12V DC Cable Length for a Security Camera: The Drop Budget

The Sum, in One Line

Voltage drop on a two-wire DC feed is V = 2 × L × I × R, where L is the one-way length in metres, I is the current in amps and R is the resistance of one conductor in ohms per metre. The factor of two is the part people leave out. Current flows out on one conductor and back on the other, so a 20 m run is 40 m of copper and the drop is doubled.

Current comes from the load, not from the system. A camera quoted at 6 W draws 0.5 A at 12 V, a camera at 12 W draws 1 A, and a camera with infrared on at night draws more than its daylight figure. Use the dark-hours number, because that is the half of the day when the run is doing the most work and the reserve is the only source.

Resistance comes from the conductor. Copper at 20 °C is close to 0.0175 ohm per metre per square millimetre of cross-section, which gives the figures below for the gauges sold as camera cable. Use the one-way length, not the total length of the spool.

The 0.6 V Budget, and Why 12 V Is Unforgiving

The usual planning convention for a low-voltage camera feed is a drop of no more than 5 percent. At 12 V nominal that is 0.6 V. It is a convention rather than a rule because the camera’s own input specification is what actually governs, and those specifications differ between models. A camera that accepts 9 to 15 V has more room than one that expects a tight 12 V, and only its datasheet settles which you have.

The volts lost do not change when the source voltage does. Drop is current multiplied by resistance, so 1 A through 1 ohm costs 1 V whether the socket sits at 12.0 V or at 13.6 V with the battery full. What changes is the percentage: the same 1 V is 8.3 percent of 12 V and 7.4 percent of 13.6 V. The calculation is therefore done on the nominal figure, where the percentage is worst.

Twelve volts is unforgiving for a structural reason. The same power at a higher voltage needs less current, and less current means less drop in the same cable. A 12 W load is 1 A at 12 V and 0.5 A at 24 V, so the same run loses half the volts. Our systems publish a DC 12 V nominal output with a maximum of 16 V, so the run has to be planned inside that envelope rather than around it. The full set of figures the socket provides is on the published 12 V output.

Three Runs, Worked Through

A 6 W camera 25 m away on 18 AWG. Current is 0.5 A and resistance is 0.021 ohm per metre, so the drop is 2 × 25 × 0.5 × 0.021 = 0.525 V, or 4.4 percent of 12 V. The camera sees about 11.5 V at the socket’s nominal figure. That run is inside the budget with a little room, and it is the shape most single-camera sites fall into.

The same cable, a 12 W camera, the same 25 m. Current doubles to 1 A and the drop doubles to 1.05 V, or 8.8 percent. The camera now sees about 10.9 V nominal. The table says 18 AWG carries 1 A about 14 m before reaching the budget, so this run needs 16 AWG, which brings the drop to 0.65 V over the same distance.

A 60 W load 10 m away on 16 AWG. Current is 5 A, the ceiling of the output, and the drop is 2 × 10 × 5 × 0.013 = 1.3 V, more than twice the budget. At the top of the current rating a 12 V feed is only good for a few metres. The 5 A figure is a socket rating, and a run of any length at 5 A belongs on a shorter path or on a second system.

Notice what the three cases have in common. Length is the input people plan around, but current moves the answer fastest, because every doubling of current halves the length the cable can carry. A site that adds a second camera to an existing run shortens the usable length of that run at the same time.

The Loss Is Energy, Not Only Volts

A long run does not just deliver less voltage, it consumes energy that came out of the reserve. The loss is I²R: at 1 A through the 1.05 ohm of a 25 m run in 18 AWG, the cable itself burns 1.05 W, which is 25 Wh a day at continuous draw. Against the smallest reserve of about 326 Wh that is 7.7 percent of the battery going into the cable, every day, before the camera is counted.

That figure belongs in the daily account rather than beside it. A camera site planned at 12 W and 288 Wh a day becomes 300 Wh a day once the run is included, and the boundary in days of zero charging shortens in proportion. If the run cannot be shortened, the reserve can be sized with the loss included, which is what the sizing order does at its second step.

The camera’s own draw is worth measuring rather than reading, because the label figure and the running figure often differ. The camera’s actual draw covers how to get the number that belongs in this arithmetic.

Where the Ratings Are Measured

The DC 12 V nominal, 60 W continuous and 5 A figures describe the DC5521 socket on the system. They are the start of the run, not the end of it, and the volts lost in the cable are lost after the socket. The 5.5 mm outside by 2.1 mm inside connector is a small contact, and a connector that runs warm under load is a connector with resistance in it rather than a normal condition.

Every system in the range uses the same output and the same connector, so the run arithmetic does not change with the model. What changes with the model is the reserve behind the socket, which is what a long run costs you in days rather than in volts. The four models and their reserves sit side by side on the product range.

The cable from the socket to the camera is the installer’s to choose and terminate. Our part ends at the DC5521 socket on its short tail.

When the Run Is Too Long

Three options, in the order worth trying. Shorten the run by moving the unit closer to the camera, which costs nothing but a decision about where the pole goes and is usually possible at the planning stage. Accept the loss and add it to the daily account, choosing a reserve that covers the extra watt-hours and a cable that keeps the drop inside the camera’s input range. Split the site into two systems, one per group of loads, which removes the long run altogether and gives each group its own reserve.

What does not work is treating the 60 W rating as available at the far end of any cable. A 60 W load is 5 A, and 5 A over a long run loses more volts than a camera will accept. Where a site genuinely needs one heavy load far from the panel, the answer is a second unit at the load rather than a heavier cable, which is the same conclusion reached for two-ended link sites in powering a bridge link.

If a site is already built and the camera behaves badly at the end of a long run, measure before changing anything. A meter at the camera terminals under load tells you whether the problem is volts arriving or something else, and the fault-finding order separates the two before parts are ordered.

Measure It Under Load

A voltage reading on an idle cable is close to useless. With no current there is no drop, so a meter at the far end of a 50 m run reads almost the same as the socket and tells you nothing about what happens when the camera starts. Take the reading with the camera connected and running, ideally with infrared on, and compare it with the reading at the socket taken at the same moment. The difference is the drop your installation actually has.

Two readings are worth recording at commissioning: the socket voltage with the load connected and the voltage at the camera. Write both on the inside of the enclosure door with the date. When the same site is visited a year later, those two numbers say whether the cable or the reserve has changed, and they cost nothing to collect.

Quick Answers

Output, recharge and model questions are collected on the answers page, and terms, minimum order and lead time for DC output orders are listed on the shipping page.