Passive POE: Need 5V @ 2A at the destination

Question

Having a bit of trouble finding the right information on google, and I'm worried I might fry some cables so I thought it best to ask some far more knowledgable than myself.

I want to connect 2 or more devices, giving them 5V/2A plus networking between them.

I figured to do so with the least amount of cables, Passive POE would be a good choice. The distance isn't far, probably maximum 20ft, as each device will be somewhere in a car, connected in to the boot/trunk area.

I'm looking at splitters and injectors, from what I've read its not as simple as just injecting 5v and it'll work? I have to worry about loss and also heat.

So my questions are, what is the best way to get 5V with max 2A current to the destinations? Do I need to send higher voltage to ensure it doesn't drop below 5, then use a regulator to ensure 5V? If I do that, the power will also be higher therefore more heat, is that ok?

Studying Wikipedia page I've read "Category 5 cable uses 24 AWG conductors, which can safely carry 360 mA at 50 V according to the latest TIA ruling"

50v * 0.36a = 18W, so if I were to send 9v * 2a = 18W. Then I could use a regulator to drop the ~9V(allowing for loss) down to 5V. Does that sound reasonable?

Many thanks

Answer 1

Let's start with a couple of straight answers: Good choice, possible, maybe, correct, yes, unless they use AWG26 or 28, which some flexible ones do, that's not how that works.

I'm betting you want more, right?

So, starting with your "biggest mistake": Power output at the end is not what a cable cares about. Cables care about current. Current creates a voltage drop caused by the wire-resistance. This voltage drop multiplied with the current creates power loss in the cable, which creates heating. Heating is bad. Bad is not good.

If you want to stick to "safe practise" you need to stay at the current level described in the standards, not the power level. So to transport your 10W, you'd need: V = P/I = 10/0.36 =~ 28V

But, that doesn't account for power loss in a conversion. If you use a buck regulator, which actively turns a given voltage into a stable 5V very close to the device, it can give you an efficiency of 85% or more, depending on which you choose. So then you are allowed to estimate using: Pin = Pout / efficiency = 10W / 0.85 =~ 11.8W.

Then you need to put V = Pin / I =~ 11.8 / 0.36 =~ 32.7V into the converter.

Because you still have wire losses you will need to put in a little more, so round up. Let's say 35V to be safe. Because of possible spikes and low-power-use moments your buck converter will need to be able to convert 28V to 38V without problem if your 35V is a clean voltage. Direct car voltages are unclean and very dangerous to shop for!

You can then create the boosted voltage from your car voltage with a boost converter that turns 9V to 34V into a nice stable 35V. Make sure it's rated for car voltages, or search for questions about "how to protect my device from car voltage spikes and load dumps".

Of course you can also just use the 48V that most early PoE systems use and get a ready made PoE to 5V converter on your device, then you just need a booster for car voltage to stable 48V.

But be very carefull in wiring all of this, make sure you look up how to transport a DC voltage over network cables, if you do it wrong your transmitter and/or receiver magnetics may melt.

Answer 2

Usually there are 8 wires that can carry Poe so 4 carry the current in and 4 more return the current. But I'd still be concerned with 2 Amps and use a higher sending voltage and use a buck convertor to locally produce 5 volts where needed. This can dramatically reduce cable current.