LED Strip Power Supply and Voltage Drop Calculator
LED Strip Power Supply and Voltage Drop Calculator
Total power, supply current and the power supply size for a run of LED strip — and the part that catches people out: the voltage the strip’s own copper loses along its length, how much light that costs at the far end, and exactly what injecting at both ends or moving to 24 V buys you.
LED strip power and voltage drop
5 m of 9.6 W/m 12 V strip fed from one end, 0.08 Ω/m per conductor, 8.7 V per LED group, 20% headroom
The strip as a distributed circuit
along the strip: dV/dx = −r·I dI/dx = −k(V − Vf) ⟹ u″ = r k u
V(x) = Vf + (V − Vf) · cosh[α(ℓ − x)] ÷ cosh(αℓ), α = √(r k)
k = (W/m ÷ V) ÷ (V − Vf) r = 2 × Ω/m per conductor
short runs: ΔV ≈ I · r · ℓ / 2
- r
- loop resistance per metre — twice the figure for one conductor, because the current goes out along one and back along the other
- k
- how much extra current one metre of strip draws per extra volt across it. It is set by the ballast resistors inside the strip
- V f
- the forward voltage of one LED group. Everything above it is what the ballast resistor sees, and the current — and so the light — is proportional to that difference
- alpha
- 1/α is the distance over which the strip’s voltage decays. A run much shorter than 1/α is fine; one much longer is hopeless whatever you do
Worked example
5 m of 9.6 W/m 12 V strip fed from one end, 0.08 Ω/m per conductor, 8.7 V per LED group, 20% headroom
Total power is 9.6 × 5 = 48.0 W, so at 12 V the nameplate current is 4.000 A and with 20% headroom the supply wants to be at least 57.6 W
The copper: 0.08 Ω/m per conductor is 0.16 Ω/m there and back, which over 5 m is 0.80 Ω of loop resistance — about 0.2155 mm² of copper in each conductor
Each metre draws 0.800 A at 12 V, and it has 3.3 V of headroom above the LED group, so k = 0.24242 A per volt per metre and α = √(r k) = 0.19695 per metre
V at the far end = 8.7 + 3.3 ÷ cosh(0.9847) = 10.863 V, a drop of 1.137 V or 9.47%
But the light is down to 65.6%, because the LED group only ever had 3.3 V to play with and has lost 1.137 V of it. A 9% voltage drop costs 34% of the brightness
Feed both ends and the drop falls from 1.137 V to 0.363 V — a factor of 3.13, not two — and the light comes back to 89.0%
The same run on a 24 V strip of the same watts per metre ends at 23.274 V: 0.726 V lost, half the volts and 3.03% against 9.47% — a quarter of the percentage
How far a run gets before the far end is visibly down
| Strip | Volts | Copper (Ω/m per conductor) | Fed from | Length for 90% at the far end | Light there |
|---|---|---|---|---|---|
| 60 LED/m 5050, 9.6 W/m | 12 V | 0.08 | one end | 5.20 m | 64% |
| the same strip | 12 V | 0.08 | both ends | 10.40 m | 64% |
| the same strip, thin copper | 12 V | 0.20 | one end | 3.29 m | 64% |
| 60 LED/m 5050, 9.6 W/m | 24 V | 0.08 | one end | 10.40 m | 64% |
| high-output 14.4 W/m | 12 V | 0.08 | one end | 4.24 m | 64% |
| high-output 14.4 W/m | 24 V | 0.08 | one end | 8.49 m | 64% |
| high-output 14.4 W/m | 24 V | 0.08 | both ends | 16.98 m | 64% |
Why the far end of a long 12 V run is dim
Sizing the power supply is the easy half: watts per metre times metres, plus headroom. Twenty per cent is the usual minimum, and more is better — a supply run at its rating runs hot and fails early. The hard half is that the strip has to carry its own current along two strips of copper foil a few millimetres wide, and those have resistance.
The copper is the input nobody can look up for you. There is no standard for it: two 9.6 W/m strips from different factories can differ by a factor of three in conductor resistance, and it is almost never on the datasheet. Two ways to get it. Measure it — put a meter on the + pad at one end of a length of strip and the + pad at the other; the LEDs will not conduct at a multimeter’s test voltage, so what you read is one conductor’s resistance for that length. Or compute it from the copper: 0.0172 Ω·mm²/m divided by the trace width times the copper thickness. The default here of 0.08 Ω/m per conductor corresponds to about 0.22 mm² — roughly a 3 mm trace in 2 oz copper, which is a decent 10 mm-wide strip. A cheap strip with 1 oz copper and narrow traces can easily be 0.2 to 0.4 Ω/m, and behaves accordingly.
