Question

The diode in the circuit shown below is ideal. The input voltage (in Volts) is given by \[ V_I = 10 \sin(100\pi t), \quad {where time} \, t \, {is in seconds.} \] The time duration (in ms, rounded off to two decimal places) for which the diode is forward biased during one period of the input is (answer in ms).

 

Hint:

The diode is forward biased during the portion of the cycle where the input voltage exceeds the threshold of 5 V. This can be determined by solving the inequality for the sine wave.

Answer 1

The input voltage \( V_I = 10 \sin(100 \pi t) \) is a sinusoidal voltage with a peak value of 10 V and a frequency of 50 Hz, since \( \omega = 100 \pi \), meaning the frequency is: \[ f = \frac{100 \pi}{2 \pi} = 50 \, {Hz} \] Thus, the period \( T \) of the input voltage is: \[ T = \frac{1}{f} = \frac{1}{50} = 0.02 \, {seconds} = 20 \, {ms} \] Step 1: Determine the condition for forward biasing of the diode
The diode will be forward biased when \( V_I \) exceeds the threshold voltage of the diode, which is 5 V in this case. We need to find when: \[ V_I = 10 \sin(100 \pi t) \geq 5 \] This condition can be rewritten as: \[ \sin(100 \pi t) \geq 0.5 \] Solving for \( t \): \[ \sin(100 \pi t) = 0.5 \quad \Rightarrow \quad 100 \pi t = \arcsin(0.5) = \frac{\pi}{6} \] Thus, \[ t = \frac{1}{100 \pi} \times \frac{\pi}{6} = \frac{1}{600} \, {seconds} = 1.67 \, {ms} \] Step 2: Determine the time duration during one period
Now we need to calculate how long the input voltage remains above 5 V during each cycle. The sine wave reaches the value of 0.5 twice during each cycle — once as the voltage rises and again when it falls. Thus, the duration during which the diode is forward biased is twice this calculated time: \[ t_{{duration}} = 2 \times 1.67 \, {ms} = 13.32 \, {ms} \] Thus, the correct answer is 13.32 ms.