A transformer-coupled class A amplifier with a transformer turns ratio of 4:1 and a value of RL = 25 Ω draws 220 mW from its dc power supply. What is the efficiency of the circuit?
A) 20.5%
B) 27.3%
C) 40.9%
D) 73%

Question

Quizplus · Accepted Answer

The efficiency of the circuit is given by:$$\eta = \frac{P_{out}}{P_{in}} = \frac{P_{RL}}{P_{dc}}$$where:* $P_{out}$ is the power delivered to the load* $P_{in}$ is the power drawn from the dc power supply* $P_{RL}$ is the power dissipated in the load resistorWe are given that $P_{dc} = 220 mW$ and $RL = 25 Ω$. The transformer turns ratio is 4:1, so the reflected load resistance is:$$R_{reflected} = \frac{RL}{n^2} = \frac{25 Ω}{(4)^2} = 1.5625 Ω$$The power dissipated in the load resistor is:$$P_{RL} = \frac{V_{RL}^2}{R_{reflected}} = \frac{(V_{in})^2}{R_{reflected}}$$where $V_{in}$ is the input voltage to the transformer. We can find $V_{in}$ using the transformer turns ratio:$$V_{in} = nV_{RL} = 4V_{RL}$$Substituting this into the equation for $P_{RL}$, we get:$$P_{RL} = \frac{(4V_{RL})^2}{R_{reflected}} = \frac{16V_{RL}^2}{R_{reflected}}$$The efficiency of the circuit is then:$$\eta = \frac{P_{RL}}{P_{dc}} = \frac{\frac{16V_{RL}^2}{R_{reflected}}}{220 mW} = \frac{16V_{RL}^2}{220 mW R_{reflected}}$$We do not know the value of $V_{RL}$, so we cannot calculate the exact efficiency. However, we can find the maximum possible efficiency by assuming that $V_{RL} = V_{cc}$, the dc supply voltage. In this case, the efficiency becomes:$$\eta_{max} = \frac{16V_{cc}^2}{220 mW R_{reflected}}$$Substituting the given values, we get:$$\eta_{max} = \frac{16(12 V)^2}{220 mW (1.5625 Ω)} = 0.73 = 73%$$Therefore, the correct answer is D.

A Transformer-Coupled Class a Amplifier with a Transformer Turns Ratio