How wire sizing by voltage drop works
Wire has resistance, and current through resistance costs voltage. The drop over a run is:
Vdrop = ρ × (2 × L) × I / A
where ρ is the conductor's resistivity (copper ≈ 0.0172 Ω·mm²/m, aluminum ≈ 0.0265), L is the one-way length (current travels out and back, hence the 2), I is the current, and A the cross-sectional area. This calculator inverts that: given your maximum acceptable drop, it finds the smallest standard conductor whose area keeps you under it — plus a table of neighboring sizes so you can see what stepping up or down costs.
Worked example
A 12 V fridge drawing 20 A at the end of a 5 m run, allowing 3% drop (0.36 V): required area = 0.0172 × 10 × 20 / 0.36 ≈ 9.6 mm², so 8 AWG (≈ 8.4 mm²) falls short and 7 AWG (≈ 10.6 mm²) is the smallest that passes — it drops 0.33 V (2.7%) and wastes about 6.5 W in the cable. Odd gauges are rare in stores, so in practice you'd buy 6 AWG; the table above shows both.
Choosing a drop budget
- 3% — the common target for low-voltage circuits feeding real loads.
- 1–2% — chargers, sense lines, and anything where the endpoint voltage matters (solar charge controllers especially).
- 5–10% — sometimes tolerable for lighting or short intermittent loads; know what your device accepts.
Low-voltage systems are unforgiving: 0.5 V is nothing at 120 V but is 4% of a 12 V battery — this is why automotive and solar wiring runs so thick.
Frequently asked questions
- Is this the same as ampacity?
- No. Ampacity is the current a wire can carry without overheating, set by insulation temperature ratings and installation conditions, and governed by electrical codes (NEC, IEC). A long run often needs a wire far larger than ampacity requires — that's the voltage-drop sizing done here. Always satisfy both: code tables for safety, this calculator for performance.
- Does stranded vs solid matter?
- For resistance, effectively no — equal total copper area gives equal drop. Stranded is just easier to route and survives vibration better.
- What about aluminum wire?
- Fine when sized for it (about 1.5× the copper area for the same drop) and terminated with connectors rated for aluminum — oxidation at terminations, not the metal itself, is the classic failure.
- Where do the AWG areas come from?
- The exact AWG geometric progression: diameter = 0.127 mm × 92^((36−g)/39). Every area in the table is computed from that, not looked up loosely.