I see that the PoE Type 2 standard assumes a maximum loop resistance of 12.5 Ω/100 m, but cat 5e is specified to have maximum loop resistance of 18.8/100 m Ω. Does this mean that PoE type 2 is not guaranteed to work over the full 100 m Ethernet range supported by Cat 5e?

  • \$\begingroup\$ You can always be less then the maximum =-D (and more then the minimum). \$\endgroup\$
    – Tyler
    Feb 19, 2016 at 17:51
  • \$\begingroup\$ What this means is that worst case scenario, you can have a compliant 5e cable that is not POE compliant, it will likely still work, but it is something to be aware of. Realistically you will have cables that don't hit either maximum. \$\endgroup\$
    – crasic
    Feb 19, 2016 at 17:58
  • \$\begingroup\$ My assumption is that the type 2 standard really does not worst-case at 100 m with cat 5e cable. I am just wondering if anyone else has made this observation. I think most people just assume PoE works to 100 m – the same reach as Ethernet. \$\endgroup\$
    – ziggle314
    Feb 19, 2016 at 17:59
  • \$\begingroup\$ Whether or not PoE+ works is highly dependent on the PSE and PD. If you assume a worst-case cable (100m, 18.8Ω), a minimally compliant PSE (50V output) and a minimally compliant PD (42.5V minimum input) sourcing maximum current, then no, it won't work. But good cable will be lower resistance, a well-designed PSE will increase the voltage at larger currents, and a well designed PD will be able to tolerate a much lower input voltage. \$\endgroup\$
    – uint128_t
    Feb 19, 2016 at 18:38
  • \$\begingroup\$ Everything I can see agrees with your statement. It appears that the type 2, 30 W standard is not guaranteed to work out to 100 m. Even Cat6, which used 23 AWG wire, is not guaranteed to work at 100 m – though it will get very close. \$\endgroup\$
    – ziggle314
    Feb 19, 2016 at 18:42

1 Answer 1


The answer is POE is a regulated power bus, the cable will burn up power but the systems should compensate for several reasons.

The POE system can source more power on the source than the device needs. It depends on the class of device, for a class 2 device 0.51W has been reserved for the "cable" at the worst case of 44V using the power equation P=I*V the current from the source would be ~160mA. The cable has only been allotted 0.51W of power to burn, so we can use another cable to find the max resistance allowed under spec, keep in mind that you have to account for both the resistance of the cable to and from the device. P=I^2*R can be rearranged to be P/I^2 = R where P=0.51W and I = 0.160A and solving for R get you 19.921 Ohms (for the total resistance of the cable). So it looks like the engineers who wrote the POE spec did their homework and added margin even for the worst case.

Thats a worst case figure, the device will usually have some margin in it (A good engineer if they needed 6.3W would probably jump to a class 2 device.

The PSE's can also source more voltage, they are required to source 44V but probably have the ability source more depending on the supply.

Don't go with cheap equipment though or you might get devices that have no margin or aren't up to the spec.

POE Explained is a great resource. I'll quote it enter image description here

Note also that these figures are upper limits to power. Power cannot be “forced” down the cable – a surprisingly common misconception. The PD simply presents a load to the cable and draws as much current as it needs. Most PoE-powered devices will draw a fixed level of power, so for example a 5 watt IP camera will advertise itself as Class 2, as it will never need more than 6.49 watts, and then draw enough current at its local voltage to provide the 5 watts it requires to operate.

The figures in the table are also based on the worst-case operating conditions for PoE. PSE voltage is nominally 48 volts, but its allowable range is between 44 and 57 volts: because efficiency increases with voltage, the losses are based on a PSE supplying the 44 volt minimum. A worst-case value for cable resistance is also assumed, and this is based on a maximum length7 of light-gauge cable, with an allowance for connectors, patch panels, or bad wiring. 7


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