2
\$\begingroup\$

I'm looking to power a 50W load approximately 140m away using 48V DC. At the far end I will use a DC-DC converter to convert this to 12VDC where it will power a small DC powered network switch I have.

I was thinking that I could use a single copper pair (+ & -) cable to do this however there might be quite a bit of power loss due to the high current running to the end load. I was working out the maths and I believe that if I were to run two copper pair cables in parallel I would both decrease the voltage drop before my load and reduce the power loss. Basically, I would have two positive cables running from my power source to my load - connected to a voltage source in parallel. I would then recombine these two cables at the far end. The same applies to the negative return path cables.

I think this because I would be halving the total current flowing through each of the cables going to and from my load. Since power dissipated is current squared times resistance I figure that I will be able to halve my power losses. Similarly, my voltage drop will also be halved as the current through each cable is halved.

I worked out that 140m of 14AWG has a resistance of around 1.16ohm. I know my load will run on 12VDC but still consumes say 50W of power so its equivalent resistance would be around 46ohm given the 1A current flowing (48V x 1A gives roughly 50W).

Just wondering if this is correct or I if have a misunderstanding somewhere.

\$\endgroup\$
7
  • \$\begingroup\$ Why are you considering two cables instead of larger diameter (smaller gauge) wire? \$\endgroup\$
    – Perry Webb
    Commented May 19, 2022 at 11:52
  • 2
    \$\begingroup\$ @JunSeo-He reducing the wire resistance does not increase dissipated power in the wire. \$\endgroup\$
    – Justme
    Commented May 19, 2022 at 11:55
  • \$\begingroup\$ And the reason is that the lowered resistance leads to a reduced voltage drop. \$\endgroup\$ Commented May 19, 2022 at 11:56
  • \$\begingroup\$ Yes you are correct.I wasnt thinking the load. \$\endgroup\$
    – Jun Seo-He
    Commented May 19, 2022 at 11:57
  • 1
    \$\begingroup\$ @BrandonKellett your reasoning is correct. Half the voltage drop and half the power loss. \$\endgroup\$ Commented May 19, 2022 at 13:51

3 Answers 3

2
\$\begingroup\$

If you are only looking at the power consumed by the wire and you know the current that will go through the wires, the power the wires consume = (I^2)*R. However, if you are trying to consume less power, consider the following.

If you use a wire resistance calculator, 140m is about 460ft. 14AWG that length is 1.162 ohms on each of the positive and negative lines. If your switch needs 50W, that is at 12V. Thus, you would need 4.16A. This gives you a voltage drop of 4.85V or each line or 9.7V total. Thus, ideally your DC-to-DC converter would have 38.3V to generate the 12V. However, the actual case needs to consider what additional current the DC-to-DC converter will draw to complete it's task.

You are probably more concerned with you total power usage rather than just the power lost on the wires. Your total power usage would be the 48V supplied times the current supplied. For that question you need to ask how does the voltage supplied to the DC-to-DC converter change the current supplied. Adding wires may not reduce the total power usage.

The efficiency of your DC-to-DC converter probably affects your power usage more than the power loss of the wire. Dropping the wire resistance could have an impact on power usage if you drop the 48V supplied to near the minimum voltage the DC-to-DC converter needs to generate the 12V.

If you have a power supply that has force and sense connections and can supply 12 V, you might consider running a pair of force lines and sense lines to the switch. Then, the power supply would raise the voltage at the source so that the switch has 12 volts, and the DC-to-DC converter is not needed.

\$\endgroup\$
0
\$\begingroup\$

Considering a DC - DC converter efficiency of 80%, the losses would be as shown below.

enter image description here

\$\endgroup\$
-3
\$\begingroup\$

You want to find the maximum power which can be consumed by the load.

schematic

simulate this circuit – Schematic created using CircuitLab

Maximum power is consumed by the load only when the impedance of the wire is the conjugate of the impedance of the load and in our case (DC) we dont have reactance ->Rw = RL.

\$\endgroup\$
5
  • 1
    \$\begingroup\$ Hi Jun Seo-He, Wire resistance is a part of the load and not of the source. Maximum power is transferred, from the source to the load, only when the load resistance is equal to the internal resistance of the source. \$\endgroup\$
    – vu2nan
    Commented May 19, 2022 at 14:24
  • \$\begingroup\$ @vu2nan how is wire resistance part of the load anywhen? \$\endgroup\$
    – Jun Seo-He
    Commented May 19, 2022 at 15:48
  • \$\begingroup\$ Hi Jun Seo-He, Wire resistance is not the internal resistance of the source. \$\endgroup\$
    – vu2nan
    Commented May 19, 2022 at 16:51
  • \$\begingroup\$ @vu2nan So what it is a useless piece of power dissipation. \$\endgroup\$
    – Jun Seo-He
    Commented May 19, 2022 at 16:52
  • \$\begingroup\$ Hi Jun Seo-He, Cable loss is a necessary evil for the OP. However, that may be reduced by increasing the cable size. The DC - DC converter loss is higher than the cable loss. \$\endgroup\$
    – vu2nan
    Commented May 21, 2022 at 7:11

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.