# Powering Cameras over Long Distances

By Brian Rhodes, Published Sep 19, 2013, 12:00am EDT (Info+)

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

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12 gage at 1.62 oms per foot ?

" For our example, we will plan to use 18 gage copper cable,

which gives us a resistance of 6.51 Ohms (per foot). "

That's a good catch! You are talking about 18 gage, not 12. But you have sharp eyes- those are 1000 foot values. Will update shortly.

Thanks!

Good article and useful info, but if you have the option, you're really better off running a higher voltage (say 48 or 55) through lighter cable out to the camera housing, and installing a local transformer or switching inverter there (such as supplied by Laird or Tygon) to generate the required 12 or 24 volts at the camera. Properly designed, this approach gives you a closely-regulated "stiff" supply that won't change output voltage even when the load changes during PTZ or heater activation. Hanging a variable load at the end of a 600 foot cable, can result in noticably higher voltage swings which may be a problem if you're operating near the edge of the tolerance window (which can be quite a bit less than +-10% in my experience).

This would be my preference as well. I had to put a switch in a junction box in the middle of a two-square-block industrial yard once, mainly as a repeater for a network feed... it required 5VDC@1A, so I just fed 24VAC down a 16/2 line from the camera power supply at the remote end, put a 3A adjustable board regulator in with the switch, and dialed in the appropriate output. Been working perfectly that way for about six years now...

Thanks for the feedback, Tom. We wanted to drill the fundamental of voltage drop calculation, and your recommendation is definitely a 'next level' insight based on best practices/experience.

You mention a PTZ, but any camera equipped with a blower/heater/defroster/wiper/illuminator is susceptible to a variable load, right?

Hi!

It seems that either you or I have missed something. The current will have to travel 660 feet to the camera and another 660 feet back to the power supply to complete the circuit. Thus the total voltage drop will be double of the figures that you have calculated.

Hello, thanks for the comment.

I do not understand your comment/questions, but keep in mind voltage drop at the camera is what we are concerned with.

Power is transmitted in a circuit - a 'pair' of both positive (red/white) and negative (black) conductors. Voltage drop is cumulative at the endpoint, in this case our camera 660 feet away.

Brian, I suspect what he means is that the negative side is the return, therefore the calculation needs to take into account round trip. If the camera is at 660 ft, the calculation needs to involve 1320 ft.

Liek it and very easily expalined. Pelase use metric meaurement for cable size in addition to AWG. Thanks, Krish

Hello, Krishnaiah:

Thanks for the feedback. I will update the chart and mentions of AWG to include metric cable sizes. Also, here's a cross reference chart:

Hi!

"Undisclosed (#0318412) Integrator" has explained it better. To state it differently, the total voltage drop will consist of the voltage drop in the outgoing wire plus the voltage drop in the return wire, because both wires will typically have equal resistance. ( I avoid using the term negative wire since that would imply that DC is being used). The net available voltage at the camera will be the power supply voltage minus these two voltage drops. To my knowledge, the resistance as stated in the chart is for a single wire and not for a pair of wires (outgoing and return) since the wire manufacturer cannot predict whether or not the ground will be used for the return path for current travelling over the wire. Since the camera is 660 ft. away, the combined length of the outgoing wire and the return wire will be 1320 ft. and the voltage drop will be double of that shown by your calculations. Please correct me if I am wrong.

You might have a valid point here in theory. Can someone validate his thinking is correct or not? In basic electronics, we draw a basic diagram with single line joining each electronic components, meaning voltage source (power supply) pumps in current, enters the resistor and out the resistor and then flow back to the voltage source. So there is a path A = entering the resistor and path B = flowing out of resistor. Because in electroncis field, path A and B is so short the internal resistance of the path is so small, it is consider to be zero or negligible. But because we dealing with a long cable wire, the internal resistance will be substantially high enough to affect the voltage source (power supply). So, I think is valid to consider both the paths A + B, as B is the returning path back and A is the entering path to the LOAD (LOAD can be anything, be it a lamp, or a PTZ camera). I reakon, to be sure of the internal resitance is a measureable value, get a cable measured for the length of say, 662 feet, cut it and pull out the red wire and black wire out, on both ends. One end is shorted, and the other end leave it open, use a multimeter and measure the internal resistance in DC current. And see what you get.

