Sorry if this is too n00b a question, I'm struggling to wrap my mind around basic EE terms for my project.

In my application I run 20 watts over a single AWG-24 wire at 5 volts. This works fine if the wire is <10 feet long; over 10 feet and my application fails; LEDs stop showing the proper color. According to those who have come before me, this is because of a "voltage drop".

I'd like to understand why something like Power over Ethernet (PoE) can send 20 watts hundreds of feet but my application cannot using a single AWG-24 wire. My intent is to roll my own solution leveraging the knowledge baked into PoE.

Suppose a PoE system uses a Cat5e cable. Cutting open this Cat5e cable reveals that it is made up of 4 pairs (cores?) of AWG-24 wires. The very same AWG-24 my application uses! I understand PoE's trick for sending 20 watts such a distance is that it steps up from 5 volts to 48 volts. Rumor has it that stepping up voltage means you can send power over longer distances.

Hold on though. I've heard there's something called Ohm's law[1] that says if you increase the voltage you either need to decrease either the current[2] or the resistance.

Since AWG-24 wires universally have the same resistance, the current must be going down in PoE! Yet PoE can still deliver 20 watts hundreds of feet away, so some kind of "combining" trickery must be happening.

What is this trickery? Is there some kind of black box that takes as input N (48 volt, M current) wire and outputs one (5 volt, NM current) wire?

Thank you kindly, community of experts, for any assistance you can offer me in my endeavor.

  1. Is Ohm's law what punked me earlier? Voltage dropped because resistance increases as distance increases?

  2. My understanding is the watts I so desperately crave are composed of volts and amps, and an amp is a way of measuring current. So many marvelous labels!


3 Answers 3


POE typically uses 48V so 20W would be less than 0.5A shared over multiple 24 AWG wires.
For your 5V system, 20W draws 4A which is why you get too much drop in the cable.

Now consider your 24AWG wire. It has a resistance per meter you can find from many sources, here's one.
Given that from this chart a single wire has approximately 0.0842 Ohms/meter, that is roughly 0.25 Ohms for 10 ft. But you have to consider the loop resistance, so your wire R value is 0.5 Ohm.
Assuming you could pass 4 A through this loop you would drop 4 * 0.5 = 2 V in doing so.


Let's say your device uses 20W of power. If you power it from 5V, you need 4A current to get 10W (5V*4A=10W). This would be quite difficult to transport over the thin UTP cable wires.

However, if you step up the voltage to 48V (standard PoE), you now only need 417mA to get 10W (48V*0.417A=20W). This is much easier to transport.

Also, PoE uses multiple wires in parallel, usually one pair for positive and one pair for negative.

According to Ohm's Law (V=R*I), voltage drop is proportional to the resistance and current. So, when you use higher voltage for the same power, you need lower current, which results in lower voltage drop. In addition, the voltage drop is an even lower part of the total voltage, resulting in higher efficiency. For example, let's say that the resistance of the cable is 2ohm:

You want 5V, 4A at the end of the cable, which means that the voltage drop is 4A*2ohm=8V. So, your power supply has to be (5+8=13V). 2oW powers your device and 32W (8V*4A) heats the cable up, resulting in ~38% efficiency (20W/(20W+32W)). If you use 50V, you now only need 0.4A, this results in 0.8V drop in the cable, so your power supply has to be 50.8V and only 0.32W are wasted as heat in the cable, resulting in 98% efficiency.

This is the reason why long distance power lines use hundreds of thousands and in some cases a bit over a million volts.

  • \$\begingroup\$ Can you explain a little more about how it uses the multiple wires in parallel? Why can't it use a single pair? What device does it use to combine those multiple pairs back into a single pair at the endpoints? (Basically trying to figure out if I need to do this multiple pairs thing in my application too) \$\endgroup\$
    – mbac32768
    Jan 7, 2017 at 18:48
  • \$\begingroup\$ Of course you could also simply select a bigger wire gauge and use a single pair of wires to get lower voltage drop... If you put multiple wires in parallel you also reduce the resistance (you are effectively making the wire bigger). You don't need a device to do this, just connect the ends together (solder or mechanical connections). In your case for example if you connected five wires each way (10 wires in total) you'd only drop 0.4 V. \$\endgroup\$ Jan 7, 2017 at 19:25

Take a closer look at Ohm's Law and the derivatives:

I=E/R, E=I*R, R=E/I


You need 20W at 5V. This means you need 4A of current (I=P/E)

The problem with your power feed is that you are getting excessive voltage drop due to the resistance of the wires (E=I*R), and while you start with 5V, you end up with something less.

You can also get 20W by using 10V and 2A. Or 20V and 1A. Or 48V and 0.41666... Amps. At 0.4A, the voltage drop through the wires will be roughly 1/10 the drop when you tried to send 4A.

POE uses the high voltage to reduce the required current, and then uses a switching regulator at the load end to efficiently drop that high voltage down to what is needed by the equipment. Switching regulators are typically 90% efficient (or better), so with 48V POE the actual current sent over the cable will be slightly higher, to compensate for both the regulator loss, and for the voltage drop in the wires. Even so, the current (and voltage drop) will be low enough that the power can be sent over the skinny wires. POE is used to send up to 25W over reasonable-length cables.

  • \$\begingroup\$ Thanks for mentioning the switching converter, that helps a lot. I think the only question mark left for me is that POE uses multiple wires in parallel. Is combining those essential to being able to send 20W long distances? If so, how does it unify them at the endpoints? \$\endgroup\$
    – mbac32768
    Jan 7, 2017 at 18:58

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