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Background

Hello, I'm building up a PCB for a PoE module based on Microchip LAN8720 IC. The recommended schematic is shown in the following figure.

recommended schematic for lan8720

I built up a prototype based on the Waveshare LAN8720 board. This Waveshare board uses an RJ-45 jack with magnetics integrated. The jack they used has a wiring as shown in the following figure. Obviously it doesn't support Power over Ethernet (PoE) with all the spare pins tight together, I had to pick a jack that supports PoE. So I made a decision to use an RJ-45 jack without magnetics and have the magnetics on the outside.

However, I made a mistake in the schematic by not connecting one of the center tap coils (ones next to the common-mode choke) together. (elaborating more on this: the two center taps should be connected to a 75 Ohm resistor each and ground through a 1000pF 3kV capacitor according to the recommended schematic in the figure above. This was the same setup the RJ45 jack with integrated magnetics have. Instead, I omitted this part by mistake and only connected them to the diode bridge for PoE). So the prototype does not work.

enter image description here

Question

I noticed this mistake while I was debugging the circuit and I found several RJ45 jacks with PoE enabled magnetics integrated. (Was too late..) However, I saw a difference in the wiring between PoE enabled jacks and the normal ones. The figure below is a jack with PoE capability and the center taps are first connected to capacitors and then to 75 Ohm resistors (C first then R). This is quiet different from the recommended (R first then C) layout.

enter image description here

My question is, is there a difference between these two topologies (C first then R) and (R first then C).

Intuitively (R first then C) topology used by the normal topology (without PoE) creates a conductive path between the two taps. Is this something to do with power delivery?

-- Update

Question in summary

What could be the difference in these two topologies marked with red boxes? Is there any underlying theory behind the difference? (Thanks @Jason for the insight on power wastage if the PoE topology had resistors connected creating a DC path)

enter image description here

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  • \$\begingroup\$ the bottom topology has extra capacitors, not reversed RC order \$\endgroup\$ – jsotola Mar 12 at 7:35
  • \$\begingroup\$ @jsotola But still, when two wires are connected across a resistor, it creates a DC connection between them right? Unlike when they are connected across capacitors.. That's my question. \$\endgroup\$ – Padmal Mar 12 at 7:49
  • \$\begingroup\$ @NoumanQaiser that's not the problem I have. It's about the topology. Think more like I need an answer from a say theoretical aspect. \$\endgroup\$ – Padmal Mar 12 at 7:51
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    \$\begingroup\$ POE is DC so a DC path inside the connector will waste power. \$\endgroup\$ – Jasen Mar 12 at 8:27
  • \$\begingroup\$ Please show the other schematic you are asking. But if you simply swap the places of the 4 capacitors with the 4 resistors, the circuit is absolutely the same and there will be no conductive path between center taps. \$\endgroup\$ – Justme Mar 12 at 8:48
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The topology that does not support PoE can have a DC path between center taps, and the resistors at the center taps are for common mode termination. There is no need for the capacitors to break the DC path. A PoE source that detects if the device supports PoE will see that the device has DC termination between pairs and will refuse to give power out. A passive PoE injector that always supplies power will damage the resistors.

The topology that supports PoE must not have a DC path between center taps. A PoE source would properly detect this and passive PoE injector will work too with this device. The resistors will still provide correct common mode termination impedance.

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  • \$\begingroup\$ This makes more sense. I forgot to add this common mode termination when I designed the schematic and most probably because of that reason, there is no communication happening. I'm going to fix it with soldering the missing parts and update this post with the results. \$\endgroup\$ – Padmal Mar 12 at 12:33

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