How to calculate NAC wire size
Start with the device minimum, the lowest available NAC terminal voltage and the complete alarm load. Set headroom deliberately; it adds to the required device voltage. This reverses a voltage-drop check: calculate each candidate gauge, then choose the smallest conductor meeting the required voltage. A lower AWG number means a larger conductor. Let L be one-way length in feet, r be resistance of one conductor in Ω/1,000 ft, I be total fixed alarm current in A, Vₛ be available NAC terminal voltage, Vₘᵢₙ be the listed device minimum and H be your headroom in V.
- Required voltage: Vᵣ = Vₘᵢₙ + H.
- Loop resistance: R = 2 × L × r ÷ 1,000.
- Voltage drop: ΔV = I × R. Last-device voltage: Vₗₐₛₜ = Vₛ − ΔV.
- Pass: Vₗₐₛₜ ≥ Vᵣ in both the source and optional battery-condition checks.
- Maximum one-way length: Lₘₐₓ = 1,000 × (Vₛ − Vᵣ) ÷ (2 × I × r), for I > 0 and Vₛ ≥ Vᵣ.
The cable-loss equations apply System Sensor's voltage-drop example, p. 4 to outgoing and return conductors. Convert meters to feet by dividing by 0.3048. Combined maximum length is the lower limit from the enabled supply checks. A source below Vᵣ means no feasible length; zero current produces no finite voltage-drop limit.
Optional point-to-point mode
For N equal-current devices equally spaced at L/N, 2L/N, …, L, each segment carries its downstream load. Our sum of these segment drops gives ΔVpoint = ΔVend × (N + 1)/(2N). Divide the end-load maximum length by the same factor. With ten devices the drop factor is 0.55; with one it is 1.
This is an algebraic extension of the I × R method: add each segment's loss using only its downstream current. Mixed currents, unequal spacing, branches or return-fed operation need an actual circuit-specific calculation.
Worked example: which gauge passes?
Take 1.5 A, a 300 ft one-way run, 24 V at the source, an enabled 20.4 V battery-condition check, a 16 V device minimum and 0 V headroom. With all load at the end, 18 AWG solid gives 20.4 − (2 × 300 × 7.765145 × 1.5 ÷ 1,000) ≈ 13.411 V on battery, so it fails. 16 AWG solid gives approximately 16.005 V and passes; its battery-limited maximum length is approximately 300.328 ft.
16 AWG stranded gives approximately 15.917 V and fails, so the smallest passing stranded option is 14 AWG, at approximately 17.581 V. Adding 0.5 V headroom raises the required voltage to 16.5 V, so 14 AWG becomes the smallest passing size for both constructions. All comparisons use full precision.
Resistance assumptions
Resistance comes from ideal annealed, uncoated copper geometry and constants in National Bureau of Standards (hosted by NIST), Copper Wire Tables, Handbook 100 (1966), §2.2 and appendices 4–5. We calculate at 75 °C and add 2% for the concentric-lay stranded estimate. Confirm actual resistance with the cable manufacturer; construction and tolerances vary. The temperature assumption does not establish cable suitability.
d(mm) = 0.127 × 92^((36 − AWG)/39); A = πd²/4; r₂₀ = (1/58) × 304.8/A; r₇₅ = r₂₀ × [1 + 0.00393 × (75 − 20)].
For comparison with your code book, use NFPA 70-2020, Chapter 9, Table 8 and its notes. Match copper coating, stranding, temperature and units. The numbers here remain NBS-derived estimates, not NEC entries. The temperature multiplier is 1.21615 from 20 °C to 75 °C; do not correct a 75 °C resistance a second time.
The 16 AWG solid example passes by only about 0.005 V. Treat that as a boundary result: a cable-specific resistance or additional loss can change the selection. Rerun the check with the specified cable data and supply condition before choosing a conductor.
Wire size is more than voltage drop
A passing voltage estimate alone does not select an approved cable. Check the panel's NAC current rating, permitted conductor sizes and terminal range, cable listing (FPL/FPLR/FPLP), mechanical and installation requirements, and the manufacturer's instructions. Consult NFPA 70-2020 Article 760 and the adopted NFPA 72 edition for your project; no standard text or tables are reproduced here.
An SLC has different equipment-specific limits. Use the NAC vs SLC guide to understand the distinction.
