Transistors: npn and pnp, how they work, and configurations
A transistor has a thin, lightly doped base between emitter and collector. In the active state, a small base current controls a much larger collector current.
npn and pnp transistors
In both, doping falls from emitter to collector to base. The collector is the largest region because it handles the most heat.
How an npn transistor works
- 1Bias it: the emitter-base junction is forward biased (VBE about 0.6 to 0.7 V for Si). The collector-base junction is reverse biased by a few volts.
- 2The forward bias pushes many electrons from the emitter into the base. This flow is the emitter current IE.
- 3The base is very thin and lightly doped, so it has few holes. Less than 5 percent of the electrons recombine there and make the small base current IB.
- 4More than 95 percent of the electrons diffuse across the base and reach the collector junction.
- 5The field of the reverse-biased collector junction sweeps them into the collector. This flow is the collector current IC.
- 6★So IE = IB + IC, and IC is almost equal to IE. A small change in IB controls a large change in IC.
★ key step
Deriving the link between α and β
α = IC/IE is the current gain with the base common. β = IC/IB is the current gain with the emitter common.
- 1Start from IE = IC + IB.
- 2Divide every term by IC: IE/IC = 1 + IB/IC.
- 3IE/IC = 1/α and IB/IC = 1/β, so 1/α = 1 + 1/β.
- 4Then 1/β = 1/α − 1 = (1 − α)/α.
- 5★So, β = α/(1 − α). Since α is just below 1, β is large.
★ key step
Three ways to connect a transistor
| Configuration | Common terminal | Input and output | Main use |
|---|---|---|---|
| Common emitter (CE) | Emitter | Input: base-emitter; output: collector-emitter | Most used: gives current, voltage and power gain |
| Common base (CB) | Base | Input: emitter-base; output: collector-base | Constant current source; current gain α is just below 1 |
| Common collector (CC) | Collector | Input: base-collector; output: emitter-collector | Impedance matching |
Worked example: α and β
In a transistor, the emitter current is 5 mA and the base current is 0.05 mA. Find IC, α and β, and check that β = α/(1 − α).
- IC = IE − IB = 5 − 0.05 = 4.95 mA.
- α = IC/IE = 4.95 ÷ 5 = 0.99.
- β = IC/IB = 4.95 ÷ 0.05 = 99.
- Check: α/(1 − α) = 0.99 ÷ 0.01 = 99, which matches.
IC = 4.95 mA, α = 0.99, β = 99
How to draw the symbols and identify the leads
- npn symbol: a short thick bar for the base B, with two slanting lines for E and C. Put the arrow on the emitter line, pointing away from the base.
- pnp symbol: the same, but the emitter arrow points towards the base.
- Biasing diagram: a small battery VBE forward biasing E-B and a larger battery VCB reverse biasing C-B. Mark IE, IB and IC with arrows.
- Leads: the lead nearest the small tab or notch on the case is the emitter.
- Codes: first letter A = germanium, B = silicon; second letter C = audio, D = power, F = radio frequency. So BC547 is a silicon audio-frequency transistor.
Transistors are the switches and amplifiers inside every mobile phone, laptop and FM radio. A phone's main chip holds billions of tiny transistors, each controlling one current with an even smaller one.
3 marks: “Explain the working of an npn transistor with a labelled biasing diagram.” E-B forward, C-B reverse. Electrons from the emitter enter the thin, lightly doped base; under 5 percent recombine (IB); over 95 percent are swept into the collector (IC). IE = IB + IC. Diagram with both batteries and the three currents.
2 marks: “Define α and β of a transistor and show that β = α/(1 − α).” α = IC/IE and β = IC/IB. From IE = IC + IB, dividing by IC gives 1/α = 1 + 1/β, so β = α/(1 − α).
2 marks: “For a transistor, α = 0.995. Find β, and the change in collector current when the emitter current changes by 2 mA.” β = α/(1 − α) = 0.995 ÷ 0.005 = 199. ΔIC = α × ΔIE = 0.995 × 2 mA = 1.99 mA.
E-B forward, C-B reverse. Doping: Emitter > Collector > Base. Size: Collector biggest, Base thinnest.