Animated Solution for Physics - Semiconductors: In an $n$-$p$-$n$ transistor circuit, the collector current is $10\text{ mA}$. If $90\%$ of the electrons emitted reach the collector

The world of electronics was revolutionized by the invention of the transistor. Among the various types, the bipolar junction transistor (BJT) stands out as a fundamental building block. In this problem, we are exploring the inner workings of an n-p-n transistor, specifically focusing on the relationship between its three terminal currents: the emitter current ($I_E$), the base current ($I_B$), and the collector current ($I_C$).

An n-p-n transistor consists of three semiconductor regions: a heavily doped n-type emitter, a very thin and lightly doped p-type base, and a moderately doped n-type collector. When the transistor is biased in its active region, the emitter-base junction is forward-biased, and the collector-base junction is reverse-biased.

Because the emitter is heavily doped, it injects a large number of electrons into the base. The base is intentionally made very thin and lightly doped so that most of these electrons do not recombine with holes there. Instead, they diffuse across the base and are swept into the collector by the strong electric field of the reverse-biased collector-base junction.

In our specific problem, we are told that the collector current is $10\text{ mA}$. We are also given a crucial piece of information: $90\%$ of the electrons emitted by the emitter reach the collector.

What happens to the other $10\%$? They recombine with holes in the p-type base region. To maintain charge neutrality, the external circuit must supply new holes to the base, which constitutes the base current ($I_B$).

The statement "$90\%$ of the electrons emitted reach the collector" can be translated directly into a mathematical relationship between the collector current and the emitter current:

We know that $I_C=10\text{ mA}$. Substituting this value into our equation gives:

To find the emitter current, we simply rearrange the equation:

For practical purposes, and to match the given options, we can round this value to $11\text{ mA}$. This tells us that the total current originating from the emitter is approximately $11\text{ mA}$.

Now that we have the emitter current and the collector current, finding the base current is straightforward. We apply Kirchhoff's Current Law (KCL) to the transistor as a whole. The transistor can be viewed as a single node where the sum of currents entering must equal the sum of currents leaving.

For an n-p-n transistor, the conventional current flows into the collector and base, and out of the emitter. Therefore, the fundamental current equation is:

We can rearrange this equation to solve for the base current:

Substituting the values we have found:

This $1\text{ mA}$ represents the small fraction of current resulting from the $10\%$ of electrons that recombined in the base.

By carefully analyzing the flow of electrons and applying fundamental circuit laws, we have determined both the emitter and base currents. The emitter current is approximately $11\text{ mA}$, and the base current is $1\text{ mA}$.

Looking at the provided options: (a) the emitter current will be $9\text{ mA}$ (Incorrect) (b) the emitter current will be $11\text{ mA}$ (Correct) (c) the base current will be $1\text{ mA}$ (Correct) (d) the base current will be $-1\text{ mA}$ (Incorrect)

Thus, the correct options are (b) and (c). This problem beautifully illustrates the delicate balance of charge carriers within a transistor and how macroscopic currents are a direct result of microscopic electron behavior.