FWR) = 0.48. Lower ripple = smoother DC output.

A half-wave rectifier uses a diode with forward resistance Rf = 20 Ω connected to a load resistance RL = 980 Ω and an AC source of peak voltage 10 V. The efficiency of rectification is approximately:

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A half-wave rectifier uses a diode with forward resistance Rf = 20 Ω connected to a load resistance RL = 980 Ω and an AC source of peak voltage 10 V. The efficiency of rectification is approximately:

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