For a half-wave rectifier having diode voltage VD and supply input of VI, the diode conducts for π – 2Θ, where Θ is given by

To solve this problem, we need to break down the information provided about the half-wave rectifier and the relationship between the diode voltage ($V_D$), supply input ($V_I$), and the conduction angle ($\pi - 2\Theta$). Here are the steps:

1. Break Down the Problem

1. Define the conduction angle in the context of a half-wave rectifier.
2. Identify the role of the diode voltage ($V_D$) and supply input ($V_I$).
3. Relate these elements to the given angle $\Theta$.

2. Relevant Concepts

1. In a half-wave rectifier, the diode will conduct when the input voltage exceeds the forward voltage drop of the diode, which can be expressed as: $V_I \sin(\omega t) > V_D$ where $\omega$ is the angular frequency of the supply.

2. The conduction angle for the diode can be derived from the inequality above, leading to: $\sin(\omega t) > \frac{V_D}{V_I}$

3. The angle at which the diode starts to conduct is given by: $\Theta = \arcsin\left(\frac{V_D}{V_I}\right)$

3. Analysis and Detail

1. The total angle of conduction for the diode is given as $\pi - 2\Theta$.

2. We already established that $\Theta = \arcsin\left(\frac{V_D}{V_I}\right)$, where:

   • $\Theta$ is the angle beyond which the diode begins to conduct.
   • The diode will stop conducting at the point where the supply voltage goes negative relative to the diode.

3. Therefore, substituting $\Theta$ into the total conduction angle: $\text{Conduction Angle} = \pi - 2 \cdot \arcsin\left(\frac{V_D}{V_I}\right)$

4. Verify and Summarize

• The derived relationship indicates how the voltage across the diode and the input supply affect the conduction angle in a half-wave rectifier.
• The calculations consistently demonstrate that as the ratio of $V_D$ to $V_I$ changes, so will the conduction angle, illustrating the diode’s behavior in AC circuits.

Final Answer

The angle $\Theta$ is given by: $\Theta = \arcsin\left(\frac{V_D}{V_I}\right)$ Thus, the total conduction angle for the diode is: $\text{Conduction Angle} = \pi - 2 \cdot \arcsin\left(\frac{V_D}{V_I}\right)$