troubleshooting
Why is my LED so dim?
Short answer: The resistor is too big for the battery. 10 kΩ on 4.5 V lets only 0.25 mA through; 220 Ω gives 11 mA and a proper glow.
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
Brightness follows current. A 5 mm LED looks properly lit from about 5 mA and is at full brightness around 20 mA. The resistor sets that current by Ohm's law, I = V ÷ R, using the voltage left over after the LED takes its 2 V:
- 4.5 V − 2.0 V = 2.5 V to spare
- 2.5 V ÷ 10 000 Ω = 0.00025 A = 0.25 mA — a faint glow you only see in the dark
- 2.5 V ÷ 220 Ω = 0.011 A = 11 mA — clearly lit
Ten thousand ohms is forty-five times more resistance than two hundred and twenty, so forty-five times less current.
The fix
Swap the resistor for 220 Ω (red-red-brown). Resistors are easy to mix up: 10 kΩ is brown-black-orange and 1 kΩ is brown-black-red — one band colour apart from each other, and both dimmer than you want.
If you want to choose the brightness, the LED resistor calculator gives the resistor for any current between 2 and 20 mA.
simulating…
Also check
- Weak battery. Every 0.5 V lost from the battery is 20 % less current in this loop. Rechargeable AAs are 1.2 V each, so three give 3.6 V, not 4.5 V — noticeably dimmer with the same resistor.
- Two LEDs in series share the spare voltage and both go dim — see why two LEDs in series are dim.
- It's a daylight thing. Compare against a known-good LED before changing anything.