Free tool
Voltage drop calculator.
Enter the supply, what the device draws, the wire and the run. See how much voltage reaches the far end, before you pull the cable.
How voltage drop works
Every wire has resistance, so some voltage is used up getting power to the far end. A longer run, a thinner wire or a bigger draw loses more.
Volts lost = amps × 2 × run length × wire resistance per foot
1.Set the power
The supply voltage and what the device draws, from its datasheet.
2.Describe the wire
Its gauge, the run one way, and how many conductors each side.
3.Read the answer
The voltage that arrives, and the wire or supply that fixes it if it is too low.
Wire resistance per 1,000 ft
| Wire | At 68 °F | At 167 °F (NEC) |
|---|---|---|
| 24 AWG (Cat5e) | 25.67 Ω | 31.2 Ω |
| 23 AWG (Cat6) | 20.36 Ω | 24.8 Ω |
| 22 AWG | 16.14 Ω | 19.6 Ω |
| 20 AWG | 10.15 Ω | 12.3 Ω |
| 18 AWG | 6.385 Ω | 7.77 Ω |
| 16 AWG | 4.016 Ω | 4.88 Ω |
| 14 AWG | 2.525 Ω | 3.07 Ω |
| 12 AWG | 1.588 Ω | 1.93 Ω |
| 10 AWG | 0.9989 Ω | 1.21 Ω |
Gauges thinner than 18 AWG are not in the code's table: their 167 °F figures are worked from the 68 °F ones, as the code does for the rest.
Worth knowing
Stay within 10%
Most 12 V and 24 V devices accept 10% low. Under 5% is comfortable.
Cameras draw the same watts
As the voltage sags they pull more current, which loses more. Too far and they never start.
Too much drop?
Thicker wire, a 24 V supply, conductors doubled up, or a power supply near the device.
Watch for CCA cable
Copper-clad aluminum has about 55% more resistance than copper.
On long runs, use 24 V
The same watts at 24 V is half the current, so the loss is a quarter as a share of the supply.
A 6 W camera on 18 AWG: 169 ft at 12 V, 676 ft at 24 V.
Where these figures come from
Wire resistance comes from the US standards body's copper tables and the electrical code itself.
Wire resistance
Solid copper at 68 °F, every gauge
18 AWG: 6.385 Ω per 1,000 ft
Copper at 167 °F, the code's Chapter 9 Table 8
18 AWG: 7.77 Ω solid, 7.95 Ω stranded
NFPA 70 (NEC)2026
No standard sets these, so they are NexVolt's working figures:
Common questions
How far can I run 18/2 for a 12 V camera?
For a typical 6 W camera on 18 AWG solid copper, about 169 ft keeps the drop within 10%; stranded 18/2 is a little shorter, about 165 ft. Power the same camera from 24 V and it reaches about 676 ft, because the same watts at twice the voltage need half the current, and 10% of 24 V is twice as many volts to lose. Check your camera's maximum watts, with IR on, on its datasheet.
Should I use 12 V or 24 V for cameras?
24 V when the runs are long. The same watts at 24 V draw half the current, so the wire loses a quarter as much as a share of the supply. A 12 W camera on 250 ft of 18 AWG will not power up at 12 V; at 24 V it gets 22.28 V and works. Check that the camera accepts 24 V (many accept 12 V DC or 24 V AC) before you change supplies.
How much voltage drop is acceptable?
Keep it within 10% of the supply: most 12 V and 24 V security devices accept that much below their rating. Under 5% is comfortable. The electrical code's 3% guideline is for branch circuits feeding equipment, not for the low-voltage side; still, the less you lose, the more margin you have for a cold night or a flat battery.
What wire size do I need for a card reader?
Readers draw little. A 3 W reader at 12 V on 22 AWG runs about 133 ft within 10%. Many readers are wired on 22 AWG composite cable for that reason. The lock is the heavy load: a 0.5 A maglock at 12 V reaches about 187 ft on 18 AWG but only about 74 ft on 22 AWG, so run 18 AWG or thicker to the lock.
Can I power a camera over Cat6 spare pairs?
For short runs. Cat6 is about 23 AWG, 20.36 ohms per 1,000 ft. A 6 W, 12 V camera on one pair (one wire out, one back) reaches about 53 ft; using two pairs (both wires of a pair on each side) doubles it to about 106 ft. For anything longer, PoE or a proper 18/2 power run is the better choice.
Why does my camera reboot at night?
Its IR lights switch on and it draws more. A 12 V camera that draws 6 W by day on 150 ft of 18 AWG gets 10.95 V; at 9 W with IR on it gets 10.33 V, 13.9% low, and it may drop out. Size the run for the night draw, use 24 V, or move the power supply closer.
Why is the answer worse for a camera than for a lock?
A camera or reader draws the same watts whatever voltage reaches it, so as the voltage sags it pulls more current, which sags it further. A lock or relay coil draws about the same current. The calculator works each one its own way: choose what the device is. On a run that is too long, a camera can fail to start at all, which the simple formula never shows.
Does voltage drop work the same on 24 V AC?
Near enough for low-voltage runs. At these lengths and currents it is the resistance of the wire that loses the voltage, and that is the same for AC and DC, so the calculator answers for 24 V AC too. Only on very long, heavy runs does AC add a little more loss.
