Question

What is the typical efficiency of half-wave rectifier?

• A. \(40.6%\)
• B. \(81.2%\)
• C. \(21.3%\)
• D. \(50.0%\)

Hint:

Important standard efficiencies: \[ \text{Half-wave rectifier} = 40.6% \] \[ \text{Full-wave rectifier} = 81.2% \] Half-wave rectifier efficiency is lower because only one half-cycle of the AC signal is utilized.

Answer 1

The Correct Option is A

Concept: Rectifier efficiency is defined as the ratio of DC output power delivered to the load to the AC input power supplied to the rectifier. Mathematically: \[ \eta=\frac{P_{dc}}{P_{ac}} \] For a half-wave rectifier:
• Only one half-cycle of the AC input is utilized.
• The other half-cycle is blocked by the diode.
• Therefore, the efficiency is comparatively low. The maximum rectification efficiency of an ideal half-wave rectifier is: \[ \eta_{max}=40.6% \] This is a standard theoretical result obtained from the ratio of DC power to AC power.

Step 1: Recall the efficiency formula for rectifiers. Efficiency is: \[ \eta=\frac{\text{DC output power}}{\text{AC input power}} \] For half-wave rectifier: \[ \eta=\frac{P_{dc}}{P_{ac}} \]

Step 2: Write the standard current expressions for half-wave rectifier. For a half-wave rectifier: \[ I_{dc}=\frac{I_m}{\pi} \] and \[ I_{rms}=\frac{I_m}{2} \] where:
• \(I_m\) is peak current
• \(I_{dc}\) is average current
• \(I_{rms}\) is RMS current

Step 3: Derive the rectification efficiency. DC output power: \[ P_{dc}=I_{dc}^2R_L \] Substituting: \[ P_{dc}=\left(\frac{I_m}{\pi}\right)^2R_L \] \[ P_{dc}=\frac{I_m^2R_L}{\pi^2} \] AC input power: \[ P_{ac}=I_{rms}^2R_L \] Substituting: \[ P_{ac}=\left(\frac{I_m}{2}\right)^2R_L \] \[ P_{ac}=\frac{I_m^2R_L}{4} \] Now: \[ \eta=\frac{P_{dc}}{P_{ac}} \] Substituting: \[ \eta=\frac{\frac{I_m^2R_L}{\pi^2}}{\frac{I_m^2R_L}{4}} \] Cancelling common terms: \[ \eta=\frac{4}{\pi^2} \] \[ \eta=\frac{4}{(3.14)^2} \] \[ \eta\approx0.406 \] Converting into percentage: \[ \eta=40.6% \]

Step 4: Write the final answer. Therefore, the typical efficiency of a half-wave rectifier is: \[ \boxed{40.6%} \] Hence, the correct option is: \[ \boxed{(A)\ 40.6%} \]