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During dead time of any DC-DC converter both the switches are turned off to avoid any shoot-through current. But in a synchronous Buck converter that would mean sudden zero current through inductor. Wouldn't that cause any problem?

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There is no problem since during the dead-time the current will flow trough the body diode of the FET discharging the inductor, very similar as a non synchronous buck. After the dead-time the FET is turned on bypassing the conducting diode. With this action you can reduce the losses since the voltage across the FET will be much lower than the forward voltage of the diode.

All in all, Synchronous Buck is all about reducing the forward losses on the Buck diode. There is no change on the operation states of the converter itself. It will work in CCM, BCM and DCM given that you have the right dead-time.

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    \$\begingroup\$ The forward recovery time of the body diode might mean that it will be slow to divert the inductor current during the dead time period, causing voltage transients. The reverse recovery time is usually not an issue as the diode will have plenty of time to recover (when the low side MOSFET passes the current). There is a fundamental change between synchronous and non-synchronous converters: a synchronous buck is inherently bidirectional and will boost the voltage on the input if the output is backfed from somewhere else. \$\endgroup\$ – jms Feb 22 '16 at 10:01
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Any inductor current is carried somewhere, for a while. There are various mechanisms to avoid this becoming a problem.

In an asynchronous buck with a switch+diode (I know you asked about synchronous, but let's start with one that you think should be OK, and show that it is less OK than you thought), the freewheel diode picks up the current, eventually. It can't do that instantaneously, as the inductance of the finite-length wires between the switch device and it is non-zero. This current first goes to charge 'unwanted' capacitances in the devices. In the event that the voltage on the switching transistor rises above its breakdown voltage, it will avalanche. Most switching devices have a 'maximum repetitive avalanche energy' specification, to cope with just this problem. Eventually, the high voltage across the wiring inductance causes the current to slew from the switch to the diode, and peace is restored.

In a high current design, great care is taken to minimise the switch/diode transfer inductance. This doesn't necessarily mean the inductance to each individually needs to be low, but it does mean they should be well coupled, so fat and close together.

How fast the current has to transfer is governed by the fall time of the switching FET. Adding a sniff of series resistance (a few ohms to the low 10s of ohms) is often used between gate driver and gate to increase the switching time a little. This is especially effective during switching as it limits the rate that the Miller capacitance charges, thus controlling the rate of voltage rise on the drain. While this increases dissipation in the off-going channel, it reduces transient overshoot which may improve EMI and other general behaviour.

So now you know what can happen in a an asynch converter, dead time does not seem so scary. First, the voltage rises on the off-going transistor, and it may avalanche. Second, current gets transferred into the rectifier device. Initially it's off, so it's first carried by the body diode. Eventually the channel comes on, which then takes up conduction.

If the wiring inductances or the current is so large that the switching transient exceeds the switch device's safe avalanche specification, then it's possible to use snubber components, typically a capacitor and resistor in series. These are best avoided if possible as they waste energy.

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    \$\begingroup\$ If the body diode is relied upon for carrying the current during the dead time period, is the forward recovery time of the body diode a significant factor? It's never stated in MOSFET datasheets. Often just the reverse recovery is mentioned. \$\endgroup\$ – jms Feb 22 '16 at 10:15
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No, because you must start the dead-time at the time where the inductor current becomes zero. This means the inductor is discharged. This means you loose no energy.

You cannot suddenly make the inductor current zero, the inductor will resist that by making the voltage such that the current will flow anyway (and your transistors might be damaged then).

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  • \$\begingroup\$ I am not sure I get your point. Current through the inductor may well never be anything near zero, but you need to switch anyway. I would say that dead time should be as short as possible, and depending if the output stage is push pull or n stacked you might not have many problems with the transistors, thanks to the body diodes. \$\endgroup\$ – Vladimir Cravero Feb 22 '16 at 8:50
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    \$\begingroup\$ @FakeMoustache : but inductor current only becomes zero in discontinuous conduction mode, not in continuous conduction mode \$\endgroup\$ – Payal Kumari Feb 22 '16 at 8:50
  • \$\begingroup\$ @VladimirCravero Relying on bulk diodes in a DCDC converter would be a very bad design choice. The point remains that you can only switch off the current through an inductor when the current reaches zero. The inductor current does become zero, it can even become negative if you leave the switch to ground on for too long. \$\endgroup\$ – Bimpelrekkie Feb 22 '16 at 9:17
  • \$\begingroup\$ "but inductor current only becomes zero in discontinuous conduction mode, not in continuous conduction mode" Correct and that is why there can be no dead time in CC mode, only in discontinous conduction mode. \$\endgroup\$ – Bimpelrekkie Feb 22 '16 at 9:20
  • \$\begingroup\$ I don't see why relying on bulk diodes is a problem. It happens, it works, it's fine, output stage is usually beefy enough... \$\endgroup\$ – Vladimir Cravero Feb 22 '16 at 9:43

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