Q.In a transistor working in the common-emitter mode, the base current is $20\ \mu\text{A}$ and the collector current is $2\ \text{mA}$. Find the current amplification factor $\beta$ and the emitter current $I_E$.

The Intuition: What Do These Gains Actually Mean?

Imagine a transistor as a tiny current valve. You send a small current into one terminal, and a much larger current flows through another. The ratio between these currents is the "gain" — how much the transistor amplifies.

But here's the catch: the transistor has three terminals — emitter, base, and collector — and you can hook it up in two fundamentally different ways. Each way gives you a different gain number, even though it's the same physical device.

Alpha ($\alpha$) is the gain when you use the transistor in common-base configuration. You send current into the emitter, and most of it flows out through the collector. A tiny bit gets lost through the base. Alpha is the fraction of emitter current that successfully reaches the collector:

Since some current always leaks through the base, $\alpha$ is always slightly less than 1 — typically 0.98 to 0.999.

Beta ($\beta$) is the gain when you use the transistor in common-emitter configuration. Here, you send a small current into the base, and that controls a much larger current flowing from collector to emitter. Beta is the ratio:

This number can be huge — 50, 100, 500 — because a tiny base current controls a large collector current.

The Simple Algebra That Connects Them

Start with the fundamental truth about transistor currents: everything that enters the emitter must leave through the base and collector.

Now write $\alpha$ and $\beta$ in terms of these currents:

From the current relation, $I_B = I_E - I_C$. Substitute into $\beta$:

Divide numerator and denominator by $I_E$:

That's it. Three lines of algebra, no magic.

What This Tells You

If $\alpha = 0.98$ (a typical value), then:

A tiny 2% loss in the emitter-to-collector current translates into a beta of 49. That's why common-emitter amplifiers are so popular — you get huge current gain from a small base signal.

The Reverse Relation

You can also solve for $\alpha$ in terms of $\beta$:

This is useful when a datasheet gives you $\beta$ (which is more common) and you need $\alpha$ for some calculation.

A Quick Check

If $\beta = 200$, then $\alpha = 200/201 \approx 0.995$. The transistor loses only 0.5% of its emitter current through the base. That's an excellent transistor.

If $\beta = 20$, then $\alpha = 20/21 \approx 0.952$. Nearly 5% of the emitter current is wasted through the base — a poor transistor by modern standards, but still usable.

The alpha-beta relationship is just a restatement of current conservation, dressed up in two different gain definitions. Understand that, and you'll never need to memorise it.

The alpha-beta relationship is a favourite short-answer question in CBSE Class 12 Physics board exams and is often searched as relation between alpha and beta transistor derivation. Because this single derivation connects two current-gain definitions used throughout Semiconductor Electronics, it is a dependable quick-revision topic for JEE Main and NEET physics too, building directly on the NCERT Class 12 curriculum.