Power Over Ethernet (PoE) Building

An Ethernet Wireless Device is a very practical approach, mainly because they can be located near the Antenna in a waterproof box and the Ethernet cable can be run a fair distance away up to 300 metres in some cases.

Ethernet connections require only two pairs of cable TX & RX, but in a standard Cat5e cable there are 4 pairs. This allows for the spare pair to used for injection of power over the Ethernet. Called Power Over Ethernet (PoE) it is common practice to use the Blue and Brown pairs for this purpose.

An issue may occur on long Ethernet runs, when this method is used in combination with some Wireless Devices that require a low voltage (5V) or a regulated power.

This is because there will be a voltage drop over the length of cable due to I²R losses. In these circumstances a higher voltage is injected into the cable and a voltage regulator is installed with the Wireless Device to drop the voltage down again to the desired voltage.

An example of a such a regulator can be found here: http://www.air-stream.org/LM2576

A useful list of PoE industry standards can be found here: http://www.air-stream.org.au/poe_standards


Injection Voltage for PoE using the 3 Amp Regulator

The 3 Amp regulator has an approximate 2V overhead required to achieve the desired voltage eg: 14V for a 12V output or 7V for a 5V output.

So we only need to calculate the resistance of the Ethernet run to determine the minimum injection voltage.

Cat5e I believe is 9 ohms per 100metres, so a 10 metre run would be about 0.9 ohm.

At 2 amps (a typical current draw) The expected drop would be calculated as:

V = I x R so this would be 2 x 0.9 = 1.8 volts.

So the minimum ejected voltage required on a ten metre run at 2 amps would be:

1.8v + 2v + 12v = 15.8 volts.

However, as the regulator has quite a high input voltage e.g. max 45V the POE injector can be anything between 15.8 and 45V. So as 24V power supplies are more common than 16V, I would tend to use a 24V power supply.

Just a side note, a switchmode regulator has similar power characteristics to that of a transformer (the current is inversely proportional to the voltage) , (e.g. more voltage less current) so you don’t need to buy a 2 amp 24V supply in this scenario.

To calculate the minimum current at 24V, convert the minimum requirement into power.

P = E x I or 15.8 x 2 = 31.6W

So the current draw at 24v will be:

P/E = I or 31.6 / 24 = 1.32 Amps

So the power supply that would be most cost effective for this scenario would be:

24V at 1.5Amps or 36VA.