200ft 12V DC Power Cable Run

[doyo 0]

Hello, all.

I need to install a camera that draws about 500mA, in about 200ft distance from the power supply. What would be the suitable amperage of the power supply? I am using 18/2.

[survtech 0]

You are pushing the envelope with 18/2 power cable. I suggest 16 gauge wire, The supply should be rated at least double of the camera requirements, or 1 amp in your case.

[doyo 0]

Thank you, survtech.

So use 16 gauge wire, and 1A power supply, correct?

Well, if 18 gauge was the only option, what would I use, 1.5A?

Can you tell me how to calculate that? I'd love to learn.

[survtech 0]

The current of the power supply is not the problem in your case, the wire gauge is. A good rule of thumb is to size your power supply at least 1-1/2 times the total draw of all cameras attached to it. Since you are only talking one camera to one supply, 1 amp is a good number.

 

For wire gauge versus distance calculations, there are plenty of them on the internet. Here is one:

 

http://www.video-insight.com/Support/Tools/Wire-Length-Calculator.aspx

[cachecreekcctv 0]

This is one reason I normally use 16/2 Belden for my cameras. You just cannot "push" more amperage through a wire. You have voltage drop, even with this lower voltage. You just might be pushing the limits with size 18 wire.

[normicgander 0]

It's not that you can't use 18AWG wire. You must adjust Vs to make up for the voltage drop (VRloop).

 

Vload = Vsupply - VRloop

 

The ampacity of 18AWG is about 2.3A for power transmission (operating temp and wire/jacket types are factors).

 

The resistance (Rc) of a single conductor of 18AWG stranded wire is about 6.9 ohms per 1000ft.

 

We have 2 conductors for a complete circuit, so:

 

The loop resistance for 200ft of 18awg is:

 

Rloop = Rc x 2 x d

------------

1000

 

Rloop = 2.76 ohms

 

or by ratios:

 

Rc= 6.9/1000 = x/200

 

Rc = 1.38 ohms x 2 conductors = 2.76 ohms

 

If It = 500mA (measure it at 12v to confirm),

 

VRloop = It x Rloop = .5A x 2.76V = 1.38V

 

The question is what is the operating voltage of the load device and are there other devices on the same supply to consider?

 

What are the operating voltages of all loads devices and how do they compare?

 

As example if you used a 12V power supply then:

 

Vload = Vs - Vrloop = 12V - 1.38V = 10.62V

 

You could use a variable power supply to compensate for voltage drop as example:

 

Vs = Vload + Vrloop = 12V + 1.38V = 13.38V

 

It's a sort of a game of numbers vs. the cable and power supplies you have available......

[CCTV_Suppliers 0]

Good observations and the formula to justify voltage drop and what it potentially need to deliver 12VDC and necessary current to fire up the DC camera......

 

What comes in mind one of such power supplies that will do this trick is made by Altronix and here is the product code: ALTV615DC44ULM - This model can give you variable voltage up to 15VDC with available current level of 4A... Even though this is for 4 camera power supply, 8 and 16 positions are also available. Very easy to use and setup...