Where can I find a basic block diagram with descriptions for a typical 10/100BASE-T PHY output driver stage?

I also wonder whether all drivers use "current-mode" outputs that lead to voltage across the external termination resistors (it seems so, because most of the time I see applications with termination resistors to VDD or GND) or are there devices that use voltage output? Why is current output the better choice? It it easier to steer current sources than voltage sources? (and therefore easier to create the different voltage levels)

  • \$\begingroup\$ Details of such chips are rarely disclosed on datasheets. Try patents or papers written on the subject. \$\endgroup\$ Commented Nov 9, 2016 at 20:45
  • \$\begingroup\$ Current output is always a better choice when you have no control over the voltage drop (for example caused by the length of cable in use). \$\endgroup\$
    – David
    Commented Nov 9, 2016 at 23:13
  • 1
    \$\begingroup\$ Agree with you that it is easier to create different voltage levels -- by turning on and off paralleled fixed current drivers. Check out "100BASE-TX Transmit" section on page 3, no block diagram but a brief description. datasheets.maximintegrated.com/en/ds/78Q2120C.pdf \$\endgroup\$
    – rioraxe
    Commented Nov 9, 2016 at 23:24
  • \$\begingroup\$ Another possible reason is that the output impedance would then simply be the external (usually pull up) resistor. \$\endgroup\$
    – rioraxe
    Commented Nov 10, 2016 at 0:24
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    \$\begingroup\$ @Junius The magnetics are a 1:1 transformer. Therefore the cable sees the same 100 ohm differential load even when the termination is on the opposite (PHY) side of the transformer. The 1:1 transformer is essentially transparent to the signal. \$\endgroup\$
    – user4574
    Commented Jan 10, 2017 at 15:37

1 Answer 1


Below is my guess of what a simplified driver can look like.

When [A1,A2] = 1,1, the output level is positive. When [A1,A2] = 0,0, the output level is negative. When [A1,A2] = 0,1 or 1,0, the output level is zero.

The output impedance is determined by the resistors, which are matched to the cable.

What is completely missing in the diagram is the slew rate control. Which I don't have much idea of how that is actually realized in an IC for this application.


simulate this circuit – Schematic created using CircuitLab

  • \$\begingroup\$ Sorry, but I don´t understand that circuit. Just for example: Let´s assume [A1,A2] = 1,1, then differential output voltage is negative (magnitude is the voltage drop via R1). Also, I don´t think that the wiring is what you wanted since there are always two transistors pulling on the same line "in parallel" to each other. Could you please check the circuit? \$\endgroup\$
    – Junius
    Commented Nov 23, 2016 at 18:41
  • \$\begingroup\$ Also, in all cases where the application notes suggest 50 Ohms pullups to VDD, also the transformer midpoint is biased to VDD. This fact would make even more troubles in the circuit above? It would for example short-circuit R1?! \$\endgroup\$
    – Junius
    Commented Nov 23, 2016 at 18:42

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