What PoE does before it puts 48 V on the cable

Plenty of things that are not PoE devices plug into an RJ45 port. That is why a PSE does not just put 48 V out: it checks who is there with a few volts, asks how much power they need, and only then switches on.

Last updated: 2026-09-10 PoEIEEE 802.3Power sequencingVoltage drop

The other end might not be a PoE device

A PoE camera, an ordinary PC, an old hub — all of them fit the same RJ45 port. Put 48 V on the pins the moment something is plugged in and you destroy the magnetics or the PHY of anything that was not expecting it.

So PoE puts two checks in front of the power. Detection asks whether this is a PoE device at all; classification asks how much power it wants. Both are done at voltages far below 48 V.

Where USB PD starts at 5 V and negotiates by messages, PoE starts by measuring the electrical character of the other end at even lower voltages. It is measurement, not protocol.

Detection: two points, and read the slope

The PSE applies a voltage somewhere in 2.7–10.1 V and measures the current, limited to 5 mA. A PD is built to look like 25 kΩ across that range; the standard accepts 23.75–26.25 kΩ.

The complication is the diodes in the path. A PD takes power either way round, so there is a bridge rectifier, which means two diodes in series are always in the way. The current does not pass through the origin.

Measure one point and divide, and the offset shows up as extra resistance. At 4 V the current is (4 − 1.4) / 25k = 104 µA, so V / I = 38.5 kΩ. A perfectly good 25 kΩ PD gets rejected.

So the PSE measures two points and takes the difference. At 4 V and 8 V the currents are 104 µA and 264 µA, and ΔV / ΔI = 4 V / 160 µA = 25.0 kΩ. The diode offset cancels in the subtraction.

That is what "stepping the voltage up gradually" actually is here. The steps are not only caution — you cannot get a slope from one point.

PoE detectionOn the left the circuit: the PSE applies 2.7 to 10.1 volts and the PD presents a 25 kilohm signature resistor behind a bridge rectifier. On the right, current against applied voltage: because of about 1.4 volts of diode offset the line does not pass through the origin. Two points, 104 microamps at 4 volts and 264 microamps at 8 volts, give 4 volts divided by 160 microamps, or 25 kilohms. Dividing a single point instead gives 38.5 kilohms and wrongly fails the device. Detection: two points in 2.7-10.1V, and the resistance comes from the slope Current is limited to 5mA. There is no 48V anywhere yet PSE applies two points, 2.7-10.1V LAN cable PD bridge rectifier 25k two diodes, about 1.4V of offset, always in the path current applied voltage 4V / 104uA 8V / 264uA slope = 1/25k assuming it passes through the origin Same PD, different answers One point: 4V / 104uA = 38.5k, which fails the check Two points: (8V - 4V) / (264uA - 104uA) = 4V / 160uA = 25.0k, which passes (23.75-26.25k)
Figure 1: detection is a two-point measurement. With the bridge diodes in the path, a single V/I does not give the right resistance.

Classification: one step up, then read the current

Once detection passes, the PSE raises the voltage to 15.5–20.5 V. The PD draws a current that encodes how much power it wants, and the PSE reads that current to decide how much to reserve.

Classification current drawn by the PD and the corresponding power
ClassPD currentReserved at the PSEAvailable at the PD
00–4 mA15.4 W12.95 W
19–12 mA4.0 W3.84 W
217–20 mA7.0 W6.49 W
326–30 mA15.4 W12.95 W
436–44 mA30 W25.5 W

Class 4 (Type 2 / PoE+) and above need either two-event classification — the same step repeated — or an LLDP negotiation after power is up. 802.3bt (Type 3 / 4) adds classes 5 to 8 and uses all four pairs, up to 71.3 W at the PD.

Classification is a reservation, not a limit. Whether the PD actually uses that power is another matter, but the PSE can no longer offer it to another port. A 48-port switch that runs out of power budget is usually the sum of these reservations.

From switch-on to switch-off

With the class settled, the PSE finally brings up the full voltage. Not instantly: it limits current so the inrush into the PD's input capacitance stays within spec, and the PD controller has its own inrush limiter.

Once power is up, the PSE keeps checking that the PD is still there. That is the MPS (Maintain Power Signature), and the PD has to keep drawing at least 10 mA.

This is where real hardware trips up. If the PD sleeps and drops below 10 mA, the PSE decides it has been unplugged and removes power. "It powers off when it goes into low-power mode" is almost always this. The fix is to keep 10 mA flowing through a dummy load while asleep, or to draw the current in pulses.

