Wire planning before buying copper
Voltage Drop Calculator
Estimate voltage drop by wire size, run length, current, and system voltage for low-voltage solar wiring.
On a 120 V circuit, losing three volts in the cable is a rounding error. On a 12 V circuit it is a quarter of the system, and it is the difference between an inverter that runs and one that shuts down on low-voltage cutoff while the battery is still half charged.
This is why voltage drop, not ampacity, usually decides wire size in low-voltage DC. Ampacity is a safety limit that asks whether the insulation survives. Voltage drop is a functionality limit that asks whether the load still works. A conductor can sit comfortably within its ampacity while starving everything downstream.
The calculation is ordinary Ohm's law. The part people get wrong is forgetting that the current has to come back.
The method
Nothing here is hidden. These are the exact expressions the calculator evaluates, in the order it evaluates them.
V_drop = 2 × L × I × RL is the ONE-WAY distance to the load. The factor of two accounts for the return conductor. Entering round-trip length here and letting the formula double it again is the most common error in this calculation.
drop% = V_drop / V_system × 100L_max = (drop_fraction × V_system) / (2 × I × R_per_foot)What each input means
| Input | What it actually is | Typical range |
|---|---|---|
| One-way run length | The distance from source to load along the actual cable route, including vertical rises and slack, not the straight-line distance. | 3–100 ft |
| Current in amps | The continuous current the circuit will carry. For an inverter feed, compute it from watts divided by efficiency divided by system voltage. | 5–200 A |
| System voltage | Nominal battery voltage — 12, 24, or 48 V. Drop percentage is relative to this. | 12, 24, 48 V |
| Wire size | AWG, which maps to a resistance per 1,000 ft from NEC Chapter 9 Table 8. Copper-clad aluminium of the same gauge has substantially higher resistance. | 14 AWG to 4/0 |
Worked example
A 12 V system feeding an 18 A load 22 feet away, checked against 12 AWG and then larger sizes.
| Step | Calculation |
|---|---|
| 12 AWG resistance | 1.93 Ω/1000 ft = 0.00193 Ω/ft |
| Drop with 12 AWG | 2 × 22 ft × 18 A × 0.00193 = 1.53 V |
| As a percentage | 1.53 V / 12 V × 100 = 12.7% |
| Retry with 8 AWG | 2 × 22 × 18 × 0.000764 = 0.605 V → 5.0% |
| Retry with 6 AWG | 2 × 22 × 18 × 0.000491 = 0.389 V → 3.2% |
| Retry with 4 AWG | 2 × 22 × 18 × 0.000308 = 0.244 V → 2.0% |
Result: 6 AWG to meet 3%, 4 AWG for comfortable margin.
Ampacity alone would have accepted 12 AWG, which is rated 25 A at 75 °C and would have run cool and legal while delivering 12.7% drop. Voltage drop forced three to four sizes larger. That gap is the entire reason this calculator exists.
Assumptions baked into the result
Every calculator makes assumptions. Most do not tell you what they are. These are ours, and if one of them does not describe your system, the answer will be wrong in a direction you can now predict.
- The conductor is copper. Aluminium has roughly 61% of copper's conductivity, and copper-clad aluminium behaves like aluminium.
- Resistance values are the NEC Chapter 9 Table 8 figures referenced at 75 °C. A hotter conductor has higher resistance and more drop than calculated.
- Termination and connection resistance is not included. Every crimp, lug, busbar, fuse holder, and switch adds a little, and a poor crimp can add a lot.
- The current is steady. Motor startup surges cause momentary drops several times larger than the steady-state figure.
What this tool will not tell you
- The result is for a single continuous conductor run. A circuit with several joints, a busbar, and a fuse holder will measure worse than this predicts.
- It does not check ampacity. A conductor must satisfy both, and ampacity is the safety-critical one.
- It does not model inrush. Size for the surge if the load is a motor, pump, or compressor.
- It assumes DC. AC circuits have reactance and power-factor considerations this does not cover.
Common questions
Should I use one-way or round-trip length?
Enter the one-way distance. The formula's factor of two already accounts for the return conductor. Entering round-trip length doubles the answer incorrectly.
What drop percentage should I target?
3% or better for anything upstream of the battery, for the battery-to-inverter run, and for any load with a low-voltage cutoff or a motor. Up to 10% is tolerable for simple resistive loads like LED lighting, where the only consequence is slightly reduced output. Note that the widely quoted NEC 3% figure appears in an informational note and is advisory, while ABYC's 3% and 10% criteria are codified.
My measured drop is worse than calculated. Why?
Almost always connection resistance. Check crimps, lug torque, and fuse holder contacts, and look for corrosion under the insulation at terminations. Also confirm the cable is genuinely pure copper and not copper-clad aluminium sold in an AWG size.
