Question

In a common emitter transistor, \( \beta = 100 \), \( R_c = 2\,\text{k}\Omega \), and the output voltage is \(2\,\text{V}\). If \( R_b = 1\,\text{k}\Omega \), find the input voltage.

• A. \(0.02\,\text{V}\)
• B. \(0.02\,\text{V}\)
• C. \(0.1\,\text{V}\)
• D. \(0.2\,\text{V}\)

Hint:

For common emitter amplifiers, remember the gain formula \( A_v = \beta \frac{R_c}{R_b} \). Then use \( A_v = \frac{V_{out}}{V_{in}} \) to find the required voltage.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
In a common emitter (CE) configuration, the transistor acts as an amplifier where a small input base current ($I_b$) controls a larger output collector current ($I_c$).
Step 2: Key Formula or Approach:
1. Output Voltage ($V_{out}$) = $I_c \times R_c$
2. Current Gain ($\beta$) = $I_c / I_b$
3. Input Voltage ($V_{in}$) = $I_b \times R_b$
Step 3: Detailed Explanation:
First, find the collector current ($I_c$): \[ 2\text{ V} = I_c \times 2000 \Omega \implies I_c = \frac{2}{2000} = 10^{-3} \text{ A} = 1 \text{ mA} \] Next, find the base current ($I_b$) using $\beta$: \[ I_b = \frac{I_c}{\beta} = \frac{10^{-3}}{100} = 10^{-5} \text{ A} \] Finally, find the input voltage ($V_{in}$): \[ V_{in} = I_b \times R_b = 10^{-5} \text{ A} \times 1000 \Omega = 10^{-2} \text{ V} \] \[ V_{in} = 0.01 \text{ V} \]
Step 4: Final Answer:
The input voltage is 0.01 V.