Question

A 'Class A' Common Emitter amplifier with \(V_{CC} = 20 \text{ V}\) draws a current I = 5A and feed a load of \(40\Omega\) through a step-up transformer N2/N1=3.16. Find the efficiency of the amplifier when it is properly matched for maximum power supply.

• A. 25 %
• B. 50 %
• C. 78.5 %
• D. 90 %

Hint:

Maximum theoretical efficiency of different Class A amplifier configurations:
Resistive load (series-fed): 25%
Transformer-coupled load: 50%
Class B push-pull: 78.5%
Class C: Can be>90% (but for tuned RF applications, not linear amplification).
"Properly matched for maximum power" usually implies operating at conditions to achieve near theoretical maximum efficiency.

Answer 1

The Correct Option is B

For a transformer-coupled Class A amplifier, the maximum theoretical efficiency is 50%. This occurs under ideal conditions with perfect impedance matching and when the amplifier delivers maximum possible AC power to the load. The question states it is "properly matched for maximum power supply" (this phrasing is a bit unusual, usually "maximum power transfer" to the load). Given it's a Class A amplifier with transformer coupling, the theoretical maximum efficiency is the key. The DC power drawn from the supply is \(P_{DC} = V_{CC} \times I_Q\), where \(I_Q\) is the quiescent collector current. The problem states "draws a current I = 5A". If this is the quiescent current \(I_Q\), then \(P_{DC} = 20V \times 5A = 100W\). For maximum efficiency in a transformer-coupled Class A amplifier, the AC power delivered to the load can be at most \(0.5 \times P_{DC}\). So, \(P_{AC,max} = 0.5 \times 100W = 50W\). Efficiency \(\eta = \frac{P_{AC,out}}{P_{DC,in}}\). Maximum efficiency \(\eta_{max} = \frac{0.5 P_{DC}}{P_{DC}} = 0.5 = 50%\). The load resistance transformation: The actual load is \(R_L = 40\Omega\). The transformer turns ratio is \(n = N2/N1 = 3.16\). The impedance seen by the primary of the transformer (at the collector of the transistor) is \(R_L' = R_L / n^2\) for a step-up transformer (N2>N1, so n>1). Wait, if N2/N1 = 3.16, it's step-up for voltage, so step-down for impedance from primary to secondary. The load \(R_L = 40\Omega\) is on the secondary. The reflected impedance to the primary is \(R_L' = (\frac{N1}{N2})^2 R_L\). Given \(N2/N1 = 3.16\), so \(N1/N2 = 1/3.16\). \(R_L' = (\frac{1}{3.16})^2 \times 40 \Omega = \frac{1}{ (3.16)^2 } \times 40 \Omega \approx \frac{1}{9.9856} \times 40 \Omega \approx \frac{1}{10} \times 40 \Omega = 4 \Omega\). For maximum AC power output in a Class A transformer-coupled amplifier, the quiescent operating point (Q-point) should be such that \(V_{CEQ} \approx V_{CC}\) (if load line allows) or \(V_{CEQ} \approx V_{CC}/2\) (for resistive load without transformer, for max symmetrical swing). With transformer coupling, the collector can swing up to \(2V_{CC}\) ideally. The condition for "properly matched for maximum power supply" refers to achieving the conditions where the efficiency approaches its theoretical maximum. The theoretical maximum efficiency for a transformer-coupled Class A amplifier is 50%. The specific component values are often given to confirm it's operating under conditions where this can be achieved, or to distract. Given that 50% is an option and is the theoretical maximum for this configuration, it's the most likely answer. \[ \boxed{50 %} \]