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Ethernet DISCUSSION
I would like my Ethernet sensor node, which needs 5 V at up to 1 A, to be powered from a standard PoE switch instead of a separate adapter. I understand that about 48 V arrives over the cable, but not how the device announces itself or which pairs carry the power.
What are the building blocks of a standards-compliant powered device, and how do I work out which power class I need?
The IEEE standards define how much power a powered device (PD) may count on. The original 802.3af provides 15.4 W at the switch port and guarantees 12.95 W at the device after cable loss, with 37 to 57 V at the device. 802.3at (PoE+) raises that to 30 W at the port and 25.5 W at the device. 802.3bt goes further, to roughly 51 W and 71 W at the device for its two higher types, using all four pairs.
The switch does not simply apply 48 V to every port. It first probes with a low voltage, and only a device presenting the correct signature gets power. That is what protects non-PoE equipment, and it is why a compliant PD needs specific circuitry rather than just a regulator.
The handshake has two stages. For detection, the switch applies between about 2.7 and 10 V and measures the current; the PD must look like a 25 kΩ resistor (the signature resistor). For classification, the switch applies roughly 15 to 20 V and the PD draws a defined current that indicates its power class: class 1 up to 3.84 W, class 2 up to 6.49 W, class 3 up to 12.95 W, with class 0 as the default for an unclassified device, treated as up to 12.95 W. Class 4 identifies a PoE+ device.
After that the switch applies the full voltage, and the PD must limit its inrush current while its input capacitor charges. It also has to keep drawing a minimum current, about 10 mA, otherwise the switch concludes that it was unplugged and removes power.
The building blocks on the device side are: a jack or transformer module with centre taps rated for PoE current; two bridge rectifiers, because power may arrive on the data pairs (through the transformer centre taps of pins 1-2 and 3-6) or on the spare pairs (4-5 and 7-8), in either polarity; a PD interface controller that provides the signature, the classification current, inrush limiting and under-voltage lockout; and a DC-DC converter from the 37 to 57 V input down to your 5 V.
Many controller ICs combine the PD interface and the converter's controller. There are also complete PD modules that take the rectified input and deliver an isolated 5 V or 12 V output, which is the quickest route for a small production run.
The class follows from your input power. 5 V at 1 A is 5 W of output. With a converter efficiency of 85 %, the input is 5 / 0.85 = 5.9 W, and the diode bridge loses a little more: two diode drops of about 0.7 V each at 5.9 W / 48 V = 0.12 A is roughly 0.2 W. About 6.1 W at the jack fits inside class 2 (6.49 W) on paper, but with very little margin, so class 3 is the sensible choice unless the load can be trimmed.
Size the converter for the whole input range. At the minimum 37 V the input current is 6.1 / 37 = 0.16 A, and the design must start and run there, not only at the nominal 48 V.
Isolation is the part people are tempted to skip. The standard requires 1500 V isolation between the Ethernet port, including the PoE supply, and any conductor a user or another device can touch: a metal case, a USB or serial connector, sensor wiring. That means an isolated converter (typically a flyback) with its feedback crossing the barrier, and a matching gap on the board. A non-isolated buck converter is only acceptable when the device is fully enclosed in insulation and has no other electrical connections at all.
Finally, do not confuse standard PoE with passive PoE, where an injector puts 24 V or 48 V on the spare pairs permanently with no detection. The two are not interchangeable: a passive injector can damage equipment that does not expect it, and a standard switch will not power a passive device that lacks the 25 kΩ signature.
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