Topic: Electronics Engineering
It is required to design a class B output stage to deliver an average output power of 12.5 W to a load of 4 $\Omega$. The power supply is selected so that $V_{CC}$ is 5 V greater than the peak output voltage. Determine the power conversion efficiency of the circuit.
- A. 48%
- B. 50%
- C. 52%
- D. 54%
Answer: C. 52%
Explanation: $V_{rms}^{2} = P \times R = 12.5~W \times 4~\Omega = 50~V^{2} \Rightarrow V_{P} = \sqrt{2 \times 50} = 10~V$.
$V_{CC} = 5 + V_{P} = 15~V$.
$I_{rms}^{2} = 12.5 / 4 \Rightarrow I_{P} = \sqrt{6.25} = 2.50~A$.
$I_{dc} = \frac{2}{\pi} \times I_{P} = \frac{2}{\pi} \times 2.5$.
$P_{dc} = V_{CC} \cdot I_{dc} = 15 \times \frac{2}{\pi} \times 2.5 \approx 23.87~W$.
$\eta = \frac{P_{ac}}{P_{dc}} \times 100 = \frac{12.5}{23.87} \times 100 = 52.3\%$.