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 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:
Show Hint
- \( D_1 \) is reverse biased, \( D_2 \) is forward biased
- \( D_1 \) and \( D_2 \) both are forward biased
- \( D_1 \) and \( D_2 \) both are reverse biased
- \( D_1 \) is forward biased, \( D_2 \) is reverse biased
The Correct Option is D
Solution and Explanation
To ascertain the diode biasing in a full-wave rectifier at \(t = 15\) ms, we first determine the input voltage at this instant. The input voltage is defined by the equation:
\[ V_{in} = 220 \sin(100 \pi t) \]
Substituting \(t = 15\) ms (equivalent to \(t = 0.015\) seconds):
\[ V_{in} = 220 \sin(100 \pi \times 0.015) \]
The argument of the sine function evaluates to:
\[ 100 \pi \times 0.015 = 1.5 \pi \]
Given that \(\sin(1.5 \pi) = -1\), the input voltage is calculated as:
\[ V_{in} = 220 \times (-1) = -220 \text{ volts} \]
In a full-wave rectifier, \(D_1\) is typically forward-biased and \(D_2\) reverse-biased during the positive half-cycle of the input. Conversely, during the negative half-cycle, the input voltage is negative, causing \(D_1\) to be reverse-biased and \(D_2\) to be forward-biased. At \(t = 15\) ms, the input voltage is \(-220\) volts, corresponding to the negative half-cycle. Therefore, \(D_1\) will be reverse-biased, and \(D_2\) will be forward-biased.
Consequently, the diode biasing is: \(D_1\) is reverse biased, \(D_2\) is forward biased.
