Constant-current drivers pin the current, but the cable drop is real. CC wire size is set by ampacity and voltage-window headroom, shown with a 40 V, 1.05 A COB example.
A constant-current driver pins the current, so many people assume "the current is fixed, any thin wire will do". On site the far end dims, flickers or drops out and the first guess is a bad driver - but the cable drop has eaten the driver's voltage window. Constant current does NOT mean zero drop; this guide does the math.
The one-sentence rule
The drop formula is identical to a constant-voltage system: ΔV = I x 2Lρ / A (both wires) - the current is fixed but the drop is just as real
A CC driver's "voltage window" (say 9-48 V) is the range it can auto-raise its output: load voltage plus drop must stay within the window or it holds the current
Once load voltage + drop > window ceiling, the driver can no longer raise voltage, so it sacrifices current - the far end dims, flickers, drops out
So CC wire size is set by two things: ampacity (no overheating) and voltage-window headroom (keep the far end inside the window); low current does not mean thin wire
The key myth: "constant current" protects the current, not the voltage - the drop still has to be calculated
Two constraints, whichever binds first
Constraint
Decides
How
Numbers in the example
Ampacity
Wire must not overheat
Look up the cable ampacity table, current <= rated
0.75 sq mm flex ~6 A, 1.5 sq mm ~10 A
Window headroom
Far end stays in window
L_max = (Vmax - Vf) / (I x 2ρ / A)
See below
For LEDs around 1 A, ampacity is usually NOT the binding limit (0.75 sq mm already carries 6 A). The real short board is almost always the voltage-window headroom - that is what to watch when sizing CC wire.
Worked example (40 V, 1.05 A COB, copper ρ = 0.0172 ohm mm2 per m)
A constant-current driver with a 9-48 V window (Vmax = 48 V) drives a COB string at Vf = 40 V, I = 1.05 A. Headroom = 48 - 40 = 8 V, so the drop may not exceed 8 V.
Drop per metre ΔV/m = I x 2ρ / A
0.75 sq mm: 1.05 x 2 x 0.0172 / 0.75 = 0.0482 V/m -> crossover L = 8 / 0.0482 ≈ 166 m
1.5 sq mm: 1.05 x 2 x 0.0172 / 1.5 = 0.0241 V/m -> crossover L = 8 / 0.0241 ≈ 332 m
Same string: moving from 0.75 to 1.5 sq mm doubles the reach from about 166 m to about 332 m. At the crossover the driver output is already at the 48 V ceiling; beyond it the COB can no longer hold 1.05 A and starts to dim.
The CC driver raises its output to Vf plus drop, but caps at 48 V; past 8 V of drop it cannot hold the current
Five field mistakes
"Constant current, any thin wire": small current tempts 0.5 sq mm bell wire; the far end drops out of the window and the driver gets blamed
Ampacity only, ignore the window: 0.75 sq mm carries 1 A easily, yet past 166 m the driver is already at the ceiling and the light dims
Load voltage near the window top: a 45 V string on a 48 V-max driver leaves only 3 V headroom, collapses within metres
Several strings on one driver: total current adds up, drop scales with total current, crossover drops sharply
Ignore terminal and joint drop: oxidised or poorly crimped joints add contact resistance that eats the headroom too, so measure the drop at joints on long runs
Acceptance checklist
Read the driver window: note Vmax (e.g. 48 V) and the load Vf (e.g. 40 V)
Compute headroom: headroom = Vmax - Vf, this is the drop budget
Compute drop from wire and distance: ΔV = I x 2Lρ / A, confirm at or below headroom
Check ampacity: I must be at or below the wire rating (0.75 sq mm flex from about 6 A)
Derive wire size from crossover: L_max = headroom / (I x 2ρ / A), thicken the wire if over
The 48 V ceiling, copper resistivity 0.0172 ohm mm2 per m and the 0.75/1.5 sq mm ratings are common values; the real window and ratings come from the driver datasheet and local wiring rules.
Send us the LED string voltage, operating current and run length. We will use the driver voltage window to tell you the wire size and the maximum reach, and suggest a matching constant-current model. Contact →