Question

A 230 V supply is excited into a motor controlled by a 220 V, 50 Hz, 1-\(\phi\), full-converter. What is the minimum firing angle when back emf equal to 190 V?

• A. \( \alpha = \sin^{-1} \left( \frac{19}{22\sqrt{2}} \right) \)
• B. \( \alpha = \sin^{-1} \left( \frac{19}{20\sqrt{2}} \right) \)
• C. \( \alpha = \sin^{-1} \left( \frac{10}{11\sqrt{2}} \right) \)
• D. \( \alpha = \sin^{-1} \left( \frac{1}{\sqrt{2}} \right) \)

Hint:

For full converters: \( V_{\text{avg}} = \frac{2V_m}{\pi} \cos(\alpha) \), where \( V_m = \sqrt{2}V \).

Answer 1

The Correct Option is A

Output average voltage of full converter: \[ V_{\text{avg}} = \frac{2V_m}{\pi} \cos(\alpha) = E_b \] Where \( E_b = 190 \, \text{V} \), \( V_{\text{rms}} = 220 \Rightarrow V_m = 220 \sqrt{2} \) Substitute: \[ 190 = \frac{2 \times 220\sqrt{2}}{\pi} \cos(\alpha) \Rightarrow \cos(\alpha) = \frac{190 \pi}{2 \cdot 220 \sqrt{2}} = \frac{19}{22\sqrt{2}} \Rightarrow \alpha = \cos^{-1} \left( \frac{19}{22\sqrt{2}} \right) \]