In a common-base mode of a transistor, the collector current is 5.488 mA for an emitter current of 5

Q: Why is $I_E$ always the largest current?

In a transistor, the emitter emits the majority charge carriers. These carriers split into two paths: a very small fraction goes to the base ($I_B$), and the vast majority reaches the collector ($I_C$). Therefore, $I_E$ is always the sum of $I_B$ and $I_C$.

Q: What is the difference between $\alpha$ and $\beta$?

$\alpha$ is the common-base current gain ($I_C/I_E$), which is always slightly less than 1. $\beta$ is the common-emitter current gain ($I_C/I_B$), which is typically much larger than 1 (often between 20 and 200).

Q: Why is the base current so small?

The base region is intentionally made very thin and lightly doped. This ensures that most of the charge carriers injected from the emitter pass straight through to the collector without recombining in the base.

Q: Can we find $\beta$ without finding $I_B$ first?

Yes! You can first calculate $\alpha = \frac{I_C}{I_E}$, and then use the relation $\beta = \frac{\alpha}{1 - \alpha}$. Both methods will yield the exact same result.

Q: Does this formula apply to both NPN and PNP transistors?

Absolutely. The fundamental current relationship $I_E = I_B + I_C$ and the definitions of $\alpha$ and $\beta$ hold true for both NPN and PNP transistors. The only difference is the actual direction of the physical current flow.

Question

In a common-base mode of a transistor, the collector current is $5.488	ext{ mA}$ for an emitter current of $5.60	ext{ mA}$. The value of the base current amplification factor ($\beta$) will be

Options:

(A) $49$

(B) $50$

(C) $51$

(D) $48$

Sigma Percentile · Accepted Answer

### Analyzing the Transistor Currents

Imagine you are looking at the inner workings of a transistor. It acts like a junction where currents meet and divide. In this problem, we are given two crucial pieces of information: the emitter current $I_E = 5.60 	ext{ mA}$ and the collector current $I_C = 5.488 	ext{ mA}$.

Our ultimate goal is to find the base current amplification factor, commonly denoted by the Greek letter $\beta$. But before we can find $\beta$, we need to understand the relationship between the currents flowing through the transistor's three terminals: the emitter, the base, and the collector.

### The Master Equation: Kirchhoff's Current Law

Think of the transistor as a single node in a circuit. According to Kirchhoff's Current Law, the total current entering a junction must equal the total current leaving it. For any transistor (whether NPN or PNP), the emitter current is always the sum of the base current and the collector current.

Mathematically, this is expressed as:

$$I_E = I_B + I_C$$

We know $I_E$ and $I_C$, but we are missing the base current $I_B$. Let's rearrange our master equation to solve for it:

$$I_B = I_E - I_C$$

Now, we carefully substitute the given values into our rearranged equation:

$$I_B = 5.60 - 5.488$$

Subtracting these values gives us the base current:

$$I_B = 0.112 	ext{ mA}$$

Notice how small the base current is compared to the emitter and collector currents! This is a fundamental characteristic of transistors, as the base region is intentionally made very thin and lightly doped.

### Calculating the Amplification Factor

Now that we have the base current, we are ready to find the base current amplification factor, $\beta$. This factor tells us how much the transistor amplifies the small base current to produce the larger collector current.

By definition, $\beta$ is the ratio of the collector current to the base current:

$$\beta = \frac{I_C}{I_B}$$

Let's substitute the values we have:

$$\beta = \frac{5.488}{0.112}$$

When we perform this division, we get a clean, whole number:

**$$\beta = 49$$**

And there we have it! The base current amplification factor is exactly 49.

### An Alternative Path

As a fun exercise, you could also solve this problem using the common-base current gain, $\alpha$. First, calculate $\alpha = \frac{I_C}{I_E}$. Once you have $\alpha$, you can find $\beta$ using the relation $\beta = \frac{\alpha}{1 - \alpha}$. If you plug in the numbers, you will find that it leads you right back to the exact same answer of 49. Physics is beautifully consistent!

Animated Solution for Physics - Semiconductors: In a common-base mode of a transistor, the collector current is $5.488 mA$ for an emitter current of $5.60 mA$ . The value of the base current amplification factor ( $β$ ) will be

Select Answer:

Visualized Solution

$I_{E}$ and $I_{C}$

Kirchhoff's Current Law

Base Current Formula

Substitution

Base Current Value

Current Amplification Factor $β$

Substituting for $β$

Final Answer

Alternative Method

The Sigma Insight: P-N Junction Diode

Analyzing the Transistor Currents

The Master Equation

Calculating the Amplification Factor

An Alternative Path

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