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Analog Electronics

BJT: The Transistor as a Switch

Think of a transistor as an electric relay: a small control current (base, IB) manages a large load current (collector, IC). The gain ratio is called β: IC = β·IB. With β=100, a 1 mA finger-tap opens a 100 mA gate.

When used as a switch, we never linger in the middle region: either fully off (IB=0 → IC=0) or fully on/saturated (VCE ≈ 0.1 V — practically a straight wire). To GUARANTEE saturation we overdrive the base by a factor of 3–10; the base resistor is simply RB = (Vcontrol−0.7)/IB.

Golden rule: with inductive loads like relays and motors, a flyback diode is mandatory. A coil resists having its current cut off abruptly; if it finds no path, it punches through the transistor with a spike of hundreds of volts. An anti-parallel diode gives that energy a safe "escape ramp".

Formulas

IC = β · IB (active region)
RB = (Vin − 0.7) / IB
Saturation: IB ≥ (3…10) · IC/β

⚡ Open this lesson in the simulator — free

Test Yourself

With a β=100 transistor, what minimum base current should you choose to switch a 100 mA load with GUARANTEED saturation?
Answer: 3–5 mA — Theoretically 1 mA; for guaranteed saturation you drive 3–5× that (3–5 mA).
Why is a flyback diode placed across a relay coil?
Answer: To quench the inductive voltage spike at turn-off — The L·di/dt spike discharges safely through the diode.
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