Hello Jayant, et al:

I've spent some time looking at this question: There is no 'standard' method of including the pair's total length or not:

For example: from Cooper's Guide (see note: 'length in feet of one wire'):

From Ed Renzi (see note: 'length of conductor in one direction, multiplied by two'):

Essentially the right answer depends on which formula or resistance values are used to calculate the voltage drop.

Many wire gage resistance charts list values including both conductor paths (a 'circuit value'), so a simple 'one way' length gives a good result. Our table, a rather ancient example from the US Bureau of Standards is constructed the way, and 'doubling up' in not needed. However, many manufacturer's charts do not, because they place no assumptions on how/to what lengths you use their products. The numbers vary based on strands vs. solids. vs. jacketing material vs. dielectric gap in terminations (screw terminal or soldered?). It is quite a comprehensive subject, with a myriad of very specific factors.

The lesson: keen eyes are needed in reviewing the table to see IF they include single-way or both-way values.

To be a CCTV installer can be tough, basically you need to be a jack of all trades, you need to know about computers, IT, networking, basic electronics (of find out the wire resistance).........etc

For those who new in dealing voltage, resistance and current. If you have a output amperage of say, 1.5A or 2A or even 3A of a DC power supply. Having 3A power supply for running @ 12V will not burn/damage/wreck a CCTV camera which works @12V @DC @1A. Although higher amperage (4A, 5A) is not recommended, because you can get seriously electrocuted if you accidentally touch the wire with your bare hands. Therefore it is best to use Alternating Current (AC) rather than DC (Direct Current) in powering up devices in long distance application.

Hi Marcus!

While the first part of your statement "having 3A power supply for 12V will not burn/damage/wreck a CCTV camera which works on 12V DC@1A" is absolutely correct, the second part is not correct. 12V DC when touched with bare hands will never cause electrocution unless your hands suffered from severe cuts and raw nerves were exposed, and at this voltage neither AC nor DC will cause electrocution. I believe that you are safe from electrocution up to about 30V or so. I believe that there are standards which require terminals carrying voltages higher than this to be covered and not left exposed to accidental touch. When you load a power supply rated at 12V 3A with 4A or 5A load, if the fuse does not blow, you will cause overheating of the internal components of the power supply which could then breakdown. If such internal breakdown of the components causes the output side to short circuit with the incoming mains side (110/220V AC), usually very unlikely in a well designed circuit layout, then only would you have a threat of electrocution.

The real reason for using AC (typically 24V for cameras) over long distances is that it gives you the freedom to step up or to step down the voltage (using a simple transformer) if really needed, and rectify the voltage to the required 12 or lower DC voltage as required by the equipment. It is a far more complicated matter to raise a low incoming DC Voltage to a higher working voltage. That would require the use of an inverter, a step up transformer, rectifier and smoothing circuit, adding greatly to the complexity, and cost while also perhaps reducing the reliability of the system.

I agree the statement you made on 2nd paragraph about using AC to power up devices over the long distance. I am have a couple of the SAMSUNG analogue CCTV camera where it comes with AC power supplier instead of DC. :-)

However, on the 1st paragraph you stating the 12VDC @ 3A denying that it will electrocute, I agree, electrocuting is a bit over exaggerating on my side, but won't you agree by touching it on the connector will give you a stinging sensation? I never dare to touch it just to be on the safe side. :-)

You might feel a little something, MAYBE... I mean, I've felt a "prick" if I'm working on the car, wrenching on a part of the body, and another part of me touches the positive battery terminal... but 12/24V isn't going to cause any harm under normal circumstances. Can't recall ever actually getting a tickle from from a camera power supply, AC or DC.

Keep in mind that the surface resistance of your dry skin in on the order of megaohms - 24V across 1meg is .0024mA. If your skin is damp/sweaty, the current conducts through the moisture, OVER your skin.

Edit: also, a higher-amperage power supply will not "burn" you where a low-amped one wouldn't. Ohm's law still applies: current flow is a factor of voltage and resistance. If your skin has a 1 meg resistance, you'll still only get .0024mA from 24V whether the power supply is capable 100mA or 100A.

Do dare touching 12 VDC power. You won't feel a thing. Hey, I don't even feel 24 volts, only 48 volts and up.

