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I can't seem to understand the point of the second MOSFET in a synchronous buck converter. The second (Q2) MOSFET has a body diode which seems to act like a normal diode in an asynchronous buck converter and when the MOSFET is conducting there is no inductor current flowing through the MOSFET, just through the diode to my understanding.

By the way the following image is not my circuit, it's just an example.

Typical synchronous buck converter with Q1 on top and Q2 on the bottom.

JRE
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littleMak
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    Although a mosfet happens to include a diode, the point of using a MOSFET instead of a diode is that if you actually switch it, it's better than a diode(resistive/smaller volt drop). The disadvantage is you must design your switching circuit timing to prevent shoot through, which is more complicated than just selecting an adequate diode. – K H Feb 10 '21 at 05:09
  • I think the body diode is a just fly back diode, "in case" you need it when driving an inductive load. Putting a diode during chip design and fabrication costs "almost nothing" in time, space or money. – tlfong01 Feb 10 '21 at 05:17
  • But then current has to be flowing from drain to source, right? @K H – littleMak Feb 10 '21 at 05:31
  • @tlfong01: The body diode is inherent to the process of making a MOSFET. They don't "add them just in case." They'd happily make them without the diode, but can't. – JRE Feb 10 '21 at 07:07
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    @tlfong01: Have a look here. – JRE Feb 10 '21 at 07:09
  • @JRE, Many thanks for pointing out ignorance. I need to Google more. Cheers. PS -Apologies to all those misled by my wrong guess. – tlfong01 Feb 10 '21 at 07:11
  • *Apologies* to all those misled by my casual wrong guess that the FET intrinsic body diode is just an option. Many thanks to @JRE for pointing out what I didn't know that I didn' know. References: [1] FETs (Field-Effect Transistors) - Alec Schmidt, Robert Nelson, 2018dec28 https://www.digikey.com/eewiki/pages/viewpage.action?pageId=49414403

    [2] The Significance of the Intrinsic Body Diodes Inside MOSFETs - Digi-Key Electronics 2016sep22 https://www.digikey.de/en/articles/the-significance-of-the-intrinsic-body-diodes-inside-mosfets

     – tlfong01 Feb 10 '21 at 07:22
  • @JRE, your reference to FET intrinsic body diode is very good, so I googled to learn more. Cheers. – tlfong01 Feb 10 '21 at 07:24
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    @K H, Your explanation of (a) using a MOSFET instead of a diode, and (b) how to design switching circuit to prevent "*shooting through*" is indeed too hard for me newbies. Anyway, I googled more to pretend that I know more than I do. :) [3] DIOFET (Diodes Schottky Integrated MOSFET) - Diodes, https://www.diodes.com/products/discrete/mosfets/diofet-diodes-schottky-integrated-mosfet/

    DIOFET is a proprietary process that monolithically integrates *a power MOSFET with a Schottky diode into a single Silicon chip*. / to continue, ...

     – tlfong01 Feb 10 '21 at 07:47
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    / cont'd ... The *integrated Schottky reduces the forward voltage drop of the body diode by almost 50% and also has a lower reverse recovery charge*.

    In application, this means the conduction and switching losses are reduced, and overall, the circuit will operate at a higher efficiency with a reduced operating temperature. In addition, the DIOFET is an avalanche rugged process and the devices have *a low gate capacitance ratio to reduce the risk of shoot-through currents*.

    The DIOFET is well suited for Point of Load (PoL) *DC-DC converters*.

     – tlfong01 Feb 10 '21 at 07:50
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    The so-called synchronous switcher can reduce power loss during continuous conduction mode of the converter. During discontinuous conduction mode, the FET may be off and it will be no different than having a diode there. – user57037 Feb 10 '21 at 08:42

2 Answers2

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and when the MOSFET is conducting there is no inductor current flowing through the MOSFET

That's not correct; it bypasses the diode and produces a more efficient flyback control so there is no power lost in the forward voltage of the diode. This leads to about 5% more power efficiency generally.

There is another use for the synchronizing MOSFET; on very light loads, the output transfer function remains the same as for continuous conduction mode (CCM) and the circuit never behaves as a converter in discontinuous mode (DCM) thus, the output is more predictable and requires less feedback to maintain stability leading to better regulation performance.

With only a diode as the flyback element, it will stop conducting as soon as the inductor current drops to zero and the converter will enter DCM. The result of this is that the output voltage is very much load dependent and requires more sophistication in the regulation and feedback circuit.

Maybe it's easier to think of the two MOSFETs acting as a hard square wave voltage source and then the LC network on the output act like a low pass filter or averager: -

enter image description here

Andy aka
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  • Aha, if I'm correct the mosfet does conduct both ways just the diode happens to be there. – littleMak Feb 10 '21 at 15:02
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    @littleMaklittleMak it does conduct both ways with pretty much the same on-resistance as the forward conduction. – Andy aka Feb 10 '21 at 15:06
  • Oh ok, that's all I was missing. Thanks!! :-) – littleMak Feb 10 '21 at 15:07
  • A nice feature of the square-wave model is that if one assumes an ideal inductor and capacitor, and a frequency high enough that inductor current doesn't change much during each cycle, the average output voltage will simply be the average voltage from the PWM, which will be the "on" voltage times the duty cycle, and the average supply current will be the continuous output current times the duty cycle. – supercat Feb 10 '21 at 19:58
  • @supe precisely. – Andy aka Feb 10 '21 at 20:09
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Even if Q2's gate and source are shorted (i.e. forcing Q2 off), yes, the circuit can operate.

But,

If Q2 is turned on, it shorts the body diode and provides a way with very-low resistance (i.e. RDS-on which can be as low as a few milliOhms) to current flow. And this significantly reduces the dissipated power and thus increases the overall efficiency. When it comes to light-load efficiency, sync-buck may not be that efficient due to operating in DCM.

Oh yes, the current will flow from Source to Drain, but it's not important. Because the intention is to close the loop when Q1 is off so that the current (induced by the energy stored by the inductor) can flow through a very-low-resistant return path.

The most important thing here is the timing of gate drive signals: If Q1 and Q2 remain on at the same time then they short the supply rails. So there should be a dead-time between them.

Rohat Kılıç
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  • Ooh, Q2 will let current flow either way so when it's on it just shorts the internal diode creating a lower voltage drop. Also sorry for being 9 hours late, it's because I posted this at midnight. – littleMak Feb 10 '21 at 15:00