Topic: Analog Circuits
For a class B amplifier providing a 20 V peak signal to 16 ohm load and a power supply of $V_{cc}=30$ V, the efficiency will be
- A. 52.3%
- B. 25.65%
- C. 75%
- D. 78.6%
Answer: A. 52.3%
Explanation: A 20-V peak signal across a 16-$\Omega$ load provides a peak load current of $I_{L}(P)=\frac{V_{L}(P)}{R_{L}}=\frac{20~V}{16~\Omega}=1.25~A$. $I_{dc}=\frac{2}{\pi}I_{L}(P)=\frac{2}{\pi}(1.25)=0.796~A$. $P_{i}(dc)=V_{cc}I_{dc}=(30V)(0.796A)=23.9W$. $P_{0}(ac)=\frac{V_{L}^{2}(P)}{2 R_{L}}=\frac{(20~V)^{2}}{2(16~\Omega)}=12.5~W$. $\%\eta=\frac{P_{0}(ac)}{P_{i}(dc)}\times100\% = \frac{12.5~W}{23.9~W}\times100\%=52.3\%$.