Question

A sinusoidal voltage of amplitude 25 V and frequency 50 Hz is applied to a half-wave rectifier using a P-N junction diode. No filter is used, and the load resistor is \( 1000\Omega \). The forward resistance \( R_f \) of the ideal diode is \( 10\Omega \). The percentage rectifier efficiency is:

• A. \( 40\% \)
• B. \( 20\% \)
• C. \( 30\% \)
• D. \( 15\% \)

Hint:

For a half-wave rectifier:
- The theoretical efficiency is \( 40.6\% \).
- Full-wave rectifiers have higher efficiency (about \( 81.2\% \)).
- The presence of a filter improves the DC component.

Answer 1

The Correct Option is A

To determine the percentage rectifier efficiency of a half-wave rectifier, we use the formula:
\[ \eta = \frac{P_{{DC}}}{P_{{AC}}} \times 100 \] where:
- \( P_{{DC}} \) is the DC power delivered to the load.
- \( P_{{AC}} \) is the AC power supplied to the rectifier.
Step 1: Given Data
- Peak voltage of AC supply: \( V_m = 25V \)
- Frequency: \( f = 50 \) Hz (not required for efficiency calculation)
- Load resistance: \( R_L = 1000\Omega \)
- Forward resistance of diode: \( R_f = 10\Omega \)
Step 2: Calculate the RMS Value of AC Input Voltage
The RMS value of the input AC voltage is given by:
\[ V_{{rms}} = \frac{V_m}{\sqrt{2}} \] Substituting the values:
\[ V_{{rms}} = \frac{25}{\sqrt{2}} = 17.68V \] Step 3: Calculate the DC Output Voltage
For a half-wave rectifier, the DC output voltage is given by:
\[ V_{{DC}} = \frac{V_m - I_{{DC}} R_f}{\pi} \] Since \( I_{{DC}} \) is unknown, we approximate \( V_{{DC}} \) as:
\[ V_{{DC}} \approx \frac{V_m}{\pi} = \frac{25}{\pi} = 7.96V \] Step 4: Calculate the DC Power Delivered to Load
The DC output current is:
\[ I_{{DC}} = \frac{V_{{DC}}}{R_L} = \frac{7.96}{1000} = 7.96 { mA} \] Thus, the DC power delivered to the load is:
\[ P_{{DC}} = V_{{DC}} I_{{DC}} = 7.96V \times 7.96 \times 10^{-3} A \] \[ = 63.37 { mW} \] Step 5: Calculate the AC Power Supplied to Rectifier
The AC power supplied to the rectifier is given by:
\[ P_{{AC}} = \frac{V_{{rms}}^2}{R_{{eq}}} \] where \( R_{{eq}} \) is the equivalent resistance:
\[ R_{{eq}} = R_L + R_f = 1000 + 10 = 1010\Omega \] \[ P_{{AC}} = \frac{(17.68)^2}{1010} \] \[ = \frac{312.64}{1010} = 0.3098 W = 309.8 { mW} \] Step 6: Calculate Efficiency
\[ \eta = \frac{P_{{DC}}}{P_{{AC}}} \times 100 \] \[ = \frac{63.37}{309.8} \times 100 \] \[ = 40\% \] Thus, the percentage rectifier efficiency is:
\[ {40\%} \]