Parallel LED Modification on a Battery Indicator
A student designs a 12 V battery-powered LED indicator. The chosen LED has a forward voltage of 2.0 V at 20 mA. The student calculates a series resistor of 500 Ω and fits the nearest preferred value, 470 Ω, rated 0.25 W. On test the LED current is 21 mA and the resistor is warm but acceptable. To make the indicator brighter for a display, the student connects a second identical LED directly in parallel with the first, keeping the same 470 Ω resistor. The battery voltage then falls to 11.2 V under load. The display must run unattended for six hours, and the battery is a sealed 12 V 1.2 Ah type. The student asks whether the parallel modification is acceptable and how long the battery will last. The LED's maximum continuous current is 25 mA.
1. Calculate the LED current with the 470 Ω resistor at 12 V, the resistor power dissipation, and the ideal battery life. State whether the 0.25 W rating is adequate.
Read the worked answer
At 12 V: I = (V_supply − V_LED)/R = (12 − 2.0)/470 = 21.3 mA. Resistor power: P = I²R = (0.0213)² × 470 ≈ 0.213 W.
A 0.25 W resistor is running at 0.213/0.25 ≈ 85% of rating. That is too close: resistor tolerance, a higher battery voltage and poor ventilation can push it over. A common design rule is to use no more than about 50–60% of the rating, so a 0.5 W resistor is the better choice.
Battery life: 1.2 Ah/0.0213 A ≈ 56 h ideal. Alkaline batteries lose usable capacity at higher currents and at end of life, so allow perhaps 70–80%: about 39–45 h. The six-hour exhibition is easily covered, even with the LED left on. As the battery voltage falls, the current also falls, so the estimate is slightly pessimistic.
2. The student connects a second identical LED directly in parallel. Calculate the total current at 11.2 V and explain why this does not double brightness. Refer to the LED I-V characteristic and current sharing.
Read the worked answer
With the second LED in parallel, the 470 Ω resistor still sets the total current. Using the measured 11.2 V and assuming V_LED ≈ 2.0 V: I_total = (11.2 − 2.0)/470 ≈ 19.6 mA. That total is shared between the two LEDs, about 9.8 mA each if they are perfectly matched. Each LED is therefore dimmer than the single LED was at 21 mA; the display does not become twice as bright.
The sharing is not guaranteed. An LED's I–V curve is steep and exponential, and V_f falls as the junction gets warmer. If one LED has a slightly lower V_f, it takes more current, heats up, and its V_f falls further, so it takes even more. With a common series resistor the total current is limited, so catastrophic runaway is unlikely, but the brightness can be very uneven and one LED may age faster. The correct fix is one resistor per LED, or the LEDs in series with a recalculated resistor.
3. Recommend a modified design for two LEDs at about 20 mA each and calculate the resistor values, power ratings and battery life. Compare a parallel and a series arrangement.
Read the worked answer
Parallel, separate resistors: each branch has its own 470 Ω resistor. Each branch current is about 21 mA, so total ≈ 42 mA. Each resistor dissipates 0.213 W, so use 0.5 W types. Ideal battery life = 1.2/0.042 ≈ 28.6 h; usable perhaps 20–23 h.
Series connection: two LEDs in series need about 4.0 V total. R = (12 − 4.0)/0.020 = 400 Ω; use the preferred value 390 Ω. Current = (12 − 4.0)/390 ≈ 20.5 mA, and each resistor (only one is needed) dissipates I²R ≈ 0.164 W, so a 0.5 W resistor is comfortable. Total battery current is only about 20.5 mA, giving 1.2/0.0205 ≈ 58 h ideal, roughly double the parallel case. At 11.2 V the current is (11.2 − 4.0)/390 ≈ 18.5 mA, still visible.
Recommendation: use the series arrangement for battery life and equal brightness, provided the supply has enough headroom. Use separate resistors if the LEDs must be independently visible or if one failing open must not extinguish the other.