CE-Transistor Amplifier Problem Overview

This problem focuses on a Common Emitter (CE) transistor amplifier and requires us to determine two crucial parameters: the input voltage and the base current. We are provided with several key details about the amplifier's operation, including the voltage across the collector resistance, the value of the collector resistance, the transistor's current amplification factor, and the input resistance (which is essentially the base resistance).

Given Values for the Transistor Amplifier Circuit

To begin our analysis, let's organize the specific values provided in the question:

Key Concepts and Essential Formulas for Transistor Amplifiers

Solving this problem relies on understanding fundamental principles of transistor operation and applying basic circuit laws:

• Ohm's Law: This fundamental law states the relationship between voltage (\(V\)), current (\(I\)), and resistance (\(R\)) as \(V = I \times R\). We will use it to calculate currents and voltages across specific resistances in the circuit.
• Current Amplification Factor (\(\beta\)): Also commonly known as the common-emitter forward current transfer ratio (\(h_{fe}\)), this factor defines how much the transistor amplifies the current. It is the ratio of the collector current (\(I_C\)) to the base current (\(I_B\)).
  \(\beta = \frac{I_C}{I_B}\)

Step-by-Step Calculation for Input Voltage and Base Current

We will calculate the required values in a sequential manner, starting with the collector current.

Collector Current Calculation

First, we need to find the collector current (\(I_C\)). We can do this by using the given audio signal voltage across the collector resistance (\(V_C\)) and the collector resistance (\(R_C\)). According to Ohm's Law:

\(V_C = I_C \times R_C\)

To find \(I_C\), we rearrange the formula:

\(I_C = \frac{V_C}{R_C}\)

Now, substituting the provided values:

\(I_C = \frac{3 \, \text{V}}{3 \times 10^3 \, \Omega} = 1 \times 10^{-3} \, \text{A} = 1 \, \text{mA}\)

Base Current Calculation

Next, we can calculate the base current (\(I_B\)) using the calculated collector current (\(I_C\)) and the given current amplification factor (\(\beta\)). The relationship is defined by the beta value:

\(\beta = \frac{I_C}{I_B}\)

Rearranging the formula to solve for \(I_B\):

\(I_B = \frac{I_C}{\beta}\)

Substituting the values we have:

\(I_B = \frac{1 \times 10^{-3} \, \text{A}}{50} = 0.02 \times 10^{-3} \, \text{A} = 20 \times 10^{-6} \, \text{A} = 20 \, \mu\text{A}\)

Input Voltage Calculation

Finally, we determine the input voltage (\(V_{in}\)). This is the voltage drop across the input resistance (\(R_{in}\)) due to the flow of the base current (\(I_B\)). We again apply Ohm's Law:

\(V_{in} = I_B \times R_{in}\)

Substitute the calculated \(I_B\) and the given \(R_{in}\):

\(V_{in} = (20 \times 10^{-6} \, \text{A}) \times (1 \times 10^3 \, \Omega)\)

\(V_{in} = 20 \times 10^{-3} \, \text{V} = 0.02 \, \text{V}\)

Summary of Results for the CE-Transistor Amplifier

Based on our detailed calculations, the determined values are:

• The input voltage for the CE-transistor amplifier is \(0.02 \, \text{V}\).
• The base current for the CE-transistor amplifier is \(20 \, \mu\text{A}\).

These values provide a comprehensive solution to the problem.