Question

A full wave rectifier circuit with diodes (\(D_1\)) and (\(D_2\)) is shown in the figure. If input supply voltage \(V_{in} = 220 \sin(100 \pi t)\) volt, then at \(t = 15\) msec: 

• A. \( D_1 \) is reverse biased, \( D_2 \) is forward biased
• B. \( D_1 \) and \( D_2 \) both are forward biased
• C. \( D_1 \) and \( D_2 \) both are reverse biased
• D. \( D_1 \) is forward biased, \( D_2 \) is reverse biased

Hint:

In a center-tapped full-wave rectifier, when the voltage across one half of the secondary transformer is positive, the corresponding diode conducts. When the input AC voltage reverses its polarity, the voltage across the other half of the secondary becomes positive, and the other diode conducts.

Answer 1

The Correct Option is D

To determine the biasing of diodes in the full wave rectifier at \(t = 15\) ms, we need to calculate the input voltage at this specific time instant. Given the input voltage equation:

\[ V_{in} = 220 \sin(100 \pi t) \]

We substitute \(t = 15\) ms (which is \(t = 0.015\) seconds) into the equation:

\[ V_{in} = 220 \sin(100 \pi \times 0.015) \]

Calculating the argument of the sine function:

\[ 100 \pi \times 0.015 = 1.5 \pi \]

Since \(\sin(1.5 \pi) = -1\), the input voltage becomes:

\[ V_{in} = 220 \times (-1) = -220 \text{ volts} \]

In a full wave rectifier, during the positive half-cycle of the input AC signal, usually \(D_1\) is forward biased and \(D_2\) is reverse biased. Conversely, during the negative half-cycle, the input voltage is negative, leading to \(D_1\) being reverse biased and \(D_2\) being forward biased. At \(t = 15\) ms, since the input voltage is \(-220\) volts (negative half-cycle), \(D_1\) becomes reverse biased, and \(D_2\) becomes forward biased.

Thus, the diodes' biasing is: \(D_1\) is forward biased, \(D_2\) is reverse biased.