Question

In the following common emitter configuration an $n p n$ transistor with current gain $\beta=100$ is used. The output voltage of the amplificr will be

• A. 10 mV
• B. 0.1 V
• C. 1.0 V
• D. 10 V

Answer 1

Answer: (C)

1.0 V

Answer 2

We know that the output voltage is given by, $v_{0}=v_{i} \times \beta \times \frac{R_{L}}{R_{B E}}$ Here, $v_{i}=1 m V, \beta=100$ $R_{L}=10 \,k \Omega, R_{B E}=1\, k \Omega$ $\therefore v_{0}=1 \times 10^{-3} \times 100 \times \frac{10}{1}=1.0\, V$