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

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Answer 1

The efficiency $η$ of half wave rectifier is given by, $η = \frac{0 . 4 0 6 R _{L}}{r _{f} + R _{L}}$

$η = \frac{0 . 4 0 6}{1 + \frac{r _{f}}{R _{L}}}$

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

$η = (\frac{0 . 4 0 6}{1 + \frac{r _{f}}{R _{L}}}) \times 100$

$η = (\frac{0 . 4 0 6}{1 + \frac{2 0}{2 0 0 0}}) \times 100$

$η = (\frac{0 . 4 0 6}{1 + 0 . 0 1}) \times 100$

$η = (\frac{0 . 4 0 6}{1 . 0 1}) \times 100$

$η = (0.4019) \times 100 = 40.2$ $%$

Answer 2

The efficiency

η

of half wave rectifier is given by,

η = \frac{0 . 4 0 6 R _{L}}{r _{f} + R _{L}}

η = \frac{0 . 4 0 6}{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

η = (\frac{0 . 4 0 6}{1 + \frac{r _{f}}{R _{L}}}) \times 100

η = (\frac{0 . 4 0 6}{1 + \frac{2 0}{2 0 0 0}}) \times 100

η = (\frac{0 . 4 0 6}{1 + 0 . 0 1}) \times 100

η = (\frac{0 . 4 0 6}{1 . 0 1}) \times 100

η = (0.4019) \times 100 = 40.2

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