In a common-emitter amplifier, a change of $0.6\ \text{mA}$ in the collector current corresponds to a change of $8\ \mu\text{A}$ in the base current. Calculate the small-signal current gain $\beta_{ac}$ of the transistor.

In a common-emitter amplifier, a change of 0.6mA in the collector current corresponds to a change of 8muA in t

Numerical · Q27

0.6\ \text{mA}

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Concept understanding — Alpha Beta Relationship

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:

\alpha = \frac{I_C}{I_E}

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:

\beta = \frac{I_C}{I_B}

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

Note

Both $\alpha$ and $\beta$ describe the same transistor, just from different wiring perspectives. They must be related — and that relation is what we're after.

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.

I_E = I_B + I_C

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

\alpha = \frac{I_C}{I_E}, \quad \beta = \frac{I_C}{I_B}

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

\beta = \frac{I_C}{I_E - I_C}

Divide numerator and denominator by $I_E$ :

\beta = \frac{I_C / I_E}{1 - I_C / I_E} = \frac{\alpha}{1 - \alpha}

\beta = \frac{\alpha}{1 - \alpha}

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

What This Tells You

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

\beta = \frac{0.98}{1 - 0.98} = \frac{0.98}{0.02} = 49

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.

Watch out

Never memorise this formula as a random equation. It's a direct consequence of $I_E = I_B + I_C$ — the most fundamental relation in transistor physics. If you forget the formula, derive it in 10 seconds.

The Reverse Relation

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

\alpha = \frac{\beta}{\beta + 1}

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

Tip

For large $\beta$ (say 100 or more), $\alpha$ is extremely close to 1. The approximation $\alpha \approx 1$ is often good enough for rough work, but never use it when precision matters — like in feedback amplifier design.

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.

Given $\Delta I_C=0.6\ \text{mA}=600\ \mu\text{A}$ and $\Delta I_B=8\ \mu\text{A}$ :

\beta_{ac}=\frac{\Delta I_C}{\Delta I_B}=\frac{600}{8}=75

✓Final answer

\beta_{ac}=75

Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2023Set ANNUAL1 markMCQ
$Q. The relation between the transistor parameters α and β is$
(a) β = α / (1 - α)
(b) α = β / (β - 1)
(c) β = (1 + α) / α
(d) α = (1 + β) / β
›Reveal solutionSolution

The current-gain relation between the common-base current gain α and common-emitter current gain β of a transistor is β = α/(1-α).

For a transistor, $\alpha = I_C/I_E$ (common-base current gain) and $\beta = I_C/I_B$ (common-emitter current gain). By Kirchhoff's current law at the transistor junction, $I_E = I_B + I_C$ , so $I_B = I_E - I_C$ .

$\beta = \dfrac{I_C}{I_E - I_C} = \dfrac{I_C/I_E}{1 - I_C/I_E} = \dfrac{\alpha}{1-\alpha}$

This is the standard relation, and it also follows that $\alpha = \beta/(1+\beta)$ .

✓Final answer
$(a) β = α / (1 - α).$
CBSE 2018Set ANNUAL1 markMCQ
$Q. If the current constant for a transistor are α & β then -$
(a) αβ = 1
(b) β > 1, α < 1
(c) α = β
(d) β < 1, α > 1
›Reveal solutionSolution

α = I_C/I_E < 1 and β = I_C/I_B ≫ 1, related by β = α/(1−α).

The common-base current gain is $\alpha = \dfrac{I_C}{I_E}$ . Since $I_E = I_C + I_B$ and the base current is small, $I_C < I_E$ , so $\alpha < 1$ (typically 0.95–0.99).

The common-emitter current gain is $\beta = \dfrac{I_C}{I_B}$ . Because the base current is very small, $\beta \gg 1$ (typically 50–300).

They are linked by $\beta = \dfrac{\alpha}{1-\alpha}$ , which confirms β > 1 when α < 1.

✓Final answer
$(b) β > 1, α < 1.$

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Q.In a common-emitter amplifier, a change of 0.6 mA0.6\ \text{mA}0.6 mA in the collector current corresponds to a change of 8 μA8\ \mu\text{A}8 μA in the base current. Calculate the small-signal current gain βac\beta_{ac}βac​ of the transistor.

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Q.In a common-emitter amplifier, a change of $0.6\ \text{mA}$ in the collector current corresponds to a change of $8\ \mu\text{A}$ in the base current. Calculate the small-signal current gain $\beta_{ac}$ of the transistor.

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