LED Strip Voltage Drop: Why the End of the Run Dims
Voltage drop explains why a 12 V strip dims after 5 m. Calculated losses at 1 oz to 3 oz copper, maximum run lengths and the three fixes that work.
HFY Applications Engineering · Published September 15, 2026 · 3 min read
Voltage drop is the loss of voltage along a strip caused by copper resistance, which makes the far end dimmer than the feed end. A 24 V strip at 12 W/m loses about 1.6 V over 5 m on a 3 oz FPC and about 4.8 V on a 1 oz FPC. Doubling the voltage quarters the loss, which is why 24 V is specified for runs above 5 m.
Key points
- Loss scales with the square of the current, so 24 V loses a quarter of what 12 V loses at the same power.
- Copper weight matters more than strip width for long runs: 3 oz roughly halves the loss of 1 oz.
- A 5% flux drop at the far end is the practical acceptance limit; 10% is visible on a plain surface.
- The three fixes are higher voltage, heavier copper, or a second feed — in that order of cost effectiveness.
The calculation, without the algebra
Voltage drop is Ohm’s law applied to a very long, very thin resistor:
V_drop = I × R
R = ρ × L / AWhere ρ for copper is 0.0172 Ω·mm²/m, L is the run length in metres (times two, because the current goes out and back), and A is the conductor cross-section.
For a 3 oz FPC at 10 mm width, one conductor is about 0.105 mm thick × 8 mm wide of copper after clearance, giving roughly 0.84 mm². Over 5 m and back, that is 10 m of conductor: R ≈ 0.20 Ω.
Measured losses by copper weight
The table below is measured on 24 V strip at 12 W/m (0.5 A per metre, 2.5 A for a 5 m run):
| Copper weight | 5 m run | 10 m run | 20 m run | Practical feed arrangement |
|---|---|---|---|---|
| 1 oz | 4.8 V | 9.6 V | 19 V | Single feed only to 5 m |
| 2 oz | 2.4 V | 4.8 V | 9.6 V | Single feed to 10 m |
| 3 oz | 1.6 V | 3.2 V | 6.4 V | Feed both ends to 15 m |
| 4 oz | 1.2 V | 2.4 V | 4.8 V | Feed both ends to 20 m |
At 12 V the same runs lose four times as much, which is why 12 V strip tops out at 5 m and 24 V at 10 m on standard copper.
What the eye actually accepts
A 5% drop in flux is the limit most lighting designers work to, because below that the gradient is not visible on a matte surface. On a glossy or mirrored surface, 3% shows because the reflection doubles the gradient.
That converts into a voltage budget: 5% of 24 V is 1.2 V. From the table, a 3 oz FPC delivers that only up to about 3.5 m on a single feed at 12 W/m — which is why long runs are always fed from both ends rather than from one.
The three fixes, ranked by cost
- Feed from both ends. Zero material cost, halves the effective run length. This solves most installations and is the first thing to try.
- Move to 24 V or 48 V. Same copper, a quarter or a sixteenth of the loss. Requires a different driver, which is a small cost at design stage and an awkward one after installation.
- Specify heavier copper. 3 oz FPC costs 8–12% more than 2 oz and is the right answer for 20 m perimeter runs where a second feed would be visible on the ceiling.
What does not work: increasing the driver voltage to compensate, or using thicker supply cable between driver and strip. The drop happens inside the strip, downstream of the cable.
Where this bites in practice
The common failure is a 20 m architectural cove specified as “24 V strip, single feed from one end”. It lights up fine on the bench and looks acceptable in daylight. At night, on a white ceiling, the gradient from 100% to 78% is unmistakable, and the fix requires a new cable run that was not in the electrical contract.
For COB LED strip and SMD LED strip we quote the recommended feed arrangement with every order, and state the predicted drop at the far end. If you are sizing the electrical side, power supply sizing covers the driver end of the same calculation.
This article describes general manufacturing practice for machined and fabricated components. It is not a substitute for a drawing-specific engineering review — tolerances, material selection and process capability depend on the individual part.