troubleshooting
Why does my LED pop on a 9 V battery?
Short answer: A 220 Ω resistor sized for 3×AA pushes about 32 mA on a 9 V battery — past the LED's 20 mA rating. On 9 V use 680 Ω for a safe 10 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
The resistor's job is to swallow the voltage left over after the LED takes its share. On a 4.5 V battery there's 2.5 V to spare; on a 9 V battery there's 7 V — nearly three times as much. The same resistor passes nearly three times the current.
Ohm's law lays it out, I = V ÷ R:
- 4.5 V, 220 Ω: (4.5 − 2.0) ÷ 220 = 11 mA ✓
- 9 V, 220 Ω: (9 − 2.0) ÷ 220 = 32 mA ✗ — a 5 mm LED is rated for 20 mA and pops around 30 mA
- 9 V, 680 Ω: (9 − 2.0) ÷ 680 = 10 mA ✓ — same brightness as the 4.5 V case
The LED doesn't care what the battery is. It still wants about 2 V across itself and about 10 mA through it. Change the battery and you have to re-pick the resistor so the current stays in that window.
The fix
Pick the resistor from the leftover voltage, not from habit:
- 9 V: 7 V spare → use 680 Ω for 10 mA, or 470 Ω for 15 mA.
- 4.5 V (3×AA): 2.5 V spare → use 220 Ω for 11 mA.
- 3 V (2×AA): 1 V spare → use 100 Ω for 10 mA.
The LED resistor calculator does the arithmetic for any battery, any LED colour, and any current you choose.
simulating…
Also check
- Weakened but not popped? 25–30 mA for a few seconds often doesn't kill a 5 mm LED outright, but it ages fast — it may be dimmer than its neighbours forever. Swap it when you fix the resistor.
- Blue or white LED on 9 V. Forward voltage ≈ 3 V, so 9 − 3 = 6 V ÷ 470 Ω ≈ 13 mA. One 470 Ω is a tidy pick for any colour at 9 V.
- Two red LEDs in series on 9 V. 2 × 2 V = 4 V dropped by the LEDs, leaving 5 V for the resistor. 470 Ω gives about 11 mA through both — the "fix" for why two LEDs in series are dim.
- Hot 9 V battery. That's a short somewhere else, not an LED problem — see why does my battery get hot.