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BJT as a switch

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Objective

Drive an LED from a 5 V logic level through an NPN switch; compare ON (saturation) and OFF (cut-off).

Procedure

Run with 5 V input: the LED glows.
Set the input battery to 0 V and run again.

Theory

With 5 V on a 4.7 kΩ base resistor Ib = 0.9 mA, enough to saturate the transistor (Vce(sat) ≈ 0.1–0.2 V) so the LED carries (5 − 1.9 − 0.15)/330 ≈ 9 mA. With 0 V in, the transistor is cut off and the LED is dark.

Questions to answer

Why do we say the transistor is 'saturated'?

Components used

2 × Battery / DC source, LED 5mm Red, 2 × Resistor 220 ohm 1/4W, Transistor BC547 (NPN)

About the model

Ideal wires; component models as listed in the catalogue (no parasitics, no temperature drift). Built without a breadboard so the schematic stays readable.