Topic: Power Electronics
A half-wave rectifier is used to charge a 12 V battery through a resistance 'R'. The input transformer is fed by 34 V AC with turns ratio 2:1. Calculate the conduction period of the diode.
- A. $136^\circ$
- B. $120^\circ$
- C. $173^\circ$
- D. $137^\circ$
Answer: B. $120^\circ$
Explanation: The RMS secondary voltage is $V_s = 34 / 2 = 17\text{ V}$. The diode conducts when the supply voltage exceeds the battery voltage: $V_s \sqrt{2} \sin \theta > 12\text{ V}$. This gives $17\sqrt{2} \sin \theta = 12$, leading to $\theta = \sin^{-1}(\frac{12}{17\sqrt{2}}) \approx 29.94^\circ$. The conduction period is $\pi - 2\theta = 180^\circ - 2(30^\circ) = 120^\circ$.