Two junctions, one thin base
An NPN bipolar junction transistor is two P/N junctions back to back: N emitter, a very thin P base, N collector. Forward-bias the base–emitter junction and the emitter injects electrons into the base. The base is so thin that almost all of them diffuse straight through and are swept into the collector by the reverse-biased collector junction. Only a few recombine in the base; they make up the small base current.
IC = β · IB, IE = IC + IB VBE ≈ 0.6–0.7 V when conducting
Output characteristics and the load line
Plot collector current against VCE for several base currents and you get a family of nearly flat curves: in the active region IC depends on IB, hardly on VCE. The collector circuit adds a load line, IC = (VCC − VCE)/RC. The operating point sits where the load line meets the curve for the present IB.
IC = (VCC − VCE) / RC
Three regions
Where Q lands on the load line decides how the transistor behaves:
- Cutoff: VBE below about 0.5 V, no base current, so no collector current. The transistor is an open switch and VCE ≈ VCC.
- Active: IC = β·IB, controlled by the base. This is the region for amplifiers.
- Saturation: the base asks for more than RC lets through. IC is capped at about (VCC − 0.2)/RC and VCE ≈ 0.2 V. The transistor is a closed switch.
The transistor as a switch
Digital and power circuits use only the two ends of the load line: cutoff for off, saturation for on. To make sure the transistor saturates, designers drive the base with two to ten times the minimum current IC,sat/β, so the switch stays on even with a low-β part.
Worked example (the simulator’s default values)
- VBB = 2 V, RB = 100 kΩ, VCC = 12 V, RC = 2.2 kΩ, β = 100.
- Base loop: IB = (2 − 0.66)/100 k = 13.4 µA.
- Collector: IC ≈ β·IB = 1.46 mA (slightly over 100 × IB, from the Early effect); VCE = 12 − 1.46 mA × 2.2 kΩ = 8.80 V: active.
- Lower RB to 22 kΩ: IB = 59 µA would ask for 5.9 mA, but RC allows only (12 − 0.2)/2.2 k = 5.36 mA. The transistor saturates at VCE = 0.19 V.
Common mistakes
- Applying IC = β·IB in saturation. There the collector circuit limits IC, and IC/IB falls below β.
- Forgetting VBE: the base current is (VBB − 0.7)/RB, not VBB/RB.
- Treating β as a precise constant. It varies widely between parts and with temperature, which is why good amplifier designs don’t depend on it.
Common questions
How does a BJT transistor work?
A small base current lets a much larger current flow from collector to emitter: IC = β·IB, with β typically 50–300. Electrons injected by the forward-biased base–emitter junction cross the thin base and are collected by the reverse-biased collector junction.
What are the cutoff, active and saturation regions?
Cutoff: no base current, transistor off. Active: IC = β·IB, used for amplifying. Saturation: IC limited by the collector circuit, VCE ≈ 0.2 V, transistor fully on.
How do you use a transistor as a switch?
Drive it between cutoff (off) and saturation (on). Choose RB so that IB is several times IC,sat/β, so the transistor saturates even with a low β.