calculator
LED resistor calculator
An LED has no brakes of its own. Pick your battery, your LED and how bright you want it, and this works out the resistor that keeps it alive — then runs the exact circuit on a breadboard below.
The answer
270 Ω — red-violet-brown
- Voltage left for the resistor: 4.5 V − 2 V = 2.5 V
- Ohm's law, R = V ÷ I: 2.5 V ÷ 0.010 A = 250 Ω
- Nearest standard value, rounding up: 270 Ω
- Real current with that resistor: 2.5 V ÷ 270 Ω ≈ 9.3 mA
- Heat in the resistor: 23 mW — a normal ¼ W (250 mW) resistor is fine.
The simulator below uses a real diode curve instead of a flat 2.0 V, so its current can differ by a milliamp or two. That is the difference between the textbook and the bench.
simulating…
Why an LED needs a resistor at all
A resistor is a part that resists current — it turns some of the push (voltage) into heat so less current flows. An LED is a diode: once the voltage across it passes its forward voltage (about 2 V for red), it stops resisting almost completely. Connect one straight across a 4.5 V battery and the extra 2.5 V has nowhere to go, so the current shoots up until the LED cooks. The resistor takes that extra voltage instead, and Ohm's law tells you exactly how much current it lets through: I = V ÷ R.
How to read the answer
The calculator rounds up to a standard value. Resistors come in fixed steps (the E12 series: 100, 120, 150, 180, 220, 270, 330, 390, 470, 560, 680, 820 and their multiples), and rounding up means slightly less current than you asked for, never more. An LED is plenty bright at 10 mA; 20 mA is the most a normal 5 mm LED should ever carry.
Common questions
Does the resistor go before or after the LED? Either. In a single loop the same current flows through every part, so the resistor limits it wherever it sits.
What if I only have a bigger resistor? Use it. A 1 kΩ resistor on 4.5 V gives about 2.5 mA: dimmer, perfectly safe. A smaller one than the answer is the only mistake.
Two LEDs in series? Add their forward voltages before subtracting: 4.5 V − 2.0 V − 2.0 V leaves 0.5 V, which is why two red LEDs in series are dim on 3×AA and fine on 9 V.
Try it for real in First Light, or explore Ohm's law.