Question

In common emitter amplifier, a change of 0.2 mA in the base current causes a change of 5 mA in the collector current. If input resistance is 2 k$\Omega$ and voltage gain is 75, the load resistance used in the circuit is

• A. 8 k$\Omega$
• B. 4 k$\Omega$
• C. 12 k$\Omega$
• D. 6 k$\Omega$

Hint:

Voltage gain can be conceptualized as $A_v = \text{Current Gain} \times \text{Resistance Gain}$. Here, the current gain is $25$. To reach a target voltage gain of $75$, the circuit needs a resistance gain of exactly $\frac{75}{25} = 3$. This means the output load resistance must be exactly 3 times larger than the input resistance: $3 \times 2\ \text{k}\Omega = 6\ \text{k}\Omega$.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The question asks for the load resistance ($R_L$) used in a common emitter (CE) transistor amplifier circuit, given the input signal resistance ($R_i$), the total voltage gain ($A_v$), and the variations in both base and collector current streams.

Step 2: Key Formula or Approach:
1. The alternating current amplification factor ($\beta$) is defined as the ratio of the change in collector current ($\Delta I_C$) to the change in base current ($\Delta I_B$):
$$\beta = \frac{\Delta I_C}{\Delta I_B}$$ 2. The voltage gain ($A_v$) of a common emitter amplifier circuit is given by the formula:
$$A_v = \beta \times \frac{R_L}{R_i}$$

Step 3: Detailed Explanation:
Let's first calculate the transistor's dynamic current gain value ($\beta$) using the given current variations:
Change in base current: $\Delta I_B = 0.2\ \text{mA}$ Change in collector current: $\Delta I_C = 5\ \text{mA}$ $$\beta = \frac{5\ \text{mA}}{0.2\ \text{mA}} = \frac{5}{0.2} = 25$$ Now, write down the other circuit values provided in the problem statement:
Input resistance parameter: $R_i = 2\ \text{k}\Omega$ Circuit voltage gain value: $A_v = 75$ Substitute $\beta$, $R_i$, and $A_v$ into the voltage gain formula to solve for the unknown load resistance $R_L$:
$$75 = 25 \times \frac{R_L}{2\ \text{k}\Omega}$$ Simplify the equation by dividing both sides by 25:
$$3 = \frac{R_L}{2\ \text{k}\Omega}$$ Isolate $R_L$ by multiplying both sides by $2\ \text{k}\Omega$:
$$R_L = 3 \times 2\ \text{k}\Omega = 6\ \text{k}\Omega$$ This calculation determines the value of the output load resistor.

Step 4: Final Answer:
The load resistance used in the circuit is 6 k$\Omega$, matching option (D).