Q127 / 180 · COMEDK 2022 · Physics
For CE transistor amplifier, the audio signal voltage across the collector resistance of $4 \mathrm{l} \Omega$ is $5 \mathrm{~V}$. If the current amplification factor of the transistor is 100 and base resistance is $2 \mathrm{k} \Omega$, then input signal voltage is
- A$75 \mathrm{mV}$
- Correct answerB$25 \mathrm{mV}$
- C$20 \mathrm{mV}$
- D$50 \mathrm{mV}$
Solution
Given the collector resistance $R_C = 4 \text{ k}\Omega = 4000 \Omega$, the output voltage $V_o = 5 \text{ V}$, and the current amplification factor $\beta = 100$.
The collector current $I_C$ is given by $I_C = \dfrac{V_o}{R_C} = \dfrac{5}{4000} \text{ A} = 1.25 \times 10^{-3} \text{ A} = 1.25 \text{ mA}$.
The base current $I_B$ is related to the collector current by $I_C = \beta I_B$. Thus, $I_B = \dfrac{I_C}{\beta} = \dfrac{1.25 \text{ mA}}{100} = 0.0125 \text{ mA} = 1.25 \times 10^{-5} \text{ A}$.
The input signal voltage $V_i$ is given by $V_i = I_B \times R_B$, where $R_B = 2 \text{ k}\Omega = 2000 \Omega$.
$V_i = (1.25 \times 10^{-5} \text{ A}) \times (2000 \Omega) = 2.5 \times 10^{-2} \text{ V} = 25 \text{ mV}$.
Answer: $25 \mathrm{mV}$
The collector current $I_C$ is given by $I_C = \dfrac{V_o}{R_C} = \dfrac{5}{4000} \text{ A} = 1.25 \times 10^{-3} \text{ A} = 1.25 \text{ mA}$.
The base current $I_B$ is related to the collector current by $I_C = \beta I_B$. Thus, $I_B = \dfrac{I_C}{\beta} = \dfrac{1.25 \text{ mA}}{100} = 0.0125 \text{ mA} = 1.25 \times 10^{-5} \text{ A}$.
The input signal voltage $V_i$ is given by $V_i = I_B \times R_B$, where $R_B = 2 \text{ k}\Omega = 2000 \Omega$.
$V_i = (1.25 \times 10^{-5} \text{ A}) \times (2000 \Omega) = 2.5 \times 10^{-2} \text{ V} = 25 \text{ mV}$.
Answer: $25 \mathrm{mV}$
Hence, the correct answer is option: 'B'.