Choosing a resistor for an LED
Light a red, green, blue or white LED from a 5 V, 9 V or 12 V supply. Work out the resistor that lets just the right current flow, and how much power the resistor turns into heat.
Core concept
An LED takes a fixed forward voltage (about 2 V for red, 3.2 V for blue and white); a resistor in series takes the rest, so by Ohm's law R = (supply − LED voltage) ÷ current
Topic: Energy and technology
An LED (light-emitting diode) glows when a small current: The flow of electricity through a wire, measured in amperes (A). All key words, about 10 to 20 mA, flows through it. Unlike a bulb it does not limit its own current: joined straight to a battery it takes too much and burns out in a flash. So we put a resistor in series. Across a lit LED there is a fixed voltage: The push that makes electricity flow, measured in volts (V). All key words, its forward voltage: about 2.0 V for red, 2.2 V for green and 3.2 V for blue and white. The resistor takes the rest of the supply voltage, and by Ohm's law (V = IR) it must be R = (supply − LED voltage) ÷ current, with the current in amperes. The resistor turns P = V × I into heat.
What you will learn
- Explain why an LED needs a series resistor.
- Use Ohm's law to find the resistor for a given supply, LED and current.
- Work out the power the resistor wastes as heat.
Before you start
- Know Ohm's law, V = IR, and that 1 mA = 0.001 A.
Materials
- LEDs, resistors and a 9 V battery (if your school has an electronics kit)
Equipment
- A multimeter (optional)
Procedure
- Record a red LED on 5 V at 20 mA.
- Change the supply to 9 V and record.
- Try a blue LED on 5 V, then a white LED on 12 V at 10 mA, and record each.
How this simulation is simplified
- Forward voltages: red 2.0 V, green 2.2 V, blue 3.2 V, white 3.2 V (typical values).