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I am trying to understand the basic operation of a synchronous switching-mode buck converter. I a continuous-mode operation the inductor current is always positive, hence from the mosfets to the load. When the pmos is off and the nmos is on -the current through the inductor is decreasing, but still positive. According to Kirchoff, the current must be sourced through the nmos device, since the pmos is off. This is what I can't understand: how can an nmos source current in this configuration? it seems to me it can only sink it! Any help would be greatly appreciated.

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  • \$\begingroup\$ "I a continuous-mode operation the inductor current is always positive" Not necessarily. Forced CCM at low output current will have negative inductor current. \$\endgroup\$
    – winny
    Sep 24 '20 at 13:35
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Well that's easy - the source is connected to the ground plane and the MOSFET is turned on, so the current comes through the MOSFET from the ground plane.

"Wait", you say, "it comes from the ground plane?"

Yes. The left end of the inductor has a voltage below the ground plane. Remember the inductor really wants to keep the current the same, and it will make whatever voltage it needs to make in order to "suck" current.

This is how a buck converter works: The current goes around in a loop, through the inductor, the lower MOSFET, and the load/smoothing capacitor, until the inductor runs out of juice (until it's fully discharged). Every so often, the controller turns on the upper MOSFET instead, and the supply "pushes" current through the inductor, recharging it.

In a non-synchronous buck converter, the lower MOSFET is replaced with a diode. When the inductor sucks current through the diode, it creates a 0.6V voltage drop across the diode which wastes energy. In a synchronous buck converter, a MOSFET is used. The MOSFET's body diode may conduct for a moment, but then the controller turns on the MOSFET which allows it to conduct with much less loss.

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  • \$\begingroup\$ So the left end of the inductor has a voltage below the ground plane. The lowest it can reach is ~-0.6V, at which point the body diode of the lower MOSFET will get forward biased. Is this correct? I guess I should pay attention to possible drain-source and gate-drain electrical stress on the upper MOSFET in this condition. Is this a correct conclusion? \$\endgroup\$
    – Itai
    Sep 24 '20 at 14:06
  • \$\begingroup\$ @Itai Note that the controller turns the MOSFET on, which creates less of a voltage drop than the body diode. If the controller doesn't turns it on fast enough then there will be a body diode voltage drop. Your upper MOSFET probably shouldn't be so close to its limit that 0.6V would push it over the edge. \$\endgroup\$
    – user253751
    Sep 24 '20 at 14:23
  • \$\begingroup\$ "If the controller doesn't turns it on fast enough then there will be a body diode voltage drop". But isn't there supposed to be a "dead-time", by design, between shutting off the the upper MOSFET and turning on the lower MOSFET to prevent a direct current path between the supply to the ground plane? \$\endgroup\$
    – Itai
    Sep 24 '20 at 14:30
  • \$\begingroup\$ @Itai should be "turn" not "turns". There will be a body diode voltage drop for a moment but then it goes away when the controller turns the MOSFET on. \$\endgroup\$
    – user253751
    Sep 24 '20 at 14:31

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