Question

A n-p-n transistor is used in common emitter mode in an amplifier it. A change of \(40 \mu \mathrm{A}\) in the base current changes the output current by \(2 \mathrm{~m} \mathrm{~A}\) and \(0.04 \mathrm{~V}\) in input voltage. The input resistance is (A) \(1 \mathrm{k} \Omega\) (B) \(10 \Omega\) (C) \(10 \mathrm{k} \Omega\) (D) \(100 \Omega\)

Short answer

The input resistance (Rin) of the n-p-n transistor operating in common emitter mode is calculated using the relationship Rin = ∆Vin / ∆Ib. Given ∆Ib = \(40 \times 10^{-6}\) A and ∆Vin = 0.04 V, we can find Rin = \( \frac{0.04}{40 \times 10^{-6}} \) = 1000 Ω or 1 kΩ. Thus, the correct option is (A) 1 kΩ.

Step-by-step solution

Understand the relationship between input voltage, base current, and input resistance in an n-p-n transistor

An n-p-n transistor operating in the common emitter mode has the following relationship between input resistance (Rin), change in input voltage (∆Vin), and change in base current (∆Ib): Rin = ∆Vin / ∆Ib We are given the values of ∆Ib and ∆Vin, and our goal is to calculate the input resistance (Rin) of the transistor.

Convert given values to standard units

The given values for the change in base current and the change in input voltage are provided in microamperes (µA) and milliamperes (mA), respectively. Convert these values to amperes (A) before proceeding with the calculation: ∆Ib = 40 µA = 40 × 10^(-6) A ∆Io = 2 mA = 2 × 10^(-3) A

Calculate the input resistance of the transistor

Plug the given values of ∆Ib and ∆Vin into the formula for input resistance: Rin = ∆Vin / ∆Ib Rin = (0.04 V) / (40 × 10^(-6) A)

Solve for Rin

Perform the division to calculate the input resistance: Rin = (0.04 V) / (40 × 10^(-6) A) = 1000 Ω = 1 kΩ

Identify the correct option

Now that we have calculated the input resistance, we can identify the correct option: The input resistance is (A) 1 kΩ.

Key concepts

Common Emitter Mode

In the realm of electronics, the common emitter mode is a popular configuration for n-p-n transistors. It’s often used in amplifiers due to its ability to amplify voltage and provide better signal gain compared to other modes. In this setup, the emitter terminal of the transistor is common to both the input and output circuits, hence the name "common emitter".
This configuration is favored because it offers:

• High voltage gain, meaning it can increase the magnitude of the input voltage significantly.
• Moderate input and output impedance levels.
• An easy setup for circuit design and analysis, especially useful for designing amplifiers.

The common emitter mode excels in applications where signal amplification is needed without a drastic reduction in input resistance.
Understanding how currents and voltages interact in this mode will give us deeper insights into designing efficient electronic circuits.

Input Resistance

Input resistance is a crucial parameter in determining how much of the input voltage is "seen" by a transistor. It essentially acts like a gatekeeper for the signal entering the device. In the common emitter mode, the input resistance (\(R_{in}\)) can be calculated using the change in input voltage (\(\Delta V_{in}\)) and the change in base current (\(\Delta I_{b}\)) as follows:
\[ R_{in} = \frac{\Delta V_{in}}{\Delta I_{b}} \]
This formula helps to determine how resistive the transistor is to incoming signals. A higher input resistance implies that most of the input signal is effectively used for amplification, making it desirable in most amplifier circuits.
In our exercise, the key was converting the base current from microamperes to amperes, and subsequently using this relationship to find the input resistance as 1 kΩ.

Input Voltage

Input voltage in the context of a transistor defines the electrical potential difference applied at the base terminal. This voltage is crucial because it directly influences the transistor's base current, thus affecting its operation and amplification levels.
In a common emitter configuration, even small changes in the input voltage can lead to significant changes in the output current. This is because the input is the control voltage that determines how much the transistor will "open" or "close" for current flow.
In this exercise, a change in input voltage of 0.04 V determines the input resistance, pointing to how these small variations are significant for the overall functionality.

Base Current

The base current (\(I_{b}\)) in an n-p-n transistor is the small current that flows into the base terminal, crucial for controlling larger currents flowing through the collector to the emitter. Despite its comparatively smaller magnitude, the base current is powerful enough to switch the transistor from an "off" to an "on" state.
Base current regulation is vital in switching applications as well as amplification operations because it dictates the amount of collector current the device allows. In an amplifier, any change in base current results in a proportionally larger change in the output current due to the transistor's gain factor.
In the given exercise, the change of 40 µA in base current was a critical parameter used to find the input resistance, illustrating how small base current inputs drive larger circuit behaviors.