Quick answer: Low-voltage lighting needs a supply matched to the lights and cable sized for their current and voltage drop. Lower voltage alone does not mean lower power consumption: power equals voltage multiplied by current.
This guide covers landscape lights and LED strips, explains what the voltage label means, and compares cable losses using assumed loads. It is a planning guide for choosing a system, with a separate calculation for the cable between the supply and the lights.
What counts as low-voltage lighting?
The everyday lighting label and the formal electrical category are different. According to Wikipedia's extra-low-voltage article, IEC 61140 / DIN EN 61140:2016 defines extra-low voltage as no more than 50 V AC or 120 V ripple-free DC, within a broader low-voltage category extending to 1,000 V AC or 1,500 V DC. These definitions do not establish a universal lighting installation rule.
For practical lighting choices, distinguish these systems:
| System | Voltage reference | What the distinction means |
|---|---|---|
| US landscape lighting | Wikipedia's landscape-lighting article lists 12 V (US) for low-voltage fixtures and 120 V (US) for line-voltage fixtures; real systems often use multi-tap transformers, so confirm against the fixture's own rating | Establish which system the fixture belongs to before choosing its supply |
| LED strip lighting | According to Wikipedia's LED-strip article, strips most commonly use 12 or 24 V DC | Match both the voltage and the DC output requirement |
| Constant-voltage supply example | The Mean Well LPV-60 datasheet lists 12 V / 5 A and 24 V / 2.5 A models, each rated 60 W | Equal wattage does not make the output voltages interchangeable |
Choose the system around the intended lights. For cabinets or shelves, start with the strip's input requirement; for a garden, start with the landscape fixture's requirement. The location alone does not determine the electrical supply.
Does lower voltage save energy?
Voltage alone cannot tell you the power draw. Using the Mean Well ratings above, P = V × I gives 12 × 5 = 60 W and 24 × 2.5 = 60 W; both outputs have the same rated power (Mean Well LPV-60 datasheet). These are output ratings, not measurements of wall-socket consumption or light output.
According to Wikipedia's extra-low-voltage article, supplying the same power at lower voltage requires higher current and causes greater resistive cable losses. That makes cable length and resistance part of the system choice. It does not support a blanket claim that either line-voltage or low-voltage lighting is always more efficient.
For a fixed power requirement, calculate current with I = P ÷ V. Then evaluate the cable at that current rather than treating the voltage label as an energy-saving feature.
How much voltage will the cable lose?
Calculate the resistance of the full circuit, including the return conductor. According to Wikipedia's voltage-drop article, DC cable drop follows Vdrop = I × R, and drop occurs in both supply and return wires.
The voltage drop calculator takes the one-way distance and doubles it internally. Its model uses Rtotal = resistance per 1,000 ft ÷ 1,000 × one-way length × 2. Entering an already doubled distance would count the return path twice.
Worked example: the same assumed load at different voltages
Assumed inputs: a 60 W nominal lighting load, a 25 ft one-way cable run, and 16 AWG copper. Compare a matched 12 V system with a matched 24 V system. Use the calculator's planning figure of 4.016 Ω per 1,000 ft for this conductor.
First calculate cable resistance:
- Round-trip length = 25 × 2 = 50 ft.
- Resistance per foot = 4.016 ÷ 1,000 = 0.004016 Ω/ft.
- Total resistance = 0.004016 × 50 = 0.2008 Ω.
