LED Strip Voltage Drop and Indicator Resistor
A school workshop builds a display using a 12 V, 2 A LED strip. The strip is mounted 20 m from the power supply and is wired with a reel of flexible cable whose datasheet gives 0.017 Ω per metre per conductor. At the supply end a fixed 12.0 V SMPS reads 12.0 V on a multimeter; at the strip end the voltage reads 10.9 V and the strip is visibly dim. The teacher also has a box of individual green LEDs marked V_F = 2.1 V at 20 mA and a selection of carbon-film resistors rated 0.25 W. A student proposes adding a 10 Ω resistor in series with the whole strip to 'stabilise' it, and another proposes setting the SMPS to 13.5 V to compensate. The SMPS has no remote-sense terminal and its output is adjustable only over 11.5–13.5 V. The strip's own current falls to about 1.6 A when its terminal voltage is 10.9 V.
1. Using the cable resistance, show whether the measured 1.1 V drop is explained, and calculate the drop at the full 2.0 A. What would a 10 Ω series resistor do at 1.6 A?
Read the worked answer
The current travels out and back, so the loop resistance is R_loop = 2 × 20 × 0.017 = 0.68 Ω. At 1.6 A the drop is 0.68 × 1.6 = 1.09 V, essentially the measured 1.1 V; the cable is a credible cause. At 2.0 A it would be 0.68 × 2.0 = 1.36 V.
A 10 Ω series resistor cannot stabilise the strip. At 1.6 A it would need 10 × 1.6 = 16 V, more than the 12 V supply, so the strip voltage and current would collapse. Even a 1 Ω resistor would drop 1.6 V. The resistor would waste power and make dimming worse.
2. The strip is later shortened to 2 m and draws 0.8 A. Calculate the terminal voltage with the SMPS at 13.5 V and decide whether 13.5 V is a safe general fix.
Read the worked answer
At 0.8 A the cable drop is 0.68 × 0.8 = 0.54 V. With the SMPS at 13.5 V the strip sees 13.5 − 0.54 = 12.96 V, about 8% above its 12 V rating. At no load it would see the full 13.5 V. At 2 A it would see 13.5 − 1.36 = 12.14 V.
So the same setting gives different strip voltages depending on current. It overvolts the strip when the load is light and only just reaches nominal at full load. The correct fix is thicker cable or moving the supply closer, not raising the source. If 13.5 V is used temporarily, it must be checked at the lightest load and against the strip rating.
3. Design the series resistor for one green LED from 12.0 V at 20 mA. Choose a standard value and power rating, and state the actual current.
Read the worked answer
R = (12.0 − 2.1) / 0.020 = 495 Ω. The nearest E24 value is 510 Ω. Actual current = (12.0 − 2.1) / 510 = 19.4 mA. Resistor power = I²R = (0.0194)² × 510 = 0.192 W.
A 0.25 W resistor would run at about 77% of rating, too close in a warm enclosure. Use a 0.5 W resistor, or two 1 kΩ 0.25 W resistors in parallel, each dissipating about 0.096 W. The LED dissipation is about 2.1 × 0.0194 = 41 mW and is not a problem.