The Corridor Night Light That Eats Batteries
A transition-year group is prototyping a night light for the school corridor. The prototype uses four white LEDs, each with its own 68 Ω series resistor, driven from a holder of three AA alkaline cells (4.5 V nominal, rated 2000 mAh). Each LED has a forward voltage of 3.1 V at 20 mA. The brief says the light must run for 8 hours a night for seven nights on one set of cells, because the caretaker will not change batteries more often. On test, the light was bright on the first night, noticeably dim by the third, and one LED is now dark while the other three still light. The group measured 3.9 V at the battery holder and a total current of 128 mA from the pack. There is a 230 V socket 1.5 m away, but any fixed wiring must be done by a contracted electrician. Materials budget is €8; the deadline is in five days.
1. Calculate the intended current per branch and the runtime the design should give. Does the design meet the brief? Show your working.
Read the worked answer
Branch current. I = (V_supply − V_f) / R = (4.5 − 3.1) / 68 = 1.4 / 68 = 0.0206 A ≈ 20.6 mA.
Total current. Four parallel branches: 4 × 20.6 = 82.4 mA.
Energy demanded by the brief. 8 h × 7 nights = 56 h. Charge needed = 82.4 mA × 56 h = 4614 mAh.
Energy available. 2000 mAh nominal, but at 82 mA the delivered capacity is well below the 2000 mAh rating (the rating assumes a much lower discharge current, roughly 20 mA). Realistically expect 1400–1600 mAh, giving 1400 / 82.4 ≈ 17 h, about two nights.
Conclusion. The design misses the requirement by a factor of about 2.5–3, and the real-world figure is closer to 3 nights than 7. The cause is structural: the resistor burns 1.4 V of the 4.5 V supply as heat (31% wasted) and the four LEDs together draw more than the cells can give. No component change can fix a battery that is too small for the duty; the topology has to change.
Useful check. Even a perfect 3.0 V battery with the same LED (V_f 3.1 V) would not light at all, which is why 3 cells were chosen - but the extra 1.4 V is being thrown away.
2. Explain the measured 128 mA total current and the dark LED, and state what hazard the fault creates.
Read the worked answer
Expected total is 82.4 mA. The measured 128 mA is 45 mA too high, and one branch is dark.
Explanation. If an LED fails short-circuit (a common failure mode after over-current or a poor solder joint), that branch becomes just the 68 Ω resistor across 4.5 V: I = 4.5 / 68 = 66 mA. It draws current but emits no light - exactly the dark LED.
Check the total. Three healthy branches at 20.6 mA plus one shorted branch at 66 mA = 61.8 + 66 = 127.8 mA ≈ 128 mA, matching the measurement. This confirms the diagnosis.
Runtime now. 2000 / 128 ≈ 15.6 h, so the pack is flat in two nights, which explains the early dimming.
Hazard. The 68 Ω resistor now dissipates P = I²R = 0.066² × 68 = 0.30 W. A standard 0.25 W resistor will overheat, discolour and can scorch the board or the plastic housing. A shorted LED in a sealed enclosure is a small but real fire risk.
Actions. Switch off, replace the LED, inspect all joints, and check the resistor's power rating: use at least 0.5 W, or better, replace the resistor dropper with a constant-current driver so a shorted LED cannot draw 3× current. Add a 100 mA polyfuse in the pack for protection.
3. Propose a redesign that meets the seven-night requirement within the €8 budget, and justify your choice against at least one alternative.
Read the worked answer
Option chosen: two LEDs at 10 mA, higher-value resistors, plus a PIR sensor.
- Resistor per branch: R = (4.5 − 3.1) / 0.010 = 140 Ω (use 150 Ω, giving 9.3 mA).
- The corridor is only occupied intermittently at night, so a PIR can cut the on-time from 8 h to an assumed 2 h per night. That assumption must be stated and tested, not guessed silently.
- Energy per week: 2 branches × 9.3 mA × 2 h × 7 = 260 mAh. Even at 6 h per night it is 780 mAh, comfortably inside a realistic 1400 mAh from the cells.
- Cost: PIR module around €3, resistors and a holder under €1, well inside €8.
Why not the alternatives.
- Bigger cells (3 × D, 12 000 mAh): works electrically but blows the budget and the caretaker still has to source D cells; it treats the symptom.
- Mains adapter on the nearby socket: a plug-in 5 V supply is electrically safe and cheap, but it needs a cable route along a corridor, which means trunking, an electrician and a cost far above €8. It also fails if the socket is switched off.
- One brighter LED at 20 mA: fewer branches but the same light output, so no real gain.
Recommendation. Build the PIR version, measure actual on-time over three nights with a data logger, and only then decide whether the mains option is worth the extra cost.