troubleshooting
Why does my LED burn out?
Short answer: Almost always: no resistor. An LED across a battery draws far more current than it can survive. A 220 Ω resistor on 3×AA holds it to about 11 mA.
See it happen
This is the broken circuit, simulated live. The readout under the board tells you what the parts are doing.
simulating…
What's going on
An LED is a diode: a one-way valve for current with a fixed "price of entry" called the forward voltage — about 2.0 V for a red one. Below that voltage almost nothing flows. Above it the LED stops resisting almost entirely, so the battery decides the current, not the LED.
A 4.5 V battery pack has about 2.5 V more than the LED wants. With nothing else in the loop that 2.5 V is dropped across the LED's own tiny resistance and the battery's, and the current climbs past 100 mA. A 5 mm LED is rated for 20 mA. It gets hot, the bond wire inside melts, and it goes dark for good — sometimes with a little pop.
A resistor is a part that turns voltage into a controlled current: I = V ÷ R (Ohm's law). Put 220 Ω in the loop and the spare 2.5 V ÷ 220 Ω = 0.011 A, or 11 mA. Bright, cool, and it lasts for years.
The fix
Put a resistor anywhere in the loop — before or after the LED makes no difference, because the same current flows through every part of a single loop.
- 3×AA (4.5 V): 220 Ω → about 11 mA.
- 2×AA (3 V): 100 Ω → about 10 mA.
- 9 V battery: 680 Ω → about 10 mA.
Any bigger resistor is also safe, just dimmer. Use the LED resistor calculator for other LEDs and batteries.
simulating…
Also check
- It died even with a resistor? Check the value: brown-black-brown is 100 Ω, red-red-brown is 220 Ω, brown-black-red is 1 kΩ. A 22 Ω (red-red-black) is too small.
- The resistor is in a different branch. If the LED has its own path from + to − that skips the resistor, the resistor isn't protecting it. Trace the loop with your finger.
- Blue and white LEDs want about 3 V and are dim on 2×AA — that's not damage, just not enough voltage.