I was looking at a schematic provided by TI for a PoE power supply, and I had questions about certain resistors. What is the purpose of R5,R6,R7, and R8 in this schematic?

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1 Answer 1


At first glance, this looks like a circuit that will do bridge rectification of a POE input signal, provides (body) diode rectification to bootstrap the system, and then allows for automatic MOSFET biasing to decrease the losses in the system.

When ADPT is pulled low, current can flow through Q6. This allows the bias voltage generators for the positive and negative side to function. Suppose PR12 is a positive voltage and PR78 is a negative voltage. There will be a current path from

  • PR12
  • D9 - D13 - Q2 (Base-Emitter)
  • Q6 - R14
  • Q10 (Base-Emitter) - D22 - D26
  • PR78

In this example, Q2 and Q10 are on. Q2 will pull down the gate voltage of PFET Q1C to about (V(PR12) - 9V), and Q10 will pull up the gate voltage of NFET Q11A to (V(PR78) + 9V). Here, C6 stabilizes the gate voltage of Q1C, reducing susceptibility to noise.

R5 is present to discharge the gate-source voltage of PMOS Q1C if the gate should no longer be pulled down. This could happen either due to driving ADPT high, or the POE power being turned off / disconnected.

C10 and R9 perform a similar function for the NPN Q2. On the bottom side, the topology is mirrored with C19 and R18 stabilizing the base voltage of the PNP Q10. R22 and C23 stabilize the gate-source voltage of NMOS Q11A

This scheme will provide significantly lower loss compared to a diode only solution, and shouldn't be speed critical as I believe PoE should be DC.

  • \$\begingroup\$ Thanks! This explanation clears up a bunch. Where does the -9V and +9V mentioned above come from? \$\endgroup\$
    – be-ee
    Aug 8, 2019 at 22:13
  • \$\begingroup\$ @be-ee D9 gives you about 8.2V when operating in zener mode, and D13 gives maybe 0.7V, and another 0.7V for the Vbe of Q2. It's a crude reference to give a voltage large enough to turn on the MOSFETs, but avoid exceeding Vgsmax, which is +/- 20V. \$\endgroup\$
    – W5VO
    Aug 9, 2019 at 0:49
  • \$\begingroup\$ But if the input of V(PR12) is 48V (typical PoE input value) won't that put the gate voltage of Q1C at 39V which is exceeds the Vgsmax? \$\endgroup\$
    – be-ee
    Aug 9, 2019 at 12:59
  • \$\begingroup\$ @be-ee Remember, it's the voltage between the gate and the source. If the input is 48V, and the gate is at 39V, then the gate-source voltage is -9V. \$\endgroup\$
    – W5VO
    Aug 9, 2019 at 14:23
  • \$\begingroup\$ Thank you. Your explanation makes full sense now. \$\endgroup\$
    – be-ee
    Aug 9, 2019 at 14:40

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