The model, and why it is a cosh. A strip is not a lump of load at the far end, and it is not a constant current either. Every cuttable segment is a few LEDs in series with a ballast resistor, so the current a segment draws is proportional to however much voltage is left above the LED group’s forward voltage. Write that down for a continuous strip and you get two coupled equations — dV/dx = −rI along the copper, dI/dx = −k(V − V_f) into the LEDs — whose solution is a hyperbolic cosine. The far-end voltage is V_f + (V − V_f)/cosh(αℓ) with α = √(rk), and that is solved exactly on this page. The familiar rule of thumb ΔV = I·r·ℓ/2 is the short-strip limit of the same expression, and it is always pessimistic: at the default settings here it over-states the drop by about 40%, because it charges the far end of the strip for current the far end is no longer drawing. Both numbers are shown so you can see the difference. The exact solution was checked against a node-by-node simulation of the strip as a 600-rung ladder, and the two agree to better than 0.02%.
Volts are not light. This is the part that surprises people. The LED group has a fixed forward voltage; only the difference between the supply and that forward voltage appears across the ballast resistor, and the current — and to a good approximation the light — is proportional to that difference. On a 12 V strip with a three-LED group at 8.7 V, the resistor only ever sees 3.3 V. Lose 1.1 V of supply and you have lost a third of the resistor’s voltage and a third of the light, for a voltage drop of only 9%. That is why a run can pass a 90%-of-nominal voltage check and still look obviously uneven. Judge it on the brightness figure, not the voltage one.
The two fixes, and what each is actually worth. Inject at both ends. Each half of the strip then carries half the total current over half the distance, and since the drop goes with the product of the two, it falls by a factor of four — not two, which is the figure usually quoted. The dimmest point moves to the middle. Use 24 V. For the same watts per metre the current halves, so the volts lost halve; but the drop that matters is the drop as a fraction of the supply, and that falls by four. This is why 24 V strip exists and why any run over about three metres should be on it. The two fixes combine: 24 V fed from both ends is sixteen times better than 12 V fed from one. A third option, splitting the run into sections each fed by its own thick wire from the supply, is really the same trick applied more times — and for sizing that wire, the voltage drop calculator handles the run from the supply to the strip, which this page does not.
For LEDs wired up individually rather than bought as a strip, the LED array calculator sizes the series resistors and the strings; the LED series resistor calculator does a single LED; and the lumens to watts converter turns the strip’s rating into the light you will actually get.
Frequently asked questions
What size power supply do I need for my LED strip?
Watts per metre times the length, plus at least 20% headroom. Five metres of 9.6 W/m at 12 V is 48 W, so a 60 W supply. Do not size it exactly: supplies run at their rating get hot, and heat is what kills them.
Why is the end of my LED strip dimmer?
Because the strip’s own copper has resistance, and all the current for the whole run has to pass through the first few centimetres of it. The voltage falls along the strip, and because each LED group sits behind a ballast resistor that only sees the difference between the supply and the LED forward voltage, a small voltage loss is a large light loss. A 9% voltage drop can cost a third of the brightness.
How long can a 12 V LED strip be?
On a decent 9.6 W/m strip fed from one end, about 5 m before the far end is at 90% of nominal — and it is already down to about two thirds brightness there. On thin copper it can be half that. Fed from both ends, or on 24 V, the same strip goes roughly twice as far.
Does injecting power at both ends halve the voltage drop?
No, it quarters it. Each half of the strip carries half the current over half the distance, and the drop is proportional to the product of those two. The commonly quoted “halves it” is wrong — the page shows the real factor for your numbers.
Is 24 V LED strip better than 12 V?
For anything longer than a couple of metres, yes. At the same watts per metre it draws half the current, so it loses half the volts in the copper — and because what matters is the loss as a fraction of the supply, the effective drop is a quarter. It also needs thinner supply wiring.
Where do I find the strip’s copper resistance?
You measure it. Put a multimeter across the + pad at one end and the + pad at the other; the LEDs do not conduct at the meter’s test voltage, so you read the conductor directly. Divide by the length. Failing that, 0.05 to 0.1 Ω/m per conductor is a good strip and 0.2 to 0.4 Ω/m is a cheap one.
Related calculators
References
- IEC 60228:2004, Conductors of insulated cables, and IEC 60028 — annealed copper at 1/58 Ω·mm²/m at 20 °C, the 0.017241 Ω·mm²/m used here to turn the strip’s measured resistance per metre into a copper cross-section, and the same constant the AWG and voltage-drop pages on this site use.
- Cree LED. XLamp XM-L LED product family data sheet, CLD-DS33 rev. 11E, relative luminous flux against forward current: 0.50 at 350 mA and 1.43 at 1,000 mA against 1.00 at the nominal 700 mA — linear to within the width of the curve up to about 1,000 mA, then flattening to 3.25 at 3,000 mA. That is why this page treats the light from an LED group as proportional to its current; above the rated current the assumption becomes optimistic.
- Glowfy. LED strip voltage drop and power injection explained. A worked example of the industry rule of thumb, quoting 3–5 m as the practical single-feed limit at 12 V and 8–10 m at 24 V, and describing both-end injection as halving the effective run length. This page agrees with the run lengths and disagrees with the arithmetic: halving the fed length quarters the drop rather than halving it, which is shown above.
- Horowitz P, Hill W. The Art of Electronics, 3rd ed. Cambridge University Press, 2015. §1.2 and §2.2 — the distributed-resistance treatment and the LED-plus-ballast behaviour that the cosh solution here rests on.
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