Question

In half wave rectifier, a p-n diode with internal resistance $20$ $\Omega$ is used. If the load resistance of $2K\Omega$ is used in the circuit, then the efficiency of this half wave rectifier is (in percentage)

• A. $80.4\%$
• B. $40.2\%$
• C. $20\%$
• D. $50\%$

Answer 1

The correct option is B $40.2\%$

The efficiency $\eta$ of half wave rectifier is given by, $\eta =\frac{0.406R_L}{r_f+R_L}$

$\eta =\frac{0.406}{1+\frac{r_f}{R_L}}$

Where, $R_L$ is the resistance of load resistor and $r_f$ is the resistance of diode in forward biased condition.

Hence, the percent efficiency is

$\eta =\left(\frac{0.406}{1+\frac{r_f}{R_L}}\right)\times 100$

$\eta =\left(\frac{0.406}{1+\frac{20}{2000}}\right)\times 100$

$\eta =\left(\frac{0.406}{1+0.01}\right)\times 100$

$\eta =\left(\frac{0.406}{1.01}\right)\times 100$

$\eta =\left(0.4019\right)\times 100=40.2$$\%$