Question

The single-phase rectifier consisting of three thyristors \(T_1, T_2, T_3\) and a diode \(D_1\) feed power to a 10 A constant current load. \(T_1\) and \(T_3\) are fired at \(\alpha = 60^\circ\) and \(T_2\) is fired at \(\alpha = 240^\circ\). The reference for \(\alpha\) is the positive zero crossing of \(V_{in}\). The average voltage \(V_O\) across the load in volts is ............ (Round off to 2 decimal places). \begin{center} \includegraphics[width=0.5\textwidth]{29.jpeg} \end{center}

Hint:

For phase-controlled rectifiers, always apply average formula \(V_{dc} = \frac{V_m}{\pi}\cos\alpha\). Adjust for firing pattern symmetry.

Answer 1

Step 1: Input waveform.
Supply: \[ v_{in} = 100 \sin(100\pi t), V_m = 100 \]

Step 2: Controlled rectifier operation.
Average load voltage: \[ V_{dc} = \frac{V_m}{2\pi} \int_{\alpha}^{\pi+\alpha} \sin\theta \, d\theta \]

Step 3: Solve integral.
\[ V_{dc} = \frac{V_m}{2\pi} \left[ -\cos\theta \right]_{\alpha}^{\pi+\alpha} = \frac{V_m}{2\pi} \left( -\cos(\pi+\alpha) + \cos(\alpha) \right) \] \[ = \frac{V_m}{2\pi} \left( -(-\cos\alpha) + \cos\alpha \right) = \frac{V_m}{2\pi} (2\cos\alpha) \] \[ V_{dc} = \frac{V_m}{\pi} \cos\alpha \]

Step 4: Substitute values.
\(V_m = 100, \; \alpha = 60^\circ\): \[ V_{dc} = \frac{100}{\pi} \cdot \cos(60^\circ) = \frac{100}{\pi} \cdot 0.5 = \frac{50}{\pi} \] \[ V_{dc} \approx 15.92 \, V \] Correction (since conduction occurs over two intervals due to T2 firing at 240°): Effective average doubled: \[ V_{dc} \approx 31.83 \, V \]

Final Answer:
\[ \boxed{31.83 \, V} \]