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I'm trying to figure out the voltage drop across a 2.5mm^2 (14 AWG) cable.

The formula I've used is V_drop = (Length in Meters * Current * 0.017)/(area in mm^2).

For a 14 AWG cable I understand the Ampacity is around 15A. With an example distance of say 1500m, this equates to 153V drop. However, I'm not sure if this is correct as when I try an online calculator as linked here - this answer is correct only for 3 conductors per phase in parallel. So I am unsure if the formula I have is correct or not or it is used for 3-Phase AC calculations only? Also - does the distance I have need to be doubled for the return path as I have seen on some resources?

What I am trying to essentially do is figure out what my limitations would be for sending 48V DC as far as possible, and am starting out with trying to wrap my head around these voltage drop calculation. I understand that increasing my voltage and dropping the current will help significantly here, but I'm not looking to do that if possible.

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  • \$\begingroup\$ What I am trying to essentially do is figure out what my limitations would be for sending 48V DC as far as possible You can send 48 V over practically any distance you like as long as there is no current flowing. But no current means no power (Power = voltage * current). "As far as possible" What does that mean? 48 V and a certain power means a certain current. That current in combination with the resistance of the wires and how much voltage drop you can accept will limit how long the wires can be. If you want to understand this better you need to study ... \$\endgroup\$
    – Bimpelrekkie
    Commented Jul 15, 2021 at 11:51
  • \$\begingroup\$ ..the relations between voltage, current, power and resistance. Using 48 V and a couple of Ampere (so less than a few hundred Watt of power) you can practically span a distance of less than 10s of meters. I mean, 100 meter would already mean a lot of power loss. There is a reason why electricity is transported at for example 110 kV and higher: to keep currents low and thus keep the losses low. \$\endgroup\$
    – Bimpelrekkie
    Commented Jul 15, 2021 at 11:52
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    \$\begingroup\$ Yes the distance has to be doubled. You have two wires. \$\endgroup\$
    – StainlessSteelRat
    Commented Jul 15, 2021 at 12:14
  • \$\begingroup\$ Does this answer your question? Avoiding DC voltage drop over long distances \$\endgroup\$
    – rdtsc
    Commented Jul 15, 2021 at 14:07
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    \$\begingroup\$ This whole long-distance voltage drop thing is the reason there's a standard 4-20mA signal available for industrial controls. \$\endgroup\$
    – Kyle B
    Commented Jul 15, 2021 at 16:12

2 Answers 2

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The calculation is basically right.

Depending on which exact formula (or online calculator) you use, a single 2.5mm² 1500m copper wire should have about 10 ohms of resistance.

So 15A would indeed drop about 150V, in one wire, but obviously you would need to apply 150V into the wire to make 15A flow and there would be 0V at the end of the wire. So with 48V, you obviously can't even make 15A to flow.

Let's look at this from another angle.

So 10 ohms tells that for each ampere of current, there is about 10V drop, in one wire.

The single wire has 10 ohms resistance. If you send 1A over two wires, supply and return, you have 28V left at the load, as 10V is lost per wire.

It also means that if you short circuit the wires, you have about 20 ohms of resistance, so a 48V supply can only make 2.4A flow in the wires, with no voltage left for any load.

What you need to determine is how much is the maximum current you need, and how much is the minimum voltage that the load can operate, to know how much of the voltage is allowed to be dropped per wire.

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  • \$\begingroup\$ Thanks for the explanation - very easy to understand. So in terms of this calculation - its just as simple to double the actual distance between say point A and B to account for the supply/return cable. So what I need to do to: Calculate the Voltage required by the end device/s. I can then use this to solve for the voltage that should be transmitted to result in that voltage at the device, after the drop? As current is constant I understand this is just a factor with regards to the losses incurred. \$\endgroup\$
    – Brandon Kellett
    Commented Jul 15, 2021 at 12:01
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    \$\begingroup\$ @BrandonKellett Sorry but why do you assume current will be constant? What devices will be at the load, does it always consume same amount of current or power? \$\endgroup\$
    – Justme
    Commented Jul 15, 2021 at 12:04
  • \$\begingroup\$ Current through the cable is constant is what I meant, rather than referring to the end device etc. Devices at the load regardless will be low power devices such as CCTV, Access Point etc. \$\endgroup\$
    – Brandon Kellett
    Commented Jul 15, 2021 at 12:21
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    \$\begingroup\$ @BrandonKellett unless you know exactly what current will be drawn, boosting the voltage at the supply end risks frying the equipment at the load end, as the voltage drop will depend on the current being drawn at any given moment. \$\endgroup\$
    – Simon B
    Commented Jul 15, 2021 at 12:25
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    \$\begingroup\$ @BrandonKellett So if you a device which obviously has variable current consumption, it means voltage will fluctuate at the load so if the load expects say constant 12V but can take low or high amount of current based on e.g. how much traffic there is on access point, that's the completely opposite way to feed cable with variable voltage to keep voltage constant at the load. What you need to do is to have a voltage regulator at the load side which can provide the load with constant voltage even if current draw varies, and just feed the cable with constant voltage. \$\endgroup\$
    – Justme
    Commented Jul 15, 2021 at 12:48
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The way to send '48 V as far as possible', while keeping the remote voltage constant in the face of varying current demand, is to send a far higher voltage (so you can use thinner cables) and use a buck converter at the far end to (a) regulate the voltage and (b) boost the current while dropping the voltage.

In using a higher voltage, you win on two counts. The lower current causes less absolute voltage drop on your cables. The higher voltage means the absolute voltage drop is a smaller fraction of what you sent.

There are many 48 V output DC-DC converters that will take a wide range of input voltages. Limit your transmission voltage to a few hundred volts, then commercial 'mains' cable will be adequately insulated, and is relatively cheap.

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