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I am designing an 10/100 Ethernet on my board and I can't understand why 3.3V should be connected to the center point of the termination. enter image description here

The schematic checklist recommends it actually and I checked several design and datasheets even gigabit ones and somewhere it is used somewhere it doesn't. As I see it is not necessary because the centerpoint of the transformer is on 3.3V which can be pulled down by the ethernet chip in case of signalling (transmission) and it would require less current without it. I will do this in the recommended way but I don't understand why which I hate very much. :)

On the other hand the receiver side would not require 3.3V at all right? (if I wouldn't use Auto-MDIX I mean)

Schematic checklist

Datasheet

I have found an interesting article in the topic from 2004. Unfortunately it doesn't explain this directly but provides and overview about line drivers of gigabit ethernet. (I think it is similar in 10/100 and this is a current-mode driver)

There is an image at the end of the article which actually shows this connection with dashed line but as I read it doesn't explain this.

Voltage-mode line drivers save on power

Quentis
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1 Answers1

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IMO it's a pull-up resistor, TXP and TXN is just a open collector output that ties the line to GND level. The total current is now the current sum from both pull-ups. One goes directly from pull-up to ground, the second goes from pull-up trough a transformer and to ground. The receive channel is the same, since it is auto-MDIX. Therefore you can use a cross over cable and the TX and RX channels got swapped automaticly.

EDIT:
I said IMO, there is no detailed description about the IC output. Here is the picture of CML output type. Do the 50 ohm pull-ups make sense now?

enter image description here

EDIT2:
The ethernet transceiver IC differs from manufacturer to manufacturer. It is necessary to use the template or demo board as recommended by the manufacturer. If some manufacturer uses centre tap tied on +Vcc and some else is floating, .... its due to the IC characteristics and can't be swapped or used an universal schematics that will fit all ICs and all magnetics.

Marko Buršič
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    Reasonable assumption, but 49.9Ω seems a tad low for a pullup! – Polynomial Dec 03 '16 at 10:37
  • The outputs are unlikely to be open collector when operating at 10 or 100 MHz! The resistors are for termination. You also didn't answer the actual question of the connection of 3.3V to the magnetics CT. – DoxyLover Dec 03 '16 at 19:28
  • @Polynomial See the edit. The CML output type needs just those 50 ohm pull-ups. – Marko Buršič Dec 04 '16 at 13:04