A sinusoidal voltage of amplitude 25 volt and frequency 50 Hz is applied to a half wave rectifier using P-n junction diode. No filter is used and the load resistor is 1000Ω. The forward resistance R_f of ideal diode is 10Ω. The percentage rectifier efficiency is

Concept:

Rectifier Efficiency:

• The rectifier efficiency η is defined as the ratio of the DC power to the AC power supplied to the rectifier.
• The DC power is calculated using the formula: P_dc = I_dc² × R_L
• The AC power is calculated using the formula: P_ac = I_rms² × (R_f + R_L)
• Rectifier Efficiency Formula: η = (P_dc / P_ac) × 100
• Given Values:
  • V_m = 25 V
  • R_L = 1000 Ω
  • R_f = 10 Ω
  • f = 50 Hz

Calculation:

Given,

V_m = 25 V, R_L = 1000 Ω, R_f = 10 Ω

 The maximum current is given by: I_m = V_m / (R_f + R_L) = 25 / (10 + 1000) = 24.75 mA

 The DC current is: I_dc = I_m / π = 24.75 / 3.14 = 7.87 mA

 The RMS current is: I_rms = I_m / 2 = 24.75 / 2 = 12.37 mA

 The DC power is: P_dc = I_dc² × R_L = (7.87 × 10^-3)² × 10³ = 61.9 mW

 The AC power is: P_ac = I_rms² × (R_f + R_L) = (12.37 × 10^-3)² × (10 + 1000) = 154.54 mW

 The rectifier efficiency is: η = (P_dc / P_ac) × 100 = (61.9 / 154.54) × 100 = 40.05%

∴ The rectifier efficiency is 40.05%.