Powering Cameras over Long Distances

In the age of Power over Ethernet, specifying external camera power is becoming a lost art. Increasingly what was once a common skill - estimating voltage drop - is an uncommonly needed exercise. However, cameras cannot always use PoE, and powering devices over long distances (e.g., hundreds of meters) requires proper calculation and sizing.

Sizing Power in 5 Steps

Ensuring your camera gets the adequate voltage and amperage when it is hung potentially hung hundreds of feet away from the supply requires more than guessing. If not sized properly, the camera may not work at all, or worse, unpredictably malfunction. Following the five steps below will ensure the right power is designed, every time:

1. Confirm Input Voltage of Camera
2. Determine Camera Amperage
3. Find Wire Resistance
4. Calculate Voltage Drop
5. Specify Correct Wire/ Power Supply

An Example Problem

Each of the above steps will be used for a theoretical example: Assume we need to power this IP PTZ camera hung 200 meters from the power supply. (To keep things simple, assume a unpowered housing.) In this example, we are using a run of coax cable and low cost EoC adapters (that do not support extended PoE ranges) for network connectivity. As a result, we need to find the right power supply to use at the head end.  We start with step one below:

1. Confirm Input Voltage of Camera: When not PoE powered, surveillance cameras are typically 12 or 24 Volt devices. However, the type of voltage varies between either direct current (DC) and alternating current (AC) specifications. Because it is safer (lower amperage) and more useful at long ranges, 24 volts AC (VAC) is a common input voltage. This value determines which type of power supply we must use, as not all power supplies (even those marketed for surveillance applications) supply all types of output voltages.

The best way to confirm this value is to check the camera's spec sheet or data page. In our example, we see our camera requires 24VAC:

Our power supply must output ~24 VAC. At this point, we note this requirement, because it impacts future steps.

2. Determine Camera Amperage: Next, we need to establish how much current our camera pulls during operation. This can be determined from the listed wattage on the spec sheet:

In our example, the camera uses 25 Watts. Because the unit is a PTZ with motorized components, this is a rather high value compared to typical 'fixed' model power consumption. While not relevant in our example, if the housing/enclosure requires power, be sure to add these values together. In any case, we must covert 'wattage' into 'amperage' using the following formula:

I = P_(W) / (PF × V )

Where I = Amps, P_(W) = Watts, PF = Power Factor (For electronic devices, use a value of 1), and V = camera voltage. Therefore:

1.04 Amps = 25 /(1 x 24)

Our example camera requires just over 1 Amp for operation. Again, note this for future use.

3. Find Wire Resistance: For the 'Wire Resistance Ohms' value, this is a standardized value taken from a list depending on conductor type. (copper vs. aluminum, stranded vs. solid conductor) These factors are commonly provided by cable manufacturers. Here's a table of resistance (per one thousand feet) in the most common copper gauges used in Low Voltage work, in US units of measure:

However, many online resources include variations adjusted for metric units of measure, or different base conductor types. For our example, we will plan to use 18 gauge copper cable, which gives us a resistance of 0.00651 Ohms (per foot).

4. Calculate Voltage Drop: So for our example, we need to run power 200 meters, or about 660 feet. At this point, we have all the needed values to calculate total voltage drop, or the amount of voltage that is dissipated during the length of run:

Voltage Drop = (Wire Resistance Ohms) / ((Total Distance)/(1000)) x (Camera Amps)

In our example, 660 feet of 18 gauge wire with a camera that draws 1.04 Amps plugs into the formula as follows:

4.46 Volts Dropped = 6.51 x (660/1000) x 1.04 Amps

During the course of the run, we will lose ~ 5 Volts due to wire resistance.

Most cameras are rated for normal operation +/- 10% of the nominal 'listed' voltage. This means a 24 VAC camera will operate normally with a 24 VAC supply if the voltage drop is around 2 volts. However, a 5 volt drop falls well outside the 'ten percent' tolerance, so we are forced to consider different options:

5. Specify Correct Wire/Power Supply: Fortunately, we have two ways to solve this problem -

Increase Voltage at Supply Source: We can select a power supply rated for a higher voltage than what we need at the camera. For example, this Altronix unit outputs 28 VAC at the source. Losing 5 Volts in the course of the run results in ~23 VAC available at the camera, which falls within the 'ten percent' tolerance:

Amperage does not diminish or change over the course of the run, and if our camera requires 1.04 amps, we can see this power supply provides us 2 amps. This particular unit will perform the task needed in our example, with both voltage drop and amperage considered. It costs about $110 USD online.

Decrease Internal Resistance of Wire: By increasing the diameter of our power cabling, we decrease the resistance value of the conduit. In our example, we used 18 gauge wire with a resistance value of 6.51 Ohm/ thousand feet. However, lets change that to run 14 gauge cable with a resistance value of 2.58 Ohm/ thousand feet. This changes our 'voltage drop' accordingly:

Voltage Drop = (Wire Resistance Ohms) / ((Total Distance)/(1000)) x (Camera Amps)

In our example, 660 feet of 14 gauge wire with a camera that draws 1.04 Amps plugs into the formula as follows:

1.77 Volts Dropped = 2.58 x (660/1000) x 1.04 Amps

During the course of the run, we will lose ~ 2 Volts due to wire resistance. In this case, our voltage loss falls within the 'ten percent' tolerance without beefing up a standard 24VAC/2A power supply.

Bigger Wire or More Juice?

The exercise above begs the question "Which option should I chose, a bigger power supply or larger wire?" In our example, our options break down like this:

• 18 AWG cable [link no longer available]: Costs ~$45 for 500', or about $0.09 per foot.
• 14 AWG cable [link no longer available]: Costs ~$60 for 500', or about $0.12 per foot.
• 24VAC/4A Power Supply: Costs ~$100 online
• 28VAC/2A Power Supply: Cost ~$110 online

For Option A, using a 28 VAC supply and 18 AWG wire, the cost is: ~($110 + $60 =) $170.

For Option B, using a 24 VAC supply and 14 AWG wire, the cost is: ~($100 + $80 =) $180.

In our example, using the larger supply and smaller wire is marginally less expensive, but the answer varies from situation to situation based on other factors. Unforseen costs of using larger wire include greater unit length weight, consuming more conduit volume, more labor to handle/install, and costing more to ship and haul to job sites.  However, larger power supplies may be less efficient, generate more heat, and may have a larger installation footprint.

The 'best answer' to this question comes must consider these other factors.  In many cases, both options - bigger wire AND more juice provides the most trouble-free and dependable result for remote and hard-to-reach cameras.

Conclusion

When it comes to specifying power, understanding and applying the 'five steps' above will guarantee proper selection of power components, every time.