Voltage drop is a term that we hear about all the time in the surveillance industry. While many people talk about it and use the charts and calculators provided by different sources, I think a smaller amount of people fully understand the physics behind it. Today I want to give a quick review of the ideas behind voltage drop and then some practical tips for installing security cameras or any other equipment.

### Equation:

Voltage drop= {2 x (length of run in ft.) x (resistance factor) x (current load in Amps)}/ 1000

The length of the run is pretty simple; it is how long the cable is between the camera and the power supply. The resistance factor is determined by the wire used and there is a chart in the NEC Chapter 9, Table 8 (rule of thumb, thinner wire will have higher resistance).

The current will be defined by the camera and will be found on the specification sheet.

Voltage Drop Example (500 mA Security Camera)

Here is a quick example: 200 ft. run, 18 Gauge wire, 500 mA cameras

VD=2* (200) * 7.7 ohm * .5A/1000

VD=1.54 V

### Voltage Range :

Cameras need a certain amount of voltage. There is normally a range of ±10% of the rated voltage, so for this example we will say that a 12V camera can work in the range of 10.8v up to 13.2v (±10% of 12v is 1.2v). If a CCTV camera gets too much voltage, the PCB would get damaged, and if too little, it won’t turn on . So if the power supply is exactly 12V at the source, then it will drop 1.54V in 200ft and supply the camera with 10.46V, which is in the range of operation.

### Voltage Drop :

One thing to remember is that the number for the voltage drop is not dependent on the supplied voltage. So if a 12V camera loses 2 volts, it wouldn’t power on because that is a 17% loss, but if a 24V camera loses 2 volts, it will work because that is only an 8% loss. So 24V cameras can run farther.

So the final test in for your setup is determining the Voltage percentage drop:

VD %=( VD/Source Voltage)*100

From our example above:

VD %=( 1.54/12)*100

VD%=7.7% (which is in the range)

### Just remember:

Longer Run=more Voltage Drop

Thinner wire (higher gauge) =more Voltage Drop

Larger Amps (more powerful camera) =more Voltage Drop

Higher source voltage=Less Voltage Percentage Drop

### For Installers :

There are times when installers have been called in because a camera is not working and they claim that the wire is supplying enough volts. The problem with that is that they unplug the camera and connect a Voltmeter to the two wires. The Voltmeter has a very low amp draw, and will show less voltage drop because of that. Once the camera is reintroduced in the circuit, the voltage drop will change, making the measurement ineffective. So we need to run the numbers on paper to determine the drop.

A primary concern when installing lengths of wire is voltage drop. The amount of voltage lost between the originating power supply and the device being powered can be significant. Improper selection of wire gauge can lead to an unacceptable voltage drop to the Camera. The following chart is designed to help calculate voltage drop per 100 feet of paired wire as a function of wire gauge and load current.

By matching load current (in AMPs) across the top of the chart with wire gauge (AWG) down the left side of the chart, one can determine voltage drop per 100 feet of paired wire run.

**NOTE:** A paired wire run represents the feed and return line to the load. Therefore, a 500 foot wire pair is equivalent to 1000 feet of total wire.

Given a load current of 1 AMP, and using 18 AWG wire, how much voltage drop can we expect at the load end for a 350 foot run of paired wire?

Using the chart, we match the row for 18 AWG and the column for 1 AMP and determine that voltage drop per 100 feet is 1.27 Volts. By dividing the paired wire length by 100, we get the factor by which we need to multiply voltage drop per 100 feet to determine total voltage drop. Therefore, 350 feet divided by 100 equals 3.5. Multiply 3.5 by 1.27 volts drop per 100 feet to get your total voltage drop. Thus the total voltage drop is 3.5 times 1.27, or 4.445 voltage drop for 350 feet.

=350/100 *1.27 = 4.45 Volts

**Example two:**

Given a camera load of 2 AMPs, that is 400 feet from the power source, which wire gauge should be selected to keep voltage drop at the camera to less than 3 volts?

To use the chart, we need to determine what the maximum voltage drop per 100 feet is. We calculate that 100 feet is 1/4 of 400 feet, thus the voltage drop allowed for 100 feet is 1/4 times 3 volts (which is the equivalent of 0.75 volts per 100 feet):voltage drop per 100 feet = 3/4 = .75 volts per 100 feet.

