0
\$\begingroup\$

Having just realised I have screwed up my board by stupidly swapping pins 4 & 5 on the magjack output, I have realised I don't properly understand the PoE power pins on the magjack. I wonder if someone could help explain before I do anything else stupid.

I am using the Kycon G8X-188S7-BP jack with integrated magnetics, the schematics below labels pins 9 & 10 as the VC+/VC- but confusingly leaves pins 4 & 5 unlabelled.

What is the purpose of combining pins 4+9 and 5+10? Why do pins 9 & 10 feed via the transformer?

Kycon G8X-188S7-BP

These are my current, hopefully now correct schematics:

[EDIT] This is for a PoE PD expecting a 24v passive mode B supply.

My schematics

[EDIT] A mode A vs B wiring schematic I was using: Mode a/B wiring

\$\endgroup\$
6
  • 1
    \$\begingroup\$ First of all, is that a device that gets powered via PoE, or the device that powers another device via PoE? At least in either case you are violating the isolation requirement, and since the pins 9 and 5 are directly connected together, accidentally using a crossover cable will burn something. They should not be directly connected, and the PoE supply must not be ground referenced. Please supply more info what you are doing so it can be answered. \$\endgroup\$
    – Justme
    Apr 20, 2022 at 17:11
  • \$\begingroup\$ Thanks for the comment! Edited to make explicit it is a Mode B PD. Why can't they be connected together given the spec. explicitly specifies 4+5 for VC+ and 7+8 for VC-? What reference should VC- take on a device? \$\endgroup\$
    – Neil
    Apr 20, 2022 at 18:11
  • 1
    \$\begingroup\$ @Neil For a Mode B ,10/100-only PD you don't need to bother the magjack since the power comes over the unused pairs. However, a proper PD is supposed to work with either Mode A or Mode B. Also, a proper PD needs to identify itself to the PSE to receive power and PoE power is 48-54 V. \$\endgroup\$
    – Zac67
    Apr 20, 2022 at 19:51
  • 1
    \$\begingroup\$ If you connect 4+9 and 5+10 you enable both Mode A and Mode B - but you really shouldn't short them but use diodes for decoupling. In any case, you're not building anything compatible with 802.3af/at/bt nor anything capable of 1000BASE-T. \$\endgroup\$
    – Zac67
    Apr 20, 2022 at 19:59
  • 1
    \$\begingroup\$ @Neil I think you are confused. Modes A and B are part of the 802.3af standard (and successors) and neither are passive. The standard also requires a PD to support both (it’s the PSE’s choice to use one or the other) along with having the necessary circuitry for identification and negotiation of power etc. Passive PoE is not standard, and just happens to use the same “unused” (in 10/100 Ethernet) pins as Mode B as this is simpler to implement. You’ll have to decide exactly what you are attempting to support: standard 802.3af PoE, custom passive, or both. \$\endgroup\$
    – jcaron
    Apr 20, 2022 at 22:39

2 Answers 2

Reset to default
1
\$\begingroup\$

It's for the two modes of PoE delivery. Mode A supplies power as a common-mode voltage on signal pins 1, 2, 3, and 6. Mode B supplies power as a voltage on unused (in 10/100) pins 4, 5, 7, and 8.

\$\endgroup\$
7
  • \$\begingroup\$ As I understand it 1000Mbs 802.3af in Mode B uses the same pins, it just use 4+5 and 7+8 for data too. But that doesn't answer the Q of why 9 & 10 in the schematic go via the magnetics. What is that doing? \$\endgroup\$
    – Neil
    Apr 20, 2022 at 18:18
  • 1
    \$\begingroup\$ It's letting you pull the common-mode voltage off those data lines. The jack is just for 10/100 so the magnetics for the other two pairs aren't installed. \$\endgroup\$
    – vir
    Apr 20, 2022 at 18:41
  • \$\begingroup\$ Apologies for being really dim here but in the schematic above, J5 and J7 both feed the common bus that pins 9 & 10 pull from. Am I misunderstanding that in 10/100 those lines are irrelevant as the CM voltage is only used for 1000BT? \$\endgroup\$
    – Neil
    Apr 20, 2022 at 20:19
  • \$\begingroup\$ Even if the PoE injector only applies power in Mode B to unused pairs, your design has the power pins of Mode A and B shorted directly together, so it energizes also the used data pairs. Depending on how the injector works, the powered data pair can end up to a device not supporting PoE at all and it can get damaged - imagine that magjack with the four capacitors replaced with short circuits which is how non-PoE devices connect them. Also, it still does not remove the problem of unisolated supply, you can't connect the PoE supply to ground. PoE must have isolated converter to have isolation. \$\endgroup\$
    – Justme
    Apr 20, 2022 at 20:27
  • 2
    \$\begingroup\$ Pins 9 and 10 are offering the ability to use mode B. The jack is wired for both and you would use the pins for the mode you want, though the standard requires that your powered device be able to use either. The problem in your schematic is that you have connected mode A and mode B together so there is the possibility of damaging a non-POE device. A proper implementation (see: ti.com/lit/ds/slvs525b/slvs525b.pdf) has two bridge rectifiers on both A and B inputs to guard against reverse polarity and to prevent power supplied on mode B pins from appearing at mode A pins. \$\endgroup\$
    – vir
    Apr 20, 2022 at 21:17
0
\$\begingroup\$

To answer and clarify based on the comments..

The circuit above was to support an "always on" or passive 24v PoE PD which I had thought was Mode B. It isn't, as jcaron pointed out, passive PoE is non-standard and just happens to use the same pins as Mode B.

My next confusion was why pins 9 & 10 were feeding off the bus bar via the magnetics. Stupidly I hadn't realised it was the other way around, pins 9 & 10 are fed via the magnetics in Mode A.

So for clarity, in this magjack schematic..

In Mode A the power is passed on Ethernet pins 1,2 (VC+) 3,6 (VC-), the power is pulled off via the magnetics and supplied to PHY pins 9 & 10

In Mode B the power is passed on Ethernet pins 4,5 (VC+) 7,8 (VC-), the power is pulled off directly and supplied to the PHY pins 4 & 5

So PHY pins 4,5 and 9,10 shouldn't be joined together in my case where I only want to support passive (non-standard) Mode B and I have no space for a full rectifier anyway.

\$\endgroup\$
1
  • \$\begingroup\$ Neil - Hi, Thanks for your post. Can you please clarify: Is this the full & final answer to your question? If so, please consider "accepting" an answer - yours or another one - to effectively mark the topic as solved. (If you want to accept your own answer, you must wait until after 11:25 UTC tomorrow as self-answers can only be "accepted" after 48 hours has elapsed since asking the question.) Thanks. \$\endgroup\$
    – SamGibson ♦
    Apr 21, 2022 at 9:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.