Question

For a common-emitter amplifier, the voltage gain is $40$. Its input and output impedances are $100\ \Omega$ and $400\ \Omega$, respectively. The power gain of the CE amplifier will be

• A. 450
• B. 400
• C. 300
• D. 500

Hint:

To solve this quickly, memorize the unified formula: $\text{Power Gain} = \frac{A_v^2 \times Z_i}{Z_o}$. Substituting the values directly gives $\frac{1600 \times 100}{400} = \frac{1600}{4} = 400$. This layout avoids splitting the calculation into two separate stages!

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The question asks for the power gain of a Common-Emitter (CE) transistor amplifier. We are given the voltage gain ($A_v = 40$), the input impedance ($Z_i = 100\ \Omega$), and the output impedance ($Z_o = 400\ \Omega$).

Step 2: Key Formula or Approach:
The power gain ($A_p$) of an amplifier can be expressed as the product of its voltage gain ($A_v$) and its alternating current gain ($\beta$ or $A_i$): $$A_p = A_v \times A_i$$ We can determine the current gain by utilizing the relationship between voltage gain, current gain, and impedances: $$A_v = A_i \times \frac{Z_o}{Z_i} \implies A_i = A_v \times \frac{Z_i}{Z_o}$$ Combining these equations yields the direct power gain formula: $$A_p = A_v^2 \times \frac{Z_i}{Z_o}$$

Step 3: Detailed Explanation:
Let's plug our given parameters into the derived current gain expression: $$A_i = 40 \times \frac{100}{400}$$ Simplify the fraction: $$A_i = 40 \times \frac{1}{4} = 10$$ Now, substitute the values of $A_v$ and $A_i$ back into our primary power gain equation: $$A_p = 40 \times 10 = 400$$

Step 4: Final Answer:
The power gain of the common-emitter amplifier is 400, which corresponds to option (B).