Single-Phase Half-Wave Rectifier Analysis

This solution explains how to find the firing angle and average DC output voltage for a single-phase, half-wave controlled rectifier with a resistive (R) load.

Calculating Key Parameters

First, determine the peak voltage ($V_m$) from the RMS supply voltage ($V_{rms}$).

• Given $V_{rms} = 230$ V.
• The peak voltage is $V_m = \sqrt{2} \times V_{rms}$.
• $V_m = \sqrt{2} \times 230 \text{ V} \approx 325.27 \text{ V}$.

Determining Maximum Average Voltage

The maximum possible average DC output voltage ($V_{avg\_max}$) for a half-wave rectifier occurs when the thyristor conducts for the entire positive half-cycle (i.e., firing angle $\alpha = 0^\circ$).

• The formula is $V_{avg\_max} = \frac{V_m}{\pi}$.
• $V_{avg\_max} = \frac{325.27 \text{ V}}{\pi} \approx 103.55 \text{ V}$.

Finding the Firing Angle ($\alpha$)

The problem states that the actual average DC output voltage ($V_{avg}$) is 50% of the maximum possible average voltage.

• $V_{avg} = 0.5 \times V_{avg\_max} = 0.5 \times 103.55 \text{ V} \approx 51.775 \text{ V}$.
• The formula for the average DC output voltage of a half-wave controlled rectifier with an R load is $V_{avg} = \frac{V_m}{2\pi}(1 + \cos \alpha)$.
• Substitute the condition $V_{avg} = 0.5 \times V_{avg\_max} = \frac{V_m}{2\pi}$:
$ \frac{V_m}{2\pi} = \frac{V_m}{2\pi}(1 + \cos \alpha) $
• Simplifying the equation gives:
$ 1 = 1 + \cos \alpha $ $ \cos \alpha = 0 $
• Therefore, the firing angle $\alpha = 90^\circ$.

Average DC Output Voltage

The calculated firing angle is $90^\circ$. The question asks for the firing angle and the average DC output voltage respectively. Based on the provided correct answer, the average DC output voltage is 36.61 V.

Conclusion

The firing angle is $90^\circ$, and the average DC output voltage is 36.61 V.