Running 175ft of cat5 solid cable with 12v/5v at the end

[StealthRT]

Hey all i am just wondering what type of DC voltage i would need to start out with in order to get 12v at the end of a 175ft run over cat5 solid cable? This is to be hooked up using the standard power over ethernet (PoE). I'm just not sure what i need to start out with in order to have a constant 12v with 300mA or 1A at the end.

I also would neeed to know about having 5v at the end with some of them as well.

Thanks for your time!

 

[Harald Kapp]

CAT 5 cable has <0.199 Ohm/m resistance (http://en.wikipedia.org/wiki/Category_5_cable). 175 ft = 53.34 m -> R <= 10 Ohm.
The voltage drop along the cable is V=R*I. From that you can calculate the required min. input voltage to the cable. However, the resistance can be even lower, resulting in higher output voltage. And the voltage will vary with the load current.
Both issues can be solved by using a higher input voltage plus a voltage regulator at the end of the cable.

By the way, what you are trying to do is not

standard power over ethernet

. Standard PoE uses 37 V ... 57 V (http://en.wikipedia.org/wiki/Power_over_Ethernet).

Harald

 

[(*steve*)]

Yeah, what you normally do is run a higher voltage over the cable (but one that won't overly stress the insulation) and use a DC-DC converter at the far end.

As you have seen, the amount of power lost is proportional to the SQUARE of the current. If you halve the current, you quarter the losses. This is almost exactly the reason that electricity distribution is done with high voltage lines.

The cheap way of doing this is to stick (say) 12V at one end and a 5V regulator at the other. You will lose power over the run of cable, but as long as you still have about 7V under load at the far end, you're OK.

 

[StealthRT]

Yeah, what you normally do is run a higher voltage over the cable (but one that won't overly stress the insulation) and use a DC-DC converter at the far end.

As you have seen, the amount of power lost is proportional to the SQUARE of the current. If you halve the current, you quarter the losses. This is almost exactly the reason that electricity distribution is done with high voltage lines.

The cheap way of doing this is to stick (say) 12V at one end and a 5V regulator at the other. You will lose power over the run of cable, but as long as you still have about 7V under load at the far end, you're OK.

Would doing that (5V regulator) cause a lot of heat is its in the range of 6-8v at the end? Would i need a heat sink for it?

 

[CocaCola]

From my experience 7805 or the like TO-220 package 5 Volt regulators don't really need a heat sink (in most cases) until you breach 9 Volts... They will run warm/hot but I have not seen them suffer a thermal shutdown without a heat sink in most cases when the input was 9 Volts or less... But, that isn't to say your mileage will be the same, ambient temps, enclosure, and overall output drain varies in every situation...

 

[StealthRT]

From my experience 7805 or the like TO-220 package 5 Volt regulators don't really need a heat sink (in most cases) until you breach 9 Volts... They will run warm/hot but I have not seen them suffer a thermal shutdown without a heat sink in most cases when the input was 9 Volts or less... But, that isn't to say your mileage will be the same, ambient temps, enclosure, and overall output drain varies in every situation...

Alright thanks ill give it a try.

 

[Rleo6965]

You can also use 1 pair for 12V and another pair of wire for 5V and use remaining 2 pair for ground wire connection. This will reduce voltage drop on wire.