How does an NPN transistor work in the active region, and what are alpha and beta?

In one line

In the active region, the forward-biased emitter-base junction injects electrons into the thin base, almost all of which are swept into the collector by the reverse-biased collector-base junction, so a small base current controls a much larger collector current.

Explained simply

Imagine a water slide with a small valve at the top. Opening the valve a little lets a big rush of water down the slide. In a transistor, a small base current is the valve and the large collector current is the rush. That is current amplification.

Technically, for active-region operation: emitter-base junction forward biased (about 0.7 V for silicon), collector-base junction reverse biased. Current gain is described by \( \alpha \) (common-base gain) and \( \beta \) (common-emitter gain).

How it works

1. Forward bias on \( J_E \) makes electrons from the heavily doped n-type emitter flow into the base: this is the emitter current \( I_E \).
2. The base is thin and lightly doped, so only a few percent of these electrons recombine with holes there. Those few make up most of the base current \( I_B \).
3. The rest diffuse across the thin base to \( J_C \), where the strong field of the reverse-biased junction sweeps them into the collector: this is \( I_C \).
4. So \( I_C \) is nearly equal to \( I_E \), and \( I_B \) is small.
5. A small reverse leakage current \( I_{CBO} \) also flows across \( J_C \); it doubles roughly every 10 °C rise in temperature.
6. The other regions: cut-off (both junctions reverse biased, transistor OFF) and saturation (both forward biased, transistor fully ON).

Key formula

\( I_E = I_B + I_C, \quad \alpha = \dfrac{I_C}{I_E}, \quad \beta = \dfrac{I_C}{I_B}, \quad \beta = \dfrac{\alpha}{1-\alpha}, \quad \alpha = \dfrac{\beta}{1+\beta} \)

\( \alpha \) = common-base current gain (slightly less than 1, e.g. 0.95–0.995), \( \beta \) (also \( h_{FE} \)) = common-emitter current gain (typically tens to a few hundred); both have no unit.

Quick example

If \( I_B = 20\,\mu \)A and \( \beta = 100 \): \( I_C = 2 \) mA, \( I_E = 2.02 \) mA, \( \alpha = 2/2.02 = 0.990 \). A microphone's tiny signal current at the base thus produces a much larger varying collector current. (Illustration only.)

Common mistakes

• Saying the collector-base junction is forward biased in the active region.
• Writing \( I_C = I_E + I_B \); the emitter current is the largest.
• Saying \( \alpha \) can be greater than 1.

How MU asks it

Numericals are not expected in this module (MU syllabus), so questions are theory with diagrams. Typical: "Explain the operation of an NPN transistor" (5 marks), or "Define \( \alpha \) and \( \beta \) and derive the relation between them" (3–5 marks). For full marks draw the biased NPN structure with electron flow and current directions, explain injection, diffusion, recombination and collection in steps, and derive \( \beta = \alpha/(1-\alpha) \) from \( I_E = I_B + I_C \).