24vdc Voltage Drop Calculator

When designing 24 VDC systems (LED strips, access control, solar, telecom, battery-fed circuits, RVs, etc.) voltage drop matters. Too much drop means dim lights, undervoltage motors/controllers, misbehavior of electronics and wasted energy. A 24 VDC Voltage Drop Calculator helps you estimate how much voltage will be lost in the conductors so you can choose the right wire gauge or the right run length.

24VDC Voltage Drop Calculator

(For round trip, this will be doubled automatically)
(Common: 1.5, 2.5, 4, 6 mm²)

The basic physics / formula

For a DC circuit the voltage drop across a conductor is Ohm’s law: Vdrop=I×RV_{drop} = I \times RVdrop​=I×R

where:

  • III = current (A)
  • RRR = total conductor resistance (Ω)

Conductor resistance for a uniform wire: R=ρLAR = \rho \frac{L}{A}R=ρAL​

where:

  • ρ\rhoρ = resistivity of the conductor (Ω·m). For copper at ~20°C, ρCu≈1.724×10−8 Ω⋅m\rho_{Cu} \approx 1.724\times10^{-8}\ \Omega\cdot mρCu​≈1.724×10−8 Ω⋅m. For aluminum ρAl≈2.82×10−8 Ω⋅m\rho_{Al} \approx 2.82\times10^{-8}\ \Omega\cdot mρAl​≈2.82×10−8 Ω⋅m.
  • LLL = length of the conductor (m) used in the resistance formula. Important: for a complete circuit use the round-trip length (out + return). If you have a single positive conductor and the return is a separate conductor of equal length, then Lround=2×LonewayL_{round}=2\times L_{oneway}Lround​=2×Loneway​.
  • AAA = cross-sectional area of the conductor (m²). For mm² conductors, AAA in m² = (mm²) × 1e-6.

So combining: Vdrop=I×ρLroundAV_{drop} = I \times \rho \frac{L_{round}}{A}Vdrop​=I×ρALround​​

If you want the percent voltage drop relative to the nominal 24 V: %  Drop=Vdrop24×100%\%\;Drop = \frac{V_{drop}}{24} \times 100\%%Drop=24Vdrop​​×100%

Rearranging: find required conductor area for a target drop

If you want to limit voltage drop to a maximum VmaxDropV_{maxDrop}VmaxDrop​ (e.g., 3% of 24 V → 0.72 V), rearrange: Arequired=ρLround×IVmaxDropA_{required} = \rho \frac{L_{round} \times I}{V_{maxDrop}}Arequired​=ρVmaxDrop​Lround​×I​

Then convert ArequiredA_{required}Arequired​ (m²) to mm² by multiplying by 10610^6106.

Practical numbers & common rules of thumb

  • Typical acceptable voltage drop for low-voltage DC feeds is often 2–3% for sensitive loads and up to 5% for less-critical runs (industry practice varies). For 24 V, 1% = 0.24 V, 3% = 0.72 V, 5% = 1.2 V.
  • Remember to use round-trip length. Many mistakes come from using the one-way length in the formula.
  • Temperature, bundling and altitude change resistance slightly — for rough sizing use 20°C resistivity; for exact engineering, apply temperature correction factors and conductor tables.

Worked examples (step-by-step)

All arithmetic shown clearly.

Example A — copper, common DIY case

System: 24 V DC, load current 10 A, one-way cable run 10 m, conductor copper, cross-section 2.5 mm².
Goal: compute voltage drop and % drop.

Step 1 — convert area to m²
A=2.5 mm2=2.5×10−6 m2A = 2.5\ \text{mm}^2 = 2.5\times10^{-6}\ \text{m}^2A=2.5 mm2=2.5×10−6 m2.

Step 2 — round-trip length
Lround=2×10 m=20 mL_{round} = 2\times10\ \text{m} = 20\ \text{m}Lround​=2×10 m=20 m.

Step 3 — copper resistivity
ρCu=1.724×10−8 Ω⋅m\rho_{Cu} = 1.724\times10^{-8}\ \Omega\cdot mρCu​=1.724×10−8 Ω⋅m.

Step 4 — resistance R=ρLroundA=1.724×10−8×202.5×10−6=0.13792 ΩR = \rho \frac{L_{round}}{A} = 1.724\times10^{-8}\times\frac{20}{2.5\times10^{-6}} = 0.13792\ \OmegaR=ρALround​​=1.724×10−8×2.5×10−620​=0.13792 Ω

Step 5 — voltage drop Vdrop=I×R=10×0.13792=1.3792 VV_{drop} = I\times R = 10\times0.13792 = 1.3792\ \text{V}Vdrop​=I×R=10×0.13792=1.3792 V

Step 6 — percent drop %Drop=1.379224×100%≈5.75%\%Drop = \frac{1.3792}{24}\times100\% \approx 5.75\%%Drop=241.3792​×100%≈5.75%

Interpretation: 2.5 mm² copper over 10 m one-way at 10 A causes ~1.38 V drop (≈5.8%), which may be too large for many 24 V electronics.

Example B — choose a conductor to meet ≤3% drop

Same scenario (10 A, 10 m one-way) but target ≤3% drop (0.72 V allowed).

Use rearranged formula: Areq=ρLround×IVmaxDropA_{req} = \rho \frac{L_{round} \times I}{V_{maxDrop}}Areq​=ρVmaxDrop​Lround​×I​

Plug numbers (copper): Areq=1.724×10−8×20×100.72=4.7889×10−6 m2A_{req} = 1.724\times10^{-8}\times\frac{20\times10}{0.72} = 4.7889\times10^{-6}\ \text{m}^2Areq​=1.724×10−8×0.7220×10​=4.7889×10−6 m2

Convert to mm²: Areq×106=4.7889 mm2 (≈ 4.8 mm2)A_{req} \times 10^6 = 4.7889\ \text{mm}^2 \ (\text{≈ 4.8 mm}^2)Areq​×106=4.7889 mm2 (≈ 4.8 mm2)

Interpretation: You need ≈4.8 mm² copper conductor (use the next common size up, e.g., 6 mm²) to keep voltage drop ≤3%.

