# Thread: Cable & Volt Drop

1. ## Cable & Volt Drop

Good day fellow sparkies!

I am currently working on a quote for a job on a private estate. A security company needs to install cameras around the perimeter (L-shaped) which is 1.2km (600m + 600m).

In total it is 24 cameras (50 meters apart). They require a plug point on each pole (inside an IP65 rated box). The power will be drawn from the security cubicle at the front gate, which has a DB.

Here are the camera's electrical specs:
Power Supply DC12V(±30%), PoE(802.3af)(Class 0)
Power Consumption DC12V:2.4W 7.1W(IR on)
PoE:3W 8.4W(IR on)

What cable and size should be used to supy the points (to compensate for volt drop)? Also, is it allowed to have all 24 points on one line? (Seeing as running multiple lines will increase the cost exponentially). The points will be mounted high up on poles, so they won't be accessible for anything else except supplying the cameras.

Please inform me on the correct regulations which may apply.

TIA

2. What voltage will the cable carry?
Will all cameras be on at the same time?
You have to work on a worse case scenario, that is how many cameras would be active at a time.

If all cameras on
Each camera point with PIR will draw 10.8W
50 points = 540W

If no cameras are on, and all PIRs are on but not active
7.1W * 50 = 355W
If you try run it at 12V, the first camera point must handle 45Amps as a worst case scenario.
If only the PIRs are on, then you need 29A at 12V.

I would use 230V, and the worst case is 3 amps at the first point.
Then use a SMPS at each camera point to power the camera.
If you require power during load shedding, then you can use an inverter, and the battery capacity will be calculated as the number of hours erquired.
In this case I would not use lead acid, but use lithium.

3. Originally Posted by Justloadit
What voltage will the cable carry?
Will all cameras be on at the same time?
You have to work on a worse case scenario, that is how many cameras would be active at a time.

If all cameras on
Each camera point with PIR will draw 10.8W
50 points = 540W

If no cameras are on, and all PIRs are on but not active
7.1W * 50 = 355W
If you try run it at 12V, the first camera point must handle 45Amps as a worst case scenario.
If only the PIRs are on, then you need 29A at 12V.

I would use 230V, and the worst case is 3 amps at the first point.
Then use a SMPS at each camera point to power the camera.
If you require power during load shedding, then you can use an inverter, and the battery capacity will be calculated as the number of hours erquired.
In this case I would not use lead acid, but use lithium.
Hi there!

The line will carry 230V. The cameras will operate at the same time. Also, they use a 12V power supply (similar to a latop) that plugs in to a normal 230V point. This power supy will be in the same IP rated box as our plug points.

So would 4mm cable suffice? Just to be on the safe side....

Sent from my SM-A515F using Tapatalk

4. I suggest you contact the cable manufacturers, and let them advise you..
4mm gives 4.6 ohms per KM.
However the amount of current/volt drop will change at each point further away from the source, because the current begins to drop as there are less units powered.

5. If you're using laptop type switch mode supplies at each camera point you will need to make allowance for their gawd awful power factor and efficiency. I'd do the cable calcs based on the camera, PIR and PoE consumption then make at least a 30% allowance on top. Even better is to actually get the supplier to lend you one camera and one power supply and actually set it up and load test it.

That said, if this is a dedicated circuit for the cameras and if the switch mode supplies have a wide operating voltage (some have a 100v-240v supply range) then in theory at least there's nothing to stop you exceeding the usually considered maximum volt drop along the entire circuit as long as the supply voltage at the far end of the circuit remains comfortably above the operating voltage of the power supplies when the circuit is under its max load. Yes, it will cost the customer money in cabling copper losses but if might save them considerable capital outlay on the cable size.

6. 24 cameras 50 meters apart = 1200m of cabling.

I would be considering a central distibution point or taping off closer distribution point ... if there is a gate ... generally there are kiosk on the property to distribute power.

or ... solar panels and backup batteries ...

or ... a ring circuit if the cables starts on one side and end s back at the same point ... that way if ther eis a problem you dont loose all the power.

There are so many factors to take into consideration ... they might add lights to the poles ... the catch with these estates ... they want the cheapest price ... then when the shyte hits the fan ... you end up with an investigation ... enegineers are called in and they try screw you for the upgrade ... engineer costs and everything else.

the best advice I can offer ... be very careful dealing with estates ... people who live in estates have short arms and deep pockets ... they want everything for the cheapest price ... until things go wrong.

7. Originally Posted by AndyD
That said, if this is a dedicated circuit for the cameras and if the switch mode supplies have a wide operating voltage (some have a 100v-240v supply range) then in theory at least there's nothing to stop you exceeding the usually considered maximum volt drop along the entire circuit as long as the supply voltage at the far end of the circuit remains comfortably above the operating voltage of the power supplies when the circuit is under its max load.
That said, lower voltage causes the current to go up as the power supply adjusts the output.
It is always Power In + Losses = Power Out.
So at 230V it can draw say 500mA, but at 115V it will draw 1Amp. Just keep this in mind as you calculate the wire size towards the end of the string.

As with any project, there is never an easy answer
Draw a diagram with each connection, and calculate the total amps from the source to the first camera, then subtract the current from this point to the next, calculate resistance of copper wire, volt drop, and calculate current again, and continue so forth.
Maybe this is going a bit too far. Use Excell to simulate your circuit, so that you can change values on the fly, and the formulas will update as you go along.

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