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:

• The efficiency of a half-wave rectifier is the ratio of the DC power delivered to the load to the AC power supplied to the rectifier.
• The formula for rectifier efficiency (η) is:
• η = P_DC / P_AC
• The current equations for a half-wave rectifier are as follows:
  • Peak current: I_m = V_m / (R_L + R_F)
  • DC current: I_DC = I_m / π
  • RMS current: I_rms = I_m / √2
• DC power: P_DC = (I_DC)² × R_L
• AC power: P_AC = (I_rms)² × (R_L + R_F)

Calculation:

Amplitude of the voltage (Vm) = 25 V

Frequency = 50 Hz

Load resistance (RL) = 1000 Ω

Forward resistance of diode (RF) = 100 Ω

Peak current:

⇒ I_m = V_m / (R_L + R_F)

⇒ I_m = 25 / (1000 + 100)

⇒ I_m = 25 / 1100

⇒ I_m = 24.75 mA

DC current:

⇒ I_DC = I_m / π

⇒ I_DC = 24.75 / 3.14

⇒ I_DC ≈ 7.87 mA

RMS current:

⇒ I_rms = I_m / √2

⇒ I_rms = 24.75 / 2

⇒ I_rms = 12.37 mA

DC power:

⇒ P_DC = (I_DC)² × R_L

⇒ P_DC = (7.87 × 10^-3)² × 10³

⇒ P_DC ≈ 61.9 mW

AC power:

⇒ P_AC = (I_rms)² × (R_L + R_F)

⇒ P_AC = (12.37 × 10^-3)² × (10 + 1000)

⇒ P_AC ≈ 154.54 mW

Efficiency:

⇒ η = P_DC / P_AC

⇒ η = (61.9 / 154.54) × 100

⇒ η ≈ 40.05%

∴ The rectifier efficiency is 40.05%.