Circuits / Getting started
LED with a series resistor
The resistor sets the LED current: I = (V − Vf) / R
Tap a part to see its voltages and currents. It is a model, not a measurement.
Predict, measure, explain
- Predict With 5 V and 330 Ω, how much current flows through the LED?
- Measure Tap the LED. Then change R1 to 1 kΩ and tap it again.
- Explain
Show the answerHide the answer
About 9.8 mA: the resistor takes the 3.2 V the LED leaves. With 1 kΩ it drops to 3.4 mA, and the LED dims.
What do you need?
- Resistor 330 ΩSearch on Amazon
- Red LEDSearch on Amazon
- Jumper wires × 1Search on Amazon
- BreadboardSearch on Amazon
How does a resistor protect an LED?
An LED holds an almost fixed voltage, so the resistor in series is what decides how much current flows. The rule is I = (V − Vf) / R: the supply voltage minus the LED's forward voltage, divided by the resistance. In this circuit, with 5 V and 330 Ω, the red LED keeps about 1.76 V and the resistor takes the other 3.24 V, so about 9.8 mA flows through both. That is well below the 20 mA most small LEDs are rated for, and still bright. Change R1 to 1 kΩ and the current drops to 3.4 mA: the LED still lights, only dimmer. Without the resistor, nothing would limit the current and the LED could burn out in a moment.
Values from the Protoboard ISO simulation: a model, not a bench measurement.
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