Example C — compare copper vs aluminum for same allowed drop

Using the same A_req as above (~4.7889e-6 m²), compute what the voltage drop would be for aluminum (rho ≈ 2.82e-8 Ω·m): VdropAl=I×ρAlLroundA=10×2.82×10−8×204.7889×10−6≈1.18 VV_{drop}^{Al} = I \times \rho_{Al}\frac{L_{round}}{A} =10\times 2.82\times10^{-8}\times\frac{20}{4.7889\times10^{-6}}\approx 1.18\ \text{V}VdropAl​=I×ρAl​ALround​​=10×2.82×10−8×4.7889×10−620​≈1.18 V

% drop ≈ 1.18/24 ×100% ≈ 4.9%. So aluminum at that area gives nearly 5% drop — worse than copper, as expected. To match copper performance you must increase aluminum conductor area.

Using the calculator in practice (steps)

  1. Decide one-way length (m or ft). Multiply by 2 for round-trip unless return uses ground plane etc.
  2. Measure or estimate load current (A). Use worst-case continuous current.
  3. Choose conductor material (copper vs aluminum) and cross-section (mm²) or AWG. Convert AWG to mm² using reference table if needed.
  4. Compute R=ρ⋅LroundAR = \rho \cdot \dfrac{L_{round}}{A}R=ρ⋅ALround​​, then Vdrop=I⋅RV_{drop} = I\cdot RVdrop​=I⋅R, then % drop = Vdrop/24×100%V_{drop}/24\times100\%Vdrop​/24×100%.
  5. If % too high, increase conductor size or shorten run length or reduce current (e.g., add local power supplies).

Quick imperial note (ohms per 1000 ft)

Electricians often use resistance per 1000 ft: R=(Rper1000ft)×length_ft1000R = (R_{per1000ft})\times\frac{length\_ft}{1000}R=(Rper1000ft​)×1000length_ft​. Copper tables give R_per1000ft for each AWG. If you use feet, be careful to use round-trip feet.

Practical tips & cautions

  • Round-trip matters. Many mistakes come from forgetting the return conductor.
  • Use continuous current for sizing; intermittent surge currents (motors) may be higher but don’t use surge to size voltage drop.
  • Temperature: conductor resistance increases with temperature. Use correction factors for hot environments.
  • Connectors & terminals add resistance. Long thin pigtails and poor crimps can add significant drop.
  • Voltage tolerance of device: some 24 V devices tolerate ±10% (i.e., 21.6–26.4 V); others have tighter requirements — check device spec.
  • Use the next standard conductor size up to give headroom.

20 FAQs (short answers)

  1. Why is voltage drop important on 24 V systems?
    Low voltage systems are sensitive; small voltage drops are a large percentage of supply voltage.
  2. Should I always use round-trip length?
    Yes — resistance is for the complete conductive path.
  3. What percent drop is acceptable?
    2–3% is good for sensitive loads; up to 5% is sometimes acceptable.
  4. How do I convert mm² to AWG?
    Use an AWG ↔ mm² conversion table (common online).
  5. Does wire temperature matter?
    Yes — resistivity increases with temperature; use corrected resistivity for high ambient temp.
  6. How do I account for multiple parallel conductors?
    Divide current among parallel conductors and compute drop for each path, or compute equivalent area.
  7. Should I size for surge currents?
    Size for continuous current; check thermal and voltage tolerance for surges.
  8. Is aluminum ok for 24 V runs?
    Yes, but needs larger cross-section than copper for same drop.
  9. Do PCB traces follow same math?
    Yes — use resistivity and cross-sectional area of the trace (thickness × width).
  10. Does insulation type affect resistance?
    Not directly; insulation affects temperature rating and allowable ampacity.
  11. Can I reduce drop by increasing voltage?
    Yes; increasing system voltage reduces percent drop, but components must accept the higher voltage.
  12. What about fuses and connectors?
    They add resistance — include them in voltage drop budgets.
  13. How precise is the resistivity constant?
    Resistivity varies slightly with alloy and temperature; the value used is typical for pure copper @20°C.
  14. Are aftermarket calculators reliable?
    Most are fine if you input correct round-trip length, current and material.
  15. How do I size for long runs in solar/RV systems?
    Use ≤2% drop for critical loads; larger cable sizes are common.
  16. What if I have multiple loads along the run?
    Do a segmental calculation: treat each segment and cumulative current for each section.
  17. Will LED strips be affected by 5% drop?
    Yes — LEDs dim and color shift; often you feed strips in multiple points to reduce drop.
  18. Do stranded & solid wire resistivity differ?
    No significant DC resistivity difference for same metal and cross-section.
  19. Is the return always the negative conductor?
    Yes in DC circuits; if chassis is used as return, check its resistance too.
  20. Can I use the calculator for AC?
    The basic resistance drop calc is the same for low-frequency AC, but AC has additional effects (skin effect at high frequencies).

Final thoughts

Voltage drop on 24 VDC systems is easy to calculate with the right formula and careful attention to round-trip length, conductor material and cross-section. Use the formulas above (or a simple 24 VDC Voltage Drop Calculator) to test scenarios and pick a conductor size that keeps voltages in the acceptable range for your equipment. When in doubt, upsizing the cable a step or two gives reliability and reduces maintenance headaches.