Using any/all Ethernet wires for power transfer?

Question

I was looking to get a small amount of power to be transmitted ~100ft, and noticed there is a standard way to move power over Ethernet cables.

We actually plan to splice the cables ourselves, and not use a standard Power over Ethernet, but wanted to confirm how much power I can dependably pass through a single pair? There will be no data transmission, no Ethernet devices. Only DC current.

Further, I wanted to know if it makes a difference which wires I use? I am not sure whether all 4 pairs (8 wires) are the same, or if only certain ones are built specially for power, while the others might be inappropriate for power, and only dependable for data?

Edit: Additional details: I have a transformer plugs into the 120V wall plug, and it says it outputs 5V 1.2A. So that is what I want the cable to support. I plan to cut the small wire coming out of the transformer and extend it using ethernet wires. If ethernet 1 pair is enough, I might do that. Or I might use 2 pairs just to keep the two transformers and end devices completely separate.

Answer 1

If you don't need compatibility with ethernet, it doesn't matter which pair you use. It doesn't even really matter much that you use pairs.

Your main concern in this endeavor will be the resistance of the cables. You can find a chart of wire resistance and from that, calculate the total resistance of your cable. For an example, cat-5 cable is typically 24-26 AWG. From a table, I see that 100 feet of 23 AWG wire has a resistance of about \$3.2\Omega\$.

Say you want to move 500mA on this wire. We can calculate the power lost in the cable:

\$ P = I^2 R = (0.5A)^2 3.2\Omega = 0.8 W \$

or the voltage you will lose on this cable:

\$ V = IR = 0.5A \cdot 3.2\Omega = 1.6 V \$

Keep in mind, this is for 100 feet of wire, but since you need a supply and a return, the total length of the wire is twice the length of the cable.

This 0.8W is not only power not doing useful work, and less power available at your load, but also heat generated in the cable. A longer cable has more area to radiate heat, so can dissapate more power. This is why most tables also include a maximum current. \$500mA\$ is just a bit too much for 23 AWG cable.

Also, the voltage drop will reduce the ability of a voltage regulator to regulate the voltage on the other end. The voltage regulator has no way to know what the voltage on the other end is, so it can't regulate it.

Another concern is the maximum voltage the insulators in the cable can withstand. Power over ethernet provides 48V, so anything below this should be safe. Anything above that, consult the datasheet for your cable.

Answer 2

If you are using the cable simply as a power cable and effectively need two conductors then using the 4 x "white striped" wires together as one conductor and the 4 solid coloured wires together as the other conductor will give you a two conductor cable with equal resistance in each conductor and lowest possible overall resistance.

My past investigations and other people's experiences showed that the resistance of the various pairs varies in cables which comply with the relevant standards and they can be significantly worse than what the standard stipulates in cable from 'less reputable suppliers'. In aquick look now I found no mention of different resistance per piar so that MAY have been CAT5 only. Maybe.

Measurement with an Ohmmeter will rapidly tell you what the situation is with the cable that you have but regardless, connecting the wires as I suggested above will give you the best result for a given cable.

Pacific Custom Cable say re Cat6 - 26 Ohms max per 100 metres per pair.

This article suggests Beldens Cat6 24AWG cable, part 1872A has a resistance of 2 Ohm per 40 feet or about 16 Ohms per 100 metres or in your case 5 Ohms loop resistance over 100 feet.
IF all pairs were the same then 4 pairs in parallel will give about 1.25 Ohm loop resistance.

Here's some "military tactical Cat6M - Kevlar armour etc.
Pretty picture of makeup
Specs - they say 23/7 AWG (23 AWG 7 strand), 75 Ohm/km - probably one leg, so 15 Ohm/100m loop - much as above for 24 AWG.
They specify 1.5A max for the RK45 connectors so 4 pair may allow 6A.

Answer 3

Yes, a bit late, but just to complete a minor issue.

One way to correct for the voltage drop is to use a power supply with voltage sense terminals and use one pair of the cable to as a voltage sensing pair to the distant end. Within reason this allows the power supply to boost the voltage at its end to compensate for the IR losses in the cable. The sense pair carries very little current, so there is no significant voltage drop. The downsides are:

1. The power supply is typically more expensive than your run-of-mill supply.
2. It leaves one less wire pair for carrying current for the end load.
3. Because of the compensation the voltage at the power supply is higher by a few volts under normal circumstances, so equipment plugged in closer to the supply will see that higher voltage.
4. You have to make sure the power supply has enough headroom to be able to correct for the voltage drop.
5. You have to make sure that the load is never powered via the sense lines, it destroys the current sense resistor(s) in the supply. So to be safe switch the AC side of the supply only, or switch both sides of the load on/off at the same time, not just the high or low side.

For more details you can search for 'remote sense power supplies EDN', there's a nice EDN article that goes into more detail on this topic including what -not- to do with this technique.