Explain, with a labelled diagram, the working of an npn transistor. Define its current gains α and β, and derive the relation between them.

1. 1 · Describe the structure. An npn transistor has a thin, lightly doped p-type base between a heavily doped n-type emitter and a larger n-type collector.
2. 2 · State the biasing. The emitter-base junction is forward biased and the collector-base junction reverse biased.
3. 3 · Explain the action. Electrons from the emitter enter the base (IE); under 5 percent recombine (IB); over 95 percent are swept into the collector (IC). So IE = IB + IC.
4. 4 · Draw the biasing diagram. [Diagram: npn transistor with a VBE battery forward biasing E-B and a larger VCB battery reverse biasing C-B; arrows for IE, IB and IC; E, B and C labelled]
5. 5 · Define the gains. α = IC/IE (just below 1); β = IC/IB (large).
6. 6 · Derive the relation. Divide IE = IC + IB by IC: 1/α = 1 + 1/β, so β = α/(1 − α).
7. 7 · Real life. This control of a big current by a small one lets transistors amplify sound in a radio.
8. 8 · Close in one line. So, a small base current controls a much larger collector current.

Scoring tip: Draw the current arrows the right way: for an npn transistor, IE comes out of the emitter while IB and IC flow in.