Then calculate current, drop, and cable power loss:
| Calculation | Assumed 12 V system | Assumed 24 V system |
|---|---|---|
| Current, I = P ÷ V | 60 ÷ 12 = 5 A | 60 ÷ 24 = 2.5 A |
| Cable drop, Vdrop = I × R | 5 × 0.2008 = 1.004 V | 2.5 × 0.2008 = 0.502 V |
| Percentage drop, Vdrop ÷ V × 100 | 1.004 ÷ 12 × 100 = 8.3667%, about 8.4% | 0.502 ÷ 24 × 100 = 2.0917%, about 2.1% |
| Voltage at load, V − Vdrop | 12 − 1.004 = 10.996 V | 24 − 0.502 = 23.498 V |
| Cable loss, P_loss = I² × R | 5² × 0.2008 = 25 × 0.2008 = 5.02 W | 2.5² × 0.2008 = 6.25 × 0.2008 = 1.255 W |
Doubling the supply voltage halves current and the drop measured in volts. Percentage drop and cable power loss each fall to one quarter: 2.0917 ÷ 8.3667 ≈ 0.25, and 1.255 ÷ 5.02 = 0.25.
The calculator's default 3% drop threshold is a planning figure, not a cited installation standard. This example exceeds that threshold at the lower voltage and falls below it at the higher voltage. It holds current at the nominal-load estimate; actual current depends on the lights' behavior as their input voltage changes. It also excludes resistance within the strip and at connections.
How do you reduce drop and choose the cable?
A thicker conductor reduces resistance and therefore voltage drop, according to Wikipedia's voltage-drop article. Shortening the cable also reduces the resistance in the calculator's length-based formula. Compare those options before replacing an otherwise suitable lighting system.
Use this planning checklist:
- Match the proposed supply voltage and output type to the lights.
- Calculate current from the load using I = P ÷ V.
- Enter the actual one-way cable distance and conductor material.
- Compare candidate conductors in the wire gauge calculator, which checks both its current-capacity table and the selected drop limit.
- Review the receiving voltage as well as the percentage drop. In the assumed example, the lower-voltage load receives 10.996 V, calculated as 12 − 1.004 V.
The calculation separates two questions: whether a conductor meets the current check and whether it keeps the voltage drop within the chosen target. Passing one does not establish the other.
What supply and strip design should you choose?
Match the output requirement first. According to Mean Well's LPV-60 datasheet, this supply family uses constant-voltage output, accepts 90–264 V AC input, and has hiccup-mode overload protection that recovers after the fault clears. Those are characteristics of this product family, not every device sold as an LED driver.
The strip's own design also matters. According to Wikipedia's LED-strip article, resistor-based constant-voltage strips are sensitive to voltage variation along long runs fed from one input, while constant-current strips use an IC in each LED group to regulate current and maintain brightness along the strip.
For a separate resistor-limited LED circuit, the LED resistor calculator applies R = (supply voltage − total LED forward voltage) ÷ current. That calculation addresses the resistor in the LED circuit; the cable-drop calculation addresses resistance between the supply and the load.
Use a qualified electrician for mains-voltage connections or alterations. A low-voltage output does not remove the mains input shown in the supply specifications.
FAQ
Is a higher-voltage system better for a long cable run?
For the same assumed power and cable, the example reduces current from 5 A to 2.5 A and cable loss from 5.02 W to 1.255 W, using I = P ÷ V and P_loss = I²R. Choose lights and a supply intended for the same voltage.
Why are the lights at the far end dimmer?
Cable drop is one explanation: according to Wikipedia's voltage-drop article, both the supply and return conductors lose voltage. The worked example calculates receiving voltage as supply voltage minus cable drop.
Do landscape lights need a transformer?
Wikipedia's landscape-lighting article lists 12 V and multi-tap transformers among the equipment used in such systems. Identify the fixture's supply requirement before choosing one.
Does low voltage mean lower electricity use?
No: voltage must be considered together with current. For the assumed systems above, P = V × I gives 12 × 5 = 24 × 2.5 = 60 W in both cases, before considering losses.
Is the calculator's drop limit a legal requirement?
The 3% default is the calculator's planning figure. It is a comparison target for the estimate, not evidence of compliance with local wiring rules.
Jack Shi
Founder & editor, LEDaskJack Shi builds and writes LEDask, an independent LED-lighting tools project operated by clooms. He designs the calculators, checks their formulas and reference values against published engineering data, and writes the guides across the site.