So, knowing that we cannot allow anything greater than a voltage drop of .75 volts per 100 feet, we can now look at the chart and select the wire gauges that will give us lower voltage drops per 100 feet at a 2 AMP load current. In this case, wire gauges of 10 (.40 V), 11 (.50 V), and 12 AWG (.64) will all suffice, with 13 AWG (.80) being a possibility.

Thus, in order to keep voltage drop at the camera to less than 3 volts given a camera load of 2 AMPs and a 400 foot paired wire run, we need to use a wire gauge in the range of 10-13 AWG.

### What is Led Driver?

An LED driver is an electrical device that regulates the power to an LED or string(s) of LEDs. What makes a driver different from conventional power supplies is that an LED driver responds to the ever-changing needs of the LED, or circuit of LEDs, by supplying a constant amount of power to the LED, as its electrical properties change with temperature.

Think of an LED driver as ‘Cruise Control’ (like in a car) for the LED, and the temperature changes of the LED are the hills and valleys it is ‘driving’ over. The power level (or ‘Speed’) of the LED is maintained constant by the driver as the electrical properties change (amount of ‘gas’ or power needed) throughout the temperature increases and decreases (or ‘hills and valleys’) seen by the LED(s). Without the proper driver, the LED may become too hot (driving too fast) and become unstable (out of control), causing poor performance (engine problems) or complete failure (crash!)

The chain is as strong as the weakest link!!

Using High quality cameras and still getting average results. It’s time to look at the power supply. Most of us tend to buy the best quality cameras (measured in TVL) but still don’t get the desired results. The reason for the following could be many:

- The first and foremost reason for average quality images would be your Power supplies. Cameras work on 12V DC and consume power starting from 100 mili amperes to 1000 Mili amperes. SMPS can be compared to the heart of the person providing oxygen to the body, similar to the human body the SMPS provides power to the CCTV system. Ensuring proper power to the camera is essential to make the CCTV system work in the most efficient manner.

- The second most important aspect to proper functioning of CCTV system would the quality of wires carrying these signals. The wires may be compared to the veins carrying this oxygen to the body. In case these veins are weak the oxygen won’t be carried in the most efficient manner. In other words the oxygen (signals) sent from the heart(SMPS) through veins (Wires) might not reach to various parts of the body (our Entire CCTV system)

**Thus – :**

**To get excellent quality picture selection of right SMPS is very important.**

**After the selection of Camera is complete, check the wattage of Camera that is mentioned in the specification sheet of Camera.**

**This figure means the Camera will draw equivalent amount of Power in Watts. The specification sheet also mentions operating voltage that is generally 12VDC or 24VAC.**

**For example if Wattage of Camera is 12Watt and if the operating voltage is 12VDC, then it means 12Watt/12V=1Amp. So under normal circumstances 12V/1A SMPS should be sufficient.**

**However this is in case Camera and SMPS is very near to each other.**

**In case cable distance is more, then there will be the voltage drop across the cable length. This will affect the picture quality.**

**The thumb rule therefore under normal circumstances is to use SMPS that will have double the current capacity. In our case 12V/2A.**

**Avoid inductive devises like tube light choke, pump, motor, florescent or sodium light near the cable between SMPS and Camera as also Camera and DVR.**

In case of AC failure, it is very important that CCTV should continue to function seamlessly as also the recording to be done by DVR. In such scenarios Battery based SMPS comes helps to achieve this. In this as soon as AC mains fail, CCTV and DVR’s start working on Battery attached to the power supply. The Battery is charged during AC mains restoration.

**The number of Battery hours is determined by – :**

Normally Battery is @12VDC. If the wattage of Camera and DVR together is say 40Watt and both are working on 12VDC. If the battery is 7AH, ten backup time calculation is as follows:

Total Wattage: 40 Watt.

Total Current: 40/12=3.3A Approximately.

By thumb rule: 100% more current so assume 6.6A current.

Battery 7AH so 7/6.6 = 1.06Hr.

So we will get more than 1Hour backup time. In case the battery is of 22AH: 22/6.6=3.3Hours. So we will get more than 3.3 hour back up.