Enter listed maximum alarm currents for the chosen settings and waveform. This fixed-current two-wire radial estimate excludes inrush, voltage-dependent current, connection and module losses, supply regulation and additional branches. Check maximum device voltage separately. Enter voltage available at NAC terminals after panel losses; raw battery voltage can overstate the available supply. The editable 20.4 V starting point comes from System Sensor's historical worked example; confirm the available output for your equipment.
Check against your code edition, procedure and the responsible inspector/engineer. This calculation aid is an engineering estimate, not design approval. Confirm project specifications and the AHJ's requirements.
For an individual gauge, use the fire alarm voltage-drop calculator. Read the voltage-drop formula guide for the method.
The ×1.02 stranded estimate can be slightly non-conservative: it can understate resistance, understate voltage drop and overstate maximum length. The 14 AWG comparison is about 0.3% below the matching NFPA 70 Chapter 9, Table 8 value. Use the specified cable resistance for design; a near-boundary pass in this tool is insufficient.
Fire alarm wire size calculator FAQ
How do I find the smallest wire gauge for an NAC circuit?
Work out the drop for each gauge, then take the smallest one that keeps the last device at or above the required voltage. The drop is ΔV = I × R, where R = 2 × L × r ÷ 1,000, L is the one-way length in feet and r is the resistance of one conductor in Ω per 1,000 ft. Take 1.5 A, a 300 ft one-way run, 24 V at the source, a 20.4 V battery-condition check and a 16 V device minimum. 18 AWG solid leaves about 13.411 V on battery and fails. 16 AWG solid leaves about 16.005 V and passes.
Why can stranded wire need a bigger gauge than solid?
The calculator adds 2% to the solid resistance as a stranded estimate. In the same example 16 AWG stranded leaves about 15.917 V on battery and fails, so the smallest passing stranded size is 14 AWG, at about 17.581 V. The 2% is an estimate and can slightly understate resistance, so use the resistance of the cable you specify.
What does headroom do?
Headroom is extra voltage you want above the device minimum. It adds to the required voltage, so a 16 V minimum plus 0.5 V of headroom needs 16.5 V at the last device. In the example 16 AWG solid then fails, and 14 AWG is the smallest passing size for both solid and stranded.
If a gauge passes, is it the right cable for the job?
Not by itself. This is a voltage estimate only. 16 AWG solid passes the example by about 0.005 V, and its battery-limited maximum length is about 300.328 ft against a 300 ft run, so a cable-specific resistance can change the pick. The panel's NAC current rating, terminal range, cable listing and installation requirements still apply. Check the manufacturer's listed data and the AHJ.
Audible output is a separate check from NAC wire size
The controls above check NAC cable loss. For a NICET Fire Alarm Systems Level II sound problem, first decide whether you are changing one source's level, combining sources at one listener, or changing the distance to one source. System Sensor's Voice Evacuation Systems Applications Guide, §3, printed pp. 3–4 explains power ratios, level addition and distance loss.
For a power ratio, ΔdB = 10 × log10(P2/P1). For the energy-sum estimate of uncorrelated sound sources measured at the same listener, with matching weighting and reference, convert each level back to a relative energy, add, then take the logarithm:
Ltotal = 10 × log10(10^(L1/10) + 10^(L2/10) + …)
Two sources each contributing 70 dB at that listener give 73.0103 dB. The increase is about 3 dB. This energy sum does not predict phase interference between coherent tones; it is not a guaranteed gain from adding another alarm appliance.
For a point-source free-field estimate at unchanged output, intensity varies as 1/r². The level change is L2 − L1 = −20 × log10(r2/r1), with both distances in the same units and measured from the source. Doubling distance gives a 6.0206 dB loss; intensity falls to one-quarter. This is the origin of the approximate 6 dB distance rule.
Apply the distance model only where free-field point-source assumptions hold. Walls, reflections and speaker directionality change a room's result. Do not use this model as a UT echo-distance correction or as proof of audibility in a building.
Look up the audibility margin separately
For NFPA 72-2022, use §18.4.4.1 for public-mode audibility, §18.4.5.1 for private-mode audibility and §18.4.6 for sleeping areas. Read the full applicable section in your own code copy before selecting the target. A calculated SPL is only an estimate to compare with that target; it does not supply the code margin or replace field measurements.