PoE Voltage Drop Calculator
Estimate how much voltage a Power-over-Ethernet run loses along the cable, and whether the powered device at the far end still gets enough voltage to work. Long runs and high-power devices can drop several volts of DC across the copper, so this tool converts device power, run length and cable resistance into the voltage and power actually reaching the device.
Enter Values
Before you rely on this: First-pass guide only. Verify safety-critical or regulated work against the relevant standards, your project requirements and a qualified professional.
How to use this calculator
- Enter the source PoE voltage (default 48 V — the nominal PoE supply voltage) and the device's power draw in watts.
- Enter the cable length in metres and the loop resistance per metre (default 0.2 Ω/m, typical of Cat5e 24AWG copper).
- Read the voltage drop, the voltage at the device and the power available there, and check the minimum-voltage warning.
How it works
The current is the device power divided by the source voltage. The cable's loop resistance is its per-metre resistance times the run length, and the voltage drop is that resistance times the current (Ohm's law). Subtracting the drop from the source gives the voltage at the device, and multiplying that by the current gives the power still available. If the device voltage falls below about 37 V — the input minimum for an 802.3af powered device — the tool flags it.
Worked example
Worked example. A 30 W device sits at the end of a 50 m Cat5e run (0.2 Ω/m) fed from 48 V. Current = 30 ÷ 48 = 0.625 A, loop resistance = 0.2 × 50 = 10 Ω, so the drop = 0.625 × 10 = 6.25 V. The device sees 48 − 6.25 = 41.75 V and about 26.09 W — safely above the 37 V minimum.
Common mistakes
- Entering the per-conductor resistance instead of the loop (out-and-back) resistance, which halves the calculated drop.
- Assuming the device receives its full rated wattage — the cable loss means the power at the far end is always lower than the power leaving the switch.
- Ignoring temperature: copper resistance rises with heat, so a run in a hot ceiling void drops more voltage than the same run measured cold.
Frequently asked questions
Why does real PoE do better than this estimate?
This simplified model assumes a single loop resistance. Real PoE (especially PoE+ and PoE++) spreads the current over two or four pairs, which lowers the effective resistance and the drop. Treat this as a conservative worst case.
What voltage does a device actually need?
It varies by standard: an 802.3af powered device is specified to work down to about 37 V at its input, while 802.3at and 802.3bt define their own minimums. Always check the device's datasheet.
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Tip: Enter any known values to calculate the remaining results.
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