The PoE power-up sequenceA graph of cable voltage against time. From 0 volts, two small detection steps at about 4 and 8 volts appear and fall back to zero. A classification pulse of about 18 volts follows and also falls back. Then the rail rises to the operating 53 volts and stays there. The detection and classification voltages are visibly far below the operating voltage. What happens, in what order, from 0V to the full rail Detection and classification happen far below the operating voltage 10V 20V 50V 57V 0 voltage on the cable 1. Detection 2.7-10.1V, under 5mA. Confirms 25k 2. Classification 15.5-20.5V. The PD current sets the class 3. Power up brought up to the full rail with inrush limited 4. Operating PSE 44-57V. The PD keeps 10mA flowing as MPS Lose the MPS and the PSE removes power and starts again from step 1. This is why low-power designs drop out
Figure 2: from 0 V to the full rail. Note how low the detection and classification voltages are compared with 53 V.

The PD minimum voltage comes from the cable drop

The standard gives different minimum voltages at each end: Type 1 is 44 V at the PSE and 37 V at the PD; Type 2 is 50 V and 42.5 V. Those are not numbers to memorise — they fall out of the cable drop.

Type 1: 350 mA maximum, channel resistance capped at 20 Ω. 44 V − 0.35 A × 20 Ω = 37 V.

Type 2: 600 mA maximum, and with four pairs the channel cap is 12.5 Ω. 50 V − 0.6 A × 12.5 Ω = 42.5 V.

Both land exactly on the specified figure. The PD minimum is simply what is left at its terminals with the worst-case cable carrying the maximum current.

So what governs a PoE design is the usual voltage drop. 24 AWG Cat5e is about 9.4 Ω per 100 m per conductor; over two pairs there and back, a 100 m run comes to roughly 9–10 Ω.

The PoE voltage budgetFor Type 1, the PSE minimum output of 44 volts less a 7.0 volt drop from 350 milliamps through 20 ohms of channel gives the PD minimum input of 37 volts. For Type 2, 50 volts less a 7.5 volt drop from 600 milliamps through 12.5 ohms gives 42.5 volts. The PD minimum input voltage comes from the cable drop Not a number to memorise, but the result of a subtraction Type 1 (802.3af, up to 350mA) 44V PSE minimum output 0.35A x 20 ohm = 7.0V channel resistance capped at 20 ohm = 37V PD minimum input Type 2 (802.3at, up to 600mA) 50V PSE minimum output 0.6A x 12.5 ohm = 7.5V four pairs bring the channel to 12.5 ohm = 42.5V PD minimum input 24 AWG Cat5e is about 9.4 ohm per 100m per conductor; over two pairs there and back, 100m comes to 9-10 ohm A lower input voltage means more current for the same power, which increases the drop again
Figure 3: the PD minimum is the PSE minimum less the cable drop. The numbers in the standard come from this subtraction.

Things that catch you out

  • Do not hang capacitance on the 25 kΩ signature. The PSE takes its two points quickly. Too much capacitance on that node and the current has not settled, so the resistance reads wrong.
  • Classification is a reservation, not a ceiling. Some PSEs will trip on overcurrent if you draw beyond your class. Declare the class that matches what you actually consume.
  • Sleep versus MPS. Low-power design and PoE fight each other directly. Either budget the 10 mA (about 0.5 W) away, or draw the current intermittently.
  • Cable length costs power. 100 m of Cat5e is around 10 Ω, which is 6–7 V at maximum current. A lower input voltage means more current for the same power, which increases the drop again.
  • "Passive PoE" exists and is not this. Some equipment simply puts 48 V on the pins with no detection or classification. A standard PD will work, but plugging in a non-PoE device will destroy it.

Frequently asked questions

Why two measurement points? Would one not do?
Because the bridge rectifier in the PD keeps the current off the origin. A single V/I carries the diode forward voltage as an error: measured at 4 V, a 25 kΩ PD looks like 38.5 kΩ and gets rejected. Taking the difference of two points leaves only the slope and cancels the offset.
My PoE device powers off when it sleeps.
Suspect the MPS (Maintain Power Signature). The PD must keep drawing at least 10 mA; below that the PSE treats it as disconnected and removes power. Keep 10 mA flowing through a dummy load while asleep, or draw current intermittently.
Why is the PD minimum 37 V in one case and 42.5 V in another?
Different cable drops. Type 1: PSE minimum 44 V, 350 mA, 20 Ω channel, so 44 − 7 = 37 V. Type 2: PSE minimum 50 V, 600 mA, 12.5 Ω channel, so 50 − 7.5 = 42.5 V. The figures in the standard are the result of that subtraction.
Is it safer to declare a higher class?
It errs on the safe side for the PD, but the PSE reserves that power. If every port on a 48-port switch over-declares, the power budget runs out and some ports get nothing. Declare the class that matches actual consumption.

Standards and references

  • IEEE 802.3 (Clause 33 / 145) — Detection, classification and MPS requirements, and the PSE / PD voltage and current ranges
  • IEEE 802.3af / 802.3at / 802.3bt — Power levels and class definitions for Types 1 to 4
  • TIA-568 / ISO-IEC 11801 — Cat5e / Cat6 conductor resistance, used for the drop estimates

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