Question

A current gain for a transistor working as CB amplifier is \(0.90\). If emitter current is \(10 \mathrm{~mA}\), then base current is (A) \(1 \mathrm{~m} \mathrm{~A}\) (B) \(2 \mathrm{~m} \mathrm{~A}\) (C) \(0.1 \mathrm{~mA}\) (D) \(0.2 \mathrm{~mA}\)

Short answer

The base current (I_B) is 1 mA, which corresponds to option (A).

Step-by-step solution

Identify given information

We are given: - Current gain (\( \alpha \)) for a transistor working as a CB amplifier = 0.90 - Emitter current (I_E) = 10 mA We need to calculate the base current (I_B).

Understand the relationship between emitter, base, and collector currents

For a transistor, the relationship between the emitter, base, and collector currents is given by: I_E = I_B + I_C In this case, we also have the current gain (\( \alpha \)), which is defined as the ratio of the collector current (I_C) to the emitter current (I_E). \( \alpha \) = I_C / I_E We can rearrange this equation to find I_C: I_C = \( \alpha \) * I_E

Calculate the collector current

Plugging in the given values: I_C = 0.90 * 10 mA = 9 mA

Calculate the base current

Now that we have the collector current, we can find the base current using the equation I_E = I_B + I_C: I_B = I_E - I_C = 10 mA - 9 mA = 1 mA So, the base current (I_B) is 1 mA, which corresponds to option (A).

Key concepts

Transistor Current Gain

In the world of electronics, transistors act as amplifiers and switches, and one crucial aspect to understand is the transistor current gain. For transistors used in common base (CB) configurations, this gain is denoted by \( \alpha \), which represents the efficiency of transferring current from the emitter to the collector.
- **Transistor Current Gain (\( \alpha \))**: It is the ratio of the collector current (\( I_C \)) to the emitter current (\( I_E \)).
\[ \alpha = \frac{I_C}{I_E} \]For a transistor in a CB configuration, \( \alpha \) is typically less than 1. This is because some current is lost through the base. In the given exercise, \( \alpha \) is 0.90, indicating that 90% of the emitter current is transferred to the collector, while 10% finds its way to the base.

Emitter Current

The emitter current, symbolized as \( I_E \), represents the total current flowing out of the transistor's emitter. It is an important quantity since it helps determine how current within the transistor is distributed between the base and collector. In transistor circuits, especially in CB amplifiers, knowing the emitter current is critical for predicting circuit behavior and performance.
- **Total Current**: The emitter current is the sum of the base current (\( I_B \)) and the collector current (\( I_C \)).
\[ I_E = I_B + I_C \]In the exercise, \( I_E \) is given as 10 mA. This means 10 mA is the total current entering the emitter that must split between the base and collector, based on the internal workings of the transistor and the specified current gain.

Base Current

Base current, denoted as \( I_B \), is one of the critical components in understanding transistor operation. Despite being small, the base current plays a major role in controlling the transistor's function as an amplifier.
- **Relationship**: Base current is calculated by subtracting the collector current from the emitter current.
\[ I_B = I_E - I_C \]In the context of the CB amplifier given in the problem statement, when the emitter current is 10 mA, and the collector current is found to be 9 mA, the resulting base current \( I_B \) is calculated as 1 mA. This illustrates how the small base current controls the larger collector current, a fundamental concept in transistor-based amplifiers.

Collector Current

The collector current, labeled as \( I_C \), is the current flowing through the collector in a transistor. This current, along with the base current, is crucial for determining how effectively a transistor functions within a circuit, especially in an amplifier configuration like the CB amplifier.
- **Calculation**: Based on the current gain and emitter current, using the formula:
\[ I_C = \alpha \times I_E \]For the given exercise, with a current gain \( \alpha \) of 0.90 and an emitter current \( I_E \) of 10 mA, the collector current \( I_C \) is calculated to be 9 mA. This collector current ensures the majority of the emitter current is channeled effectively and is responsible for the amplification process inherent to CB amplifiers.