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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
Odisha ChseOdisha CHSE +2 Science Board Exam 2023MCQ· 1mImportance★★★★★
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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 (α\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:

α=ICIE\alpha = \frac{I_C}{I_E}

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

Beta (β\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:

β=ICIB\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 α\alpha and β\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.

IE=IB+ICI_E = I_B + I_C

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

α=ICIE,β=ICIB\alpha = \frac{I_C}{I_E}, \quad \beta = \frac{I_C}{I_B}

From the current relation, IB=IE−ICI_B = I_E - I_C. Substitute into β\beta:

β=ICIE−IC\beta = \frac{I_C}{I_E - I_C}

Divide numerator and denominator by IEI_E:

β=IC/IE1−IC/IE=α1−α\beta = \frac{I_C / I_E}{1 - I_C / I_E} = \frac{\alpha}{1 - \alpha}

β=α1−α\beta = \frac{\alpha}{1 - \alpha}

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

What This Tells You

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

β=0.981−0.98=0.980.02=49\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 IE=IB+ICI_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 α\alpha in terms of β\beta:

α=ββ+1\alpha = \frac{\beta}{\beta + 1} …

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