Describe transistor action in an n-p-n transistor in terms of the motion of majority and minority charge carriers across its two junctions.

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Accepted Answer

Concept understanding — Transistor Action

Transistor Action — The Intuition First

Imagine a narrow hallway with a door at each end. The first door (emitter-base junction) is wide open — people can pour in easily. The second door (collector-base junction) is almost closed, but it opens inward only, so once someone passes through the hallway they can leave but cannot come back. The hallway itself is very short and thin (the base). Most people who enter through the first door will walk straight through and exit the second door, because the hallway is too narrow for them to linger or turn around.

That is the core picture of a bipolar junction transistor (BJT) in active mode.

The emitter is heavily doped — it has a huge supply of charge carriers (electrons for an n-p-n, holes for a p-n-p). The base is very thin and lightly doped. The collector is moderately doped and physically large. When you forward-bias the emitter-base junction, carriers flood from the emitter into the base. Once inside the base, these carriers are minority carriers — they are not supposed to be there. The base region is so thin that most of them diffuse across it before they have a chance to recombine with majority carriers. When they reach the collector-base junction, the reverse bias there sweeps them into the collector because the electric field of the depletion region pulls them across.

That is transistor action: a small current in the emitter-base loop controls a much larger current in the collector-base loop, because nearly all the injected carriers make it to the collector.

The Precise Statement

Important

Transistor action is the phenomenon in which, under active-mode biasing (EB junction forward biased, CB junction reverse biased), the majority of charge carriers injected from the emitter into the base diffuse across the thin base region and are collected by the reverse-biased collector-base junction, producing a collector current that is nearly equal to the emitter current.

Mathematically, the collector current $I_C$ is related to the emitter current $I_E$ by the common-base current gain $\alpha$ :

I_C = \alpha I_E

where $\alpha$ is typically $0.95$ to $0.999$ . The small fraction that is lost — carriers that recombine in the base or are injected back into the emitter — constitutes the base current $I_B$ :

I_E = I_B + I_C

The more familiar common-emitter current gain $\beta$ is defined as:

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

A typical $\beta$ of 100 means that a base current of $10\ \mu\text{A}$ can control a collector current of $1\ \text{mA}$ — that is the amplification.

Why It Works — The Three Conditions

Emitter heavily doped — ensures a huge supply of carriers to inject. The emitter current is almost entirely due to these injected carriers, not due to majority carriers from the base.
Base very thin ( $\ll$ diffusion length) — injected carriers reach the collector before they recombine. The base width $W_B$ is much smaller than the minority-carrier diffusion length $L_n$ (for n-p-n). Recombination loss is minimal.
Collector reverse biased — the strong electric field at the CB junction pulls carriers across, preventing them from accumulating at the base-collector boundary. It also prevents injection of carriers from collector into base. …

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The emitter-base junction is forward biased, so the heavily doped n-type emitter injects a large number of electrons into the thin, lightly doped p-type base. As minority carriers there, these electrons could recombine with the base's holes, but because the base is so thin and lightly doped, only a small fraction actually does -- forming the small base current $I_B$ . The great majority survive to reach the base-collector junction, where the reverse-bias field there sweeps them across into the collector, forming the large collector current $I_C$ , with $I_ …

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

CBSE 2024Set ANNUAL1 mark
$Q. In a transistor, base is made very thin and doped with little impurity atoms, why? OR Which one you prefer to use among common-base or common-emitter transistors as amplifier and why?$
›Reveal solutionSolution
A thin, lightly-doped base minimises electron-hole recombination inside the base region, so nearly all carriers emitted from the emitter successfully reach the collector — this is exactly what makes transistor action (current amplification) possible.

In a transistor (say n-p-n), the emitter is heavily doped (to inject a large number of majority carriers, electrons, into the base) and the base is deliberately made:

VERY THIN (its width is much smaller than the diffusion length of the minority carriers) — so that the carriers injected from the emitter cross it quickly, in a very short transit time, before they have a chance to recombine with the base's own majority carriers (holes).

LIGHTLY DOPED — so the base has a low concentration of holes (majority carriers there); with fewer available holes, the probability that an emitter-injected electron recombines with a hole while crossing the base is small.

…
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CBSE 2018Set ANNUAL1 markMCQ
$0.505\,mA$
(a) $0.25\,mA$
(b) $0.5\,mA$
(c) $0.52\,mA$
(d) $0.55\,mA$
›Reveal solutionSolution

Transistor current relation $I_E = I_B+I_C$ directly gives $I_C = I_E - I_B$ .

For any transistor, Kirchhoff's current law applied to the three terminals gives the fundamental relation:

$I_E = I_B + I_C$

…

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CBSE 2017Set ANNUAL1 markMCQ
$Q. The emitter base junction of a given transistor is forward biased and its collector base junction is reverse biased. If the base current is increased, then its :$
(a) $V_{CE}$ will increase
(b) $I_C$ will decrease
(c) $I_C$ will increase
(d) $V_{CC}$ will increase
›Reveal solutionSolution

In active-mode operation ( $I_C=\beta I_B$ ), raising the base current proportionally raises the collector current.

The bias condition described — emitter-base (EB) junction forward biased, collector-base (CB) junction reverse biased — is precisely the definition of a bipolar junction transistor (BJT) operating in its active region, the normal mode used for amplification. In this mode, a small current $I_B$ injected into the thin, lightly doped base controls a much larger current $I_C$ flowing from collector to emitter, related by the transistor's common-emitter current gain $\beta$ (or $h_{FE}$ ):

$I_C = \beta I_B.$

Physically, forward-biasing the EB junction injects majority carriers (electrons, in an npn transistor) from the emitter into the thin base region; because the base is thin and lightly doped, most of these carriers diffuse across it without recombining and are then swept into the collector by the strong reverse-bias field at the CB junction. The fraction of base current that must be supplied is fixed (mostly) by the transistor's construction (as $\beta$ ), so increasing the base current $I_B$ directly and proportionally increases the collector current $I_C$ that this active-mode relation permits.

…

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Q.Describe transistor action in an n-p-n transistor in terms of the motion of majority and minority charge carriers across its two junctions.

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