Question

A transistor is connected in common emitter configuration. The collector supply is \(8 \mathrm{~V}\) and the voltage drop across a resistor of 800 in the collector circuit is \(0.5 \mathrm{~V}\). If the current gain factor \(\alpha\) is \(0.96\), then base current will be (A) \(27 \mu \mathrm{A}\) (B) \(26 \mu \mathrm{A}\) (C) \(25 \mu \mathrm{A}\) (D) \(24 \mu \mathrm{A}\)

Short answer

The short answer is (B) \(26 \mu \mathrm{A}\).

Step-by-step solution

Find the current flowing through the resistor

We are given the voltage drop across the resistor as 0.5 V, and the resistance as 800 Ω. We can use Ohm's Law to find the current flowing through the resistor: \(I = \frac{V}{R}\) Where I is the current, V is the voltage drop, and R is the resistance. Plug in the given values: \(I = \frac{0.5}{800} = 0.000625 \mathrm{~A}\)

Calculate the collector current

In a common emitter configuration, the current flowing through the resistor is the same as the collector current (Ic). Therefore, \(I_c = 0.000625 \mathrm{~A}\)

Use the current gain factor to find the base current

We are given the current gain factor α as 0.96. The relationship between α, base current (Ib), and collector current (Ic) can be expressed as: \(\alpha = \frac{I_c}{I_c + I_b}\) We can rearrange the equation to find the base current (Ib): \(I_b = \frac{I_c}{\alpha} - I_c\) Now, plug in the given values: \(I_b = \frac{0.000625}{0.96} - 0.000625 = 2.604 \times 10^{-5} \mathrm{~A}\)

Convert base current to microamperes and check the options

To convert the base current from amperes to microamperes, multiply by 10^6: \(I_b = 2.604 \times 10^{-5} \mathrm{~A} \times 10^6 = 26.04 \mu \mathrm{A}\) Checking the given options, we find that the closest answer to our calculation is: (B) \(26 \mu \mathrm{A}\)

Key concepts

Ohm's Law

Ohm's Law is fundamental in understanding electric circuits. It helps determine the relationship between voltage, current, and resistance. The law states that the current flowing through a conductor between two points is directly proportional to the voltage across the two points and inversely proportional to the resistance. It is mathematically expressed as:
\[ I = \frac{V}{R} \]
where:

• \(I\) represents the current (in amperes)
• \(V\) is the voltage drop (in volts)
• \(R\) is the resistance (in ohms)

In the context of the transistor problem, Ohm's Law allows us to find the current flowing through a known resistor with a specific voltage drop across it. This current is vital for further calculations in the circuit, especially when dealing with transistors in common emitter configuration.

Current Gain Factor

The current gain factor, usually denoted by \(\alpha\) or \(\beta\), is a key concept in transistor operation. It describes how effectively a transistor amplifies the input current. In this given problem, the factor \(\alpha\) represents the ratio of collector current \(I_c\) to the emitter current \(I_e\) in common emitter configuration with the formula:
\[ \alpha = \frac{I_c}{I_e} \]
For the exercise, \(\alpha\) is provided as 0.96. This indicates that 96% of the emitter current is converted into collector current, with the remaining 4% being the base current. Understanding \(\alpha\) helps in finding out the relationship between the different currents in a transistor, which is crucial to solving for unknown values like the base current \(I_b\).
Knowing \(\alpha\) is essential to calculate and confirm the effective amplification capabilities of a transistor operated in this configuration.

Base Current Calculation

Base current calculation is an integral part of analyzing transistor circuits. It involves understanding how different circuit parameters relate and influence each other. In problem-solving, like in this exercise, one often needs to express the base current \(I_b\) in terms of other known quantities using the given current gain factor.
The formula that relates \(\alpha, I_c\), and \(I_b\) is:
\[ \alpha = \frac{I_c}{I_c + I_b} \]
This can be rearranged to find \(I_b\) by using the known values of \(\alpha\) and \(I_c\):
\[ I_b = \frac{I_c}{\alpha} - I_c \]
In the exercise, after computing the collector current \(I_c = 0.000625 \) A, we use \(\alpha = 0.96\) to find the base current. Solving this gives us \(I_b = 26.04 \mu \mathrm{A}\). Understanding base current calculations helps achieve precise transistor-based circuit design, optimizing the desired current output and signal processing.