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.