But, if you work outside barefoot in a puddle, that's a totally different ball game. I recently read instructions of an outdoor 12 VDC motion detector and they do warn of danger of working with even such low power under wet environment. I've never tested that, and won't. Water and electricity don't mix.

Nothing will happen standing barefoot in a puddle, either, unless your power source has some sort of reference to an earth ground. Not saying to run out and try it just to see what happens, just saying... don't be paranoid about statements that make no logical sense.

Warnings like that are more likely added either for legal ass-covering, or by someone who doesn't understand how electricity works.

Matt, not sure how you define power source with reference to earth ground. I thought all power were reference to ground.

Brian,

Thanks for the clarification. Somehow, I have always used charts that give the resistance for a single wire.

As I said in the beginning one of us has missed something. One lives and learns.

Thanks again,

I'm not 100% sure this observation is always correct, but:

'Electronics discipline' charts (small values, low voltages, shorter distances, etc) tend to NOT include 'circuit-value' resistance. Circuit distances are quite short, and component loss is a much bigger source of drop.

However:

'Electrician discipline' charts (larger values, higher voltages, longer distances, etc) tend to included doubled values. Circuit distances can be vast, covering hundreds of feet.

Being that security equipment is indeed low voltage, but can cover vast distances, we end up in a confusing grey area.

Instead of working out the calculations using Ohm's Law (V = I x R), in paper which will be hard to those newcomers in the CCTV field, would it work? For example, when I get a 662feet of power cable, cut it so that, I exposed 2 ends of the cable, grab the Red wire and Black wire to be short-ciruited on one end and then use a multimeter to measure the DC resistance on the other end of the cable and assume the DC and AC resistance is roughly the same?

Marcus,

In theory you are correct. In practice, it would be better to do a theoretical calculation and make sure that you are using the right wire and power supply etc. than to cut the wire and then discover that you have the wrong gauge of wire or the wrong voltage power supply. This may prove to be very expensive both in terms of time to complete the job and in terms of costs involved in replacing the inadequate wire or power supply with the correct ones.

Yeah you right. Why waste money and time when I almost have none \$\$\$ left in my wallet. :-). Gosh! I wish I can remember to do this calculation on powering up remote device, when I am in the 50's and perhpas 60's. Pardon me for not reading the whole discussion on this interesting stuff. So....I am going to ask a dumb question here, for calculating the CORRECT voltage drop on the cabling, do I need to factor in the resistance as X2 or just X1

As Brian has already stated, it depends on the wire tables that you are using. From all the tables that I have seen, the value of 6.51 ohms per 1000 ft for 18 gauge wire is correct for a single run only, and you will need to x2 to get the total resistance for the complete run. Use that (6.51 ohms per 1000 ft for 18 awg wire) as a reference point to check for the particular gauge and length that you plan to use and you should not go wrong. Always remember that there may be minor variations in the value (6.51 ohms) from one manufacturer to another.

After reading Brian's explanation on his findings about the factor of x2 the resistance, I am still lost in the bush, am I dumb or what? :-) So, indeed it is true that you need to double up the resistance of the wire if such application arise. Then, there will be 3 voltage drops applicable, 1st voltage drop will occur at the 1st run/path of the wiring (current entering) to the load (PTZ, CCTV camera, case blower), the 2nd voltage drop will occur at the load itself, and the 3rd/last voltage drop will occur at the wiring (current existing) from the load back to the voltage source (power supply unit):-)

The answer: Yes. Fundamentally the resistance need to be doubled.

Some charts assume this, while others do not because they double the value in the subsequent formula. For the formula used in the post, the resistance values either need to be 'circuit values', or doubled for the return leg.

Instead of charts and slide rules, make it easy on yourself and use any of a number of online calculators... like this one (already plugged the numbers in for you :), or this one, or if you're dealing specifically with Cat5e (baluns over UTP, for example), I find this one handy.

Marcus,

Further to my earlier observation, it may be wiser to just assume that the resistance as seen from the charts is only for a single run, use double the value for the complete run and go ahead. At worst you would have left a greater safety margin and incurred substantially lower costs, than if you had found out your error later and then had to replace the wire and/or the power supply. Always better to err on the safe side.

Brian great article... was the sole basis for using coax vs ethernet due to distance limitation on ethernet? if so what are the possibilities of using an utp with ethernet extenders and and secondary power (non poe)...