Question

A n-p-n transistor is used in common emitter made in an amplifier it. A change of \(40 \mu \mathrm{A}\) in the base current changes the output current by $2 \mathrm{~mA}\( and \)0.04 \mathrm{~V}$ in input voltage. An amplifier has voltage gain \(A_{V}=1000\). The voltage gain in \(\mathrm{dB}\) is (A) \(20 \mathrm{~dB}\) (B) \(30 \mathrm{~dB}\) (C) \(3 \mathrm{~dB}\) (D) \(60 \mathrm{~dB}\)

Short answer

The voltage gain in dB is 60 dB, which corresponds to option (D).

Step-by-step solution

Identify the given information

We are given the following information about the n-p-n transistor and the amplifier: - Voltage gain, \(A_V = 1000\) - The voltage gain in dB needs to be found - Options for the voltage gain in dB are 20 dB, 30 dB, 3 dB, and 60 dB.

Apply the formula for voltage gain in dB

The formula for voltage gain in dB is: \[Gain_{dB} = 20 \times \log_{10}(A_V)\] Where Gain_{dB} is the voltage gain in decibels and \(A_V\) is the voltage gain. In this case, the voltage gain \(A_V = 1000\).

Calculate the voltage gain in dB

Now we substitute the given value of \(A_V\) into the formula and solve for Gain_{dB}: \[ Gain_{dB} = 20 \times \log_{10}(1000) \\ Gain_{dB} = 20 \times 3 \\ Gain_{dB} = 60 \, dB \]

Match the calculated Gain_{dB} with the given options

The calculated voltage gain in dB is 60 dB. Comparing this with the given options, we find that it matches option (D). Therefore, the correct answer is option (D), 60 dB.