What PoE does before it puts 48 V on the cable

The other end might not be a PoE device

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

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

Detection: two points, and read the slope

The PSE applies 2.7–10.1 V and measures the current, limited to 5 mA. A PD is built to look like 25 kΩ there (accepted: 23.75–26.25 kΩ).

A PD's bridge rectifier always puts two diodes in series, so the current does not pass through the origin. Measure one point and divide: at 4 V the current is (4 − 1.4) / 25k = 104 µA, so V / I = 38.5 kΩ, and a 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. You cannot get a slope from one point, hence the steps.

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 the power it wants, and the PSE reserves accordingly.

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

Classification is a reservation: the PSE can no longer offer that power to another port. A 48-port switch running out of budget is usually the sum of these.

From power-up to normal operation

With the class settled, the PSE brings up the full voltage, but not instantly: it limits current so the inrush into the PD's input capacitance stays within spec.

When the MPS drops, the power goes

The PSE keeps checking that the PD is still there. That is the MPS (Maintain Power Signature): the PD has to keep drawing at least 10 mA. If the PD sleeps and drops below that, the PSE decides it has been unplugged and removes power. Powering off in low-power mode is almost always this.

Keep 10 mA (about 0.5 W) flowing through a dummy load while asleep, or draw it in pulses.

The PD minimum voltage comes from the cable drop

The minimum voltages differ at each end because of the cable drop. Type 1: 350 mA maximum, channel resistance capped at 20 Ω, so 44 V − 0.35 A × 20 Ω = 37 V. Type 2: 600 mA maximum, and with four pairs the cap is 12.5 Ω, so 50 V − 0.6 A × 12.5 Ω = 42.5 V. The PD minimum is what is left 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.

Things that catch you out

• Do not hang capacitance on the 25 kΩ signature: the PSE takes its two points quickly, and the current has not settled, so the resistance reads wrong.
• Some PSEs will trip on overcurrent if you draw beyond your class. Declare the class that matches what you actually consume.
• 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" puts 48 V on the pins with no detection or classification. Plugging in a non-PoE device will destroy it.